192266681 Solution of Physics for Scientists and Engineers 3rd Edition by Douglas C Giancoli

Chapter 1 CHAPTER 1 - Introduction, Measurement, Estimating 1.

(a) Assuming one significant figure, we have 11010 yr. 10 billion yr = 10109 yr = 10 7 17 310 s. (b) (110 yr)(310 s/yr) =

2.

(a) (b) (c) (d) (e) (f) (g)

4 significant figures. Because the zero is not needed for placement, we have 4 significant figures. 3 significant figures. Because the zeros are for placement only, we have 1 significant figure. Because the zeros are for placement only, we have 2 significant figures. 4 significant figures. 2, 3, or 4 significant figures, depending on the significance of the zeros.

3.

(a) (b) (c) (d) (e) (f)

1,156 = 1.156103. 21.8 = 2.18101. 0.0068 = 6.810–3. 27.635 = 2.7635101. 0.219 = 2.1910–1. 22 = 2.2101.

4.

(a) (b) (c) (d) (e)

8.69104 = 7.1103 = 6.610–1 = 8.76102 = 8.6210–5 =

5.

% uncertainty = [(0.25 m)/(3.26 m)] 100 = 7.7%. Because the uncertainty has 2 significant figures, the % uncertainty has 2 significant figures.

6.

We assume an uncertainty of 1 in the last place, i. e., 0.01, so we have % uncertainty = [(0.01 m)/(1.28 m)] 100 = 0.8%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.

7.

We assume an uncertainty of 0.2 s. (a) % uncertainty = [(0.2 s)/(5 s)] 100 = 4%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. (b) % uncertainty = [(0.2 s)/(50 s)] 100 = 0.4%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. (c) % uncertainty = [(0.2 s)/(5 min)(60 s/min)] 100 = 0.07%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.

8.

For multiplication, the number of significant figures in the result is the least number from the multipliers; in this case 2 from the second value. (2.079102 m)(0.07210–1) = 0.15101 m = 1.5 m.

9.

To add, we make all of the exponents the same: 9.2103 s + 8.3104 s + 0.008106 s = 0.92104 s + 8.3104 s + 0.8104 s 1.0105 s. = 10.02104 s = Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the third value.

86,900. 7,100. 0.66. 876. 0.000 086 2.

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Chapter 1 10. We assume an uncertainty of 0.1104 cm. We compare the area for the specified radius to the area for the extreme radius. A1 = pR12 = p(3.8104 cm)2 = 4.54109 cm2 ; A2 = pR22 = p[(3.8 + 0.1)104 cm]2 = 4.78109 cm2 , so the uncertainty in the area is ?A =A2 – A1 = 0.24109 cm2 = 0.2109 cm2. We write the area as A = (4.5 ± 0.2)109 cm2. We could also treat the change as a differential: dA =2pR dR = 2p(3.8104 cm)(0.1104 cm) = 2108 cm2. 11. We compare the volume with the specified radius to the volume for the extreme radius. V1 = )pR13 = )p(2.86 m)3 = 98.0 m3; V2 = )pR23 = )p(2.86 m + 0.08 m)3 = 106.45 m3, so the uncertainty in the volume is ?V =V2 –V1 = 8.5 m3; and the % uncertainty is % uncertainty = [(8.5 m3)/(98.0 m3)] 100 = 9%. 12. (a) (b) (c) (d) (e)

106 volts = 1 megavolt = 1 Mvolt. 10–6 meters = 1 micrometer = 1 µm. 6103 days = 6 kilodays = 6 kdays. 18102 bucks = 18 hectobucks = 1.8 kbucks. 810–9 pieces = 8 nanopieces = 8 npieces.

13. (a) (b) (c) (d) (e) (f)

286.6 mm = 286.610–3 m = 0.286 6 m. 85 µV = 8510–6 V = 0.000 085 V. 760 mg = 76010–3 g = 76010–6 g = 0.000 760 kg. This assumes that the last zero is significant. 60.0 picoseconds = 60.010–12 s = 0.000 000 000 060 0 s. 22.5 femtometers = 22.510–15 m = 0.000 000 000 000 022 5 m. 2.50 gigavolts = 2.50109 volts = 2,500,000,000 volts.

14. 50 hectocars = 50102 cars = 5,000 cars. 1 megabuck/yr = 1106 bucks/yr = 1,000,000 bucks/yr 15. If we assume a height of 5 ft 10 in, we have height = 5 ft 10 in = 70 in = (70 in)[(1 m)/(39.37 in)] =

(millionaire). 1.8 m.

16. (a) 93 million mi = 93106 mi = (93106 mi)[(1610 m)/(1 mi)] = 150Gm. (b) 1.51011 m = 150109 m =

1.51011 m.

17. (a) 1 ft2 = (1 ft2)[(1 yd)/(3 ft)]2 = 0.111 yd2. 2 2 2 10.76 ft2. (b) 1 m = (1 m )[(3.28 ft)/(1 m)] = 18. We find the time from time = distance/speed = [(1.0 mi)(1.61 km/mi)]/[2300 km/h)/(3600 s/h)] =

2.5 s.

19. (a) 1.010–10 m = (1.010–10 m)[(1 in)/(2.54 cm)][(100 cm)/(1 m)] = 3.910–9 in. (b) We let the units lead us to the answer: 1.0108 atoms. (1.0 cm)[(1 m)/(100 cm)][(1 atom)/(1.010–10 m)] = 20. To add, we make all of the units the same: 2.00 m + 142.5 cm + 7.24105 µm = 2.00 m + 1.425 m + 0.724 m = 4.149m = 4.15m. Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the first value.

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Chapter 1 21. (a) 1 km/h = (1 km/h)[(0.621 mi)/(1 km)] = 0.621 mi/h. (b) 1 m/s = (1 m/s)[(1 ft)/(0.305 m)] = 3.28 ft/s. (c) 1 km/h = (1 km/h)[(1000 m)/(1 km)][(1 h)/(3600 s)] = 0.278 m/s. A useful alternative is 1 km/h = (1 km/h)[(1 h)/(3.600 ks)] = 0.278 m/s. 22. For the length of a one-mile race in m, we have 1 mi = (1 mi)[(1610 m)/(1 mi)] = 1610 m; so the difference is 110 m. The % difference is % difference = [(110 m)/(1500 m)] 100 = 7.3%. = (2.998108 m/s)(1.00 yr)[(365.25 days)/(1 yr)][(24 h)/(1 day)][(3600 s)/(1 h)] = 9.461015 m. 6.31104 AU. (b) 1.00 ly = (9.461015 m)[(1 AU)/(1.50108 km)][(1 km)/(1000 m)] = 8 8 (c) speed of light = (2.99810 m/s)[(1 AU)/(1.5010 km)][(1 km)/(1000 m)][(3600 s)/(1 h)] = 7.20 AU/h.

23. (a) 1.00 ly

24. For the surface area of a sphere, we have AMoon = 4prMoon2 = 4p[!(3.48106 m)]2 = 3.801013 m2. We compare this to the area of the earth by finding the ratio: AMoon /AEarth = 4prMoon2 /4prEarth2 = (rMoon/rEarth)2 = [(1.74103 km)/(6.38103 km)]2 = 7.44 x 10–2. Thus we have AMoon = 7.44 x 10–2 AEarth. 25. (a) (b) (c) (d)

2800 = 2.8103 ˜ 1103 = 103. 86.30102 ˜ 100102 = 104. 0.0076 = 0.7610–2 ˜ 10–2. 15.0108 = 1.50109 ˜ 109.

26. We assume that a good runner can run 6 mi/h (equivalent to a 10-min mile) for 5 h/day. Using 3000 mi for the distance across the United States, we have time = (3000 mi)/(6 mi/h)(5 h/day) ˜ 100 days. 27. We assume a rectangular house 40 ft30 ft, 8 ft high; so the total wall area is Atotal = [2(40 ft) + 2(30 ft)](8 ft) ˜ 1000 ft2. If we assume there are 12 windows with dimensions 3 ft5 ft, the window area is Awindow = 12(3 ft)(5 ft) ˜ 200 ft2. Thus we have % window area = [Awindow/Atotal ](100) = [(200 ft2)/(1000 ft2)](100) ˜ 20%. 28. If we take an average lifetime to be 70 years and the average pulse to be 60 beats/min, we have N = (60 beats/min)(70 yr)(365 day/yr)(24 h/day)(60 min/h) ˜ 2109 beats. 29. If we approximate the body as a box with dimensions 6 ft, 1 ft, and 8 in, we have volume = (72 in)(12 in)(8 in)(2.54 cm/in)3 ˜ 1105 cm3.

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Chapter 1 30. We assume the distance from Beijing to Paris is 10,000 mi. (a) If we assume that today’s race car can travel for an extended period at an average speed of 40 mi/h, we have time = [(10,000 mi)/(40 mi/h)](1 day/24 h) ˜ 10 days. (b) If we assume that in 1906 a race car could travel for an extended period at an average speed of 5 mi/h, we have time = [(10,000 mi)/(5 mi/h)](1 day/24 h) ˜ 80 days. 31. We assume that each dentist sees 10 patients/day, 5 days/wk for 48 wk/yr, for a total number of visits of Nvisits = (10 visits/day)(5 days/wk)(48 wk/yr) ˜ 2400 visits/yr. We assume that each person sees a dentist 2 times/yr. (a) We assume that the population of San Francisco is 700,000. We let the units lead us to the answer: N = (700,000)(2 visits/yr)/(2400 visits/yr) ˜ 600 dentists. (b) Left to the reader to estimate the population. 32. If we assume that the person can mow at a speed of 1 m/s and the width of the mower cut is 0.5 m, the rate at which the field is mown is (1 m/s)(0.5 m) = 0.5 m2/s. If we take the dimensions of the field to be 110 m by 50 m, we have time = [(110 m)(50 m)/(0.5 m2/s)]/(3600 s/h) ˜ 3 h. 33. We assume an average time of 3 yr for the tire to wear d = 1 cm and the tire has a radius r = 30 cm and a width w = 10 cm. Thus the volume of rubber lost by a tire in 3 yr is V = wd2pr. If we assume there are 100 million vehicles, each with 4 tires, we have m = (100106 vehicles)(4 tires/vehicle)(0.1 m)(0.01 m)2p(0.3 m/tire)(1200 kg/m3)/(3 yr) ˜ 3108 kg/yr. 34. We assume that one-third of the floor space can be used for shelves and there are five shelves, each with a depth of 25 cm, in a stack. If the average book has a width of 4.0 cm, we have N = [@(1500 m2)(5 shelves)/(4.0 cm/book)(25 cm/shelf)](100 cm/m)2 = 2.5105 books. 35. For the right triangle shown on the diagram we have r2 + d2 = (r + h)2; d2 = 2rh + h2 = 2(6.4106 m)(200 m) + (200 m)2, which gives d = 5.1104 m = 51 km.

36. Because every term must have the same dimensions, we have v = At3 – Bt; [L/T] = [A][T]3 – [B][T]. Thus we get [A] = [L/T]/[T3] = [L/T4], and [B] = [L/T]/[T] = [L/T2]. 37. For the units we have A = [L/T4] = m/s4,

and

B = [L/T2] = m/s2.

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Chapter 1

38. We test to see if each term has the same dimensions. (a) x = vt2 + 2at; [L] =? [L/T][T]2 + [L/T2][T]; [L] ? [LT] + [L/T]; therefore, this is not correct. 2 (b) x = v0t + !at ; [L] =? [L/T][T] + [L/T2][T]2; [L] = [L] + [L]; therefore, this may be correct. (c) x = v0t + 2at2; [L] =? [L/T][T] + [L/T2][T]2; [L] = [L] + [L]; therefore, this may be correct. 39. (a) (b) (c) (d)

1.0 Å = (1.010–10 m)/(10–9 m/nm) = 0.10 nm. 1.0 Å = (1.010–10 m)/(10–15 m/fm) = 1.0105 fm. 1.0 m = (1.0 m)/(10–10 m/Å) = 1.01010 Å. From the result for Problem 23, we have 9.51025 Å. 1.0 ly = (9.51015 m)/(10–10 m/Å) =

40. We use the values for the masses from Table 1–3. (a) Nbacterium = (10–15 kg)/(10–27 kg/proton) = 1012 protons (or neutrons). –17 –27 10 (b) NDNA = (10 kg)/(10 kg/proton) = 10 protons (or neutrons). 2 –27 (c) Nhuman = (10 kg)/(10 kg/proton) = 1029 protons (or neutrons). 41 –27 (d) Ngalaxy = (10 kg)/(10 kg/proton) = 1068 protons (or neutrons). 41. (a) 1.00 yr = (365.25 days)(24 h/day)(3600 s/h) = 3.16107 s 7 –9 16 3.1610 ns. (b) 1.00 yr = (3.1610 s)/(10 s/ns) = 3.1710–8 yr. (c) 1.00 s = (1.00 s)/(3.16107 s/yr) = 42. 1 hectare = (104 m2)[(3.28 ft)/(1 m)]2[(1 acre)/(4104 ft2)] =

˜ p107 s.

2.69 acres.

43. (a) The maximum number of buses is needed during rush hour. If we assume that at any time there are 40,000 persons commuting by bus and each bus has 30 passengers, we have N = (40,000 commuters)/(30 passengers/bus) ˜ 1000 buses ˜ 1,000 drivers. (b) Left to the reader. 44. If we ignore any loss of material from the slicing, we find the number of wafers from (30 cm)(10 mm/cm)/(0.60 mm/wafer) = 500 wafers. For the maximum number of chips, we have (500 wafers)(100 chips/wafer) = 50,000 chips. 45. If we assume there is 1 automobile for 2 persons and a U. S. population of 250 million, we have 125 million automobiles. We estimate that each automobile travels 15,000 miles in a year and averages 20 mi/gal. Thus we have N = (125106 automobiles)(15,000 mi/yr/automobile)/(20 mi/gal) ˜ 11011 gal/yr. 46. We let D represent the diameter of a gumball. Because there are air gaps around the gumballs, we estimate the volume occupied by a gumball as a cube with volume D3 . The machine has a square crosssection with sides equivalent to 10 gumballs and is about 14 gumballs high, so we have N = volume of machine/volume of gumball = (14D)(10D)2/D3 ˜ 1.4103 gumballs.

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Chapter 1

47. We will convert all units to meters. The volume used in one year is V = [(40,000 persons)/(4 persons/family)](1200 L/family · day)  (365 days/yr)(10–3 m3/L) = 4.4106 m3/yr.           If we let d represent the loss in depth, we have d = V/area = (4.4106 m3/yr)/(50 km2)(103 m/km)2 ˜ 0.09 m ˜ 9 cm/yr. 48. For the volume of a 1-ton rock, we have V = (2000 lb)/(3)(62 lb/ft3) ˜ 11 ft3. If we assume the rock is a sphere, we find the radius from 11 ft3 = )pr3, which gives r ˜ 1.4 ft, so the diameter would be

˜ 3 ft ˜ 1 m.

49. We find the amount of water from its volume: m = (5 km)(8 km)(1.0 cm)(105 cm/km)2(10–3 kg/cm3)/(103 kg/t) = 40104 t ˜

4105 t.

50. We will use a pencil with a diameter of 5 mm and assume that it is held 0.5 m from the eye. Because the triangles AOD and BOC are similar, we can equate the ratio of distances: BC/AD = OQ/OP; BC/(0.005 m) = (3.8105 km)/(0.5 m), which gives BC ˜ 4103 km.

51. We assume that we can walk an average of 15 miles a day. If we ignore the impossibility of walking on water and travel around the equator, the time required is time = 2prEarth/speed = 2p(6103 km)(0.621 mi/km)/(15 mi/day)(365 days/yr) ˜ 4 yr. 52. If we use 0.5 m for the cubit, for the dimensions we have 150 m long, 25 m wide, and 15 m high. 53. From the diagram we see that d/L = tan ; (120 yd)/L = tan 30°, which gives L = 2.1102 yd = (2.1102 yd)(3 ft/yd)/(3.28 ft/m) =

1.9102 m.

54. We assume the oil slick is circular with diameter D and thickness d. For the volume we have V = d(pr2) = #dpD2; (1 L)(1000 cm3/L)/(1102 cm/m)3 = #(210–10 m)pD2, which gives D ˜ 3103 m.

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Chapter 1

55. The trigonometric function is not a linear function of the angle, so we can find the uncertainty in the sine by calculating two values. 3%. (a) Percent uncertainty in  = (0.5°/15.0°) 100 = We find the percent uncertainty in the sine from sin 15.0° = 0.2588; sin 15.5° = 0.2672; Percent uncertainty in sin  = [(0.2672 – 0.2588)/0.2588] 100 = 3%. 0.7%. (b) Percent uncertainty in  = (0.5°/75.0°) 100 = We find the percent uncertainty in the sine from sin 75.0° = 0.9659; sin 75.5° = 0.9681; Percent uncertainty in sin  = [(0.9681 – 0.9659)/0.9659] 100 = 0.2%. Note that it is possible to approximate uncertainties with differential quantities, if the angle is not very small (i. e., sin  is not small). We have d(sin ) = cos  d; d(sin )/(sin ) = d/(tan ). The angle must be in radians, so we get d(sin )/(sin ) = [(0.5°)(p/180°)]/tan 15.0° = 0.03 = 3%; d(sin )/(sin ) = [(0.5°)(p/180°)]/tan 75.0° = 0.002 = 0.2%. 56. Because you are lying on the sand, your line of sight is tangent to the surface of the Earth. If the height of the hull is h, from the triangle on the diagram we have r2 + d2 = (r + h)2; d2 = 2rh + h2 = 2(6.38106 m)(2.5 m) + (2.5 m)2, which gives d = 5.6103 m = 5.6 km.

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Chapter 2

CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1.

We find the time from average speed = d/t; 15 km/h = (75 km)/t , which gives

t = 5.0 h.

2.

We find the average speed from average speed = d/t = (280 km)/(3.2 h) =

3.

We find the distance traveled from average speed = d/t; (110 km/h)/(3600 s/h) = d/(2.0 s), which gives

88 km/h.

d= 6.110–2 km = 61 m.

4.

We find the average velocity from æ = (x2 – x1)/(t2 – t1) = (– 4.2 cm – 3.4 cm)/(6.1 s – 3.0 s) =

5.

We find the average velocity from æ = (x2 – x1)/(t2 – t1) = (8.5 cm – 3.4 cm)/[4.5 s – (– 2.0 s)] = 0.78 cm/s (toward + x). Because we do not know the total distance traveled, we cannot calculate the average speed.

6.

(a) We find the elapsed time before the speed change from speed = d1/t1 ; 65 mi/h = (130 mi)/t1 , which gives t1 = 2.0 h. Thus the time at the lower speed is t2 = T – t1 = 3.33 h – 2.0 h = 1.33 h. We find the distance traveled at the lower speed from speed = d2/t2 ; 55 mi/h = d2/(1.33 h), which gives d2 = 73 mi. The total distance traveled is D = d1 + d2 = 130 mi + 73 mi = 203 mi. (b) We find the average speed from average speed = d/t = (203 mi)/(3.33 h) = 61 mi/h. Note that the average speed is not !(65 mi/h + 55 mi/h). The two speeds were not maintained for equal times.

7.

Because there is no elapsed time when the light arrives, the sound travels one mile in 5 seconds. We find the speed of sound from speed = d/t = (1 mi)(1610 m/1 mi)/(5 s) ˜ 300 m/s.

8.

(a) We find the average speed from average speed = d/t = 8(0.25 mi)(1.61103 m/mi)/(12.5 min)(60 s/min) = (b) Because the person finishes at the starting point, there is no displacement; thus the average velocity is æ = x/t = 0.

9.

(a) (b)

– 2.5 cm/s (toward – x).

4.29 m/s.

We find the average speed from average speed = d/t = (160 m + 80 m)/(17.0 s + 6.8 s) = 10.1 m/s. The displacement away from the trainer is 160 m – 80 m = 80 m; thus the average velocity is æ = x/t = (80 m)/(17.0 s + 6.8 s) = + 3.4 m/s, away from trainer.

10. Because the two locomotives are traveling with equal speeds in opposite directions, each locomotive will travel half the distance, 4.25 km. We find the elapsed time from Page 1

Chapter 2

speed1 = d1/t ; (95 km/h)/(60 min/h) = (4.25 km)/t, which gives

t = 2.7 min.

x (m)

11. (a) We find the instantaneous velocity from the slope of the straight line from t = 0 to t = 10.0 s: 20 v10 = ?x/?t = (2.8 m – 0)/(10.0 s – 0) = 0.28 m/s. (b) We find the instantaneous velocity from the slope 10 of a tangent to the line at t = 30.0 s: v30 = ?x/?t = (22 m – 10 m)/(35 s – 25 s) = 1.2 m/s. 0 0 (c) The velocity is constant for the first 17 s (a straight line), so the velocity is the same as the velocity at t = 10 s: æ05 = 0.28 m/s. (d) For the average velocity we have æ2530 = ?x/?t = (16 m – 8 m)/(30.0 s – 25.0 s) = 1.6 m/s. (e) For the average velocity we have æ4050 = ?x/?t = (10 m – 20 m)/(50.0 s – 40.0 s) = – 1.0 m/s.

10

20

30

12. (a) (b) (c) (d)

Constant velocity is indicated by a straight line, which occurs from t = 0 to 17 s. The maximum velocity is when the slope is greatest: t = 28 s. Zero velocity is indicated by a zero slope. The tangent is horizontal at t = 38 s. Because the curve has both positive and negative slopes, the motion is in both directions.

13. (a)

We find the average speed from 13.4 m/s. average speed = d/t = (100 m + 50 m)/[8.4 s + @(8.4 s)] = The displacement away from the trainer is 100 m – 50 m = 50 m; thus the average velocity is æ = x/t = (50 m)/[8.4 s + @(8.4 s)] = + 4.5 m/s, away from master.

(b)

40

t (s)

14. (a) We find the position from the dependence of x on t: x = 2.0 m – (4.6 m/s)t + (1.1 m/s2)t2. x1 = 2.0 m – (4.6 m/s)(1.0 s) + (1.1 m/s2)(1.0 s)2 = – 1.5 m; x2 = 2.0 m – (4.6 m/s)(2.0 s) + (1.1 m/s2)(2.0 s)2 = – 2.8 m; 2 2 x3 = 2.0 m – (4.6 m/s)(3.0 s) + (1.1 m/s )(3.0 s) = – 1.9 m. (b) For the average velocity we have æ13 = ?x/?t = [(– 1.9 m) – (– 1.5 m)]/(3.0 s – 1.0 s) = – 0.2 m/s (toward – x). (c) We find the instantaneous velocity by differentiating: v = dx/dt = – (4.6 m/s) + (2.2 m/s2)t; v2 = – (4.6 m/s) + (2.2 m/s2)(2.0 s) = – 0.2 m/s (toward – x); v3 = – (4.6 m/s) + (2.2 m/s2)(3.0 s) = + 2.0 m/s (toward + x). 15. Because the velocities are constant, we can use the relative speed of the car to find the time: t = d/vrel = [(0.100 km)/(90 km/h – 75 km/h)](60 min/h) = 0.40 min = 24 s. 16. We find the total time for the trip by adding the times for each leg: T = t1 + t2 = (d1/v1) + (d2/v2) = [(2100 km)/(800 km/h)] + [(1800 km)/(1000 km/h)] = 4.43 h. We find the average speed from average speed = (d1 + d2)/T = (2100 km + 1800 km)/(4.43 h) = 881 km/h. Note that the average speed is not !(800 km/h + 1000 km/h). The two speeds were not maintained for equal times.

Page 2

50

Chapter 2

17. We find the time for the outgoing 200 km from t1 = d1/v1 = (200 km)/(90 km/h) = 2.22 h. We find the time for the return 200 km from t2 = d2/v2 = (200 km)/(50 km/h) = 4.00 h. We find the average speed from average speed = (d1 + d2)/(t1 + tlunch + t2) = (200 km + 200 km)/(2.22 h + 1.00 h + 4.00 h) = 55 km/h. Because the trip finishes at the starting point, there is no displacement; thus the average velocity is æ = x/t = 0. 18. If

vAG is the velocity of the automobile with respect to the ground, vTG the velocity of the train with respect to the ground, and vAT the velocity of the automobile with respect to the train, then vAT = vAG – vTG . If we use a coordinate system in the reference frame of the train with the origin at the back of the train, we find the time to pass from x1 = vATt1 ; 1.10 km = (90 km/h – 80 km/h)t1 , which gives t1 = 0.11 h = 6.6 min. With respect to the ground, the automobile will have traveled x1G = vAGt1 = (90 km/h)(0.11 h) = 9.9 km. If the automobile is traveling toward the train, we find the time to pass from x2 = vATt2 ; 1.10 km = [90 km/h – (– 80 km/h)]t2 , which gives t2 = 0.00647 h = 23.3 s. With respect to the ground, the automobile will have traveled x2G = vAGt2 = (90 km/h)(0.00647 h) = 0.58 km.

19. We find the time for the sound to travel the length of the lane from tsound = d/vsound = (16.5 m)/(340 m/s) = 0.0485 s. We find the speed of the ball from v = d/(T – tsound) = (16.5 m)/(2.50 s – 0.0485 s) = 6.73 m/s. 20. We find the average acceleration from Æ = v/t = [(95 km/h)(1 h/3.6 ks) – 0]/(6.2 s) =

4.3 m/s2.

21. We find the time from Æ = v/t ; 1.6 m/s2 = (110 km/h – 80 km/h)(1 h/3.6 ks)/t, which gives

Page 3

t =

5.2 s.

Chapter 2

22.

v (m/s)

40 30 20 10 0

0

10

20

30

40

50

60

70

80

90 100 110 120

t (s) (a) The maximum velocity is indicated by the highest point, which occurs at (b) Constant velocity is indicated by a horizontal slope, which occurs from (c) Constant acceleration is indicated by a straight line, which occurs from t = 0 to 30 s, and t = 90 s to 107 s. (d) The maximum acceleration is when the slope is greatest: t = 75 s. 23. We find the acceleration (assumed to be constant) from v2 = v02 + 2a(x2 – x1); 0 = [(100 km/h)/(3.6 ks/h)]2 + 2a(55 m), which gives The number of g’s is N =  a/g = (7.0 m/s2)/(9.80 m/s2) = 0.72.

t = 50 s. t = 90 s to 107 s.

a = – 7.0 m/s2.



24. For the average acceleration we have Æ2 = ?v/?t = (24 m/s – 14 m/s)/(8 s – 4 s) = 2.5 m/s2; Æ4 = ?v/?t = (44 m/s – 37 m/s)/(27 s – 16 s) = 0.6 m/s2.

50 5th gear 4th gear

v (m/s)

40 3rd gear

30 20

2nd gear

10 0

1st gear 0

10

20

t (s)

25. (a) For the average acceleration we have Æ1 = ?v/?t = (14 m/s – 0)/(3 s – 0) = 4.7 m/s2. (b) For the average acceleration we have Æ3 = ?v/?t = (37 m/s – 24 m/s)/(14 s – 8 s) = 2.2 m/s2. (c) For the average acceleration we have Æ5 = ?v/?t = (52 m/s – 44 m/s)/(50 s – 27 s) = 0.3 m/s2. (d) For the average acceleration we have Page 4

30

40

Chapter 2

Æ14 = ?v/?t = (44 m/s – 0)/(27 s – 0) = 1.6 m/s2. Note that we cannot add the four average accelerations and divide by 4.

27. We find the velocity and acceleration by differentiating x = (6.0 m/s)t + (8.5 m/s2)t2: v = dx/dt = (6.0 m/s) + (17 m/s2)t; a = dv/dt = 17 m/s2. 28. The position is given by x = At + 6Bt3. (a) All terms must give the same units, so we have A ~ x/t = m/s; and B ~ x/t3 = m/s3. (b) We find the velocity and acceleration by differentiating: v = dx/dt = A + 18Bt2; a = dv/dt = 36Bt. Page 5

6 5 4 3 2 1 0

a (m/s 2 )

v (m/s)

26. (a) We take the average velocity during a time interval as the instantaneous velocity at the midpoint of the time interval: vmidpoint = æ = x/t. Thus for the first interval we have v0.125 s = (0.11 m – 0)/(0.25 s – 0) = 0.44 m/s. (b) We take the average acceleration during a time interval as the instantaneous acceleration at the midpoint of the time interval: amidpoint = Æ = v/t. Thus for the first interval in the velocity column we have a0.25 s = (1.4 m/s – 0.44 m/s)/(0.375 s – 0.125 s) = 3.8 m/s2. The results are presented in the following table and graph. t(s) x(m) t(s) v(m/s) t(s) a(m/s2) 0.0 0.0 0.0 0.0 acceleration 0.125 0.44 0.25 0.11 0.25 3.8 0.375 1.4 0.50 0.46 0.50 4.0 30 0.625 2.4 0.75 1.06 0.75 4.5 0.875 3.5 1.00 1.94 1.06 4.9 20 1.25 5.36 1.50 4.62 1.50 5.0 1.75 7.85 2.00 8.55 2.00 5.2 2.25 10.5 10 2.5013.79 2.50 5.3 velocity 2.75 13.1 3.0020.36 3.00 5.5 3.25 15.9 3.50 28.31 3.50 5.6 0 3.75 18.7 0 1 2 3 4 5 6 4.00 37.65 4.00 5.5 t (s) 4.25 21.4 4.5048.37 4.50 4.8 4.75 23.9 5.0060.30 5.00 4.1 5.25 25.9 5.5073.26 5.50 3.8 5.75 27.8 6.0087.16 Note that we do not know the acceleration at t = 0.

Chapter 2

(c) For the given time we have v = dx/dt = A + 18Bt2 = A + 18B(5.0 s)2 = A + (450 s2)B; a = dv/dt = 36Bt = 36B(5.0 s) = (180 s)B. (d) We find the velocity by differentiating: v = dx/dt = A – 3Bt–4.

29. We find the acceleration from v = v0 + a(t – t0); 21 m/s = 12 m/s + a(6.0 s), which gives We find the distance traveled from x = !(v + v0)t = !(21 m/s + 12 m/s)(6.0 s) =

a = 1.5 m/s2.

99 m.

30. We find the acceleration (assumed constant) from v2 = v02 + 2a(x2 – x1); 0 = (25 m/s)2 + 2a(75 m), which gives a = – 4.2 m/s2. 31. We find the length of the runway from v2 = v02 + 2aL; (32 m/s)2 = 0 + 2(3.0 m/s2)L, which gives 32. We find the average acceleration from v2 = v02 + 2Æ(x2 – x1); (44 m/s)2 = 0 + 2Æ(3.5 m), which gives

Æ = 2.8102 m/s2.

33. We find the average acceleration from v2 = v02 + 2Æ(x2 – x1); (11.5 m/s)2 = 0 + 2Æ(15.0 m), which gives We find the time required from x = !(v + v0)t ; 15.0 m = !(11.5 m/s + 0), which gives

L = 1.7102 m.

Æ = 4.41 m/s2.

t = 2.61 s.

34. The average velocity is æ = (x – x0)/(t – t0), so we must find the position as a function of time. Because the acceleration is a function of time, we find the velocity by integrating a = dv/dt : ? dv = ? a dt = ? (A + Bt) dt; v = At + (Bt2/2) + C. If v = v0 when t = 0, C = v0; so we have v = At + (Bt2/2) + v0. We find the position by integrating v = dx/dt : ? dx = ? v dt = ? [At + (Bt2/2) + v0] dt; x = (At2/2)+ (Bt3/6) + v0t + D. If x = x0 when t = 0, D = x0; so we have x = (At2/2)+ (Bt3/6) + v0t + x0. Thus the average velocity is æ = (x – x0)/(t – t0) = {[(At2/2)+ (Bt3/6) + v0t + x0] – x0}/(t – 0) = (At/2)+ (Bt2/6) + v0. If we evaluate (v + v0)/2, we get (v + v0)/2 = {[At + (Bt2/2) + v0] + v0}/2 = (At/2)+ (Bt2/4) + v0 , which is not the average velocity. Page 6

Chapter 2

35. For the constant acceleration the average speed is !(v + v0), thus x = !(v + v0)t: = !(0 + 22.0 m/s)(5.00 s) =

55.0 m.

36. We find the speed of the car from v2 = v02 + 2a(x1 – x0); 0 = v02 + 2(– 7.00 m/s2)(75 m), which gives

v0 = 32 m/s.

37. We convert the units for the speed: (55 km/h)/(3.6 ks/h) = 15.3 m/s. (a) We find the distance the car travels before stopping from v2 = v02 + 2a(x1 – x0); 0 = (15.3 m/s)2 + 2(– 0.50 m/s2)(x1 – x0), which gives x1 – x0 = 2.3102 m. (b) We find the time it takes to stop the car from v = v0 + at ; 0 = 15.3 m/s + (– 0.50 m/s2)t, which gives t = 31 s. (c) With the origin at the beginning of the coast, we find the position at a time t from x = v0t + !at2. Thus we find x1 = (15.3 m/s)(1.0 s) + !(– 0.50 m/s2)(1.0 s)2 = 15 m; x4 = (15.3 m/s)(4.0 s) + !(– 0.50 m/s2)(4.0 s)2 = 57 m; x5 = (15.3 m/s)(5.0 s) + !(– 0.50 m/s2)(5.0 s)2 = 70 m. During the first second the car travels 15 m – 0 = 15 m. During the fifth second the car travels 70 m – 57 m = 13 m. 38. We find the average acceleration from v2 = v02 + 2Æ(x2 – x1); 0 = [(95 km/h)/(3.6 ks/h)]2 + 2Æ(0.80 m), which gives The number of g’s is Æ = (4.4102 m/s2)/[(9.80 m/s2)/g] = 44g.

Æ = – 4.4102 m/s2.

39. We convert the units for the speed: (90 km/h)/(3.6 ks/h) = 25 m/s. With the origin at the beginning of the reaction, the location when the brakes are applied is x0 = v0t = (25 m/s)(1.0 s) = 25 m. (a) We find the location of the car after the brakes are applied from v2 = v02 + 2a1(x1 – x0); 0 = (25 m/s)2 + 2(– 4.0 m/s2)(x1 – 25 m), which gives x1 = 103 m. (b) We repeat the calculation for the new acceleration: v2 = v02 + 2a2(x2 – x0); 0 = (25 m/s)2 + 2(– 8.0 m/s2)(x2 – 25 m), which gives x2 = 64 m. 40. We find the acceleration of the space vehicle from v = v0 + at; 162 m/s = 65 m/s + a(10.0 s – 0.0 s), which gives a = 9.7 m/s2. We find the positions at the two times from x = x0 + v0t + !at2; x2 = x0 + (65 m/s)(2.0 s) + ! (9.7 m/s2)(2.0 s)2 = x0 + 149 m; Page 7

Chapter 2

x6 = x0 + (65 m/s)(6.0 s) + ! (9.7 m/s2)(6.0 s)2 = x0 + 565 m. Thus the distance moved is x6 – x2 = 565 m – 149 m = 416 m = 4.2102 m. 41. We use a coordinate system with the origin at the a v0 = 0 initial position of the front of the train. We can TRAIN find the acceleration of the train from the motion up to the point where the front of the train passes D L the worker: y=0 v12 = v02 + 2a(D – 0); 2 (25 m/s) = 0 + 2a(140 m – 0), which gives a = 2.23 m/s2. Now we consider the motion of the last car, which starts at – L, to the point where it passes the worker: v22 = v02 + 2a[D – (– L)] = 0 + 2(2.23 m/s2)(140 m + 75 m), which gives v2 = 31 m/s. 42. With the origin at the beginning of the reaction, the location when the brakes are applied is d0 = v0tR . We find the location of the car after the brakes are applied, which is the total stopping distance, from v2 = 0 = v02 + 2a(dS – d0), which gives dS = v0tR – v02/(2a). Note that a is negative. 43. (a)

We assume constant velocity of v0 through the intersection. The time to travel at this speed is t = (dS + dI)/v0 = tR – (v0/2a) + (dI/v0). (b) For the two speeds we have t1 = tR – (v01/2a) + (dI/v01) = 0.500 s – [(30.0 km/h)/(3.60 ks/h)2(– 4.00 m/s2)] + [(14.4 m)(3.60 ks/h)/(30.0 km/h)] = 3.27 s. t2 = tR – (v02/2a) + (dI/v02) = 0.500 s – [(60.0 km/h)/(3.60 ks/h)2(– 4.00 m/s2)] + [(14.4 m)(3.60 ks/h)/(60.0 km/h)] = 3.45 s. Thus the chosen time is 3.45 s.

44. We convert the units: (95 km/h)/(3.6 ks/h) = 26.4 m/s. (140 km/h/s)/(3.6 ks/h) = 38.9 m/s. We use a coordinate system with the origin where the motorist passes the police officer, as shown in the diagram. The location of the speeding motorist is given by xm = x0 + vmt = 0 + (38.9 m/s)t. The location of the police officer is given by xp = x0 + v0p(1.00 s) + v0p(t – 1.00 s) + !ap(t – 1.00 s)2

v 0p

= 0 + (26.4 m/s)t + !(2.00 m/s2)(t – 1.00 s)2. The officer will reach the speeder when these locations coincide, so we have x m = x p;

ap

vm

x=0 t=0

t = 1.00 s

(38.9 m/s)t = (26.4 m/s)t + !(2.00 m/s2)(t – 1.00 s)2. The solutions to this quadratic equation are 0.07 s and 14.4 s. Because the time must be greater than 1.00 s, the result is t = 14.4 s. 45. If the police car accelerates for 6.0 s, the time from when the speeder passed the police car is 7.0 s. From the analysis of Problem 44 we have Page 8

Chapter 2

x m = x p; vmt = v0pt + !ap(t – 1.00 s)2 ; vm(7.0 s) = (26.4 m/s)(7.0 s) + !(2.00 m/s2)(6.0 s)2, which gives vm =

32 m/s (110 km/h).

46. We find the assumed constant speed for the first 27.0 min from v0 = ?x/?t = (10,000 m – 1100 m)/(27.0 min)(60 s/min) = 5.49 m/s. The runner must cover the last 1100 m in 3.0 min (180 s). If the runner accelerates for t s, the new speed will be v = v0 + at = 5.49 m/s + (0.20 m/s2)t; and the distance covered during the acceleration will be x1 = v0t + !at2 = (5.49 m/s)t + !(0.20 m/s2)t2. The remaining distance will be run at the new speed, so we have 1100 m – x1 = v(180 s – t); or 1100 m – (5.49 m/s)t – !(0.20 m/s2)t2 = [5.49 m/s + (0.20 m/s2)t](180 s – t). This is a quadratic equation: 0.10 t2 – 36 t + 111.8 = 0, with the solutions t = + 363 s, + 3.1 s. Because we want a total time less than 3 minutes, the physical answer is t = 3.1 s. 47. We use a coordinate system with the origin at the top of the cliff and down positive. To find the time for the object to acquire the velocity, we have v = v0 + at ; (100 km/h)/(3.6 ks/h) = 0 + (9.80 m/s2)t, which gives t = 2.83 s. 48. We use a coordinate system with the origin at the top of the cliff and down positive. To find the height of the cliff, we have y = y0 + v0t + !at2 = 0 + 0 + !(9.80 m/s2)(2.75 s)2 =

37.1 m.

49. We use a coordinate system with the origin at the top of the building and down positive. (a) To find the time of fall, we have y = y0 + v0t + !at2; (b)

t = 8.81 s. 380 m= 0 + 0 + !(9.80 m/s2)t2, which gives We find the velocity just before landing from v = v0 + at = 0 + (9.80 m/s2)(8.81 s) = 86.3 m/s.

50. We use a coordinate system with the origin at the ground and up positive. (a) At the top of the motion the velocity is zero, so we find the height h from v2 = v02 + 2ah; 0 = (20 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 20 m. (b) When the ball returns to the ground, its displacement is zero, so we have y = y0 + v0t + !at2 0 = 0 + (20 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the ball starts up), and

t = 4.1 s.

51. We use a coordinate system with the origin at the ground and up positive. We can find the initial velocity from the maximum height (where the velocity is zero): v2 = v02 + 2ah; 0 = v02 + 2(– 9.80 m/s2)(2.55 m), which gives v0 = 7.07 m/s. When the kangaroo returns to the ground, its displacement is zero. For the entire jump we have Page 9

Chapter 2

y = y0 + v0t + !at2; 0 = 0 + (7.07 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the kangaroo jumps), and

t = 1.44 s.

52. We use a coordinate system with the origin at the ground and up positive. When the ball returns to the ground, its displacement is zero, so we have y = y0 + v0t + !at2; v0 = 15 m/s. 0 = 0 + v0(3.1 s) + !(– 9.80 m/s2)(3.1 s)2, which gives At the top of the motion the velocity is zero, so we find the height h from v2 = v02 + 2ah; 0 = (15 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 12 m. 53. We use a coordinate system with the origin at the ground and up positive. We assume you can throw the object 4 stories high, which is about 12 m. We can find the initial speed from the maximum height (where the velocity is zero): v2 = v02 + 2ah; 0 = v02 + 2(– 9.80 m/s2)(12 m), which gives v0 = 15 m/s.

54. We use a coordinate system with the origin at the ground and up positive. (a) We can find the initial velocity from the maximum height (where the velocity is zero): v2 = v02 + 2ah; 0 = v02 + 2(– 9.80 m/s2)(1.20 m), which gives v0 = 4.85 m/s. (b) When the player returns to the ground, the displacement is zero. For the entire jump we have y = y0 + v0t + !at2; 0 = 0 + (4.85 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the player jumps), and

t = 0.990 s.

55. We use a coordinate system with the origin at the ground and up positive. When the package returns to the ground, its displacement is zero, so we have y = y0 + v0t + !at2; 0 = 115 m + (5.60 m/s)t + !(– 9.80 m/s2)t2. The solutions of this quadratic equation are t = – 4.31 s, and t = 5.44 s. Because the package is released at t = 0, the positive answer is the physical answer: 5.44 s.

+

v0 y0

a

H y=0

56. We use a coordinate system with the origin at the release point and down positive. Because the object starts from rest, v0 = 0. The position of the object is given by y = y0 + v0t + !at2 = 0 + 0 + !gt2. The positions at one-second intervals are y0 = 0; y1 = !g(1 s)2 = (1 s2)!g; y2 = !g(2 s)2 = (4 s2)!g; y3 = !g(3 s)2 = (9 s2)!g; … . The distances traveled during each second are Page 10

Chapter 2

d1 = y1 – y0 = (1 s2)!g; d2 = y2 – y1 = (4 s2 – 1 s2)!g = (3 s2)(!g) = 3 d1; d3 = y3 – y2 = (9 s2 – 4 s2)!g = (5 s2)(!g) = 5 d1; … . 57. We use a coordinate system with the origin at the ground and up positive. Without air resistance, the acceleration is constant, so we have v2 = v02 + 2a(y – y0); v2 = v02 + 2(– 9.8 m/s2)(0 – 0) = v02, which gives v = ± v0. The two signs represent the two directions of the velocity at the ground. The magnitudes, and thus the speeds, are the same. 58. We use a coordinate system with the origin at the ground and up positive. (a) We find the velocity from v2 = v02 + 2a(y – y0); v2 = (23.0 m/s)2 + 2(– 9.8 m/s2)(12.0 m – 0), which gives v = ± 17.1 m/s. The stone reaches this height on the way up (the positive sign) and on the way down (the negative sign). (b) We find the time to reach the height from v = v0 + at; ± 17.1 m/s = 23.0 m/s + (– 9.80 m/s2)t, which gives t = 0.602 s, 4.09 s. (c) There are two answers because the stone reaches this height on the way up (t = 0.602 s) and on the way down (t = 4.09 s).

59. We use a coordinate system with the origin at the release point and down positive. On paper the apple measures 8 mm, which we will call 8 mmp. If its true diameter is 10 cm, the conversion is 0.10 m/8 mmp. The images of the apple immediately after release overlap. We will use the first clear image which is 10 mmp below the release point. The final image is 63 mmp below the release point, and there are 7 intervals between these two images. The position of the apple is given by y = y0 + v0t + !at2 = 0 + 0 + !gt2. For the two selected images we have y1 = !gt12; (10 mmp)(0.10 m/8 mmp) = !(9.8 m/s2)t12, which gives t1 = 0.159 s; y2 = !gt22; (63 mmp)(0.10 m/8 mmp) = !(9.8 m/s2)t22, which gives t2 = 0.401 s. Thus the time interval between successive images is ?t = (t2 – t1)/7 = (0.401 s – 0.159 s)/7 = 0.035 s. 60. We use a coordinate system with the origin at the ground and up positive. (a) We find the velocity when the rocket runs out of fuel from v12 = v02 + 2a(y1 – y0); v12 = 0+ 2(3.2 m/s2)(1200 m – 0), which gives v1 = 87.6 m/s = 88 m/s. (b) We find the time to reach 1200 m from v1 = v0 + at1; 87.6 m/s = 0 + (3.2 m/s2)t1, which gives t1 = 27.4 s = 27 s. (c) After the rocket runs out of fuel, the acceleration is – g. We find the maximum altitude (where the velocity is zero) from v22 = v12 + 2(– g)(h – y1); 0 = (87.6 m/s)2 + 2(– 9.80 m/s2)(h – 1200 m), which gives h = 1590 m. (d) We find the time from v2 = v1 + (– g)(t2 – t1) Page 11

Chapter 2

(e)

(f)

0 = 87.6 m/s + (– 9.80 m/s2)(t2 – 27.4 s), which gives t2 = 36 s. We consider the motion after the rocket runs out of fuel: v32 = v12 + 2(– g)(y3 – y1); v32 = (87.6 m/s)2 + 2(– 9.80 m/s2)(0 – 1200 m), which gives v3 = – 177 m/s = We find the time from v3 = v1 + (– g)(t3 – t1) – 177 m/s = 87.6 m/s + (– 9.80 m/s2)(t3 – 27.4 s), which gives t3 = 54 s.

– 1.8102 m/s.

61. We use a coordinate system with the origin at the top of the window and down positive. We can find the velocity at the top of the window from the motion past the window: y = y0 + v0t + !at2; 2.2 m = 0 + v0(0.30 s) + !(9.80 m/s2)(0.30 s)2, which gives v0 = 5.86 m/s. For the motion from the release point to the top of the window, we have v02 = vrelease2 + 2g(y0 – yrelease); (5.86 m/s)2 = 0 + 2(9.80 m/s2)(0 – yrelease), which gives yrelease = – 1.8 m. The stone was released 1.8 m above the top of the window.

v=0 a

y=0

v0 H

62. We use a coordinate system with the origin at the nozzle and up positive. For the motion of the water from the nozzle to the ground, we have y = y0 + v0t + !at2; – 1.5 m = 0 + v0(2.0 s) + !(– 9.80 m/s2)(2.0 s)2, which gives

v0 = 9.1 m/s.

63. If the height of the cliff is H, the time for the sound to travel from the ocean to the top is tsound = H/vsound. The time of fall for the rock is T – tsound. We use a coordinate system with the origin at the top of the cliff and down positive. For the falling motion we have y = y0 + v0t + !at2; H = 0 + 0 + !a(T – tsound)2 = !(9.80 m/s2)[3.4 s – H/(340 m/s)]2 . This is a quadratic equation for H: 4.2410–5 H2 – 1.098H + 56.64 = 0, with H in m; which has the solutions H = 52 m, 2.58104 m. The larger result corresponds to tsound greater than 3.4 s, so the height of the cliff is 52 m. 64. We use a coordinate system with the origin at the ground, up positive, and t = 0 when the first object is thrown. (a) For the motion of the rock we have y1 = y0 + v01t + !at2 = 0 + (12.0 m/s)t + !(– 9.80 m/s2)t2. For the motion of the ball we have y2 = y0 + v02(t – 1.00 s) + !a(t – 1.00 s)2 = 0 + (20.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2. When the two meet we have y1 = y2;

(b) (c)

(12.0 m/s)t + !(– 9.80 m/s2)t2 = (20.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2, which gives t = 1.40 s. We find the height from y1 = y0 + v01t + !at2 = 0 + (12.0 m/s)(1.40 s) + !(– 9.80 m/s2)(1.40 s)2 = 7.18 m. If we reverse the order we have Page 12

+

Chapter 2

y1 = y2; (20.0 m/s)t + !(– 9.80 m/s2)t2 = (12.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2, which gives t = 9.38 s. We find the height from y1 = y0 + v01t + !at2 = 0 + (20.0 m/s)(9.38 s) + !(– 9.80 m/s2)(9.38 s)2 = – 244 m. This means they never collide. The rock, thrown later, returns to the ground before the ball does. To confirm this we find the time when the rock strikes the ground: y2 = y0 + v02(t – 1.00 s) + !a(t – 1.00 s)2 0 = 0 + (12.0 m/s)(t – 1.00 s) + !(– 9.80 m/s2)(t – 1.00 s)2, which gives t = 3.45 s. At this time the position of the ball is y1 = y0 + v01t + !at2 = 0 + (20.0 m/s)(3.45 s) + !(– 9.80 m/s2)(3.45 s)2 = 10.7 m. 65. We use a coordinate system with the origin at the ground, up positive, with t1 the time when the rocket reaches the bottom of the window and t2 = t1 + 0.15 s the time when the rocket reaches the top of the window. A very quick burn means we can assume that the rocket has an initial velocity at the ground. The position of the rocket is given by y = v0t + !at2. For the positions at the bottom and top of the window we have 10.0 m = v0t1 + !(– 9.80 m/s2)t12; 12.0 m = v0t2 + !(– 9.80 m/s2)t22 = v0(t1 + 0.15 s) + !(– 9.80 m/s2)(t1 + 0.15 s)2. Thus we have two equations for the two unknowns: v0 and t1 . The results of combining the equations are t1 = 0.590 s (where we have taken the positive time) and v0 = 19.8 m/s. We find the maximum height from v2 = v02 + 2ah; 0 = (19.8 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 20.0 m.

66. We find the total displacement by integration:

x=

t2 t1

v dt =

t2 t1

2

25 m/ s + 18 m / s t dt = 25 m / s t 2 – t1 + 9.0 m/ s

= 25 m / s 3.5 s – 1.5 s + 9.0 m / s 2

3.5 s

2

– 1.5 s

2

=

2

2

2

t2 – t1

140 m.

67. (a) If we make the suggested change of variable, we have u = g – kv, and du = – k dv, and a = g – kv = u. Thus from the definition of acceleration, we have a = dv/dt; u = – du/k dt, or du/u = – k dt. When we integrate, we get g – kv g

d u = t – k d t , which gives ln g – kv = – kt, or g – kv = g e – kt. u g 0

Thus the velocity as a function of time is v = (g/k)(1 – e – kt). (b) When the falling body reaches its terminal velocity, the acceleration will be zero, so we have a = g – kvterm = 0, or vterm = g/k.

Note that this is the terminal velocity from the velocity expression because as t  8, e – kt  0. Page 13

Chapter 2

68. (a) We find the speed by integration: v v0

dv =

t 0

a dt =

t 0

At 1/ 2 d t, wh i chgi ves

v – v 0 = 23 At 3/ 2 – 0, or v = v 0 + 23 At 3/ 2 = (b) We find the displacement by integration: x 0

dx =

t

v dt =

0 2 2 3 5

5/ 2

t 0

10 m/ s +

2 3

2.0 m / s 5/ 2 t 3/ 2 .

v 0 + 23 At 3/ 2 dt, wh i chgi ves 5/ 2

4 4 x = v 0t + A t , or x = v 0 t + 15 At = 10 m / s t + 15 2.0 m / s 5/ 2 t (c) For the given time we have a = (2.0 m/s5/2)(5.0 s)1/2 = 4.5 m/s2; 5/2 v = (10 m/s) + %(2.0 m/s )(5.0 s)3/2 = 25 m/s; x = (10 m/s)(5.0 s) + (4/15)(2.0 m/s5/2)(5.0 s)5/2 = 80 m.

5/ 2

.

69. The height reached is determined by the initial velocity. We assume the same initial velocity of the object on the Moon and Earth. With a vertical velocity of 0 at the highest point, we have v2 = v02 + 2ah; 0 = v02 + 2(– g)h, so we get v02 = 2gEarthhEarth = 2gMoonhMoon , or hMoon = (gEarth/gMoon)hEarth = 6hEarth.

70. For the falling motion, we use a coordinate system with the origin y 01 = 0 at the fourth-story window and down positive. For the stopping motion in the net, we use a coordinate system with the origin at the original position of the net and down positive. (a) We find the velocity of the person at the unstretched net + (which is the initial velocity for the stretching of the net) H from the free fall: v022 = v012 + 2a1(y1 – y01) = 0 + 2(9.80 m/s2)(15.0 m – 0), which gives v02 = 17.1 m/s. We find the acceleration during the stretching of the net from v22 = v022 + 2a2(y2 – y02); y 02 = 0 0 = (17.1 m/s)2 + 2a2(1.0 m – 0), which gives a2 = – 1.5102 m/s2. (b) To produce the same velocity change with a smaller acceleration requires a greater displacement. Thus the net should be loosened.

v 01 = 0

a1

a2

v 02 = v 1

71. We assume that the seat belt keeps the occupant fixed with respect to the car. The distance the occupant moves with respect to the front end is the distance the front end collapses, so we have v2 = v02 + 2a(x – x0); Page 14

Chapter 2

0 = [(100 km/h)/(3.6 ks/h)]2 + 2(– 30)(9.80 m/s2)(x – 0), which gives x = 72. If the lap distance is D, the time for the first 9 laps is t1 = 9D/(199 km/h), the time for the last lap is t2 = D/æ, and the time for the entire trial is T = 10D/(200 km/h). Thus we have T = t1 + t2 ; 10D/(200 km/h) = 9D/(199 km/h) + D/æ, which gives

1.3 m.

æ = 209.5 km/h.

73. We use a coordinate system with the origin at the release point and down positive. (a) The speed at the end of the fall is found from v2 = v02 + 2a(x – x0) = 0 + 2g(H– 0), which gives v = (2gH)1/2. (b) To achieve a speed of 50 km/h, we have v = (2gH)1/2; (50 km/h)/(3.6 ks/h) = [2(9.80 m/s2)H50]1/2, which gives H50 = 9.8 m. (c) To achieve a speed of 100 km/h, we have v = (2gH)1/2; (100 km/h)/(3.6 ks/h) = [2(9.80 m/s2)H100]1/2, which gives H100 = 39 m.

Page 15

30 A B

25 20

x (m)

74. For the motion from A to B, (a) The object is moving in the negative direction. The slope (the instantaneous velocity) is negative; the x-value is decreasing. (b) Because the slope is becoming more negative (greater magnitude of the velocity), the object is speeding up. (c) Because the velocity is becoming more negative, the acceleration is negative. For the motion from D to E, (d) The object is moving in the positive direction. The slope (the instantaneous velocity) is positive; the x-value is increasing. (e) Because the slope is becoming more positive (greater magnitude of the velocity), the object is speeding up. (f) Because the velocity is becoming more positive, the acceleration is positive. (g) The position is constant, so the object is not moving, the velocity and the acceleration are zero.

E

15 10 5 0

D

C

0

1

2

3

t (s)

4

5

6

Chapter 2

75. (a)

(b)

(c)

At the top of the motion the velocity is zero, so we find the maximum height of the second child from v2 = v022 + 2ah2; 0 = (5.0 m/s)2 + 2(– 9.80 m/s2)h2 , which gives h2 = 1.28 m = 1.3 m. If the first child reaches a height h1 = 1.5h2 , we find the initial speed from v2 = v012 + 2ah1 = v012 + 2a(1.5h2) = v012 – 1.5v022 = 0; v012 = (1.5)(5.0 m/s)2, which gives v01 = 6.1 m/s. We find the time for the first child from y = y0 + v0t + !at2 0 = 0 + (6.1 m/s)t + !(– 9.80 m/s2)t2, which gives t = 0 (when the child starts up), and

t = 1.2 s.

76. We use a coordinate system with the origin at the a initial position of the front of the train. We can v0 = 0 find the acceleration of the train from the motion TRAIN up to the point where the front of the train passes D the worker: L v12 = v02 + 2a(D – 0); y=0 (20 m/s)2 = 0 + 2a(180 m – 0), which gives a = 1.11 m/s2. Now we consider the motion of the last car, which starts at – L, to the point where it passes the worker: v22 = v02 + 2a[D – (– L)] = 0 + 2(1.11 m/s2)(180 m + 90 m), which gives v2 = 24 m/s.

77. We use a coordinate system with the origin at the roof of the building and down positive, and call the height of the building H. (a) For the first stone, we have y1 = y01 + v01t1 + !at12; H = 0 + 0 + !(g)t12, or H = !gt12. For the second stone, we have y2 = y02 + v02t2 + !at22;

t1

+

= (30.0 m/s)(t1 – 2.00 s) + !(g)(t1 – 2.00 s)2 = (30.0 m/s)t1 – 60.0 m + !gt12 – (2.00 s)gt1 + (2.00 s2)g. When we eliminate H from the two equations, we get 0 = (30.0 m/s)t1 – 60.0 m – (2.00 s)gt1 + (2.00 s2)g, which gives t1 = 3.88 s. We use the motion of the first stone to find the height of the building: Page 16

v 02

y = 0 v 01 = 0

H = 0 + (30.0 m/s)t2 + !(g)t22

(b)

t2 = t1 – 2.00 s

a

H

Chapter 2

78. We convert the units: (110 km/h)/(3.6 ks/h) = 30.6 m/s. We use a coordinate system with the origin where the motorist passes the police officer, as shown in the diagram. (b) The location of the speeding motorist is given by xm = x0 + vmt, which we use to find the time required: 700 m = (30.6 m/s)t, which gives t = 22.9 s. (c) The location of the police car is given by xp = x0 + v0pt + !apt2 = 0 + 0 + !apt2, which we use to find the acceleration: ap = 2.67 m/s2. 700 m = !ap(22.9 s)2, which gives (d) We find the speed of the officer from vp = v0p + apt; = 0 + (2.67 m/s2)(22.9 s) = 61.1 m/s = 220 km/h

48.4 m/s. (a)

x

(c)

H = !gt12 = !(9.80 m/s2)(3.88 s)2 = 73.9 m. We find the speeds from v1 = v01 + at1 = 0 + (9.80 m/s2)(3.88 s) = 38.0 m/s; v2 = v02 + at2 = 30.0 m/s + (9.80 m/s2)(3.88 s – 2.00 s) =

Speeder

Police officer

t

(about 135 mi/h!).

79. We convert the maximum speed units: vmax = (90 km/h)/(3.6 ks/h) = 25 m/s. (a) There are (36 km)/0.80 km) = 45 trip segments, which means 46 stations (with 44 intermediate stations). In each segment there are three motions. a3 v max a1 Motion 1 is the acceleration to vmax. Station We find the time for this motion from vmax = v01 + a1t1; L1 L2 L3 25 m/s = 0 + (1.1 m/s2)t1 , which gives t1 = 22.7 s. x = 0 x1 = 0 x2 = 0 3 We find the distance for this motion from 2 x1 = x01 + v01t + !a1t1 ; L1 = 0 + 0 + !(1.1 m/s2)(22.7 s)2 = 284 m. Motion 2 is the constant speed of vmax , for which we have Page 17

t1

Station

Chapter 2

x2 = x02 + vmaxt2 ; L2 = 0 + vmaxt2 . Motion 3 is the acceleration from vmax to 0. We find the time for this motion from 0 = vmax + a3t3; 0 = 25 m/s + (– 2.0 m/s2)t3 , which gives t3 = 12.5 s. We find the distance for this motion from x3 = x03 + vmaxt + !a3t32 ;

(b)

L3 = 0 + (25 m/s)(12.5 s) + !(– 2.0 m/s2)(12.5 s)2 = 156 m. The distance for Motion 2 is L2 = 800 m – L1 – L3 = 800 m – 284 m – 156 m = 360 m, so the time for Motion 2 is t2 = L2/vmax = (360 m)/(25 m/s) = 14.4 s. Thus the total time for the 45 segments and 44 stops is T = 45(t1 + t2 + t3) + 44(20 s) = 45(22.7 s + 14.4 s + 12.5 s) + 44(20 s) = 3112 s = 52 min. There are (36 km)/3.0 km) = 12 trip segments, which means 13 stations (with 11 intermediate stations.) The results for Motion 1 and Motion 3 are the same: t1 = 22.7 s, L1 = 284 m, t3 = 12.5 s, L3 = 156 m. The distance for Motion 2 is L2 = 3000 m – L1 – L3 = 3000 m – 284 m – 156 m = 2560 m, so the time for Motion 2 is t2 = L2/vmax = (2560 m)/(25 m/s) = 102 s. Thus the total time for the 12 segments and 11 stops is T = 12(t1 + t2 + t3) + 11(20 s) = 12(22.7 s + 102 s + 12.5 s) + 11(20 s) = 1870 s = 31 min. This means there is a higher average speed for stations farther apart.

80. d2

d3

d4

a2

v1

d1

x=0

Page 18

d5

d6

Chapter 2

We convert the units for the speed limit: (50 km/h)/(3.6 ks/h) = 13.9 m/s. (a) If we assume that we are traveling at the speed limit, the time to pass through the farthest intersection is t1 = (d1 + d2 + d3 + d4 + d5 + d6)/v1 = (10 m + 15 m + 50 m + 15 m + 70 m + 15 m)/(13.9 m/s) = 12.6 s. Because this is less than the time while the lights are green, yes, you can make it through. (b) We find the time required for the second car to reach the speed limit: vmax = v02 + a2t2a; 13.9 m/s = 0 + (2.0 m/s2)t2a , which gives t2a = 6.95 s. In this time the second car will have traveled x2a = v02t1 + !a2t2a2 = 0 + !(2.0 m/s2)(6.95 s)2 = 48 m. The time to travel the remaining distance at constant speed is t2b = (d2 + d3 + d4 + d5 + d6 – x2a)/vmax = (15 m + 50 m + 15 m + 70 m + 15 m – 48 m)/(13.9 m/s) = 8.42 s. Thus the total time is ttotal = t2a + t2b = 6.95 s + 8.42 s = 15.4 s. No, the second car will not clear all the lights. 81. We use a coordinate system with the origin at the ground and up positive. (a) We find the initial speed from the motion to the window: v12 = v02 + 2a(y1 – y0); (14 m/s)2 = v02 + 2(– 9.80 m/s2)(25 m – 0), which gives v0 = 26 m/s. (b) We find the maximum altitude from v22 = v02 + 2a(y2 – y0); 0 = (26.2 m/s)2 + 2(– 9.80 m/s2)(y2 – 0), which gives y2 = 35 m. (c) We find the time from the motion to the window: v1 = v0 + at1 14 m/s = 26.2 m/s + (– 9.80 m/s2)t1 , which gives t1 = 1.2 s. Thus it was thrown 1.2 s before passing the window. (d) We find the time to reach the street from y = y0 + v0t + !at2;

+

y2

v2 = 0

y1

v1 H

y=0

0 = 0 + (26.2 m/s)t + !(– 9.80 m/s2)t2. This is a quadratic equation for t, which has the solutions t = 0 (the initial throw), 5.3 s. Thus the time after the baseball passed the window is 5.3 s – 1.2 s = 4.1 s.

82. (a)

We find the time required for the fugitive to reach his maximum speed: vmax = v0f + aftf1; 8.0 m/s = 0 + (4.0 m/s2)tf1 , which gives tf1 = 2.0 s. In this time the fugitive will have traveled xf1 = v0ftf1 + !aftf12 = 0 + !(4.0 m/s2)(2.0 s)2 = 8.0 m. From this time the fugitive will run at constant speed. When he reaches the box car, we have xtrain = xf ; vtraint = xf1 + vmax(t – tf1); Page 19

v0

a

Chapter 2

(b)

(6.0 m/s)t = 8.0 m + (8.0 m/s)(t – 2.0 s), which gives t = We can find the distance from the motion of the train: xtrain = vtraint = (6.0 m/s)(4.0 s) = 24 m.

4.0 s.

83. We use a coordinate system with the origin at the top of the cliff and up positive. (a) For the motion of the stone from the top of the cliff to the ground, we have y = y0 + v0t + !at2;

(b)

(c)

– 65.0 m = 0 + (10.0 m/s)t + !(– 9.80 m/s2)t2. This is a quadratic equation for t, which has the solutions t = – 2.76 s, 4.80 s. Because the stone starts at t = 0, the time is 4.80 s. We find the speed from v = v0 + at = 10.0 m/s + (– 9.80 m/s2)(4.80 s) = – 37.0 m/s. The negative sign indicates the downward direction, so the speed is 37.0 m/s. The total distance includes the distance up to the maximum height, down to the top of the cliff, and down to the bottom. We find the maximum height from v2 = v02 + 2ah; 0 = (10.0 m/s)2 + 2(– 9.80 m/s2)h, which gives h = 5.10 m. The total distance traveled is d = 5.10 m + 5.10 m + 65.0 m = 75.2 m.

84. The instantaneous velocity is the slope of the x vs. t graph:

1.5

v (m/s)

20

x (m)

2.0

10

1.0 0.5

0

0

10

20

30

40

50

0

t (s) –0.5 –1.0 –1.5 –2.0

85. We use a coordinate system with the origin where the

Page 20

10

20

30

t (s)

40

50

Chapter 2

initial action takes place, as shown in the diagram. x=0 The initial speed is (50 km/h)/(3.6 ks/h) = 13.9 m/s. t=0 If she decides to stop, we find the minimum stopping distance v0 from v12 = v02 + 2a1(x1 – x0); L1 L2 2 2 0 = (13.9 m/s) + 2(– 6.0 m/s )x1 , which gives x1 = 16 m. v=0 x1 a1 Because this is less than L1 , the distance to the intersection, she can safely stop in time. x2 a2 If she decides to increase her speed, we find the acceleration from the time to go from 50 km/h to 70 km/h (19.4 m/s): v = v0 + a2t ; 19.4 m/s = 13.9 m/s + a2(6.0 s), which gives a2 = 0.917 m/s2. We find her location when the light turns red from x2 = x0 + v0t2 + !a2t22 = 0 + (13.9 m/s)(2.0 s) + !(0.917 m/s2)(2.0 s)2 = 30 m. Because this is L1 , she is at the beginning of the intersection, but moving at high speed. She should decide to stop! 86. We use a coordinate system with the origin at the water and up positive. We find the time for the pelican to reach the water from y1 = y0 + v0t + !at12; 0 = 16.0 m + 0 + ! (– 9.80 m/s2)t2, which gives t1 = 1.81 s. This means that the fish must spot the pelican 1.81 s – 0.20 s = 1.61 s after the pelican starts its dive. We find the height of the pelican at this time from y2 = y0 + v0t + !at22; = 16.0 + 0 + !(– 9.80 m/s2)(1.61 s)2 =

3.3 m.

87. In each case we use a coordinate system with the origin at the beginning of the putt and the positive direction in the direction of the putt. The limits on the putting distance are 6.0 m < x < 8.0 m. For the downhill putt we have: v2 = v0down2 + 2adown(x – x0); 0 = v0down2 + 2(– 2.0 m/s2)x. When we use the limits for x, we get 4.9 m/s < v0down < 5.7 m/s, or ?v0down = 0.8 m/s. For the uphill putt we have: v2 = v0up2 + 2aup(x – x0); 0 = v0up2 + 2(– 3.0 m/s2)x. When we use the limits for x, we get 6.0 m/s < v0up < 6.9 m/s, or ?v0up = 0.9 m/s. The smaller spread in allowable initial velocities makes the downhill putt more difficult.

Page 21

Chapter 2

88. We use a coordinate system with the origin at the initial position of the car. The passing car’s position is given by L x1 = x01 + v0t + !a1t2 = 0 + v0t + !a1t2. v0 v0 The truck’s position is given by 1 xtruck = x0truck + v0t = D + v0t. The oncoming car’s position is given by D x2 = x02 – v0t = L – v0t. x=0 For the car to be safely past the truck, we must have x1 – xtruck = 10 m;

v0 2

v0t + !a1t2 – (D + v0t) = !a1t2 – D = 10 m, which allows us to find the time required for passing: !(1.0 m/s2)t2 – 30 m = 10 m, which gives t = 8.94 s. At this time the car’s location will be x1 = v0t + !a1t2 = (25 m/s)(8.94 s) + !(1.0 m/s2)(8.94 s)2 = 264 m from the origin. At this time the oncoming car’s location will be x2 = L – v0t = 400 m – (25 m/s)(8.94 s) = 176 m from the origin. Because this is closer to the origin, the two cars will have collided, so the passing attempt should not be made. 89. We use a coordinate system with the origin at the roof of the building and down positive. We find the time of fall for the second stone from v2 = v02 + at2 ; 12.0 m/s = 0 + (9.80 m/s2)t2 , which gives t2 = 1.22 s. During this time, the second stone fell y2 = y02 + v02t2 + !at22 = 0 + 0 + !(9.80 m/s2)(1.22 s)2 = 7.29 m. The time of fall for the first stone is t1 = t2 + 1.50 s = 1.22 s + 1.50 s = 2.72 s. During this time, the first stone fell y1 = y01 + v01t1 + !at12 = 0 + 0 + !(9.80 m/s2)(2.72 s)2 = 36.3 m. Thus the distance between the two stones is y1 – y2 = 36.3 m – 7.29 m = 29.0 m. 90. For the vertical motion of James Bond we use a coordinate system with the origin at the ground and up positive. We can find the time for his fall to the level of the truck bed from y = y0 + v0t + !at2;

y0 v truck

1.5 m = 10 m + 0 + !(– 9.80 m/s2)t2, which gives t = 1.32 s. D During this time the distance the truck will travel is x = x0 + vtruckt = 0 + (30 m/s)(1.32 s) = 39.6 m. Because the poles are 20 m apart, he should jump when the truck is there is a pole at the bridge.

Page 22

y=0

2 poles

away, assuming that

Chapter 3

CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1.

2.

We choose the west and south coordinate system shown. D 2x For the components of the resultant we have W RW = D1 + D2 cos 45° D 2y D 2 = (200 km) + (80 km) cos 45° = 257 km; RS = 0 + D2 = 0 + (80 km) sin 45° = 57 km. We find the resultant displacement from R = (RW2 + RS2)1/2 = [(257 km)2 + (57 km)2]1/2 = 263 km; tan  = RS/RW = (57 km)/(257 km) = 0.222, which gives  = 13° S of W. We choose the north and east coordinate system shown. For the components of the resultant we have RE = D2 = 10 blocks; RN = D1 – D3 = 18 blocks – 16 blocks = 2 blocks. We find the resultant displacement from R = (RE2 + RN2)1/2 = [(10 blocks)2 + (2 blocks)2]1/2 = 10 blocks; tan  = RN/RE = (2 blocks)/(10 blocks) = 0.20, which gives  = 11° N of E.

3.

From Fig. 3–6c, if we write the equivalent vector addition, we have V1 + Vwrong = V2 , or Vwrong = V2 – V1.

4.

We find the vector from V = (Vx2 + Vy2)1/2 = [(8.80)2 + (– 6.40)2]1/2 = 10.9; tan  = Vy/Vx = (– 6.40)/(8.80) = – 0.727, which gives  = 36.0° below the x-axis.

The resultant is

13.6 m, 18° N of E. North

V2

V1 V3

VR



East

Page 1



Ry

R Rx S

N D2 D1 D3



R

E

y Vx

x

 V

5.

D1

Vy

Chapter 3

6.

(a) V1x = – 6.0, V1y = 0; y V2x = V2 cos 45° = 4.5 cos 45° = 3.18 = 3.2, R V2y = V2 sin 45° = 4.5 sin 45° = 3.18 = 3.2. V2 (b) For the components of the sum we have   Rx = V1x + V2x = – 6.0 + 3.18 = – 2.82; x V1 Ry = V1y + V2y = 0 + 3.18 = 3.18. We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 2.82)2 + (3.18)2]1/2 = 4.3; tan  = Ry/Rx = (3.18)/(2.82) = 1.13, which gives  = 48° above – x-axis. Note that we have used the magnitude of Rx for the angle indicated on the diagram.

7.

(a) y (b) For the components of the vector we have Vx = – V cos  = – 14.3 cos 34.8° = – 11.7; Vy = V sin  = 14.3 sin 34.8° = 8.16. V (c) We find the vector from Vy 2 2 1/2 2 2 1/2 V = (Vx + Vy ) = [(– 11.7) + (8.16) ]  = 14.3; x tan  = Vy/Vx = (8.16)/(11.7) = 1.42, which gives Vx  = 34.9° above – x-axis. This is within significant figures. Note that we have used the magnitude of Vx for the angle indicated on the diagram.

8.

Because the vectors are parallel, the direction can be indicated by the sign. (a) C = A + B = 6.8 + (– 5.5) = 1.3 in the + x-direction.

(a) C = A + B C

B

x

A (b) C = A – B

(b) C = A – B = 6.8 – (– 5.5) = 12.3 in the + x-direction.

C A

(c) C = B – A = – 5.5 – (6.8) = – 12.3 in the + x-direction or

(c)

C=B–A

12.3 in the – x-direction.

C –A

9.

(a) Using the given angle, we find the components from VN = V cos 38.5° = (635 km/h) cos 41.5° = 476 km/h; VW = V sin 38.5° = (635 km/h) sin 41.5° = 421 km/h. (b) We use the velocity components to find the displacement components: dN = VN t = (476 km/h)(3.00 h) = 1.43103 km; dW = VW t = (421 km/h)(3.00 h) = 1.26103 km.

Page 2

x –B

B

x

Chapter 3

10. The vectors are V1 = – 6.0i + 8.0j, V2 = 4.5i – 5.0j. (a) For the magnitude of V1 we have  V1 = (V1x2 + V1y2)1/2 = [(– 6.0)2 + (8.0)2]1/2 = 10.0. We find the direction from tan 1 = V1y/V1x = (8.0)/(– 6.0) = – 1.33. From the signs of the components, we have 1 = 53° above – x-axis. (b) For the magnitude of V2 we have  V2 = (V2x2 + V2y2)1/2 = [(4.5)2 + (– 5.0)2]1/2 = 6.7. We find the direction from tan 2 = V2y/V2x = (– 5.0)/(4.5) = – 1.11. From the signs of the components, we have 2 = 48° below + x-axis. (c) For the sum V1 + V2 we have V1 + V2 = – 1.5i + 3.0j. For the magnitude of V1 + V2 we have  V1 + V2 = [(– 1.5)2 + (3.0 )2]1/2 = 3.4. We find the direction from tan 1+2 = (3.0 )/(– 1.5) = – 2.0. From the signs of the components, we have 1+2 = 63° above – x-axis. (d) For the difference V2 – V1 we have V2 – V1 = 10.5i – 13.0j. For the magnitude of V1 + V2 we have  V2 – V1 = [(10.5)2 + (– 13.0 )2]1/2 = 16.7. We find the direction from tan 2–1 = (– 13.0 )/(10.5) = – 1.24. From the signs of the components, we have 2–1 = 51° below + x-axis. 







11. The vectors are V1 = 4i – 8j, V2 = i + j, V3 = – 2i + 4j. (a) For the sum V1 + V2 + V3 we have V1 + V2 + V3 = 3i – 3j. For the magnitude of V1 + V2 + V3 we have  V1 + V2 + V3 = [(3)2 + (– 3 )2]1/2 = 4.2. We find the direction from tan a = (– 3)/(3) = – 1.0. a = 45° below + x-axis. From the signs of the components, we have (b) For V1 – V2 + V3 we have V1 – V2 + V3 = i – 5j. For the magnitude of V1 – V2 + V3 we have  V1 – V2 + V3 = [(1)2 + (– 5 )2]1/2 = 5.1. We find the direction from tan b = (– 5)/(1) = – 5.0. b = 79° below + x-axis. From the signs of the components, we have 



Page 3

Chapter 3

12. (a) For the components we have Rx = Ax + Bx + Cx = 44.0 cos 28.0° – 26.5 cos 56.0° + 0 = 24.0; Ry = Ay + By + Cy = 44.0 sin 28.0° + 26.5 sin 56.0° – 31.0 = 11.6. (b) We find the resultant from R = (Rx2 + Ry2)1/2 = [(24.0)2 + (11.6)2]1/2 = 26.7; tan  = Ry/Rx = (11.6)/(24.0) = 0.483, which gives  = 25.8° above + x-axis.

y R

 

C

B

x

A

 13. (a) For the components we have y Rx = Bx – Ax –A B = – 26.5 cos 56.0° – 44.0 cos 28.0° = – 53.7; Ry = By – Ay R = 26.5 sin 56.0° – 44.0 sin 28.0° = 1.3.  We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 53.7)2 + (1.3)2]1/2 = 53.7; tan  = Ry/Rx = (1.3)/(53.7) = 0.0245, which gives  = 1.40° above – x-axis. Note that we have used the magnitude of Rx for the angle indicated on the diagram. (b) For the components we have y Rx = Ax – Bx A = 44.0 cos 28.0° – (– 26.5 cos 56.0°) = 53.7; –B Ry = Ay – By = 44.0 sin 28.0° – 26.5 sin 56.0° = – 1.3. We find the resultant from  R R = (Rx2 + Ry2)1/2 = [(53.7)2 + (– 1.3)2]1/2 = 53.7; tan  = Ry/Rx = (1.3)/(53.7) = 0.0245, which gives  = 1.40° below + x-axis, which is opposite to the result from (a).

Page 4

x

x

Chapter 3

14. (a) For the components we have Rx = Ax – Bx + Cx = 44.0 cos 28.0° – (– 26.5 cos 56.0°) + 0 = 53.7; Ry = Ay – By + Cy = 44.0 sin 28.0° – 26.5 sin 56.0° – 31.0 = – 32.3. We find the resultant from R = (Rx2 + Ry2)1/2 = [(53.7)2 + (– 32.3)2]1/2 = 62.7; tan  = Ry/Rx = (32.3)/(53.7) = 0.602, which gives  = 31.0° below + x-axis. Note that we have used the magnitude of Ry for the angle indicated on the diagram. (b) For the components we have Rx = Ax + Bx – Cx = 44.0 cos 28.0° + (– 26.5 cos 56.0°) – 0 = 24.0; Ry = Ay + By – Cy = 44.0 sin 28.0° + 26.5 sin 56.0° – (– 31.0) = 73.6. We find the resultant from R = (Rx2 + Ry2)1/2 = [(24.0)2 + (73.6)2]1/2 = 77.4; tan  = Ry/Rx = (73.6)/(24.0) = 3.07, which gives  = 71.9° above + x-axis. (c) For the components we have Rx = Cx – Ax – Bx = 0 – 44.0 cos 28.0° – (– 26.5 cos 56.0°) = – 24.0; Ry = Cy – Ay – By = – 31.0 – 44.0 sin 28.0° – 26.5 sin 56.0° = – 73.6. We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 24.0)2 + (– 73.6)2]1/2 = 77.4; tan  = Ry/Rx = (73.6)/(24.0) = 3.07, which gives  = 71.9° below – x-axis. Page 5

y A

–B





x

 C R

y

–C

R

B

 

A



x

y x

 –A –B 



R

C

Chapter 3

15. (a) For the components we have Rx = Bx – 2Ax = – 26.5 cos 56.0° – 2(44.0 cos 28.0° ) = – 92.5; Ry = By – 2Ay = 26.5 sin 56.0° – 2(44.0 sin 28.0°) = – 19.3. We find the resultant from R = (Rx2 + Ry2)1/2 = [(– 92.5)2 + (– 19.3)2]1/2 = 94.5; tan  = Ry/Rx = (19.3)/(92.5) = 0.209, which gives  = 11.8° below – x-axis. (b) For the components we have Rx = 2Ax – 3Bx + 2Cx = 2(44.0 cos 28.0°) – 3(– 26.5 cos 56.0°) + 2(0) = 122.2; Ry = 2Ay – 3By + 2Cy = 2(44.0 sin 28.0° ) – 3(26.5 sin 56.0°) + 2(– 31.0) = – 86.6. We find the resultant from R = (Rx2 + Ry2)1/2 = [(122.2)2 + (– 86.6)2]1/2 = 150; tan  = Ry/Rx = (86.6)/(122.2) = 0.709, which gives  = 35.3° below + x-axis.

Page 6

y –A

B



Chapter 3

R

z V

y

A

D

y North

H

–B

 A –B

 

x

x

East

17. (a) We find the x-component from A2 = Ax2 + Ay2; y R Ax = ± 82.9. (90.0)2 = Ax2 + (– 35.0)2; which gives (b) If we call the new vector B, we have Rx = Ax + Bx ; Ay A2 – 80.0 = + 82.9 + Bx , which gives Bx = – 162.9; Ry = Ay + By ; 0 = – 35.0 + By , which gives By = + 35.0. We find the resultant from B = (Bx2 + By2)1/2 = [(– 162.9)2 + (+ 35.0)2]1/2 = 166.6;  = 12.1° above – x-axis. tan  = By/Bx = (35.0)/(162.9) = 0.215, which gives



–B C

x C A1

18. We find the velocity and acceleration by differentiating: r = (7.60 m/s)ti + (8.85 m)j – (1.00 m/s2)t2k; v = dr/dt = (7.60 m/s)i – (2.00 m/s2)tk; a = dv/dt = – (2.00 m/s2)k. 19. The positions of the particle at the two times are r1 = (7.60 m/s)(1.00 s)i + (8.85 m)j – (1.00 m/s2)(1.00 s)2k = (7.60 m)i + (8.85 m)j – (1.00 m)k; r3 = (7.60 m/s)(3.00 s)i + (8.85 m)j – (1.00 m/s2)(3.00 s)2k = (22.8 m)i + (8.85 m)j – (9.00 m)k. The average velocity is vav = ?r/?t = [(15.2 m)i – (8.00 m)k]/(3.00 s – 1.00 s) = (7.60 m/s)i – (4.00 m/s)k. The instantaneous velocity at the midpoint of the interval is v2 = (7.60 m/s)i – (2.00 m/s2)(2.00 s)k = (7.60 m/s)i – (4.00 m/s)k. Note that this is the same as the average velocity because the acceleration is constant. The magnitude is v2 = [(7.60 m/s)2 + (4.00 m/s)2]1/2 = 8.59 m/s. 20. The y-position is constant. The x- and z-components are x = (7.60 m/s)t, and z = – (1.00 m/s2)t2. If we eliminate t, we get z = – (1.00 m)x2/(7.60 m)2, which is a parabola. Thus the path is parabola centered on the – z-axis at y = 8.85 m. 21. (a) Because we do not know the displacement over the given time interval, the average velocity is unknown. (b) The average acceleration is aav = ?v/?t = [(27.5 m/s)i – (– 18.0 m/s)j]/(8.00 s) = (3.44 m/s2)i + (2.25 m/s2)k. The magnitude is Page 7

x



–A

16. For the components we have Dx = Hx = – 4580 sin 32.4° = – 2454 m; Dy = Hy = + 4580 cos 32.4° = + 3867 m; Dz = V = + 2450 m. By extending the Pythagorean theorem, we find the magnitude from D = (Dx2 + Dy2 + Dz2)1/2 = [(– 2454 m)2 + (3867 m)2 + (2450 m)2]1/2 = 5194 m.



Chapter 3

 aav = [(3.44 m/s2)2 + (2.25 m/s2)2]1/2 = 4.11 m/s2. We find the direction from tan  = (2.25 m/s2)/(3.44 m/s2) = 0.654, which gives  = 33.2° north of east. (c) Because we do not know the distance traveled, the average speed is unknown. 

22. (a) For the vertical component we have aV = (3.80 m/s2) sin 30.0° = 1.90 m/s2 down. (b) Because the elevation change is the vertical displacement, we find the time from the vertical motion, taking down as the positive direction: y = v0yt + !aVt2; 250 m = 0 + !(1.90 m/s2)t2, which gives

t = 16.2 s.

23. The acceleration is a = (4.0 m/s2)i + (3.0 m/s2)j. (a) We find the velocity by integrating: v

dv =

0

v=

t 0

t 0

a dt ;

(4.0 m / s 2)i + (3.0 m/ s 2)j dt =

(4.0 m / s 2)t i + (3.0 m / s 2)tj.

(b) The speed of the particle is  v = {[(4.0 m/s2)t]2 + [(3.0 m/s2)t]2}1/2 = (c) We find the position by integrating: 

r 0

dr =

r=

t 0

t 0

(5.0 m/s2)t.

v dt; 2

2

2

2

2

2

(4.0 m / s )ti + (3.0 m/ s )t dt = 12 (4.0 m / s )t i + 12 (3.0 m / s )t j 2

2

= (2.0 m / s2 )t i + (1.5 m / s 2)t j. (d) For the given time we have v = (4.0 m/s2)ti + (3.0 m/s2)tj = (4.0 m/s2)(2.0 s)i + (3.0 m/s2)(2.0 s)j = (8.0 m/s)i + (6.0 m/s)j.  v = (5.0 m/s2)t = (5.0 m/s2)(2.0 s) = 10.0 m/s. r = (2.0 m/s2)t2i + (1.5 m/s2)t2j = (2.0 m/s2)(2.0 s)2i + (1.5 m/s2)(2.0 s)2j = (8.0 m)i + (6.0 m)j. 

Page 8

Chapter 3

24. The acceleration is a = (– 3.0 m/s2)i + (4.5 m/s2)j. We find the velocity by integrating: v v0

t

dv =

0

a dt ;

v – (5.0 m/ s )i =

t 0

(– 3.0 m / s 2)i + (4.5 m/ s 2)j dt = (– 3.0 m/ s 2)ti + (4.5 m / s 2)t j, or

v = [(– 3.0 m/s2)t + 5.0 m/s]i + (4.5 m/s2)tj. We find the position by integrating: r 0

dr =

t 0

v d t;

t

(– 3.0 m/ s 2)t + 5.0 m/ s i + (4.5 m / s 2)t d t =

r=

1 2 (–

2

0 2

2

= (– 1.5 m/ s 2)t + (5.0 m/ s)t i + (2.25 m / s2 )t j. To find the time at which the particle reaches its maximum x-coordinate, we set dx/dt = 0: dx/dt = (– 3.0 m/s2)t + 5.0 m/s = 0, which gives t = 1.67 s. The velocity is v = [(– 3.0 m/s2)t + 5.0 m/s]i + (4.5 m/s2)tj = 0i + (4.5 m/s2)(1.67 s)j = (7.5 m/s)j. The position is r = [(– 1.5 m/s2)t2 + (5.0 m/s)t]i + (2.25 m/s2)t2j = [(– 1.5 m/s2)(1.67 s)2 + (5.0 m/s)(1.67 s)]i + (2.25 m/s2)(1.67 s)2j = (4.2 m)i + (6.3 m)j.

25. The position is r = (6.0 m) cos (3.0 s–1)ti + (6.0 m) sin (3.0 s–1)tj. (a) We find the velocity by differentiating: v = dr/dt = – (6.0 m)(3.0 s–1) sin (3.0 s–1)ti + (6.0 m)(3.0 s–1) cos (3.0 s–1)tj = – (18.0 m/s) sin (3.0 s–1)ti + (18.0 m/s) cos (3.0 s–1)tj. (b) We find the acceleration by differentiating: a = dv/dt = – (18.0 m/s)(3.0 s–1) cos (3.0 s–1)ti – (18.0 m/s)(3.0 s–1) sin (3.0 s–1)tj = – (54.0 m/s2) cos (3.0 s–1)ti – (54.0 m/s2) sin (3.0 s–1)tj. (c) The magnitude of r is  r= {(6.0 m) cos (3.0 s–1)t]2 + [(6.0 m) sin (3.0 s–1)t]2]}1/2 = 6.0 m. Thus the particle is always 6.0 m from the origin, so it is traveling in a circle. (d) We see that a = – (9.0 s–2)r, so we have a = (9.0 s–2)r, with the angle between the vectors being 180°, that is, in opposite directions. (e) We see that  v = {[(18.0 m/s) sin (3.0 s–1)t]2 + [(18.0 m/s) cos (3.0 s–1)t]2}1/2 = 18.0 m/s, so v = (3.0 s–1)r, and a = (9.0 s–2)r = v2/r. 



Page 9

2

3.0 m/ s 2)t + (5.0 m/ s )t i + 12 (4.5 m / s 2)t j

Chapter 3

26. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction down. We find the time for the tiger to reach the ground from its vertical motion: y = y0 + v0yt + !ayt2; 6.5 m = 0 + 0 + !(9.80 m/s2)t2, which gives t = 1.15 s. The horizontal motion will have constant velocity. We find the distance from the base of the rock from x = x0 + v0xt; x = 0 + (4.0 m/s)(1.15 s) = 4.6 m.

O

v0

x

h

y

27. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction down. We find the height of the cliff from the vertical displacement: y = y0 + v0yt + !ayt2; y = 0 + 0 + !(9.80 m/s2)(3.0 s)2 = 44 m. The horizontal motion will have constant velocity. We find the distance from the base of the cliff from x = x0 + v0xt; x = 0 + (2.1 m/s)(3.0 s) = 6.3 m. 28. Because the initial and final locations are at the same level, we can use the expression for the horizontal range. The horizontal range on Earth is given by R = v02 sin(20)/g, whereas on the Moon it is RMoon = v02 sin(20)/gMoon. Because we have the same v0 and 0 , when we divide the two equations, we get RMoon/R = g/gMoon , or RMoon = (g/gMoon)R = [g/(g/6)]R = 6R, so a person could jump six times as far.

29. Because the water returns to the same level, we can use the y v0 expression for the horizontal range: 2 R = v0 sin(2)/g; 3.0 m = (5.5 m/s)2 sin(2)/(9.80 m/s2), which gives  x sin(2) = 0.971, or 2 = 76° and 104°, O R so the angles are 38° and 52°. At the larger angle the water has a smaller horizontal velocity but spends more time in the air, because of the larger initial vertical velocity. Thus the horizontal displacement is the same for the two angles.

Page 10

Chapter 3

30. We choose a coordinate system with the origin at the release point, with x horizontal and y vertical, with the positive direction down. We find the time of fall from the vertical displacement: y = y0 + v0yt + !ayt2;

O

9.0 m = 0 + 0 + !(9.80 m/s2)t2, which gives t = 1.35 s. The horizontal motion will have constant velocity. We find the initial speed from x = x0 + v0xt; 8.5 m = 0 + v0(1.35 s), which gives v0 = 6.3 m/s.

v0

x

h

y

31. We find the time of flight from the vertical displacement: y = y0 + v0yt + !ayt2; 0 = 0 + (18.0 m/s)(sin 32.0°)t + !(– 9.80 m/s2)t2, which gives t = 0, 1.95 s. The ball is kicked at t = 0, so the football hits the ground 1.95 s later. 32. We choose a coordinate system with the origin at the release point, with x horizontal and y vertical, with the positive direction down. The horizontal motion will have constant velocity. We find the time required for the fall from x = x0 + v0xt; 36.0 m = 0 + (22.2 m/s)t, which gives t = 1.62 s. We find the height from the vertical motion: y = y0 + v0yt + !ayt2; h = 0 + 0 + !(9.80 m/s2)(1.62 s)2 =

12.9 m.

33. We choose a coordinate system with the origin at the release point, with x horizontal and y vertical, with the positive direction up. We find the time required for the fall from the vertical motion: y = y0 + v0yt + !ayt2; – 2.2 m = 0 + (14 m/s)(sin 40°)t + !(– 9.80 m/s2)t2. The solutions of this quadratic equation are t = – 0.22 s, 2.06 s. Because the shot is released at t = 0, the physical answer is 2.06 s. We find the horizontal distance from x = x0 + v0xt; x = 0 + (14 m/s)(cos 40°)(2.06 s) = 22 m.

y v0 

O h

x

34. We choose a coordinate system with the origin at the release point, with y vertical and the positive direction up. At the highest point the vertical velocity will be zero, so we find the time to reach the highest point from vy = v0y + aytup; 0 = v0y + (– g)tup , which gives tup = v0y/g. We find the elevation h at the highest point from vy2 = v0y2 + 2ay(y– y0); 0 = v0y2 + 2(– g)h, which gives h = v0y2/2g. Page 11

Chapter 3

We find the time to fall from the highest point from y = y0 + v0yt + !aytdown2; 0 = h + 0 + !(– g)tdown2 , which gives tdown = (2h/g)1/2 = [2(v0y2/2g)/g]1/2 = v0y/g, which is the same as tup. 35. To plot the trajectory, we need a relationship between x and y, which can be obtained by eliminating t from the equations for the two components of the motion: x = v0xt = v0 (cos )t; y = y0 + v0yt + !ayt2 = 0 + v0 (sin )t + !(– The relationship is y = (tan )x – !g(x/v0 cos )2. g)t2.

y (m) 50 90°

40

75° 60°

30 20

30°

10

45°

15° 0

0

20

40

60

80

100

x (m)

36. The arrow will hit the apple at the same elevation, so we can use the expression for the horizontal range: R = v02 sin(20)/g; 25.0 m = (22.5 m/s)2 sin(20)/(9.80 m/s2), which gives sin(20) = 0.484, or 20 = 28.9° , 0 = 14.5°. Note that the other angle, 75.5°, is not realistic because the arrow will strike the apple at this angle. 37. We choose a coordinate system with the origin at the release point, with x horizontal and y vertical, with the positive direction down. Because the horizontal velocity of the package is constant at the horizontal velocity of the airplane, the package is always directly under the airplane. The time interval is the time required for the fall, which we find from the vertical motion: y = y0 + v0yt + !ayt2; 160 m = 0 + 0 + !(9.80 m/s2)t2, which gives

t = 5.71 s.

38. (a) Because the athlete lands at the same level, we can use the expression for the horizontal range: R = v02 sin(20)/g; 7.80 m = v02 sin[2(33.0°)]/(9.80 m/s2), which gives v0 = 9.15 m/s. (b) For an increase of 5%, the initial speed becomes v0 = (1 + 0.05)v0 = (1.05)v0 , and the new range is R = v02 sin(20)/g = (1.05)2v02 sin(20)/g = 1.10R. Thus the increase in the length of the jump is R – R = (1.10 – 1)R = 0.10(7.80 m) = 0.78 m.

39. (a) At the highest point, the vertical velocity vy = 0. We find the maximum height h from vy2 = v0y2 + 2ay(y – y0); 0 = [(51.2 m/s) sin 44.5°]2 + 2(– 9.80 m/s2)(h – 0), which gives h = 65.7 m. (b) Because the projectile returns to the same elevation, we have y = y0 + v0yt + !ayt2; 0 = 0 + (51.2 m/s)(sin 44.5°)t + !(– 9.80 m/s2)t2, which gives t = 0, and 7.32 s. Because t = 0 was the launch time, the total time in the air was 7.32 s. Page 12

Chapter 3

(c) We find the horizontal distance from x = v0xt = (51.2 m/s)(cos 44.5°)(7.32 s) = 267 m. (d) The horizontal velocity will be constant: vx = v0x = (51.2 m/s) cos 44.5° = 36.5 m/s. We find the vertical velocity from vy = v0y + ayt = (51.2 m/s) sin 44.5° + (– 9.80 m/s2)(1.50 s) = 21.2 m/s. The magnitude of the velocity is v = (vx2 + vy2)1/2 = [(36.5 m/s)2 + (21.2 m/s)2]1/2 = 42.2 m/s. We find the angle from  = 30.1° above the horizontal. tan  = vy/vx = (21.2 m/s)/(36.5 m/s) = 0.581, which gives

y

40. (a) We choose a coordinate system with the origin at the base of the cliff, with x horizontal and y vertical, with the positive direction up. We find the time required for the fall from the vertical motion: y = y0 + v0yt + !ayt2;

(b)

(c)

(d) (e)

v0



0 = 125 m + (65.0 m/s)(sin 37.0°)t + !(– 9.80 m/s2)t2, h which gives t = – 2.45, 10.4 s. Because the projectile starts at t = 0, we have t = 10.4 s. We find the range from the horizontal motion: X = v0xt = (65.0 m/s)(cos 37.0°)(10.4 s) P x = 540 m . O X For the velocity components, we have vx = v0x = (65.0 m/s) cos 37.0° = 51.9 m/s. vy = v0y + ayt = (65.0 m/s) sin 37.0° + (– 9.80 m/s2)(10.4 s) = – 62.8 m/s. When we combine these components, we get v = (vx2 + vy2)1/2 = [(51.9 m/s)2 + (– 62.8 m/s)2]1/2 = 81.5 m/s. We find the angle from  = 50.4° below the horizontal. tan  = vy/vx = (62.8 m/s)/(51.9 m/s) = 1.21, which gives

41. We use the coordinate system shown in the diagram. To see if the water balloon hits the boy, we will find the location of the water balloon and the boy when the water balloon passes the vertical line that the boy follows. We find the time at which this occurs from the horizontal motion of the water balloon: d = v0xt = (v0 cos 0)t , which gives t =d/(v0 cos 0). At this time the location of the boy is yboy = y0boy + v0yboyt + !ayt2 = h – !g[d/(v0 cos = h + 0 + !(– g)[d/(v0 cos The vertical position of the water balloon will be yballoon = y0balloon + v0yballoont + !ayt2

0)]2

y

boy

h

v0

0 0)]2.

O

water balloon

d

= 0 + (v0 sin 0)[d/(v0 cos 0)] + !(– g)[d/(v0 cos 0)]2 = d sin 0/cos 0 – !g[d/(v0 cos 0)]2 = d tan 0 – !g[d/(v0 cos 0)]2 . Because h = d tan 0 , we have yballoon = yboy.

42. The horizontal range is R = v02 sin(20)/g = 2v02 sin 0 cos 0/g. At the maximum height the vertical velocity will be zero. We find the maximum height from vy2 = v0y2 + 2ay; Page 13

x

Chapter 3

0 = (v02 sin 0)2 + 2(– g)hmax , which gives hmax = v02 sin2 0/2g. When we equate this to the range, we get v02 sin2 0/2g = 2v02 sin 0 cos 0/g, which gives tan 0 = 4, 0 = 76.0°. 43. The ball passes the goal posts when it has traveled the horizontal distance of 36.0 m. From this we can find the time when it passes the goal posts: x = v0xt; 36.0 m = (20.0 m/s) cos 37.0° t, which gives t = 2.25 s. To see if the kick is successful, we must find the height of the ball at this time: y = y0 + v0yt + !ayt2 = 0 + (20.0 m/s) sin 37.0° (2.25 s) + !(– 9.80 m/s2)(2.25 s)2 = 2.24 m. Thus the kick is unsuccessful because it passes 0.76 m below the bar. To have a successful kick, the ball must pass the goal posts with an elevation of at least 3.00 m. We find the time when the ball has this height from y = y0 + v0yt + !ayt2; 3.00 m = 0 + (20.0 m/s) sin 37.0° t + !(– 9.80 m/s2)t2. The two solutions of this quadratic equation are t = 0.28 s, 2.17 s. The horizontal distance traveled by the ball is found from x = v0xt = (20.0 m/s) cos 37.0° t. For the two times, we get x = 4.5 m, 34.7 m. Thus the kick must be made no farther than 34.7 m from the goal posts (and no nearer than 4.5 m). If the vertical velocity is found at these two times from vy = v0y + ayt = (20.0 m/s) sin 37.0° + (– 9.80 m/s2)t = + 9.3 m/s, – 9.3 m/s, we see that the ball is falling at the goal posts for a kick from 34.7 m and rising at the goal posts for a kick from 4.5 m. 44. (a) For the velocity of the projectile motion we have v = v0 cos 0 i + (v0 sin 0 – gt) j = v0 cos 0 i + [v0 sin 0 – (9.80 m/s2)(3.0 s)] j = (7.6 m/s) i + (4.8 m/s) j. This gives us two equations for the two unknowns, v0 and 0: v0 cos 0 = 7.6 m/s; v0 sin 0 – 29.4 m/s = 4.8 m/s. When we solve them we get v0 = 35 m/s, 0 = 77°. The horizontal range is R = v02 sin(20)/g = (35 m/s)2 sin[2(77°)]/(9.80 m/s2) = 53 m. (b) At the maximum height the vertical velocity will be zero. We find the maximum height from vy2 = v0y2 + 2ay; 0 = (v0 sin 0)2 + 2(– g)hmax ; hmax = v02 sin2 0/2g = (35 m/s)2 sin2 (77°)/2(9.80 m/s2) = 60 m. (c) From the symmetry of the motion, for the velocity components we have vx = v0x = 7.6 m/s; vy = – v0y = – (35 m/s) sin 77° = – 34 m/s. The speed is v = (vx2 + vy2)1/2 = [(7.6 m/s)2 + (– 34 m/s)2]1/2 = 35 m/s. We find the angle from  = 77° below the horizontal. tan  = vy/vx = (– 34 m/s)/(7.6 m/s) = – 4.47, which gives

Page 14

Chapter 3

45. (a) We choose a coordinate system with the origin at the jump point, with x horizontal and y vertical, with the positive direction up. The horizontal motion will have constant velocity: x = x0 + v0xt; L = 0 + v0 cos 0t; 3.0 m = (v0 cos 0)(1.3 s), which gives v0 cos 0 = 2.31 m/s. For the vertical motion we have y = y0 + v0yt + !ayt2;

y v0 0

O h

L

– h = 0 + (v0 sin 0)t + !(– g)t2; – 5.0 m = (v0 sin 0)(1.3 s) – !(9.80 m/s2)(1.3 s)2, which gives v0 sin 0 = 2.52 m/s. When we divide the two equations, we get tan 0 = 1.093, 0 = 47.5°. Thus we find the magnitude of v0 from v0 sin 47.5° = 2.52 m/s, which gives v0 = 3.42 m/s, so v0 = 3.42 m/s, 47.5° above the horizontal. (b) At the maximum height the vertical velocity will be zero. We find the maximum height from vy2 = v0y2 + 2ay; 0 = (v0 sin 0)2 + 2(– g)hmax ; hmax = v02 sin2 0/2g = (3.42 m/s)2 sin2 (47.5°)/2(9.80 m/s2) = 0.32 m, or 5.32 m above the water. (c) The horizontal velocity is constant. We find the vertical velocity from vfy2 = v0y2 + 2ay; vfy2 = (3.42 m/s)2 sin2 (47.5°) – 2(9.80 m/s2)(– 5.0 m), which gives vfy = – 10.2 m/s. We find the direction from tan  = vfy/vfx = (– 10.2 m/s)/(2.31 m/s) = – 4.42, which gives  = 77° below the horizontal. We find the magnitude of v from vf sin 77° = 10.2 m/s, which gives vf = 10.5 m/s, so vf = 10.5 m/s, 77° below the horizontal.

46. (a) We choose a coordinate system with the origin at the jump point, with x horizontal and y vertical, with the positive direction up. The horizontal motion will have constant velocity. We find the time required for the fall from x = x0 + v0xt; L = 0 + v0t, which gives t = L/v0. For the vertical motion we have y = y0 + v0yt + !ayt2; – h = 0 + 0 + !(– g)t2, so h = !g(L/v0)2; 36 m/s (130 km/h). 1.5 m = !(9.80 m/s2)(20 m)2/v02, which gives v0 = (b) If the ramp makes an angle 0 with the horizontal, we have x = x0 + v0xt; L = 0 + (v0 cos 0)t, which gives t = L/v0 cos 0 . For the vertical motion we have y = y0 + v0yt + !ayt2; Page 15

x

Chapter 3

– h = 0 + (v0 sin 0)t + !(– g)t2, so h = – (v0 sin 0)(L/v0 cos 0) + !g(L/v0 cos 0)2; 1.5 m = – (20 m) tan 10° + !(9.80 m/s2)(20 m)2/v02 cos2 10°, which gives v0 =

20 m/s (72 km/h).

47. We will take down as the positive direction. The direction of motion is the direction of the velocity. For the velocity components, we have vx = v0x = v0 . vy = v0y + ayt = 0 + gt = gt. We find the angle that the velocity vector makes with the horizontal from tan  = vy/vx = gt/v0 , or  = tan–1(gt/v0) below the horizontal.

y

48. We use the coordinate system shown in the diagram. For the horizontal motion we have x = v0xt; R = (v0 cos 0)t; which gives t = R/(v0 cos 0). For the vertical motion we have y = y0 + v0yt + !ayt2 ;

v0

0 O

h R

h = 0 + (v0 sin 0)t + !(– g)t2 = (v0 sin 0)[R/(v0 cos 0)] – !g[R/(v0 cos 0)]2. This is a quadratic equation for R: R2 – (2v02 sin 0 cos 0/g)R + 2hv02 cos2 0/g = 0, with solutions R = (v02 sin 0 cos 0/g){1 ± [1 – (2gh/v02 sin2 0)]1/2} = !R0{1 ± [1 – (2gh/v02 sin2 0)]1/2}, where R0 is the horizontal range. We use the + sign for h < 0, and h > 0 if the projectile has passed the highest point. 49. We choose a coordinate system with the origin at the base of the hill, with x horizontal and y vertical, with the positive direction up. When the object lands on the hill, y = x tan . The distance up the hill is given by d2 = x2 + y2 = x2(1 + tan2 ). Thus to maximize d, we can maximize x. For the horizontal and vertical motions we have x = x0 + v0xt = 0 + (v0 cos )t = (v0 cos )t;

y v0



d



O

y = y0 + v0yt + !ayt2 = 0 + (v0 sin )t + !(– g)t2 = (v0 sin )t – !gt2. We combine these equations to get x a function of : y = x tan  = (v0 sin )(x/v0 cos ) – !g(x/v0 cos )2, which gives x = (2v02 cos /g)(sin  – cos tan ). We can simplify this by using trigonometric identities to get functions of 2: x = (v02/g)[sin 2 – (1 + cos 2tan ]. To find the value of  for maximum x, we set dx/d = 0: dx/d = (v02/g)[2 cos 2 – (– 2 sin 2) tan ] = 0, which gives

 = ! tan–1 (– cot ). tan 2 = – cot , or Note that the negative sign means using the angle greater than 90° for the inverse tangent.

Page 16

x

x

Chapter 3

50. We find the time of flight from the vertical motion: y = y0 + v0yt + !ayt2; 0 = 0 + (v0 sin 0)t + !(– g)t2, which gives t = 2v0 sin 0/g = 2(32 m/s) sin 55°/(9.80 m/s2) = 5.35 s. The horizontal distance the ball travels is x = x0 + v0xt = 0 + (v0 cos )t = (v0 cos )t = (32 m/s) cos 55° (5.35 s) = 98.2 m. From the top view of the positions as the outfielder runs from A to B, we see that d1 = x sin  = (98.2 m) sin 22° = 36.8 m; d2 = L – x cos  = 85 m – (98.2 m) cos 22° = – 6.0 m. Thus the angle from the line to the plate that the outfielder must run is given by tan  = d1/d2 = (36.8 m)/(– 6.0 m) = – 6.13,  = 99.3°. The distance is d = (d12 + d22)1/2 = [(36.8 m)2 + (– 6.0 m)2]1/2 = 37.3 m, so his speed is v = d/t = (37.3 m)/(5.35 s) = 7.0 m/s. Thus the outfielder must run 7.0 m/s , 99° from the line to home plate. 51. The centripetal acceleration is aR = v2/r = (500 m/s)2/(3.50103 m)(9.80 m/s2/g) = 52. The centripetal acceleration is aR = v2/r = (0.85 m/s)2/(3.6 m) =

y B

v0





d



A

L

O Top view B x

d1 



L

d2

O

7.29g up.

0.20 m/s2 toward the center.

53. The centripetal acceleration of the Earth is aR = v2/r = (2pr/T)2/r = 4p2r/T2 = 4p2(1.51011 m)/(3.16107 s)2 =

5.910–3 m/s2 toward the Sun.

54. The speed of the speck is v= 2pr/T = 2prf. Thus the centripetal acceleration is aR = v2/r = (2prf)2/r = 4p2rf2 = 4p2(0.15 m)[(45 min–1)/(60 s/min)]2 =

3.3 m/s2.

55. To complete an orbit in time T, the speed of the shuttle must be v= 2pr/T. Thus the centripetal acceleration in terms of g is aR/g = v2/rg = (2pr/T)2/rg = 4p2r/gT2 = 4p2(6.38106 m + 0.40106 m)/(9.80 m/s2)[(90 min)(60 s/min)]2, which gives 56. An object at the equator is moving with the rotational speed of the surface of the Earth: v = 2prE/T = 2p(6.38106 m)/(86,400 s) = 464 m/s. The acceleration of gravity is reduced by the radial acceleration: Thus ?g = – v2/rE = – (464 m/s)2/(6.38106 m) = – 0.0337 m/s2 (0.343% of g). 57. We check the form of aR = v2/r by using the dimensions of each variable: [aR] = [v/t] = [d/t2] = [L/T2]; [v2] = [(d/t)2] = [(L/T)2] = [L2/T2]; [r] = [d] = [L]. Thus we have [v2/r] = [L2/T2]/[L] = [L/T2], which are the dimensions of aR. Page 17

aR = 0.94g.

A

Chapter 3

58. The position is r = (2.0 m) cos (3.0 s–1)ti + (2.0 m) sin (3.0 s–1)tj. (a) The magnitude of r is  r = {[(2.0 m) cos (3.0 s–1)t]2 + [(2.0 m) sin (3.0 s–1)t]2}1/2 = 2.0 m. Thus the particle is always 2.0 m from the origin, so it is traveling in a circle. (b) We find the velocity by differentiating: v = dr/dt = – (2.0 m)(3.0 s–1) sin (3.0 s–1)ti + (2.0 m)(3.0 s–1) cos (3.0 s–1)tj = – (6.0 m/s) sin (3.0 s–1)ti + (6.0 m/s) cos (3.0 s–1)tj. We find the acceleration by differentiating: a = dv/dt = – (6.0 m)(3.0 s–1) cos (3.0 s–1)ti – (6.0 m)(3.0 s–1) sin (3.0 s–1)tj = – (18.0 m/s2) cos (3.0 s–1)ti – (18.0 m/s2) sin (3.0 s–1)tj. (c) The magnitude of v is  v = {[(6.0 m/s) cos (3.0 s–1)t]2 + [(6.0 m/s) sin (3.0 s–1)t]2}1/2 = 6.0 m/s, The magnitude of a is  a = {[(18.0 m/s2) cos (3.0 s–1)t]2 + [(18.0 m/s2) sin (3.0 s–1)t]2}1/2 = 18.0 m/s2, (d) We see that v2/r = [(6.0 m/s)2/(2.0 m)] = 18.0 m/s2 = a. (e) If we compare r and a we see that a = – (9.0 s–2)r, so the acceleration vector is always opposite to the direction of r and thus points toward the center of the circle. 





59. If

vHR is the velocity of Huck with respect to the raft, v HB vHB the velocity of Huck with respect to the bank, and vRB the velocity of the raft with respect to the bank, then vHB = vHR + vRB , as shown in the diagram.  From the diagram we get v RB vHB = (vHR2 + vRB2)1/2 = [(1.0 m/s)2 + (2.5 m/s)2]1/2 = 2.7 m/s. We find the angle from tan  = vHR/vRB = (1.0 m/s)/(2.5 m/s) = 0.40, which gives  = 22° from the river bank.

v HR

60. (a) If

vBS is the velocity of the boat with respect to the shore, vBW the velocity of the boat with respect to the water, and vWS the velocity of the water with respect to the shore, then vBS = vBW + vWS , as shown in the diagram. From the diagram we get vBS = (vBW2 + vWS2)1/2 = [(2.20 m/s)2 + (1.20 m/s)2]1/2 = 2.51 m/s. We find the angle from tan  = vBW/vWS = (2.20 m/s)/(1.20 m/s) = 1.83, which gives  = 61.4° from the shore. (b) Because the boat will move with constant velocity, the displacement will be d = vBSt = (2.51 m/s)(3.00 s) = 7.52 m at 61.4° to the shore.

v BS

61. Because the planes are approaching along the same line, for the relative velocity we have v = v1 – v2 = 780 km/h – (– 780 km/h) = 1560 km/h. We find the time before they would meet from t = d/v = (10.0 km)/(1560 km/h) = 0.00641 h = 23.1 s.

Page 18

v BW

 v WS

Chapter 3

62. If

vPA is the velocity of the airplane with respect to the air, vPG the velocity of the airplane with respect to the ground, and vAG the velocity of the air(wind) with respect to the ground, then vPG = vPA + vAG , as shown in the diagram. (a) From the diagram we find the two components of vPG:  vPGE = vAG cos 45° = (90.0 km/h) cos 45° = 63.6 km/h; v PG v PA vPGS = vPA – vAG sin 45° = 550 km/h – (90.0 km/h) sin 45° = 486 km/h. For the magnitude we have vPG = (vPGE2 + vPGS2)1/2 = [(63.6 km/h)2 + (486 km/h)2]1/2 = 490 km/h. v AG 45° We find the angle from East tan  = vPGE/vPGS = (63.6 km/h)/(486 km/h) = 0.131, which gives  = 7.45° east of south. South (b) Because the pilot is expecting to move south, we find the easterly distance from this line from d = vPGEt = (63.6 km/h)(12.0 min)/(60 min/h) = 12.7 km. Of course the airplane will also not be as far south as it would be without the wind.

63. From the vector diagram of Example 3–13, we have vBW2 = vBS2 + vWS2 ; (1.85 m/s)2 = vBS2 + (1.20 m/s)2 , which gives

vBS = 1.41 m/s.

64. If

vPW is the velocity of the passenger with respect to the water, y vPB the velocity of the passenger with respect to the boat, and vBW the velocity of the boat with respect to the water, then vPW = vPB + vBW , as shown in the diagram. From the diagram we find the two components of vPW: vPWx = vPB cos 45° + vBW = (0.60 m/s) cos 45° + 1.80 m/s = 2.22 m/s. vPWy = vPB sin 45° = (0.60 m/s) sin 45° = 0.424 m/s. For the magnitude we have vPW = (vPWx2 + vPWy2)1/2 = [(2.22 m/s)2 + (0.424 m/s)2]1/2 = 2.26 m/s. We find the angle from tan  = vPWy/vPWx = (0.424 m/s)/(2.22 m/s) = 0.191, which gives  = 11° above the water.

v PW 

v BW

v WS

65. If

vBS is the velocity of the boat with respect to the shore, vBW the velocity of the boat with respect to the water, and vWS the velocity of the water with respect to the shore, then vBS = vBW + vWS , as shown in the diagram. (a) From the diagram we have vWS = vBW sin  = (3.70 m/s) sin 29.5° = 1.82 m/s. (b) vBS = vBW cos  = (3.70 m/s) cos 29.5° = 3.22 m/s.

Page 19

v BS



v BW

v PB

45° x

Chapter 3

66. If

vBS is the velocity of the boat with respect to the shore, vBW the velocity of the boat with respect to the water, and vWS the velocity of the water with respect to the shore, then vBS = vBW + vWS , as shown in the diagram. We find the angle of the boat’s motion with respect to the shore from the distances: tan  = dshore/driver = (120 m)/(280 m) = 0.429, which gives  = 23.2°. The y-component of vBS is also the y-component of vBW: vBSy = vBWy = (2.40 m/s) sin 45° = 1.70 m/s. We find the x-component from vBSx = vBSy tan  = (1.70 m/s) tan 23.2° = 0.727 m/s. For the x-component of the relative velocity, we use the diagram to get  vWS= vBWx – vBSx = (2.40 m/s) cos 45° – 0.727 m/s = 0.97 m/s.

y v WS

v BS



v BW 45° x



67. If

vSB is the velocity of the swimmer with respect to the bank, vSW the velocity of the swimmer with respect to the water, and vWB the velocity of the water with respect to the bank, then vSB = vSW + vWB , as shown in the diagram. (a) We find the angle from tan  = vWB/vSW = (0.80 m/s)/(1.00 m/s) = 0.80, which gives  = 38.7°. Because the swimmer travels in a straight line, we have tan  = dshore/driver ; 0.80 = dshore/(75 m), which gives dshore = 60 m. (b) We can find how long it takes by using the components across the river: t = driver/vSW = (75 m)/(1.00 m/s) = 75 s.

68. We have a new diagram, as shown. From the diagram, we have sin  = vWB/vSW = (0.80 m/s)/(1.00 m/s) = 0.80, which gives  = 53°.

v WB

v SW

v WB

v SW

69. The velocities are shown in the diagram. For the relative velocity of car 1 with respect to car 2, we have v12 = v1g – v2g . For the magnitude we have v12 = (v1g2 + v2g2)1/2 = [(30 km/h)2 + (50 km/h)2]1/2 = 58 km/h. We find the angle from  = 31°. tan  = v1g/v2g = (30 km/h)/(50 km/h) = 0.60, which gives For the relative velocity of car 2 with respect to car 1, we have v21 = v2g – v1g = – (v1g – v2g) = – v12 = 58 km/h opposite to v12.

Page 20

v SB





v SB

v 12 v 1g

 v 2g

Chapter 3

70. If

vPW is the velocity of the airplane with respect to the wind, vPG the velocity of the airplane with respect to the ground, and vWG the velocity of the wind with respect to the ground, then vPG = vPW + vWG , as shown in the diagram. We have two unknowns: vPG and  (or ). If we use the law of sines for the vector triangle, we have vPW/sin (90° +  ) = vWG/sin  , or sin  = (vWG/vPW)sin (90° +  ) = [(120 km/h)/(680 km/h)] sin (90.0° + 35.0°) = 0.145, or  = 8.31°. 43.3° N of E. Thus we have =  +  = 8.31° + 35.0° =

N v WG

v PW v PG











71. If

vCG is the velocity of the car with respect to the ground, vMG the velocity of the motorcycle with respect to the ground, and vMC the velocity of the motorcycle with respect to the car, then vMC = vMG – vCG . Because the motion is in one dimension, for the initial relative velocity we have vMC = vMG – vCG = (95.0 km/h – 75.0 km/h)/(3.6 ks/h) = 5.56 m/s. For the linear motion, in the reference frame of the car we have x = x0 + v0t + !at2; a = 0.0889 m/s2. 60.0 m = 0 + (5.56 m/s)(10.0 s) + !a(10.0 s)2, which gives Note that this is also the acceleration in the reference frame of the ground.

72. (a) For the magnitude of the resultant to be equal to the sum of the magnitudes, the two vectors must be parallel. (b) The expression is the one we use when we find the magnitude of a vector from its rectangular components. Thus the two vectors must be perpendicular. (c) The only way to have the sum and difference of two magnitudes be equal is for V2 = – V2, or V2 = 0. Only a zero vector has zero magnitude: V2 = 0.

73. The displacement is shown in the diagram. For the components, we have Dx = 60 m, Dy = – 35 m, Dz = – 12 m. To find the magnitude we extend the process for two dimensions: D = (Dx2 + Dy2 + Dz2)1/2 = [(60 m)2 + (– 35 m)2 + (– 12 m)2]1/2 = 70 m. The direction is specified by the two angles shown, which we find from tan h = Dy/Dx = (35 m)/(60 m) = 0.583, h = 30° from the x-axis toward the – y-axis; which gives tan v = Dz/(Dx2 + Dy2)1/2 = (12 m)/[(60 m)2 + (– 35 m)2]1/2 = 0.173, which gives

z up

y North

D1 h

D2

v

D

D3

v = 9.8° below the horizontal.

74. If we assume constant acceleration along the slope, we have v = v0 + at; 0 = [(120 km/h)/(3.6 ks/h)] + a(12 s), which gives a = – 2.78 m/s2 along the slope. For the components we have ahorizontal = a cos 30° = (– 2.78 m/s2) cos 30° = – 2.4 m/s2 (opposite to the truck’s motion). Page 21

x East

E

Chapter 3

avertical = a sin 30° = (– 2.78 m/s2) sin 30° =

– 1.4 m/s2 (down).

75. The horizontal velocity is constant, and the vertical velocity will y be zero when the pebbles hit the window. Using the coordinate system shown, we find the vertical component of the initial v0 velocity from vy2 = v0y2 + 2ay(h – y0) ; O 0 = v0y2 + 2(– 9.80 m/s2)(8.0 m – 0), which gives v0y = 12.5 m/s. (We choose the positive square root because we know that the pebbles are thrown upward.) We find the time for the pebbles to hit the window from the vertical motion: vy = v0y + ayt; 0 = 12.5 m/s + (– 9.80 m/s2)t, which gives t = 1.28 s. For the horizontal motion we have x = x0 + v0xt; 9.0 m = 0 + v0x(1.28 s), which gives v0x = 7.0 m/s. Because the pebbles are traveling horizontally when they hit the window, this is their speed.

h x

76. (a) We choose a coordinate system with the origin at the release point, with x horizontal and y vertical, with the positive direction down. We find the time of flight from the horizontal motion: x = x0 + v0xt; 65.0 m = 0 + (145 m/s)t, which gives t = 0.448 s. We find the distance the bullet falls from y = y0 + v0yt + !ayt2; y = 0 + 0 + !(9.80 m/s2)(0.448 s)2 = 0.985 m. (b) The bullet will hit the target at the same elevation, so we can use the expression for the horizontal range: R = v02 sin(20)/g; 65.0 m = (145 m/s)2 sin(20)/(9.80 m/s2), which gives sin(20) = 0.0303, or 20 = 1.74° , 0 = 0.87°. The larger angle of 89.1° is unrealistic.

77. We see from the diagram that cos  = Ax/A = (46.4)/(52.8) = 0.879, or  = ± 28.5°. For the y-component, we have Ay = A sin  = (52.8) sin (± 28.5°) = ± 25.2. The second vector (below the x-axis) is not shown in the diagram.

y A Ay

 x Ax

Page 22

Chapter 3

78. The velocity of the rain with respect to the train is vRT = vR – vT , where vR is the velocity of the rain with respect to the ground and vT is the velocity of the train with respect to the ground. From the diagram we have tan  = vT/vR , or vR = vT/tan .

v RT

 vR

vT

79. If

vPW is the velocity of the airplane with respect to the wind, N vPG the velocity of the airplane with respect to the ground, and vWG the velocity of the wind with respect to the ground, then vPG = vPW + vWG , as shown in the diagram. v PG 45° Because the plane has covered 180 km in 1.00 hour, vPG = 180 km/h. v PW We use the diagram to write the component equations: vWGE = vPGE = vPG sin 45° = (180 km/h) sin 45° = 127 km/h; v vWGN = vPGN – vPWN = – vPG cos 45° – vPW  WG = – (180 km/h) cos 45° – (– 240 km/h) = 113 km/h. For the magnitude we have vWG = (vWGE2 + vWGN2)1/2 = [(127 km/h)2 + (113 km/h)2]1/2 = 170 km/h. We find the angle from tan  = vWGN/vWGE = (113 km/h)/(127 km/h) = 0.886, which gives  = 41.5° N of E.

E

80. We can find the time for the jump from the horizontal motion: x = vxt; t = x/vx = (8.0 m)/(9.2 m/s) = 0.87 s. It takes half this time for the jumper to reach the maximum height or to fall from the maximum height. If we consider the latter, we have y = y0 + v0yt + !ayt2, or 0 = hmax + 0 + !(– 9.80 m/s2)(0.435 s)2, which gives hmax = 0.93 m. 81. Because the golf ball returns to the same elevation, we can use the expression for the horizontal range. The horizontal range on Earth is given by R = v02 sin(20)/g, whereas on the Moon it is RMoon = v02 sin(20)/gMoon. Because we assume the same v0 and 0 , when we divide the two equations, we get RMoon/R = g/gMoon , or gMoon = (R/RMoon)g = [(30 m)/(180 m)](9.80 m/s2) = 1.6 m/s2.

82. We choose a coordinate system with the origin at home plate, x horizontal and y up, as shown in the diagram. The minimum speed of the ball is that which will have the ball just clear the fence. The horizontal motion is x = v0xt; 92 m = v0 cos 40° t, which gives v0t = 120 m. The vertical motion is y = y0 + v0yt + !ayt2; Page 23

y

v0

 O

h x

Chapter 3

12 m = 1.0 m + v0 sin 40° t + !(– 9.80 m/s2)t2. We can use the first equation to eliminate v0t from the second and solve for t, which gives t = 3.67 s. When this value is used in the first equation, we get v0 = 33 m/s. 83. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction down. We find the time for the diver to reach the water from the vertical motion: y = y0 + v0yt + !ayt2; t = 2.7 s. 35 m = 0 + 0 + !(9.80 m/s2)t2, which gives The horizontal motion will have constant velocity. We find the minimum horizontal initial velocity needed to land beyond the rocky outcrop from x = x0 + v0xt; 5.0 m = 0 + v0(2.7 s), which gives v0 = 1.9 m/s. 84. We use the coordinate system shown in the diagram. We find the time for the ball to reach the net from the vertical motion: y = y0 + v0yt + !ayt2; 0.90 m = 2.50 m + 0 + !(– 9.80 m/s2)t2, which gives t = 0.571 s. We find the initial velocity from the horizontal motion: x = v0xt; 15.0 m = v0 (0.571 s), which gives v0 = 26.3 m/s. We find the time for the ball to reach the ground from the vertical motion: y = y0 + v0yt2 + !ayt22;

y v0

h O

L

x

t2 = 0.714 s. 0 = 2.50 m + 0 + !(– 9.80 m/s2)t22, which gives We find where it lands from the horizontal motion: x2 = v0t2 = (26.3 m/s)(0.714 s) = 18.8 m. Because this is 18.8 m – 15.0 m = 3.8 m beyond the net, which is less than 7.0 m, the serve is good. 85. (a) When the boat moves upstream the speed with respect to the bank is v – u. When the boat moves downstream the speed with respect to the bank is v + u. For each leg the distance traveled is !D, so the total time is t = [!D/(v – u)] + [!D/(v + u)] = !D(v + u + v – u)/(v – u)(v + u) = Dv/(v2 – u2). (b) To move directly across the river the boat must head at an angle , as shown in the diagram. From the diagram we see that the speed with respect to the shore is vBS = (v2 – u2)1/2, v  during both legs of the trip. Thus the total time is t = [!D/(v2 – u2)1/2] + [!D/(v2 – u2)1/2] = D/(v2 – u2)1/2. We must assume that u < v, otherwise the boat will be swept downstream and never make it across the river. This appears in our answers as a negative time in (a) and the square root of a negative number in (b). u

Page 24

v BS

Chapter 3

86. If

vPA is the velocity of the airplane with respect to the air, vPG the velocity of the airplane with respect to the ground, and vAG the velocity of the air(wind) with respect to the ground, then vPG = vPA + vAG , as shown in the diagram. From the diagram we find the two components of vPG: vPGW = vAG sin 45° = (100 km/h) sin 45° = 70.7 km/h; vPGN = vPA – vAG cos 45° = 200 km/h – (100 km/h) cos 45° = 129.3 km/h. For the magnitude we have vPG = (vPGW2 + vPGN2)1/2 = [(70.7 km/h)2 + (129.3 km/h)2]1/2 = 147 km/h. We find the angle from tan  = vPGW/vPGN = (70.7 km/h)/(129.3 km/h) = 0.547, which gives  = 28.7° west of north.

v AG 45° v PA v PG

 East South

87. If

vAG is the velocity of the automobile with respect to the ground, vHG the velocity of the helicopter with respect to the ground, and vHA the velocity of the helicopter with respect to the automobile, then vHA = vHG – vAG . For the horizontal relative velocity we have vHA = vHG – vAG = (200 km/h – 150 km/h)/(3.6 ks/h) = 13.9 m/s. This is the initial (horizontal) velocity of the parcel, so we can find the time of fall from y = y0 + v0yt + !ayt2; 78.0 m = 0 + 0 + !(+ 9.80 m/s2)t2, which gives t = 3.99 s. During this time, we find the horizontal distance the parcel travels with respect to the car from x = vHAt = (13.9 m/s)(3.99 s) = 55.4 m. Because the helicopter is always directly above the parcel, this is how far behind the automobile the helicopter must be when the parcel is dropped. Thus we find the angle from tan  = y/x = (78.0 m)/(55.4 m) = 1.41, which gives  = 54.6° below the horizontal.

88. We use the coordinate system shown in the diagram, with up positive. For the horizontal and vertical motions we have x = x0 + v0xt, L = 0 + (v0 cos )t = (v0 cos )t; y = y0 + v0yt + !ayt2;

y

v0

h = 0 + (v0 sin )t + !(– g)t2 = (v0 sin )t – !gt2.  We combine these equations to get O h = (v0 sin )(L/v0 cos ) – !g(L/v0 cos )2, which gives v02 = gL2/2(L tan  – h) cos2 . From this we can find the speed required to hit the center of the basket: L v02 = (9.80 m/s2)(11.00 m)2/2[(11.00 m) tan 38.0° – (2.60 m – 2.10 m)] cos2 38.0°, which gives v0 = 10.86 m/s. If we treat the small change in L as a differential, we find the small change in v0 by differentiating: 2v0 dv0 = (g/2 cos2 ){[2L/(L tan  – h)] + L2(– tan )/(L tan  – h)2} dL = [gL2/2(L tan  – h) cos2 ]{[tan  – (2h/L)] /(L tan  – h)} dL = v02{[tan  – (2h/L)] /(L tan  – h)} dL, so

dv0 = !v0{[tan  – (2h/L)] /(L tan  – h)} dL

= !(10.86 m/s){[(tan 38.0° – 2(0.50 m)/(11.00 m)]/[(11.00 m) tan 38.0° – 0.50 m]} (± 0.22 m) Page 25

h x

Chapter 3

= ± 0.10 m/s. Thus the range of initial speeds is

10.76 m/s < v0 < 10.96 m/s.

89. (a) The rotational speed at the equator is v0 = 2prE/T = 2p(6.38106 m)/(86,400 s) = 464 m/s. (b) At a latitude , the radius of rotation (distance to the axis of rotation) is rE cos . The rotational speed is v = 2prE cos /T = v0 cos  = (464 m/s) cos 40° = 355 m/s. 90. We use the coordinate system shown in the diagram, with up positive. For the horizontal motion, we have x = v0xt; L = (v0 cos )t; 195 m = (v0 cos )(7.6 s), which gives v0 cos  = 25.7 m/s. For the vertical motion, we have y = y0 + v0yt + !ayt2;

y v0 H

 O

x

L

H = 0 + (v0 sin )t + !(– g)t2;

155 m = (v0 sin )(7.6 s) + !(– 9.80 m/s2)(7.6 s)2, which gives v0 sin  = 57.6 m/s. We can find the initial angle  by dividing the two results: tan  = (v0 sin  )/(v0 cos  ) = (57.6 m/s)/(25.7 m/s) = 2.24, which gives  = 66.0°. Now we can use one of the previous results to find the initial velocity: v0 = (25.7 m/s)/cos  = (25.7 m/s)/cos 66.0° = 63 m/s. Thus the initial velocity is 63 m/s, 66° above the horizontal. 91. We use the coordinate system shown in the diagram. If Agent Logan heads downstream at speed vB at an y angle , the time required to cross the river is t1 = D/vB cos . The distance he will travel in the y-direction in this time is y = (vB sin + vW)t1 = (D/vB)[(vB sin + vW)/cos ]. Because he will be below the point directly across the river, he must run this distance, which, at speed vL , will take a time t2 = y/vL = (D/vBvL)[(vB sin + vW)/cos ]. Thus the total time is t = t1 + t2 = (D/vBvL)[(vL + vB sin + vW)/cos ]. To find the angle that produces the minimum time, we set dt/d = 0: dt/d = (D/vBvL){vB – [(vL + vB sin  + vW)(– sin )/cos2 ]}

vB

vL vW



x D

= (D/vBvL){[vB cos2  + (vL + vW) sin  + vB sin2 ]/cos2 } = (D/vBvL cos2 )[vB + (vL + vW) sin ] = 0, which gives sin  = – vB/(vL + vW) = – (1.50 m/s)/( 3.00 m/s + 0.80 m/s) = – 0.395,  = – 23°. The time to reach the shore is t1 = D/vB cos  = (1600 m)/(1.50 m/s) cos (– 23°) = 1160 s. The distance he must run is y = (vB sin + vW)t1 = [(1.50 m/s) sin (– 23°) + 0.80 m/s](1160 s) = 243 m. The running time is t2 = y/vL = (243 m)/(3.00 m/s) = 81 s, so the total time is t = t1 + t2 = 1160 s + 81 s = 1240 s = 20.7 min. Thus Agent Logan must row at an angle of 23° upstream and run 243 m in a total time of 20.7 min. Note that if the initial angle is chosen upstream, it is very difficult to keep the signs straight, because the Page 26

Chapter 3

landing point could be above or below the point directly across the river, that is, positive or negative. This is also a good problem for a numerical solution on a spreadsheet, where absolute values can be used.

92. If

vSW is the velocity of the ship with respect to the water, vJS the velocity of the jogger with respect to the ship, and vJW the velocity of the jogger with respect to the water, then vJW = vJS + vSW . We choose the direction of the ship as the positive direction. Because all vectors are parallel, in each case the motion is one-dimensional. When the jogger is moving toward the bow, we have vJW = vJS + vSW = 2.0 m/s + 8.5 m/s = 10.5 m/s in the direction of the ship's motion. When the jogger is moving toward the stern, we have vJW = – vJS + vSW = – 2.0 m/s + 8.5 m/s = 6.5 m/s in the direction of the ship's motion.

93. The centripetal acceleration must equal g: g = v2/R = (2pR/T)2/R = 4p2R/T2; 9.80 m/s2 = 4p2(0.55103 m)/T2, which gives T = 47.1 s. For the rotation speed we have revolutions/day = (86,400 s/day)/(47.1 s/rev) = 1.8103 rev/day. 94. The centripetal acceleration is aR = v2/r, or r = v2/aR . Thus r is minimal when aR is maximal: rmin = [(700 km/h)/(3.6 ks/h)]2/(6.0)(9.80 m/s2) = 6.4102 m =

Page 27

0.64 km.

Chapter 4

1

CHAPTER 4 - Dynamics: Newton’s Laws of Motion 1.

We convert the units: # lb = (0.25 lb)(4.45 N/lb) ˜ 1 N.

2.

If we select the bike and rider as the object, we apply Newton’s second law to find the mass: ?F = ma; m = 116 kg. 255 N = m(2.20 m/s2), which gives

3.

We apply Newton’s second law to the object: ?F = ma; F = (7.010–3 kg)(10,000)(9.80 m/s2) = 6.9102 N.

4.

Without friction, the only horizontal force is the tension. We apply Newton’s second law to the car: ?F = ma; FT = (1250 kg)(1.30 m/s2) = 1.63103 N.

5.

We find the weight from the value of g. 5.7102 N. (a) Earth: FG = mg = (58 kg)(9.80 m/s2) = 2 (b) Moon: FG = mg = (58 kg)(1.7 m/s ) = 99 N. (c) Mars: FG = mg = (58 kg)(3.7 m/s2) = 2.1102 N. 2 (d) Space: FG = mg = (58 kg)(0 m/s ) = 0.

6.

The acceleration can be found from the car’s one-dimensional motion: v = v0 + at; 0 = [(90 km/h)/(3.6 ks/h)] + a(7.0 s), which gives a = – 3.57 m/s2. We apply Newton’s second law to find the required average force ?F = ma; F = (1050 kg)(– 3.57 m/s2) = – 3.8103 N. The negative sign indicates that the force is opposite to the velocity.

7.

The required average acceleration can be found from the one-dimensional motion: v2 = v02 + 2a(x – x0); (155 m/s)2 = 0 + 2a(0.700 m – 0), which gives a = 1.72104 m/s2. We apply Newton’s second law to find the required average force ?F = ma; F = (6.2510–3 kg)(1.72104 m/s2) = 107 N.

8.

(a) The weight of the box depends on the value of g: FG = m1g = (30.0 kg)(9.80 m/s2) = 294 N. We find the normal force from ?Fy = may; FN – m1g = 0, which gives FN = m1g = 294 N. (b) We select both blocks as the object and apply Newton’s second law: ?Fy = may; FN1 – m1g – m2g = 0, which gives FN1 = (m1 + m2)g = (30.0 kg + 20.0 kg)(9.80 m/s2) = If we select the top block as the object, we have ?Fy = may; FN2 – m2g = 0, which gives FN2 = m2g = (20.0 kg)(9.80 m/s2) = 196 N.

Page 1

(a) FN

y

m 1g

x

(b) 490 N.

m 2g

m 1g

FN1

FN2 m 1g

Chapter 4

9.

2

The required average acceleration can be found from the one-dimensional motion: v2 = v02 + 2a(x – x0); 0 = [(100 km/h)/(3.6 ks/h)]2 + 2a(150 m – 0), which gives a = – 2.57 m/s2. We apply Newton’s second law to find the required force ?F = ma; F = (3.6105 kg)(– 2.57 m/s2) = – 9.3105 N. The weight of the train is mg = (3.6105 kg)(9.80 m/s2) = 3.5106 N, so Superman must apply a force that is 25% of the weight of the train.

10. The acceleration of the first box is a1 = F/m1 , so the speed after a time t is v1 = v0 + a1t = a1t. For the second box we have a2 = F/m2 = F/2m1 = !a1 , so the speed after a time t is v2 = v0 + a2t = !a1t, or 11. Because the line snapped, the tension FT > 25 N. We write ?F = ma from the force diagram for the fish: y-component: FT – mg = ma, or FT = m(a + g). We find the minimum mass from the minimum tension: 25 N = mmin(3.5 m/s2 + 9.80 m/s2), which gives mmin = 1.9 kg. Thus we can say m > 1.9 kg.

v2 = !v1.

+y FT

mg

12. The average acceleration of the ball can be found from the one-dimensional motion: v2 = v02 + 2a(x – x0); 0 = (41.0 m/s)2 + 2a(0.120 m – 0), which gives a = – 7.00103 m/s2. We apply Newton’s second law to find the required average force applied to the ball: ?F = ma; F = (0.140 kg)(– 7.00103 m/s2) = – 9.80102 N. The force on the glove has the same magnitude but the opposite direction: 9.80102 N in the direction of the ball’s motion. 13. The average acceleration of the shot can be found from the one-dimensional motion: v2 = v02 + 2a(x – x0); (13 m/s)2 = 0 + 2a(2.8 m – 0), which gives a = 30.2 m/s2. We apply Newton’s second law to find the required average force applied to the shot: ?F = ma; F = (7.0 kg)(30.2 m/s2) = 2.1102 N. 14. We write ?F = ma from the force diagram for the car: y-component: FT – mg = ma, or FT = m(a + g) = (1200 kg)(0.80 m/s2 + 9.80 m/s2) =

+y 1.3104

N.

FT

mg

Page 2

Chapter 4

3

15. We write ?F = ma from the force diagram for the bucket: y-component: FT – mg = ma; 63.0 N – (7.50 kg)(9.80 m/s2) = (7.50 kg)a, a = – 1.40 m/s2 (down). which gives

+y

FT

mg

16. The maximum tension will be exerted by the motor when the elevator is accelerating upward. We write ?F = ma from the force diagram for the elevator: y-component: FTmax – mg = ma, or FTmax = m(a + g) = (4125 kg)(0.0600 + 1)(9.80 m/s2) = 4.29104 N. The minimum tension will be exerted by the motor when the elevator is accelerating downward. We write ?F = ma from the force diagram for the car: y-component: FTmin – mg = ma, or FTmin = m(a + g) = (4125 kg)(– 0.0600 + 1)(9.80 m/s2) = 3.80104 N. 17. To have the tension less than the weight, the thief must have a downward acceleration so that the tension FT = meffectiveg. We write ?F = ma from the force diagram for the thief: y-component: mg – FT= ma. We find the minimum acceleration from the maximum tension: mg – meffectiveg= mamin ; (65 kg – 57 kg)(9.80 m/s2)= (65 kg)amin , which gives amin = 1.2 m/s2. Thus we can say a (downward) = 1.2 m/s2.

18. The scale reads the force the person exerts on the scale. From Newton’s third law, this is also the magnitude of the normal force acting on the person. We write ?F = ma from the force diagram for the person: y-component: FN – mg = ma, or 0.75mg – mg = ma, which gives a = (0.75 – 1)(9.80 m/s2) =

+y

FT

mg

FT

+y

a

mg

+y

FN – 2.5 m/s2 (down).

19. The maximum tension will be exerted by the motor when the elevator has the maximum acceleration. We write ?F = ma from the force diagram for the elevator: y-component: FTmax – mg = mamax ; 21,750 N – (2100 kg)(9.80 m/s2) = (2100 kg)amax , which gives amax = 0.557 m/s2.

+y

mg

FT

mg 20. We can find the constant acceleration from Page 3

Chapter 4

4

x = v0t + !at2; 2.00g. 402 m = 0 + !a (6.40 s)2, which gives a = 19.6 m/s2 = We find the required horizontal force from Fhorizontal = ma = (280 kg)(19.6 m/s2) = 5.49103 N. 21. From Newton’s third law, the gases will exert a force on the rocket that is equal and opposite to the force the rocket exerts on the gases. (a) With up positive, we write ?F = ma from the force diagram for the rocket: Fgases – mg = ma; 33106 N – (2.75106 kg)(9.80 m/s2) = (2.75106 kg)a, a = 2.2 m/s2. which gives (b) If we ignore the mass of the gas expelled and any change in g, we can assume a constant acceleration. We find the velocity from v = v0 + at = 0 + (2.2 m/s2)(8.0 s) = 18 m/s. (c) We find the time to achieve the height from y = y0 + v0t + !at2; 9500 m = 0 + 0 + !(2.2 m/s2)t2, which gives

+y mg

Fgases

t = 93 s.

22. With down positive, we write ?F = ma from the force diagram for the skydivers: mg – FR = ma; (a) Before the parachute opens, we have mg – #mg = ma, which gives a = &g = 7.4 m/s2 (down). (b) Falling at constant speed means the acceleration is zero, so we have mg – FR = ma = 0, which gives FR = mg = (120.0 kg)(9.80 m/s2) = 1176 N.

FR

mg +y

23. We find the velocity necessary for the jump from the motion when the +y person leaves the ground to the highest point, where the velocity is zero: v2 = vjump2 + 2(– g)h; mg 0 = vjump2 + 2(– 9.80 m/s2)(0.80 m), which gives vjump = 3.96 m/s. F N We can find the acceleration required to achieve this velocity during the crouch from vjump2 = v02 + 2a(y – y0); (3.96 m/s)2 = 0 + 2a(0.20 m – 0), which gives a = 39.2 m/s2. Using the force diagram for the person during the crouch, we can write ?F = ma: FN – mg = ma; FN – (61 kg)(9.80 m/s2) = (61 kg)(39.2 m/s2), which gives FN = 3.0103 N. From Newton’s third law, the person will exert an equal and opposite force on the ground: 3.0103 N downward. 24. (a) We find the velocity just before striking the ground from v12 = v02 + 2gh; v12 = 0 + 2(9.80 m/s2)(3.5 m), which gives v1 = 8.3 m/s. (b) We can find the average acceleration required to bring the person to rest from v2 = v12 + 2a(y – y0); 0 = (8.3 m/s)2 + 2a(0.70 m – 0), which gives a = – 49 m/s2. Using the force diagram for the person during the crouch, we can write ?F = ma: mg – Flegs = ma; Page 4

mg

Flegs +y

Chapter 4

5

(43 kg)(9.80 m/s2) – Flegs = (43 kg)(– 49 m/s2), which gives

Flegs = 2.5103 N up.

25. (a) If we assume that he accelerates for a time t1 over the first 45 m and reaches a top speed of v, we have x1 = !(v0 + v)t1 = !vt1 , or t1 = 2x1/v = 2(45 m)/v = (90 m)/v. Because he maintains this top speed for the last 55 m, we have t2 = (55 m)/v. Thus the total time is T = t1 + t2 = (90 m)/v+ (55 m)/v = 10.0 s. When we solve for v, we get v = 14.5 m/s; so the acceleration time is t1 = (90 m)/(14.5 m/s) = 6.21 s. We find the constant acceleration for the first 45 m from a = ?v/?t = (14.5 m/s – 0)/(6.21 s) = 2.34 m/s2. We find the horizontal force component that will produce this acceleration from F = ma = (62 kg)(2.34 m/s2) = 1.4102 N. (b) As we found in part (a): v = 14.5 m/s. 26. If the box does not move, the acceleration is zero. FT FT y Using the force diagrams, we write ?F = ma. For the hanging weight, we have FT – FG2 = 0, or FT = FG2 . For the box on the floor, we have FT – FG1 + FN = 0, or FN = FG1 – FT = FG1 – FG2 . (a) FN = 85 N – 30 N = 55 N. FG2 FN FG1 (b) FN = 85 N – 60 N = 25 N. (c) FN = 85 N – 90 N = – 5 N. Because the floor can only push up on the box, it is not possible to have a negative normal force. Thus the normal force is 0; the box will leave the floor and accelerate upward. 27. In order for the resultant to have no northerly component, the second force must have a northerly component equal in magnitude to that of the first force but pointing toward the south. Because the magnitudes of both the forces are equal and their northerly components are equal, the westerly components must also be equal. From the symmetry, the second force must be in the southwesterly direction.

28.

(a)

(b)

mg

mg FN

Page 5

N

F1

F2 W

R

Chapter 4

6

29.

(b)

(a) Fbat

mg

mg

Note that we ignore air resistance.

30. (a) We find the horizontal acceleration from the horizontal component of the force exerted on the sprinter, which is the reaction to the force the sprinter exerts on the block: F cos  = ma; 1.3 m/s2. (80 N) cos 22° = (57 kg)a, which gives a = (b) For the motion of the sprinter we can write v = v0 + at = 0 + (1.3 m/s2)(0.34 s) = 0.44 m/s. 31. (a) For the components of the net force we have (a) Fax = – F1 = – 20.2 N; Fay = – F2 = – 26.0 N. F1  We find the magnitude from Fa2 = Fax2 + Fay2 = (– 20.2 N)2 + (– 26.0 N)2, Fa Fa = 32.9 N. which gives We find the direction from tan  =  Fay/ Fax= (26.0 N)/(20.2 N) = 1.29,  = 52.2° below – x-axis. which gives The acceleration will be in the direction of the net force: aa = Fa/m = (32.9 N)/(29.0 kg) = 1.13 m/s2, 52.2° below – x-axis. y (b) (b) For the components of the net force we have Fbx = F1 cos  = (20.2 N) cos 30° = 17.5 N; Fby = F2 – F1 sin  = 26.0 N – (20.2 N) sin 30° = 15.9 N. F2 Fb We find the magnitude from  Fb2 = Fbx2 + Fby2 = (17.5 N)2 + (15.9 N)2,  Fb = 23.6 N. which gives F1 We find the direction from tan  =  Fby/ Fbx = (15.9 N)/(17.5 N) = 0.909,  = 42.3° above + x-axis. which gives The acceleration will be in the direction of the net force: ab = Fb/m = (23.6 N)/(29.0 kg) = 0.814 m/s2, 42.3° above + x-axis. 



32.

y x F2





(b) Because the velocity is constant, the acceleration is zero. We write ?F = ma from the force diagram for the mower: x-component: F cos  – FR = ma = 0, which gives FR = (78.0 N) cos 45° = 55.2 N. (c) y-component: FN – mg – F sin  = ma = 0, which gives FN = (13.0 kg)(9.80 m/s2) + (78.0 N) sin 45° = 183 N. (d) We can find the acceleration from the motion of the mower: a = ?v/?t = (1.2 m/s – 0)/(2.0 s) = 0.60 m/s2. For the x-component of ?F = ma we now have F cos  – FR = ma; F cos 45° – 55.2 N = (13.0 kg)(0.60 m/s2), which gives F = Page 6

(a)

y

F



x

FR FN

89 N.

mg

x

Chapter 4

7

33. (a) With down positive, we write ?F = ma from the force diagram for m1: m1g – FT = m1a. With up positive, we write ?F = ma from the force diagram for m2: FT – m2g = m2a; (b) When we add the two equations, we get (m1 – m2)g = (m1 + m2)a, or a = (m1 – m2)g/(m1 + m2) = (1150 kg – 1000 kg)(9.80 m/s2)/(1150 kg + 1000 kg) = 0.68 2 m/s . Using one of the equations, we find the tension: m1g – FT = m1a, or FT = m1(g – a) = (1150 kg )(9.80 m/s2 – 0.68 m/s2) = 10,500 N. 34. (a) We select the helicopter and the car as the system. We write ?F = ma from the force diagram: Fair Fair– mhg – mcg = (mh + mc)a, which gives Fair = (mh + mc)(a + g) = (7500 kg + 1200 kg)(0.52 m/s2 + 9.80 m/s2) = 8.98104 N. m hg (b) We select the car as the system. We write ?F = ma from the force diagram: FT– mcg = mca, which gives FT = mc(a + g) m cg = (1200 kg)(0.52 m/s2 + 9.80 m/s2) = 1.24104 N. 35. (a) Because the buckets are at rest, the acceleration is zero. We write ?F = ma from the force diagram for each bucket: lower bucket: FT2 – m2g = m2a = 0, which gives FT2 = m2g = (3.5 kg)(9.80 m/s2) = 34 N. upper bucket: FT1 – FT2 – m1g = m1a = 0, which gives FT1 = FT2 + m1g = 34 N + (3.5 kg)(9.80 m/s2) = 68 N. (b) The two buckets must have the same acceleration. We write ?F = ma from the force diagram for each bucket: lower bucket: FT2 – m2g = m2a, which gives FT2 = m2(g + a) = (3.5 kg)(9.80 m/s2 + 1.60 m/s2) = 40 N. upper bucket: FT1 – FT2 – m1g = m1a, which gives FT1 = FT2 + m1(g + a) = 40 N + (3.5 kg)(9.80 m/s2 + 1.60 m/s2) = 80 N. Note that we could have taken both buckets as the system to find the tension in the upper cord. 36. (a) Because the speed is constant, the acceleration is zero. We write ?F = ma from the force diagram: FT + FT – mg = ma = 0, which gives FT = !mg = !(58 kg)(9.80 m/s2) = (b) Now we have: FT + FT – mg = ma;

2.8102 N.

2(1.10)(!mg) – mg = ma, which gives a = 0.10g = 0.10(9.80 m/s2) = 0.98 m/s2.

Page 7

a

+y a

+y m 1g

+y

FT

m cg

FT1

y

FT2

m 1g

x FT2

m 2g

FT +y

FT

FT

FT

mg

m 2g

Chapter 4

8

37. There is no acceleration. From the symmetry of the force diagram, we see that the tension is the same on each side of Arlene. We write ?F = ma from the force diagram: ?Fy = may: 2FT sin  – mg = 0, which gives FTmin = mg/2 sin max = (50.0 kg)(9.80 m/s2)/2 sin 10° = 1.4103 N.

y

x



FT

FT



mg

38. There is no acceleration. From symmetry we have 3 pairs of wires, with the vertical components equal to one-sixth the weight. We write ?F = ma from the force diagram: ?Fy = may; FT1 = mg /6 = (70.0 kg)(9.80 m/s2)/6 = 114 N; FT2 cos 30° = mg/6 = (70.0 kg)(9.80 m/s2)/6, which gives FT2 = 132 N. FT3 cos 60° = mg/6 = (70.0 kg)(9.80 m/s2)/6, which gives FT3 = 229 N.

39. There is no acceleration perpendicular to the line L. We write ?F = ma from the force diagram: ?Fx = max; FB sin B – FA sin A = 0; FB sin 30° – (4500 N) sin 50° = 0, which gives FB = 6890 N. The resultant force is in the y-direction: ?Fy = FB cos B + FA cos A = (6890 N) cos 30° + (4500 N) cos 50° =

L A

B

FA 8860 N.

y

x

40. (a) From the force diagram for the block, we have ?F = ma: x-component: mg sin  = ma; y-component: FN – mg cos  = 0. From the x-equation we find the acceleration: a = g sin  = (9.80 m/s2) sin 22.0° = 3.67 m/s2. (b) For the motion of the block, we find the speed from v2 = v02 + 2a(x – x0); v2 = 0 + 2(3.67 m/s2)(12.0 m – 0), which gives v = 9.39 m/s.

Page 8

FN

 y

 x

mg

FB

Chapter 4

9

41. We choose the origin for x at the bottom of the plane. Note that down the plane (the direction of the acceleration) is positive. FN (a) From the force diagram for the block, we have ?F = ma: x-component: mg sin  = ma;  y-component: FN – mg cos  = 0. y From the x-equation we find the acceleration: a = g sin  = (9.80 m/s2) sin 22° = 3.67 m/s2.  For the motion of the block, we find the distance until it momentarily stops from x mg v2 = v02 + 2a(x – x0); 0 = (– 4.0 m/s)2 + 2(3.67 m/s2)(x – 0), which gives x = – 2.2 m. Thus the block travels 2.2 m up the plane. (b) We find the time from the start to return to the bottom from x = x0 + v0t + !at2; 0 = 0 + (– 4.0 m/s)t + !(3.67 m/s2)t2, which gives t = 0 (the start), and

42. We can find the angle the wire makes with the vertical from sin  = d/L = (0.10 m)/(4.0 m) = 0.025, which gives  = 1.43°. (a) We write ?F = ma from the force diagram for the chandelier: x-component: FT sin  – F = 0, or F = FT sin ; y-component: FT cos  – mg = 0, or mg = FT cos .    If we divide the two equations, we get F = mg tan  = (27 kg)(9.80 m/s2) tan 1.43° = 6.6 N. (b) From the y-equation we have FT = mg/cos  = (27 kg)(9.80 m/s2)/cos 1.43° = 2.6102 N.

t = 2.2 s.

y

L 

F

FT

d

x

mg

43. We find the two accelerations: a1 = F0/m; a2 = 2F0/m. We choose the origin of coordinates at the initial position. For the motion up to time t0 we have x1 = x01 + v01t + !a1t2 = 0 + 0 + !(F0/m)t02 = !(F0/m)t02; and v1 = v01 + a1t0 = 0 + (F0/m)t0 = (F0/m)t0. These values become the initial ones for the motion after t0: x2 = x02 + v02t + !a2t2 = !(F0/m)t02 + (F0/m)t0(2t0 – t0) + !(2F0/m)(2t0 – t0)2 =

Page 9

((F0/m)t02.

Chapter 4

10

44. The two buckets and the cords must have the same acceleration. We write ?F = ma from the force diagram for the lower bucket: FTbottomlow – m2g = m2a, which gives FTbottomlow = m2(g + a) = (3.5 kg)(9.80 m/s2 + 1.60 m/s2) = 40 N. The tensions shown in the diagram are exerted by the cords. If we consider the lower cord, which now has weight wbottom , as the system, the tensions will be in the opposite directions. FTbottomhigh – FTbottomlow – wbottom = (wbottom/g)a, which gives FTbottomhigh = FTbottomlow + wbottom + (wbottom/g)a, = 40 N + 2.0 N + [(2.0 N)/(9.80 m/s2)](1.60 m/s2) = 42.3 N. We write ?F = ma from the force diagram for the upper bucket: FTtoplow – FTbottomhigh – m1g = m1a, which gives FTtoplow = FTbottomhigh + m1(g + a) = 42.3 N + (3.5 kg)(9.80 m/s2 + 1.60 m/s2) = 82.2 N. If we consider the upper cord, which now has weight wtop , as the system, the tensions will be in the opposite directions. FTtophigh – FTtoplow – wtop = (wtop/g)a, which gives FTtophigh = FTtoplow + wtop + (wtop/g)a, = 82.2 N + 2.0 N + [(2.0 N)/(9.80 m/s2)](1.60 m/s2) = 84.5 N. Thus we have top cord: FTtophigh = 85 N; FTtoplow = 82 N; bottom cord: FTbottomhigh = 42 N; FTbottomlow = 40 N. 45. If we select the first and second cars as the system, the only horizontal force is the tension in the coupling between the locomotive and the first car. From the force diagram, we have ?Fx = (m1 + m2)ax , or F1 = (m + m)a = 2ma. If we select the second car as the system, the only horizontal force is the tension in the coupling between the first car and the second car. From the force diagram, we have ?Fx = m2ax , or F2 = ma. Thus we have F1/F2 = 2ma/ma = 2, for any nonzero acceleration. 46.

(b) If we select all three blocks as the system, we have ?Fx = max: F = (m1 + m2 + m3)a, which gives a = F/(m1 + m2 + m3). (c) For the three blocks individually, for ?Fx = max we have Fnet1 = m1a = m1F/(m1 + m2 + m3); Fnet2 = m2a = m2F/(m1 + m2 + m3); Fnet3 = m3a = m3F/(m1 + m2 + m3). (d) From the force diagram for block 1 we have Fnet1 = F – F12 = m1a, which gives F12 = F – m1a = F – m1F/(m1 + m2 + m3) = F(m2 + m3)/(m1 + m2 + m3). This is also F21 (Newton’s third law). From the force diagram for block 2 we have Fnet2 = F21 – F23 = m2a, which gives F23 = F21 – m2a = F – m1a – m2a = F – (m1 + m2)F/(m1 + m2 + m3) = Fm3/(m1 + m2 + m3). Page 10

(a)

FTtophigh y

FTtoplow

a m1g

FTbottomhigh

FT2

FTbottomlow

m 2g

locomotive 2

1 x

2

1

F1

F2

2

F

m1 x

m2

m3

FN1

F

F12

m 1g F21

FN2 m 2g

F32

FN3

m 3g

F23

Chapter 4

11

This is also F32 (Newton’s third law). (e) When we use the given values, we get a = F/(m1 + m2 + m3) = (96.0 N)/(12.0 kg + 12.0 kg + 12.0 kg) = 2.67 m/s2. Fnet1 = m1a = (12.0 kg)(2.67 m/s2) = 32.0 N. Fnet2 = m2a = (12.0 kg)(2.67 m/s2) = 32.0 N. Fnet3 = m3a = (12.0 kg)(2.67 m/s2) = 32.0 N. Because the blocks have the same mass and the same acceleration, we expect Fnet1 = Fnet2 = Fnet3 = 32.0 N. For the forces between the blocks we have F21 = F12 = F – m1a = 96.0 N – (12.0 kg)(2.67 m/s2) = 64.0 N. 2 F32 = F23 = F – m1a – m2a = 96.0 N – (12.0 kg)(2.67 m/s ) – (12.0 kg)(2.67 m/s2) = 32.0 N. 47.

(a) (b) If the system is released from rest, the blocks will have the same acceleration in the directions indicated on the diagram. We write ?F = ma from the force diagram for each block: x-component (m1): FT = m1a; y-component (m1): FN – m1g = 0; y-component (m2): m2g – FT = m2a. When we add the first and third equations, we get m2g = (m1 + m2)a, which gives a = m2g/(m1 + m2). When we use this result in the first equation, we get FT = m1m2g/(m1 + m2).

y

FN

FT x a

m 1g

FT a y

m2 g

48. (a) From the result for Problem 47 we have a = m2 g/(m1 + m2) = (6.0 kg)(9.80 m/s2)/(13.0 kg + 6.0 kg) = (b) For the motion of the top block we have x1 = x01 + v01t + !a1t2;

3.1 m/s2.

0.90 s. 1.250 m = 0 + 0 + !(3.12 m/s2)t2 , which gives t = (c) From the result for Problem 47 we have a = m2 g/(m1 + m2); (1/100)g = (1.0 kg)g/(m1 + 1.0 kg), which gives m1 = 99.0 kg. 49. If the blocks are released from rest, they will have the same acceleration with the directions indicated on the diagram. Because the cord has mass, the tension will vary along the cord. We call the magnitude of the tension at the pulley FT3. With the upper block and the horizontal piece of the cord as the system, the tension at the pulley will be to the right. We write ?F = ma: x-component: FT3 = [m1 + mC(¬1/¬)]a. With the lower block and the vertical piece of the cord as the system, the tension at the pulley will be up. We write ?F = ma: y-component: [m2 + mC(¬2/¬)]g – FT3 = [m2 + mC(¬2/¬)]a. When we add the two equations, we get a = [m2 + mC(¬2/¬)]g/(m1 + m2 + mC).

Page 11

FN

m 1g

FT3

¬1 a

¬2

y x

FT3

a m 2g

y

Chapter 4

12

50. Forces are drawn for each of the blocks. Because the string doesn’t stretch, the tension is the same at each end of the string, and the accelerations of the blocks have the same magnitude. Note that we take the positive direction in the direction of the acceleration for each block. We write ?F = ma from the force diagram for each block: y-component (block 1): FT – m1g = m1a; y-component (block 2): m2g – FT = m2a. y By adding the equations, we find the acceleration: FT a = (m2 – m1)g/(m1 + m2) FT 2 = (3.2 kg – 2.2 kg)(9.80 m/s )/(3.2 kg + 2.2 kg) a a = 1.81 m/s2 for both blocks. For the motion of block 1 we take the origin at the ground y and up positive. When block 2 hits the ground, we have m 1g v12 = v012 + 2a(y1 – y01) m 2g 2 = 0 + 2(1.81 m/s ) (3.60 m – 1.80 m), which gives v1 = 2.56 m/s. Once block 2 hits the ground, FT 0 and block 1 will have the downward acceleration of g. For this motion of block 1 up to the highest point reached, we have v2 = v12 + 2a(h – y1) 0 = (2.56 m/s)2 + 2(– 9.80 m/s2) (h – 3.60 m), which gives h = 3.93 m. 51.

y x

m2

FT2

FT2

FT1 mc

FT1

FP m1

The blocks and the cord will have the same acceleration. If we select the two blocks and cord as the system, we have ?Fx = max: FP = (m1 + m2 + mC)a, which gives a = FP/(m1 + m2 + mC) = (40.0 N)/(10.0 kg + 12.0 kg + 1.0 kg) = 1.74 m/s2. For block 1 we have ?Fx = max: FP – FT1 = m1a; 40.0 N – FT1 = (10.0 kg)(1.74 m/s2), which gives FT1 = 22.6 N. For block 2 we have ?Fx = max: FT2 = m2a; FT2 = (12.0 kg)(1.74 m/s2) = 20.9 N. Note that we can see if these agree with the analysis of ?Fx = max for the cord: FT1 – FT2 = mCa: 22.6 N – 20.9 N = (1.0 kg)a, which gives a = 1.7 m/s2.

Page 12

Chapter 4

13

52. Because the cord connecting the masses does not stretch and we can ignore the mass of the pulley, the tension is the same at each end of this cord, and the accelerations of the blocks have the same magnitude. Forces are drawn for each of the blocks. Note that we take the positive direction in the direction of the acceleration for each block. We write ?F = ma from the force diagram for each block: FT y-component (block 1): FT – m1g = m1a; y-component (block 2): m2g – FT = m2a. y By adding the equations, we find the acceleration: FT a = (m2 – m1)g/(m1 + m2). We use this in the equation for block 1 to find the tension: a FT = m1(m2 – m1)g/(m1 + m2) + m1g = 2m1m2g/(m1 + m2). There is no acceleration of the pulley. We write ?F = ma from the force diagram for the pulley: m 1g FTC – FT – FT = 0, or FTC = 2FT = 4m1m2g/(m1 + m2) = 4(1.2 kg)(3.2 kg)(9.80 m/s2)/(1.2 kg + 3.2 kg) = 34 N. 53. If m does not move on the incline, both blocks must have the same horizontal acceleration. From the force diagram for the block of mass m, we have ?F = ma: x-component: FN1 sin  = ma; y-component: FN1 cos  – mg = 0. When we combine these two equations, we get: a = g tan , and FN1 = mg/cos . From the force diagram for the block of mass M, we have ?F = ma: x-component: F – FN1 sin  = Ma; y-component: FN2 –FN1 cos  – Mg = 0. We use the previous results in the first equation to get F = mg tan  + Mg tan  = (m + M)g tan .

Page 13

FTC

FT FT a y m 2g

y

a x

FN2





FN1

FN1

F

 Mg

mg

Chapter 4

14

54. If there is no motion of m1 relative to m3 , y FN1 all blocks must have the same horizontal FT acceleration. Thus the tension in the cord x acting on m2 must have a horizontal a component as shown in the diagram. m 1g For the mass m2 , we have ?F = ma: x-component: FT sin  = m2a; FT  y-component: FT cos  – m2g = 0. FT When we combine these two equations, FT  we get: FN3 F FN1 a = g tan , and FT = m2g/cos . For the mass m1 , we have ?F = ma: m2 g x-component: FT = m1a; y-component: FN1 – m1g = 0. We can combine the two x-equations to m 3g get sin  = m2/m1. If we put this in a triangle, as shown, we can find tan . Thus the acceleration is m1 a = g tan  = m2g/(m12 – m22)1/2. m2 For the block m3 (including the pulley), we have ?F = ma:  x-component: F – FT sin  – FT = m3a; (m 1 2 – m 2 2 )1/2 y-component: FN3 – FN1 – FT cos  – m3g = 0. When we use the previous results in the x-equation, we get F = FT sin  + FT + m3a = (m2g/cos ) sin  + m1a + m3a = (m2 + m1 + m3)a = (m1 + m2 + m3)m2g/(m12 – m22)1/2. Note that we would obtain this result directly if we had chosen the three blocks and the pulley as the system. The only horizontal force would be F, which would produce the acceleration of the three blocks. 55. The force diagrams for each of the masses and the movable pulley are shown. Note that we take down as positive and the indicated accelerations are relative to the fixed pulley. A downward acceleration of m3 means an upward acceleration of the movable pulley. If we call ar the (downward) acceleration of m1 with respect to the movable pulley, we have a1 = ar – a3 and a2 = – ar – a3 , FT3 because the acceleration of m2 with respect to the pulley must be the negative of m1’s acceleration with respect to a3 the pulley. If the mass of the pulley is negligible, for the movable pulley we write ?Fy = may: FT1 FT1 FT3 +y 2FT1 – FT3 = (0)(– a3), so 2FT1 = FT3. For each of the masses, for ?Fy = may we get m1g – FT1 = m1a1 = m1(ar – a3), mass m1: mass m2: m2g – FT1 = m2a2 = m2(– ar – a3), mass m3: m3g – FT3 = m3a3. We have four equations for the four unknowns: FT1 , FT3 , ar , and a3. After some careful algebra, we get a3 = [(m1m3 + m2m3 – 4m1m2)/(m1m3 + m2m3 + 4m1m2)]g; ar = [2(m1m3 – m2m3)/(m1m3 + m2m3 + 4m1m2)]g; FT1 = [4m1m2m3/(m1m3 + m2m3 + 4m1m2)]g; and FT3 = [8m1m2m3/(m1m3 + m2m3 + 4m1m2)]g. Page 14

FT1

a3 FT1 a2

a1

m 1g

m 2g

m 3g

Chapter 4

15

We can now find the other accelerations: a1 = [(m1m3 – 3m2m3 + 4m1m2)/(m1m3 + m2m3 + 4m1m2)]g; a2 = [(– 3m1m3 + m2m3 + 4m1m2)/(m1m3 + m2m3 + 4m1m2)]g. 56. Because the force produces the acceleration, we have ?F = ma: x-component: Ct2 = ma = m dv/dt. We integrate this to find the velocity: v

t

t

2

3

Ct dt , wh i ch gi ves v = Ct . m 3m 0 0 0 Because v = dx/dt, we integrate this to find the displacement: x 0

dv =

dx =

t 0

adt =

v dt =

t

Ct 3 d t, wh i ch gi v es 0 3m

4 x = Ct . 12m

57. From the result of Problem 49 we have the acceleration of each mass and the cord: a = [m2 + mC(¬2/¬)]g/(m1 + m2 + mC). To simplify the symbol, we let y (= ¬2) represent the vertical displacement of the hanging mass. Thus we have a = dv/dt = [m2 + mC(y/¬)]g/(m1 + m2 + mC). This contains three variables, v, y, and t, so we cannot integrate directly. They are not independent, because v = dy/dt . We use the chain rule to eliminate the variable t: dv/dt = (dv/dy)(dy/dt) = v(dv/dy) = [m2 + mC(y/¬)]g/(m1 + m2 + mC), or [m2 + mC(y/¬)]g dy = (m1 + m2 + mC)v dv. When we integrate this, we get y

0

m 2 + (m C y/ ) g d y =

v

m 1 + m 2 + m C v d v , which gives

0

[m2y + mC(y2/2¬)]g = (m1 + m2 + mC)v2/2, or

v = {[2m2¬2 + mC(¬22/¬)]g/(m1 + m2 + mC)}1/2.

58. (a) We assume the directions indicated in the diagram and call the magnitude of the tension in the cable at the pulley FT. FT FT The only other forces are those due to gravity. For a uniform cable, the mass of each side is proportional to the length of cable on that side. Note that we take the positive direction in the a a direction of the acceleration on each side. We write ?F = ma from the force diagram for each side: FT – (M/L)(L – y)g = (M/L)(L – y)a; L–y y (M/L)yg – FT = (M/L)ya. By adding the equations, we find the acceleration: a = (2y – L)g/L. (b) We assume that y0 > L/2, so the cable accelerates in the direction we assumed in part (a). We have a = dv/dt = (2y – L)g/L. This contains three variables, v, y, and t, so we cannot integrate directly. They are not independent, because v = dy/dt . We use the chain rule to eliminate the variable t: dv/dt = (dv/dy)(dy/dt) = v(dv/dy) = (2y – L)g/L, or [(2y – L)g/L] dy = v dv. When we integrate this, we get L y0

(2 y/ L) – 1 g d y =

( y 2g/ L)

L y0

– gy

L y0

vf 0

v d v , which gives

= gL – ( y02g/ L) – gL + gy 0 = v f2 / 2, or

vf = [2gy0(1 – y0/L)]1/2. Page 15

Chapter 4

16

(c) When y0 = %L, we get vf = [2g%L(1 – %)]1/2 =

%(gL)1/2.

59. The acceleration can be found from the blood’s one-dimensional motion: v = v0 + at; 0.35 m/s = (0.25 m/s) + a(0.10 s), which gives a = 1.00 m/s2. We apply Newton’s second law to find the required force ?F = ma; F = (2010–3 kg)(1.00 m/s2) = 2.010–2 N. 60. We apply Newton’s second law to the person: ?F = ma; F = (70 kg)(– 30)(9.80 m/s2) = – 2.1104 N (opposite to motion). We find the distance from v2 = v02 + 2a (x – xo) 0 = [(90 km/h)/(3.6 ks/h)]2 + 2(– 30)(9.80 m/s2)(x – xo), which gives x – xo =

1.1 m.

61. We find the downward acceleration from the motion: v2 = v02 + 2a(y – y0); (29 m/s)2 = 0 + 2a(55 m – 0), which gives a = 7.64 m/s2. The forces are gravity downward and the air resistance upward. We can write ?F = ma: mg – FR = ma; (2.0 kg)(9.80 m/s2) – FR = (2.0 kg)(7.64 m/s2), which gives FR = 4.3 N. 62. We write ?F = ma from the force diagram for the fish: y-component: FT – mg = ma, or FT = m(a + g). (a) At constant speed, a = 0, so we have FTmax = mmaxg; 45 N = mmax(9.80 m/s2), which gives mmax = 4.6 kg. (b) For an upward acceleration, we have FTmax = mmax(a + g); 45 N = mmax(2.0 m/s2 + 9.80 m/s2), which gives mmax = 3.8 kg. Note that we have ignored any force from the water. 63. For the motion until the elevator stops, we have v2 = v02 + 2a(x – x0); 0 = (3.5 m/s)2 + 2a(3.0 m), which gives a = – 2.04 m/s2. We write ?F = ma from the force diagram for the elevator: mg – FT = ma; or (1300 kg)(9.80 m/s2) – FT = (1300 kg)(– 2.04 m/s2), which gives FT = 1.5104 N.

Page 16

+y FT

mg

FT +y

mg

Chapter 4

17

64. The tension in the cables supporting the load must have a horizontal component as shown in the diagram. For the load, we have ?F = ma: x-component: FT sin  = ma; y-component: FT cos  – mg = 0. When we combine these two equations, we get: a = g tan  = (9.80 m/s2) tan 5.0° = 0.86 m/s2.

a

 FT

y x

mg

65. From the force diagram for the soap, we have ?F = ma: x-component: mg sin  = ma; y-component: FN – mg cos  = 0. Thus the acceleration is a = g sin . We find the time to reach the bottom from x = x0 + v0t + !at2;

FN



x

 mg

t = 1.9 s. 3.0 m = 0 + 0 + !(9.80 m/s2)(sin 9.5°)t2, which gives Because the acceleration is independent of the mass, there would be no change in the time for a heavier mass. 66. (a) The forces and coordinate systems are shown in the diagram. From the force diagram, with the block m2 as the system, we can write ?F = ma: y-component: m2g – FT = m2a. From the force diagram, with the block m1 as the system, we can write ?F = ma: x-component: FT – m1g sin  = m1a. When we eliminate FT between these two equations, we get a = (m2 – m1 sin )g/(m1 + m2). (b) Because up the plane is our positive direction, we have a down (negative) requires m2 < m1 sin . a up (positive) requires m2 > m1 sin .

FT

FN

FT



x

y

y

 m1g

m 2g

67. (a) The forces and coordinate systems are shown in the diagram. Note that we take up the incline as the FT positive direction. From the force diagram, with the block m2 as the system, we can write ?F = ma: y-component: m2g – FT = m2a. From the force diagram, with the block m1 as the system, y we can write ?F = ma: x-component: FT – m1g sin  = m1a. m 2g When we eliminate FT between these two equations, we get a = (m2 – m1 sin )g/(m1 + m2). = [1.00 kg – (1.00 kg)(sin 30°)](9.80 m/s2)/(1.00 kg + 1.00 kg) = 2.45 m/s2 (up the incline). (b) If the system remains at rest, the acceleration is zero, so we have a = (m2 – m1 sin )g/(m1 + m2) = 0, or m2 = m1 sin  = (1.00 kg)(sin 30°) = 0.50 kg. (c) From the force equations, we have Page 17

y

FT

FN



x

 m1g

y

Chapter 4

18

FT = m2(g – a). For part (a), we get FT = (1.00 kg)(9.80 m/s2 – 2.45 m/s2) = For part (b), we get FT = (0.50 kg)(9.80 m/s2 – 0) = 4.9 N.

7.35 N.

y

68. (a) Forces are drawn for each of the blocks. Note that we use two different coordinate systems. We write ?F = ma from the FT force diagram for each block: FN 1 For block 1: x-component: m1g sin 1 – FT = m1a. For block 2: x-component: FT – m2 g sin 2 = m2a. m 1g By adding the equations, we have a = (m1 sin 1 – m2 sin 2)g/(m1 + m2). (b) If the system remains at rest, the acceleration is zero, so we have a = (m1 sin 1 – m2 sin 2)g/(m1 + m2) = 0, or m1 sin 1 = m2 sin 2; (5.0 kg)(sin 30°) = m2 sin 20°, which gives m2 = 7.3 kg. From the force equations, we have FT = m1(g sin 1 – a) = (5.0 kg)[(9.80 m/s2 )(sin 30°) – 0] =

y x

x

FT

m 2g

Page 18

2

25 N.

69. Both motions have constant velocity, so the acceleration is zero. From the force diagram for the motion coasting down the hill, we can write ?F = ma: x-component: mg sin  – FR = 0, or FR = mg sin .  From the force diagram for the motion climbing up the hill, we can write ?F = ma: x-component: F – mg sin  – FR = 0, so F = FR + mg sin  = 2mg sin  = 2(65 kg)(9.80 m/s2) sin 6.0° = 1.3102 N. 70. We take the positive direction upward. The scale reads the force the person exerts on the scale. From Newton’s third law, this is also the magnitude of the normal force acting on the person. The effective mass on the scale is mscale = FN/g. We write ?F = ma from the force diagram for the person: y-component: FN – mg = ma, or mscale = FN/g = m(a + g)/g. (a) When the elevator is at rest, a = 0: mscale = m(0 + g)/g = m = 75.0 kg (735 N).

FN 2

1

FN

FR



mg

+y

FN mg

x

Chapter 4

19

(b) When the elevator is climbing at constant speed, a = 0: mscale = m(0 + g)/g = m = 75.0 kg (735 N). (c) When the elevator is falling at constant speed, a = 0: mscale = m(0 + g)/g = m = 75.0 kg (735 N). (d) When the elevator is accelerating upward, a is positive: mscale = m(a + g)/g = (75.0 kg)(3.0 m/s2 + 9.80 m/s2)/(9.80 m/s2) = 98.0 kg (960 N). (e) When the elevator is accelerating downward, a is negative: mscale = m(a + g)/g = (75.0 kg)(– 3.0 m/s2 + 9.80 m/s2)/(9.80 m/s2) = 52.0 kg (510 N).

71. We find the acceleration of the car on the level from y v = v0 + at; x (21 m/s) = 0 + a(14.0 s), which gives a = 1.5 m/s2. This acceleration is produced by the net force: Fnet = ma = (1100 kg)(1.5 m/s2) = 1650 N. If we assume the same net force on the hill, with no acceleration on the steepest hill, from the force  diagram we have x-component: Fnet – mg sin  = 0; 1650 N – (1100 kg)(9.80 m/s2) sin  = 0, which gives sin  = 0.153, or

FN Fnet mg

 = 8.8°.

72. The velocity is constant, so the acceleration is zero. (a) From the force diagram for the bicycle, we can write ?F = ma: x-component: mg sin  – Fair = 0, or mg sin  = cv; (80 kg)(9.80 m/s2) sin 5.0° = c(6.0 km/h)/(3.6 ks/h), which gives c = 41 kg/s. (b) We have an additional force in ?F = ma: x-component: F + mg sin  – Fair = 0, so F = cv – mg sin  = [(41 kg/s)(20.0 km/h)/(3.6 ks/h)] – (80 kg)(9.80 m/s2) sin 5.0° = 73. From the force diagram for the watch, we have ?F = ma: x-component: FT sin  = ma; y-component: FT cos  – mg = 0, or FT cos  = mg. If we divide the two equations, we find the acceleration: a = g tan  = (9.80 m/s2) tan 25° = 4.56 m/s2. For the motion of the aircraft, we find the takeoff speed from v = v0 + at = 0 + (4.56 m/s2)(18 s) = 82 m/s (300 km/h).

FN



mg

1.6102 N.

y

 FT x a

Page 19

Fair

mg

x

Chapter 4

20

74. We assume the cart is pushed up the ramp at constant speed. From the force diagram for the cart, we have ?F = ma: x-component: F – mg sin  = ma = 0; y-component: FN – mg cos  = 0. The maximum angle is determined by the maximum applied force: Fmax = mg sin max; 20 N = (20 kg)[(9.80 m/s2) sin max , which gives max = 5.9°.

y x

FN v

 F

 mg

75. There are only three different tensions. The tension in the rope that goes around both pulleys is constant: FT1 = FT2 = F. We choose up positive and assume that the piano is lifted with no acceleration. If the masses of the pulleys are negligible, we can write ?Fy = may. (a) If we select the piano and bottom pulley as the system, we have FT2 + FT1 – Mg = 0, which gives

FT3

y FT4

F = !Mg. 2FT1 = Mg, or FT1 = (b) For the individual elements, we have piano: FT4 – Mg = 0, which gives FT4 = Mg. bottom pulley: FT2 + FT1 – FT4 = 0, which gives

FT2

FT1 F

FT2 FT4 Mg

Bottom pulley

FT1 Top pulley

FT1 = FT2 = !Mg. 2FT1 = FT4 = Mg, or top pulley: FT3 – FT1 – FT2 – F = 0, which gives FT3 = 3FT1 = *Mg. 76. From the force diagram for the pilot, we have ?F = ma: Fseat x-component: Fseat cos  – mg sin  = ma, or Fseat cos  = mg sin  + ma; x y-component: Fseat sin  – mg cos  = 0, or Fseat sin  = mg cos .  If we divide the two equations, we find the angle of the force from the seat: tan  = (g cos )/(g sin  + a) = (9.80 m/s2)(cos 45°)/[(9.80 m/s2)(sin 45°) + 4.5 m/s2] = 0.607, so  = 31.2°.  We can now find the magnitude of the force from the y-equation: Fseat sin 31.2° = (75 kg)(9.80 m/s2) cos 45°, which gives Fseat = 1.00103 N, 31° above the direction of the plane, or 76° above the horizontal. 77. We find the acceleration of the automobile from v = v0 + at; 0 = (50 km/h)/(3.6 ks/h) + a(0.20 s), which gives a = – 69.4 m/s2. Page 20

a

mg

y

Chapter 4

21

We assume that the automobile is on a level surface and there is no relative motion of the chair with respect to the seat back or the seat back with respect to the frame of the automobile. If we treat the child as a particle, this horizontal acceleration is produced by the net force of the straps: Fnet = ma = (12 kg)(– 69.4 m/s2) = – 8.3102 N. 78. We select the frame as the system and write ?F = ma from the force diagram: FT– mg = ma, which gives FT = m(a + g) = (600 kg)(0.15 + 1) (9.80 m/s2) = 6.76103 N.

+y

FT

a

mg

79. There will be two horizontal forces on the train: the force exerted against the track, Ftrack , and the drag force from the air, Fdrag , which depends on the speed. We select the train as the system and write ?F = ma: Ftrack– Fdrag = ma. (a) The maximum acceleration will occur when the force on the track is maximum and the drag force is minimum. This occurs when the speed is zero, that is, when the train is starting to move. Thus we have 400103 N – 0 = (660,000 kg)amax , which gives amax = 0.606 m/s2. (b) At top speed the acceleration is zero, so we have 150 kN. 150103 N – Fdrag = 0, which gives Fdrag = 150103 N =

Page 21

Chapter 5

CHAPTER 5 - Further Applications of Newton’s Laws 1.



2.



3.

The friction is kinetic, so Ffr = kFN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write ?F = ma: x-component: F – kFN = 0; y-component: FN – Mg = 0. Thus FN = Mg, and   F = kFN = kMg = (0.30)(12.0 kg)(9.80 m/s2) = 35 N. If k = 0, there is no force required to maintain constant speed.

FN

y

F x Ffr

Mg

(a) In general, static friction is given by Ffr = sFN. Immediately FN before the box starts to move, the static friction force reaches its maximum value: Ffr,max = sFN. For the instant before the box F starts to move, the acceleration is zero. Using the force diagram for the box, we can write ?F = ma: x-component: F – sFN = 0; Ffr Mg y-component: FN – Mg = 0. Thus FN = Mg, and    F = sFN = sMg ; 25.0 N = s(6.0 kg)(9.80 m/s2), which gives s = 0.43. (b) When the box accelerates and the friction changes to kinetic, we have F – kFN = Ma; 25.0 N – k(6.0 kg)(9.80 m/s2) = (6.0 kg)(0.50 m/s2), which gives k = 0.37. (a)

(b)

FN

Ffr



(c)

FN

Ffr

FN





Ffr mg

mg

mg

In (a) the friction is static and opposes the impending motion down the plane. In (b) the friction is kinetic and opposes the motion down the plane. In (c) the friction is kinetic and opposes the motion up the plane. 4.

If we simplify the forces so that there is one normal force, we have the diagram shown. We can write ?F = ma: x-component: – Ffr + mg sin  = 0; y-component: FN – mg cos  = 0. When we combine the two equations, we have tan  = Ffr/FN = s. Thus we have tan max = s = 0.8, max = 39°.

Page 1

y

FN

Ffr

x mg



y x

Chapter 5

5.

6.

If we simplify the forces so that there is one normal force, we have the diagram shown. The friction force provides the acceleration. We can write ?F = ma: x-component: Ffr = Ma; y-component: FN – Mg = 0. Thus we have a = Ffr/M = sFN/M = sg. The minimum value of s is s,min = a/g = 0.20.

FN y x a Ffr

Mg

If we simplify the forces so that there is one normal force, we have the diagram shown. The friction force provides the acceleration. We can write ?F = ma: x-component: Ffr = Ma; y-component: FN – Mg = 0. Thus we have a = Ffr/M = sFN/M = sg. The maximum value of a is amax = sg = (0.80)(9.80 m/s2) = 7.8 m/s2.

FN

x a Ffr

Mg

7.

While the box is sliding down, friction will be up the FN plane, opposing the motion. From the force diagram for the box, we have ?F = ma:  x-component: mg sin  – Ffr = ma; y Ffr y-component: FN – mg cos  = 0. From the x-equation, we have  Ffr = mg sin  – ma = m(g sin  – a ) x mg = (15.0 kg)[(9.80 m/s2) sin 30° – (0.30 m/s2)] v = 69 N. Because the friction is kinetic, we have Ffr = kFN = kmg cos ; 69 N = k(15.0 kg)(9.80 m/s2) cos 30°, which gives k = 0.54.

8.

On the horizontal the only force is the friction force, which provides the acceleration: – Ffr = ma1. On the hill, we have x-component: – mg sin  – Ffr = ma2 , or a2 = – g sin  + a1 = – (9.80 m/s2) sin 13° + (– 4.80 m/s2) = – 7.00 m/s2.

Page 2

v

FN y

y



x Ffr mg

Chapter 5

9.

For the hanging box we can write ?F = ma: y-component: mIIg – FT = 0. For the box on the table we can write ?F = ma: x-component: FT – Ffr = 0; y-component: FN – mIg = 0. When we combine the equations, we have Ffr = FT = sFN = smIg. Thus we have mI = mII/s = (2.0 kg)/(0.25) = 8.0 kg.

FT x FT

Ffr

10. From the force diagram for the soap, we have ?F = ma: x-component: mg sin  – Ffr = ma; y-component: FN – mg cos  = 0, or FN = mg cos . The friction is kinetic, so Ffr = kFN = kmg cos . From the x-equation we find the acceleration: a = g sin  – kg cos  = g(sin  – k cos  ) = (9.80 m/s2)[sin 8.0° – (0.060) cos 8.0°] = 0.78 m/s2. For the motion of the soap, we find the time from x = x0 + v0t + !at2; 9.0 m = 0 + 0 + !(0.78 m/s2)t2, which gives

y

FN

y

m Ig

m IIg

FN

 y

 x

mg

t = 4.8 s.

11. The friction is kinetic, so Ffr = kFN. Because the push is no longer there, the only horizontal force is the friction force. Using the force diagram for the box, we can write ?F = ma: x-component: – kFN = Ma; y-component: FN – Mg = 0. Thus we have a = – kg. For the motion with constant acceleration, we have v2 = v02 + 2ax = v02 + 2(– kg)x; 0 = (2.5 m/s)2 + 2(– 0.25)(9.80 m/s2)x, which gives x = 1.3 m. 12. The boxes and the cord have the same acceleration with the directions indicated on the diagram. The length of FN the horizontal piece of the cord is ¬1, so the vertical piece Ffr has length ¬2 = ¬ – ¬1. Because the cord has mass, the tension will vary along the cord. We call the magnitude m Ig of the tension at the pulley FT3. y With the upper box and the horizontal piece of the cord as the system, the tension at the pulley will be to the right. We write ?F = ma: x-component: FT3 – kFN = [mI + mC(¬1/¬)]a; y-component: FN – mIg = 0, so FT3 – kmIg = [mI + mC(¬1/¬)]a. With the lower block and the vertical piece of the cord as the system, the tension at the pulley will be up. We write ?F = ma: y-component: [mII + mC(¬2/¬)]g – FT3 = [mII + mC(¬2/¬)]a. When we add the two equations, we get a = [mII + mC(¬2/¬) – kmI]g/(mI + mII + mC). Page 3

Ffr

FN

y x

Ffr

Mg

FT3

¬1 a

x

FT3 ¬2

a m IIg

y

Chapter 5

Note that as the cord moves over the pulley, the acceleration will not be constant.

13. From the force diagram for the car we have y x-component: F – FT = mcara; y-component : FN,car – mcarg = 0. FN, trailer From the force diagram for the trailer we x have FT FT x-component: FT – kFN,trailer = mtrailera; y-component: FN,trailer – mtrailerg = 0, Ffr or FN,trailer = mtrailerg. m car g m trailer g If we add the two x-equations, we get F – kmtrailerg = mcara + mtrailera; 3.5103 N – (0.15)(350 kg)(9.80 m/s2) = (1000 kg + 350 kg)a, which gives a = 2.21 m/s2. We can find the force on the trailer from the x-equation for the trailer: FT – kmtrailerg = mtrailera; FT – (0.15)(350 kg)(9.80 m/s2) = (350 kg)(2.21 m/s2), which gives FT = 1.3103 N.

FN, car

14. (a) If the automobile does not skid, the friction is static, with Ffr = sFN. On a level road, the normal force is FN = mg. The static friction force is the only force slowing the automobile and will be maximum in order to produce the minimum stopping distance. We find the acceleration from ?Fx = max: – smg = ma, which gives a = – sg . For the motion until the automobile stops, we have vfinal2 = v02 + 2a(x – x0); 0 = v2 + 2(– sg)(xmin), which gives xmin = v2/2sg . (b) For the given data we have xmin = [(95 km/h)/(3.6 ks/h)]2/2(0.75)(9.80 m/s2) = 47 m. (c) The only change is in the value of g: xmin = [(95 km/h)/(3.6 ks/h)]2/2(0.75)(1.63 m/s2) = 2.8102 m. 15. (a) We write ?F = ma from the force diagram for the snow while it is stationary on the roof:

Page 4

F

Chapter 5



x-component: mg sin – Ffr = 0; y y-component: FN – mg cos  = 0.  When we combine the equations, we get FN Ffr = mg sin = sFN = smg cos . x Thus we have s = tan = tan 30° = 0.58. Ffr (b) We write ?F = ma from the force diagram for the  snow while it is sliding on the roof: x-component: mg sin – kFN = ma;  y-component: FN – mg cos  = 0.  mg  Thus a = g(sin – k cos ) = (9.80 m/s2)[sin 30°– (0.20) cos 30°] = 3.20 m/s2. For the motion on the roof, we have v12 = v02 + 2a(x1 – x0) = 0 + 2(3.20 m/s2)(5.0 m), which gives v1 = 5.7 m/s. (c) The motion when the snow leaves the roof is projectile motion, with an initial velocity of v1 = 5.7 m/s at 30° below the horizontal. If we use the new coordinate system shown, we have vx = v1 cos  = (5.7 m/s) cos 30° = 4.9 m/s; vy2 = (– v1 sin  )2 + 2gh = [– (5.7 m/s) sin 30°]2 + 2(– 9.80 m/s2)(– 10.0 m), which gives vy = – 14.3 m/s. The speed of the snow is v = (vx2 + vy2)1/2 = [(4.9 m/s)2 + (– 14.3 m/s)2]1/2 = 15 m/s.

y v1 h

16. On a level road, the normal force is FN = mg. The kinetic friction force is the only force slowing the automobile. We find the acceleration from the horizontal component of ?F = ma: – kmg = ma, which gives a = – kg = – (0.80)(9.80 m/s2) = – 7.84 m/s2. For the motion until the automobile stops, we have v2 = v02 + 2a(x – x0); 0 = v02 + 2(– 7.84 m/s2)(80 m), which gives v0 = 35 m/s (130 km/h). 17. (a) The two crates must have the same acceleration. F From the force diagram for crate 1 we have m2 m1 x-component: F – F12 – kFN1 = m1a; x y-component: FN1 – m1g = 0, or FN1 = m1g. FN1 From the force diagram for crate 2 we have F12 F x-component: F12 – kFN2 = m2a; y-component: FN2 – m2g = 0, or FN2 = m2g. Ffr1 m 1g If we add the two x-equations, we get F – km1g – km2g = m1a + m2a; FN2 F21 = – F 12 F – k(m1 + m2)g = (m1 + m2)a; 750 N – (0.12)(80 kg + 210 kg)(9.80 m/s2) = Ffr2 m2g 1.4 m/s2. (80 kg + 210 kg)a, which gives a = (b) We can find the force between the crates from the x-equation for crate 2: F12 – km2g = m2a; F12 – (0.12)(210 kg)(9.80 m/s2) = (210 kg)(1.41 m/s2), which gives F12 = 5.4102 N. Page 5

x

Chapter 5

(c) If the crates are reversed, the acceleration will be the same: a = 1.4 m/s2. For the force between the crates, we have F12 – km1g = m1a; F12 – (0.12)(80 kg)(9.80 m/s2) = (80 kg)(1.41 m/s2), which gives F12 = 2.1102 N. 18. (a) While the block is sliding down, friction will be up the plane, opposing the motion. From the force diagram for the block, we have ?F = ma: x-component: mg sin  – kFN = ma. y-component: FN – mg cos  = 0. When we combine these, we have a = g(sin – k cos ) = (9.80 m/s2)[sin 22.0°– (0.17) cos 22.0°] = 2.1 m/s2. (b) For the motion of the block, we have v2 = v02 + 2a(x – x0) = 0 + 2(2.13 m/s2)(9.3 m), which gives v= 6.3 m/s.

FN

 y

Ffr

 x

mg v

19. We choose the origin for x at the bottom of the plane. FN Note that down the plane (the direction of the acceleration) is positive. (a) From the force diagram for the block, we have ?F = ma:  x-component: mg sin  + kFN = ma. y y-component: FN – mg cos  = 0. Ffr When we combine these, we have  a = g(sin + k cos ) x mg = (9.80 m/s2)[sin 22.0°+ (0.17) cos 22.0°] = 5.22 m/s2. v For the motion on the block until it stops, we have v2 = v02 + 2a(x – x0); 0 = (– 3.0 m/s)2 + 2(5.22 m/s2)(x – 0), which gives x = – 0.86 m. Thus the block travels 86 cm up the plane. (b) We find the time to reach the highest point from v = v0 + auptup ; 0 = (– 3.0 m/s) + (5.22 m/s2)tup , which gives tup = 0.575 s. When the block slides down the plane, the friction force will reverse, so the acceleration is adown = g(sin – k cos ) = (9.80 m/s2)[sin 22.0°– (0.17) cos 22.0°] = 2.13 m/s2. We find the time to slide down from x = x0 + vt + !adowntdown2; 0 = – 0.862 m + 0 + !(2.13 m/s2)tdown2, which gives tdown = 0.900 s. Thus the total time is t = tup + tdown = 0.575 s + 0.900 s = 1.5 s. Page 6

Chapter 5

20. We find the angle each cable makes with the ground from sin  = (18 m)/(30 m) = 0.600,  = 36.9°. From the force diagram for the concrete block, we have ?F = ma: x-component: FT cos  – Ffr = 0 y-component: FN – mg + FT sin  = 0. When we combine these, we have FT cos  = Ffr = sFN = s(mg – FT sin ), or FT = smg/(cos + s sin ) = (0.80)(1600 N)/[cos 36.8° + (0.80) sin 36.8°]. so FT = 1.0103 N. 21. We choose the coordinate system shown in the force diagram and assume the cord is taut. (a) From the force diagram for each block, we have ?F = ma: y-component (1) FN1 – m1g cos  = 0, or FN1 = m1g cos ; x-component (1): m1g sin  – 1FN1 – FT = m1a; v (m1 sin  – 1m1 cos )g – FT = m1a; y-component (2) FN2 – m2g cos  = 0, or FN2 = m2g cos ; x-component (2): m2g sin  – 2FN2 + FT = m2a; (m2 sin  – 2m2 cos )g + FT = m2a. When we add the two x-equations, we have (m1 + m2)a = [(m1 + m2) sin – (1m1 + 2m2) cos ]g;

FT

 y FN x Ffr

mg

y

x FN1

FT

FN2 FT

Ffr2



m 2g

Ffr1

 m 1g

(5.0 kg + 5.0 kg)a = {[(5.0 kg + 5.0 kg) sin 30°– [(0.20)(5.0 kg) + (0.30)(5.0 kg)] cos 30°}(9.80 m/s2) which gives a = 2.8 m/s2. (b) We find the tension from FT = m1[(sin  – 1 cos )g – a]

2.1 N. = (5.0 kg){[sin 30° – (0.20) cos 30°](9.80 m/s2) – (2.78 m/s2)} = Note that the positive result justifies our assumption that the cord is taut. 22. We choose the coordinate system shown in the force diagram. From the force diagram for each block, we have ?F = ma: y y-component (1): FN1 – m1g cos  = 0, or FN2 FN1 = m1g cos ; x-component (1): m1g sin  – 1FN1 – FT = m1a1; x FT? (m1 sin  – 1m1 cos )g – FT = m1a1; Ffr2 FN1 FT? y-component (2): FN2 – m2g cos  = 0, or v FN2 = m2g cos ;  m 2g x-component (2): m2g sin  – 2FN2 + FT = m2a2; Ffr1 (m2 sin  – 2m2 cos )g + FT = m2a2.  The motion will depend on whether there is a tension in m g the cord. If there is no tension, the blocks will have 1 different accelerations: a1 = (sin  – 1 cos )g; a2 = (sin  – 2 cos )g. (a) If there is no tension and 1 < 2 , we see that a1 > a2 . The lower block would move away from the upper block, creating a tension in the cord. Thus the two blocks must slide with the same acceleration. (b) If there is no tension and 1 > 2 , we see that a1 < a2 . The upper block would move toward from the lower block, and the tension would remain zero. Thus each block will slide with its own acceleration until the upper block contacts the lower block. Page 7

Chapter 5

(c) In part (a) the two blocks have the same acceleration. When we add the two x-equations, we have (m1 + m2)a = [(m1 + m2) sin – (1m1 + 2m2) cos ]g, or

a = {sin  – [(1m1 + 2m2)/(m1 + m2)] cos }g. We find the tension from FT = m1[(sin  – 1 cos )g – a] = m1g{sin  – 1 cos  – sin  + [(1m1 + 2m2)/(m1 + m2)] cos }; FT = [m1m2(2 – 1)/(m1 + m2)]g cos .

Note that if 1 > 2 , we would get a negative tension, which the cord cannot provide. In part (b), FT = 0; the two blocks have different accelerations: a1 = (sin  – 1 cos )g; a2 = (sin  – 2 cos )g. 23. We choose the coordinate system shown in the force diagram. If the blocks are connected by a rod, which can support a tension or a compression, they must have the same acceleration. From the force diagram for each block, we have ?F = ma: y-component (1): FN1 – m1g cos  = 0, or FN1 = m1g cos ; x-component (1): m1g sin  – 1FN1 – FT = m1a; (m1 sin  – 1m1 cos )g – FT = m1a; y-component (2): FN2 – m2g cos  = 0, or FN2 = m2g cos ; x-component (2): m2g sin  – 2FN2 + FT = m2a; (m2 sin  – 2m2 cos )g + FT = m2a. When we add the two x-equations, we have (m1 + m2)a = [(m1 + m2) sin – (1m1 + 2m2) cos ]g, or

y

FN2

x FN1

v

FT

FT

Ffr2



m 2g

Ffr1

 m 1g

a = {sin  – [(1m1 + 2m2)/(m1 + m2)] cos }g. We find the tension from FT = m1[(sin  – 1 cos )g – a] = m1g{sin  – 1 cos  – sin  + [(1m1 + 2m2)/(m1 + m2)] cos }; FT = [m1m2(2 – 1)/(m1 + m2)]g cos .

If 1

< 2 , there will be tension in the rod; if 1 > 2 , there will be compression in the rod.

24. (a) For the hanging box we can write ?F = ma: y-component: m2g – FT = 0. For the box on the table we can write ?F = ma: x-component: FT – Ffr = 0; y-component: FN – m1g = 0. When we combine the equations, we have Ffr = FT = m2g = sFN = sm1g. Thus we have m1 = m2/s = (2.0 kg)/(0.40) = 5.0 kg.  (b) The acceleration is zero, so the only change is that the friction is kinetic: m1 = m2/k = (2.0 kg)/(0.30) = 6.7 kg.

Page 8

y

FN

x

FT

FT Ffr

m1g

y m 2g

Chapter 5

25. (a) We select the origin at the bottom of the ramp, with y up positive. We find the acceleration from the motion up FN the ramp: 2 2 v = v0 + 2a(x – x0); v 0 = v02 + 2a(d – 0), which gives a = – v02/2d. When the block slides up the ramp, kinetic friction will Ffr be down, opposing the motion. From the force diagram for the block, we have ?F = ma:  x-component: – mg sin  – kFN = ma; mg y-component: FN – mg cos  = 0. When we eliminate FN from the two equations and use the result for a, we get

x



– mg sin  – kmg cos  = m(– v02/2d), which gives k = (v02/2gd cos ) – tan . (b) Once the block stops, the friction becomes static and will be up the plane, to oppose the impending motion down. If the block remains at rest, the acceleration is zero. The static friction force must be = sFN and we have x-component: – mg sin  + Ffr = 0, or Ffr = mg sin  = smg cos . Thus we know that s = tan .

26. We simplify the forces to the three shown in the diagram. y If the car does not skid, the friction is static, with Ffr = sFN. The static friction force will be maximum just before the car x slips. We write ?F = ma from the force diagram: x-component: mg sin max– sFN = 0; y-component: FN – mg cos max = 0. When we combine these, we get tan max = s = 0.15, or max = 8.5°. Thus a car will slip on any driveway with an incline greater than 8.5°. The only driveway safe to park in is Bonnie’s.

27. The direction of the kinetic friction force is determined by the direction of the velocity, not the direction of the acceleration. We assume the block on the plane is moving up, so the friction force is down. The forces and coordinate systems are shown in the diagram. From the force diagram, with the block m2 as the system, we can write ?F = ma: y-component: m2g – FT = m2a. From the force diagram, with the block m1 as the system, we can write ?F = ma: x-component: FT – Ffr – m1g sin  = m1a; y-component: FN – m1g cos  = 0; with Ffr = kFN. Page 9

FN

mg



FN y

FT FT



x Ffr

Ffr

 m1g

y m 2g

Chapter 5

When we eliminate FT between these two equations, we get a = (m2 – m1 sin  – km1 cos )g/(m1 + m2). (a) For a mass m1 = 5.0 kg, we have a = [5.0 kg – (5.0 kg) sin 30° – (0.10)(5.0 kg) cos 30°](9.80 m/s2)/(5.0 kg + 5.0 kg) = 2.0 m/s2 up the plane. The acceleration is up the plane because the answer is positive. This agrees with our assumption for the direction of motion. (b) For a mass m1 = 2.0 kg, we have a = [5.0 kg – (2.0 kg) sin 30° – (0.10)(2.0 kg) cos 30°](9.80 m/s2)/(2.0 kg + 5.0 kg) = 5.4 m/s2 up the plane. The acceleration is up the plane because the answer is positive. This agrees with our assumption for the direction of motion. 28. If we assume the block is moving up the plane, we find the mass required for zero acceleration from a = (m2 – m1up sin  – km1up cos )g/(m1up + m2); 0 = [5.0 kg – m1up sin 30° – (0.50)m1up cos 30°](9.80 m/s2)/(m1up + 5.0 kg). which gives m1up = 5.4 kg. If we assume the block is moving down the plane, the direction of the friction force reverses. We find the mass required for zero acceleration from a = (m2 – m1down sin  + km1down cos )g/(m1down + m2); 0 = [5.0 kg – m1down sin 30° + (0.50)m1down cos 30°](9.80 m/s2)/(m1down + 5.0 kg). which gives m1down = 75 kg. Because s = k , if the blocks are at rest, they will remain at rest if 5.4 kg = m1 = 75 kg. 29. The kinetic friction force will be up the slide to oppose the motion. We choose the positive direction in the direction of the acceleration. From the force diagram for the child, we have ?F = ma: x-component: mg sin  – Ffr = ma; y-component: FN – mg cos  = 0. When we combine these, we get a = g sin  – kg cos  = g(sin  – k cos ). We can use this for the frictionless slide if we set k = 0. For the motion of the child, we have v2 = v02 + 2a(x – x0) = 0 + 2ad, where d is the distance along the slide. If we form the ratio for the two slides, we get (vfriction/vnone)2 = afriction/anone = (sin  – k cos )/sin ; (!)2 = (sin 28° – k cos 28°)/sin 28°, which gives

FN Ffr

x mg

k = 0.40.

30. On a level belt, the normal force is FN = mg. The static friction force provides the acceleration and reverses direction for the second half of the trip. We find the acceleration from the horizontal component of ?F = ma: Ffr = ma, or a = Ffr/m = sg. From the symmetry of the motion, for the first half of the distance we have x = v0t + !at2; Page 10

y

FN

y x

mg

Ffr

a

Chapter 5

!D = 0 + !a(!t)2, or t2 = 4D/a.

We see that the minimum time will be achieved with the maximum acceleration: amax = sg: tmin2 = 4(10 m)/(0.60)(9.80 m/s2), which gives tmin = 2.6 s. 31. The velocity is constant, so the acceleration is zero. (a) From the force diagram for the bicycle, we can write ?F = ma: x-component: mg sin  – FD = 0, or mg sin  = cv2; (80.0 kg)(9.80 m/s2) sin 7.0° = c[(9.5 km/h)/(3.6 ks/h)]2, which gives c = 14 kg/m. (b) We have an additional force in ?F = ma: x-component: F + mg sin  – FD = 0, so F = cv2 – mg sin  = (13.7 kg/m)[(25 km/h)/(3.6 ks/h)]2 – (80 kg)(9.80 m/s2) sin 7.0° = 32. (a) The two blocks must have the same acceleration. From the force diagram for the top block we have x-component: Ffr1 = M1a; y-component: FN1 – M1g = 0, or FN1 = M1g. For the static friction force we have Ffr1 = M1a = sFN1 = M1g. Thus we have  = a/g = (5.2 m/s2)/(9.80 m/s2) = 0.53. (b) If the coefficient of friction is less, the top block will slide. Because the friction force is one-half the force in part (a), the acceleration is also reduced by half: a1 = !a = !(5.2 m/s2) = 2.6 m/s2. (c) For the relative acceleration we have a12 = a1 – a = 2.6 m/s2 – 5.2 m/s2 = – 2.6 m/s2. (d) From the force diagram for the bottom block we have x-component: F – Ffr1 = M2a; y-component: FN2 – FN1 – M2g = 0, or FN2 = M1g + M2g. For part (a), if we add the two x-equations, we get Fa = M1a + M2a = (4.0 kg + 12.0 kg)(5.2 m/s2) = 83 N. For part (b), if we add the two x-equations, we get Fb = M1a1 + M2a = (4.0 kg)(2.6 m/s2) + (12.0 kg)(5.2 m/s2) = 73 N.

33. If the triangular block is pushed so that the small block tends to move up the incline, the static friction force on the small block will be down the incline, as shown. We choose a coordinate system with the x-axis in the direction of the acceleration a of the triangular block. From the force diagram for the small block we have Page 11

FN

mg

Ffr



5.7102 N.

FN1 y M 1g

Ffr1 a

FN1 Ffr1

x

FN2 F

M 2g

x

FN Chapter 5

Ffr x-component: FN sin  + Ffr cos  = max;  Ffr y-component: FN cos  – mg – Ffr sin  = may. mg y a The top block will not slide until Ffr > FN. As long as FN2 F FN this is not true, ax = a, and ay = 0. Thus we find the x limiting acceleration by using these conditions:  FN cos  – mg – FN sin  = 0, or FN = mg/(cos  –  sin ); FN sin  + FN cos  = mamax , or Mg amax = FN(sin  +  cos )/m = g(sin  +  cos )/(cos  –  sin ). From the force diagram for the triangular block we have x-component: F – FN sin  – Ffr cos  = Ma. Thus the maximum force F for the small block to not slide, and thus the minimum force to make the small block slide, is Fmin = FN sin  + Ffr cos  + Mamax = (m + M)amax = (m + M)g(sin  +  cos )/(cos  – s sin ). 34. The centripetal acceleration of the Earth is aR = v2/r = (2pr/T)2/r = 4p2r/T2 = 4p2(1.501011 m)/(3.16107 s)2 = 5.9310–3 m/s2 toward the Sun. The net force that produces this acceleration is Fnet = mEaR = (5.981024 kg)(5.9310–3 m/s2) = 3.551022 N toward the Sun. This force is the gravitational attraction from the Sun. 35. If the car does not skid, the friction is static, with Ffr = sFN. This friction force provides the centripetal acceleration. We take a coordinate system with the x-axis in the direction of the centripetal acceleration. We write ?F = ma from the force diagram for the auto: x-component: Ffr = maR = mv2/R; y-component: FN – mg = 0. The speed is maximum when Ffr = Ffr,max = sFN. When we combine the equations, the mass cancels, and we get sg = vmax2/R; (0.55)(9.80 m/s2) = vmax2/(80.0 m), which gives vmax = 21 m/s. The mass canceled, so the result is independent of the mass.

y FN aR Ffr mg

36. The force on the discus produces the centripetal acceleration: F = maR = mv2/r; 60.0 N = (2.00 kg)v2/(1.00 m), which gives v = 5.48 m/s. 37. (a) The centripetal acceleration is aR = v2/r = (1.50 m/s)2/(9.0 m) = 0.25 m/s2 toward the center. (b) The net horizontal force that produces this acceleration is Fnet = maR = (25.0 kg)(0.25 m/s2) = 6.3 N toward the center. 38. The centripetal acceleration is aR = v2/r = 2r = (2pf)2r; (100,000)(9.80 m/s2) = (2pf)2(0.090 m), which gives f = 525 rev/s =

Page 12

3.2104 rpm.

x

Chapter 5

39. Yes. If the bucket is traveling fast enough at the top of the circle, in addition to the weight of the water a force from the bucket, similar to a normal force, is required to provide the necessary centripetal acceleration to make the water go in the circle. From the force diagram, we write FN + mg = ma = mvtop2/R. The minimum speed is that for which the normal force is zero: 0 + mg = mvtop,min2/R, or vtop,min = (gR)1/2.

FN

mg

R

40. At the top of the trip, both the normal force and the weight are downward, which we take for the positive direction. We write ?F = ma from the force diagram for the passenger: y-component: FN + mg = mv2/R. The speed v will be minimum when the normal force is minimum. The normal force can only push away from the seat, that is, with our coordinate system it must be positive, so FNmin = 0. Thus we have vmin2 = gR, or vmin = (gR)1/2 = [(9.80 m/s2)(8.0 m)]1/2 = 8.9 m/s.

41. The static friction force provides the centripetal acceleration. We write ?F = ma from the force diagram for the coin: x-component: Ffr = mv2/R; y-component: FN – mg = 0. The highest speed without sliding requires Ffr,max = sFN. The maximum speed before sliding is vmax = 2pR/Tmin = 2pRfmax = 2p(0.120 m)(50 /min)/(60 s/min) = 0.628 m/s. Thus we have smg = mvmax2/R s(9.80 m/s2) = (0.628 m/s)2/(0.120 m), which gives s = 42. The car will leave the road if the normal force becomes zero. For the radial direction the net force provides the radial acceleration: mg cos  – FN = mv2/R, or FN = mg cos  – mv2/R. For the car to stay on the road, FN > 0, so we have mg cos  > mv2/R, or R > v2/g cos . We see that we must use this restriction at the largest angle: Rmin = v2/g cos max Page 13

FN

mg

R

y FN Ffr Mg

x R

0.34.

FN



v

mg

Chapter 5

= [(90 km/h)/(3.6 ks/h)]2/(9.80 m/s2) cos 22° =

69 m.

43. We find the speed of the skaters from the period of rotation: v = 2pr/T = 2p(0.80 m)/(3.0 s) = 1.68 m/s. The pull or tension in their arms provides the centripetal acceleration: FT = mv2/R; = (60.0 kg)(1.68 m/s)2/(0.80 m) = 2.1102 N. 44. The net force on Tarzan will provide his centripetal acceleration, which we take as the positive direction. We write ?F = ma from the force diagram for Tarzan: FT – mg = ma = mv2/R. The maximum speed will require the maximum tension that Tarzan can create: 1400 N – (80 kg)(9.80 m/s2) = (80 kg)v2/(4.8 m), which gives v = 6.1 m/s.

R

FT

v

45. The mass moves in a circle of radius r and has a centripetal acceleration. We write ?F = ma from the force diagram for the mass: x-component: FT cos = mv2/r; y-component: FT sin – mg = 0. Combining these, we get rg = v2 tan ; (0.600 m)(9.80 m/s2) (7.54 m/s)2 tan , which gives  = 5.91°. tan  = 0.103, or We find the tension from FT= mg/sin  = (0.150 kg)(9.80 m/s2)/ sin 5.91° = 14.3 N.

mg

y FT



x

mg

46. For the rotating ball, the tension provides the centripetal acceleration, ?FR = MaR: FT = Mv2/R. We see that the tension increases if the speed increases, so the maximum tension determines the maximum speed: FTmax = Mvmax2/R; 80 N = (0.35 kg)vmax2/(1.0 m), which gives vmax = 15 m/s. If there were friction, it would be kinetic opposing the motion of the ball around the circle. Because this is perpendicular to the radius and the tension, it would have no effect on the maximum speed. 47. At the top of the hill, the normal force is upward and the weight is downward, which we select as the positive direction. (a) We write ?F = ma from the force diagram for the car: mcarg – FNcar = mv2/R; (1000 kg)(9.80 m/s2) – FNcar = (1000 kg)(20 m/s)2/(100 m), which gives FNcar = 5.8103 N. (b) When we apply a similar analysis to the driver, we have (70 kg)(9.80 m/s2) – FNdriver = (70 kg)(20 m/s)2/(100 m), which gives FNdriver = 4.1102 N. (c) For the normal force to be equal to zero, we have (1000 kg)(9.80 m/s2) – 0 = (1000 kg)v2/(100 m), which gives v = 31 m/s (110 km/h or 70 mi/h).

Page 14

FN

R

mg

Chapter 5

48. The masses will have different velocities: v1 = 2pr1/T = 2pr1 f; v2 = 2pr2/T = 2pr2 f. We choose the positive direction toward the center of the circle. For each mass we write ?Fr = mar: m1: FT1 – FT2 = m1v12/r1 = 4p2m1r1 f 2; m2: FT2 = m2v22/r2 = 4p2m2r2 f 2. When we use this in the first equation, we get FT1 = FT2 + 4p2m1r1 f 2; thus FT1 = 4p2f 2(m1r1 + m2r2); FT2 = 4p2f 2m2r2.

r2 r1 FT1

Because the angle is small, we have sin !(d) ˜ !(d), and get 2FT !(d) = (m /2p) d (2pRf )2/R, which gives FT = 2pmRf 2.

v1

R

49. We consider a small segment of the hoop which subtends an angle d, so it has a mass dm = (m/2pR)R d = (m/2p) d. The speed of the segment is v = 2pRf. The net radial force from the two tensions provides the centripetal acceleration: 2FT sin !(d) = dm v2/R.

m1

FT2

FT2 m 2 v2

d /2

FT dm d

r FT

50. We convert the speeds: FN (70 km/h)/(3.6 ks/h) = 19.4 m/s; y (100 km/h)/(3.6 ks/h) = 27.8 m/s. aR At the speed for which the curve is banked perfectly, there is no need for a friction force. We take the x-axis in the direction of the centripetal acceleration. x Ffr We write ?F = ma from the force diagram for the car:  x-component: FN1 sin  = ma1 = mv12/R;  y-component: FN1 cos  – mg = 0. mg Combining these, we get v12 = gR tan . (19.4 m/s)2 (9.80 m/s2) (65 m) tan , which gives  = 30.6°. tan  = 0.591, or At a higher speed, there is need for a friction force, which will be down the incline. If the automobile does not skid, the friction is static, with Ffr = sFN. We write ?F = ma from the force diagram for the car: x-component: FN2 sin  + Ffr cos  = ma2 = mv22/R; y-component: FN2 cos  – Ffr sin  – mg = 0. We eliminate Ffr by multiplying the x-equation by sin , the y-equation by cos , and adding: FN2 = m{[(v22/R) sin ] + g cos }. By reversing the trig multipliers and subtracting, we eliminate FN2 to get Ffr = m{[(v22/R) cos ] – g sin }. If the automobile does not skid, the friction is static, with Ffr = sFN: m{[(v22/R) cos ] – g sin } = sm{[(v22/R) sin ] + g cos }, or Page 15

Chapter 5

s = {[(v22/R) cos ] – g sin }/{[(v22/R) sin ] + g cos } = [(v22/gR)] – tan ]/{[(v22/gR) tan ] + 1}. When we express tan  in terms of the design speed, we get s = [(v22/gR) – (v12/gR)]/{[(v22/gR)(v12/gR)] + 1} = (1/gR)(v22 – v12)/[(v1v2/gR)2 + 1] = [1/(9.80 m/s2)(65 m)][(27.8 m/s)2 – (19.4 m/s)2]/{[(19.4 m/s)(27.8 m/s)/(9.80 m/s2)(65 m)]2 + 1} = 0.36. 51. We convert the speed: (90 km/h)/(3.6 ks/h) = 25.0 m/s. FN y At the speed for which the curve is banked perfectly, a there is no need for a friction force. We take the x-axis in the direction of the centripetal acceleration. We write ?F = ma from the force diagram for the car: x Ffr x-component: FN1 sin  = ma1 = mv12/R;  y-component: FN1 cos  – mg = 0.  Combining these, we get mg tan  = v12/gR 2 2 = (25.0 m/s) /(9.80 m/s )(60 m) = 1.062, or  = 46.7°. At a higher speed, there is need for a friction force, which will be down the incline to help provide the greater centripetal acceleration. If the automobile does not skid, the friction is static, with Ffr = sFN. At the maximum speed, Ffr = sFN. We write ?F = ma from the force diagram for the car: x-component: FN2 sin  + sFN2 cos  = ma2 = mvmax2/R; y-component: FN2 cos  – sFN2 sin  – mg = 0, or FN2(cos  – s sin ) = mg. When we eliminate FN2 by dividing the equations, we get vmax2 = gR(sin  + s cos )/(cos  – s sin ) = gR(tan  + s)/(1 – s tan  ) = gR(v12 + sgR)/(gR – sv12)] = (9.80 m/s2)(60 m)[(25.0 m/s)2 + 0.30(9.80 m/s2)(60 m)]/[(9.80 m/s2)(60 m) – 0.30(25.0 2 m/s) ], which gives vmax = 34.3 m/s = 123 km/h. At a lower speed, there is need for a friction force, which will be up the incline to prevent the car from sliding down the incline. If the automobile does not skid, the friction is static, with Ffr = sFN. At the minimum speed, Ffr = sFN. The reversal of the direction of Ffr can be incorporated in the above equations by changing the sign of s , so we have vmin2 = gR[(sin  – s cos )/(cos  + s sin )] = gR(v12 – sgR)/(gR + sv12)] = (9.80 m/s2)(60 m)[(25.0 m/s)2 – 0.30(9.80 m/s2)(60 m)]/[(9.80 m/s2)(60 m) + 0.30(25.0 2 m/s) ], which gives vmin = 18.4 m/s = 66 km/h. Thus the range of permissible speeds is 66 km/h < v < 123 km/h. 52. (a) Because the tangential acceleration has a constant magnitude,

Page 16

Chapter 5

we have ætan = !(v0 + v2) = #(2pr)/t;

y t=0

!(0 + v2) = p(2.0 m)/2(2.0 s),

which gives v2 = 3.1 m/s. (b) The average velocity during the interval is vav = ?r/?t = [(2.0 m)i – (2.0 m)j]/(2.0 s) = (1.0 m/s)i – (1.0 m/s)j = 1.4 m/s, 45° below x-axis. (c) The average acceleration during the interval is aav = ?v/?t = [(3.14 m/s)(– j) – 0]/(2.0 s) = – (1.6 m/s2)j = 1.6 m/s2, down.

r = 2.0 m t = 2.0 s x v2

53. The constant magnitude of the tangential acceleration is y atan = ?v/?t = (3.14 m/s – 0)/(2.0 s) = 1.6 m/s2. atan0 t=0 The magnitude of the radial acceleration is t = 1.0 s aR = v2/r. atan1 (a) At t = 0.0 s, we have 2 atan0 = (1.6 m/s )i; r = 2.0 m aR1 aR0 = 0. t = 2.0 s Thus the acceleration is a0 = (1.6 m/s2)i. x (b) We find the distance traveled along the circle from aR2 2 2 2 d1 = v0t1 + !atant1 = 0 + !(1.57 m/s )(1.0 s) = 0.785 m. atan2 The angle traveled in this time is 1 = d1/r = (0.785 m)/(2.0 m) = 0.393 rad = 22.5°. The speed at this time is v1 = v0 + atant1 = 0 + (1.57 m/s2)(1.0 s) = 1.57 m/s. Thus we have atan1 = 1.6 m/s2, 22.5° below x-axis; aR1 = (1.57 m/s)2/(2.0 m) = 1.2 m/s2, 112.5° below x-axis. Thus the acceleration is a1 = [(1.57 m/s2) cos 22.5°i – (1.57 m/s2) sin 22.5°j] + [– (1.23 m/s2) sin 22.5°i – (1.23 m/s2) cos 22.5°j] = (0.98 m/s2)i – (1.7 m/s2)j. (c) At t = 2.0 s, the particle is on the x-axis. We have v2 = v0 + atant2 = 0 + (1.57 m/s2)(2.0 s) = 3.14 m/s. atan2 = 1.6 m/s2, in – y-direction; aR2 = (3.14 m/s)2/(2.0 m) = 4.93 m/s2, in – x-direction. Thus the acceleration is a2 = – (4.9 m/s2)i – (1.6 m/s2)j.

Page 17

Chapter 5

54. (a) We find the speed from the radial component of the acceleration: aR = a sin  = v12/R ; (0.210)(9.80 m/s2) sin 28.0° = v12/(3.60 m), which gives v1 = 1.86 m/s.  (b) Assuming constant tangential acceleration, we find the speed from v2 = v1 + atant = (1.86 m/s) + (0.210)(9.80 m/s2)(cos 28.0°)(2.00 s) = 5.50 m/s.

55. (a) We find the tangential acceleration from atan = dv/dt = d[3.6 m/s + (1.5 m/s3)t2]/dt = (3.0 m/s3)t = (3.0 m/s3)(3.0 s) = (b) We find the radial acceleration from aR = v2/r = [3.6 m/s + (1.5 m/s3)(3.0 s)2]2/(20 m) = 15 m/s2.

atan

 a

aR

9.0 m/s2.

56. The tangential component of the force produces the tangential acceleration: Ftan = matan = m(b + ct2). We find the tangential speed by integrating: v

v0

dv =

t

0

atan d t =

t

b + ct 2 d t ;

0

v – v0 = bt + @ct3, or v = v0 + bt + @ct3. Thus the radial component of the force is FR = maR = mv2/r; FR = m(v0 + bt + @ct3)2/r. 57. We want to find the powers of m and b, so we assume they are  and , respectively, and put in their dimensions, where b has the dimensions of F/v:  = mb; [T] = [M] [MLT–2/LT–1] = [M] +  [T]– . Equating exponents of each dimension, we get  = – 1;  +  = 0, so  = –  = + 1. Thus we have  = m/b. 58. (a) With a drag force proportional to v, the terminal velocity is vT = mg/b; 9 m/s = (310–5 kg)(9.80 m/s2)/b, which gives b = 3.310–5 kg/s. (b) The time to reach 63% of the terminal velocity is  = m/b = (310–5 kg)/(3.310–5 kg/s) = 0.92 s. 59. (a) For an initial velocity downward, we take the positive direction downward. The drag force will be upward, so we have mg – bv = m dv/dt, or dv/[v – (mg/b)] = – (b/m) dt. When we integrate, we get v v0

dv = v – (mg/ b)

t 0

b d t; –m

ln{[v – (mg/b)]/[v0 – (mg/b]} = – bt/m, or v = (mg/b) + [v0 – (mg/b)] e – bt/m. (b) For an initial velocity upward, we take the positive direction upward. Page 18

Chapter 5

The drag force and gravity will be downward, so we have – mg – bv = m dv/dt, or dv/[v + (mg/b)] = – (b/m) dt. This is the same as in part (a), with g replaced by – g. Thus we have v = – (mg/b) + [v0 + (mg/b)] e –bt/m, v = 0. For the motion after the object comes momentarily to rest, we use the result from part (a) with v0 = 0. 60. (a) For a falling object, we have mg – bv2 = m dv/dt. When it reaches its terminal velocity, we get mg – bvT2 = m dv/dt = 0; or vT = (mg/b)1/2. (b) We find the value of b from vT = (mg/b)1/2; 60 m/s = [(75 kg)(9.80 m/s2)/b]1/2, which gives b = 0.20 kg/m. (c) For the same terminal velocity, the curve for FD  v2 lies above the curve for FD  v. Terminal velocity is reached sooner because the resisting force increases more rapidly. Note that to have the same terminal velocity, the coefficient b will be different.

v FD = – bv2

vT

FD = – bv

t

0

61. (a) We find the coefficient for the velocity-dependent drag force from FD2 = bv2; 1.0 N = b(2.2 m/s)2, which gives b = 0.21 kg/m. The total drag force is FD = FD1 + FD2 = 4.0 N + (0.21 kg/m)v2. (b) To coast down the slope, the net force along the slope must be zero: mg sin  – FD = ma = 0; (80 kg)(9.80 m/s2) sin  = 4.0 N + (0.21 kg/m)(10 m/s)2, which gives

 = 1.8°.

62. From Example 5–15, we have mg – bv = ma, or a = g – (b/m)v, and v = (mg/b)(1 – e –bt/m). When we use the expression for v in that for a, we get a = g – g(1 – e –bt/m) = ge –bt/m. From the definition of v = dy/dt, we have dy/dt = (mg/b)(1 – e –bt/m), or (b/mg) dy = (1 – e –bt/m) dt. When we integrate, we get y 0

(b/ mg) d y =

t 0

1 – e– bt/ m dt;

(b/mg)y = t + (m/b)(e –bt/m – 1), which gives

y = (mg/b)[t – (m/b)(1 – e –bt/m)].

63. When the engine is shut off, the only force on the boat is the drag force of the water, so we have FD = – bv = m dv/dt, or dv/v = – b dt/m. When we integrate, we get v v0

dv = v

t 0

– (b/ m) d t;

ln (v/v0) = – bt/m, or v = v0 e –bt/m. We find the value of b/m from the change in velocity in 3.0 s: ln (!) = – (b/m)(3.0 s), which gives b/m = 0.231 /s. From the definition of v = dx/dt, we have Page 19

Chapter 5

dx/dt = v0 e –bt/m, or dx = v0 e –bt/m dt. When we integrate, we get x

0

dx =

t

0

v 0 e– bt / m dt;

= (– mv0/b)(e –bt/m – 1) = (mv0/b)(1 – e –bt/m) = [(2.4 m/s)/(0.231 /s)][1 – e –(0.231 /s)(8)] =

x

10 m.

64. The drag force produces the acceleration, so we have FD = – bv1/2 = m dv/dt, or dv/v1/2 = – b dt/m. When we integrate, we get v v0

t

dv = v 1/2

0

– (b/ m) d t;

2(v1/2 – v01/2) = – bt/m, or v = v0 – (bv01/2/m)t + (b2/4m2)t2. From the definition of v = dx/dt, we have dx/dt = v0 –(bv01/2/m)t + (b2/4m2)t2, or dx = [v0 –(bv01/2/m)t + (b2/4m2)t2] dt. When we integrate, we get x 0

dx =

t 0

v0 –(bv 01/2 / m)t + (b2 /4 m 2)t2 d t;

x = v0t – (bv01/2/2m)t2 + (b2/12m2)t3. 65. The drawer will suddenly open when the resisting static friction force reaches its maximum value: Ffr,max = sFN. Frequently drawers are stuck from pressure on the sides and top of the drawer. Here we assume that the friction force is produced only by the normal force on the bottom of the drawer. For ?F = ma we have x-component: F – sFN = 0; y-component: FN – Mg = 0. Thus FN = Mg, and    F = sFN = sMg; 8.0 N = s(2.0 kg)(9.80 m/s2), which gives s = 0.41.

66. If an object with one of the surfaces is placed on an inclined FN plane of the other surface, and the object remains stationary, Ffr for ?F = ma we have  y x-component: mg sin  – Ffr = 0; y-component: FN – mg cos  = 0. x  Thus we have tan  = Ffr/FN. mg If we increase the angle until the object just begins to move, we know that the static friction force is maximum, so we have tan  = sFN/FN = s . Thus we can determine the coefficient of static friction by measuring the angle at which the object starts to slide. 67. We can find the required acceleration, assumed constant, from x = v0t + !at2; (0.250 mi)(1610 m/mi) = 0 + !a (6.0 s)2, which gives a = 22.4 m/s2. If we assume that the tires are just on the verge of slipping, Ffr,max = sFN , so we have x-component: sFN = ma; y-component: FN – mg = 0. Page 20

Chapter 5

Thus we have s = ma/mg = a/g = (22.4 m/s2)/(9.80 m/s2) =

2.3.

68. We find the maximum permissible deceleration from the motion until the automobile stops: v = v0 + at; 0 = [(45 km/h)/(3.6 ks/h)] + amax(3.5 s), which gives amax = – 3.57 m/s2. The minimum time for deceleration without the cup sliding means that the static friction force, which is the force producing the deceleration of the cup, is maximum. On a level road, the normal force is FN = mg. The maximum static friction force is Ffr,max = sFN. For the horizontal component of ?F = ma, we have – smg = mamax , which gives s = – amax/g = – (– 3.57 m/s2)/(9.80 m/s2) = 0.36. 69. While the box is sliding down, friction will be up the plane, opposing the motion. From the force diagram for the box, we have ?F = ma: x-component: mg sin  – Ffr = ma. y-component: FN – mg cos  = 0. From the x-equation, we have Ffr = mg sin  – ma = m(g sin  – a ) = (18.0 kg)[(9.80 m/s2) sin 37.0° – (0.270 m/s2)] = 101 N. Because the friction is kinetic, we have Ffr = kFN = kmg cos ; 101.3 N = k(18.0 kg)(9.80 m/s2) cos 37.0°, which gives

FN

 y

Ffr

 x

mg v

k = 0.719.

70. If the crate does not slide, it must have the same acceleration as the truck. The friction is static, with Ffr = sFN. On a level road, the normal force is FN = mg. If we consider the crate as the system, the static friction force will be opposite to the direction of motion (to oppose the impending motion of the crate toward the front of the truck), is the only force providing the acceleration, and will be maximum in order to produce the maximum acceleration. We find the acceleration from the horizontal component of ?F = ma: smg = ma, which gives a = sg = – (0.75)(9.80 m/s2) = – 7.4 m/s2.

71. (a) For the object to move with the ground, the static friction force must provide the same acceleration. With the usual coordinate system, for ?F = ma we have x-component: Ffr = ma; y-component: FN – mg = 0. For static friction, Ffr = sFN , or ma = smg ; thus s = a/g. In order not to slide when the acceleration is maximum, the minimum required coefficient of static friction is s = amax/g. (b) For the greatest acceleration, the minimum required coefficient is s = amax/g =(4.0 m/s2)/(9.80 m/s2) = 0.41. Because this is greater than 0.25, the chair will slide.

Page 21

Chapter 5

72. While the roller coaster is sliding down, friction will be up FN the hill, opposing the motion. From the force diagram for the roller coaster, we have ?F = ma:  x-component: mg sin  – Ffr = mg sin  – kFN = ma. y y-component: FN – mg cos  = 0. Ffr We combine these equations to find the acceleration:  a = g sin  – kg cos  = g(sin  – k cos ) x mg = (9.80 m/s2)[sin 45° – (0.12) cos 45°] = 6.10 m/s2. v For the motion of the roller coaster, we find the speed from v2 = v02 + 2a(x – x0); v2 = [(6.0 km/h)/(3.6 ks/h)]2 + 2(6.10 m/s2)(45.0 m – 0), which gives v = 23 m/s (85 km/h). 73. On a level road, the normal force is FN = mg. The kinetic friction force is the only force slowing the motorcycle. We find the acceleration from the horizontal component of ?F = ma: – kmg = ma, which gives a = – kg = – (0.80)(9.80 m/s2) = – 7.84 m/s2. For the motion through the sandy stretch, we have v2 = v02 + 2a(x – x0); v2 = (20.0 m/s)2 + 2(– 7.84 m/s2)(15 m), which gives v = ± 12.8 m/s. The negative sign corresponds to the motorcycle going beyond the sandy stretch and returning, assuming the same negative acceleration after the motorcycle comes to rest. This will not occur, so the motorcycle emerges with a speed of 13 m/s. If the motorcycle did not emerge, we would get a negative value for v2, indicating that there is no real value for v. 74. We write ?F = ma from the force diagram for the stationary hanging mass, with down positive: mg – FT = ma = 0; which gives FT = mg. For the rotating puck, the tension provides the centripetal acceleration, ?FR = MaR: FT = Mv2/R. When we combine the two equations, we have Mv2/R = mg, which gives v = (mgR/M)1/2.

FN FT FT Mg

R y

puck

mg

75. The horizontal force on the astronaut produces the centripetal acceleration: F = maR = mv2/r; (7.75) m (9.80 m/s2) = mv2/(10.0 m), which gives v = 27.6 m/s. The rotation rate is Rate = v/2pr = (27.6 m/s)/2p(10.0 m) = 0.439 rev/s. Note that the results are independent of mass, and thus are the same for all astronauts. 76. The velocity of the people is v = 2pR/T = 2pRf = 2p(5.0 m)(0.50 /s) = 15.7 m/s. The force that prevents slipping is an upward friction force. The normal force provides the centripetal acceleration. We write ?F = ma from the force diagram for the person: x-component: FN = mv2/R; y-component: Ffr – mg = 0. Because the friction is static, we have Ffr = sFN , or mg = smv2/R. Page 22

R

Ffr

FN mg

y x

Chapter 5

Thus we have s = gR/v2 = (9.80 m/s2)(5.0 m)/(15.7 m/s)2 = 0.20. There is no force pressing the people against the wall. They feel the normal force and thus are applying the reaction to this, which is an outward force on the wall. There is no horizontal force on the people except the normal force. 77. The assumed constant magnitude of the tangential acceleration is atan = ?v/?t = (30 m/s – 0)/(9.0 s) = 3.33 m/s2. For the components of the net force we have ?Ftan = matan = (1000 kg)(3.33 m/s2) = 3.3103 N; ?FR = maR = mv2/r = (1000 kg)(30 m/s)2/(450 m) = 2.0103 N. 78. We convert the speed: (80 km/h)/(3.6 ks/h) = 22.2 m/s. FN We take the x-axis in the direction of the centripetal y acceleration. We find the speed when there is no need a for a friction force. We write ?F = ma from the force diagram for the car: x Ffr x-component: FN1 sin  = ma1 = mv12/R; y-component: FN1 cos  – mg = 0.  Combining these, we get  v12 = gR tan  = (9.80 m/s2)(80 m) tan 14°, mg which gives v1 = 14.0 m/s. Because the speed is greater than this, a friction force is required. Because the car will tend to slide up the slope, the friction force will be down the slope.We write ?F = ma from the force diagram for the car: x-component: FN2 sin  + Ffr cos  = ma2 = mv22/R; y-component: FN2 cos  – Ffr sin  – mg = 0. We eliminate FN2 by multiplying the x-equation by cos , the y-equation by sin , and subtracting: Ffr = m{[(v22/R ) cos  ] – g sin }

= (1000 kg)({[(22.2 m/s)2/(80 m)] cos 14°} – (9.80 m/s2) sin 14°) =

Page 23

3.6103 N down the slope.

Chapter 5

79. (a) There will be two forces in the vertical direction. We write ?F = ma from the force diagram. At point A the radial acceleration will be down, so we take FN that as the positive direction: mg – FNA = maA = mv2/R, so FNA = m(g – v2/R). At point B the radial acceleration is zero, so we have mg – FNB = maB = 0, so FNB = mg. At point C the radial acceleration will be up, so we take that as the positive direction: FNC – mg = maC = mv2/R, so FNC = m(g + v2/R). mg Thus we see that FNC > FNB > FNA. (b) The force diagram for the driver would look the same and there will be similar normal forces on the driver to provide the radial accelerations, so the drive will feel heaviest at C, and lightest at A. (c) Because the normal force cannot be negative, the car will lose contact at A when mg = mv2/R. Thus the maximum speed at A without losing contact is vAmax = (gR)1/2. 80. We find the constant tangential acceleration from the motion around the turn: vtan2 = v02 + 2atan(xtan – x0) [(320 km/h)/(3.6 ks/h)]2 = 0 + 2atan[p(200 m) – 0], which gives atan = 6.29 m/s2. The centripetal acceleration depends on the speed, so it will increase around the turn. We find the speed at the halfway point from v12 = v02 + 2atan(x1 – x0) = 0 + 2(6.29 m/s2)[p(100 m) – 0], which gives v1 = 62.8 m/s. The radial acceleration is aR = v12/R = (62.8 m/s)2/(200 m) = 19.7 m/s2. The magnitude of the acceleration is a = (atan2 + aR2)1/2 = [(6.29 m/s2)2 + (19.7 m/s2)2]1/2 = 20.7 m/s2. On a flat surface, FN = Mg; and the friction force must provide the acceleration: Ffr = Ma. With no slipping the friction is static, so we have Ffr = sFN , or Ma = sMg. Thus we have s = a/g = (20.7 m/s2)/(9.80 m/s2) = 2.11. 81. (a) We find the speed from the radial component of the acceleration: aR = a sin  = v12/R ; (1.05 m/s2) sin 32.0° = v12/(2.70 m), which gives v1 = 1.23 m/s.  (b) Assuming constant tangential acceleration, we find the speed from v2 = v1 + atant = (1.23 m/s) + (1.05 m/s2)(cos 32.0°)(2.00 s) = 3.01 m/s.

atan

 a

aR

Page 24

Chapter 5

82. For each object we take the direction of the acceleration as y the positive direction. The kinetic friction from the table will oppose the motion of the bowl. (a) From the force diagrams, we have ?F = ma: FN FT x-component(bowl):FT – Ffr = mbowla; x Ffr y-component(bowl):FN – mbowlg = 0; y-component(cat): mcatg – FT = mcata. With Ffr = kFN , we have m bowl g FT = mbowla + kmbowlg. When we eliminate FT , we get (mcat – kmbowl)g = (mcat + mbowl)a; [(5.0 kg) – (0.44)(11 kg)](9.80 m/s2) = (5.0 kg + 11 kg)a, which gives a = 0.098 m/s2. (b) We find the time for the bowl to reach the edge of the table from x = x0 + v0t + !at2; 0.90 m = 0 + 0 + !(0.098 m/s2)t2, which gives

y

m cat g

t = 4.3 s.

83. Before the mass slides, the friction is static, with Ffr = sFN. The static friction force will be maximum just before the mass slides. We write ?F = ma from the force diagram: x-component: mg sin – sFN = 0; y-component: FN – mg cos  = 0. When we combine these, we get tan  = s = 0.60, or  = 31°.

FN

 y

Ffr

 mg

x

84. We assume there is no tension in the rope and simplify the forces to those shown. From the force diagram, we have ?F = ma: x-component: FNshoes – FNwall = 0, Ffrshoes FNwall so the two normal forces are equal: FNshoes = FNwall = FN; y-component: Ffrshoes + Ffrwall – mg = 0. y For a static friction force, we know that Fsfr = sFN. FNshoes The minimum normal force will be exerted when the static x friction forces are at the limit: mg sshoesFNshoes + swallFNwall = mg; (0.80 + 0.60)FN = (70 kg)(9.80 m/s2), which gives FN = 4.9102 N. 85. (a) We can define the radius of curvature as the radius at any instant of the circle such that the acceleration perpendicular to the path corresponds to the radial acceleration: mg cos  = mv2/r, or r = v2/g cos , where  is the angle of the baseball’s motion below the horizontal.  (b) At the initial release point,  = 0°, so we have r0 = v02/g = (30 m/s)2/(9.80 m/s2) = 92 m.

Page 25

FT

Ffrwall

v0



v



mg

Chapter 5

86. (a) If motion is just about to begin, the static friction y force on the block will be maximum: Ffr,max = sFN , and the acceleration will be zero. We write ?F = ma from the force diagram for each object: FN FT x-component (block): FT – sFN = 0; x FT Ffr y-component (block): FN – m1g = 0; y-component (bucket): (m2 + msand)g – FT = 0. m 1g When we combine these equations, we get (m2 + msand)g = sm1g, which gives y (m 2 + msand )g msand = sm1 – m2 = (0.450)(28.0 kg) – 1.00 kg = 11.6 kg. (b) When the system starts moving, the friction becomes kinetic, and the tension changes. The force equations become x-component (block): FT2 – kFN = m1a; y-component (block): FN – m1g = 0; y-component (bucket): (m2 + msand)g – FT2 = (m2 + msand)a. When we combine these equations, we get (m2 + msand – km1)g = (m1 + m2 + msand)a, which gives a = [1.00 kg + 11.6 kg – (0.320)(28.0 kg)](9.80 m/s2)/(28.0 kg + 1.00 kg + 11.6 kg) = 0.879 m/s2. 87. (a) We can find the radius of the halfcircle from the radial acceleration. We choose the x-direction in the radial direction. We write ?F = ma from the force diagram: Flift cos  – mg = 0; Flift sin  = mv2/r. When we combine these, we get tan  = v2/gr; tan 38° = [(520 km/h)/(3.6 ks/h)]2/(9.80 m/s2)r, which gives r = 2.7103 m. We find the time to complete the halfcircle from t = pr/v = p(2.7103 m)/[(520 km/h)/(3.6 ks/h)] = 59 s. (b) The passengers will feel a greater normal force from the seat.

Page 26

y

Flift

x 

aR

mg

Chapter 5

88. At the speed for which the curve is banked perfectly, FN y there is no need for a friction force. We take the x-axis a in the direction of the centripetal acceleration. We write ?F = ma from the force diagram for the car: x-component: FN0 sin  = ma0 = mv02/R; x Ffr y-component: FN0 cos  – mg = 0.  Combining these, we get 2  tan  = v0 /gR. mg At a higher speed, there is need for a friction force, which will be down the incline to help provide the greater centripetal acceleration. If the automobile does not skid, the friction is static, with Ffr = sFN. At the maximum speed, Ffr = sFN. We write ?F = ma from the force diagram for the car: x-component: FN sin  + sFN cos  = ma2 = mvmax2/R; y-component: FN cos  – sFN sin  – mg = 0, or FN(cos  – s sin ) = mg. When we eliminate FN by dividing the equations, we get vmax2 = gR(sin  + s cos )/(cos  – s sin ) = gR(tan  + s)/(1 – stan ) = gR(v02 + sgR)/(gR – sv02)]. At a lower speed, there is need for a friction force, which will be up the incline to prevent the car from sliding down the incline. If the automobile does not skid, the friction is static, with Ffr = sFN. At the minimum speed, Ffr = sFN. The reversal of the direction of Ffr can be incorporated in the above equations by changing the sign of s , so we have vmin2 = gR[(sin  – s cos )/(cos  + s sin )]= gR(v02 – sgR)/(gR + sv02)]. Thus the range of permissible speeds is {gR(v02 – sgR)/(gR + sv02)]}1/2 < v < {gR(v02 + sgR)/(gR – sv02)]}1/2. 89. The pendulum swings out until the tension in the suspension provides the centripetal acceleration, which is the centripetal acceleration of the train. The forces are shown in the diagram. We write ?F = ma from the force diagram for the pendulum: x-component: FT sin  = mv2/r; y-component: FT cos  – mg = 0. When these equations are combined, we get tan  = v2/rg; 29.2 m/s. tan 17.5° = v2/(275 m)(9.80 m/s2), which gives v =

Page 27

y

 FT

aR x

mg

Chapter 5

90. For the falling object we have mg – kv2 = m dv/dt, or dv/[(mg/k) – v2] = (k/m) dt. When we integrate, we get v 0

dv = (mg/ k) – v 2

t 0

k m dt ; v

–1 1/ 2 –1 1/2 1 1 v/ (mg/ k) = v/ (mg/ k) = kt 1/ 2 tanh 1/ 2 tanh m. 0 (mg/ k) (mg/ k) When we rearrange this for v, we get mg 1/ 2 1/ 2 v= tanh kg/ m t . k We integrate again to get x:

x 0

dx =

t 0

mg k

1/ 2

tanh

kg/ m

1/ 2

t dt =

mg k

1/2

m kg

1/2

t 0

sinh kg/ m

1/2

cosh kg/ m

1/2

t t

kg/ m

1/2

dt ;

1/ 2 1/ 2 x = m ln cosh kg/ m t – ln (1) = m ln cosh kg/ m t . k k For the given data of m = 75 kg, and k = 0.22 kg/m, we get t, s v, m/s x, m 0.0 0.0 0 2.0 18.9 19 4.0 34.1 73 6.0 44.4 152 8.0 50.6 248 10.0 54.0 353 12.0 55.9 463 14.0 56.8 576 16.0 57.3 690 18.0 57.5 805 20.0 57.7 920 We see from the expression for v, or from the original equation of motion when we set dv/dt = 0, that the terminal speed is vT = (mg/k)1/2 = [(75 kg)(9.80 m/s2)/(0.22 kg/m)]1/2 = 57.8 m/s. The initial acceleration is g. The terminal speed is reached because the force of air resistance increases until it balances the force of gravity.

91. For the rising rocket we have – mg – kv2 = m dv/dt. We can use the chain rule to replace the variable t with the variable y: – mg – kv2 = m dv/dt = m (dv/dy)(dy/dt) = mv dv/dy, or v dv/[v2 + (mg/k)] = – (k/m) dy. When we integrate from the initial speed up to the highest point, we get

Page 28

Chapter 5 0

h

2v dv = – 2k m d y; 2 0 v 0 v + (mg/ k) ln v 2 + (mg/ k)

0

= ln

(mg/ k)

= – 2kh m ;

v0 v 02 + (mg/ k) (250 kg)(9.80 m / s 2)/ (0.65 kg /

2(0.65 kg / m)h m) =– , which gives h = (250 kg ) (120 m/ s) + (250 kg )(9.80 m/ (0.65 kg / m ) If there were no air resistance, a = – g, and we find the maximum height from v2 = v02 + 2ay; ln

2

s 2 )/

0 = (120 m/s)2 + 2(– 9.80 m/s2)h, which gives

h = 735 m, almost 2.5.

92. (a) If the bead does not move along the hoop, it must have only the radial acceleration moving in a circle of radius R = r sin . We write ?F = ma from the force diagram: x-component: FN sin  = maR = mv2/R; FN sin  = mR(2pf )2 = 4p2f 2mr sin ; y-component: FN cos  – mg = 0, or FN = mg /cos . Combining these, we get cos  = g/4p2f 2r. (b) For the given data we get cos  = (9.80 m/s2)/4p2(4.0 rev/s)2(0.20 m) = 0.0776,  = 86°. (c) Because there is no friction, the bead cannot ride as high as the center. There would be no force to balance the downward force of gravity. 93. The speed of the mass is v = r2pf. If the mass does not move along the cone, it must have only the radial acceleration moving in a circle of radius r. The static friction force could be either way along the cone. We write ?F = ma from the force diagram: x-component: FN cos  ± Ffr sin  = mv2/r = 4p2f 2mr; y-component: FN sin  — Ffr cos  – mg = 0. For static friction we have Ffr = sFN. The extreme positions occur when Ffr = sFN, so we have FN (cos  ± s sin ) = 4p2f 2mr; FN (sin  — s cos  ) = mg, which gives r = g(cos  ± s sin )/4p2f2(sin  — s cos  ). We can write this as g(1 – s tan )/4p2f 2(tan  + s) < r < g(1 + s tan )/4p2f 2(tan  – s).

Page 29

302 m.

y

FN

r  x

mg

y FN

 r

x

Ffr

aR



mg

Chapter 5

94. We write ?F = ma from the force diagram: tangential: mg cos  = matan; radial: FT – mg sin  = maR. From the tangential equation we get atan = g cos  = (9.80 m/s2) cos 30° = 8.5 m/s2. The radial acceleration is aR = v2/r = (6.0 m/s)2/(0.80 m) = 45 m/s2. From the radial equation we get FT = m(aR + g sin ) = (1.0 kg)[45 m/s2 + (9.80 m/s2) sin 30°] = 50 N.

r 

FT v



Page 30

mg

Chapter 6

CHAPTER 6 - Gravitation and Newton’s Synthesis 1.

Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.6710–11 N · m2/kg2)(5.981024 kg)(1400 kg)/[3(6.38106 m)]2 = 1.52103 N.

2.

The acceleration due to gravity on the surface of a planet is g = F/M = GMplanet/r2. For the Moon we have gMoon = (6.6710–11 N · m2/kg2)(7.351022 kg)/(1.74106 m)2 =

3.

4.

1.62 m/s2.

The acceleration due to gravity on the surface of a planet is g = F/M = GMplanet/r2. If we form the ratio of the two accelerations, we have gplanet/gEarth = (Mplanet/MEarth)/(rplanet/rEarth)2, or gplanet = gEarth(Mplanet/MEarth)/(rplanet/rEarth)2 = (9.80 m/s2)(1)/(2.5)2 =

1.6 m/s2.

The acceleration due to gravity on the surface of a planet is g = F/M = GMplanet/r2. If we form the ratio of the two accelerations, we have gplanet/gEarth = (Mplanet/MEarth)/(rplanet/rEarth)2, or gplanet = gEarth(Mplanet/MEarth)/(rplanet/rEarth)2 = (9.80 m/s2)(3.0)/(1)2 =

29 m/s2.

5.

The acceleration due to gravity at a distance r from the center of the Earth is g = F/M = GmEarth/r2. If we form the ratio of the two accelerations for the different distances, we have gh/gsurface = [(rEarth)/(rEarth + h)]2 = [(6400 km)/(6400 km + 300 km)]2 which gives gh = 0.91gsurface.

6.

The acceleration due to gravity at a distance r from the center of the Earth is g = F/M = GmEarth/r2. If we form the ratio of the two accelerations for the different distances, we have g/gsurface = [(rEarth)/(rEarth + h)]2 ; (a) g = (9.80 m/s2)[(6400 km)/(6400 km + 3.20 km)]2 = 9.80 m/s2. 2 2 4.36 m/s2. (b) g = (9.80 m/s )[(6400 km)/(6400 km + 3200 km)] =

7.

We choose the coordinate system shown in the figure and y find the force on the mass in the lower left corner. L Because the masses are equal, for the magnitudes of the 1 forces from the other corners we have F1 = F3 = Gmm/r12 = (6.6710–11 N · m2/kg2)(8.5 kg)(8.5 kg)/(0.70 m)2 = 9.8310–9 N; F2 = Gmm/r22 F1 = (6.6710–11 N · m2/kg2)(8.5 kg)(8.5 kg)/[(0.70 m)/cos 45°]2 F2 = 4.9210–9 N. From the symmetry of the forces we see that the resultant will be 4 F3 along the diagonal. The resultant force is F = 2F1 cos 45° + F2 = 2(9.8310–9 N) cos 45° + 4.9210–9 N = 1.910–8 N toward center of the square.

Page 1

2

L

3

x

Chapter 6

8.

For the magnitude of each attractive gravitational force, we have Sun Venus Earth Jupiter Saturn FV = GmEmV/rV2 = GfVmE2/rV2 = (6.6710–11 N · m2/kg2)(0.815)(5.981024 kg)2/[(108 – 150)109 m]2 = 1.101018 N; FJ = GmEmJ/rJ2 = GfJmE2/rJ2 = (6.6710–11 N · m2/kg2)(318)(5.981024 kg)2/[(778 – 150)109 m]2 = 1.921018 N; FSa = GmEmSa/rSa2 = GfSamE2/rSa2 = (6.6710–11 N · m2/kg2)(95.1)(5.981024 kg)2/[(1430 – 150)109 m]2 = 1.381017 N. The force from Venus is toward the Sun; the forces from Jupiter and Saturn are away from the Sun. For the net force we have Fnet = FJ + FSa – FV = 1.921018 N + 1.381017 N – 1.101018 N = 9.61017 N away from the Sun.

9.

For the magnitudes of the forces on the mass m from the other masses we have F2 = Gm(2m)/x02 = 2Gm2/x02; F3 = Gm(3m)/(x02 + y02) = 3Gm2/(x02 + y02); F4 = Gm(4m)/y02 = 4Gm2/y02. The force F3 is at an angle  above the x-axis, with sin  = y0/(x02 + y02)1/2, and cos  = x0/(x02 + y02)1/2. Thus the resultant force is F = (F2 + F3 cos )i + (F3 sin  + F4)j = Gm2{(2/x02) +[3/(x02 + y02)][x0/(x02 + y02)1/2]}i +

y 3m

4m y0 F4

F3

 m

x0

F2

x

2m

Gm2{[3/(x02 + y02)][y0/(x02 + y02)1/2] + (4/y02)}j

=

Gm2{(2/x02) +[3x0/(x02 + y02)3/2]}i + Gm2{[3y0/(x02 + y02)3/2] + (4/y02)}j.

10. At each planet the gravitational force from the other will be negligible, so we have WEarth = MgEarth = (70 kg)(9.80 m/s2) = 690 N; WMars = MgMars = (70 kg)(3.7 m/s2) = 260 N. We assume the planets are at their closest, so their separation is D = 228109 m – 150109 m = 78109 m. If we let d represent the distance from the Earth’s center, the net gravitational force (toward the Earth) is Fnet = M{[GmEarth/d2] – [GmMars/(D – d)2]}. If we let d = xD, so x varies from a small number to almost 1, and use g = Gm/r2, we can write this as Fnet = M{[GmEarth/r2] – [GmMars/(D – r)2]}.

Fnet 670 N

= (M/D2){gEarth(rEarth/x)2 – gMars[rMars/(1 – x)2]}

Fnet = 0 Mars 0 Earth

0.25

0.50

0.75

x = d/ D

= (MrEarth2/D2){(gEarth/x2) – [gMars(rMars/rEarth)2/(1 – x)2]}. The net force will be zero when (gEarth/x2) = [gMars(rMars/rEarth)2/(1 – x)2], or x(rMars/rEarth)(gMars/gEarth)1/2 = 1 – x; x[(3400 km)/6380 km)][(3.7 m/s2)/(9.8 m/s2)]1/2 = 1 – x, which gives x = 0.75, (about 75% of the total distance). The net force will be very small except when near a planet’s surface, so we have the graph shown. Page 2

– 270 N

Chapter 6

Note that the negative value is an indication of the direction change.

11. The weight of objects is determined by g, which depends on the mass and radius of the Earth: g = GmE/rE2. If the density remains constant, the mass will be proportional to the radius cubed: mE/mE = (rE/rE)3. If we form the ration of g’s, we have g/g = (mE/mE)/(rE/rE)2 = (mE/mE)/(mE/mE)2/3 = (mE/mE)1/3 = 21/3 = 1.26. 12. The acceleration due to gravity on the surface of Mars is gMars = GmMars/rMars2; 3.7 m/s2 = (6.6710–11 N · m2/kg2)mMars/(3.4106 m)2, which gives mMars =

6.41023 kg.

13. Because the gravitational force is always attractive, the two forces will be in opposite directions. If we call the distance from the Earth to the Moon D and let x be the distance from the Earth where the magnitudes of the forces are equal, we have GmMM/(D – x)2 = GmEM/x2, which becomes mMx2 = mE(D – x)2. (7.351022 kg)x2 = (5.981024 kg)[(3.84108 m) – x]2, which gives x= 3.46108 m from Earth’s center. 14. We relate the speed of the Earth to the period of its orbit from v = 2pR/T. The gravitational attraction from the Sun must provide the centripetal acceleration for the circular orbit: GmEmS/R2 = mEv2/R = mE(2pR/T)2/R = mE4p2R/T2, so we have GmS = 4p2R3/T2; (6.6710–11 N · m2/kg2)mS = 4p2(1.501011 m)3/(3.16107 s)2, which gives mS = 2.01030 kg. This is the same as found in Example 6–9. 15. (a) The acceleration due to gravity at a distance r = rE + ?r from the center is g = Gm/r2 = Gm/(rE + ?r)2. If we use the binomial expansion and keep only the first two terms, we get g = (Gm/rE2)[1 + (?r/rE)]–2 = g[1 – 2(?r/rE) + 3(?r/rE)2 + ] ˜ g – 2g(?r/rE). Thus we have g – g = ?g ˜ – 2g(?r/rE). Note that this could also be obtained by treating the changes as differentials: dg = – 2(Gm/r3) dr = – 2g dr/r.  (b) The negative sign means that g decreases with an increase in height.  (c) At a height of 100 km we get ?g = – 2g(?r/rE) = – 2(9.80 m/s2)(100 km)/(6.38103 km) = – 0.307 m/s2. Thus we have g = g + ?g = 9.800 m/s2 – 0.307 m/s2 = 9.493 m/s2. If we use Eq. 6–1, we get g = GmE/(rE + ?r)2 = (6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m + 100103 m)2 = 9.499 m/s2.

Page 3

Chapter 6

16. At a latitude angle , the radius of the circular motion is R = rE cos  and the speed of a point on the surface is v = R, so the centripetal acceleration is a = v2/R = R2 = rE2 cos . If we think of the effective g measured by a spring, we have the forces indicated on the diagram. With the coordinate system shown, for ?F = ma we have x-component: mgeff cos  – mg cos  = – mrE2 cos , or geff cos  = g cos  – rE2 cos ; y-component: mgeff sin  – mg sin  = 0, or geff sin  = g sin . When we divide the two equations, we get tan  = (g sin )/(g cos  – rE2 cos )

y



mg eff

mR 2



R 

x

mg  rE

= (9.800 m/s2)(sin 45°)/{(9.800 m/s2)(cos 45°) – (6.38106 m)[(2p/(24 h)(3600 s/h)]2 cos 45} = 1.0035, so  = 45.10°. Thus for geff we have geff = (g sin )/sin  = (9.80 m/s2)(sin 45°)/sin 45.10° = 9.78 m/s2, 0.10° from the “vertical.”

17. For a stationary object at the equator, we know that the effective weight is w0 = mg – mve2/rE , where ve = 2prE f is the speed of a point on the equator. When the ship moves with respect to the surface at speed v, the effective speed will be ve ± v, depending on the direction. The positive sign is used for a ship traveling eastward; the negative sign for a ship traveling westward. Thus the apparent weight is w = mg – m(ve ± v)2/rE = mg – (mve2/rE)[1 ± (v/ve)]2 = mg – (mve2/rE)[1 ± 2(v/ve) + (v/ve)2]. Because v « ve , we can drop the last term to get w = mg – (mve2/rE) — 2(mvev/rE). If we divide this by the effective weight for a stationary object, we have w/w0 = [mg – (mve2/rE) — 2(mvev/rE)]/(mg – mve2/rE) = 1 — 2(mvev/rE)/(mg – mve2/rE) = 1 — 2(vev/grE)/(1 – ve2/grE).

Because ve2/grE « 1, we get w/w0 ˜ 1 — 2(vev/grE) ˜ 1 ± 2(vev/grE) = 1 ± 4pfv/g, if we reverse the meaning of the sign for the direction of the ship: positive sign is used for a ship traveling westward; the negative sign for a ship traveling eastward.

18. The gravitational attraction must provide the centripetal acceleration for the circular orbit: GmEM/r2 = Mv2/r, or v2 = GmE/(rE + h) = (6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m + 5.20106 m), 5.89103 m/s. which gives v = 19. The required centripetal acceleration of the circular orbit is provided by the gravitational attraction: GmEM/r2 = Mv2/r, so we have v2 = GmE/(rE + h) = (6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m + 0.60106 m), 7.56103 m/s. which gives v = Page 4

Chapter 6

20. The greater tension will occur when the elevator is accelerating upward, which we take as the positive direction. We write ?F = ma from the force diagram for the monkey: FT – mg = ma; 200 N – (16.0 kg)(9.80 m/s2) = (16.0 kg)a, which gives a= 2.70 m/s2 upward. Because the rope broke, the tension was greater than 200 N, so this was the minimum acceleration.

y FT a mg

21. We relate the speed to the period of revolution from v = 2pr/T. The required centripetal acceleration is provided by the gravitational attraction: GmMM/r2 = Mv2/r = M(2pr/T)2/r = M4p2r/T2, so we have GmM = 4p2(rM + h)3/T2; (6.6710–11 N · m2/kg2)(7.41022 kg) = 4p2(1.74106 m + 1.00105 m)3/T2, 2.0 h. which gives T = 7.06103 s = 22. We relate the speed to the period of revolution from v = 2pr/T. For the required centripetal acceleration, provided by gravity, we have aR = v2/r = (2pr/T)2/r = 4p2r/T2; 9.80 m/s2 = 4p2(6.38106 m)/T2, which gives T = 5.07103 s (1.41 h). The result is independent of the mass of the satellite. 23. We take the positive direction upward. The spring scale reads the normal force expressed as an effective mass: FN/g. +y We write ?F = ma from the force diagram: FN – mg = ma, or meffective = FN/g = m(1 + a/g). (a) For a constant speed, there is no acceleration, so we have FN meffective = m(1 + a/g) = m = 56 kg. (b) For a constant speed, there is no acceleration, so we have mg meffective = m(1 + a/g) = m = 56 kg. (c) For the upward (positive) acceleration, we have meffective = m(1 + a/g) = m(1 + 0.33g/g) = 1.33(56 kg) = 75 kg. (d) For the downward (negative) acceleration, we have meffective = m(1 + a/g) = m(1 – 0.33g/g) = 0.67(56 kg) = 38 kg. (e) In free fall the acceleration is – g, so we have meffective = m(1 + a/g) = m(1 – g/g) = 0.

Page 5

Chapter 6

24. The centripetal acceleration has a magnitude of aR = v2/R = (2pR/T)2/R = 4p2R/T2 = 4p2[!(27.5 m)]/(10.5 s)2 = 4.92 m/s2. At each position we take the positive direction in the direction of the acceleration. Because the seat swings, the normal force from the seat is upward and the weight is downward. The apparent weight is measured by the normal force. (a) At the top, we write ?F = ma from the force diagram: – FNtop + mg = maR , or FNtop = mg(1 – aR/g). For the fractional change we have Fractional change = (FNtop – mg)/mg = – aR/g = – (4.92 m/s2)/(9.80 m/s2) = – 0.502 (– 50.2%). (b) At the bottom, we write ?F = ma from the force diagram: FNbottom – mg = maR , or FNbottom = mg(1 + aR/g). For the fractional change we have Fractional change = (FNbottom – mg)/mg = + aR/g = + (4.92 m/s2)/(9.80 m/s2) = + 0.502 (+ 50.2%).

FNtop mg

R

FNbottom

mg

25. The acceleration due to gravity is g = Fgrav/M = Gm/r2 = (6.6710–11 N · m2/kg2)(7.41022 kg)/(4.1106 m)2 = 0.29 m/s2. We take the positive direction toward the Moon. The apparent weight is measured by the normal force. We write ?F = Ma from the force diagram: – FN + Mg = Ma, (a) For a constant velocity, there is no acceleration, so we have – FN + Mg = 0, or FN = Mg = (75 kg)(0.29 m/s2) = 22 N (toward the Moon). (b) For an acceleration toward the Moon, we have – FN + Mg = Ma, or FN = M(g – a) = (75 kg)(0.29 m/s2 – 2.6 m/s2) = – 1.7102 N (away from the Moon). 26. We relate the speed to the period from v = 2prE/T. To be apparently weightless, the acceleration of gravity must be the required centripetal acceleration, so we have aR = g = v2/rE = (2prE/T)2/rE = 4p2rE/T2; 9.80 m/s2 = 4p2(6.38106 m)/T2, which gives T = 5.07103 s (1.41 h). 27. (a) The attractive gravitational force between the stars is providing the required centripetal acceleration for the circular motion. (b) We relate the orbital speed to the period of revolution from v = 2pr/T, where r is the distance to the midpoint. The gravitational attraction provides the centripetal acceleration: Gmm/(2r)2 = mv2/r = m(2pr/T)2/r = m4p2r/T2, so we have m = 16p2r3/GT2 = 16p2(4.01010 m)3/(6.6710–11 N · m2/kg2)[(12.6 yr)(3.16107 s/yr)]2 = 9.61026 kg. Page 6

Chapter 6

28. (a) We relate the speed to the period of revolution from v = 2pr/T. We know that the gravitational attraction of a mass M provides the centripetal acceleration: GmplanetM/r2 = Mv2/r = M(2pr/T)2/r = M4p2r/T2, so we have mplanet = 4p2r3/GT2. Thus the density is  = mplanet/V = [4p2r3/GT2]/)pr3 = 3p/GT2. (b) For the Earth we have  = 3p/GT2 = 3p/(6.6710–11 N · m2/kg2)[(90 min)(60 s/min)]2 = 4.8103 kg/m3. 3 3 Note that the density of iron is 7.810 kg/m . 29. The gravitational attraction must provide the centripetal acceleration for the circular orbit: GmEM/r2 = Mv2/r, or v2 = GmE/(rE + h) = (6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m + 8.8103 m), 7.9103 m/s. which gives v = 30. (a) The gravitational attraction must provide the centripetal acceleration for the circular orbit: GmEM/r2 = Mv2/r, or v = (GmE/r)1/2. Because the change in radius is small compared to r, we can approximate the changes as differentials, so we differentiate the expression for v: dv = – !(GmE/r3)1/2 dr. We assume that the height of the orbit is small compared to the Earth’s radius, which we use for r: dv = – ![(6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m)3]1/2(– 1.0103 m) = + 0.62 m/s. Thus the time required to catch up is ?t = ?x/dv = (30103 m)/(0.62 m/s) = 4.8104 s = 13 h. (b) To catch up in 8.0 hours, the increase in speed must be dv = ?x/?t = (30103 m)/(8.0 h)(3600 s/h) = 1.05 m/s. We find the decrease in radius from dv = – !(GmE/r3)1/2 dr; 1.05 m/s = – ![(6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m)3]1/2 dr, – 1.7 km. which gives dr = – 1.7103 m = 31. Each body will attract one of the other bodies with a force given by F = Gmm/L2. From the symmetry we see that the net force on a body will be toward the center of the triangle and will provide the centripetal acceleration required for the circular orbit: Fnet = 2(Gmm/L2) cos 30° = mv2/r = mv2/(L/2 cos 30°), which gives v = (Gm/L)1/2.

M L F 30°

M

Page 7

L

F

L

M

Chapter 6

32. If we take the upward acceleration of the elevator as positive, the effective value of gravity in the elevator is geff = g + aelev. The acceleration of the mass relative to the plane must be along the surface of the plane. From the diagram we see that FN mgeff sin  = marel , so arel = (g + aelev) sin . (a) For an upward acceleration of the elevator, we get arel = (g + aelev) sin  = (g + 0.50g) sin 30° arel = 0.75g (down the plane). (b) For a downward acceleration of the elevator, we get  arel = (g + aelev) sin  = (g – 0.50g) sin 30°  = 0.25g (down the plane). mg eff (c) For free fall aelev = – g, so we have geff = 0, and we get arel = (g + aelev) sin  = 0. (d) For constant speed aelev = 0, so we have geff = g, and we get arel = (g + aelev) sin  = g sin 30° = 0.50g (down the plane). 33. From Kepler’s third law, T2 = 4p2r3/GmE , we can relate the periods of the satellite and the Moon: (T/TMoon)2 = (r/rMoon)3; (T/27.4 d)2 = [(6.38106 m)/(3.84108 m)]3, which gives T = 0.0587 days (1.41 h). 34. From Kepler’s third law, T2 = 4p2r3/GmS , we can relate the periods of Icarus and the Earth: (TIcarus/TEarth)2 = (rIcarus/rEarth)3; (410 d/365 d)2 = [rIcarus/(1.501011 m)]3, which gives rIcarus = 1.621011 m. 35. From Kepler’s third law, T2 = 4p2r3/GmS , we can relate the periods of the Earth and Neptune: (TNeptune/TEarth)2 = (rNeptune/rEarth)3; (TNeptune/1 yr)2 = [(4.51012 m)/(1.501011 m)]3, which gives TNeptune = 1.6102 yr. 36. We use Kepler’s third law, T2 = 4p2r3/GmE , for the motion of the Moon around the Earth: T2 = 4p2r3/GmE ; [(27.4 d)(86,400 s/d)]2 = 4p2(3.84108 m)3/(6.6710–11 N · m2/kg2)mE , which gives mE = 5.981024 kg. 37. We use Kepler’s third law, T2 = 4p2r3/GmE , for the motion of the Sun: T2 = 4p2r3/GmGalaxy; = 4p2[(3104 ly)(3108 m/s)(3.16107 s/yr)]3/(6.6710–11 N · m2/kg2)(41041 kg), which gives T = 61015 s = 2108 yr. 38. From Kepler’s third law, T2 = 4p2r3/GmJupiter , we have mJupiter = 4p2r3/GT2. Page 8

Chapter 6

(a) mJupiter = 4p2rIo3/GTIo2 = 4p2(422106 m)3/(6.6710–11 N · m2/kg2)[(1.77 d)(86,400 s/d)]2 = (b) mJupiter = 4p2rEuropa3/GTEuropa2 = 4p2(671106 m)3/(6.6710–11 N · m2/kg2)[(3.55 d)(86,400 s/d)]2 = mJupiter = 4p2rGanymede3/GTGanymede2 = 4p2(1070106 m)3/(6.6710–11 N · m2/kg2)[(7.16 d)(86,400 s/d)]2 = mJupiter = 4p2rCallisto3/GTCallisto2 = 4p2(1883106 m)3/(6.6710–11 N · m2/kg2)[(16.7 d)(86,400 s/d)]2 = The results are consistent.

1.901027 kg. 1.901027 kg; 1.891027 kg; 1.901027 kg.

39. From Kepler’s third law, T2 = 4p2r3/GmJupiter , we can relate the distances of the moons: (r/rIo)3 = (T/TIo)2. Thus we have (rEuropa/422103 km)3 = (3.55 d/1.77 d)2, which gives rEuropa = 6.71105 km. (rGanymede/422103 km)3 = (7.16 d/1.77 d)2, which gives rGanymede = 1.07106 km. 3 3 2 (rCallisto/42210 km) = (16.7 d/1.77 d) , which gives rCallisto = 1.88106 km. All values agree with the table. 40. (a) From Kepler’s third law, T2 = 4p2r3/GmS , we can relate the periods of the assumed planet and the Earth: (Tplanet/TEarth)2 = (rplanet/rEarth)3; (Tplanet/1 yr)2 = (3)3, which gives Tplanet = 5.2 yr. (b) No, the radius and period are independent of the mass of the orbiting body. 41. An apparent gravity of one g means that the normal force from the band Fgrav is mg, where g = GmE/rE2. FN The normal force and the gravitational attraction from the Sun provide r the centripetal acceleration: 2 2 2 GmSm/rSE + mGmE/rE = mv /rSE , or v2 = G[(mS/rSE) + (mErSE/rE2)] = (6.6710–11 N · m2/kg2)[(1.991030 kg)/(1.501011 m) + (5.981024 kg)(1.501011 m)/(6.38106 m)2], which gives v = 1.2106 m/s. For the period of revolution we have T = 2prSE/v = [2p(1.501011 m)/(1.2106 m/s)]/(86,400 s/day) = 9.0 Earth-days. 42. (a) In a short time interval ?t, the planet will travel a distance v ?t along its orbit. This distance has been exaggerated on the diagram. Kepler’s second dN dF law states that the area swept out by a line from vN ² t the Sun to the planet in time ?t is the same Sun anywhere on the orbit. If we take the areas swept out at thenearest and farthest points, as shown on the diagram, and approximate the areas as triangles (which is a good approximation for very small ?t), we have !dN(vN ?t) = !dF(vF ?t), which gives vN/vF = dF/dN. (b) For the average velocity we have æ = 2p[!(dN + dF)]/T = p(1.471011 m + 1.521011 m)/(3.16107 s) = 2.97104 m/s. From the result for part (a), we have vN/vF = dF/dN = 1.52/1.47 = 1.034, or vN is 3.4% greater than vF . For this small change, we can take each of the extreme velocities to be ± 1.7% from the average. Thus we have Page 9

vF ² t

Chapter 6

vN = 1.017(2.97104 m/s) = vF = 0.983(2.97104 m/s) =

3.02104 m/s; 2.92104 m/s.

43. (a) From Kepler’s third law, T2 = 4p2r3/GmS , we can relate the periods of Hale-Bopp and the Earth: (THB/TE)2 = (rHB/rE)3; [(3000 y)/(1 yr)]2 = (rHB/(1 A.U.)]3, which gives rHB = 2.1102 A.U. (3.11013 m). (b) We find the farthest distance from rHB = !(dN + dF); 4.2102 A.U. 2.1102 A.U. = !(1.0 A.U. + dF), which gives dF = (c) From Problem 42 we have vN/vF = dF/dN = (4.2102 A.U.)/(1.0 A.U.) = 4.2102. 44. We call D the separation of the Earth and Moon and take the positive direction toward the Earth. Because the gravitational fields are in opposite directions, we have g = gE – gM = [GmE/(D/2)2] – [GmM/(D/2)2] = (4G/D2)(mE – mM) = 4[(6.6710–11 N · m2/kg2)/(3.84108 m)2](5.981024 kg – 7.41022 kg) = 1.0710–2 N/kg toward Earth. 45. (a) The gravitational field is g = GmS/DSE2 = (6.6710–11 N · m2/kg2)(2.01030 kg)/(1.501011 m)2 = 5.910–3 N/kg. (b) Because the field due to the Sun is about 1000 times smaller than the field from the Earth, the effect is not significant. y 46. (a) From the symmetry, the magnitudes of the gravitational field from each particle on the y-axis are equal: g1 = g2 = Gm/r2 = Gm/(x02 + y2). g1 g2 The symmetry means that the total field will be toward – y  y r and is given by m m g = 2(Gm/r2) cos  (– j) = – 2(Gmy/r3) j = – [2Gmy/(x02 + y2)3/2] + x0 – x0 j. (b) We find the locations of the maximum magnitude by setting the first derivative equal to zero: 2Gm 2Gm dg 2Gmy 2y =– + 32 = – x 20 – y 2 + 3 y 2 = 0. 5/ 2 dy 2 2 3/ 2 2 2 5/ 2 2 x +y x + y2 x +y 0

0

0

Thus we have y = ± x0/v2. 2y2 = x02, or Note that these are maximum points, because the field is zero at the origin and very far away. The maximum values are gmax = – 2{Gm(± x0/v2)/[x02 + (± x0/v2)2]3/2} j = — (4Gm/33/2x02) j = — (0.77 Gm/x02) j. 47. The acceleration due to gravity at a distance r from the center of the Earth is g = F/M = GmEarth/r2. If we form the ratio of the two accelerations for the different distances, we have g/gsurface = [rEarth/(rEarth + h)]2; 1/2 = [(6400 km)/(6400 km + h)]2 , which gives h = 2.7103 km. 48. (a) The mass does not depend on the gravitational force, so it is (b) For the weights we have wEarth = mgEarth = (3.0 kg)(9.80 m/s2) = 29 N; Page 10

3.0 kg on both.

x

Chapter 6

wplanet = mgplanet = (3.0 kg)(12.0 m/s2) =

36 N.

49. The acceleration due to gravity on the surface of the neutron star is g = F/M = Gm/r2 = (6.6710–11 N · m2/kg2)(5)(2.01030 kg)/(10103 m)2 =

6.71012 m/s2.

50. The acceleration due to gravity at a distance r from the center of the Earth is g = F/M = GmEarth/r2. If we form the ratio of the two accelerations for the different distances, we have g/gsurface = (rEarth/r)2; 1/10 = [(6400 km)/r]2 , which gives r = 2.0104 km. 51. The acceleration due to gravity on the surface of the white dwarf star is g = F/M = Gm/r2 = (6.6710–11 N · m2/kg2)(2.01030 kg)/(1.74106 m)2 =

4.4107 m/s2.

52. The time for one revolution about the Moon is T = 2pR/v. The required centripetal acceleration is provided by the gravitational attraction: GmMm/R2 = mv2/R = m(2pR/T)2/R = m4p2R/T2, so we have GmM = 4p2(rM + h)3/T2; (6.6710–11 N · m2/kg2)(7.41022 kg) = 4p2(1.74106 m + 1.00105 m)3/T2, 2.0 h. which gives T = 7.06103 s = 53. We assume a separation of 1 m between two persons of mass 60 kg. If we assume we can sense a force of 0.5 N, we have F = Gmm/r2 ; G = 110–4 N · m2/kg2 ˜ 106 G. 0.5 N = G(60 kg)(60 kg)/(1 m)2, which gives 54. (a) The two forces are FS = GmSmE/rSE2 = (6.6710–11 N · m2/kg2)(2.01030 kg)(5.981024 kg)/(1.51011 m)2 = 3.51022 N; 2 FM = GmMmE/rME = (6.6710–11 N · m2/kg2)(7.41022 kg)(5.981024 kg)/(3.84108 m)2 = 2.01020 N. If we form the ratio, we get FS/FM = 3.51022 N/2.01020 N = 1.8102. (b) The direction of the Moon’s gravitational field changes more frequently, with a period slightly less than 28 days; and the relative change in the gravitational field from one side of the Earth to the other is greater for the Moon. 55. We relate the orbital speed to the period of revolution from v = 2pR/T. The required centripetal acceleration is provided by the gravitational attraction: GmSm/R2 = mv2/R = m(2pR/T)2/R = m4p2R/T2, so we have GmS = 4p2R3/T2. For the two extreme orbits we have (6.6710–11 N · m2/kg2)(5.691026 kg) = 4p2(7.3107 m)3/Tinner2, which gives Tinner = 2.0104 s = 5 h 35 min; –11 2 2 26 (6.6710 N · m /kg )(5.6910 kg) = 4p2(17107 m)3/Touter2, which gives Touter = 7.1104 s = 19 h 50 min. Because the mean rotation period of Saturn is between the two results, with respect to a point on the surface of Saturn, the edges of the rings are moving in opposite directions. 56. (a) The gravitational attraction must provide the centripetal acceleration for the circular orbit: Page 11

Chapter 6

GmEm/r2 = mv2/r, or v2 = GmE/(rE + h) = (6.6710–11 N · m2/kg2)(5.981024 kg)/[6.38106 m + (11,000)(1.852103 m)], 3.9103 m/s. which gives v = (b) For the period of revolution we have T = 2pr/v = 2p[6.38106 m + (11,000)(1.852103 m)]/(3.9103 m/s.) 12 h. = 4.3104 s =

57. (a) If we assume Eros is a cylinder, its mass is mEros = pr2h = (2.3103 kg/m3)p(3103 m)2(40103 m) = 2.61015 kg. We assume NEAR orbits around the narrow waist of Eros and use Kepler’s third law: T2 = 4p2r3/GmEros = 4p2[(15103 m) + (3103 m)]3/(6.6710–11 N · m2/kg2)(2.61015 kg), which gives T = 3.6104 s = 10 h. (b) With the same mass and density, the sphere must have the same volume: )pr03 = (40 km)p(3 km)2, which gives r0 = 6.5 km. (c) The acceleration due to gravity on the surface of a spherical Eros is g = GmEros/r02 = (6.6710–11 N · m2/kg2)(2.61015 kg)/(6.5103 m)2 = 4.210–3 m/s2. 58. We relate the speed to the period of revolution about the midpoint from v = 2p(r/2)/T, where r is the separation. The gravitational attraction provides the centripetal acceleration: Gmm/r2 = mv2/(r/2) = 2m[2p(r/2)/T]2/r = 2mp2r/T2, so we have m = 2p2r3/GT2 = 2p2(360109 m)3/(6.6710–11 N · m2/kg2)[(5.0 yr)(3.16107 s/yr)]2 =

5.51029 kg.

59. From Kepler’s third law, T2 = 4p2R3/GmS , we can relate the periods of Halley’s comet and the Earth to find the mean distance of the comet from the Sun: (THalley/TEarth)2 = (RHalley/REarth)3; (76 yr/1 yr)2 = [RHalley/(1.501011 m)]3, which gives RHalley = 2.681012 m. This mean distance is half the sum of the nearest and farthest distances. If we take the nearest distance to the Sun as zero, the farthest distance is d = 2RHalley = 2(2.681012 m) = 5.41012 m. It is still orbiting the Sun and thus is in the Solar System. The planet nearest it is Pluto. 60. We relate the speed to the period of revolution from v = 2pr/T, where r is the distance to the center of the Milky Way. The gravitational attraction provides the centripetal acceleration: GmgalaxymS/r2 = mSv2/r = mS(2pr/T)2/r = mS4p2r/T2, so we have mgalaxy = 4p2r3/GT2 = 4p2[(30,000 ly)(9.51015 m/ly)]3/ (6.6710–11 N · m2/kg2)[(200106 yr)(3.16107 s/yr)]2 = The number of stars (“Suns”) is Page 12

3.41041 kg.

Chapter 6

(3.41041 kg)/(2.01030 kg) =

1.71011.

61. The acceleration due to gravity on the surface of a planet is gP = Fgrav/M = GmP/R2. If we form the ratio of the expressions for Jupiter and the Earth, we have gJupiter/gEarth = (mJupiter/mEarth)(REarth/RJupiter)2; gJupiter/gEarth = [(1.91027 kg)/(6.01024 kg)][(6.38106 m)/(7.1107 m)]2, which gives gJupiter = 2.56gEarth . This has not taken into account the centripetal acceleration. We ignore the small contribution on Earth. The centripetal acceleration on the equator of Jupiter is aR = v2/R = (2pR/T)2/R = 4p2R/T2 = 4p2(7.1107 m)/[(595 min)(60 s/min)]2 = 2.2 m/s2 = 0.22gEarth . The centripetal acceleration reduces the effective value of g: gJupiter = gJupiter – aR = 2.56gEarth – 0.22gEarth = 2.3gEarth.

62. The gravitational attraction from the core must provide the centripetal acceleration for the orbiting stars: Gmstarmcore/R2 = mstarv2/R, so we have mcore = v2R/G = (780103 m/s)2(5.71017 m)/(6.6710–11 N · m2/kg2) = 5.21039 kg. If we compare this to our Sun, we get mcore/mSun = (5.21039 kg)/(2.01030 kg) = 2.6109 . 63. The acceleration due to gravity on the surface of a planet is gP = F/M = GmP/r2, so we have mP = gPr2/G. 64. (a) The attractive gravitational force on the plumb bob is FM = GmmM/DM2. Because ?F = 0, we see from the force diagram: tan = FM/mg = (GmmM/DM2)/(mGmE/RE2), where we have used GME/RE2 for g. Thus we have = tan–1 (mMRE2/mEDM2). (b) For the mass of a cone with apex half-angle , we have mM = V = @ph3 tan2 

FT



FT mg  FM

FM

mg

= (3103 kg/m3)@p(4103 m)3 tan2 30° = 71013 kg. (c) Using the result from part (a) for the angle, we have tan  = mMRE2/mEDM2 = (71013 kg)(6.4106 m)2/(6.01024 kg)(5103 m)2 = 210–5, ˜ (110–3)°. which gives 65. Your weight at a distance r from the center of the Earth is w = GmmE/r2. The rate of change of r is dr/dt = v, so the rate of change of your weight is dw/dt = GmmE(– 2/r3)(dr/dt) = – 2GmmEv/r3. 66. The contribution to g of the oil deposit (with density oil) is g = Gmoil/D2 = GoilV/D2, where V is the volume of oil and D is the distance of the oil from the measurement site. We assume the change in g from average is caused by the substitution of oil for rock, that is, the difference between the Page 13

Chapter 6

contribution of the oil and what would be contributed by the same volume of rock. Thus we have ?g = G ?m/D2 = G(?)V/D2 = G(oil – rock)V/D2; (– 210–7)(9.80 m/s2) = (6.6710–11 N · m2/kg2)(1000 kg/m3 – 3000 kg/m3)V/(2000 m)2, V = 5.8107 m3. which gives The mass of the oil is moil = oilV = (1000 kg/m3 )(5.8107 m3) = 5.81010 kg. 67. The particle falls along a radial line. We take the positive direction upward. During the fall the gravitational attraction provides the acceleration: – GmEm/r2 = ma = m dv/dt = m (dv/dr)(dr/dt) = mv (dv/dr), which we can write – (GmE/r2) dr = v dv. We find the velocity by integration: rE

2rE

– (Gm E/ r 2) d r =

vf

0

v d v , which gives (Gm E/ r)

rE

2 rE

= 12 v f2, or

vf2 = 2GmE[(1/rE) – (1/2rE)] = GmE/rE = (6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m), which gives vf =

Page 14

7.9103 m/s.

Chapter 7

CHAPTER 7 - Work and Energy 1.

The displacement is in the direction of the gravitational force, thus W = Fh cos 0° = mgh = (250 kg)(9.80 m/s2)(2.80 m) = 6.86103 J.

2.

The displacement is opposite to the direction of the retarding force, thus W = Fx cos 180° = (535 N)(1.25103 m)(– 1) = – 6.69105 J.

3.

Because there is no acceleration, the contact force must have the same magnitude as the weight. The displacement in the direction of this force is the vertical displacement. Thus, W = F ?y = (mg) ?y = (65.0 kg)(9.80 m/s2)(20.0 m) = 1.27104 J.

4.

(a) Because there is no acceleration, the horizontal applied force must have the same magnitude as the friction force. Thus, W = F ?x = (230 N)(4.0 m) = 9.2102 J. (b) Because there is no acceleration, the vertical applied force must have the same magnitude as the weight. Thus, W = F y = mg y = (1200 N)(4.0 m) = 4.8103 J.

5.

Because there is no acceleration, from the force diagram we see that FN = mg, and F = Ffr = kmg. Thus, W = F x cos 0° = kmg x cos 0° = (0.50)(160 kg)(9.80 m/s2)(10.3 m)(1) = 8.1103 J.

y F

FN N x f fr

mg

²x

6.

Because the speed is zero before the throw and when the rock reaches the highest point, the positive work of the throw and the (negative) work done by the (downward) weight must add to zero. Thus, Wnet = Wthrow + mgh cos 180° = 0, or h = – Wthrow/mg cos 180° = – (80.0 J)/(1.85 kg)(9.80 m/s2)(–1) = 4.41 m.

7.

1 J = (1 kg · m/s2)(1 m)(1000 g/kg)(100 cm/m)2 = 1107 g · cm/s2 = 1 J = (1 N · m)(0.225 lb/N)(3.28 ft/m) = 0.738 ft · lb.

8.

The maximum amount of work that the hammer can do is the work that was done by the weight as the hammer fell: Wmax = mgh cos 0° = (2.0 kg)(9.80 m/s2)(0.50 m)(1) = 9.8 J. People add their own force to the hammer as it falls in order that additional work is done before the hammer hits the nail, and thus more work can be done on the nail.

9.

If we assume the width of a cut is 0.30 m, and we cut the lawn parallel to the long side, the number of cuts required is N = (10.0 m)/(0.30m/cut) = 33 cuts, each 20 m long. Thus the total work (ignoring turn-arounds) is W = Fx = (15 N)(33)(20 m) = 1.0104 J.

Page 1

1107 erg.

Chapter 7

10. The minimum work is needed when there is no acceleration. y (a) From the force diagram, we write F = ma: y-component: FN – mg cos  = 0; x-component: Fmin – mg sin  = 0. For a distance d along the incline, we have Wmin = Fmind cos 0° = mgd sin  (1) f fr = (950 kg)(9.80 m/s2)(310 m) sin 9.0° = 4.5105 J. (b) When there is friction, we have x-component: Fmin – mg sin  – kFN = 0, or Fmin = mg sin  + kmg cos , For a distance d along the incline, we have Wmin = Fmind cos 0° = mgd (sin  + k cos )(1) = (950 kg)(9.80 m/s2)(310 m)(sin 9.0° + 0.25 cos 9.0°) = 11. We assume that the input force is exerted perpendicular to the lever. When the lever rotates through a small angle , the distance through which the input force acts is ¬I, and the distance the output force acts is ¬O. If the output work is equal to the input work, we have WI = FI¬I = WO = FO¬O, or FO/FI = ¬I/¬O .

x

FN

F



d

mg

1.2106 J.

FI

¬I FO

¬O

12. Because the net work must be zero, the work to stack the books will have the same magnitude as the work done by gravity. For each book the work is mg times the distance the center is raised (zero for the first book, one book-height for the second book, etc.). W1 = 0, W2 = mgh, W3 = mg2h; … . Thus for eight books, we have W = W1 + W2 + W3 + … + W8 = mgh(0 + 1 + 2 + … + 7) = (1.7 kg)(9.80 m/s2)(0.043 m)(28) = 20 J. 13. (a) From the force diagram, we write F = ma: y-component: FN – mg cos  = 0; x-component: – F – kFN + mg sin  = 0. Thus we have  F = – kFN + mg sin  = – kmg cos  + mg sin  = mg (sin  – k cos ) = (380 kg)(9.80 m/s2)(sin 27° – 0.40 cos 27°) = 3.6102 N. (b) Because the piano is sliding down the incline, we have WF = F d cos 180° = (3.6102 J)(3.5 m)(– 1) = – 1.3103 J. (c) For the friction force, we have Wfr = kmg cos d cos 180° = (0.40)(380 kg)(9.80 m/s2)(cos 27°)(3.5 m)(– 1) = – 4.6103 J. (d) For the force of gravity, we have Wgrav = mg d cos 63° = (380 kg)(9.80 m/s2)(3.5 m)(cos 63°) = 5.9103 J. (e) Because the normal force does no work, we have Wnet = Wgrav + WF + Wfr + WN = 5.9103 J– 1.3103 J– 4.6103 J 0 = 0. Page 2

v y

FN Ffr F



mg

x

Chapter 7

14. (a) From the plot of the force exerted by the catapult, we see that the takeoff distance is 90 m. The displacement is in the direction of the force from the engine, thus Wengine = Fengineh cos 0° = (130103 N)(90 m) = 1.2107 J. (b) The work done by the catapult is the area under the F vs. x graph: W = !(F1 + F2)(x2 – x1) = !(1100103 N + 65103 N)(90 m) = 5.2107 J.

F(kN) 1100

65 0

15. Because the motion is in the x-direction, we see that the weight and normal forces do no work: WFN = Wmg = 0. From the force diagram, we write F = ma: x-component: FP cos  – Ffr = 0, or Ffr = FP cos . For the work by these two forces, we have WFP = FP ?x cos  = (14 N)(15 m) cos 20° = 2.0102 J. Wfr = FP cos  ?x cos 180°= (14 N) cos 20° (15 m)(– 1) = – 2.0102 J. As expected, the total work is zero: WFP = – Wfr = 2.0102 J. 16. (a) To find the required force, we use the force diagram to write Fy = may: F – Mg = Ma, so we have F = M(a + g) = M(0.10g + g) = 1.10Mg. (b) For the work, we have WF = Fh cos 0° = 1.10Mgh.



FP

y N FN x Ffr

F

mg

+y

a

Mg

17. We take the scalar product of each unit vector with the vector: i · V = Vx i · i + Vy i · j + Vz i · k = Vx(1) + Vy(0) + Vz(0) = Vx ; j · V = Vx j · i + Vy j · j + Vz j · k = Vx(0) + Vy(1) + Vz(0) = Vy ; k · V = Vx k · i + Vy k · j + Vz k · k = Vx(0) + Vy(0) + Vz(1) = Vz . 18. The magnitudes of the vectors are A = (Ax2 + Ay2 + Az2)1/2 = [(6.8)2 + (4.6)2 + (6.2)2]1/2 = 10.3. B = (Bx2 + By2 + Bz2)1/2 = [(8.2)2 + (2.3)2 + (– 7.0)2]1/2 = 11.0. We find the angle from the scalar product: A · B = AxBx + AyBy + AzBz = AB cos ; (6.8)(8.2) + (4.6)(2.3) + (6.2)(– 7.0) = (10.3)(11.0) cos , which gives cos  = 0.203, 19. A · (– B) = Ax(– Bx)+ Ay(– By) + Az(– Bz) = – (AxBx + AyBy + AzBz) = – A · B. 20. We write the two vectors using unit vectors: V1 = V1 k; V2 = V2 cos i + V2 sin k. For the scalar product we have Page 3

90

x(m)

 = 78°.

Chapter 7

V1 · V2 = V1xV2x + V1yV2y + V1zV2z = 0 + 0 + (V1)(V2 sin ) = (75)(50)[sin(–48°)] =

– 2.8103.

21. (a) A · (B + C) = (7.0 i – 8.5 j) · [(– 8.0 + 6.8) i + (8.1 – 7.2) j + 4.2 k] = (7.0)(– 1.2) + (– 8.5)(0.9) = – 16.1. (b) (A + C) · B = [(7.0 + 6.8) i + (– 8.5 – 7.2) j + 0 k] · (– 8.0 i + 8.1 j + 4.2 k) = (13.8)(– 8.0) + (– 15.7)(8.1) + 0 = – 238. (c) (B + A) · C = [(– 8.0 + 7.0) i + (8.1 – 8.5) j + (4.2 + 0) k] · (6.8 i – 7.2 j) = (– 1.0)(6.8) + (– 0.4)(– 7.2) + (4.2)(0) = – 3.9. 22. The scalar product is A · B = A · (Bx i + By j + Bz k). If we use the distributive law, we get A · B = (A · i)Bx + (A · j)By + (A · k)Bz = [(Ax i + Ay j + Az k) · i]Bx + [(Ax i + Ay j + Az k) · j]By + [(Ax i + Ay j + Az k) · k]Bz If we use the distributive law again, we get A · B = (Ax i · i + Ay j · i + Az k · i)Bx + (Ax i · j + Ay j · j + Az k · j)By + (Ax i · k + Ay j · k + Az k · k)Bz = AxBx + AyBy + AzBz. 23. Because C lies in the xy-plane and is perpendicular to B, we have B · C = BxCx + ByCy = 9.6Cx + 6.7Cy = 0. For the scalar product of A and C we have A · C = AxCx + AyCy = – 4.8Cx + 7.8Cy = 20.0. When we solve these two equations for the two unknowns, we get Cx = –1.25, and Cy = 1.79. C = – 1.3i + 1.8j. Thus 24. For the sum and difference vectors, we have Csum = A + B, Cdiff = A – B. If we form the scalar product, we get Csum · Cdiff = (A + B) · (A – B) = A · A + B · A – A · B – B · B = A2 – B2 = 0, because A and B have the same magnitude. Because the scalar product is zero, the sum and difference vectors are perpendicular. 25. The magnitude of the vector is V = (Vx2 + Vy2 + Vz2)1/2 = [(20.0)2 + (12.0)2 + (– 14.0)2]1/2 = 27.2. We find the angle this vector makes with an axis by taking the appropriate scalar product with the unit vector: V · i = Vx = V cos x , or cos x = Vx/V = 20.0/27.2 = 0.732, x = 42.7°; V · j = Vy = V cos y , or cos y = Vy/V = 12.0/27.2 = 0.441, y = 63.8°; V · k = Vz = V cos z , or cos z = Vz/V = – 14.0/27.2 = – 0.515, z = 121°. 26. From the triangle we see that a + c = b, or c = b – a. If we take the scalar product of c with itself, we have c · c = (b – a) · (b – a); c2 = b2 – a · b – b · a + a2 = a2 + b2 – 2ab cos .

Page 4

b

c

 a

Chapter 7

27. We find the angle between A and B by taking the scalar product: A · B = AxBx + AyBy = AB cos ; 20.0 = (12.0)(4.0) cos , which gives cos  = 0.417, so  = 65.4°, – 65.4°. Thus the two angles that B may make with the x-axis are 95°, – 35° from x-axis.

y A 30° x

28. We use the two expressions for the scalar product: A · B = AB cos  = AxBx + AyBy ; AB cos ( – ) = (A cos )(B cos ) + (A sin )(B sin ); cos ( – ) = cos  cos  + sin  sin .

y A 



B x

29. We have shown the vector sum of B and C on the diagram. The scalar product can be considered as the magnitude of one vector times the component of the other vector parallel to the first. If we let B|| be the component of B parallel to A, etc., then we want to show that A · (B + C) = A · B + A · C , or A(B + C) = A B + A C = A(B + C). We see from the diagram that (B + C) = B + C , so the scalar product is distributive.

y B+C C B

C

A C||

B||

x

(B+C)||

30. Although the pedal travels around the circumference of the circle that the pedal makes, we can find the work done by taking the component of the displacement parallel to the force, which is the diameter: W = Fd = (470 N)(0.36 m) = 1.7102 J. 31. We obtain the forces at the beginning and end of the motion: at x1 = 0.030 m, F1 = kx1 = (84 N/m)(0.030 m) = 2.52 N; at x2 = 0.055 m, F2 = kx2 = (84 N/m)(0.055 m) = 4.62 N. From the graph the work done in stretching the object is the area under the F vs. x graph: W = !(F1 + F2)(x2 – x1) = !(2.52 N + 4.62 N)(0.055 m – 0.030 m) = 0.089 J.

F(x)

0

Page 5

x1

x2

x

Chapter 7

32. For an irregular path, we find the work by considering the path to be an infinite number of differential steps. We find the (differential) work for each step and add (integrate):

W G =

d W G = FG · d s = –

y

mg cos  d s.

W=–

h 0



F

From the diagram we see that cos  ds = dy, so we have

h ds

mg d y = – mgh,

which is the result of Example 7–2.

33. The work done in moving the object is the area under the Fx vs. x graph. For the motion from 0.0 m to 11.0 m, we find the area of two triangles and one rectangle: W = !(300 N)(3.0 m – 0.0 m) + (300 N)(7.0 m – 3.0 m) + !(300 N)(11.0 m – 7.0 m) = 2.3103 J.

Fx (N) 300

0

3.0

11.0

10

15

x(m)

Fx (N)

34. The work done in moving the object is the area 400 under the Fx vs. x graph. 300 (a) For the motion from 0.0 m to 10.0 m, we find 200 the area of two triangles and one rectangle: 100 W = !(400 N)(3.0 m – 0.0 m) 0 + (400 N)(7.0 m – 3.0 m) + –100 !(400 N)(10.0 m – 7.0 m) –200 = 2.8103 J. (b) For the motion from 0.0 m to 15.0 m, we add the negative area of two triangles and one rectangle: W = 2.8103 J – !(200 N)(11.7 m – 10.0 m) – (200 N)(13.7 m – 11.7 m) – !(200 N)(15.0 m – 13.7 m) =

7.0

5

x(m)

2.1103 J.

35. We consider the area under the curve to be seven rectangles, each with a length on the distance axis of ?x = (30.0 m – 10.0 m)/7 = (20.0 m)/7. We estimate the height of each segment from the graph to get W = ? Fi ?x = (?Fi) ?x = (185 N + 175 N + 150 N + 120 N + 110 N + 100 N + 100 N)(20.0 m)/7 = 2.7103 J. 36. The resisting force opposes the penetration. If we assume no acceleration, the applied force must be equal to this in the direction of the penetration. For a variable force, we find the work by integration:

W=

F · ds =

d 0

5 kx 4 dx = kx 5

d 0

=

5

kd . 5

37. For a variable force, we find the work by integration: Page 6

Chapter 7 X

W =

F · ds = 2

0

3

4

kx + ax + bx dx

4

= kx + ax + bx 2 4 5

X

5

0

2

4

5

kX + aX + bX . 2 4 5

=

38. If we let y be the length of chain hanging over the edge and use  for linear weight density, the weight hanging is w = y. As the next differential length dy comes over the edge, the weight w will fall a distance dy, during which the force of gravity will do work dW = w dy. If L1 is the initial hanging length and L is the total length of the chain, we integrate this from L1 to L:

W=

L L1

y d y=

y 2 2

L

= L1

 2 2 (20 N/ m ) 2 2 L – L1 = (3.0 m ) – (1 .0 m ) = 2 2

39. (a) If we use r as the displacement, the force of gravity is F(r) negative and ?r is negative. Thus we can plot the force as positive with a positive change in r, as shown. If we approximate the area as two rectangles, the average forces for the two are at rE + #h = 6.38106 m + #(3.0106 m) = 7.13106 m, F1 = GMEm/r2 = (6.6710–11 N · m2/kg2)(5.981024 kg)  (2500 kg)/(7.13106 m)2 = 1.96104 N. at rE + &h = 6.38106 m + &(3.0106 m) = 8.63106 m, 0 F2 = GMEm/rE2 = (6.6710–11 N · m2/kg2)(5.981024 kg)(2500 kg)/ (8.63106 m)2 = 1.34104 N. From the graph the work done is the area under the F vs. r graph: W = (F1 + F2)!h = !(1.96104 N + 1.34104 N)(3.0106 m) = 4.951010 J = (b) To find the work by integration, we have re  Gm em Gm em re W = F dr = – dr = r = Gme m r1 – 1 2 e re + h r r +h r +h e

=

Gm em

re

1–

y w

80 J.

rE

rE + h

5.01010 J .

e

re re = mg re 1 – re + h re + h 6

6.38  10 m 10 = 5.00  10 J. 6 9.38  10 m Thus we see that our approximation in (a) is within 3%. 2

6

= (2500 kg )(9.80 m/ s )(6.38  10 m) 1 –

40. We find the speed from K = !mv2; 6.2110–21 J = !(5.3110–26 kg)v2, which gives v =

484 m/s.

41. (a) K2 = !mv22 = 3K1 = 3(!mv12), which gives v2 = v1v3, so the speed increases by a factor of v3. 2 2 2 (b) K2 = !mv2 = !m(!v1) = #(!mv1 ) = #K1 , so the kinetic energy changes by a factor of #.

Page 7

+y

r

Chapter 7

42. The work done on the electron decreases its kinetic energy: W = ?K = !mv2 – !mv02 = 0 – !(9.1110–31 kg)(1.70106 m/s)2 = 43. The work done on the car decreases its kinetic energy: W = ?K = !mv2 – !mv02 = 0 – !(1300 kg)[(100 km/h)/(3.6 ks/h)]2 = 44. The work done on the arrow increases its kinetic energy: W = Fd = ?K = !mv2 – !mv02; (105 N)(0.80 m) = !(0.085 kg)v2 – 0, which gives v =

– 1.3210–18 J.

– 5.02105 J.

44 m/s.

45. The work done by the force of the glove decreases the kinetic energy of the ball: W = Fd = ?K = !mv2 – !mv02; F(0.25 m) = 0 – !(0.145 kg)(32 m/s)2 , which gives F = – 3.0102 N. The force by the ball on the glove is the reaction to this force: 3.0102 N in the direction of the motion of the ball. 46. The work done by the braking force decreases the kinetic energy of the car: W = ?K; – Fd = !mv2 – !mv02 = 0 – !mv02. Assuming the same braking force, we form the ratio: d2/d1 = (v02/v01)2 = (1.50)2 = 2.25. 47. On a level road, the normal force is mg, so the kinetic friction force is kmg. Because it is the (negative) work of the friction force that stops the car, we have W = ?K; – kmg d = !mv2 – !mv02; – (0.38)m(9.80 m/s2)(78 m) = – !mv02, which gives v0 = 24 m/s (87 km/h or 54 mi/h). Because every term contains the mass, it cancels. 48. With m1 = 2m2 , for the initial condition we have K1 = !K2 ; !m1v12 = !(!m2v22), or 2m2 v12 = !m2v22, which gives v1 = !v2. After a speed increase of ?v, we have K1 = K2; !m1(v1 + ?v)2 = !m2(v2 + ?v)2; 2m2(!v2 + 7.0 m/s)2 = m2(v2 + 7.0 m/s)2. When we take the square root of both sides, we get v2(!v2 + 7.0 m/s) = ± (v2 + 7.0 m/s), which gives a positive result of v2 = For the other speed we have v1 = !v2 = 4.9 m/s.

9.9 m/s.

49. On the horizontal FN = mg, so the friction force is Ffr = mg. (a) The work by the applied force is WF = Fd = (6.0 N)(12 m) = 72 J. (b) The work by friction is Wfr = – FNd = – kmgd = – (0.30)(1.0 kg)(9.80 m/s2)(12 m) = – 35 J. (c) The normal force and the weight do no work. The net work increases the kinetic energy of the mass: WF + Wfr = ?K; 72 J – 35 J = Kf – 0, which gives Kf = 37 J. 50. The force from the spring is opposite to the displacement, so the work it does is negative. The work changes the kinetic energy of the car: Page 8

Chapter 7

Wspring = ?K; – !kx2 = 0 – !mv2; – !k(2.2 m)2 = – !(1200 kg)[(60 km/h)/(3.6 ks/h)]2 , which gives k =

6.9104 N/m.

51. On the level the normal force is FN = mg, so the friction force is Ffr = kmg. The normal force and the weight do no work. The net work increases the kinetic energy of the mass: Wnet = ?K = !mvf2 – !mvi2; F(L1 + L2) – kmgL2 = !mvf2 – 0; (225 N)(11.0 m + 10.0 m) – (0.20)(66.0 kg)(9.80 m/s2)(10.0 m) = !(66.0 kg)vf2, which gives vf = 10.2 m/s. 52. Because the force F holds the spring a distance x, the spring constant is k = F/x. The work done by the spring changes the kinetic energy of the mass: Wspring = ?K; – (!kxf2 – !kxi2) = !mvf2 – !mvi2. (a) When the spring returns to xf = 0, we have – (0 – !kx2) = !mvf2 – 0, or vf = (k/m)1/2x = (Fx/m)1/2. (b) When the spring returns to xf = !x, we have – [!k(!x)2 – !kx2] = !mvf2 – 0, or vf = (3k/4m)1/2x = (3Fx/4m)1/2. 53. There will be an additional (negative) work done by the friction force. The net work increases the kinetic energy of the mass: – (!kxf2 – !kxi2) – kmg ?x = !mvf2 – !mvi2; – (0 – !kx2) – kmgx = 0 – 0, which gives k = kx/2mg = F/2mg. 54. (a) From the force diagram we write Fy = may: FT – mg = ma; FT – (355 kg)(9.80 m/s2) = (355 kg)(0.15)(9.80 m/s2), which gives FT = 4.00103 N. (b) The net work is done by the net force: Wnet = Fneth = (FT – mg)h = [4.00103 N – (355 kg)(9.80 m/s2)](33.0 m) = 1.72104 J. (c) The work done by the cable is Wcable = FTh = (4.00103 N)(33.0 m) = 1.32105 J. (d) The work done by gravity is Wgrav = – mgh = – (355 kg)(9.80 m/s2)(33.0 m) = – 1.15105 J. Note that Wnet = Wcable + Wgrav. (e) The net work done on the load increases its kinetic energy: Wnet = ?K = !mv2 – !mv02 ; 1.72104 J = !(355 kg)v2 – 0, which gives v = 9.84 m/s. 55. (a) For the work done by the applied force we have WP = FP d cos  = (150 N) cos 30° (5.0 m) = 6.5102 J. (b) For the force of gravity, we have Wgrav = (– mg sin  )d = – (20 kg)(9.80 m/s2)(5.0 m) sin 30° = – 4.9102 J. (c) Because the normal force is perpendicular to the displacement, it does no work: WN = 0. (d) The net work done on the block increases its kinetic energy: Page 9

y

FT

+y

mg

x

FN

FP

 mg



a

Chapter 7

Wnet = WP + Wgrav + WN = !mv2 – !mv02 ; 6.5102 J – 4.9102 J + 0 = !(20 kg)v2 – 0, which gives v =

4.0 m/s.

56. (a) The work done by the applied force is the same: WP = 6.5102 J. y (b) The work done by gravity is the same: Wgrav = – 4.9102 J. (c) The work done by the normal force is the same: WN = 0. FP (d) From the force diagram, we write F = ma: y-component: FN – mg cos  – FP sin  = 0, or  FN = mg cos  + FP sin  = (20 kg)(9.80 m/s2) cos 30° + (150 N) sin 30° = 245 N. The friction force is Ffr = kFN , so the work done by friction is Wfr = – kFNd = – (0.10)(245 N)(5.0 m) = – 1.2102 J. The net work done on the block increases its kinetic energy: Wnet = WP + Wgrav + WN + Wfr = !mv2 – !mv02 ; 6.5102 J – 4.9102 J + 0 – 1.2102 J = !(20 kg)v2 – 0, which gives v =

57. (a) The work done by gravity is the decrease in the potential energy: Wgrav = – mg(hf – hi) = – (755 kg)(9.80 m/s2)(0 – 22.5 m) = 1.66105 J. (b) The work done by gravity increases the kinetic energy: Wgrav = ?K; 1.66105 J = !(755 kg)v2 – 0, which gives v = 21.0 m/s. (c) For the motion from the break point to the maximum compression of the spring, we have Wspring + Wgrav = ?K; – (!kxf2 – !kxi2) – mg(hf – hi) = !mvf2 – !mvi2; – [!(8.00104 N/m)x2 – 0] – (755 kg)(9.80 m/s2)(– x – 22.5 m) = 0 – 0. This is a quadratic equation for x, which has the solutions x = – 1.95 m, 2.13 m. Because x must be positive, the spring compresses 2.13 m.

58. We consider an instance when the spring has a stretched length D and the mass has a velocity v, as shown in the diagram. If we take a differential segment of the spring at position x with a length dx, the mass of the segment is dM = (MS/D) dx and the velocity of the segment is vx = (x/D)v. When we add the kinetic energies of the segments of the spring by integrating, we get Page 10

x

FN

Ffr mg



1.9 m/s.

v=0 +y

h

y=0 x

v

D

m x

dx

Chapter 7 D

MS v 2 D 2 =1 x dx 2 D3 0 0 MS v 2 D3 1 MS v 2 = 1 = . 2 D3 2 3 3 When we add the kinetic energy of the mass m, we have K = !mv2 + !(@MS)v2 = !(m + @MS)v2, so the effective mass is M = m + @MS. K spring =

xv D

1 MS d x 2 D

2

59. For the proton we have 1 W p = Kp = m pc2 –1 ; 2 2 1 – (vp / c ) – 13

3.2  10

– 27

J = (1.67  10

8

2

kg )(3.0  10 m / s)

1 – 1, 2 2 1 – (vp / c )

which gives vp/c = 0.0651, or vp = 2.0107 m/s. Using the classical formula, we get Wp = !mpvpc2; 3.210–13 J = !(1.6710–27 kg)vpc2, which gives vpc = 2.0107 m/s, the same as the relativistic formula. For the electron we have 1 W e = Ke = m e c2 – 1; 1 – (ve2/ c2) 2 – 13 – 31 8 1 3.2  10 J = (9.1  10 kg)(3.0  10 m / s) 2 2 – 1, 1 – (v e / c ) which gives ve/c = 0.98, or ve = 2.9108 m/s. Using the classical formula, we get We = !mevec2; 3.210–13 J = !(9.110–31 kg)vec2, which gives vec = 8.4108 m/s, not only much greater than the relativistic formula, but greater than c, which is impossible. 60. (a) For the relativistic calculation we have 1 K a = mc2 –1 1 – (va2/ c 2)

1 2 2 –1 = 8 1 – (3.00  10 m/ s) / (3.00  10 m/ s) Using the classical formula, we get Ka = !mvac2 = !m(3.00104 m/s)2 = (4.50108 J/kg)m, the same. 8

= m(3.00  10 m/ s)

2

4

8

(4.50  10 J/ kg)m.

(b) For the relativistic calculation we have 1 K b = mc 2 –1 1 – (v b2/ c2 )

1 2 2 –1 = 8 1 – (3.00  10 m / s) / (3.00  10 m / s) Using the classical formula, we get Kb = !mvbc2 = !m(3.00106 m/s)2 = (4.501012 J/kg)m, the same. (c) For the relativistic calculation we have 1 K c = mc 2 –1 1 – (v c2/ c 2) 2 8 1 = m(3.00  10 m / s) 2 2 – 1 = 7 8 1 – (3.00  10 m / s) / (3.00  10 m / s) Using the classical formula, we get 8

= m(3.00  10 m/ s )

2

6

Page 11

12

(4.50  10 J/ kg)m.

14

(4.53  10 J/ kg)m.

Chapter 7

Kc = !mvcc2 = !m(3.00107 m/s)2 = (4.501014 J/kg)m, slightly less. 61. We let N represent the number of books of mass m that can be placed on a shelf. Because the initial and final kinetic energies are zero, for each book the net work is zero: W – mg(vertical distance center of mass of book is raised) = 0. From the diagram we see that the work required to fill the nth shelf is Wn = Nmg[D + !h + (n – 1)H]. Thus for the five shelves, we have W = W1 + W2 + W3 + W4 + W5 = Nmg[5(D + !h) + H + 2H + 3H + 4H] = Nmg[5(D + !h) + 10H] = (25)(1.60 kg)(9.80 m/s2){5[0.120 m + !(0.220 m)] + 10(0.330 m)} =

H

+y

H

h

D

y =0

1.74103 J.

62. (a) The kinetic energy of the locust is K = !mv2 = !(3.010–3 kg)(3.0 m/s)2 = 1.410–2 J. (b) If 40% of the energy is turned into kinetic energy, we have E = K/0.40 = (1.410–2 J)/0.40 = 3.410–2 J.

63. (a) The initial kinetic energy of the block is y Ki = !mvi2 = !(6.0 kg)(2.2 m/s)2 = 14.5 J = 15 J. (b) The work done by the applied force is FP WP = FP cos  d = (75 N) cos 37° (7.0 m) = 4.2102 J. (c) The work done by friction is  Wfr = – Ffrd = – (25 N)(7.0 m) = – 1.75102 J = – 1.8102 J. (d) The work done by gravity is Wgrav = (– mg sin  )d = – (6.0 kg)(9.80 m/s2) sin 37° (7.0 m) = – 2.5102 J. (e) Because the normal force is perpendicular to the displacement, it does no work: WN = 0. (f) The net work done on the block changes its kinetic energy: Wnet = WP + Wgrav + WN + Wfr = Kf – Ki ; 4.2102 J – 2.5102 J + 0 – 1.75102 J = Kf – 14.5 J, which gives Kf =

x

FN

Ffr mg



10 J.

64. (a) On a horizontal surface, FN = mg, and both forces do no work. The tension in the cord is always perpendicular to the displacement, so it does no work. Thus work is done only by the friction force. From the work-energy principle, for one revolution we have Wnet = Wgrav + WN + WT + Wfr1 = !mvf2 – !mvi2 ; 0 + 0 + 0 + Wfr1 = !m(0.75v0)2 – !mv02, which gives Wfr1 = – 0.22mv02. (b) Because Ffr = kmg, the work done in one revolution is Page 12

Chapter 7

Wfr1 = – 0.22mv02 = – kmg(2pR), which gives k = 0.035v02/gR. (c) Because the friction force has a constant magnitude, the work done in N revolutions is N times the work done in one revolution. From the work-energy principle we have Wfr = NWfr1 = !mvf2 – !mvi2; N(– 0.22mv02) = 0 – !mv02, which gives N = 2.3 rev. 65. (a) The work done by the two forces is W = W1 + W2 = F1 · d + F2 · d = (F1 + F2) · d = (1.50 N – 0.70 N)(8.0 m) + (– 0.80 N + 1.20 N)(6.0 m) + (0.70 N)(5.0 m) = 12 J. (b) The work done by the friction force is Wfr = Ffr · d = – 0.20[(1.50 N)(8.0 m) + (– 0.80 N)(6.0 m) + (0.70 N)(5.0 m)] = – 2.1 J. Thus the net work is Wnet = W + Wfr = 12.3 J – 2.1 J = 10 J. (c) As found in part (b), Wfr = – 2.1 J. 66. For an assumed constant force acting the length of the barrel, from the work-energy principle we have W = Fd = !mvf2 – !mvi2; F(15 m) = !(1250 kg)(750 m/s)2, which gives F = 2.3107 N (5.2106 lb). 67. The work done by the force is W = F · d = Fxdx + Fydy + Fzdz = (10.0 kN)(5.0 m) + (9.0 kN)(4.0 m) + (12.0 kN)(0) = 86 kJ. If we use the other expression for the scalar product, we have W = F · d = Fd cos ; 86 kJ = [(10.0 kN)2 + (9.0 kN)2 + (12.0 kN)2]1/2[(5.0 m)2 + (4.0 m)2]1/2 cos . which gives  = 42°. cos  = 0.746,

68. The two vectors are shown in the diagram. Their lengths are A = [(0.230 nm)2 + (0.133 nm)2]1/2 = 0.266 nm; B = [(0.077 nm)2 + (0.133 nm)2 + (0.247 nm)2]1/2 = 0.291 nm. We find the angle between the vectors from the scalar product: A · B = AxBx + AyBy + AzBz = AB cos ; (0.230 nm)(0.077 nm) + (0.133 nm)(0.133 nm) + (0)(0.247 nm) = (0.266 nm)(0.291 nm) cos , which gives cos  = 0.457,  = 62.8°.

z 3 (0.077, 0.133,0.247)

B 1

y

(0,0,0)

A x

69. For a variable force, we find the work by integration:

W=

F · ds =



d

Ae – kx dx = – A e – kx k

 d

= 0 – – A e – kd = k

A e – (0.10m)k. k

70. We find the work required to compress the spring by integration:

Page 13

2

(0.230,0.133,0)

Chapter 7

W = =

F · ds =

2.0 m 0

(150 N/ m )x + (12 N / m 3)x 3 dx

(150 N/ m)x 2 (12 N/ m 3 )x 4 + 2 4

2.0 m 0

2

4

= (75 N/ m)(2.0 m ) + (3.0 N/ m 3)(2.0 m) = 348 J.

When the spring is released, it will do this work on the ball. From the work-energy principle we have W = !mvf2 – !mvi2; 348 J = !(3.0 kg)vf2 – 0, which gives vf = 15 m/s. 71. The work done by the air resistance decreases the kinetic energy of the ball: W = – Faird = ?K = !mv2 – !mv02 = !m(0.90v0)2 – !mv02 = !mv02[(0.90)2 – 1]; – Fair(15 m) = !(0.25 kg)[(110 km/h)/(3.6 ks/h)]2[(0.90)2 – 1], which gives Fair =

1.5 N.

72. For the motion during the impact until the car comes momentarily to rest, we use the work-energy principle: Wspring = ?K; – (!kxf2 – !kxi2) = !mvf2 – !mvi2; – [!k(0.015 m)2 – 0] = 0 – !(1150 kg)[(8 km/h)/(3.6 ks/h)]2, which gives k = 2.5107 N/m. 73. The maximum acceleration will occur at the maximum compression of the spring: kxmax = mamax = m(5.0g), which gives xmax = 5.0mg/k. For the motion from reaching the spring to the maximum compression of the spring, we use the work-energy principle: Wspring = ?K; – (!kxf2 – !kxi2) = !mvf2 – !mvi2; – (!kxmax2 – 0) = 0 – !mv2. When we use the previous result, we get ! mv2 = !k(5.0mg/k)2, which gives k = 25mg2/v2 = 25(1300 kg)(9.80 m/s2)2/[(90 km/h)/(3.6 ks/h)]2 = 5.0103 N/m.

74. (a) The net work decreases the kinetic energy: Wsnow + Wgrav = Wsnow + mgd = ?K = !mvf2 – !mvi2; Wsnow + (80 kg)(9.80 m/s2)(1.1 m) = !(80 kg)[0 – (50 m/s)2] = – 1.0105 J. (b) We find the average force from F = W/d = (– 1.0105 J)/(1.1 m) = – 9.1104 N. (c) With air resistance during the fall we have Wair + Wgrav = Wair + mgh = ?K= (!mvf2 – !mvi2) Wair + (80 kg)(9.80 m/s2)(370 m) = !(80 kg)[(50 m/s)2 – 0], which gives Wair =

– 1.9105 J.

75. (a) Because the rider exerts a force on each side of the pedals, when the front sprocket has turned through Nfront revolutions, the work done will be Wrider = Nfront 2FDfront = 2Nfront (0.90mg)Dfront = 1.80mgNfrontDfront . The number of points that the chain passes over must be the same for the front and back sprockets: Nfront(42 points/rev) = Nback(19 points/rev), so the number of revolutions of the rear wheel is Nback = (42/19)Nfront . Page 14

Chapter 7

After Nback revolutions, the bike travels a distance d = Nback2pRwheel = (42/19)Nfront2pRwheel . For the work-energy principle we have Wrider + Wgrav = ?K = 0, or 1.80mgNfrontDfront – (m + mbike)gd sin  = 0, or 1.80mgNfrontDfront = (m + mbike)g(42/19)Nfront2pRwheel sin ; (1.80)(60 kg)gNfront(0.36 m) = (60 kg + 12 kg)g(42/19)Nfront2p(0.34 m) sin , which gives sin  = 0.114,  = 6.6°. (b) If the force is applied tangential to the pedal motion, the work done is Wrider = Nfront F2pRfront = Nfront (0.90mg)2pRfront = 1.80pmgNfrontRfront . For the work-energy principle we have Wrider + Wgrav = ?K = 0, or 1.80pmgNfrontRfront – (m + mbike)gd sin  = 0, or 1.80pmgNfrontRfront = (m + mbike)g(42/19)Nfront2pRwheel sin ; (1.80)p(60 kg)gNfront(0.18 m) = (60 kg + 12 kg)g(42/19)Nfront2p(0.34 m) sin , which gives sin  = 0.180,  = 10.3°. 76. (a) Because the acceleration is essentially zero, we have FT cos  = mg; FT sin  = F, which gives F = mg tan . Because the force is variable, we find the work done by integration. The displacement for a differential change in angle, ds = L d, is tangent to the path of the pendulum, so we have

W = =

0

F · ds = 0 0

0

L



FT

ds

(F cos  ) L d 

 F

mgL t a n  cos  d  = mgL

0 0

si n  d  = – mgL cos 

0 0

mg

= mgL(1 – cos  0 ). (b) Because the angle between the force of gravity and the displacement is changing, we find the work done by integration: W grav =

mg · d s =

= – mgL

0 0

0 0

(– mg sin ) L d 

sin  d  = + mgL cos 

0 0

=

mgL(cos  0 – 1).

Note that this is – mgh, where h is the height the pendulum has risen. Because the tension in the cord is always perpendicular to the displacement, no work is done: WT = 0. Note that the net force of all three forces is zero, as it must be since there is no change in kinetic energy.

Page 15

Chapter 8

CHAPTER 8 - Conservation of Energy 1.

The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = ½kxf2 – 0; 35.0 J = ½(82.0 N/m)xf2 – 0, which gives xf =

2.

0.924 m.

For the potential energy change we have ?U = mg ?y = (5.0 kg)(9.80 m/s2)(1.5 m) =

74 J.

3.

For the potential energy change we have ?U = mg ?y = (58 kg)(9.80 m/s2)(3.8 m) =

2.2103 J.

4.

(a) With the reference level at the ground, for the potential energy change we have 7.56105 J. ?U = mg ?y = (66.5 kg)(9.80 m/s2)(2660 m – 1500 m) = (b) The minimum work would be equal to the change in potential energy: Wmin = ?U = 7.56105 J. (c) Yes, the actual work will be more than this. There will be additional work required for any kinetic energy change, and to overcome retarding forces, such as air resistance and ground deformation.

5.

(a) With the reference level at the ground, for the potential energy we have Ua = mgya = (2.20 kg)(9.80 m/s2)(2.40 m) = 51.7 J. (b) With the reference level at the top of the head, for the potential energy we have Ub = mg(yb – h)= (2.20 kg)(9.80 m/s2)(2.40 m – 1.70 m) = 15.1 J. (c) Because the person lifted the book from the reference level in part (a), the potential energy is equal to the work done: 51.7 J. In part (b) the initial potential energy was negative, so the final potential energy is not the work done, which was still 51.7 J.

6.

For the potential energy U = 3x2 + 2xy + 4y2z, we find the components of the force from Fx = – ?U/?x = – 6x – 2y – 0 = – (6x + 2y); Fy = – ?U/?y = – 0 – 2x – 8yz = – (2x + 8yz); Fz = – ?U/?z = – 0 – 0 – 4y2 = – (4y2). Thus the force is F = – (6x + 2y)i – (2x + 8yz)j – (4y2)k.

7.

(a) Because the force F = (– kx + ax3 + bx4)i is a function only of position, it is (b) We find the form of the potential energy function from U = – ? F · dr = – ? (– kx + ax3 + bx4)i · (dxi + dyj + dzk) ½kx2 – ¼ax4 – $bx5 + constant. = – ? (– kx + ax3 + bx4)dx =

8.

This force is not conservative. If the object moves along a path that returns to the starting point, the direction of the motion (the direction of the velocity) changes. Because the direction of the force changes with the change in direction of the motion, the net work done by the force is not zero.

9.

The potential energy of the spring is zero when the spring is not stretched or compressed (x= 0). (a) For the change in potential energy, we have ½k(x2 – x02). ?U = ½kx2 – ½kx02 = (b) If we call compressing positive, we have ?Ucompression = ½k(+ x0)2 – 0 = ½kx02; ?Ustretching = ½k(– x0)2 – 0 = ½kx02. The change in potential energy is the

same.

Page 1

conservative.

Chapter 8

10. We choose the potential energy to be zero at the ground (y = 0). Because the tension in the vine does no work, energy is conserved, so we have E = K0 + U0 = Kf + Uf ;

½mv02 + mgy0 = ½mvf2 + mgyf ; ½m(5.0 m/s)2 + m(9.80 m/s2)(0) = ½m(0)2 + m(9.80 m/s2)h,

L

which gives h = 1.3 m. No, the length of the vine does not affect the height; it affects the angle. v0



h

y=0

11. We choose the potential energy to be zero at the bottom (y = 0). Because there is no friction and the normal force does no work, energy is conserved, so we have E = K1 + U1 = K2 + U2 ;

½mv12 + mgy1 = ½mv22 + mgy2 ; ½m(0)2 + m(9.80 m/s2)(105 m) = ½mv22 + m(9.80 m/s2)(0), which gives v2 =

45.4 m/s.

This is 160 km/h½ It is a good thing there is friction on the ski slopes.

12. We choose the potential energy to be zero at the bottom (y = 0). Because there is no friction and the normal force does no work, energy is conserved, so we have E = K1 + U1 = K2 + U2 ;

½mv12 + mgy1 = ½mv22 + mgy2 ; ½mv12 + m(9.80 m/s2)(0) = ½m(0)2 + m(9.80 m/s2)(1.22 m), which gives v1 =

4.89 m/s.

13. We choose the potential energy to be zero at the level of the center of mass before the jump (y = 0). We find the minimum speed by ignoring any frictional forces. Energy is conserved, so we have E = K1 + U1 = K2 + U2 ;

½mv12 + mgy1 = ½mv22 + mgy2 ; ½mv12 + m(9.80 m/s2)(0) = ½m(0.70 m/s)2 + m(9.80 m/s2)(2.10 m), which gives v1 =

6.5 m/s. Note that the initial velocity will not be horizontal, but will have a horizontal component of 0.70 m/s. 14. We choose y = 0 at the level of the trampoline. (a) We apply conservation of energy for the jump from the top of the platform to the trampoline: E = K1 + U1 = K2 + U2 ;

½mv12 + mgH = ½mv22 + 0; ½m(5.0 m/s)2 + m(9.80 m/s2)(2.0 m) = ½mv22,

which gives v2 = 8.0 m/s. (b) We apply conservation of energy from the landing on the trampoline to the maximum depression of the trampoline. If we ignore the small change in gravitational potential energy, we have E = K2 + U2 = K3 + U3 ;

½mv22 + 0 = 0 + ½kx2; ½(75 kg)(8.0 m/s)2 = ½(5.2104 N/m)x2,

v1

+y

y =0

which gives x = 0.30 m. This will increase slightly if the gravitational potential energy is taken into account.

Page 2

H v2

Chapter 8

15. (a) For the motion from the bridge to the lowest point, we use energy conservation: Ki + Ugravi + Ucordi = Kf + Ugravf + Ucordf ;

y=0 L0

0 + 0 + 0 = 0 + mg(– h) + ½k(h – L0)2 ;

+y

0 = – (60 kg)(9.80 m/s2)(31 m) + ½k(31 m – 12 m)2, 1.0102 N/m. which gives k = (b) The maximum acceleration will occur at the lowest point, where the upward restoring force in the cord is maximum: kxmax – mg = mamax ; (1.0102 N/m)(31 m – 12 m) – (60 kg)(9.80 m/s2) = (60 kg)amax , which gives amax = 22 m/s2.

16. We choose y = 0 at point B. With no friction, energy is conserved. The initial (and constant) energy is E = EA = mghA + ½mvA2 = m(9.8 m/s2)(30 m) + 0 = (294 J/kg)m . At point B we have E = mghB + ½mvB2; m/s2)(0)

(294 J/kg)m = m(9.8 which gives vB = 24 m/s. At point C we have E = mghC + ½mvC2;

+½

mvB2,

h x

v=0

A C

hA

D

hC

hD

B

(294 J/kg)m = m(9.8 m/s2)(25 m) + ½mvC2, which gives vC = 9.9 m/s. At point D we have E = mghD + ½mvD2; (294 J/kg)m = m(9.8 m/s2)(12 m) + ½mvD2, which gives vD = 19 m/s. 17. We choose the potential energy to be zero at the compressed position (y = 0). (a) For the motion from the release point to where the ball leaves the spring, we use energy conservation: Ki + Ugravi + Uspringi = Kf + Ugravf + Uspringf ; 0 + 0 + ½kx2 = ½mv2 + mgx + 0; ½(900 N/m)(0.150 m)2 = ½(0.300 kg)v2 + (0.300 kg)(9.80 m/s2)(0.150 m), which gives v = 8.03 m/s. (b) For the motion from the release point to the highest point, we use energy conservation: Ki + Ugravi + Uspringi = Kf + Ugravf + Uspringf ; 0 + 0 + ½kx2 = 0 + mgh + 0; 0 + 0 + ½(900 N/m)(0.150 m)2 = (0.300 kg)(9.80 m/s2)h, which gives h = 3.44 m. Page 3

v=0

h

+y

v x y=0

Chapter 8

18. (a) Because the horizontal speed of the ball does not change, the speed at the highest point is v2 = v1 cos  = (10 m/s) cos 30° = 8.7 m/s. (b) We choose the potential energy to be zero at the ground (y = 0). For the motion from the release point to the highest point, we use energy conservation: K1 + U1 = K2 + U2 ;

½mv12 + 0 = ½mv22 + mgh; ½(10 m/s)2 = ½(8.67 m/s)2 + (9.80 m/s2)h, which gives h =

1.3 m.

19. The potential energy is zero at x = 0. (a) Because energy is conserved, the maximum speed occurs at the minimum potential energy: Ki + Ui = Kf + Uf ;

½mv02 + ½kx02 = ½mvmax2 + 0, which gives

vmax = [v02 + (kx02/m)]1/2. (b) The maximum stretch occurs at the maximum potential energy or the minimum kinetic energy: Ki + Ui = Kf + Uf ;

½mv02 + ½kx02 = 0 + ½kxmax2, which gives

xmax = [x02 + (mv02/k)]1/2.

20. (a) The work done against gravity is the increase in the potential energy: W = mgh = (75.0 kg)(9.80 m/s2)(92.0 m) = 6.76104 J. (b) If this work is done by the force on the pedals, we need to find the distance that the force acts over one revolution of the pedals and the number of revolutions to climb the hill. We find the number of revolutions from the distance along the incline: N = (h/ sin )/(5.10 m/revolution) = [(92.0 m)/ sin 9.50°]/(5.10 m/revolution) = 109 revolutions. Because the force is always tangent to the circular path, in each revolution the force acts over a distance equal to the circumference of the path: pD. Thus we have W = NFpD; 547 N. 6.76104 J = (109 revolutions)Fp(0.360 m), which gives F =

Page 4

Chapter 8

21. We choose y = 0 at the lowest point of the swing. (a) We apply conservation of energy from the release point to the lowest point: E = K1 + U1 = K2 + U2 ; 0 + mgL(1 – cos 0) = ½mva + 0; 2

0 + (9.80 m/s2)(2.00 m)(1 – cos 30.0°) = ½va2, 2.29 m/s. which gives va = (b) We apply conservation of energy from the release point to the given point: E = K1 + U1 = K3 + U3 ;

L



h

y=0

va

0 + mgL(1 – cos 0) = ½mvb2 + mgL(1 – cos b); 0 + (9.80 m/s2)(2.00 m)(1 – cos 30.0°) = ½vb2 + (9.80 m/s2)(2.00 m)(1 – cos 15.0°), 1.98 m/s. which gives vb = (c) We apply conservation of energy from the release point to the given point: E = K1 + U1 = K4 + U4 ; 0 + mgL(1 – cos 0) = ½mvc2 + mgL(1 – cos c); 0 + (9.80 m/s2)(2.00 m)(1 – cos 30.0°) = ½vc2 + (9.80 m/s2)(2.00 m)[1 – cos (– 15.0°)], 1.98 m/s. which gives vc = Because this is the same elevation as in part (b), the answer is the same. (d) The net force along the cord must provide the radial acceleration: FT – mg cos  = mv2/L, or FT = m[(v2/L) + g cos ]. Thus we have FTa = m[(va2/L) + g cos a] = (0.070 kg){[(2.29 m/s)2/(2.00 m)] + (9.80 m/s2) cos 0°} =

FTb = m[(vb2/L) + g cos b] = (0.070 kg){(1.98 m/s)2/(2.00 m)] + (9.80 m/s2) cos 15.0°} =

0.87 N. 0.80 N.

FTc = m[(vc2/L) + g cos c] = (0.070 kg){[(1.98 m/s)2/(2.00 m)] + (9.80 m/s2) cos (– 15.0°) } = (e) With an initial kinetic energy, conservation of energy from the release point to the given point becomes E = K1 + U1 = K4 + U4 ;

0.80 N.

½mv02 + mgL(1 – cos 0) = ½mv2 + mgL(1 – cos ).

Thus we have ½(1.20 m/s)2 + (9.80 m/s2)(2.00 m)(1 – cos 30.0°) = ½va2 + (9.80 m/s2)(2.00 m)(1 – cos 0°), 2.59 m/s; which gives va =

½(1.20 m/s)2 + (9.80 m/s2)(2.00 m)(1 – cos 30.0°) = ½vb2 + (9.80 m/s2)(2.00 m)(1 – cos 15.0°),

which gives vb =

½(1.20

2.31 m/s;

+ (9.80 m/s2)(2.00 m)(1 – cos 30.0°) = ½vc2 + (9.80 m/s2)(2.00 m)[1 – cos (– 15.0°)], 2.31 m/s. which gives vc = m/s)2

22. The maximum acceleration will occur at the maximum compression of the spring: kxmax = Mamax = M(5.0g), which gives xmax = 5.0Mg/k. For the motion from reaching the spring to the maximum compression of the spring, we use energy conservation: Page 5

Chapter 8

Ki + Uspringi = Kf + Uspringf ;

½Mv2 + 0 = 0 + ½kxmax2.

When we use the previous result, we get ½Mv2 = ½k(5.0Mg/k)2, which gives k = 25Mg2/v2 = 25(1200 kg)(9.80 m/s2)2/[(100 km/h)/(3.6 ks/h)]2 =

23. The maximum acceleration will occur at the lowest point, where the upward restoring force in the spring is maximum: kxmax – Mg = Mamax = M(5.0g), which gives xmax = 6.0Mg/k. With y = 0 at the initial position of the top of the spring, for the motion from the break point to the maximum compression of the spring, we use energy conservation: Ki + Ugravi + Uspringi = Kf + Ugravf + Uspringf ; 0 + Mgh + 0 = 0 + Mg(– xmax) + ½kxmax2. When we use the previous result, we get Mgh = – [6.0(Mg)2/k] + ½k(6.0Mg/k)2, which gives

24. (a) The net force toward the center of the sphere provides the radial acceleration: mg cos  – FN = mv2/r. The skier will leave the sphere when the normal force becomes zero, or va2 = gr cos a. With the reference level at the center of the sphere, we apply conservation of energy from the top to the point where the skier leaves: E = K1 + U1 = K2 + U2 ;

3.7103 N/m.

v=0 +y

h

y=0 k = 12Mg/h.

x

FN v

 r

mg

0 + mgr = ½mva2 + mgr cos a; gr = ½gr cos a + gr cos a , which gives cos a = %, a = 48.2° . (b) The condition for the normal force to become zero is vb2 = gr cos b. There will be negative work done by the friction force, so we have Wf = ?K + ?U = (½mvb2 – 0) + mgr(cos b – 1) = ½mgr cos b + mgr cos b – mgr, which we write cos b = %[1 + (Wf/mgr)]. Because Wf is negative, this means that cos b < cos a , or b > a . Page 6

Chapter 8

25. The thermal energy is equal to the loss in kinetic energy: Ethermal = – ?K = ½mvi2 – ½mvf2 = ½(2)(6500 kg)[(95 km/h)/(3.6 ks/h)]2 – 0 =

4.5106 J.

26. We choose the bottom of the slide for the gravitational potential energy reference level. The thermal energy is the negative of the change in kinetic and potential energy: Ethermal = – (?K + ?U) = ½mvi2 – ½mvf2 + mg(hi – hf) = 0 – ½(16.0 kg)(2.25 m/s)2 + (16.0 kg)(9.80 m/s2)(2.50 m – 0) =

27. (a) We find the normal force from the force diagram for the ski: y-component: FN1 = mg cos ; which gives the friction force: Ffr1 = kmg cos . For the work-energy principle, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); – kmg cos  L = (½mvf2 – 0) + mg(0 – L sin ); – (0.090)(9.80 m/s2) cos 20° (100 m) = ½vf2 – (9.80 m/s2)(100 m) sin 20°, which gives vf = 22 m/s. (b) On the level the normal force is FN2 = mg, so the friction force is Ffr2 = kmg. For the work-energy principle, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi);

FN1

352 J.

L

FN2

Ffr1 mg y=0



– kmg D = (0 – ½mvi2) + mg(0 – 0); – (0.090)(9.80 m/s2)D = – ½(22.5 m/s)2 , 2.9102 m. which gives D = 28. We choose the reference level for the gravitational potential energy at the ground. (a) With no air resistance during the fall we have 0 = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi), or

½(vf2 – 0) = – (9.80 m/s2)(0 – 12.0 m), which gives vf =

15.3 m/s.

(b) With air resistance during the fall we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi);

Fair(12.0 m) = ½(0.145 kg)[(8.00 m/s)2 – 0] + (0.145 kg)(9.80 m/s2)(0 – 12.0 m), which gives Fair = – 1.03 N. 29. On the level the normal force is FN = mg, so the friction force is Ffr = kmg. For the work-energy principle, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); F(L1 + L2) – kmg L2 = (½mvf2 – 0) + mg(0 – 0); (350 N)(15 m + 15 m) – (0.25)(90 kg)(9.80 m/s2)(15 m) = ½(90 kg)vf2, which gives vf = 13 m/s.

Page 7

Ffr2

mg

Chapter 8 A

30. We choose y = 0 at point B. For the work-energy principle applied to the motion from A to B, we have WNC = ?K + ?U = (½mvB2 – ½mvA2) + mg(hB – hA);

C

– 0.20mgL = (½mvB2 – ½mvA2) + mg(0 – hA);

hA

– 0.20(9.80 m/s2)(45.0 m) = ½vB2 – ½(1.70 m/s)2 – (9.80 2 m/s )(30 m), which gives vB = 20 m/s.

L FN Ffr

– kmg cos  L = (0 – ½mvi2) + mg(L sin  – 0); – k(9.80 m/s2) cos 17° (12 m) =

mg

– ½(11 m/s)2 + (9.80 m/s2)(12 m) sin 17°, 0.23. which gives k =

0 = ½mvB2 – 0 + 0 – mgr;

A

D hD

B

31. We find the normal force from the force diagram for the skier: y-component: FN = mg cos ; which gives the friction force: Ffr = kmg cos . For the work-energy principle for the motion up the incline, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi);

32. (a) For the motion from A to B, there is no friction and the normal force does no work. For the work-energy principle, we have WNC = (½mvB2 – ½mvA2) + mg(hB – hA);

hC



y=0

m r

y=0

C

B

D L

0 = ½vB2 – (9.80 m/s2)(2.0 m), which gives vB = 6.3 m/s. (b) On the level the normal force is FN = mg, so the friction force is Ffr = kmg. The work done by this force from B to C is WNC = – kmgL = – (0.25)(1.0 kg)(9.80 m/s2)(3.0 m) = – 7.4 J. (c) For the motion from B to C, the normal force does no work. For the work-energy principle, we have WNC = (½mvC2 – ½mvB2) + mg(hC – hB); – 7.4 J = ½(1.0 kg)vC2 – ½(1.0 kg)(6.3 m/s)2 + 0 – 0, which gives vC = 4.9 m/s. (d) For the motion from C to the point where the block momentarily comes to rest, there is no friction. For the work-energy principle, we have WNC = (½mv2 – ½mvC2) + mg(h – hC) + (½kx2 – ½kxC2); 0 = 0 – ½(1.0 kg)(4.9 m/s)2 + 0 – 0 + ½k(0.20 m)2 – 0, 6.1102 N/m. which gives k = 33. On the level the normal force is FN = mg, so the friction force is Ffr = kmg. The block is at rest at the release point and where it momentarily stops before turning back. For the work-energy principle, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + (½kxf2 – ½kxi2); – kmg L = (0 – 0) + ½k(xf2 – xi2); – k(0.620 kg)(9.80 m/s2)(0.050 m + 0.023 m) = ½(180 N/m)[(0.023 m)2 – (– 0.050 m)2], Page 8

Chapter 8

which gives k =

0.40.

34. We find the spring constant from the force required to compress the spring: k = F/xi = ( – 22 N)/(– 0.18 m) = 122 N/m. On the level the normal force is FN = mg, so the friction force is Ffr = kmg. The block is at rest at the release point and where it momentarily stops before turning back. For the work-energy principle, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + (½kxf2 – ½kxi2); – kmg L = (0 – 0) + ½k(xf2 – xi2); – (0.30)(0.180 kg)(9.80 m/s2)(0.18 m + xf) = ½(122 N/m)[xf2 – (– 0.18 m)2]. This reduces to the quadratic equation 61xf2 + 0.529xf – 1.88 = 0, which has the solutions xf = 0.17 m, – 0.18 m. The negative solution corresponds to no motion, so the physical result is xf =

35. (a) On the level the normal force is FN = mg, so the friction force is Ffr = kmg. For the motion from the impact point to where the block stops, for the work-energy principle we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + (½kxf2 – ½kxi2);

0.17 m.

v1

– kmgx = (0 – ½mv12) + ½k(x2 – 0);

x – (0.30)(2.0 kg)(9.80 m/s2)x = – ½(2.0 kg)(1.3 m/s)2 + ½(120 v =0 N/m)x2. This reduces to the quadratic equation 60x2 + 5.88x – 1.69 = 0, which has the solutions x = 0.126 m, – 0.22 m. The negative solution corresponds to positive friction work, so the physical result is x = 0.13 m. (b) For the block to remain at rest at the maximum compressed position, the magnitude of the restoring force in the spring must equal the magnitude of the static friction force: kx = Fs = smg, or s = kx/mg = (120 N/m)(0.126 m)/(2.0 kg)(9.80 m/s2) = 0.77. (c) Before the spring reaches its natural length, it is pushing on the block. At the natural length, the force goes to zero. Beyond the natural length the spring would want to pull on the block, but it is not attached; therefore the block leaves the spring. If we consider the motion from the initial impact to the point where the block leaves the spring, for the work-energy principle, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + (½kxf2 – ½kxi2); – kmg2x = (½mvf2 – ½mv12) + (0 – 0); – (0.30)(9.80 m/s2)2(0.126 m) = ½vf2 – ½(1.3 m/s)2, which gives vf =

0.46 m/s.

36. We choose the potential energy to be zero at the ground (y = 0). We convert the speeds: (500 km/h)/(3.6 ks/h) = 139 m/s; (200 km/h)/(3.6 ks/h) = 55.6 m/s. (a) If there were no air resistance, energy would be conserved: Page 9

Chapter 8

0 = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); 0 = ½(1000 kg)[vf2 – (139 m/s)2] + (1000 kg)(9.80 m/s2)(0 – 3500 m), which gives vf = 297 m/s = 1.07103 km/h. (b) With air resistance we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); – F(hi/ sin ) = ½m(vf2 – vi2) + mg(0 – hi); – F(3500 m)/sin 10° = ½(1000 kg)[(55.6 m/s)2 – (139 m/s)2] + (1000 kg)(9.80 m/s2)(– 3500 m) 2.1103 N. which gives F = 37. (a) The gravitational force provides the radial acceleration: FG = GMEmS//rS2 = mSv2//rS , or v2 = GME/rS. Thus the kinetic energy is K = ½mSv2 = GMEmS/2rS. (b) With U = 0 at infinity, the potential energy is U = – GMEmS/rS. (c) For the ratio we have K/U = (GMEmS/2rS)/(– GMEmS/rS) = – ½. 38. Because energy is conserved, we have K1 + U1 = K2 + U2 ;

½mv12 – GMEm/rE = 0 – GMEm/r, or v12/2GME = (1/rE) – (1/r); (850 m/s)2/2(6.6710–11 N · m2/kg2)(5.981024 kg) = (1/6.38106 m) – (1/r), which gives r = 6.42106 m. Thus the height above the surface of the Earth is h = r – rE = 6.42106 m – 6.38106 m = 4104 m.

39. The escape velocity from a mass M, as determined by energy conservation, is found from vesc2 = 2GM/r. (a) To escape from the Sun’s surface, we have vesc2 = 2GMS/rS = 2(6.6710–11 N · m2/kg2)(2.01030 kg)/(7.0108 m), 6.2105 m/s. which gives vesc = (b) To escape from the Sun when at the Earth’s location, we have vesc2 = 2GMS/r = 2(6.6710–11 N · m2/kg2)(2.01030 kg)/(1.501011 m), 4.2104 m/s. which gives vesc = Because the gravitational attraction provides the radial acceleration of the Earth, we have GMSME/r2 = MEvorbit2/r, or vorbit2 = GMS/r. For the ratio we get vesc2/vorbit2 = (2GMS/r )/(GMS/r), or vesc/vorbit = v2. 40. The change in gravitational potential energy is ?U = – (GmME)[(1/r2) – (1/r1)]. In terms of the height above the surface of the Earth, y, which is small compared to rE , we have 1/r1 = 1/(rE + y1) = 1/{rE[1 + (y1/rE)]} ˜ (1/rE)[1 – (y1/rE)];

1/r2 = 1/(rE + y2) = 1/{rE[1 + (y2/rE)]} ˜ (1/rE)[1 – (y2/rE)];

where we have used 1/(1 + x) ˜ 1 – x, when x « 1. When we use these in the expression for ?U, we get Page 10

Chapter 8

?U ˜ – (GmME)(1/rE){[1 – (y2 /rE)] – [1 – (y1 /rE)]} = (GmME/rE2)(y2 – y1) = mg(y2 – y1). 41. The change in gravitational potential energy is ?U = – (GmME)[(1/r2) – (1/r1)]. In terms of the height above the surface of the Earth, h, we have 1 – 1 U = – GmM E rE + h rE 1 Gm ME 1 – Gm M E 1 + h/ rE – 1 = h/ rE = 1 + 1 + h/ r E rE rE h/ rE h mgh Gm ME Gm M E = = = . rE 1 + h/ rE r 2E 1 + h/ rE 1 + h/ r E 42. (a) Because the gravitational attraction provides the radial acceleration of the satellite, we have GmME/r2 = mv2/r, or v2 = GME/r. The total energy of the satellite is E = K + U = ½mv2 – GMEm/r = ½mGME/r – GMEm/r = – ½GmME/r. (b) From the expression for the total energy, we see that a decrease in E (E becomes more negative) means a decrease in r. Because the kinetic energy is positive, a decrease in r means an increase in kinetic energy. 43. The escape velocity for a mass m is the speed required so the mass can get infinitely far away with essentially zero velocity. From energy conservation we have K1 + U1 = K2 + U2 ;

½mvesc2 – GMEm/r = 0 – GMEm/8, or vesc2 = 2GME/r.

Because the gravitational attraction provides the radial acceleration of the satellite, we have GmME/r2 = mvorbit2/r, or vorbit2 = GME/r. For the ratio we get vesc2/vorbit2 = (2GME/r )/(GME/r), or vesc/vorbit = v2.

44. (a) The change in gravitational potential energy is ?U = – (GMSME)[(1/r2) – (1/r1)] = – (6.6710–11 N · m2/kg2)(2.01030 kg)(5.981024 kg)[(1/1.5211011 m) – (1/1.4711011 m)] = 1.81032 J. (b) Because the total energy is constant, the change in kinetic energy is – 1.81032 J. ?K = – ?U = Note that the orbit is not circular so the gravitational attraction is not always perpendicular to the velocity. Consequently it does not just provide the radial acceleration. (c) Because energy is conserved for the Earth–Sun system, the change in the total energy is zero. 45. The velocity of a point on the surface of the Earth is vsurface = rE = (6.38106 m)(2p rad/day)/(86,400 s/day) = 464 m/s to the East. The required escape speed is found from vesc2 = 2GME/rE = 2(6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m), which gives vesc = 1.12104 m/s. (a) If the speed with respect to the Earth is v0 , when the rocket is fired eastward, we have vesc = v0 + vsurface; 1.12104 m/s = v0 + 464 m/s, which gives v0 = 1.07104 m/s. (b) If the speed with respect to the Earth is v0 , when the rocket is fired westward, we have vesc = v0 – vsurface; Page 11

Chapter 8

1.12104 m/s = v0 – 464 m/s, which gives v0 = 1.17104 m/s. (c) If the speed with respect to the Earth is v0 , when the rocket is fired vertically upward, we have vesc2 = v02 + vsurface2; (1.12104 m/s)2 = v02 + (464 m/s)2, which gives v0 = 1.12104 m/s. 46. (a) Because the speed is zero at the maximum height, from energy conservation we have K1 + U1 = K2 + U2 ;

½mv02 – GMEm/rE = 0 – GMEm/(rE + h), or ½v02 = GME{(1/rE) – [1/(rE + h)]}= GME[h/rE(rE + h)].

When we solve this for h we get h = v02rE2/(2GME – v02rE) = rE/[(2GME/v02rE) – 1]. (b) For the given speed we have h = (6.38106 m)/{[2(6.6710–11 N · m2/kg2)(5.981024 kg)/(8.2103 m/s)2(6.38106 m)] – 1} = 7.4106 m = 7.4103 km. 47. (a) The escape velocity at a distance r from the center of the Earth is vesc = (2GME/r)1/2. We find the rate at which vesc changes with a change in r by differentiating: dvesc/dr = – ½(2GME/r3)1/2 = – vesc/2r. (b) The escape speed from the surface of the Earth is found from vesc02 = 2GME/rE = 2(6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m), which gives vesc0 = 1.12104 m/s. The approximate change at a height of 300 km is ?vesc ˜ (– vesc0/2rE) ?r = [– (1.12104 m/s)/2(6.38106 m)](300103 m) = – 263 m/s. The new escape velocity is vesc = vesc0 + ?vesc = 1.12104 m/s – 263 m/s = 1.09104 m/s.

48. (a) For energy conservation for the motion until just before the meteor strikes the sand, we have K1 + U1 = K2 + U2 ;

½mv12 – GMEm/(rE + h) = ½mv22 – GMEm/rE , or v22 = v12 + 2GME{(1/rE) – [1/(rE + h)]}; v22 = (90.0 m/s)2 + 2(6.6710–11 N · m2/kg2)(5.981024 kg) 

{(1/6.38106 m) – [1/(6.38106 m + 0.800106 m)]}

which gives v2 = 3.73103 m/s. (b) Because the change in potential energy is very small, the work changes the kinetic energy: WNC = ?K = 0 – ½mv12 = – ½(575 kg)(3.73103 m/s)2 = – 4.00109 J. (c) We find the average force from WNC = – Favd; – 4.01109 J = – Fav(3.25 m), which gives Fav = 1.23109 N. (d) The work done by the sand produces the thermal energy: Ethermal = – WNC = 4.00109 J. 49. The total energy of the satellite in an orbit of radius r is E = K + U = ½mv2 – GMEm/r = ½mGME/r – GMEm/r = – ½mGME/r. Page 12

Chapter 8

Thus the work required is W = ?E = – ½mGME[(1/r2) – (1/r1)] = – ½mGME[(1/3rE) – (1/2rE)] =

GmME/12rE.

50. (a) We can find the total work needed by adding the works required to bring the masses in one at a time. Each work is found from the change in potential energy. The initial potential energy of each mass is zero. Because the first mass is still infinitely far from the others, we have W1 = 0. When we bring in the second mass, we have W2 = – Gm1m2/r12 – 0 = – Gm1m2/r12. When we bring in the third mass, we have W3 = (– Gm3m1/r13 – Gm3m2/r23) – 0 = – G[(m3m1/r13) + (m3m2/r23)]. For the total work, we get W = W1 + W2 + W3 = 0 – (Gm1m2/r12) – G[(m3m1/r13) + (m3m2/r23)] = – G[(m1m2/r12) + (m1m3/r13) + (m2m3/r23)]. Note that the negative sign is an indication that the masses attract each other, so the energy is less than the value at infinity, which is zero. (b) This is the energy stored in the system, and thus is the potential energy of the system. It is not the potential energy of one or two of the bodies. A change in any one of the bodies will change the potential energy. (c) The negative of W can be considered the binding energy of the system. This positive work would have to be done to separate the masses. 51. If we neglect friction, we can apply conservation of energy: K1 + U1 = K2 + U2 ;

½mv12 – GMEm/r1 = ½mv22 – GMEm/r2 , or v22 = v12 + 2GME[(1/r2) – (1/r1)]; v22 = (600 m/s)2 + 2(6.6710–11 N · m2/kg2)(5.981024 kg)[(1/6.38106 m) – (1/5.0109 m)] which gives v2 = 1.1104 m/s.

52. We can consider the uniform spherical shell to be a solid sphere of radius r1 of the same density and a smaller sphere of radius r2 with a negative density of the same magnitude. If the additional mass in the smaller sphere is M, the mass with negative density is – M. With our reference level of zero potential energy at infinity, the gravitational potential energy of the mass m is the sum of the two contributions: U = – G(M + M)m/r – G(– M)m/r = – GmM/r .

m r1

r2

53. The escape speed from the surface of the Earth, ignoring the Sun, is found from vE2 = 2GME/rE = 2(6.6710–11 N · m2/kg2)(5.981024 kg)/(6.38106 m), which gives vE = 1.12104 m/s = 11.2 km/s. The escape speed from the Sun when at the Earth’s orbit, ignoring the Earth, is found from Page 13

r

Chapter 8

vS2 = 2GMS/rSE = 2(6.6710–11 N · m2/kg2)(2.01030 kg)/(1.501011 m), which gives vS = 4.22104 m/s = 42.2 km/s. The orbital speed of the Earth is vO = rSE = (1.501011 m)(2p rad/yr)/(3.17107 s/yr) = 2.98104 m/s = 29.8 km/s. (a) In the reference frame of the Earth, if the spacecraft leaves the surface of the Earth with speed v, we find the speed v at a distance where the gravitational attraction is negligible from energy conservation: K1 + U1 = K2 + U2 ;

½mv2 – GMEm/rE = ½mv2 – 0, or v2 = v2 + 2GME/rE = v2 + vE2. The reference frame of the Earth is orbiting the Sun with speed vO . In the reference frame of the Sun, the speed far from the Earth is vS = v + vO . When we use this in the previous result, we get v2 = (vS – vO)2 + vE2, or v = [vE2 + (vS – vO)2]1/2 = [(11.2 km/s)2 + (42.2 km/s – 29.8 km/s)2]1/2 = 16.7 km/s. (b) The required kinetic energy is K = ½mv2, so we have K/m = ½v2 = ½(1.67104 m/s)2 = 1.40108 J/kg. 54. The amount of work required is the increase in potential energy: W = mg y. We find the time from P = W/t = mg y/t; 25.5 s. 1750 W = (285 kg)(9.80 m/s2)(16.0 m)/t, which gives t = 55. We find the equivalent force exerted by the engine from P = Fv; (18 hp)(746 W/hp) = F(90 km/h)/(3.6 ks/h), which gives F = 5.4102 N. At constant speed, this force is balanced by the average retarding force, which must be

5.4102 N.

56. (a) 1 hp = (550 ft·lb/s)(4.45 N/lb)/(3.281 ft/m) = 746 W. (b) P = (100 W)/(746 W/hp) = 0.134 hp. 57. (a) The original kinetic energy is K = ½mv2 = ½(80 kg)(5.0 m/s)2 = 1.0103 J. (b) Work equal to the change in kinetic energy must be done. Thus the average power is Pav = ?K/t = (1.0103 J)/(1.0 s) = 1.0103 W. 58. On the level the normal force is FN = mg, so the friction force is Ffr = kmg. To keep the box moving at constant speed, the applied force must balance the friction force. Thus the minimum horsepower required is P = Fv = Ffrv = kmgv = (0.45)(300 kg)(9.80 m/s2)(1.20 m/s)/(746 W/hp) = 2.1 hp. 59. We find the average resistance force from the acceleration: FR = ma = m ?v/?t = (1000 kg)(70 km/h – 90 km/h)/(3.6 ks/h)(6.0 s) = – 926 N. If we assume that this is the resistance force at 80 km/h, the engine must provide an equal and opposite force to maintain a constant speed. We find the power required from P = Fv = (926 N)(80 km/h)/(3.6 ks/h) = 2.1104 W = (2.1104 W)/(746 W/hp) = 28 hp. 60. We find the work from W = Pt = (3.0 hp)(746 W/hp)(1 h)(3600 s/h) = Page 14

8.1106 J.

Chapter 8

61. The work done by the shot-putter increases the kinetic energy of the shot. We find the power from P = W/t = ?K/t = (½mvf2 – ½mvi2)/t = ½(7.3 kg)[(14 m/s)2 – 0]/(1.5 s) =

4.8102 W

(about 0.6 hp).

62. The work done by the pump increases the potential energy of the water. We find the power from P = W/t = ?U/t = mg(hf – hi)/t = (m/t)g(hf – hi) = [(18.0 kg/min)/(60 s/min)](9.80 m/s2)(3.50 m – 0) = 10.3 W. 63. The work done increases the potential energy of the player. We find the power from P = W/t = ?U/t = mg(hf – hi)/t = (105 kg)(9.80 m/s2)[(140 m) sin 30° – 0]/(61 s) = 1.2103 W (about 1.6 hp). 64. From the force diagram for the car, at constant speed we have: x-component: F – Ffr = mg sin . Because the power output is P = Fv, we have (P/v) – Ffr = mg sin . The maximum power determines the maximum angle: (Pmax/v) – Ffr = mg sin max ; (120 hp)(746 W/hp)/[(70 km/h)/(3.6 ks/h)] – 600 N = (1000 kg)(9.80 m/s2) sin max , which gives sin max = 0.409, or max = 24°.

y

FN

x

F

f fr



v mg

65. The work done by the lifts increases the potential energy of the people. We assume an average mass of 70 kg and find the power from P = W/t = ?U/t = mg(hf – hi)/t = (m/t)g(hf – hi) = [(47,000 people/h)(70 kg/person)/(3600 s/h)](9.80 m/s2)(200 m – 0) (about 2.4103 hp). = 1.8106 W 66. For the work-energy principle applied to coasting down the hill a distance L, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); – FfrL = (½mv2 – ½mv2) + mg(0 – L sin ), which gives Ffr = mg sin . Because the climb is at the same speed, we assume the resisting force is the same. For the work-energy principle applied to climbing the hill a distance L, we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); FL – FfrL = (½mv2 – ½mv2) + mg(L sin  – 0); (P/v) – mg sin  = mg sin , which gives P = 2mgv sin  = 2(75 kg)(9.80 m/s2)(5.0 m/s) sin 7.0° =

9.0102 W

(about 1.2 hp).

67. Because the rate of work is P = Fv and the applied force produces the acceleration, we find the velocity and acceleration as a function of time: x = (5.0 m/s3)t3 – (8.0 m/s2)t2 – (30 m/s)t; v = dx/dt = (15.0 m/s3)t2 – (16.0 m/s2)t – (30 m/s); a = dv/dt = (30.0 m/s3)t – (16.0 m/s2). Thus the rate of work is P = Fv = mav = m[(30.0 m/s3)t – (16.0 m/s2)][(15.0 m/s3)t2 – (16.0 m/s2)t – (30 m/s)]. (a) At t = 2.0 s, we have P = (0.280 kg)[(30.0 m/s3)(2.0 s) – (16.0 m/s2)][(15.0 m/s3)(2.0 s)2 – (16.0 m/s2)(2.0 s) – (30 m/s)] = – 25 W. Note that the negative sign means there are times when the applied force is opposite to the motion. (b) At t = 4.0 s, we have Page 15

Chapter 8

= (0.280 kg)[(30.0 m/s3)(4.0 s) – (16.0 m/s2)][(15.0 m/s3)(4.0 s)2 – (16.0 m/s2)(4.0 s) – (30 m/s)] = + 4.3103 W. (c) Over a time interval, the average net power produces the change in kinetic energy: P = W/?t = ?K/?t = (½mvf2 – ½mvi2)/?t = ½m(vf2 – vi2)/?t. We find the velocities at the three times: v0 = (15.0 m/s3)(0)2 – (16.0 m/s2)(0) – (30 m/s) = – 30 m/s; v2 = (15.0 m/s3)(2.0 s)2 – (16.0 m/s2)(2.0 s) – (30 m/s) = – 2.0 m/s; v4 = (15.0 m/s3)(4.0 s)2 – (16.0 m/s2)(4.0 s) – (30 m/s) = 146 m/s. From t = 0 to t = 2.0 s, we have P = ½(0.280 kg)[(– 2.0 m/s)2 – (– 30 m/s)2]/(2.0 s – 0) = – 63 W. From t = 2.0 s to t = 4.0 s, we have P = ½(0.280 kg)[(146 m/s)2 – (– 2.0 m/s)2]/(4.0 s – 2.0 s) = + 1.5103 W. P

68. The potential energy of the mass is ½kx2. At the release point the kinetic energy is zero, so the total energy is E = ½kx02. The mass will move toward x = 0. From the diagram, we see that its kinetic energy will be the difference between E and U and will be maximum at x = 0. As the mass moves past x = 0, its kinetic energy will decrease until the mass reaches x = – x0 , where the motion will reverse.

U(x) E K 0

x0

69. (a) Because the mass has no initial kinetic energy, the total energy is the initial potential energy: E = U0 = ½kx02 = ½(160 N/m)(1.0 m)2 = 80 J. (b) Because energy is conserved, we have E = K + U = K + ½kx2; 60 J. 80 J = K + ½(160 N/m)[½(1.0 m)]2, which gives K = (c) The maximum kinetic energy occurs when the potential energy is zero at the equilibrium position: E = Kmax = 80 J. (d) We find the maximum speed from E = Kmax = ½mvmax2; 5.7 m/s at x = 0. 80 J = ½(5.0 kg)vmax2, which gives vmax = (e) The magnitude of the acceleration is maximum at the endpoints of the motion: Fmax = kx0 = mamax ; (160 N/m)(1.0 m) = (5.0 kg)amax , which gives amax = 32 m/s2 at x = ± x0.

Page 16

x

Chapter 8

70. (a) To find where U is a minimum, we set the (c) first derivative equal to zero: U(r) = (– a/r6) + (b/r12); dU/dr = (+ 6a/r7) – (12b/r13) = 0, rmin = (2b/a)1/6. which gives That this is a minimum can be seen from the plot. It can be confirmed by finding the sign of the second derivative at this value of r: = (– 42a/r8) + (156b/r14) d2U/dr2 = (6/r2)[(– 7a/r6) + (36b/r12)]. At rmin the value of the bracket is [(– 7a/rmin6) + (36b/rmin12)] = [(– 7a2/2b) + (18a2/2b)] > 0, so U(rmin) is a minimum.

(b) (d)

(e)

(f)

U(r)

(b/ a) 1/6

r

(2b/ a) 1/6 The value of U goes to 8 as r goes to zero, so the maximum is at rmax = 0. When we set U(r) = 0, we get U(r) = (– a/r6) + (b/r12) = 0, which gives r = (b/a)1/6, and 8. The kinetic energy of the system is K = E – U, and it must be positive, that is, E > U. From the plot we see that, if E < 0, the system is bound and the atom must oscillate between the turning points where E = U. From the plot we see that, if E > 0, the system is unbound and the atoms can reach an infinite separation. From the plot we see that F > 0 (away from the other atom) when the slope is negative: F > 0, 0 < r < (2b/a)1/6. From the plot we see that F < 0 (toward the other atom) when the slope is positive: F < 0, (2b/a)1/6 < r < 8. From the plot we see that F = 0 when the slope is zero: F = 0, r = (2b/a)1/6. We find F from F(r) = – dU/dr = (– 6a/r7) + (12b/r13).

71. For the binding energy, we have Ebinding = U(8) – U(rmin) = 0 – [(– a/rmin6) + (b/rmin12)] = [a/(2b/a)] – [b/(2b/a)2] =

a2/4b.

72. We choose the potential energy to be zero at the ground (y = 0). Energy is conserved, so we have E = Ki + Ui = Kf + Uf ;

½mvi2 + mgyi = ½mvf2 + mgyf ; ½m(180 m/s)2 + m(9.80 m/s2)(165 m) = ½mvf2 + m(9.80 m/s2)(0), which gives vf =

189 m/s.

Note that we have not found the direction of the velocity.

73. We choose the potential energy to be zero at the initial level of the center of mass (y = 0). We find the minimum speed by ignoring any frictional forces. Energy is conserved, so we have E = Ki + Ui = Kf + Uf ;

½mvi2 + mgyi = ½mvf2 + mgyf ; ½mvi2 + m(9.80 m/s2)(0) = ½m(6.5 m/s)2 + m(9.80 m/s2)(1.1 m), which gives vi =

8.0 m/s. Note that the initial velocity will not be horizontal, but will have a horizontal component of 6.5 m/s. 74. The work done increases the potential energy of the cyclist. We find the speed from P = W/t = ?U/t = mg(hf – hi)/t = mg(L sin – 0)/t = mgv sin  (0.20 hp)(746 W/hp) = (85 kg)(9.80 m/s2)v sin 12°, which gives v = 0.86 m/s.

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Chapter 8

75. The work done increases the potential energy of the elevator. We find the power output from P = W/t = ?U/t = mg(hf – hi)/t = (850 kg)(9.80 m/s2)(32.0 m)/(11.0 s)(746 W/hp) = 32.5 hp. 76. We choose the reference level for the gravitational potential energy at the ground. (a) With no air resistance during the fall we have 0 = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi), or

½(vf2 – 0) = – (9.80 m/s2)(0 – 18 m), which gives vf =

19 m/s.

(b) With air resistance during the fall we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi);

Fair(18 m) = ½(0.20 kg)[(10.0 m/s)2 – 0] + (0.20 kg)(9.80 m/s2)(0 – 18 m), which gives Fair = – 1.4 N. 77. We choose the reference level for the gravitational potential energy indicated on the diagram. (a) Because there is no friction, we apply conservation of energy from the top A to the point where the skier leaves the ramp B: E = KA + UA = KB + UB ;

A vB

H

0 + mgH = ½mvB2 + mgh;

y=0

h B

(9.80 m/s2)(45 m) = ½vB2 + (9.80 m/s2)(4.4 m), which gives vB = 28 m/s. (b) For the motion from B to C, the horizontal velocity is constant, so we have x = s cos  = vBt, or t = (s cos )/vB . For the vertical motion, we have y = yB + 0 + ½(– g)t2;

s

C 

– s sin  = h – ½g[(s cos )/vB ]2; – s sin 30° = 4.4 m – ½(9.80 m/s2)[(s cos 30°)/(28 m/s)]2. This reduces to a quadratic equation for s: s2 – (108 m)s – 952 m2 = 0, which has a positive solution of s =

1.2102 m.

78. Because the jump occurs at the end of the ramp, the horizontal velocity at B will be the same as before. Thus we still have t = (s cos )/vB . For the vertical motion, we have y = yB + vyt + ½(– g)t2; – s sin  = h + (3.0 m/s)[(s cos )/vB] – ½g[(s cos )/vB ]2; – s sin 30° = 4.4 m + (3.0 m/s)[(s cos 30°)/(28 m/s)] – ½(9.80 m/s2)[(s cos 30°)/(28 m/s)]2. This reduces to a quadratic equation for s: s2 – (128 m)s – 952 m2 = 0, which has a positive solution of s = 1.4102 m.

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Chapter 8

79. We choose the reference level for the gravitational potential energy at the lowest point. The tension in the cord is always perpendicular to the displacement and thus does no work. (a) With no air resistance during the fall, we have 0 = ?K + ?U = (½mv12 – ½mv02) + mg(h1 – h0), or

y L h

½(v12 – 0) = – g(0 – L), which gives v1 =

(2gL)1/2. (b) For the motion from release to the rise around the peg, we have 0 = ?K + ?U = (½mv22 – ½mv02) + mg(h2 – h0), or

FT

peg

y=0

½(v22 – 0) = – g[2(L – h) – L] = g(2h – L) = 0.60gL, which gives v2 =

mg

(1.2gL)1/2.

80. From Problem 79, at the top of the circular motion about the peg, we have v22 = 2g(2h – L). The tension in the cord and the weight provide the radial acceleration: FT + mg = mv22/r = m(2g)(2h – L)/(L – h). Because the cord cannot push, for the ball to make a complete circle, FT = 0, or m(2g)(2h – L)/(L – h) = mg; 4h – 2L = L – h, or h = 0.60L. 81. (a) The work done against gravity is the increase in the potential energy: W = ?U = mg(hf – hi) = (65 kg)(9.80 m/s2)(3900 m – 2200 m) = 1.1106 J. (b) We find the power from P = W/t = (1.1106 J)/(5.0 h)(3600 s/h) = 60 W = 0.081 hp. (c) We find the power input from Pinput = P/efficiency = (60 W)/(0.15) = 4.0102 W = 0.54 hp. 82. (a) With y = 0 at the bottom of the circle, we call the start point A, the bottom of the circle B, and the top of the circle C. At the top of the circle we have the forces mg and FNtop , both downward, that provide the centripetal acceleration: mg + FNtop = mvC2/r. The minimum value of FNtop is zero, so the minimum speed at C is found from vCmin2 = gr. From energy conservation for the motion from A to C we have KA + UA = KC + UC ;

A FNtop C h

0 + mgh = ½mvC2 + mg(2r), thus the minimum height is found from ghmin = ½vCmin2 + 2gr = ½gr + 2gr, which gives hmin = 2.5r. (b) From energy conservation for the motion from A to B we have KA + UA = KB + UB ; 0 + mg2h = 5mgr = ½mvB2 + 0, which gives vB2 = 10gr. At the bottom of the circle we have the forces mg down and FNbottom up that provide the centripetal acceleration: – mg + FNbottom = mvB2/r. If we use the previous result, we get FNbottom = mvB2/r + mg = 11mg. (c) From energy conservation for the motion from A to C we have KA + UA = KC + UC ; Page 19

mg

r FNbottom

mg

B

Chapter 8

0 + mg2h = 5mgr = ½mvC2 + mg(2r), which gives vC2 = 6gr. At the top of the circle we have the forces mg and FNtop , both down, that provide the centripetal acceleration: mg + FNtop = mvC2/r. If we use the previous result, we get FNtop = mvC2/r – mg = 5mg. (d) On the horizontal section we have FN = mg.

83. We choose the reference level for the gravitational potential energy at the lowest point. (a) With no air resistance during the fall, we have 0 = ?K + ?U = (½mv2 – ½mv02) + mg(h – h0), or

½(v2 – 0) = – g(0 – H), which gives

v = (2gH)1/2 = [2(9.80 m/s2)(80 m)] = 40 m/s. (b) If 60% of the kinetic energy of the water is transferred, we have P = (0.60)½mv2/t = (0.60)½(m/t)v2 = (0.60)½(550 kg/s)(40 m/s)2 = 2.6105 W (about 350 hp). 84. We convert the speeds: (10 km/h)/(3.6 ks/h) = 2.78 m/s; (30 km/h)/(3.6 ks/h) = 8.33 m/s. We use the work-energy principle applied to coasting down the hill a distance L to find b: WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); – bv12L = (½mv12 – ½mv12) + mg(0 – L sin ), which gives b = (mg/ v12) sin  = [(75 kg)(9.80 m/s2)/(2.78 m/s)2] sin 4.0° = 6.63 kg/m. For the work-energy principle applied to speeding down the hill a distance L, the cyclist must provide a force so we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); F2L – bv22L = (½mv22 – ½mv22) + mg(0 – L sin ), which gives F2 = bv22 – mg sin . The power supplied by the cyclist is P = F2v2 = [(6.63 kg/m)(8.33 m/s)2 – (75 kg)(9.80 m/s2) sin 4.0°](8.33 m/s) = 3.41103 W. For the work-energy principle applied to climbing the hill a distance L, the cyclist will provide a force F3 = P/v3 , so we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); (P/v3)L – bv32L = (½mv32 – ½mv32) + mg(L sin  – 0), which gives [(3.41103 W)/v3] – (6.63 kg/m)v32 = (75 kg)(9.80 m/s2) sin 4.0°. This is a cubic equation for v3 , which has one real solution: v3 = 7.69 m/s. The speed is (7.69 m/s)(3.6 ks/h) = 28 km/h.

Page 20

Chapter 8

85. We choose the reference level for the gravitational potential energy at the bottom. From energy conservation for the motion from top to bottom, we have Ktop + Utop = Kbottom + Ubottom ;

½mvtop2 + mg2R = ½mvbottom2 + 0, which gives

v top mg

FNtop vbottom2 = vtop2 + 4gR. At the bottom of the circle we have the forces mg down and FNbottom up that provide the centripetal acceleration: R – mg + FNbottom = mvbottom2/R, which gives FNbottom = (mvbottom2/R) + mg. At the top of the circle we have the forces mg and FNtop , both down, that provide the centripetal acceleration: FNbottom mg + FNtop = mvtop2/R, which gives 2 FNtop = (mvtop /R) – mg. If we subtract the two equations, we get mg FNbottom – FNtop = (mvbottom2/R) + mg – [(mvtop2/R) – mg] v bottom = (m/R)(vbottom2 – vtop2) + 2mg = 4mg + 2mg = 6mg. The speed must be above the minimum at the top so the roller coaster does not leave the track. From Problem 82, we know that we must have h > 2.5R. The result we found does not depend on the radius or speed.

86. We choose y = 0 at the scale. We find the spring constant from the force (your weight) required to compress the spring: k = F1/x1 = ( – 700 N)/(– 0.5010–3 m) = 1.4106 N/m. We apply conservation of energy for the jump to the scale. Both the initial and final kinetic energy are zero. If we ignore the small change in gravitational potential energy when the scale compresses, we have Ki + Ui = Kf + Uf ; 0 + mgH = 0 + ½kx22; (700 N)(1.0 m) = ½(1.4106 N/m)x22, which gives x2 = 0.032 m. The reading of the scale is F2 = kx2 = (1.40106 N/m)(0.032 m) = 4.4104 N. 87. We choose the potential energy to be zero at the lowest point (y = 0). (a) Because the tension in the rope does no work, energy is conserved, so we have Ki + Ui = Kf + Uf ;

½mv02 + 0 = 0 + mgh = mg(L – L cos ) = mgL(1 – cos ); ½m(5.0 m/s)2 = m(9.80 m/s2)(10.0 m)(1 – cos )



L FT

29°. which gives cos  = 0.872, or  = h y=0 (b) The velocity is zero just before he releases, so there is no centripetal acceleration. There is a tangential acceleration which has been mg v0 decreasing his tangential velocity. For the radial direction we have FT – mg cos  = 0; or FT = mg cos  = (75 kg)(9.80 m/s2)(0.872) = 6.4102 N. (c) The velocity and thus the centripetal acceleration is maximum at the bottom, so the tension will be maximum there. For the radial direction we have FT – mg = mv02/L, or Page 21

Chapter 8

FT = mg + mv02/L = (75 kg)[(9.80 m/s2) + (5.0 m/s)2/(10.0 m)] =

9.2102 N.

88. We choose the potential energy to be zero at the floor. The work done increases the potential energy of the athlete. We find the power from P = W/t = ?U/t = mg(hf – hi)/t = (70 kg)(9.80 m/s2)(5.0 m – 0)/(9.0 s) = 3.8102 W (about 0.5 hp).

r 89. (a) For the Yukawa potential U r = – U 0 r0 e – r/ r0 , we find F(r) from r r F r = – d U = – U 0 02 e – r/ r 0 + U 0 0 – r1 e – r/ r0 = dr r 0 r (b) For the ratio of the two forces, we have F 3r 0 r e– 3 = 0 (1/ 3) + 1 – 1 = 2 e – 2 = 0.030. 1+1 3r0 9 F r0 e

–

U 0 r0 + 1 e – r/ r0. r r

(c) From the potential of the charged particles U r = – C r , we find F(r) from F r = – d U = – C2 . dr r For the ratio of the two forces, we have 2 F 3r 0 r = 0 2=1= 0.11. 9 F r0 3r0 We see that the force decreases more rapidly for the Yukawa potential. The exponential factor makes the Yukawa force drop off more rapidly with distance, so it is a “short-range” force.

90. We choose the potential energy to be zero at the bottom of the incline (y = 0). (a) Because there is no acceleration perpendicular to the incline, we have FN = mg cos , so the friction force is Ffr = kmg cos . For the motion from the bottom of the incline to where the sled stops a distance s up the incline, for the work-energy principle we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); – kmg cos  s = (0 – ½mv12) + mg(s sin );

s FN Ffr mg y=0



0.41 m. – (0.25)(9.80 m/s2)s cos 30° = – ½(2.4 m/s)2 + (9.80 m/s2)(s sin 30°), which gives s = (b) When the sled stops, the friction becomes static. For the sled to not get stuck, we have mg sin  > smg cos , or s < tan 30° = 0.58. (c) On the way down the friction force changes direction to always oppose the motion. For the motion from the bottom of the incline up and then back down to the bottom, for the work-energy principle we have WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi); – kmg cos  (2s) = (½mv22 – ½mv12) + 0; – (0.25)(9.80 m/s2) cos 30° [2(0.41 m)] = ½v22 – ½(2.4 m/s)2 , which gives v2 =

1.5 m/s.

91. If we consider a small length L of the circular stream with radius R that leaves the hose, the mass of this much water is m = pR2L. If the water leaves the hose with speed v, the time to leave the hose is t = L/v. If we neglect air resistance and apply conservation of energy from leaving the hose to the highest point, we have K1 + U1 = K2 + U2 ; Page 22

Chapter 8

½mv2 + 0 = 0 + mgh, or v = (2gh)1/2 = [2(9.80 m/s2)(30 m)]1/2 = 24.2 m/s.

The minimal supplied power produces the kinetic energy of the water as it leaves the hose: P = W/t = ?K/t = ½mv2/t = ½pR2Lv2/t = ½pR2v3 = ½(1.00103 kg/m3)p(1.510–2 m)2(24.2 m/s)3/(746 W/hp) = 6.7 hp. 92. We can find the required non-conservative work that is required, and thus the thermal energy dissipated, by using the work-energy principle: WNC = ?K + ?U = (½mvf2 – ½mvi2) + mg(hf – hi) = m[(½vf2 – ½vi2) + g(hf – hi)] m) sin 20°}

= (1500 kg){½[(30 km/h)/(3.6 ks/h)]2 – ½[(90 km/h)/(3.6 ks/h)]2 + (9.80 m/s2)(– 0.30103

= – 1.9106 J. Thus the thermal energy dissipated is

1.9106 J.

93. We choose the surface of the moon as our reference level. We can apply conservation of energy from the moment of engine shut off to the landing: K1 + U1 = K2 + U2 ;

½mv12 + mgh1 = ½mv22 + mgh2 ; ½mv12 + mgh1 = ½mv22 + 0, or h1max = ½(v2max2 – v12)/g.

(a) If the initial velocity is zero, we have hamax = ½(v2max2 – v12)/g = ½[(3.0 m/s)2 – 0]/(1.62 m/s2) = 2.8 m. (b) If the initial velocity is downward, we have hbmax = ½(v2max2 – v12)/g = ½[(3.0 m/s)2 – (– 2.0 m/s)2]/(1.62 m/s2) = 1.5 m. (c) If the initial velocity is upward, we have hcmax = ½(v2max2 – v12)/g = ½[(3.0 m/s)2 – (2.0 m/s)2]/(1.62 m/s2) = 1.5 m. Note that this must be the same as in part (b), because when the lander goes up and returns to the initial elevation, it will have the same speed.

94. The output comes from the decrease of the potential energy of the water. Thus we have P = ?U/t = mg(hf – hi)/t = Vg(hf – hi)/t; 100106 W = (1.00103 kg/m3)V(9.80 m/s2)(500 m)/(3600 s), which gives V = 7.3104 m3. This assumes 100% efficiency. 95. If 80% of the electrical power is used to increase the potential energy of the water, we have 0.80P = mg(hf – hi)/t; 0.80P = (1.00106 kg/day)(9.80 m/s2)(400 m)/(24 h/day)(3600 s/h), 76 hp. which gives P = 5.7104 W = 96. (a) We find the energy required to place the satellite into orbit by finding the change in total energy of the satellite. Before the satellite is launched, it has kinetic energy because it has the surface speed of a point on the equator: v1 = 2prE/t = 2p(6.38106 m)/(24 h)(3600 s/h) = 464 m/s; K1 = ½mv12 = ½(12,000 kg)(464 m/s)2 = 1.29109 J. With the reference level at infinity, the initial potential energy is U1 = – GmME/rE = – (6.6710–11 N · m2/kg2)(12,000 kg)(5.981024 kg)/(6.38106 m) = – 7.501011 J. Thus the initial total energy is E1 = K1 + U1 = 1.29109 J – 7.501011 J = – 7.491011 J. We find the speed of the satellite in orbit from the required radial acceleration: GmME/r22 = mv22/r2 , which gives v22 = GME/r2. Page 23

Chapter 8

The total energy in orbit is E2 = K2 + U2 = ½mv22 – GmME/r2 = ½GmME/r2 – GmME/r2 = – ½GmME/r2 = – ½(6.6710–11 N · m2/kg2)(12,000 kg)(5.981024 kg)/(6.38106 m + 1.00106 m) = – 3.241011 J. Thus the energy required is W = E2 – E1 = – 3.241011 J – (– 7.491011 J) = 4.251011 J. (b) If the satellite is launched from the north pole, its initial kinetic energy is zero, so we have W = E2 – U1 = – 3.241011 J – (– 7.501011 J) = 4.261011 J. 97. (a) At point B the distance of the satellite from the center of the Earth is rB = [(½a)2 + b2]1/2 = [(8,000 km)2 + (13,900 km)2]1/2 = 16,000 km. For energy conservation for the motion from A to B, we have KA + UA = KB + UB ;

B

b

A

C

½mvA2 – GMEm/rA = ½mvB2 – GMEm/rB , or vB2 = vA2 + 2GME(1/rB – 1/rA); vB2 = (8650 m/s)2 + 2(6.6710–11 N · m2/kg2)  (5.981024 kg)[(1/16.0106 m) – (1/8.0106 m)] 5.00103 m/s. which gives vB = (b) For energy conservation for the motion from A to C, we have KA + UA = KC + UC ;

a

a/2

a/2

½mvA2 – GMEm/rA = ½mvC2 – GMEm/rC , or vC2 = vA2 + 2GME[1/(*a) – 1/(½a)]; vC2 = (8650 m/s)2 + 2(6.6710–11 N · m2/kg2)(5.981024 kg)[(1/24.0106 m) – (1/8.0106 m)] which gives vC = 2.89103 m/s.

98. (a)

U(r)/ U 0 0.3 0.2 0.1 0

r 4

8

12

16

– 0.1 – 0.2 When we set U(r) = 0, we get U(r) = U0[(2/r2) – (1/r)] = 0, which gives r = To find where U is a minimum, we set the first derivative equal to zero: Page 24

2.

20

Chapter 8

U(r) = U0[(2/r2) – (1/r)]; dU/dr = U0[(– 4/r3) + (1/r2) = 0, which gives rmin = 4. (b) The minimum potential energy at rmin is U(rmin) = U0[(2/rmin2) – (1/rmin)] = U0[(2/42) – (1/4)] = – U0/8. When we plot the constant energy of – 0.050U0 , we see that the turning points are ˜ 2.5 and 18. The kinetic energy of the system is K = E – U, so the maximum K occurs at minimum U: Kmax = E – Umin = – 0.050U0 – (– 0.125U0) = 0.075U0 at r = 4.

Page 25

Chapter 9

CHAPTER 9 - Linear Momentum and Collisions 1.

We find the force on the expelled gases from F = ?p/?t = (?m/?t)v = (1200 kg/s)(50,000 m/s) = 6.0107 N. An equal, but opposite, force will be exerted on the rocket: 6.0107 N, up.

2.

For the momentum p = 4.8t2i – 8.0j – 8.9tk, we find the force from F = dp/dt = 9.6ti – 8.9k, in SI units.

3.

(a) p = mv = (0.030 kg)(12 m/s) = 0.36 kg · m/s. (b) The force, opposite the direction of the velocity, changes the momentum: F = ?p/?t; 0.12 kg · m/s. – 2.010–2 N = (p2 – 0.36 kg · m/s)/(12 s), which gives p2 =

4.

The change in momentum is ?p = p2 – p1 = mvj – mvi = (0.145 kg)(30 m/s)j – (0.145 kg)(30 m/s)i =

5.

The force changes the momentum: F = (26 N)i – (12 N/s2)t2j = dp/dt. Because F is a variable force, we integrate to find the momentum change:

dp = p =

F dt; 2.0 s 1.0 s

2

(26 N)i – (12 N/ s 2)t j d t 2

3

= (26 N)t i – (4.0 N/ s )t j 6.

– (4.35 kg · m/s)i + (4.35 kg · m/s)j.

2.0 s 1.0 s

=

(26 N s)i – (28 N s)j.

If M is the initial mass of the rocket and m2 is the mass of the expelled gases, the final mass of the rocket is m1 = M – m2. Because the gas is expelled perpendicular to the rocket in the rocket’s frame, it will still have the initial forward velocity, so the velocity of the rocket in the original direction will not change. We find the y-component of the rocket’s velocity after firing from v1 = v0 tan  = (120 m/s) tan 23.0° = 50.9 m/s. Using the coordinate system shown, for momentum conservation in the y-direction we have 0 + 0 = m1v1 – m2v2 , or (M – m2)v1 = m2v2 ; (4200 kg – m2)(50.9 m/s) = m2(2200 m/s), which gives m2 =

Before



v0

y

gas x



Page 1

After

95 kg.

v2

v1

Chapter 9

7.

(a) We choose downward as positive. For the fall we have y = y0 + v0t1 + !at12; h = 0 + 0 + !gt12, which gives t1 = (2h/g)1/2. To reach the same height on the rebound, the upward motion must be a reversal of the downward motion. Thus the time to rise will be the same, so the total time is ttotal = 2t1 = 2(2h/g)1/2 = (8h/g)1/2. (b) We find the speed from v = v0 + at1 = 0 + g(2h/g)1/2 = (2gh)1/2. (c) To reach the same height on the rebound, the upward speed at the floor must be the same as the speed striking the floor. Thus the change in momentum is ?p = m(– v) – mv = – 2m(2gh)1/2 = = – (8m2gh)1/2 (up). (d) For the average force on the ball we have F = ?p/?t = – (8m2gh)1/2/(8h/g)1/2 = – mg (up). Thus the average force on the floor is mg (down), a surprising result.

8.

We convert the speed: (100 km/h)/(3.6 ks/h) = 27.8 m/s. In a time ?t all of the air within a distance v ?t will strike the building; so the mass being brought to rest is ?m = Av ?t. Thus the average force on the air is F = ?p/?t = (?m/?t) ?v = Av(0 – v) = – Av2 = – (1.3 kg/m3)(40 m)(60 m)(27.8 m/s)2 = – 2.4106 N. The average force on the building is the reaction to this: 2.4106 N.

9.

For the one-dimensional motion, we take the direction of the first car for the positive direction. For this perfectly inelastic collision, we use momentum conservation: M1v1 + M2v2 = (M1 + M2)V; (9700 kg)(18 m/s) + 0 = (9700 kg + M2)(4.0 m/s), which gives M2 = 3.4104 kg.

10. For this one-dimensional motion, we take the direction of the halfback for the positive direction. For this perfectly inelastic collision, we use momentum conservation: M1v1 + M2v2 = (M1 + M2)V; (90 kg)(5.0 m/s) + (130 kg)(– 2.5 m/s) = (90 kg + 130 kg)V, which gives V= 0.57 m/s (direction of halfback). 11. The new nucleus and the alpha particle will recoil in opposite directions. Momentum conservation gives us 0 = MV – mv , 0 = (57m)V – m(2.5105 m/s), which gives V= 4.4103 m/s. 

12. For the horizontal motion, we take the direction of the car for the positive direction. The load initially has no horizontal velocity. For this perfectly inelastic collision, we use momentum conservation: M1v1 + M2v2 = (M1 + M2)V; (10,500 kg)(15.0 m/s) + 0 = (10,500 kg + 6350 kg)V, which gives V= 9.35 m/s. 13. During the throwing we use momentum conservation for the one-dimensional motion: 0 = (mboat + mchild)vboat + mpackagevpackage ; 0 = (55.0 kg + 26.0 kg)vboat + (5.40 kg)(10.0 m/s), which gives vboat = – 0.667 m/s (opposite to the direction of the package). 14. Momentum conservation gives us mv1 + Mv2 = mv1 + Mv2, (0.012 kg)(190 m/s) + 0 = (0.012 kg)(150 m/s) + (2.0 kg)v2, which gives v2= 0.24 m/s. Page 2

v1 m

v2 = 0 M

v 2 M

v 1 m

Chapter 9

15. Momentum conservation gives 0 = m1v1 + m2v2, or v2/v1 = – m1/m2 . The ratio of kinetic energies is K2/K1 = !m2v22/!m1v12 = (m2/m1)(v2/v1)2 = 2. When we use the result from momentum, we get (m2/m1)(– m1/m2)2 = 2, which gives m1/m2 = 2. The fragment with the lesser kinetic energy has the greater mass. 16. For the collision we use momentum conservation: x-direction: m1v1 + 0 = m1v1 cos 1 + m2v2 cos 2 ; m(17 m/s) = mv1 cos 45° + mv2 cos 30°; y-direction: 0 + 0 = – m1v1 sin 1 + m2v2 sin 2 ; 0 = – mv1 sin 45° + mv2 sin 30°. The mass cancels, so we solve these two equations for the two unknowns: v1 = 8.8 m/s; v2 = 12.4 m/s.

After v 2

Before

2 v1

y

1 x

v 1

17. For momentum conservation we have mv0i = %m v + @m(2v0)j, which gives v = *v0 i – v0 j. The rocket’s forward speed increases because the fuel is shot backward relative to the rocket. 18. For momentum conservation of the decay of a stationary neutron into the three particles, we have 0 = m1v1 + m2v2 + m3v3 , or v3 = – [(m1/m3)v1 + (m2/m3)v2]. Because (m1/m3)v1 + (m2/m3)v2 defines a plane containing v1 and v2, the third velocity must lie in the same plane. 19. Because the initial momentum is zero, the momenta of the three products of the decay must add to zero. If we draw the vector pneutrino  diagram, we see that pnucleus = (pelectron2 + pneutrino2)1/2 = [(8.610–23 kg · m/s)2 + (6.210–23 kg · m/s)2]1/2 = 1.110–22 kg · m/s. pelectron We find the angle from tan  = pneutrino/pelectron = (6.210–23 kg · m/s)/(8.610–23 kg · m/s) = 0.721, so the angle is 36° from the direction opposite to the electron’s.

pnucleus



20. (a) With respect to the Earth after the explosion, one section will have a speed v1 and the other will have a speed v2 = v1 + vrelative. Momentum conservation gives us mv = !mv1 + !mv2, or v = !v1 + !(v1 + vrelative) = v1 + !vrelative; 5.10103 m/s. 6.50103 m/s = v1 + !(2.80103 m/s), which gives v1 = The other section will have v2 = v1 + vrelative = 5.10103 m/s + 2.80103 m/s = 7.90103 m/s. (b) The energy supplied by the explosion increases the kinetic energy: E = ?K = [!(!m)v12 + !(!m)v22] – !mv2 = ![!(900 kg)(5.10103 m/s)2] + ![!(900 kg)(7.90103 m/s)2] – !(900 kg)(6.50103 m/s)2 = 8.82108 J. Page 3

Chapter 9

21. (a) For the initial projectile motion, the horizontal velocity is constant, so the velocity at the highest point, before the breakup, is v = vhi = v0 cos 60° i = (100 m/s) cos 60° i = (50 m/s)i. During the explosion, momentum is conserved: mv = m1v1 + m2v2 + m3v3 ; mvh i = – @mvh j + @mvh i + @mv3 ; which gives v3 = 2vh i + vh j = 2(50 m/s)i + (50 m/s)j = (100 m/s)i + (50 m/s)j . (b) The energy supplied by the explosion increases the kinetic energy: E = ?K = !(@m)vh2 + !(@m)vh2 + !(@m)[(2vh)2 + vh2)] – !mvh2 = %mvh2 = %(200 kg)(50 m/s)2 =

3.3105 J.

22. We find the average force on the ball from F = ?p/?t = m ?v/?t =(0.145 kg)[(56.0 m/s) – (– 35.0 m/s)]/(5.0010–3 s) =

2.64103 N.

23. We find the average force on the ball from F = ?p/?t = m ?v/?t = (0.0600 kg)[(65.0 m/s) – 0]/(0.0300 s) = 130 N. Because the weight of a 60-kg person is ˜ 600 N, this force is not large enough. 24. The momentum parallel to the wall does not change, therefore the impulse will be perpendicular to the wall. With the positive direction toward the wall, we find the impulse on the ball from Impulse = ?p = m ?v = m[(– v sin ) – (v sin )] = – 2mv sin  = – 2(0.060 kg)(28 m/s) sin 45° = – 2.4 N · s. The impulse on the wall is in the opposite direction: 2.4 N · s.

25. We find the average force on the water from F = ?p/?t = (?m/?t) ?v = (60 kg/s)(10 m/s)[(– 0.75) – 1] = – 1.1103 N. The average force on the turbine blades is the reaction to this:

v

v



1.1103 N.

26. In the reference frame of the capsule before the push, we take the positive direction in the direction the capsule will move. (a) Momentum conservation gives us mvastronaut + Mvsatellite = mvastronaut + Mvsatellite , 0 + 0 = (140 kg)(– 2.50 m/s) + (1800 kg) vsatellite , which gives vsatellite = 0.194 m/s. (b) We find the force on the satellite from Fsatellite = ?psatellite/?t = msatellite ?vsatellite/?t = (1800 kg)(0.194 m/s – 0)/(0.500 s) = 700 N. There will be an equal but opposite force on the astronaut. (c) The kinetic energies are Kastronaut = !m vastronaut2 = !(140 kg)(2.50 m/s)2 = 438 J. 



Ksatellite = !M vsatellite2 = !(1800 kg)(0.194 m/s)2 = 27. (a) We find the average force on the molecule from F = ?p/?t = m ?v/?t Page 4



33.9 J.

Chapter 9

= m[(– v) – (+ v)]/?t = – 2mv/?t. The average force on the wall is the reaction to this: 2mv/?t. (b) If t is the average time between collisions, the number of collisions in time T is N = T/t. Thus in the time T the total momentum change is N(2mv), so the average force on the wall is Fwall = N(2mv)/T = 2mv/t. 28. (a) The impulse is the area under the F vs. t curve. The value of each block on the graph is 1 block = (50 N)(0.01 s) = 0.50 N · s. We estimate there are 10 blocks under the curve, so the impulse is Impulse = (10 blocks)(0.50 N · s/block) ˜ 5.0 N · s. (b) We find the final velocity of the ball from Impulse = ?p = m ?v; 5.0 N · s = (0.060 kg)(v – 0), which gives v = 83 m/s.

29.

(a)

(b) The force would become zero at t = 780/(2.6105) = 3.010–3 s. The impulse is the area under the F vs. t curve, which is a triangle: J = !(780 N)(3.010–3 s) = 1.2 N · s. (c) We integrate the variable force to find the impulse:

J=

F dt =

3.0  10

–3

0

s

0

3.0

t (ms)

(780 N) – (2 .6  10 5 N/ s)t d t –3

= (780 N)t – (1.3  105 N/ s)t 2 –3

F (N) 780

3.0  10

s

0 5

–3

2

= (780 N)(3.0  10 s) – (1.3  10 N/ s) (3.0  10 s) = 1.2 N · s. (d) We find the mass of the bullet from Impulse = ?p = m ?v; 3.9 g. 1.17 N · s = m(300 m/s – 0), which gives m = 3.910–3 kg = 30. The maximum force that each leg can exert without breaking is (170106 N/m2)(2.510–4 m2) = 4.25104 N, so, if there is an even landing with both feet, the maximum force allowed on the body is 8.50104 N. We use the work-energy principle for the fall to find the landing speed: 0 = ?K + ?U; 0 = !mvland2 – 0 + (0 – mghmax), or vland2 = 2ghmax . The impulse from the maximum force changes the momentum on landing. If we take down as the positive direction and assume the landing lasts for a time t, we have – Fmaxt = m ?v = m(0 – vland), or t = mvland/Fmax . We have assumed a constant force, so the acceleration will be constant. For the landing we have y = vlandt + !at2 = vland(mvland/Fmax) + !(– Fmax/m)(mvland/Fmax)2 = !mvland2/Fmax = mghmax/Fmax ; 0.60 m = (75 kg)(9.80 m/s2)hmax/(8.50104 N), which gives hmax = 69 m.

Page 5

Chapter 9

31. We choose the upward direction as positive. (a) If we consider a mass ?m of water falling to the pan, we can find the speed just before hitting the pan from energy conservation: 0 = ?K + ?U; 0 = !(?m)v2 – 0 + [0 – (?m)gh], or v2 = 2gh. We find the average force required to stop the water from F = ?p/?t = (?m/?t) ?v = (?m/?t)[0 – (– v)] = (?m/?t)(2gh)1/2. The average force on the pan is the reaction to this: – (?m/?t)(2gh)1/2 (down). The scale reading is the increased normal force. After a time t, the water in the pan has a mass m = (?m/?t)t. If the acceleration of the water in the pan is negligible, we have FN – (?m/?t)(2gh)1/2 – mg = 0, or FN = (?m/?t)(2gh)1/2 + (?m/?t)gt = (?m/?t)[(2gh)1/2 + gt]

(0.84 N) + (1.2 N/s)t. = (0.12 kg/s){[2(9.80 m/s2)(2.5 m)]1/2 + (9.80 m/s2)t} = (b) After 15 s we have FN = (0.84 N) + (1.18 N/s)(15 s) = 18.5 N. (c) After a time t, the water in the pan has a mass m = (?m/?t)t. The height of this water in the cylinder will be h = m/A = (?m/?t)t/A = (0.12 kg/s)t/(1.0103 kg/m3)(2010–4 m2) = (6.010–2 m/s)t. The water falling into the cylinder at time t will have fallen a distance h – h. Thus the speed of the water will be given by v2 = 2g(h – h), and the impact force on the pan will be F = – (?m/?t)[2g(h – h)]1/2 (down). If the acceleration of the water in the pan is negligible, we have FN – (?m/?t)[2g(h – h)]1/2 – mg = 0, or FN = (?m/?t)[2g(h – h)]1/2 + (?m/?t)gt = (?m/?t){[2g(h – h)]1/2+ gt}

= (0.12 kg/s)({2(9.80 m/s2)[(2.5 m) – (6.010–2 m/s)t]}1/2 + (9.80 m/s2)t)

= (0.12 kg/s){[(49 m2/s2) – (1.18 m2/s3)t]1/2 + (9.80 m/s2)t}. After 15 s we have FN = (0.12 kg/s){(49 m2/s2) – (1.18 m2/s3)t]}1/2 + (9.80 m/s2)t}

= (0.12 kg/s){[(49 m2/s2) – (1.18 m2/s3)(15 s)]1/2 + (9.80 m/s2)(15 s)} =

18.3 N.

32. For the elastic collision of the two balls, we use momentum conservation for this one-dimensional motion: m1v1 + m2v2 = m1v1 + m2v2; (0.540 kg)(3.90 m/s) + (0.320 kg)(0) = (0.540 kg)v1 + (0.320 kg)v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 3.90 m/s – 0 = v2 – v1. Combining these two equations, we get v1 = 0.998 m/s, and v2 = 4.89 m/s. 33. For the elastic collision of the two pucks, we use momentum conservation for this one-dimensional motion: m1v1 + m2v2 = m1v1 + m2v2; (0.450 kg)(4.20 m/s) + (0.900 kg)(0) = (0.450 kg)v1 + (0.900 kg)v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 4.20 m/s – 0 = v2 – v1. Page 6

Chapter 9

Combining these two equations, we get v1 = – 1.40 m/s (rebound), and

v2 = 2.80 m/s.

34. For the elastic collision of the two balls, we use momentum conservation for this one-dimensional motion: m1v1 + m2v2 = m1v1 + m2v2; (0.060 kg)(7.50 m/s) + (0.090 kg)(3.00 m/s) = (0.060 kg)v1 + (0.090 kg)v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 7.50 m/s – 3.00 m/s = v2 – v1. Combining these two equations, we get v1 = 2.10 m/s, and v2 = 6.60 m/s. 35. (a) For the elastic collision of the two balls, we use momentum conservation for this one-dimensional motion: m1v1 + m2v2 = m1v1 + m2v2; (0.220 kg)(6.5 m/s) + m2(0) = (0.220 kg)(– 3.8 m/s) + m2v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 6.5 m/s – 0 = v2 – (– 3.8 m/s), which gives v2 = 2.7 m/s. (b) Using the result for v2 in the momentum equation, we get m2 = 0.84 kg. 36. (a) For the elastic collision of the two balls, we use momentum conservation: m1v1 + m2v2 = m1v1 + m2v2; (0.280 kg)v1 + m2(0) = (0.280 kg)v1 + m2(!v1). Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or v1 – 0 = !v1 – v1, which gives v1 = – !v1. Using this result in the momentum equation, we get m2 = 0.840 kg. (b) The fraction transferred is fraction = ?K2/K1 = !m2(v22 – v22)/!m1v12 = !m2[(!v1)2 – 0]/!m1v12 = #m2/m1 = #(0.840 kg)/(0.280 kg) =

0.750.

37. Because mass is conserved, the mass of the new nucleus is M2 = 222 u – 4.0 u = 218 u. Momentum conservation gives us M1V1 = M2V2 + mv , (222 u)(500 m/s) = (218 u)(450 m/s) + (4.0 u)v , which gives v= 3.2103 m/s. 



38. For the elastic collision of the two balls, we use momentum conservation: m1v1 + m2v2 = m1v1 + m2v2; mv1 + 0 = m(– v1/4) + m2v2, or m2v2 = 5mv1/4. Because the collision is elastic, the relative speed does not change: v1 – 0 = – (v1 – v2); v1 = v2 – (– v1/4), or v2 = 3v1/4. Combining these two equations, we get m2 = 5m/3.

Page 7

Chapter 9

39. For the elastic collision, we use momentum conservation: m1v1 + m2v2 = m1v1 + m2v2; m1v1 + 0 = m1v1 + m2v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or v1 – 0 = v2 – v1. When we combine the two equations, we get v1 = (m1 – m2)v1/(m1 + m2). The fraction of kinetic energy lost by the neutron is ?K1/K1 = (!m1v12 – !m1v12)/!m1v12 = 1 – [(m1 – m2)/(m1 + m2)]2 = 4m1m2/(m1 + m2)2. (a) For m2 = 1.01 u, we get ?K1/K1 = 4m1m2/(m1 + m2)2 = 4(1.01 u)(1.01 u)/(1.01 u + 1.01 u)2 = 1.00. (b) For m2 = 2.01 u, we get ?K1/K1 = 4m1m2/(m1 + m2)2 = 4(1.01 u)(2.01 u)/(1.01 u + 2.01 u)2 = 0.89. (c) For m2 = 12.00 u, we get ?K1/K1 = 4m1m2/(m1 + m2)2 = 4(1.01 u)(12.00 u)/(1.01 u + 12.00 u)2 = 0.29. (d) For m2 = 208 u, we get ?K1/K1 = 4m1m2/(m1 + m2)2 = 4(1.01 u)(208 u)/(1.01 u + 208 u)2 = 0.019. 40. For the elastic collision, we use momentum conservation for this one-dimensional motion: m1v1 + m2v2 = m1v1 + m2v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or v1 = v2 – v1 + v2. If we put this in the momentum equation, we get m1v1 + m2v2 = m1v2 – m1v1 + m1v2 + m2v2, or v2 = [2m1/(m1 + m2)]v1 + [(m2 – m1)/(m1 + m2)]v2 . When we use this in the relative speed equation, we get v1 = v2 – v1 + v2 = v2 – v1 + [2m1/(m1 + m2)]v1 + [(m2 – m1)/(m1 + m2)]v2 = [(m1 – m2)/(m1 + m2)]v1 + [2m2/(m1 + m2)]v2. Note that this is just an interchange of 1 and 2 in the previous result. 41. (a) At the maximum compression of the spring, there will be no relative motion of the two blocks. Because there is no friction, we can use momentum conservation: m1v1 + m2v2 = m1v1 + m2v2; m1v1 + 0 = (m1 + m2)V, or V = m1v1/(m1 + m2). Energy is also conserved, so we have !m1v12 + 0 = !(m1 + m2)V2 + !kx2;

!m1v12 = !(m1 + m2)[m1v1/(m1 + m2)]2 + !kx2, or x2 = m1m2v12/k(m1 + m2) = (2.0 kg)(4.5 kg)(8.0 m/s)2/(850 N/m)(2.0 kg + 4.5 kg), which gives x = 0.32 m. (b) From the initial motion of the first block to the final separation, all horizontal forces are internal to the system of the two blocks and are conservative. For momentum conservation we have m1v1 + m2v2 = m1v1 + m2v2; Page 8

Chapter 9

m1v1 + 0 = m1v1 + m2v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or v2 = v1 – 0 + v1. When we combine the equations, we get v1 = (m1 – m2)v1/(m1 + m1) = (2.0 kg – 4.5 kg)(8.0 m/s)/(2.0 kg + 4.5 kg) = – 3.1 m/s (rebound). For v2 we get v2 = v1 + v1 = 8.0 m/s + (– 3.1 m/s) = 4.9 m/s. (c) Yes, because the force by the spring is conservative, the collision is elastic.

42. We let V be the speed of the block and bullet immediately after the collision and before the pendulum swings. For this perfectly inelastic collision, we use momentum conservation: mv + 0 = (M + m)V; (0.018 kg)(180 m/s) = (0.018 kg + 3.6 kg)V, which gives V = 0.896 m/s. Because the tension does no work, we can use energy conservation for the swing: !(M + m)V2 = (M + m)gh, or V2 = 2gh; (0.896 m/s)2 = 2(9.80 m/s2)h, which gives h = 0.0409 m. We find the horizontal displacement from the triangle: L2 = (L – h)2 + x2; 0.48 m. (2.8 m)2 = (2.8 m – 0.0409 m)2 + x2, which gives x =

L

43. (a) The velocity of the block and projectile after the collision is v = mv1/(m + M). The fraction of kinetic energy lost is fraction lost = – ?K/K = – [!(m + M)v2 – !mv12]/!mv12 = – {(m + M)[mv1/(m + M)]2 – mv12}/mv12 = – [m/(m + M)] + 1 = + M/(m + M). (b) For the data given we have fraction lost = M/(m + M) = (380 g)/(14.0 g + 380 g) =

0.964.

44. Momentum conservation gives 0 = m1v1 + m2v2; 0 = m1v1 + 1.5m1v2, or v1 = – 1.5v2. The kinetic energy of each piece is K2 = !m2v22; K1 = !m1v12 = !(m2/1.5)(– 1.5v2)2 = (1.5)!m2v22 = 1.5K2 . The energy supplied by the explosion produces the kinetic energy: E = K1 + K2 = 2.5K2; 17,500 J = 2.5K2 , which gives K2 = 7000 J. For the other piece we have K1 = E – K2 = 17,500 J – 7000 J = 10,500 J. Thus K(heavier) = 7000 J; K(lighter) = 10,500 J. 45. Using the coordinate system shown, for momentum conservation we have

Page 9

x

v m

L



h M

m+M m+M V

Chapter 9

y-momentum: – mv sin 1 + mv sin 2 = 0, or 1 = 2 . x-momentum: mv cos 1 + mv cos 2 = 2mv2; 2mv cos 1 = 2mv/3; cos 1 = @, or 1 = 70.5° = 2 . The angle between their initial directions is  = 1 + 2 = 2(70.5°) = 141°.

y

1 2

p1

x p

p2

46. On the horizontal surface after the collision, the normal force on the joined cars is FN = (m + M)g. We find the common speed of the joined cars immediately after the collision by using the work-energy principle for the sliding motion: Wfr = ?K; – µk(m + M)gd = 0 – !(M + m)V2; (0.40)(9.80 m/s2)(4.8 m) = !V2, which gives V = 6.13 m/s. For the collision, we use momentum conservation: mv + 0 = (m + M)V; (0.95103 kg)v = (0.95103 kg + 2.2103 kg)(6.13 m/s), which gives

v = 20 m/s

(73 km/h).

47. (a) For a perfectly elastic collision, we use momentum conservation: m1v1 + m2v2 = m1v1 + m2v2, or m1(v1 – v1) = m2(v2 – v2). Kinetic energy is conserved, so we have !m1v12 + !m2v22 = !m1v12 + !m2v22, or m1(v12 – v12) = m2(v22 – v22), which can be written as m1(v1 – v1)(v1 + v1) = m2(v2 – v2)(v2 + v2). When we divide this by the momentum result, we get v1 + v1 = v2 + v2 , or v1 – v2 = v2 – v1 . If we use this in the definition of the coefficient of restitution, we get e = (v1 – v2)/(v2 – v1) = (v2 – v1)/(v2 – v1) = 1. For a completely inelastic collision, the two objects move together, so we have v1 = v2, which gives e = 0. (b) We find the speed after falling a height h from energy conservation: !mv12 = mgh, or v1 = (2gh)1/2. The same expression holds for the height reached by an object moving upward: v1 = (2gh)1/2. Because the steel plate does not move, when we take into account the directions we have e = (v1 – v2)/(v2 – v1 ) = [(2gh)1/2 – 0]/{0 – [– (2gh)1/2]}, so e = (h/h)1/2. 48. We assume the explosion creates equal and opposite forces on the two pieces. Momentum conservation for the explosion gives us 0 = m1v1 + m2v2; 0 = m1v1 + 3m1v2, or v1 = – 3v2. On the horizontal surface after the collision, the normal force on a block is FN = mg. We relate the speed of a block immediately after the collision to the distance it slides from the work-energy principle for the sliding motion: Page 10

Chapter 9

Wfr = ?K; – µkmgd = 0 – !mv2, or d = !v2/µkg. The coefficient of friction is the same for each piece. If we form the ratio of the distances, we get d1/d2 = (v1/v2)2 = (– 3)2 = 9, with the lighter block traveling farther.

49. For the momentum conservation of this one-dimensional collision, we have m1v1 + m2v2 = m1v1 + m2v2. (a) If the bodies stick together, v1 = v2 = V: (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg + 3.0 kg)V, which gives V = v1 = v2 = 1.9 m/s. (b) If the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 5.5 m/s – (– 4.0 m/s) = 9.5 m/s = v2 – v1. The momentum equation is (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg)v1 + (3.0 kg)v2, or (5.0 kg)v1 + (3.0 kg)v2 = 15.5 kg · m/s. When we combine these two equations, we get v1 = – 1.6 m/s, v2 = 7.9 m/s. (c) If m1 comes to rest, v1 = 0. (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = 0 + (3.0 kg)v2, which gives v1 = 0, v2 = 5.2 m/s. (d) If m2 comes to rest, v2 = 0. (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg)v1 + 0, which gives v1 = 3.1 m/s, v2 = 0. (e) The momentum equation is (5.0 kg)(5.5 m/s) + (3.0 kg)(– 4.0 m/s) = (5.0 kg)(– 4.0 m/s) + (3.0 kg)v2, which gives v1 = – 4.0 m/s, v2 = 12 m/s. The result for (c) is reasonable. The 3.0-kg body rebounds. The result for (d) is not reasonable. The 5.0-kg body would have to pass through the 3.0-kg body. To check the result for (e) we find the change in kinetic energy: ?K = (!m1v12 + !m2v22) – (!m1v12 + !m2v22) = ![(5.0 kg)(– 4.0 m/s)2 + (3.0 kg)(12 m/s)2] – ![(5.0 kg)(5.5 m/s)2 + (3.0 kg)(– 4.0 m/s)2] = + 156 J. Because the kinetic energy cannot increase in a simple collision, the result for (e) is not reasonable.

Page 11

Chapter 9

50. (a) We let dM be the distance the block moves during the collision. We estimate the average velocity of each object as if the acceleration were constant, vav = (vi + vf)/2. Because the velocity of the bullet relative to the block after the collision is zero, if d is the distance the bullet L moves relative to the block, we have ?t = d/[(v1 + 0)/2] = 2d/v1 , where we have used v1 for the initial velocity of the bullet. M m The distance the block has moved is dM = [(0 + v)/2] ?t = vd/v1 . v1 (b) During the collision, the block will swing a small angle  :   ˜ sin   = dM/L = vd/Lv1 , which means cos   ˜ 1 –  2/2. We estimate the magnitude of the tension during the collision: FT cos   = (m + M)g, or FT ˜ (m + M)g. The horizontal impulse during the collision is created by the tension: – FT (sin  )av ?t = (m + M)v – mv1 ; – FT ( /2) ?t = (m + M)v – mv1 ; – (m + M)g(vd/2Lv1)(2d/v1) = (m + M)v – mv1 , which gives v = [mv1/(m + M)]/[1 + (gd2/Lv12)]. If (gd2/Lv12) « 1, we have v ˜ [1 – (gd2/Lv12)]mv1/(m + M). We find ?p from the impulse: ?p = – (m + M)g(vd2/Lv12) = – (m + M)(gd2/Lv12)[1 – (gd2/Lv12)]mv1/(m + M) = – (mgd2/Lv1)[1 – (gd2/Lv12)] ˜ – mgd2/Lv1. (c) During the collision, the block will rise a height h = L(1 – cos  ) ˜ L[1 – (1 –  2/2)] = L(vd/Lv1)2/2. When we use the previous result for v, we get h = [1 – (gd2/Lv12)]2[m/(m + M)]2d2/2L ˜ [m/(m + M)]2d2/2L. After the collision, the block will rise an additional height h – h. From energy conservation, we have Page 12



L m+M

h dM

d

v

Chapter 9

= [2g(h – h)]1/2 = (2gh)1/2[1 – (h/h)]1/2 ˜ (2gh)1/2[1 – (h/2h)] ˜ (2gh)1/2{1 – [m/(m + M)]2d2/4hL}. When we use the result for v from part (b), we have [mv1/(m + M)]/[1 + (gd2/Lv12)] = (2gh)1/2{1 – [m/(m + M)]2d2/4hL}, or v

v1 = [(m + M)/m](2gh)1/2{1 – [m/(m + M)]2d2/4hL}[1 + (gd2/Lv12)]. If we assume that we can use the result from Example 9–11, v = [(m + M)/m](2gh)1/2, for the first factor and for v1 in the last term and we keep only terms in d2 when we multiply the factors, we get

v1 ˜ v{1 – [m/(m + M)]2d2/4hL + (gd2/Lv2)}. Thus the fractional error is (v1 – v)/v = gd2/Lv2 – [m/(m + M)]2d2/4hL

= (d2/L){[m2/(m + M)22h] – [m/(m + M)]2/4h} =

d2m2/2(m + M)2hL.

51. For the collision we use momentum conservation: x-direction: m1v1 + 0 = (m1 + m2)v cos ; (3.3 kg)(7.8 m/s) = (3.3 kg + 4.6 kg)v cos , which gives v cos  = 3.26 m/s. y-direction: 0 + m2v2 = (m1 + m2)v sin ; (4.6 kg)(10.2 m/s) = (3.3 kg + 4.6 kg)v sin , which gives v sin  = 5.94 m/s. We find the direction by dividing the equations: tan  = (5.94 m/s)/(3.26 m/s) = 1.82, 61° from first eagles’s direction. so  = We find the magnitude by squaring and adding the equations: v = [(5.94 m/s)2 + (3.26 m/s)2]1/2 = 6.8 m/s.

52. (a) Using the coordinate system shown, for momentum conservation we have x-momentum: mAvA + 0 = mAvA cos A + mBvB cos B; Before y-momentum: 0 + 0 = mAvA sin A – mBvB sin B. (b) With the given data, we have x: (0.400 kg)(1.80 m/s) = vA (0.400 kg)(1.10 m/s) cos 30° + (0.500 kg)vB cos B, which gives vB cos B = 0.678 m/s; y: 0 = (0.400 kg)(1.10 m/s) sin 30° – (0.500 kg)vB sin B , which gives vB sin B = 0.440 m/s. We find the magnitude by squaring and adding the equations: vB = [(0.440 m/s)2 + (0.678 m/s)2]1/2 = 0.808 m/s. We find the direction by dividing the equations: tan B = (0.440 m/s)/(0.678 m/s) = 0.649, so B = 33.0°.

Page 13

y p



p1

x p2

After v A

A y

B x

v B

Chapter 9

53. (a) Using the coordinate system shown, for momentum conservation we have x-momentum: mv + 0 = 0 + 2mv2 cos , or 2v2 cos  = v; y-momentum: 0 + 0 = – mv1 + 2mv2 sin , or 2v2 sin  = v1. If we square and add these two equations, we get v2 + v12 = 4v22. For the conservation of kinetic energy, we have !mv2 + 0 = !mv12 + !(2m)v22, or v2 – v12 = 2v22. When we add this to the previous result, we get v2 = 3v22. Using this in the x-momentum equation, we get 30°. cos  = v3/2, or  = (b) From part (a) we have v2 = v/v3. Using the energy result, we get v12 = v2 – 2v22 = v2 – 2v2/3 = @v2, or v1 = v/v3. (c) The fraction of the kinetic energy transferred is fraction = ?K1/K1 = K2/K1 = !(2m)v22/!mv2 = m(v2/3)/!mv2 =

y

p2 



p1

x

p1

%.

54. Using the coordinate system shown, for momentum conservation we have x: 0 + mv2 = mv1 cos  + 0; 3.7 m/s = v1 cos ; y: mv1 + 0 = mv1 sin  + mv2; 2.0 m/s = v1 sin  + v2, or v1 sin  = 2.0 m/s – v2. For the conservation of kinetic energy, we have mv 2 !mv12 + !mv22 = !mv12 + !mv22; (2.0 m/s)2 + (3.7 m/s)2 = v12 + v22. We have three equations in three unknowns:  , v1, v2. We eliminate  by squaring and adding the two momentum results, and then combine this with the energy equation, with the results:  = 0°, v1 = 3.7 m/s, v2 = 2.0 m/s. The two billiard balls exchange velocities. 55. Using the coordinate system shown, for momentum conservation we have x: mnvn + 0 = mnvn cos 1 + mHevHe cos 2; mn(6.2105 m/s) = mnvn cos 1 + 4mnvHe cos 45°, or vn cos 1 = (6.2105 m/s) – 4vHe cos 45°. y: 0 + 0 = + mnvn sin 1 – mHevHe sin 2; 0 = + mnvn sin 1 – 4mnvHe sin 45°, or vn sin 1 = 4vHe sin 45°. For the conservation of kinetic energy, we have !mnvn2 + 0 = !mnvn2 + !mHevHe2; Page 14

y

mv 2 

mv 1 



x

mv 1

y pn

1 

pn

2  p He 

x

Chapter 9

mn(6.2105 m/s)2 = mnvn2 + 4mnvHe2, or vn2 + 4vHe2 = 3.841011 m2/s2. We have three equations in three unknowns: 1, vn, vHe. We eliminate 1 by squaring and adding the two momentum results, and then combine this with the energy equation, with the results: 1 = 76°, vn = 5.1105 m/s, vHe = 1.8105 m/s. 56. Using the coordinate system shown, for momentum conservation we have x-momentum: mNevNe + 0 = mNevNe cos Ne + mxvx cos x ; y-momentum: 0 + 0 = mNevNe sin Ne – mxvx sin x . From the second equation we get mxvx = mNevNe (sin Ne)/sin x . When we use this in the first equation, we get mNevNe = mNevNe cos Ne + [mNevNe (sin Ne)/sin x ] cos x = mNevNe [cos Ne + (sin Ne)/tan x ]. For the perfectly elastic collision, kinetic energy is conserved: !mNevNe2 + 0 = !mNevNe2 + !mxvx2;

y

m Ne v Ne 

Ne

m Ne v Ne

x

x

mx m xv x

!mNevNe2[cos Ne + (sin Ne)/tan x]2 = !mNevNe2 + !mNevNe2 (mNe/mx)(sin2 Ne)/sin2 x ;

[cos Ne + (sin Ne)/tan x]2 = 1 + (mNe/mx)(sin2 Ne)/sin2 x ; [cos 55.6° + (sin 55.6°)/tan 50.0°]2 = 1 + (mNe/mx)(sin2 55.6°)/sin2 50.0°, which gives mNe/mx = 0.501, so mx = (20 u)/(0.501) = 40 u.

57. Using the coordinate system shown, for momentum conservation we have x: mv1 + 0 = mv1 cos 1 + mv2 cos 2 , or y v1 = v1 cos 1 + v2 cos 2 ; y: 0 + 0 = mv1 sin 1 – mv2 sin 2 , or 0 = v1 sin 1 – v2 sin 2 . For the conservation of kinetic energy, we have mv 1 !mv12 + 0 = !mv12 + !mv22; m v12 = v12 + v22. We square each of the momentum equations: v12 = v12 cos2 1 + 2v1v2 cos 1 cos 2 + v22 cos2 2 ; 0 = v12 sin2 1 – 2v1v2 sin 1 sin 2 + v22 sin2 2 . If we add these two equations and use sin2  + cos2  = 1, we get v12 = v12 + 2v1v2(cos 1 cos 2 – sin 1 sin 2) + v22 . If we subtract the energy equation, we get 0 = 2v1v2(cos 1 cos 2 – sin 1 sin 2), or cos 1 cos 2 – sin 1 sin 2 = 0. We reduce this with a trigonometric identity: cos 1 cos 2 – sin 1 sin 2 = cos(1 + 2) = 0, which means that 1 + 2 = 90°. 58. Using the coordinate system shown, for momentum conservation we have

Page 15

mv 1 

1 2 mv 2 

x

Chapter 9

x: m1v1 + 0 = m1v1 cos 1 + m2v2 cos 2; y: 0 + 0 = m1v1 sin 1 – m2v2 sin 2. Before We eliminate 2 by using sin2 2 + cos2 2 = 1, to get m1 (m2v2)2 = (m1v1)2 – 2m12v1v1 cos 1 + (m1v1)2. For the conservation of kinetic energy, we have v1 m2 !m1v12 + 0 = !m1v12 + !m2v22. or (m2v2)2 = m1m2(v12 – v12). We see from this, as expected, that v1 = v1 . When we combine the energy and momentum results, we get 2m1v1v1 cos 1 = (m1 + m2)v12 + (m1 – m2)v12, or

After v 1 m1 y

1 

m2 x

2  v 2

cos 1 = ![(1 + m2/m1)(v1/v1) + (1 – m2/m1)(v1/v1)], with 0 = v1 = v1 . (a) For m1 < m2 , we consider the extreme values for v1. If v1 = v1, we get cos 1 = ![(1 + m2/m1) + (1 – m2/m1)] = 1, so 1 = 0°. If v1 = 0, we get cos 1 = ![0 + (1 – m2/m1)(v1/0)] = – 8.

Of course the value of the cosine cannot have a value of – 8, but this means that its value can start at + 1 and become negative, so there is a value of v1 > 0 where the cosine is – 1. Thus we have 0° = 1 = 180°. (b) For m1 > m2 , we consider the extreme values for v1. If v1 = v1, we get cos 1 = ![(1 + m2/m1) + (1 – m2/m1)] = 1, so 1 = 0°. If v1 = 0, we get cos 1 = ![0 + (1 – m2/m1)(v1/0)] = + 8.

Of course the value of the cosine cannot have a value of + 8, but this means that its value remains positive. This implies that there is some angle for which the cosine has a minimum positive value. To find this angle we set the first derivative of the cosine function equal to zero: d(cos 1)/dv1 = ![(1 + m2/m1)(1/v1) – (1 – m2/m1)(v1/v12)] = 0, which gives v1 = v1[(1 – m2/m1)/(1 + m2/m1)]1/2. Thus the minimum value of cos 1 is

(cos 1)min = !{(1 + m2/m1)[(1 – m2/m1)/(1 + m2/m1)]1/2 + (1 – m2/m1)[(1 + m2/m1)/(1 – m2/m1)]1/2} = [1 – (m2/m1)2]1/2, which determines the maximum angle . Thus we have 0° = 1 = , where cos2  = 1 – (m2/m1)2. Note that we could not use the derivative in part (a) because the cosine does not have a mathematical minimum before cos 1 reaches – 8. 59. We choose the origin at the carbon atom. The center of mass will lie along the line joining the atoms: xCM = (mCxC + mOxO)/(mC + mO) = [0 + (16 u)(1.1310–10 m)]/(12 u + 16 u) = 6.510–11 m from the carbon atom. 60. We choose the origin at the front of the car: xCM = (mcarxcar + mfrontxfront + mbackxback)/(mcar + mfront + mback) = [(1150 kg)(2.50 m) + (140 kg)(2.80 m) + (210 kg)(3.90 m)]/(1150 kg + 140 kg + 210 kg) = 2.72 m from the front of the car.

Page 16

Chapter 9

61. From the symmetry of the hydrogen triangle, we know that the center of mass will be perpendicular to the plane of the hydrogen atoms along a line from the center of the triangle. We find the height above the triangle from zCM = [3mH(0) + mNzN]/(3mH + mN) = [0+ (14 u)(0.037 nm)]/[(3(1 u) + 14 u] = 0.030 nm above center of H triangle.

62. Because the cubes are made of the same material, their masses will be proportional to the volumes: m1 , m2 = 23m1 = 8m1 , m3 = 33m1 = 27m1 . From symmetry we see that yCM = 0. We choose the x-origin at the outside edge of the small cube: xCM = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3)

z N H H

H

y

3¬0

¬0

2¬0 x

= {m1(!¬0) + 8m1[¬0 + !(2¬0)] +

27m1[¬0 + 2¬0 + !(3¬0)]}/(m1 + 8m1 + 27m1) = 138¬0/36 = 3.83¬0 from the outer edge of the small cube. 63. We choose the origin at the center of the raft, which is the center of mass of the raft: xCM = (Mxraft + m1x1 + m2x2 + m3x3)/(M + m1 + m2 + m3) = [0 + (1200 kg)(9.0 m) + (1200 kg)(9.0 m) + (1200 kg)(– 9.0 m)]/[6200 kg + 3(1200 kg)] = 1.10 m (East). yCM = (Myraft + m1y1 + m2y2 + m3y3)/(M + m1 + m2 + m3) = [0 + (1200 kg)(9.0 m) + (1200 kg)(– 9.0 m) + (1200 kg)(– 9.0 m)]/[6200 kg + 3(1200 kg)] = – 1.10 m (South).

Page 17

y

m1 M

m3

x m2

Chapter 9

64. We know from the symmetry that the center of mass lies on a line containing the center of the plate and the center of the hole. We choose the center of the plate as origin and x along the line joining the centers. Then yCM = 0. A uniform circle has its center of mass at its center. 2R We can treat the system as two circles: a circle of radius 2R, density  and mass p(2R)2 with x1 = 0; CM a circle of radius R, density –  and mass – pR2 with x2 = 0.80R. We find the center of mass from xCM = (m1x1 + m2x2)/(m1 + m2) = [4pR2 (0) – pR2 (0.80R)]/(4pR2 – pR2) = – 0.27R. The center of mass is along the line joining the centers 0.07R outside the hole. 65. We know from the symmetry that the center of mass lies on the y-axis: xCM = 0. We choose a differential length of the wire r d, shown in the diagram, which has a mass dm = (m/pr)r d. We integrate over the wire to find yCM: 

 y CM =

y dm

0

=

0

= r (– cos  )

 0

R C C´

x

0.80R

y dm d

r sin  (m/ r) r d  

dm

y

r



x

C

(m/ r) r d 

= r [– ( – 1) + (1)] = 2r .

Thus the center of mass is at

xCM = 0, yCM = 2r/p.

y 66. We know from the symmetry that the center of mass lies on the y-axis: xCM = 0. We treat the plate as an infinite number of semicircular wires. For the wire with radius r and thickness dr, we choose a differential length r d, shown in the diagram, which has a mass dm = (2m/pR2)r ddr. We integrate over the plate to find yCM:



r sin  (2m/ R ) r d  d r 2

y dm

y CM =

dm

=

=

m

3

2 R 2

 2 R (– cos  ) = 2R [– ( – 1) + (1)] = 4R . 2 3 3 R 3 0 Thus the center of mass is at xCM = 0, yCM = 4R/3p.

=

Page 18

R d

dr r

 C R 0

2

r dr

 0

sin  d 

x

Chapter 9

z

67. We know from the symmetry that the center of mass lies on the z-axis: xCM = 0, yCM = 0. For a differential element we use a circle at a height z, thickness dz, and radius r = (R/h)z. If  is the mass density of the cone, the mass of this element is dm = pr2 dz. We integrate over the cone to find zCM:



h

z dm

z CM =

=

0 h

dm

0

h

 z r2 dz =

 r2 dz

Thus the center of mass is at

R

r

dz

z

2

z (R/ h) z 2 dz

0 h

2

z (R/ h) z 2 dz

0

y

4

=

(h / 4) 3h = . 3 4 (h / 3)

x

xCM = 0, yCM = 0, zCM = 3h/4 above the point.

68. We know from the symmetry that the center of mass lies on the z-axis: xCM = 0, yCM = 0. For a differential element we use a square at a height z, thickness dz, and side L = (s/h)(h – z). If  is the mass density of the pyramid, the mass of this element is dm = L2 dz. We relate h to s by representing the edge of the pyramid in terms of its components: s2 = h2 + (s/2)2 + (s/2)2, which gives h = s/v2. We integrate over the plate to find zCM:



s

h 0 h 0

dz

h L

y s s x

dm

=

z

z dm

z CM =

h

h

 z (s/ h) 2 (h – z) 2 dz = 2

2

 (s/ h) (h – z) dz

0

h 0

2

(h z – 2hz 2 + z 3 ) d z (h 2 – 2hz + z 2 ) d z

[h 2(h 2 / 2) – 2h(h3 / 3) + (h 4 / 4)] h = = s . 2 2 3 4 4 2 [h (h) – 2h(h / 2) + (h / 3)] Thus the center of mass is at xCM = 0, yCM = 0, zCM = s/4v2 above the base. =

69. (a) If we choose the origin at the center of the Earth, we have xCM = (mEarthxEarth + mMoonxMoon)/(mEarth + mMoon) = [0 + (7.351022 kg)(3.84108 m)]/(5.981024 kg + 7.351022 kg) = 4.66106 m. Note that this is less than the radius of the Earth and thus is inside the Earth. (b) The CM found in part (a) will move around the Sun on an elliptical path. The Earth and Moon will revolve about the CM. Because this is near the center of the Earth, the Earth will essentially be on the elliptical path around the Sun. The motion of the Moon about the Sun is more complicated. 70. We find the velocity of the center of mass from vCM = (m1v1 + m2v2)/(m1 + m2)

= {(35 kg)[(12 m/s)i – (16 m/s)j] + (35 kg)[(– 20 m/s)i + (14 m/s)j]}/(35 kg + 35 kg) = (– 4 m/s)i + (– 1 m/s)j.

Page 19

Chapter 9

71. We choose the origin of our coordinate system at the man. (a) For their center of mass we have xCM = (mwomanxwoman + mmanxman)/(mwoman + mman) = [(50 kg)(11.0 m) + 0]/(50 kg + 70 kg) = 4.6 m. (b) Because the center of mass will not move, we find the location of the woman from xCM = (mwomanxwoman + mmanxman)/(mwoman + mman) 4.6 m = [(50 kg)xwoman + (70 kg)(2.8 m)]/(50 kg + 70 kg), which gives xwoman = 7.1 m. The separation of the two will be 7.1 m – 2.8 m = 4.3 m. (c) The two will meet at the center of mass, so he will have moved 4.6 m. 72. The center of mass will land at the same point, 2d from the launch site. If part I is still stopped by the explosion, it will fall straight down, as before. (a) We find the location of part II from the center of mass: xCM = (mIxI + mIIxII)/(mI + mII) 2d = [mId + 3mIxII]/(mI + 3mI), which gives xII = 7d/3, or 2d/3 closer to the launch site. (b) For the new mass distribution, we have xCM = (mIxI + mIIxII)/(mI + mII) 2d = [3mIId + mIIxII]/(3mII + mII), which gives xII = 5d, or 2d farther from the launch site. 73. The forces on the balloon, gondola, and passenger are balanced, so the center of mass does not move relative to the Earth. As the passenger moves down at a speed v relative to the balloon, the balloon will move up. If the speed of the balloon is v relative to the Earth, the passenger will move down at a speed v – v relative to the Earth. We choose the location of the center of mass as the origin and determine the positions after a time t: xCM = (mballoonxballoon + mpassengerxpassenger)/(mballoon + mpassenger) 0 = [Mvt – m(v – v)t]/(M + m), which gives v = mv/(M + m) up. If the passenger stops, the gondola and the balloon will also stop. There will be equal and opposite impulses acting when the passenger grabs the rope to stop. 74. We choose the origin of our coordinate system at the initial position of the 75-kg person. For the location of the center of mass of the system we have xCM = (m1x1 + m2x2 + mboatxboat)/(m1 + m2 + mboat) = [(75 kg)(0) + (55 kg)(3.0 m) + (80 kg)(1.5 m)]/ (75 kg + 55 kg + 80 kg) = 1.36 m. Thus the center of mass will be 3.0 m – 1.36 m = 1.64 m from the 55-kg person. When the two people exchange seats, the center of mass will not move. The end where the 75-kg person started, which was 1.36 m from the center of mass, is now 1.64 m from the center of mass, that is, the boat must have moved 1.64 m – 1.36 m = 0.28 m toward the initial position of the 75-kg person.

Page 20

m2

m1 m boat CM

x L

m2

m1 m boat L

CM

x

Chapter 9

75. We find the time for the man to reach the other end from t = L/vrel = (25 m)/(2.0 m/s) = 12.5 s. If we let v2 be the speed of the flatcar while the man is walking, the speed of the man is v1 = v2 + vrel. Because the velocity of the center of mass of the system of flatcar and man does not change, we have vCM = (m1v1 + m2v2)/(m1 + m2) 5.0 m/s = [(90 kg)(v2 + 2.0 m/s) + (200 kg)v2]/(90 kg + 200 kg), which gives v2 = 4.38 m/s. The flatcar will have moved x = v2t = (4.38 m/s)(12.5 s) = 55 m. 76. The external gravitational force and the thrust produce the acceleration of the rocket: Fext + vrel dM/dt = Ma; – Mg + vrel dM/dt = M(3.0 g), or vrel(– 30 kg/s) = 4.0(2500 kg)(9.80 m/s2), which gives vrel = – 3.3103 m/s (down). Thus the exhaust speed is 3.3103 m/s. 77. We can consider the gravel dropping on the belt as a collision. The velocity of the gravel relative to the belt is the negative of the velocity of the belt. External forces on the belt are from the motor and friction, so we have Fext + vrel dM/dt = Ma; Fmotor – Ffr + vrel dM/dt = 0; Fmotor – 140 N + (– 2.20 m/s)(75.0 kg/s) = 0, which gives Fmotor = 305 N. The required output power of the motor is Pmotor = Fmotorv = (305 N)(2.20 m/s)/(746 W/hp) = 0.899 hp. When the gravel drops from the conveyor belt, it does not stop but still has the same horizontal velocity. Thus there is no additional interaction with the belt. The gravel dropping on the belt must still be accelerated, so the power required does not change. 78. The value of g at the altitude is g = GME/r2 = (6.6710–11 N · m2/kg2)(5.981024 kg)/[(6.40 + 6.38)106 m]2 = 2.44 m/s2. The external gravitational force and the thrust produce the acceleration of the rocket: Fext + vrel dM/dt = Ma; – Mg + vrel dM/dt = Ma, or vrel dM/dt = M(a + g) (– 1200 m/s)(dM/dt) = (25,000 kg)(1.7 m/s2 + 2.44 m/s2), which gives dM/dt = – 86 kg/s. 79. (a) Because chemical changes do not change the mass, the thrust from the ejected fuel is Ffuel = vrel dMfuel/dt = (– 550 m/s)(– 4.2 kg/s) = 2.3103 N. (b) The air is collected by the airplane at the speed of the airplane and ejected with the speed of the fuel, so the net thrust is Fair = vrel1 dMair/dt + vrel2 dMair/dt = (– 270 m/s)(100 kg/s) + (– 550 m/s)(– 100 kg/s) = 2.8104 N. (c) The power delivered by the two thrusts is P = (Ffuel + Fair)v = (2.3103 N + 2.8104 N)(270 m/s)/(746 W/hp) = 1.1104 hp.

Page 21

Chapter 9

80. Because the sand is leaking from the hole, there will be no relative velocity of the leaking sand with respect to the sled. Thus there is no thrust imparted to the sled. If we take the sled and remaining sand at a time when its mass is m, we have the forces indicated in the diagram. The acceleration will be g sin  down the slope, independent of the mass. We find the time from x = !at2; 7.0 s. 120 m = !(9.80 m/s2) sin 30° t2, which gives t = 81. From the result of Problem 57, the angle between the two final directions will be 90° for an elastic collision. We take the initial direction of the cue ball to be parallel to the side of the table. The angle for the struck ball to hit the pocket after the collision is tan 1 = 1.0/v3.0, which gives 1 = 30°. Thus the angle for the cue ball will be 90° – 30° = 60°. From the diagram we see that tan 2 = (4.0 – 1.0)/v3.0, which gives 2 = 60°. Because this is the angle the cue ball will have, this will be a “scratch shot.”

FN

 y

 x

mg

4.0 1.0

¦3.0

1

2

82. We find the force on the person from the magnitude of the force required to change the momentum of the air: F = ?p/?t = (?m/?t)v = (40 kg/s · m2)(1.50 m)(0.50 m)(100 km/h)/(3.6 ks/h) = 8.3102 N. The maximum friction force will be Ffr = µmg ˜ (1.0)(70 kg)(9.80 m/s2) = 6.9102 N, so the forces are about the same. 83. We find the speed after being hit from the height h using energy conservation: !mv2 = mgh, or v = (2gh)1/2 = [2(9.80 m/s2)(55.6 m)]1/2 = 33.0 m/s. We see from the diagram that the magnitude of the change in momentum is ?p = m(v2 + v2)1/2 = (0.145 kg)[(35.0 m/s)2 + (33.0 m/s)2]1/2 = 6.98 kg · m/s. We find the force from F ?t = ?p; F(0.5010–3 s) = 6.98 kg · m/s, which gives F = 1.4 104 N. We find the direction of the force from 43.3°. tan  = v/v = (33.0 m/s)/(35.0 m/s) = 0.943,  = 84. Using the coordinate system shown, for momentum conservation we have

Page 22

Chapter 9

y-momentum: mv1 + 0 = mv1 cos  + Mv2 cos 2 ; 5M(12.0 m/s) = 5Mv1 cos  + Mv2 cos 80°, or 5v1 cos  = – v2 cos 80° + 60.0 m/s. x-momentum: 0 = – mv1 sin  + Mv2 sin 2 ; 0 = – 5Mv1 sin  + Mv2 sin 80°, or 5v1 sin  = v2 sin 80°. For the conservation of kinetic energy, we have !mv12 + 0 = !mv12 + !Mv22; 5M(12.0 m/s)2 = 5Mv12 + Mv22, or 5v12 + v22 = 720 m2/s2. We have three equations in three unknowns: , v1, v2. We eliminate  by squaring and adding the two momentum results, and then combine this with the energy equation, with the results: (a) v2 = 3.47 m/s. (b) v1 = 11.9 m/s. (c)  = 3.29°.

85. On the horizontal surface after the collision, the normal force is FN = (m + M)g. We find the common speed of the block and bullet immediately after the embedding by using the work-energy principle for the sliding motion: Wfr = ?K;

y

 mv 1 

Mv 2 

m) = ! which gives V = 6.82 m/s. 0.25(9.80 For the collision, we use momentum conservation: mv + 0 = (M + m)V; (0.015 kg)v = (0.015 kg + 1.10 kg)(6.82 m/s), which gives

x

M mv 1

m = 5M

V

FN

m v

M Ffr

– µk(m + M)gd = 0 – !(M + m)V2; m/s2)(9.5

2

(m+M )g

V2,

v = 5.1102 m/s.

86. We let V be the speed of the block and bullet immediately after the embedding and before the two start to rise. For this perfectly inelastic collision, we use momentum conservation: mv + 0 = (M + m)V; (0.021 kg)(310 m/s) = (0.021 kg + 1.40 kg)V, which gives V = 4.58 m/s. For the rising motion we use energy conservation, with the potential energy reference level at the sheet: Ki + Ui = Kf + Uf ;

!(M + m)V 2 + 0 = 0 + (m + M)gh, or

h = V 2/2g = (4.58 m/s)2/2(9.80 m/s2) =

1.07 m.

87. For the elastic collision of the two balls, we use momentum conservation: mv1 + m2v2 = mv1 + m2v2; mv1 + 0 = m(– 0.600v1) + m2v2, or m2v2 = 1.600mv1 . Because the collision is elastic, the relative speed does not change: v1 – 0 = – (v1 – v2); v1 = v2 – (– 0.600v1), or v2 = 0.400v1 . Combining these two equations, we get m2 = 4.00m.

Page 23

V

M v

m

m+M

Chapter 9

88. On the horizontal surface, the normal force on a car is FN = mg. We find the speed of a car immediately after the collision by using the work-energy principle for the succeeding sliding motion: Wfr = ?K; – µkmgd = 0 – !mv2. We use this to find the speeds of the cars after the collision: 0.60(9.80 m/s2)(15 m) = !vA2, which gives vA = 13.3 m/s; 0.60(9.80 m/s2)(30 m) = !vB2, which gives vB = 18.8 m/s. For the collision, we use momentum conservation: mAvA + mBvB = mAvA + mBvB ; (2000 kg)vA + 0 = (2000 kg)(13.3 m/s) + (1000 kg)(18.8 m/s), which gives vA = 22.7 m/s. We find the speed of car A before the brakes were applied by using the work-energy principle for the preceding sliding motion: Wfr = ?K; – µkmAgd = !mAvA2 – !mAvA02; – 0.60(9.80 m/s2)(15 m) = ![(22.7 m/s)2 – vA02], which gives vA0 = 26.3 m/s = 94.7 km/h =

59 mi/h.

89. The energy transformed to thermal or other forms of energy is the loss in kinetic energy of the cars: = – [!(m1 + m2)v2 – !m1v12]/!m1v12 – ?K/K1 = 1 – [(m1 + m2)/m1](v/v1)2 = 1 – [(10,000 kg + 10,000 kg)/(10,000kg)](1/2)2 = 0.50 = 50%.

90. Because mass is conserved, the mass of the third piece must be m. The initial momentum is zero. For momentum conservation we have x-direction: 0 = (2m)(2v) – mv3 cos , or v3 cos  = 4v; y-direction: 0 = mv – mv3 sin , or v3 sin  = v. When we divide the two equations, we get tan  = 0.25,  = 14°. From the first equation, we get v3 = 4v/cos 14° = 4.1v, 104° from the direction of the first piece.

y

mv 2m(2v ) x

 mv 3

91. (a) (b)

No. The spring exerts equal but opposite forces on the blocks. For the system these forces are internal forces. Because there are no external forces on the system of the two blocks, momentum is conserved: 0 = m1v1 + m2v2 , which gives v1/v2 = – m2/m1. For the ratio of kinetic energies, we have K1/K2 = !m1v12/!m2v22 = (m1/m2)(v1/v2)2 = (m1/m2)(– m2/m1)2 = m2/m1. The center of mass is initially at rest. Because there are no external forces on the system of the two blocks, the center of mass does not move. The two friction forces will be in opposite directions, but they need not be equal. Thus there will be a net external force on the system. Momentum is not conserved, and the center of mass will move. 

(c) (d) (e)

Page 24

Chapter 9

92. We choose the coordinate system shown. There are 10 cases. xCM = (5mx1 + 3mx2 + 2mx3)/(10m) = [5(!¬) + 3(¬ + !¬) + 2(2¬ + !¬)]/(10) = 1.2¬. yCM = (7my1 + 2my2 + my3)/(10m) = [7(!¬) + 2(¬ + !¬) + (2¬ + !¬)]/(10) = 0.9¬. The center of mass is 1.2¬ from the left, and 0.9¬ from the back of the pallet.

x

¬ y

93. For the system of railroad car and snow, the horizontal momentum will be constant. For the horizontal motion, we take the direction of the car for the positive direction. The snow initially has no horizontal velocity. For this perfectly inelastic collision, we use momentum conservation: M1v1 + M2v2 = (M1 + M2)V; (5800 kg)(8.60 m/s) + 0 = [5800 kg + (3.50 kg/min)(60.0 min)]V, which gives V= 8.29 m/s. Note that there is a vertical impulse, so the vertical momentum is not constant.

94. (a) There are no horizontal external forces acting on the car and snow. If v is the velocity of the car, the relative velocity of the snow is – v. Thus we have Fext + vrel dM/dt = M dv/dt; 0 – v dM/dt = M dv/dt, or dv/v = – dM/M. We integrate this, with the initial velocity v0 and the initial mass M0: v v0

dv = – v

M M0

dM; M

v ln v – ln v 0 = – ln M + ln M 0 , or ln v 0 = ln

M0 . M

Thus we see that v = v0(M0/M). The mass is increasing at a uniform rate dM/dt, so the mass after a time t is M = M0 + (dM/dt)t. Thus we have v = v0{M0/[M0 + (dM/dt)t]}. (b) After a time of 60.0 min, we have v = (8.60 m/s){(5800 kg)/[5800 kg + (3.50 kg/min)(60.0 min)]} = 8.29 m/s, in agreement. 95. (a) We take the direction of the meteor for the positive direction. For this perfectly inelastic collision, we use momentum conservation: Mmeteorvmeteor + MEarthvEarth = (Mmeteor + MEarth)V; (108 kg)(15 103 m/s) + 0 = (108 kg + 6.01024 kg)V, which gives Page 25

V= 2.510–13 m/s.

Chapter 9

(b) The fraction transformed was fraction = ?KEarth/Kmeteor = !mEarthV 2/!mmeteorvmeteor2 = (6.01024 kg)(2.510–13 m/s)2/(108 kg)(15 103 m/s)2 = (c) The change in the Earth’s kinetic energy was ?KEarth = !mEarthV 2 = !(6.01024 kg)(2.510–13 m/s)2 =

1.710–17.

0.19 J.

96. We find the speed for falling or rising through a height h from energy conservation: !mv2 = mgh, or v2 = 2gh. y (a) The speed of the first block after sliding down the incline m and just before the collision is v1 = [2(9.80 m/s2)(3.60 m)]1/2 = 8.40 m/s. h For the elastic collision of the two blocks, we use momentum  conservation: mv1 + Mv2 = mv1 + Mv2; (2.20 kg)(8.40 m/s) + (7.00 kg)(0) = (2.20 kg)v1 + (7.00 kg)v2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 8.40 m/s – 0 = v2 – v1. Combining these two equations, we get v1 = – 4.38 m/s, v2 = 4.02 m/s. (b) We find the height of the rebound from v12 = 2gh; (– 4.38 m/s)2 = 2(9.80 m/s2)h, which gives h = 0.979 m. The distance along the incline is d = h/sin  = (0.979 m)/sin 30° = 1.96 m.

M x

97. Because energy is conserved for the motion up and down the incline, mass m will return to the level with the speed – v1. For a second collision to occur, mass m must be moving faster than mass M: – v1= v2. In the first collision, the relative speed does not change: v1 – 0 = – (v1 – v2), or – v1 = v1 – v2, so the condition becomes v1 – v2 = v2, or v1 = 2v2. For the first collision, we use momentum conservation: mv1 + 0 = mv1 + Mv2, or v1 – v1 = (M/m)v2. When we use the two versions of the condition, we get v1 – v1 = 3v2, so we need (M/m) = 3, or m = M/3. 98. (a) If the skeet were not hit by the pellet, the horizontal distance it would travel can be found from the range expression for projectile motion: R = (v02/g) sin 2 = [(30 m/s)2/(9.80 m/s2)] sin 2(30°) = 79.5 m. At the collision the skeet will have the x-component of the initial velocity: v1 = v0 cos  = (30 m/s) cos 30° = 26.0 m/s. We use energy conservation to find the height Page 26

y v1

v0



v2 x total

h

x

h

² x

Chapter 9

attained by the skeet when the collision occurs: !Mv02 = !Mv12 + Mgh;

!(30 m/s)2 = !(26.0 m/s)2 + (9.80 m/s2)h, which gives h = 11.5 m.

Using the coordinate system shown, for momentum conservation of the collision we have x: Mv1 + 0 = (M + m)Vx ; (250 g)(26.0 m/s) = (250 g + 15 g)Vx , which gives Vx = 24.5 m/s; y: 0 + mv2 = (M + m)Vy ; (15 g)(200 m/s) = (250 g + 15 g)Vy , which gives Vy = 11.3 m/s. We use energy conservation to find the additional height attained by the skeet after the collision: !(M + m)(Vx2 + Vy2)= !(M + m)Vx2 + (M + m)gh;

![(24.5 m/s)2 + (11.3 m/s)2]= !(24.5 m/s)2 + (9.80 m/s2)h, which gives

h = 6.54 m. (b) We find the time after the collision for the skeet to reach the ground from the vertical motion: y = y0 + Vy t + !(– g)t2; – 11.5 m = 0 + (11.3 m/s)t – !(9.80 m/s2)t2. The positive solution to this quadratic equation is t = 3.07 s. The horizontal distance from the collision is x = Vx t = (24.5 m/s)(3.07 s) = 75 m. The total horizontal distance covered is xtotal = !R + x = !(79.5 m) + 75 m = 115 m. Because of the collision, the skeet will have traveled an additional distance of ?x = xtotal – R = 115 m – 79.5 m = 35 m. 99. Obviously the spacecraft will have negligible effect on the motion of Saturn. In the reference frame of Saturn, we can treat this as the equivalent of a small mass “bouncing off” a massive object. The relative velocity of the spacecraft in this reference frame will be reversed. The initial relative velocity of the spacecraft is vSpS = vSp – vS = 10.4 km/s – (– 9.6 km/s) = 20.0 km/s. so the final relative velocity is vSpS = – 20.0 km/s. Therefore, we find the final velocity of the spacecraft from vSpS = vSp – vS ; – 20.0 km/s = vSp – (– 9.6 km/s), which gives vSp = – 29.6 km/s, so the final speed of the spacecraft is 29.6 km/s.

y 100. Because the two segments of the mallet are uniform, we know that the center of mass of each segment is at its midpoint. L We choose the origin at the bottom of the handle. The mallet will spin m about the CM, which is the point that will follow a parabolic trajectory: d xCM = (md + ML)/(m + M) = [(0.500 kg)(12.0 cm) + (2.00 kg)(24.0 cm + 4.00 cm)]/(0.500 kg + 2.00 kg) = 24.8 cm. 101. (a) We find the impulse on the ball from J = ?p = m ?v =(0.045 kg)(50 m/s – 0) = 2.25 N · s = (b) The average force is F = J/?t = (2.25 N · s)/(5.010–3 s) = 4.5102 N.

2.3 N · s.

102. (a) The given velocity is relative to the shuttle. If the shuttle gets a final speed in the – z-direction of vf , the final speed of the satellite will be v – vf . Momentum conservation gives Page 27

M

x

Chapter 9

0 = msatellite(v – vf) – mshuttlevf , or vf = msatellitev/(msatellite + mshuttle) = (800 kg)(0.30 m/s)/(800 kg + 90,000 kg) = (b) We find the average force on the satellite from F = ?p/?t = [msatellite(v – vf) – 0]/?t = (800 kg)[0.30 m/s – 2.610–3 m/s]/(4.0 s) = 59 N.

2.610–3 m/s.

103. (a) There is an obvious loss of kinetic energy, so this is an inelastic collision. (b) If we assume a constant acceleration, we can find the time from x = !(v0 + v)t; 0.10 s. 0.70 m = !{[(50 km/h)/(3.6 ks/h) + 0}t, which gives t = (c) We find the average impulsive force from F = ?p/?t = [m(v – v0)]/?t = (1000 kg)[0 – (50 km/h)/(3.6 ks/h)]/(0.10 s) = – 1.4105 N.

104. For the system of the two blocks, the spring force is internal. Because there is no friction, we can use momentum conservation: m1v1 + m2v2 = m1v1 – m2v2; 0 + 0 = mv1 – 3mv2, which gives v1 = 3v2. Kinetic energy is also conserved, so we have 0 + 0 + !kD2 = !m1v12 + !m2v22; kD2 = m(3v2)2 + 3mv22, which gives v2 = (k/12m)1/2D, and v1 = (3k/4m)1/2D.

105. We find the speed after falling a height h from energy conservation: !Mv2 = Mgh, or v = (2gh)1/2. The speed of the first cube after sliding down the incline and just before the collision is v1 = [2(9.80 m/s2)(0.20 m)]1/2 = 1.98 m/s. For the elastic collision of the two cubes, we use momentum conservation: Mv1 + mv2 = Mv1 + mv2;

h

M(1.98 m/s) + !M(0) = Mv1 + !Mv2. Because the collision is elastic, the relative speed does not change: v1 – v2 = – (v1 – v2), or 1.98 m/s – 0 = v2 – v1. Combining these two equations, we get Page 28

v

H

Chapter 9

v1 = 0.660 m/s, and v2 = 2.64 m/s. Because both cubes leave the table with a horizontal velocity, they will fall to the floor in the same time, which we find from H = !gt2; 0.90 m = !(9.80 m/s2)t2, which gives t = 0.429 s. Because the horizontal motion has constant velocity, we have x1 = v1t = (0.660 m/s)(0.429 s) = 0.28 m; x2 = v2t = (2.64 m/s)(0.429 s) = 1.1 m.

Page 29

Chapter 10 1

CHAPTER 10 - Rotational Motion About a Fixed Axis 1.

(a) (b) (c) (d) (e)

2.

The subtended angle in radians is the size of the object divided by the distance to the object:  = 2rSun/r; (0.5°)(p rad/180°) = 2rSun/(150106 km), which gives rSun ˜ 6.5105 km.

3.

We find the distance from  = h/r; (7.5°)(p rad/180°) = (300 m)/r; which gives r =

4.

30° = (30°)( rad/180°) = p/6 rad = 0.524 rad; 57° = (57°)(p rad/180°) = 19p/60 = 0.995 rad; 90° = (90°)(p rad/180°) = p/2 = 1.571 rad; 360° = (360°)(p rad/180°) = 2p = 6.283 rad; 420° = (420°)(p rad/180°) = 7p/3 = 7.330 rad.

2.3103 m.

From the definition of angular acceleration, we have  = ?/?t = [(20,000 rev/min)(2p rad/rev)/(60 s/min) – 0]/(5.0 min)(60 s/min) =

7.0 rad/s2.

5.

From the definition of angular velocity, we have  /t , and we use the time for each hand to turn through a complete circle, 2p rad. (a) second /t = (2p rad)/(60 s) = 0.105 rad/s. (b) minute /t = (2p rad)/(60 min)(60 s/min) = 1.75 10–3 rad/s. /t (c) hour = (2p rad)/(12 h)(60 min/h)(60 s/min) = 1.45 10–4 rad/s. (d) For each case, the angular velocity is constant, so the angular acceleration is zero.

6.

(a) The Earth moves one revolution around the Sun in one year, so we have orbit /t = (2p rad)/(1 yr)(3.16 107 s/yr) = 1.99 10–7 rad/s. (b) The Earth rotates one revolution in one day, so we have rotation /t = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 10–5 rad/s.

7.

All points will have the angular speed of the Earth:  = /t = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 10–5 rad/s. Their linear speed will depend on the distance from the rotation axis. (a) On the equator we have v = rEarth = (6.38106 m)(7.27 10–5 rad/s) = 464 m/s. (b) At a latitude of 66.5° the distance is rEarth cos 66.5°, so we have v = rEarth cos 66.5°  = (6.38106 m)(cos 66.5°)(7.27 10–5 rad/s) = (c) At a latitude of 40.0° the distance is rEarth cos 40.0°, so we have v = rEarth cos 40.0°  = (6.38106 m)(cos 40.0°)(7.27 10–5 rad/s) =

8.

185 m/s. 355 m/s.

The subtended angle in radians is the size of the object divided by the distance to the object. A pencil Page 1

Chapter 10 2

with a diameter of 6 mm will block out the Moon if it is held about 60 cm from the eye. For the angle subtended we have Moon = Dpencil/rpencil ˜ (0.6 cm)/(60 cm) ˜ 0.01 rad. We estimate the diameter of the Moon from Moon = DMoon/rMoon; 0.01 rad = DMoon/(3.8 105 km), which gives DMoon ˜ 4 103 km.

9.

(a)  = (2500 rev/min)(2p rad/rev)/(60 s/min) = 262 rad/s. (b) The linear speed of the point on the edge is the tangential speed: v = r = (0.175 m)(262 rad/s) = 46 m/s. Because the speed is constant, the tangential acceleration is zero. There will be a radial acceleration: aR = 2R = (262 rad/s)2(0.175 m) = 1.2 104 m/s2 radial.

10. (a) The angular speed of the merry-go-round is  = (1 rev)(2p rad/rev)/(4.0 s) = 1.57 rad/s. The linear speed of the child is the tangential speed: v = r = (1.2 m)(1.57 rad/s) = 1.9 m/s. (b) The child will have a radial acceleration: aR = 2R = (1.57 rad/s)2(1.2 m) = 3.0 m/s2 radial. 11. In each revolution the ball rolls a distance equal to its circumference, so we have L = N(pD); 3.5 m = (15.0)pD, which gives D = 0.074 m = 7.4 cm. 12. For a constant acceleration, we find the total angle the wheel turns from  = !(0 + )t = ![(210 rev/min + 380 rev/min)][(2p rad/rev)/(60 s/min)](6.5 s) = 201 rad = 32.0 rev. For each revolution the point on the edge will travel one circumference, so the total distance traveled is d = pD = (32.0 rev)p(0.60 m) = 60 m.  13. The final angular speed is  = (1 rpm)(2p rad/rev)/(60 s/min) = 0.105 rad/s. (a) We find the angular acceleration from  = ?/?t = (0.105 rad/s – 0)/(10.0 min)(60 s/min) = 1.75 10–4 rad/s2. (b) We find the angular speed after 5.0 min:  = 0 + t = 0 + (1.75 10–4 rad/s2)(5.0 min)(60 s/min) = 5.25 10–2 rad/s. At this time the radial acceleration of a point on the skin is aR = 2r = (5.25 10–2 rad/s)2(4.25 m) = 1.17 10–2 m/s2. The tangential acceleration is atan = r = (1.75 10–4 rad/s2)(4.25 m) = 7.44 10–4 m/s2.

14. We use the initial conditions of t = 0, 0 = 0, and 0. If we write the definition of angular acceleration as Page 2

Chapter 10 3

d =  dt, we integrate to find :  t t d =  d t =  d t; 0

0

0

 – 0 = t, or  = 0 + t; Eq. (10–9a). If we write the definition of angular velocity as d =  dt, we integrate to find :  0

 d =

t 0

 dt =

t 0

(0 + t) dt;

 – 0 = 0t + !t2 – 0, or  = 0t + !t2;

Eq. (10–9b).

15. (a) If there is no slipping, the linear tangential acceleration of the pottery wheel and the rubber wheel at the contact point must be the same: atan = R11 = R22 ; (2.0 cm)(7.2 rad/s2) = (25.0 cm)2 , which gives 2 = 0.58 rad/s2. (b) We find the time from  = 0 + t; (65 rev/min)(2p rad/rev)/(60 s/min) = 0 + (0.58 rad/s2)t, which gives t = 12 s. 16. (a) We find the instantaneous angular velocity by differentiating:  = d/dt = d[(6.0 rad/s)t – (8.0 rad/s2)t2 + (4.5 rad/s4)t4]/dt = (6.0 rad/s) – (16.0 rad/s2)t + (18.0 rad/s4)t3. (b) We find the instantaneous angular acceleration by differentiating:  = d/dt = d[(6.0 rad/s) – (16.0 rad/s2)t + (18.0 rad/s4)t3]/dt = – (16.0 rad/s2) + (54.0 rad/s4)t2. (c) At t = 3.0 s, we have  = (6.0 rad/s) – (16.0 rad/s2)(3.0 s) + (18.0 rad/s4)(3.0 s)3 = 444 rad/s = 4.4 102 rad/s;  = – (16.0 rad/s2) + (54.0 rad/s4)(3.0 s)2 = 470 rad/s2 = 4.7102 rad/s2. (d) The angular positions at the two times are  = (6.0 rad/s)(2.0 s) – (8.0 rad/s2)(2.0 s)2 + (4.5 rad/s4)(2.0 s)4 = 52 rad;  = (6.0 rad/s)(3.0 s) – (8.0 rad/s2)(3.0 s)2 + (4.5 rad/s4)(3.0 s)4 = 311 rad. We find the average angular velocity from av = /t = (311 rad – 52 rad)/(3.0 s – 2.0 s) = 2.6 102 rad/s. (e) The angular velocities at the two times are  = (6.0 rad/s) – (16.0 rad/s2)(2.0 s) + (18.0 rad/s4)(2.0 s)3 = 118 rad/s;  = 444 rad/s from part (c). We find the average angular acceleration from av = /t = (444 rad/s – 118 rad/s)/(3.0 s – 2.0 s) = 3.3 102 rad/s2. Note that the average angular velocity in part (d) is not !( + ). The angular acceleration is not constant!

17. (a) If we write the definition of angular acceleration as d =  dt, we integrate to find : Page 3

Chapter 10 4  0

d =

t 0

 dt =

t 0

2

(5.0 rad / s4)t – (3.5 rad / s 3)t dt ;

 = @(5.0 rad/s4)t3 – !(3.5 rad/s3)t2 = (1.67 rad/s4)t3 – (1.75 rad/s3)t2. (b) If we write the definition of angular velocity as d =  dt, we integrate to find :  t t 3 2 d  =  dt = (1.67 rad / s4 )t – (1.75 rad / s3 )t dt; 0

0

0

 = #(1.67 – @(1.75 rad/s3)t3 = (0.418 rad/s4)t4 – (0.583 rad/s3)t3. (c) At t = 2.0 s, we have  = (1.67 rad/s4)(2.0 s)3 – (1.75 rad/s3)(2.0 s)2 = 6.4 rad/s;  = (0.418 rad/s4)(2.0 s)4 – (0.583 rad/s3)(2.0 s)3 = 2.0 rad. rad/s4)t4

18. We find the initial and final angular velocities of the wheel from the rolling condition: 0 = v0/r = [(90.0 km/h)/(3.6 ks/h)]/(0.45 m) = 55.6 rad/s;  = v/r = [(60.0 km/h)/(3.6 ks/h)]/(0.45 m) = 37.0 rad/s. (a) We find the angular acceleration from 2 = 02 + 2; (37.0 rad/s)2 = (55.6 rad/s)2 + 2 (85 rev)(2p rad/rev), which gives = – 1.6 rad/s2. (b) We find the additional time from final =  + t; 0 = 37.0 rad/s + (– 1.6 rad/s2)t, which gives t = 23 s. 19. (a) The direction of 1 is along the axle. At the instant shown, z 1 is in the – x-direction.  The direction of 2 is up. At any time 2 is in the + z-direction. 2 (b) At the instant shown, we have the vector diagram shown.  We find the magnitude from x 1 2 = 12 + 22 = (50.0 rad/s)2 + (35.0 rad/s)2, which gives  = 61.0 rad/s. We find the angle from tan  = 2/1 = (35.0 rad/s)/(50.0 rad/s) = 0.700, so  = 35.0°. Thus the resultant angular velocity is  = 61.0 rad/s, 35.0° above – x-axis. (c) Because 2 is constant, only 1 will be changing. It will rotate at 2 in the xy-plane. If we let t = 0 at the instant shown, we have 1 = 1[– cos (2t)i – sin(2t)j]. From the definition of the angular acceleration we have  = d/dt = d(1 + 2)/dt = d1/dt = 12[+ sin (2t)i – cos(2t)j]. At t= 0, we get  = – 12 j = – (50.0 rad/s)(35.0 rad/s)j = – (1.75 103 rad/s2)j. 20. If  is the angle between the force and the surface of the door, we have 36 m · N. (a)  = LF sin  = (0.96 m)(38 N) sin 90° = 32 m · N. (b)  = LF sin  = (0.96 m)(38 N) sin 60.0° =

y L



Page 4

x F

Chapter 10 5

21. We use the force diagram shown. For the torques, we have 35 m · N. (a)  = Fd sin  = (700 N)(0.050 m) sin 90° = 30 m · N. (b)  = Fd sin  = (700 N)(0.050 m) sin 60° =

y

F x

d



22. Because the force and the moment arm are in the xy-plane, the torque will be along the z-axis. From the diagram, we see that the torque is  = FR = FyR = (+ 38.6 N)!(0.270 m) = 5.21 m · N CCW. If we curl our fingers counterclockwise, we see that our thumb points in the z-direction. Thus the torque is  = (5.21 m · N)k.

y

F

x

R

23. We assume clockwise motion, so the frictional torque is counterclockwise. If we take the clockwise direction as positive, we have net = rF1 – RF2 + RF3 – fr = (0.10 m)(35 N) – (0.20 m)(30 N) + (0.20 m)(20 N) – 0.30 m · N = 1.2 m · N (clockwise).

F1 F2 135°

r

fr R

F3 24. (a) We let L be the length of the beam and take clockwise as the positive direction. For the net torque about point C, we have C = !L(F1 sin 1) – !L(F3 sin 3)



= !(2.0 m)(50 N) sin 30° – !(2.0 m)(50 N) sin 60° = – 18 m · N (CCW). (b) For the net torque about point P, we have P = L(F1 sin 1) – !L(F2 sin 2)

1

F2

F1

2 C

= (2.0 m)(50 N) sin 30° – !(2.0 m)(60 N) sin 45° = 7.6 m · N (CW).

P

F3

3

Page 5

Chapter 10 6

25. The force to produce the required torque is Fwrench = /L = (90 m · N)/(0.26 m) = 3.5 102 N. Because this torque is balanced by the torque produced by the bolt on the wrench, an equal torque is produced on the bolt. Because there are six points where a force is applied to the bolt, we have Fbolt = (/r)/6 = (90 m · N)/6(0.0075 m) = 2.0 103 N. 26. The moment of inertia of a sphere about an axis through its center is I = (2/5)MR2 = (2/5)(12.0 kg)(0.80 m)2 = 3.1 kg · m2. 27. The torque produces the angular acceleration of the grindstone: motor = I = !MR2 ?/?t = !(1.4 kg)(0.20 m)2[(1800 rev/s)(2p rad/rev) – 0]/(6.0 s) =

53 m · N.

28. If M is the total mass and D is the effective separation, each atom has a mass !M and is !D from the axis. We find the distance D from I = 2(!M)(!D)2 = #MD2; 1.2 10–10 m. 1.9 10–46 kg · m2 = #(5.3 10–26 kg)D2, which gives D = 29. (a) Because we can ignore the mass of the rod, for the moment of inertia we have I = mballR2 = (2.4 kg)(1.2 m)2 = 3.5 kg · m2. (b) To produce constant angular velocity, the net torque must be zero: net = applied – friction = 0, or applied = FfrR = (0.020 N)(1.2 m) = 0.024 m · N. 30. (a) For the moment of inertia about the y-axis, we have Ia = ?miRi2 = md12 + Md12 + m(d2 – d1)2 + M(d2 – d1)2 = (1.8 kg)(0.50 m)2 + (3.1 kg)(0.50 m)2 + 6.1 kg · m2. (1.8 kg)(1.00 m)2 + (3.1 kg)(1.00 m)2 = (b) For the moment of inertia about the x-axis, all the masses are the same distance from the axis, so we have Ib = ?miRi2 = (2m + 2M)(!h)2 0.61 kg · m2. = [2(1.8 kg) + 2(3.1 kg)](0.25 m)2 = It will be harder to accelerate the array around the y-axis, because the moment of inertia is greater.

y m

d1

(180 rev)(2p rad/rev) = 0 + !(15.0 s)2, which gives  = 10.1 rad/s2. This acceleration is produced by the torque:  = I = ^MR2 ; 15 kg. 10.8 m · N = ^M(0.42 m)2(10.1 rad/s2), which gives M = 33. The torque produces the angular acceleration of the merry-go-round: net = I = !MR2 ?/?t = !(31,000 kg)(7.0 m)2(3.0 rad/s – 0)/(24 s) = Page 6

9.5 104 m · N.

m x

h M

31. For the moment of inertia of the rotor blades we have I = 3(@mbladeL2) = mbladeL2 = (160 kg)(3.75 m)2 = 2.25 103 kg · m2. We find the required torque from  = I = I( – 0)/t = (2.25 10–3 kg · m2)[(5.0 rev/s)(2p rad/rev) – 0]/(8.0 s) = 8.8 103 m · N. 32. We find the angular acceleration for the motion:  = 0t + !t2;

d2

M

Chapter 10 7

34. (a) For the moment of inertia about the axis, we have Ia = ?miRi2 = M(0) + M¬2 + M(2¬)2 + M(3¬)2 = 14M¬2 (b) The force will be minimum when the moment arm is maximum, which is when the force is perpendicular to the rod. Thus we have  = I; F(3¬) = (14M¬2), which gives F = 14M¬/3. (c) This minimum force will be perpendicular to the rod and to the axis. 35. Because the force varies with time, the angular acceleration will vary with time:  = FTR0 – fr = I; [(3.00 N/s)t – (0.20 N/s2)t2](0.330 m) – 1.10 m · N = (0.385 kg · m2), which gives  = – (2.86 rad/s2) + (2.57 rad/s3)t – (0.171 rad/s4)t2. We integrate this variable acceleration to find the angular velocity:  0

d =

t 0

 dt =

t 0

2

– (2.86 rad / s 2) + (2.57 rad / s 3)t – (0.171 rad / s 4)t dt;



= – (2.86 rad/s2)t + (1.29 rad/s3)t2 – (0.057 rad/s4)t3 = – (2.86 rad/s2)(8.0 s) + (1.29 rad/s3)(8.0 s)2 – (0.057 rad/s4)(8.0 s)3 = 30.2 rad. The linear speed of a point on the rim is the tangential speed: v = R0 = (0.330 m)(30.2 rad) = 10 m/s. 36. We find the acceleration from friction = I = !MR2; – 1.00 m · N = !(4.70 kg)(0.0780 m)2, which gives  = – 69.9 rad/s2. We find the angle turned through from 2 = 02 + 2; 0 = [(10,000 rpm)(2p rad/rev)/(60 s/min)]2 + 2(– 69.9 rad/s2), which gives = 7.84 103 rad = 1.25 103 rev. We can find the time from  = 0 + t; 0 = (10,000 rev/min)(2p rad/rev)/(60 s/min) + (– 69.9 rad/s2)t; which gives t = 37.



(a) + (b) We take the positive direction as the direction of the acceleration FN 1 FT1 for each block and clockwise for the pulley. We apply ?F = ma to each block to find the tensions: FT1 – m1g sin 1 = m1a; FT1 – (8.0 kg)(9.80 m/s2) sin 30° = 1 m 1g (8.0 kg)(1.00 m/s2), which gives FT1 = 47 N. m2g sin 2 – FT2 = m2a; (10.0 kg)(9.80 m/s2) sin 60° – FT2 = (10.0 kg)(1.00 m/s2), which gives FT2 = 75 N. (c) The net torque acting on the pulley is net = FT2R – FT1R = (75 N)(0.25 m) – (47 N)(0.25 m) = 7.0 m · N. We apply ? = Ito find the moment of inertia: net = I = I(a/R)   7.0 m · N = I(1.00 m/s2)/(0.25 m), which gives I = 1.7 kg · m2.

Page 7

15.0 s.

FT1

FT2 FT2

2

+ FN 2

m 2g

Chapter 10 8

38. (a) We take the positive directions indicated on the diagram. The linear acceleration of the blocks and the angular acceleration of the pulley are related: a = R0. We apply ?F = ma to each block: m1g – FT1 = m1a; FT2 – m2g = m2a. If we add the equations, we get FT2 – FT1 = (m1 + m2)a – (m1 – m2)g. We apply ? = Ito the pulley: net = I ;

 out of page

R0 FT1

FT2

FT1

FT1R0 – FT2R0 = !MR02 = !MR0a. When we use the result from the force equations, we get – (m1 + m2)a + (m1 – m2)g = !Ma;

FT2

+y m 1g

– (3.80 kg + 3.40 kg)a + (3.80 kg – 3.40 kg)(9.80 m/s2) = !(0.80 kg)a, 0.52 m/s2. which gives a = (b) We find the acceleration from the motion of the block: v = v0 + at; 0 = (0.20 m/s) + a(6.2 s), which gives a = – 0.0323 m/s2. We have an additional (negative) torque on the pulley from friction: FT1R0 – FT2R0 – fr = !MR02 = !MR0a.

m 2g

– (m1 + m2)a + (m1 – m2)g – (fr/R0) = !Ma; – (3.80 kg + 3.40 kg)(– 0.0323 m/s2) + (3.80 kg – 3.40 kg)(9.80 m/s2) – [fr/(0.030 m)] = which gives f =

!(0.80 kg)(– 0.0323 m/s2), 0.12 m · N.

39. Thin hoop (through center): Mk2 = MR02, which gives k = R0; Thin hoop (through diameter): Mk2 = (MR02/2) + (Mw2)/12, which gives k = [(R02/2) + (w2)/12]1/2; 2 2 Solid cylinder (through center): Mk = MR /2, which gives k = R/v2; k = [(R12 + R22)/2]1/2; Hollow cylinder (through center): Mk2 = M(R12 + R22)/2, which gives Uniform sphere (through center): Mk2 = 2Mr02/5, which gives k = (2r02/5]1/2; 2 2 Rod (through center): Mk = M¬ /12, which gives k = ¬/v12; k = ¬/v3; Rod (through end): Mk2 = M¬2/3, which gives k = [(¬2 + w2)/12]1/2. Plate (through center): Mk2 = M(¬2 + w2)/12, which gives

Page 8

+y

Chapter 10 9

40. (a) Although there are forces at the point A at the tabletop, they create no torque about point A. For the angular motion we have A = IA; mg!¬ cos  = @m¬2 = @m¬2(d/dt), or d/dt = (3g/2¬) cos . This equation contains three variables, but we can eliminate one from the definition of angular velocity:  = d/dt = – d/dt. If we use each side as a multiplier of the previous equation, we get  d/dt = (3g/2¬)(– d/dt) cos , or  d = – (3g/2¬) cos  d. We integrate to get ():  0

 d  = – (3g/ 2 )

 2/ 2

 0

 / 2

 ¬



 / 2

A

Ffr

cos  d  ;

= – (3g/ 2 ) sin 

FN

mg

= – (3g/ 2 )(sin  – 1),

which gives 2 = (3g/¬)(1 – sin ). (b) At the tabletop,  = 0, so we have 2 = (3g/¬)(1 – 0), or  = (3g/¬)1/2. The speed of the tip is v = ¬ = ¬(3g/¬)1/2 = (3g¬)1/2. 41. (a) The final angular velocity of the hammer is  = v/r = (29.0 m/s)/(2.00 m) = 14.5 rad/s. atan We find the angular acceleration from 2 = 02 + 2; a (14.5 rad/s)2 = 0 + 2 (4 rev)(2p rad/rev),  4.18 rad/s2. which gives = aR (b) For the tangential acceleration we have atan = r = (4.18 rad/s2)(2.00 m) = 8.37 m/s2. r (c) For the radial acceleration at release we have aR = 2r = (14.5 rad/s)2(2.00 m) = 421 m/s2. (d) The magnitude of the resultant acceleration of the hammer is a = (atan2 + aR2)1/2 = [(8.37 m/s2)2 + (421 m/s2)2]1/2 = 421 m/s2. This acceleration is provided by the net force exerted on the hammer, so we have Fnet = ma = (7.30 kg)(421 m/s2) = 3.07103 N. (e) We find the angle from 1.14° . tan  = atan/aR = (8.37 m/s2)/(421 m/s2) = 0.0199, which gives  =



P

42. From the parallel-axis theorem, we have I = ICM + Mh2 = M¬2/12 + M(¬/2)2 = M¬2/3. 43. (a) Because the two diagonals are equivalent and perpendicular, we can relate the moment of inertia about an axis through the center perpendicular to the plate to the moments of inertia about the diagonals: Ic = Ia + Ia ; M(s2 + s2)/12 = 2Ia , or Ia = Ms2/12. (b) Because the two axes parallel to the sides are equivalent and perpendicular, we can relate the moment of inertia about an axis through the center perpendicular to the plate to the moments of inertia about the b axes: Page 9

a

b

s

a

b

s

Chapter 10 10

Ic = Ib + Ib ; M(s2 + s2)/12 = 2Ib , or

Ib = Ms2/12.

44. We can consider the door to be made of a large number of horizontal rods. Because each rod is rotating about an axis perpendicular to its end, the moment of inertia of the door is I = @Mw2 = @(25 kg)(1.0 m)2 = 8.3 kg · m2. 45. (a) We use the parallel-axis theorem: I = 2(ICM + Mh2) = 2[(2MR02/5) + M(3R0/2)2] = (b) If we treat the spheres as point masses, we get I = 2[M(3R0/2)2] = 4.50MR02. The error is error = (I – I)/I = (4.50 – 5.30)/(5.30) = – 0.15 =

M

5.30MR02.

M R0

R0

R0

– 15%.

46. (a) If we treat the sphere as a point mass, we get Ia = MR02. (b) We use the parallel-axis theorem: Ib = ICM + Mh2 = (2MR12/5) + MR02. (c) The error is error = (Ia – Ib)/Ib = [MR02 – (2MR12/5) – MR02]/[(2MR12/5) + MR02] = [– (2R12/5)]/[(2R12/5) + R02] = – (0.40)(0.10 m)2/[(0.40)(0.10 m)2 + (1.0 m)2] = – 0.0040 =

M R0 R1

– 0.40%.

47. (a) We use the parallel-axis theorem: Ia = ICM + Mh2 = (MR02/2) + M(R0/4)2 = 9MR02/16. (b) If we consider two horizontal axes, they must have the same moment of inertia. We use the perpendicular-axis theorem: ICM = Ib + Ib ; MR02/2 = 2Ib , or Ib = MR02/4. (c) We can use the result from part (b) and the parallel-axis theorem: Ic = Ib + Mh2 = (MR02/4) + MR02 = 5MR02/4. 48. (a) The center of mass will be along the line from the center of M the wheel to the weight: xCM = md/(M + m) d x = (1.50 kg)(0.22 m)/(7.0 kg + 1.50 kg) m = 0.0388 m = 3.9 cm from the center. R1 (b) The moment of inertia is I = (MR02/2) + MxCM2 + m(d – xCM)2 = [(7.0 kg)(0.32 m)2/2] + (7.0 kg)(0.0388 m)2 + (1.50 kg)(0.22 m – 0.0388 m)2 = 0.42 kg · m2.

Page 10

Chapter 10 11

49. The density of the sphere is  = M/(4pR03/3) = 3M/4pR03. We consider a disk of thickness dx a distance x from the center perpendicular to the x-axis. The radius of this disk is y, where x2 + y2 = R02. The mass of the disk is dm =  py2 dx, so the moment of inertia of the disk is dI = (dm)y2/2. We integrate from x = – R0 to x = R0 to find the moment of inertia of the sphere:

 I=1 2 = =

y2 d m =

 2

R0 – R0

 2

y4 d x =

 2

R 40 – 2R02x 2 + x 4 d x =

R0 –R0

y x dx

x

2

2

R0 – x 2 d x

 4 R x – 2 R02x 3 + 1 x 5 3 5 2 0

R0 – R0

3 M  2R 50 – 4 R50 + 2 R 50 = 3 M 16R 50 = 2 MR 20. 3 5 5 4R 30 2 8R03 15

50. We select a differential element of the rod of length dx a distance x from the center of the rod. The element is equivalent to a point mass with a mass of dm = (M/¬) dx. We integrate from x = – ¬/2 to x = ¬/2 to find the moment of inertia of the rod:

I=

R0

x2

dm = M

/2 – /2

x2

dx = M x 3 3

/2 – /2

= 2M 3 2

3

/2 – / 2

x 2 dx

w/ 2 – w/ 2

dy+

/2 – / 2

dx

w/ 2 – w/ 2

¬

x dx

x

2

= M . 12

51. (a) We select a differential element of the plate with sides dx and dy at the location x, y. The element is equivalent to a point mass with a mass of dm = (M/¬w) dx dy. We integrate from x = – ¬/2 to x = ¬/2 and y = – w/2 to y = w/2 to find the moment of inertia of the plate: I = r2 dm = M x 2 + y2 d x d y w

= M w

y

y dy

¬

y w

x x

dx

y2 d y

/2 w/ 2 w/ 2 3 / 2 y3 = M x y +x 3 – w/ 2 w 3 – / 2 – w/ 2 – / 2 3

3 2 = M 2 w + 2 w = M + w2 . 8 8 12 3w (b) For an axis along the edge parallel to the y-axis, we can consider the plate to be an infinite number of rods of length ¬ and width dy with a mass (M/w) dy. The moment of inertia of each rod is dI = (¬2/12) dm = (M/w)(¬2/12) dx. When we add (integrate), we get I = (M/w)(¬2/12)w = M¬2/12. I = Mw2/12. Similarly, for the other edge we get

52. (a) The angular momentum is L = I = !MR02 = ! (2.8 kg)(0.18 m)2(1500 rev/min)(2p rad/rev)/(60 s/min) = (b) We find the required torque from Page 11

7.1 kg · m2/s.

Chapter 10 12

= ?L/?t = (0 – 7.1 kg · m2/s)/(7.0 s) =

– 1.0 m · N.

53. Because the diver in the air is an isolated system, for the conservation of angular momentum we have L = I11 = I22 , or I2/I1 = 1/2 ; 1/3.5 = 1/(2 rev/1.5 s), which gives 1 = 0.38 rev/s. 54. Because the skater is an isolated system, for the conservation of angular momentum we have L = I11 = I22 ; (4.6 kg · m2)(1.0 rev/2.0 s) = I2(3.0 rev/s), which gives I2 = 0.77 kg · m2. She accomplishes this by pulling her arms closer to her body. 55. (a) As the arms are raised some of the person’s mass is farther from the axis of rotation, so the moment of inertia has increased. For the isolated system of platform and person, the angular momentum is conserved. As the moment of inertia increases, the angular velocity must decrease. (b) If the mass and thus the moment of inertia of the platform can be neglected, for the conservation of angular momentum, we have L = I11 = I22 , or I2/I1 = 1/2 = (1.30 rad/s)/(0.80 rad/s) = 1.6. 56. (a) We approximate the mass distribution as a solid cylinder. The angular momentum is L = I = !mR2 = !(55 kg)(0.15 m)2[(3.5 rev/s)(2p rad/rev)] = 14 kg · m2/s. (b) If the arms do not move, the moment of inertia will not change. We find the torque from the change in angular momentum:  = L/?t = (0 – 14 kg · m2/s)/(5.0 s) = – 2.7 m · N. 57. (a) The Earth rotates one revolution in one day, so we have rotation /t = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 10–5 rad/s. If we assume the Earth is a uniform sphere, the angular momentum is Lrotation = Irotationrotation = (^Mr2)rotation = (^)(6.0 1024 kg)(6.4 106 m)2(7.27 10–5 rad/s) = 7.11033 kg · m2/s. (b) The Earth moves one revolution around the Sun in one year, so we have orbit /t = (2p rad)/(1 yr)(3.16 107 s/yr) = 1.99 10–7 rad/s. The angular momentum is Lrevolution = Irevolutionrevolution = (MR2)revolution = (6.0 1024 kg)(1.5 1011 m)2(1.99 10–7 rad/s) = 2.71040 kg · m2/s. 58. The angular momentum is L = I = [(M¬2/12) + 2m(¬/2)2] =

[(M/12) + (m/2)]¬2 .

59. When the people step onto the merry-go-round, they have no initial angular momentum. For the system of merry-go-round and people, angular momentum is conserved: Imerry-go-round0 + Ipeoplei = (Imerry-go-round + Ipeople) ; Imerry-go-round0 + 4mR2(0) = (Imerry-go-round + 4mR2) ; (1950 kg · m2)(0.80 rad/s) = [(1950 kg · m2) + 4(65 kg)(2.4 m)2], which gives  = 0.45 rad/s. If the people jump off in a radial direction with respect to the merry-go-round, they have the tangential velocity of the merry-go-round: v = R0. For the system of merry-go-round and people, angular momentum is conserved: (Imerry-go-round + Ipeople)0 = Imerry-go-round + Ipeople0 , which gives 0 = 0.80 rad/s. The angular speed of the merry-go-round does not change. Note that the angular momentum of the people will change when contact is made with the ground. Page 12

Chapter 10 13

60. (a) Because the woman is walking toward the center of the platform, her position is given by r = R – vt, and she will have the same angular velocity as the platform. Because friction of rotation is negligible, angular momentum will be conserved: L = (Iplatform + Iperson1)0 = (Iplatform + Iperson2) ; (!MR2 + mR2)0 = [!MR2 + m(R – vt)2], which gives

 = (!M + m)R20/[(!M + m)R2 – 2mRvt + mv2t2]. (b) When the woman gets to the center, conservation of angular momentum gives us L = (Iplatform + Iperson1)0 = (Iplatform + Iperson2)f ; (!MR2 + mR2)0 = [!MR2 + 0]f , which gives f = (!MR2 + mR2)0/!MR2 = (M + 2m)0/M. Note that this is the result we obtain if we put the time to reach the center, tf = R/v, in the result from part (a). 61. The work done increases the kinetic energy of the rotor: W = ?K = !I2 – 0 = !(4.25 10–2 kg · m2)[(10,000 rev/min)(2p rad/rev)/(60 s/min)]2 = 62. We find the power of the engine from P =  = (280 m · N)(4000 rev/min)(2p rad/rev)/(60 s/min)(746 W/hp) =

2.33104 J. 157 hp.

63. Conservation of angular momentum gives us Iii = Iff , or f/i = Ii/If . The ratio of kinetic energies is Kf/Ki = !Iff2/!Iii2 = (If/Ii)(f/i)2 = (If/Ii)(Ii/If)2 = Ii/If = (Ri/Rf)2 = [(7105 km)/(10 km)]2 = 5109. The increased kinetic energy came from the loss of gravitational potential energy. 64. We can use the slowing motion to find the frictional torque acting on the platform: fr = I = I ?/?t = !MR2(0 – 0)/t = – !(280 kg)(5.5 m)2(3.8 rev/s)(2p rad/rev)/(18 s) = – 5.62103 m · N. To maintain a constant angular speed, the motor must supply an equal and opposite torque. The power required is P =  = (5.62103 m · N)(3.8 rev/s)(2p rad/rev)/(746 W/hp) = 180 hp.

Page 13

Chapter 10 14

65. For the system of the two blocks and pulley, no work will be done by nonconservative forces. The rope ensures that each block has the same speed v and the angular speed of the pulley is  = v/R0. We choose the reference level for gravitational potential energy at the floor. The rotational inertia of the pulley is I = !MR02. For the work-energy principle we have Wnet = ?K + ?U; 0 = [(!

m1v2

+!

m2v2

+!

I2)

m2

– 0] + m1g(h – 0) + m2g(0 – h);

! +! + !( ! = (m2 – m1)gh; 2 ![m1 + m2 + !M]v = (m2 – m1)gh; ![35.0 kg + 38.0 kg + !(4.8 kg)]v2 = m1v2

v=

m2v2

R0

M

MR02)(v/R0)2

(38.0 kg – 35.0 kg)(9.80 1.4 m/s.

m/s2)(2.5

h

m), which gives

h=0

m1

66. Because friction can be ignored, the energy of the rod is conserved: Ki + Ui = Kf + Uf; 0 + Mghi = !I2 + Mghi ;

h=0 ¬



0 + (– Mg!¬ cos ) = !(@M¬2)2 + (– Mg!¬), which gives  = [(3g/¬)(1 – cos )]1/2. The speed of the free end is v = ¬ = [(3g¬)(1 – cos )]1/2.

Mg

67. (a) When we use the parallel-axis theorem for ¬ ¬ R the arms, we get m Ia = (MR2/2) + 2[(m¬2/12) + m(R + ¬/2)2] m = [(60 kg)(0.12 m)2/2] + M 2(5.0 kg)[(0.60)2/12 + (0.12 m + 0.30 m)2] 2 = 2.5 kg · m . (b) When the arms are at the sides, all of their mass is the same distance from the axis, so we get (a) Ib = (MR2/2) + 2mR2 = [(M/2) + 2m]R2 = [(60 kg)/2 + 2(5.0 kg)](0.12 m)2 = 0.58 kg · m2.  (c) From the conservation of angular momentum about the vertical axis, we have Iaa = Ibb ; 0.35 s. (2.5 kg · m2)(1 rev/1.5 s) = (0.58 kg · m2)(1 rev/tb), which givestb = (d) The change in kinetic energy is ?K = !Iaa2 – !Ibb2

R ¬

¬ M

m

(b)

= !(2.5 kg · m2)(2p rad/1.5 s)2 – !(0.58 kg · m2)(2p rad/0.35 s)2 = – 72 J. (e) Because the kinetic energy decreases, it will be easier to lift your arms when rotating. In the rotating system, the arms tend to move away from the center of rotation. 68. (a) The kinetic energy of the system is K = !I2 = !(m1r12 + m2r22)2 = ![(4.00 kg)(0.250 = 14.0 J.

m)2



+ (3.00 kg)(0.250 Page 14

m)2](8.00

rad/s)2

m1

r1

r2

m2

m

Chapter 10 15



(b) The net horizontal forces produce the radial acceleration: F = m1r12 = (4.00 kg)(0.250 m)(8.00 rad/s)2 = 64.0 N; F = m2r22 = (3.00 kg)(0.250 m)(8.00 rad/s)2 = 48.0 N. Note that because these forces are not equal, there will be a horizontal force on the axle. The force of gravity on each mass is balanced by a vertical force from the rod. (c) With the origin at m1 , the position of the center of mass is xCM = m2(r1 + r2)/(m1 + m2) = (3.00 kg)(0.500 m)/(4.00 kg + 3.00 kg) = 0.214 m. The kinetic energy is now K = !I2 = !(m1r12 + m2r22)2 = 2 = The net horizontal forces are F = m1r12 = (4.00 kg)(0.214 m)(8.00 rad/s)2 = 54.9 N; 2 F = m2r2 = (3.00 kg)(0.500 m – 0.214 m)(8.00 rad/s)2 = 54.9 N. Note that there will not be a horizontal force on the axle.

13.7 J.

69. If the cylinder rolls without slipping, the speed of the center of mass is v = R. Because energy is conserved, from the top to the bottom of the incline we have 0 = ?K + ?U; 0 = [(!mv2 + !I2) – 0] + (0 – mgh), or !mv2 + !(!mR2)(v/R)2 = mgh, which gives v = ()gh)1/2 = [)(9.80 m/s2)(11.8 m)]1/2 = 12.4 m/s. 70. (a) The Earth rotates one revolution in one day, so we have rotation /t = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 10–5 rad/s. The kinetic energy of rotation is Krotation = !Irotationrotation2 = !(^Mr2)rotation2 = !(^)(6.0 1024 kg)(6.4 106 m)2(7.27 10–5 rad/s)2 = 2.61029 J. (b) The Earth moves one revolution around the Sun in one year, so we have orbit /t = (2p rad)/(1 yr)(3.16 107 s/yr) = 1.99 10–7 rad/s. The kinetic energy of revolution is Krevolution = !Irevolutionrevolution2 = !(MR2)revolution2 = !(6.0 1024 kg)(1.5 1011 m)2(1.99 10–7 rad/s)2 = 2.71033 J. We see that the kinetic energy of revolution is much greater than that of rotation, so the total energy is Ktotal = 2.71033 J. 71. The angular speed of the rolling ball is  = v/R. The total kinetic energy will have a translational term for the center of mass and a term for the rotational energy about the center of mass: Ktotal = Ktrans + Krot = !Mv2 + !I2 = !Mv2 + !(^MR2)(v/R)2 = 7Mv2/10 = 7(7.3 kg)(5.3 m/s)2/10 = 1.4 102 J.

72. (a) Because there is no acceleration of the center of mass, the net force must be zero, so we have FT = Mg. (b) You do work because the tension force you apply moves a distance. This work increases the kinetic energy of the spool: W = ?K + ?U; W = (!I2 – 0) + 0 = !(!mR2)2 = #mR22.

Page 15

FT Chapter 10 16

+y

R

Mg

73. (a) If the pipe rolls without slipping, the speed of the center of mass is v = R. Because energy is conserved, from the top to the bottom of the incline we have 0 = ?K + ?U; FN D R 0 = [(!mv2 + !I2) – 0] + (0 – mgD sin ), or !mv2 + !(mR2)(v/R)2 = mv2 = mgD sin , which gives Ffr v = (gD sin )1/2 = [(9.80 m/s2)(5.60 m) sin 21.5°]1/2 = 4.48 m/s. mg  y=0 (b) The kinetic energy is K = !mv2 + !I2 = !mv2 + !(mR2)(v/R)2 = mv2 = (0.0600 kg)(4.48 m/s)2 = 1.21 J.  (c) For the rolling pipe, the acceleration of the center of mass and the angular acceleration are related: a = R. From the force diagram we have FN = mg cos , and mg sin  – Ffr = ma. For the angular acceleration around the center of mass, we have FfrR = I = (mR2)(a/R), or Ffr = ma. When we use this in the force equation, we get Ffr = !mg sin . If the pipe is not to slip, the friction force must be less than the maximum static friction: Ffr = sFN;

!mg sin  = smg cos ; ! tan 21.5° = s, which gives

s = 0.197.

Page 16

Chapter 10 17

74. (a) We use the accelerating reference frame of the car. The ball is rolling, so aball = r. In the accelerating frame, we must add a fictitious force – ma to the center of mass of the ball. Thus for forces and torques, we have Ffr – ma = maball ; – Ffrr = (2mr2/5) = 2mraball/5. When we combine these, we get – 2maball/5 – ma = maball , which gives aball = – 5a/7 (backward). (b) Relative to the ground we have aball = aball + a = (– 5a/7) + a = 2a/7 (forward).

75. If the ball rolls without slipping, the speed of the center of mass is v = r0. Because energy is conserved, for the motion from A to B we have 0 = ?K + ?U; 0 = [(!mv2 + !I2) – 0] + (mgr0 – mgR0), or

+ a – ma FN

r

Ffr

Mg

A R0 C

!mv2 + !(^mr02)(v/r0)2 = mg(R0 – r0), which gives v = [10g(R0 – r0)/7]1/2.

y=0

76. (a) If the ball rolls without slipping, the speed of the center of mass is v = r0. Because energy is conserved, for the motion from A to C we have 0 = ?K + ?U; 0 = (!mvC2 + !IC2) – 0 + mg[R0 – (R0 – r0) cos ] – mgR0 , or

!mvC2 + !(^mr02)(vC/r0)2 = mg(R0 – r0) cos ,

B

A



R0

C y=0

B

D

x=0

which gives vC = [10g(R0 – r0)(cos )/7]1/2 = [10(9.80 m/s2)(0.260 m – 0.015 m)(cos 45°)/7]1/2 = 1.56 m/s. (b) The initial location of the ball when it leaves the track is x0 = (R0 – r0) sin  = (0.260 m – 0.015 m) sin 45° = 0.173 m; y0 = R0 – (R0 – r0) cos  = 0.260 m – (0.260 m – 0.015 m) cos 45° = 0.0868 m; The ball hits the ground when its center is r0 above the ground. For the projectile motion , we have y = y0 + (vC cos )t + !(–g)t2; 0.015 m = 0.0868 m + (1.56 m/s)(cos 45°)t – !(9.80 m/s2)t2, or (4.90 m/s2)t2 – (1.10 m/s)t – 0.0718 m = 0. This quadratic equation has the positive solution t = 0.278 s. For the horizontal motion, we have D = x0 + (vC sin )t = 0.173 m + (1.56 m/s)(sin 45°)(0.278 s) = 0.48 m. Page 17

Chapter 10 18

77. We let M be the mass of the car without the wheels and m the mass of a wheel of radius r. (a) We assume the wheels are rolling, so v = r. The total kinetic energy is the sum of the translational energy of the car and the rotational energy of the wheels: K = !Mv2 + 4(!mv2 + !I2) = ![(M + 4m)v2 + 4(!mr2)(v/r)2] = !(M + 4m + 2m)v2 = ![(1100 kg) + 2(35 kg)][(100 km/h)/(3.6 ks/h)]2 = 4.5105 J. (b) The fraction of the kinetic energy in the wheels is fraction = 4(!mv2 + !I2)/K = [2mv2 + 4(!mr2)(v/r)2]/K = 3mv2/K = 3(35 kg)[(100 km/h)/(3.6 ks/h)]2/(4.5105 J) = 0.18 (18%). (c) When the car is towed, while the car is accelerating, there will be a friction force opposing the linear motion of a wheel, causing the angular acceleration that starts the wheel turning. Assuming no slipping, for the angular motion of a wheel, we have  = I = !mr2(a/r); Ffrr = !mra, or Ffr = !ma. For the linear motion of the car, we have Ftow – 4Ffr = (M + 4m)a; Ftow – 4(!ma) = (M + 4m)a; 1.71 m/s2. 2000 N – 4(!)(35 kg)a = (1100 kg)a, which gives a = (d) If we ignore the rotation of the tires (set I = 0), for the linear motion of the car we have Ftow = (M + 4m)a; 2000 N = (1100 kg)a, which gives a = 1.82 m/s2. Thus the error is error = (1.82 m/s2 – 1.71 m/s2)(100)/(1.71 m/s2) = 6.4%. 78. From Example 10–23, we see that the acceleration of the center of mass is %g and is constant. Thus we can find the speed from v2 = v02 + 2ah; v2 = 0 + 2(%g)h, which gives v =

()gh)1/2.

79. (a) From Example 10–24, we see that while the ball is slipping, the acceleration of the center of mass is – kg and is constant. The ball slips for a time T = 2v0/7kg, so we find the distance from xslip = v0t + !at2 = v0(2v0/7kg) + !(– kg)(2v0/7kg)2 = 12v02/49kg. (b) Once the ball starts rolling at time T, the linear speed is constant v = v0 + at = v0 + (– kg)(2v0/7kg) = 5v0/7. We can find the angular speed from  = 0 + t, or from  = v/R = (5v0/7)/R = 5v0/7R. 80. (a) Because the surface of the wheel that initially touches the ground is moving backward, the friction force on the wheel is forward to oppose the slipping. (b) For the linear motion of the center of mass, we have ?F = ma; kMg = Ma, which gives a = kg. Thus the speed is v = v0 + at = 0 + (kg)t = kgt. For the angular motion about the center of mass, we have ? = ICMCM; – kMgR = MR2, which gives  = – 2kg/R. Page 18

Chapter 10 19

Thus the angular speed is  = 0 + t = 0 + (– 2kg/R)t. We see that v increases and  decreases. The wheel will start rolling when v = R; kgT = [0 + (– 2kg/R)T]R, which gives T = R0/3kg. (c) Once the ball starts rolling at time T, the linear speed is constant v = v0 + at = 0 + (kg)(R0/3kg) = R0/3.

81. (a) For the linear motion of the center of mass, we have ?F = ma; Mg sin  – kMg cos  = Ma, which gives a = (sin  – kcos )g = [sin 33.0° – (0.10) cos 33.0°](9.80 m/s2) 4.5 m/s2. = 4.52 m/s2= For the angular motion about the center of mass, we have ? = ICMCM;

 FN H

Ffr

Mg (kMg cos )R0 = ^MR02, which gives  = 5(kg cos )/2R0 = 5(0.10)(9.80 m/s2)(cos 33.0°)/2(0.110 m) 19 rad/s2. = 18.7 rad/s2 = (b) The acceleration of the center of mass is constant. Thus we can find the speed from v2 = v02 + 2ax; v2 = 0 + 2(4.52 m/s2)(2.0 m)/sin 33.0°, which gives v = 5.76 m/s = 5.8 m/s. (c) We find the time to reach the bottom of the incline from v = v0 + at; 5.76 m/s = 0 + (4.52 m/s2)t, which gives t = 1.28 s. At this time the angular speed is  = 0 + t = 0 + (18.7 rad/s2)(1.28 s) = 23.9 rad/s. Page 19

a



Chapter 10 20

The total kinetic energy at the bottom is K = !Mv2 + !ICM2 = !(0.850 kg)(5.76 m/s)2 + 2 = 15.3 J. (d) The loss in mechanical energy is – ?E = – ?K – ?U = Ki – Kf + Ui – Uf = (0 – 15.3 J) + (0.850 kg)(9.80 m/s2)(2.0 m) – 0 = 1.4 J. (e) When there is no slipping, we use the results from Example 10-21: a = 5g(sin )/7 = 5(9.80 m/s2)(sin 33.0°)/7 = 3.81 m/s2 = 3.8 m/s2. v = (10gH/7)1/2 = [10(9.80 m/s2)(2.0 m)/7]1/2 = 5.29 m/s = 5.3 m/s.  = v/R0 = (5.29 m/s)/(0.110 m) = 48.1 rad/s. = !Mv2 + !ICM2 = !(0.850 kg)(5.29 m/s)2 + 2 = 16.7 J. – ?E = – ?K – ?U = Ki – Kf + Ui – Uf = (0 – 16.7 J) + (0.850 kg)(9.80 m/s2)(2.0 m) – 0 = 0. As expected, there is no loss in mechanical energy when rolling. (f) For a box sliding down the incline, the linear equations will be the same as for the slipping sphere. Thus the acceleration and speed will be the same: a = 4.5 m/s2; v = 5.8 m/s. There will be no rotational kinetic energy, so the kinetic energy will be less: K = !Mv2 = !(0.850 kg)(5.76 m/s)2 = 14.1 J. K

82. We take the positive direction in the direction of motion. For the linear motion of the center of mass, we have ?F = ma; – Ffr = Ma. For the angular motion about the center of mass, we have ? = ICMCM; – ¬FN + Ffrr0 = ^Mr02. When we use the rolling condition, a = r0, and the result from the linear motion, we get – ¬FN + Ffrr0 = ^Mr02 = ^Mr0a = – ^Ffrr0, which gives ¬FN = 7Ffrr0/5, or N = 7fr/5.

+ FN ¬ r0 Ffr

83. The subtended angle in radians is the size of the object divided by the distance to the object: Sun = 2RSun/rSun = 2(6.96105 km)/(149.6106 km) = 9.3010–3 rad (0.53°); 3 3 Moon = 2RMoon/rMoon = 2(1.7410 km)/(38410 km) = 9.0610–3 rad (0.52°). These are almost equal, so eclipses can occur. 84. (a) We choose the reference level for gravitational potential energy at the initial position at the bottom of the incline. The kinetic energy will be the translational energy of the center of mass and the rotational energy about the center of mass. Because there is no work done by friction while the cylinder is rolling, for the work-energy principle we have Wnet = ?K + ?U;

d



R

0 = [0 – (!Mv2 + !I2)] + Mg(d sin  – 0). Because the cylinder is rolling, v = R. For a hoop the moment of inertia is MR2. Thus we get !Mv2 + !(MR2)(v2/R2) = Mv2 = Mgd sin ; 1.23 m. (2.10 m/s)2 = (9.80 m/s2)d sin 21.5°, which gives d = (b) We find the time to go up the incline from the linear motion (which has constant acceleration): d = !(v + 0)t; Page 20

y=0

Chapter 10 21

1.23 m = !(2.10 m/s)t, which gives t = 1.17 s. Because there are no losses to friction, the time to go up the incline will be the same as the time to return. The total time will be T = 2t = 2.34 s. 85. If there is no slipping, the tangential speed of the outer edge of the wheel is the tangential speed of the outer edge of the roller: v = R11 = R22 , which gives 1/2 = R2/R1. 86. For the same side of the Moon to always face the Earth, the angular velocity of the orbital motion and the angular velocity of the spinning motion must be the same. We use r for the radius of the Moon and R for the distance from the Earth to the Moon. For the ratio of angular momenta, we have Lspin/Lorbital = ^Mr2/MR2 = ^(r/R)2 = ^[(1.74 106 m)/(3.84 108 m)]2 =

8.21 10–6.

87. Because the spool is rolling, vCM = R. The velocity of the rope at the top of the spool, which is also the velocity of the person, is v = R + vCM = 2R = 2vCM. Thus in the time it takes for the person to walk a distance ¬, the center of mass will move a distance ¬/2. Therefore, the length of rope that unwinds is ¬rope = ¬ – (¬/2) = ¬/2. The center of mass will move a distance ¬/2.

 v CM R



88. After 3.0 s the velocity of the CM will be vCM = 0 + aCMt = (1.00 m/s2)(3.0 s) = 3.0 m/s. Because the wheel is rolling, vCM = R. The velocity at the top of the wheel is v = R + vCM = 2R = 2vCM = 2(3.0 m/s) = 6.0 m/s.

v

v v CM R

89. Initially there is no angular momentum about the vertical axis. Because there are no torques about this vertical axis for the system of platform and wheel, the angular momentum about the vertical axis is zero and conserved. We choose up for the positive direction. (a) From the conservation of angular momentum about the vertical axis, we have L = 0 = IPP + IWW , which givesP = – (IW/IP)W (down).  (b) From the conservation of angular momentum about the vertical axis, we have L = 0 = IPP + IWW cos 60°, which givesP = – (IW/2IP)W (down). (c) From the conservation of angular momentum about the vertical axis, we have L = 0 = IPP + IW(– W) , which givesP = (IW/IP)W (up). (d) Because the total angular momentum is zero, when the wheel stops, the platform and person must also stop. P = 0. Thus 90. We select a differential element of the rod of length dx a distance x from the center of the rod. Because the mass per unit length changes uniformly from 0 to 20 over the length of the rod, its rate of change is 0/¬. At x = 0 the mass per unit length is *0 , so the mass per unit length Page 21

y

0

20

¬

x x

dx

Chapter 10 22

as a function of x is  = (0/¬)x + *0 , and the mass of the element is dm =  dx. The total mass of the rod is ¬/ 2  ¬/ 2 M=  dx = ( 0 / ¬)x + 32  0 d x = 32  0¬. –¬/ 2 –¬/ 2 We integrate from x = – ¬/2 to x = ¬/2 to find the moment of inertia of the rod:

I=

¬/ 2 –¬/ 2

x 2 d x =

4 = 0 x 4¬

¬/ 2

/2 – /2

3 +3 x 2 3 – ¬/ 2

(  0 / ¬)x + 32  0 x 2 d x ¬/ 2 –¬/ 2

= 0 0 +

3

¬ = 8

 0¬ 3 M¬ 2 = . 8 12

91. (a) If we let d represent the spacing of the teeth, which is the same on both sprockets, we can relate the number of teeth to the radius for each wheel: NFd = 2pRF , and NRd = 2pRR , which gives NF/NR = RF/RR. The linear speed of the chain is the tangential speed for each socket: v = RFF = RRR. Thus we have R/F = RF/RR = NF/NR. (b) For the given data we have R/F = 52/13 = 4.0. (c) For the given data we have R/F = 42/28 = 1.5.

92. (a) The angular acceleration of the ball-arm system is

 = atan/d1 = (7.0 m/s2)/(0.30 m) = 23.3 rad/s2. Because we ignore the mass of the arm, for the moment of inertia we have I = mballd12 = (1.00 kg)(0.30 m)2 = 0.090 kg · m2. Thus we find the required torque from  = I  = (0.090 kg · m2)(23.3 rad/s2) = 2.1 m · N. (b) Because the force from the triceps muscle is perpendicular to the line from the axis, we find the force from F = /d2 = (2.1 m · N)/(0.025 m) = 84 N.

Page 22

Chapter 10 23

93. (a) The final angular velocity of the arm and ball is  = v/d1 = (10.0 m/s)/(0.30 m) = 33.3 rad/s. We find the angular acceleration from  = 0 + t; 33.3 rad/s = 0 + (0.22 s), which gives = 1.5 102 rad/s2. (b) For the moment of inertia of the ball and arm we have I = mballd12 + @marmd12 = (1.00 kg)(0.30 + @(3.4 kg)(0.30 = 0.192 Because the force from the triceps muscle is perpendicular to the line from the axis, we find the force from F = /d2 = I /d2 = (0.192 kg · m2)(1.52 102 rad/s2)/(0.025 m) = m)2

m)2

m ball

d1

m arm A

d2

F

kg · m2.

1.2 103 N.

94. For the accelerating motion, we have applied = I = I( – 0)/t = @mL2( – 0)/t



= @(2.2 kg)(0.95 m)2[(3.0 rev/s)(2p rad/rev) – 0]/(0.20 s) =

62 m · N.

95. (a) We find the angular acceleration from  = 0t + !t2; 

(20 rev)(2p rad/rev) = 0 + ![(1 min)(60 s/min)]2, which gives (b) We find the final angular speed from  = 0 + t = 0 + (0.070 rad/s2)(60 s) = 4.2 rad/s = 40 rpm.

 = 0.070 rad/s2.

96. If there are no external torques, angular momentum will be conserved: L = Idisk1 + Irod(0) = (Idisk + Irod)2 ;

(Mr2/2)1 = [(Mr2/2) + (ML2/12)]2 = {(Mr2/2) + [M(2r)2/12]}2 , which gives 2 = (3/5)1 = (3/5)(7.0 rev/s) = 4.2 rev/s.

97. We find the required constant angular acceleration from  = 0 + t; (30 rev/min)(2p rad/rev)(1 min/60 s) = 0 + (5.0 min)(60 s/min), which gives  = 0.0105 rad/s2. The moment of inertia of the solid cylinder is !MR2. Because we have four forces creating the torque that produces the required acceleration, we have  = I; 4FR = !MR2, or F = MR/8 = (2000 kg)(3.0 m)(0.0105 rad/s2)/8 = 7.9 N.

98. We choose the reference level for gravitational potential energy at the bottom. The kinetic energy will be the translational energy of the center of mass and the rotational energy about the center of mass. (a) Because there is no work done by friction while the sphere is rolling, for the work-energy principle we have Wnet = ?K + ?U; 0 = (!Mv2 + !I2 – 0) + Mg(0 – d sin ). Because the sphere is rolling, v = R. The rotational inertia is ^MR2. Thus we get !Mv2 + !(^MR2)(v/R)2 = Mgd sin , which gives v = (10gd sin /7)1/2, and  = (10gd sin /7)1/2/R. Page 23

d R

y=0



Chapter 10 24

When we use the given data, we get v = [10(9.80 m/s2)(10.0 m) sin 30°/7]1/2 = 8.37 m/s,  = v/R = (8.37 m/s)/(0.200 m) = 41.8 rad/s. (b) For the ratio of kinetic energies we have Ktrans/Krot = !Mv2/!I2

and

2.50. = Mv2/(^MR2)(v/R2)2 = (c) None of the answers depends on the mass; the rotational speed depends on the radius. 99. We convert the speed: (90 km/h)/(3.6 ks/h) = 25 m/s. (a) We assume that the linear kinetic energy that the automobile acquires during each acceleration is not regained when the automobile slows down. For the work-energy principle applied to the 300-km trip we have Wnet = ?K + ?U; – FfrD = [20(!Mv2) – Kflywheel] + 0, or Kflywheel = (20)!(1400 kg)(25 m/s)2 + (500 N)(300 103 m) = 1.6 108 J. (b) We find the angular velocity of the flywheel from Kflywheel = !I2 = !(!mR2)2; 2.2 103 rad/s. 1.6 108 J = #(240 kg)(0.75 m)22, which gives  = (c) We find the time from t = Kflywheel /P = (1.6 108 J)/(150 hp)(746 W/hp) = 1.43 103 s = 24 min. 100. Each wheel supports one-quarter of the weight. For the wheels to spin, the applied torque must be greater than the maximum frictional torque produced by the static friction from the pavement: applied = Ffrr = sFNr = (0.75)#(1250 kg)(9.80 m/s2)(0.33 m) = 7.6 102 m · N.

applied R

Mg /4

FN

101. We choose the clockwise direction as positive. (a) With the force acting, we write  = I about the axis from the force diagram for the roll: Fr – fr = I1 ; (3.2 N)(0.076 m) – 0.11 m · N = (2.9 10–3 kg · m2)1 , which gives 1 = 45.9 rad/s2. We find the angle turned while the force is acting from 1 = 0t + !1t12 = 0 + !(45.9 rad/s2)(1.3 s)2 = 38.8 rad. The length of paper that unrolls during this time is Page 24

fr r

F

Ffr

Chapter 10 25

s1 = r1 = (0.076 m)(38.8 rad) = 2.9 m. (b) With no force acting, we write  = I about the axis from the force diagram for the roll: – fr = I2 ; – 0.11 m · N = (2.9 10–3 kg · m2)2 , which gives 2 = – 37.9 rad/s2. The initial velocity for this motion is the final velocity from part (a): 1 = 0 + 1t1 = 0 + (45.9 rad/s2)(1.3 s) = 59.7 rad/s. We find the angle turned after the force is removed from 22 = 12 + 222 ; 0 = (59.7 rad/s)2 + 2(– 37.9 rad/s2)2 , which gives 2 = 47.0 rad. The length of paper that unrolls during this time is s2 = r2 = (0.076 m)(47.0 rad) = 3.6 m.

102. We assume that m2 > m1 and choose the coordinates shown on the force diagrams. Note that we take the positive direction in the direction of the acceleration for each object. Because the linear acceleration of the masses is the tangential acceleration of the rim of the pulley, we have a = atan = R0. We write Fy = may for m2: m2g – FT2 = m2a. We write Fy = may for m1: FT1 – m1g = m1a. We write  = I for the pulley about its axle: FT2R0 – FT1R0 = I = Ia/R0 , or FT2 – FT1 = Ia/R02. If we add the two force equations, we get FT1 – FT2 = (m1 + m2)a + (m1 – m2)g. When we add these two equations, we get a = (m2 – m1)g/(m1 + m2 + I/R02). If the moment of inertia of the pulley is ignored, from the torque equation, we see that the two tensions will be equal. For the acceleration, we set I = 0 and get a0 = (m2 – m1)g/(m1 + m2). Thus we see that a0 > a.

 into page

R FT1

FT2

FT1

FT2

+y m 1g

m 2g

+y

103. (a) By walking to the edge, the moment of inertia of the person changes. Because the system of person and platform is isolated, angular momentum will be conserved: L = (Iplatform + Iperson1)1 = (Iplatform + Iperson2)2 ; [1000 kg · m2 + (75 kg)(0)2](2.0 rad/s) = [1000 kg · m2 + (75 kg)(3.0 m)2]2 , which gives 2 = 1.2 rad/s. (b) For the kinetic energies, we have K1 = !(Iplatform + Iperson1)12 = !(1000 kg · m2 + 0)(2.0 rad/s)2 = 2.0 103 J; K2 = !(Iplatform + Iperson2)22 = 2 = Thus there is a loss of 8.0 102 J, a decrease of 40%. Page 25

1.2 103 J.

Chapter 10 26

104. The wheel will roll about the contact point A. We write  = I about the point A: F(R – h) + FN1[R2 – (R – h)2]1/2 – Mg[R2 – (R – h)2]1/2 = IA. When the cylinder does roll over the curb, contact with the ground is lost and FN1 = 0. Thus we get

F

F = {IA + Mg[R2 – (R – h)2]1/2}/(R – h) = [IA/(R – h)] + [Mg(2Rh – h2)1/2/(R – h)]. The minimum force occurs when  = 0: Fmin = Mg[h(2R – h)]1/2/(R – h).

105. (a) The speed of the block is the tangential speed of the cylinder, so v = R. For the system of block and cylinder, if there is no friction, energy is conserved. After the block has moved a distance D, we have 0 = ?K + ?U; 0 = [(!mv2 + !I12) – 0] + (0 – mgD sin );

! + !( ! = mgD sin ; !mv2 + #Mv2 = mgD sin ; [!(3.0 kg) + #(30 kg)]v2 = mv2

MR2)(v/R)2

FN 2

R FN 1

A

h

Mg

FN 1 FT

+

R

Ffr1 Mg

FT

FN 2

Ffr2 mg

m/s2)(1.80



m) sin 30°, (3.0 kg)(9.80 which gives v= 1.7 m/s. (b) From the force diagram for the block we see that Ffr2 = FN2 = mg cos . For the cylinder, we assume a single vertical normal force. Because there is no linear acceleration of the center of mass, we have FN1 = Mg + FT sin . Because the block accelerates down the plane, FT < mg sin , and m = 0.1M. Thus we ignore the contribution of FT sin  (and we are justified in ignoring the horizontal component of FN1) to get FN1 ˜ Mg, and Ffr1 = FN1 = Mg. When the block moves a distance D, the surface of the rotating cylinder will move a distance D through the depression. For the work-energy theorem we have Wfr = ?K + ?U; – Ffr1D – Ffr2D = [(!mv2 + !I12) – 0] + (0 – mgD sin ); – (M + m cos )gD = !mv2 + #Mv2 – mgD sin ; – (0.035)[30 kg + (3.0 kg) cos 30°](9.80 m/s2)(1.80 m) = [!(3.0 kg) + #(30 kg)]v2 – (3.0 kg)(9.80 m/s2)(1.80 m) sin 30°, which gives v= 0.84 m/s.

Page 26

Chapter 10 27

106. The mass removed from the hole is mhole = (M/pR02)pR12 = (R1/R0)2M. When we subtract the moment of inertia of the hole about the center (found using the parallel-axis theorem), the moment of inertia about the center is I = Isolid – Ihole = !MR02 – (!mholeR12 + mholeh2) = !MR02 – M(R1/R0)2(!R12 + h2) =

R0 C h

!M[R02 – (R14/R02) – (2R12h2/R02)].

R1

107. (a) At the top of the loop, if the marble stays on the track, the normal force and the weight provide the radial acceleration: FN + mg = mv2/R0. The minimum value of the normal force is zero, so we find the minimum speed at the top from mg = mvmin2/R0 , or vmin2 = gR0. Because the marble is rolling, the corresponding angular velocity at the top is min = vmin/r0 , so the minimum kinetic energy at the top is Kmin = !mvmin2 + !Imin2 = !mvmin2 + !(^mr02)(vmin/r0)2 = 7mvmin2/10 =

7mgR0/10. If there are no frictional losses, we use energy conservation from the release point to the highest point of the loop: Ki + Ui = Kf + Uf; 0 + mghmin = Kmin + mg2R0 = 7mgR0/10 + 2mgR0 , which gives hmin = 2.7R0. (b) At the top of the loop, if the marble stays on the track, the normal force and the weight provide the radial acceleration: FN + mg = mv2/(R0 – r0). The minimum value of the normal force is zero, so we find the minimum speed at the top from mg = mvmin2/(R0 – r0), or vmin2 = g(R0 – r0). Because the marble is rolling, the corresponding angular velocity at the top is min = vmin/r0 , so the minimum kinetic energy at the top is Kmin = !mvmin2 + !Imin2 = !mvmin2 + !(^mr02)(vmin/r0)2 = 7mvmin2/10 = 7mg(R0 – r0)/10. If there are no frictional losses, we use energy conservation from the release point to the highest point of the loop: Ki + Ui = Kf + Uf; 0 + mghmin = Kmin + mg(2R0 – r0) = 7mg(R0 – r0)/10 + mg(2R0 – r0), which gives hmin = 2.7R0 – 1.7r0 .

Page 27

Chapter 10 28

108. (a) The disk will move in the direction of the force. (b) For the linear motion of the center of mass, we have ?F = ma; F = Ma, which gives a = F/M = (30 N)/(21.0 kg) = 1.43 m/s2. The acceleration of the center of mass is constant. Thus we can find the speed from v2 = v02 + 2ax; v2 = 0 + 2(1.43 m/s2)(9.0 m), which gives v = 5.1 m/s. (c) For the angular motion about the center of mass, we have ? = ICMCM;

v



R

F

FR = !MR2, which gives  = 2F/MR = 2(30 N)/(21.0 kg)(0.850 m) = 3.36 rad/s2. We can find the time for the motion from x = v0t + !at2; 9.0 m = 0 + !(1.43 m/s2)t2, which gives t = 3.55 s. At this time the angular speed is  = 0 + t = 0 + (3.36 rad/s2)(3.55 s) = 11.9 rad/s = 12 rad/s. (Note that v ? R, because the disk is not rolling.) (d) The angle that the disk has turned through in this time is  = 0t + !t2 = 0 + !(3.36 rad/s2)(3.55 s)2 = 21.2 rad. The length of string that has unwrapped is the distance the rim of the disk has rotated: s = R = (0.850 m)(21.2 rad) = 18 m. 109. (a) The yo-yo is considered as three cylinders, with a total mass of M = 2Mdisk + Mhub = 2(0.050 kg) + 0.0050 kg = 0.105 kg, and a moment of inertia of the yo-yo about its axis of I = 2(!MdiskRdisk2) + !MhubRhub2 = (0.050 kg)[!(0.075 m)]2 + !(0.0050 kg)[!(0.010 m)]2 = 7.0410–5 kg · m2. Because the yo-yo is rolling about a point on the rim of the hub, vCM = Rhub. The kinetic energy of the yo-yo is the translational kinetic energy of the CM and the rotational kinetic energy about the CM. Because the top of the string does not move, the tension in the string does no work. Thus energy is conserved: Ki + Ui = Kf + Uf ;

FT

y = 0 at top

R hub

+y

R disk

(2m disk + m hub )g

0 + 0 = !MvCM2 + !I2 + Mg(– L);

!MvCM2 + !I(vCM/Rhub)2 = MgL; !{0.105 kg + [(7.0410–5 kg · m2)/(0.0050 m)2]}vCM2 = (0.105 kg)(9.80 m/s2)(1.0 m),

which gives vCM = 0.84 m/s. (b) For the fraction of the kinetic energy that is rotational, we have Krot/(Ktrans + Krot) = !I2 /(!MvCM2 + !I2 ) = I2/[M(Rhub)2 + I2] = 1/[(MRhub2/I) + 1]

= 1/{[(0.105 kg)(0.0050 m)2/(7.0410–5 kg · m2)] + 1} =

Page 28

0.96.

Chapter 11

CHAPTER 11 – General Rotation 1.

z (a) For the magnitudes of the vector products we have i  i = ii sin 0° = 0; k j  j = jj sin 0° = 0; j k  k = kk sin 0° = 0. y (b) For the magnitudes of the vector products we have i  j = ij sin 90° = (1)(1)(1) = 1; i i  k = ik sin 90° = (1)(1)(1) = 1; x j  k  = jk sin 90° = (1)(1)(1) = 1. From the right hand rule, if we rotate our fingers from i into j, our thumb points in the direction of k. Thus i  j = k. Similarly, when we rotate i into k, our thumb points along – j. Thus i  k = – j. When we rotate j into k, our thumb points along i. Thus j  k = i.

2.

(a) We have A = – Ai and B = Bk. For the direction of A  B we have the positive y-axis. – i  k = – (– j) = j, (b) For the direction of B  A we have k  (– i) = – (k  i) = – (j) = – j, the negative y-axis. (c) For the magnitude of A  B we have A  B = AB sin 90° = AB. For the magnitude of B  A we have B  A = BA sin 90° = AB. This is expected, because B  A = – A  B.

3.

The magnitude of the tangential acceleration is atan = r. From the diagram we see that , r and atan are all perpendicular, and rotating  into r gives a vector in the direction of atan . Thus we have atan =   r. The magnitude of the radial acceleration is aR = 2r = r = v. From the diagram we see that , v and aR are all perpendicular, and rotating  into v gives a vector in the direction of aR. Thus we have aR =   v.

axis

atan



v

 r

aR

4.

When we use the component forms for the vectors, we have A  (B + C) = [Ay(Bz + Cz) – Az(By + Cy)]i + [Az(Bx + Cx) – Ax(Bz + Cz)]j + [Ax(By + Cy) – Ay(Bx + Cx)]k = (AyBz – AzBy)i + (AzBx – AxBz)j + (AxBy – AyBx)k + (AyCz – AzCy)i + (AzCx – AxCz)j + (AxCy – AyCx)k = A  B + A  C.

5.

For the limiting process we have d (A  B) (A  B) [(A + A)  (B + B)] – (A  B) = lim = lim t  0 t t  0 t dt [(A  B) + (A  B) + (A  B) + (A  B)] – (A  B) = lim t  0 t B A A  B = A  d B + d A  B. = tlim A  t + t  B + 0 t dt dt The last term is dropped because it is the product of two differentials.

Page 1

Chapter 11

6.

(a) When we use the component forms for the vectors, we have A  B = (Axi + Ayj + Azk)  (Bxi + Byj + Bzk) = AxBx(i  i) + AxBy(i  j) + AxBz(i  k) + AyBx(j  i) + AyBy(j  j) + AyBz(j  k) + AzBx(k  i) + AzBy(k  j) + AzBz(k  k) = 0 + AxBy(k) + AxBz(– j) + AyBx(– k) + 0 + AyBz(i) + AzBx(j) + AzBy(– i) + 0 = (AyBz – AzBy)i + (AzBx – AxBz)j + (AxBy – AyBx)k. (b) When we use the rules for evaluating a determinant, we get i j k A x A y A z = i A y B z – A zB y – j A x B z – A zB x + k A x B y – A y B x Bx By B z

= i A y B z – A zB y + j A z Bx – A x Bz + k A x By – A y Bx = A  B. 7.

(a) For the vectors A = 7.0i – 3.5j and B = – 8.5i + 7.0j + 2.0k, we have i j k A  B = 7.0 – 3.5 0 – 8.5 7.0 2.0

= i (– 3.5)(2.0) – (0)(7.0) – j (7.0)(2.0) – (0)(– 8.5) + k (7.0)(7.0) – (– 3.5)(– 8.5) = – 7.0i – 14.0j + 19.3k. (b) The magnitudes of the vectors are z A = [(7.0)2 + (– 3.5)2]1/2 = 7.83; B = [(– 8.5)2 + (7.0)2 + (2.0)2]1/2 = 11.2; A  B = [(– 7.0)2 + (– 14.0)2 + (19.3)2]1/2 = 24.8. We find the angle between A and B from A A  B = AB sin  ; 24.8 = (7.83)(11.2) sin , which gives sin  = 0.283, 164°. which gives  = 16°, 164°. From the diagram, we see that  = x

B

8.

We choose the z-axis as the axis of rotation. If we choose t = 0 z when the position vector lies in the xz-plane, we have  = k;  r = r[(cos t)i + (sin t)j] + zk. Thus the velocity is r y v = dr/dt = r[(– sin t)i + (cos t)j]. t For the vector product of  and r we have x   r = k  {r[(cos t)i + (sin t)j] + zk} = r[(cos t)j – (sin t)i] = v. We see that O can be anywhere on the rotation axis. If O is not on the axis, we can write the position vector as r = r – b, where b is the position of O relative to the previous origin. Thus we have v = dr/dt = dr/dt = v; but   r =   (r – b) = v – (  b) ? v. Thus O must be on the axis.

9.

When we use the component forms for the vectors, we have A · (B  C) = Ax(ByCz – BzCy) + Ay(BzCx – BxCz) + Az(BxCy – ByCx) = AxByCz – AxBzCy + AyBzCx – AyBxCz + AzBxCy – AzByCx. B · (C  A) = Bx(CyAz – CzAy) + By(CzAx – CxAz) + Bz(CxAy – CyAx) = BxCyAz – BxCzAy + ByCzAx – ByCxAz + BzCxAy – BzCyAx = A · (B  C). C · (A  B) = Cx(AyBz – AzBy) + Cy(AzBx – AxBz) + Cz(AxBy – AyBx) = CxAyBz – CxAzBy + CyAzBx – CyAxBz + CzAxBy – CzAyBx = A · (B  C). Note that if the three vectors lie in a plane, B  C will be perpendicular to the plane and thus to A, so A · (B  C) = 0. Page 2

y

Chapter 11

10. For the torque we have  = r  F = [(4.0i + 8.0j + 6.0k) m]  [(16.0j – 4.0k) N] = {[(8.0)(– 4.0) – (6.0)(16.0)]i + [0 – (4.0)(– 4.0)]j + [(4.0)(16.0) – 0]k} m · N = (– 128i + 16j + 64k) m · N. 11. We write the force as F = F[(cos  )i + (sin  )j] = (188 N)(cos 33.0° i + sin 33.0° j). For the torque we have  = r  F = [(0.220i + 0.335j) m]  [(188 N)(cos 33.0° i + sin 33.0° j)] = (188 N)[(0.220 m)(sin 33.0°) – (0.335 m)(cos 33.0°)]k = – (30.3 m · N)k (in – z-direction). 12. (a) For the torque we have     = r  F = [(0.215 m)i]  [(22.8 N)j – (21.6 N)k] = [0 – 0]i + [0 – (0.215 m)(– 21.6 N)]j + [(0.215 m)(22.8 N) – 0]k = (4.64 m · N)j + (4.90 m · N)k. For the magnitude we have  = [(4.64 m · N)2 + (4.90 m · N)2]1/2 = 6.75 m · N. The torque lies in the yz-plane. We find the angle from the y-axis from tan  = z/y = (4.90 m · N)/(4.64 m · N) = 1.056, which gives  = 46.6°. Thus we have  = 6.75 m · N in yz-plane 46.6° from y-axis. (b) The angular acceleration must be along the z-axis:  = k. Thus we see that the torque is not parallel to . The net torque must be along the z-axis, so the constraint of the axis means there is an additional torque provided by the bearings. 13. For the torque we have  = r  F = [(8.0 m)j + (6.0 m)k]  [(± 2.4 kN)i – (3.0 kN) j] = [0 – (6.0 m)(– 3.0 kN)]i + [(6.0 m)(± 2.4 kN) – 0]j + [0 – (8.0 m)(± 2.4 kN)]k = (18 m · kN)i ± (14 m · kN)j — (19 m · kN)k. 14. For the angular momentum we have ¬ = r  p = (xi + yj + zk)  (pxi + pyj + pzk) =

(ypz – zpy)i + (zpx – xpz)j + (xpy – ypx)k.

15. For a particle moving in a circle the magnitude of the angular momentum is ¬ = mvr, and the moment of inertia is I = mr2.. The kinetic energy is K = mv2/2 = m(¬/mr)2/2 = ¬2/2mr2 = ¬2/2I. 16. (a) About the origin O we have ¬ = r  p = r  mv = rmv into page = (b) About the origin O we have ¬ = r  p = r  mv = rmv sin 180° =

r

mvd into page. 0.

m

r

v

d O

O

17. We let d be the displacement of the second particle relative to the first particle. Thus r1 = r2 + d. For the angular momentum we have ¬ = r1  p1 + r2  p2 = (r2 + d)  p + r2  (– p) = d  p, which is independent of the choice of origin.

Page 3

Chapter 11

18. For the angular momentum we have ¬ = r  mv = m(r  v) = (0.060 kg)[(7.0 m)i + (– 6.0 m)j]  [(2.0 m/s)i – (8.0 m/s) k] = (0.060 kg){[(– 6.0 m)(– 8.0 m/s) – 0]i + [0 – (7.0 m)(– 8.0 m/s)]j + [0 – (– 6.0 m)(2.0 m/s)]k} = (2.9i + 3.4j + 0.72k) kg · m2/s. 19. For the angular momentum we have i j k = r  mv = (7.6 kg) 1.0 2.0 3.0 (m 2/ s) – 5.0 – 4.5 – 3.1

= i (2.0)(–3.1) – (3.0)(–4.5) – j (1.0)(–3.1) – (3.0)(–5.0) + k (1.0)(– 4.5) – (2.0)(– 5.0) = (55i – 90j + 42k) kg m 2/ s .

(7.6 kg  m 2/ s)

20. (a) The force on the falling particle is mg. If we use the other particle as the origin, the magnitude of the torque acting on m is  = rmg = mgd (constant). The direction is horizontal, perpendicular to the table edge. (b) The particle falls with a constant acceleration from rest, so the velocity is v = gt (down). For the magnitude of the angular momentum we have ¬ = rmv = mgtd. The direction is horizontal, perpendicular to the table edge. (c) The rate of change of the angular momentum is d¬/dt = mgd = . 21. (a) The moment of inertia of the system about one end is I = m(¬/3)2 + m(2¬/3)2 + m¬2 + (M¬2/3) = [(14m/9) + (M/3)]¬2. The kinetic energy is K = I2/2 = [(14m/9) + (M/3)]¬22/2 = [(7m/9) + (M/6)]¬22.  (b) For the angular momentum we have L = I = [(14m/9) + (M/3)]¬2.

Page 4

Chapter 11

22. (a) We choose the coordinates shown on the force diagrams. Note that we take the positive direction in the direction of the acceleration for each object. Because the linear acceleration of the masses is the tangential acceleration of the rim of the pulley, we have a = atan = R0. We write Fy = may for m2: m2g – FT2 = m2a. We write Fy = may for m1: FT1 – m1g = m1a. We write  = I for the pulley about its axle: FT2R0 – FT1R0 = I = !MR02a/R0 , or FT2 – FT1 = !Ma. If we add the two force equations, we get FT1 – FT2 = (m1 + m2)a + (m1 – m2)g. When we add these two equations, we get a = (m2 – m1)g/(m1 + m2 + !M)



 into page

R0 FT1 FT1

+y

FT2 m 1g

m 2g

= (8.8 kg – 7.0 kg)(9.80 m/s2)/[7.0 kg + 8.8 kg + !(0.80 kg)] = 1.1 m/s2. (b) If the moment of inertia of the pulley is ignored, from the torque equation, we see that the two tensions will be equal. If we add the force equations, we get a0 = (m2 – m1)g/(m1 + m2). Thus the error is (a0 – a)/a = (a0/a) – 1 = [(m1 + m2 + !M)/(m1 + m2)] – 1 = !M/(m1 + m2) = !(0.80 kg)/(7.0 kg + 8.8 kg) = 0.025 =

3%.

23. With the positive direction CCW, for the angular momentum about the axis of the pulley we have L = R0mv + I = R0mv + I(v/R0) = [R0m + (I/R0)]v. The net torque is produced by mg and the frictional torque:  = dL/dt; mgR0 – fr = [R0m + (I/R0)] dv/dt, which gives a = (mgR0 – fr)/[R0m + (I/R0)] = [(1.53 kg)(9.80 m/s2)(0.330 m) – 1.10 m · N]/ {(0.330 m)(1.53 kg) + [(0.385 kg · m2)/(0.330 m)]} = 2.30 m/s2.

Page 5

FT2

R0

v mg

+y

Chapter 11

24. The linear momentum for each mass will be tangent to its circular motion with magnitude p = mv = md = (0.600 kg)(30 rad/s)(0.20 m) = 3.6 kg · m/s. (a) For the component of angular momentum along the axle we have Laxle = d1p1 + d2p2 = 2d1p1 = 2(0.20 m)(3.6 kg · m/s) = 1.44 kg · m2/s. (b) We choose the point of the axle midway between the particles for the axis. Each angular momentum will be perpendicular to r and p as shown on the diagram. We see that the two angular momenta are parallel, so the total angular momentum will also make an angle  with the horizontal. From the symmetry of the distances,  = 45°, so the angular momentum makes an angle of 90 – 45° = 45° with the axle.

 L1



p out

d1

m1 r1

O r2

 d2

25. (a) With the positive direction CCW, for the angular momentum about the axis of the pulley we have L = R0M1v + R0M2v + I = R0M1v + R0M2v + I(v/R0) = [R0M1 + R0M2 + (I/R0)]v.  (b) Because M1g is balanced by the normal force on the horizontal surface, the net torque is from M2g only:  = dL/dt; M2gR0 = [R0M1 + R0M2 + (I/R0)] dv/dt, which gives a = M2g/[M1 + M2 + (I/R02)].

L2 p in m2

v FN

R0

M 1g

v M 2g

26. (a) With the positive direction CCW, we have the same FN v formula for the angular momentum: L = [R0M1 + R0M2 + (I/R0)]v. A  (b) Because M1g is balanced by the normal force on the horizontal R FT1 d 0 surface, the friction force is Ffr = kM1g. If we consider only the block M1 , which does not rotate, the net torque must be zero, Ffr M 1g which means that the normal force and the weight are separated  by a distance d. The torque of this couple must be balanced by the torque from the tension and friction. If we take the point A as the axis, we have kM1gR0 – M1gd = 0, or – M1gd = – kM1gR0. For the entire system the tension is an internal force. Because the line of the friction force passes through the center of the pulley, the net torque is produced by M2g and the couple from FN and M1g:  = dL/dt; M2gR0 – kM1gR0 = [R0M1 + R0M2 + (I/R0)] dv/dt, which gives a = (M2 – kM1)g/[M1 + M2 + (I/R02)]. 27. The impulse changes the linear momentum of the center of mass: J = ?p = mvCM; 8.5 10–3 N · s = (0.040 kg)vCM , which gives vCM = 0.213 m/s. The moment of the impulse about the center of mass changes the angular momentum: rJ = ?L = ICMCM; [!(0.070 m) – 0.020 m](8.5 10–3 N · s) = [(0.040 kg)(0.070 m)2/12]CM , Page 6

¬ J

d

Chapter 11

which gives CM = 7.8 rad/s. The rod rotates at 7.8 rad/s about the center of mass, which moves with constant velocity of 0.21 m/s.

28. We select a differential element of the rod a distance r from the center, which has a mass dm = (M/¬)dr. The angular momentum of this element is dL = r  v dm, where v has magnitude (r sin ). From the diagram, we see that for the lower half of the rod, r and v reverse direction. Thus the direction of dL will be the same for all the elements, and L will be perpendicular to the rod, making an angle 90° –  with the axis. We integrate to find the magnitude of the angular momentum:

L=

dL =

r(r si n )  d m =

= ( M/ ¬) si n  r3 / 3 =

¬/ 2 – ¬/ 2

¬/ 2 – ¬/ 2

 dL

 dm

 r O

r(r si n ) (M/ ¬) dr

= ( M/ ¬) si n  2/ 3 ¬/ 2

z

3

2

1/ 12 M¬  si n  .

29. In the inertial reference frame the position of the center of mass is rCM = ?miri/?mi = ?miri/M. In the center of mass frame the position of the ith particle is ri* = ri – rCM, so vi* = vi – vCM. When we combine these, we see that ?miri* = ?miri – ?mi(?miri/M) = 0, and ?mivi* = ?mivi – ?mi(?mivi/M) = 0, which of course defines the center of mass frame. When we use the above relations in the definition of the angular momentum, we get L = ?ri  pi = ?ri  mivi = ?[(rCM + ri*)  (mivCM + mivi*)] = ?(rCM  mivCM) + ?(ri*  mivCM) + ?(rCM  mivi*) + ?(ri*  mivi*) = rCM  MvCM + ?(miri*)  vCM + rCM  (?mivi*) + ?(ri*  mivi*) = rCM  MvCM + 0 + 0 + ?(ri*  mivi*) = L* + rCM  MvCM . 30. The net torque to maintain the rotation is supplied by the forces at the bearings. If the bearings are a distance d from the center, we have net = 2Fd = I2/tan  = [(m1r12 + m2r22)sin2 ]2/tan , which gives F = (m1r12 + m2r22)2 sin  cos /2d. Page 7

y

Chapter 11

31. If we choose the center of the circle that m1 makes, O, as origin the angular momentum will be parallel to . It will have a magnitude L = (r sin )m1v1 = m1r2 sin2 . L is constant in both magnitude and direction, so we have net = dL/dt; m1 F1(d – r cos ) – F2(d + r cos ) = 0, or F2 = [(d – r cos )/(d + r cos )]F1. The radial acceleration of the mass must be produced by a radial force in the rod: Frod = m1v12/(r sin ) = m1r2 sin . Because the rod has no mass, the net force on the rod is zero. For the net horizontal force on the axle we have F1 + F2 – Frod = 0, or F1 + F2 = m1r2 sin . When we combine this with the result from the torque equation, we have F1 + [(d – r cos )/(d + r cos )]F1 = m1r2 sin , which gives F1 = [(d + r cos )/2d]m1r2 sin ; F2 = [(d – r cos )/2d]m1r2 sin .



F1 O



Frod

r

32. We use the result from Problem 30: F = (m1r12 + m2r22)2 sin  cos /2d = 2m1r122 sin  cos /2d = 2(0.60 kg)(0.30 m)2(11.0 rad/s)2 sin 23.0° cos 23.0°/(0.20 m) =

O

Frod

d

F2

24 N.

33. We use the result from Problem 31: F1 = [(d + r cos )/2d]m1r2 sin  = [0.10 m + (0.30 m) cos 23.0°](0.60 kg)(0.30 m)(11.0 rad/s)2 sin 23°/(0.20 m) = F2 = [(d – r cos )/2d]m1r2 sin  = [0.10 m – (0.30 m) cos 23.0°](0.60 kg)(0.30 m)(11.0 rad/s)2 sin 23°/(0.20 m) =

Page 8

16 N; – 7.5 N.

Chapter 11

34. (a) From the symmetry we see that F1 = F2 = F. The two forces from the bearings produce the radial acceleration of the center of mass: F1 + F2 = 2F = Mr2;



F1

F = !(11.8 kg)(0.0100 m)[(11.2 rev/s)(2p = 292 N. The reaction to this force will be exerted on each bearing: 292 N. (b) To balance the wheel, we want the center of mass to be at the axle. The added mass will be on the line joining the axle and the center of mass of the wheel. Thus we have md + Mr = 0; (1.00 kg)d + (11.8 kg)(0.0100 m), which gives d = – 0.118 m. Thus the mass should be placed 11.8 cm from the center of the wheel opposite to the axle. rad/rev)]2

r R

O

F2

35. In the collision, during which we ignore any motion of the rod, angular momentum about the pivot point will be conserved: Li = Lf ; mv(!¬) + 0 = Itotal = [@M¬2 + m(!¬)2], which gives  = 6mv/(4M + 3m)¬. For the rotation about the pivot after the collision, energy will be conserved. If the center of the rod reaches a height h, the bottom of the rod will swing to a height H = 2h, so we have Ki + Ui = Kf + Uf ;

M m

¬ v

h

!Itotal2 + 0 = 0 + (m + M)gh; ![@M¬2 + m(!¬)2][6mv/(4M + 3m)¬]2 = (m + M)gH/2, which gives

H = 3m2v2/g(3m + 4M)(m + M).

36. For the system of stick and bullet during the collision, angular momentum about the center of mass is conserved: Li = Lf ; mvi(#¬) + 0 = mvi(#¬) + Irod ; (3.0 g)(250 m/s)#(1.0 m) = (3.0 g)(160 m/s)#(1.0 m) + [(300 g)(1.0 m)2/12)], which gives = 2.7 rad/s. 37. The initial angular speed of the Earth is E  (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 10–5 rad/s. We approximate the Earth as a uniform sphere, with IE = ^MErE2. For the system of meteor and Earth, angular momentum is conserved – mvrE sin  + IEarthE = (IEarth + Imeteor). Note that after the collision, Imeteor = mrE2. This gives  = (IEarthE – mvrE sin )/(IEarth + Imeteor) = [(1 – mvrE sin /^MErE2E)/(1 + mrE2/^MErE2)]E Page 9

m

v

 rE

Earth

Chapter 11

= [(1 – 5mv sin /2MErEE)/(1 + 5m/2ME)]E. Because the mass of the Earth is much greater than the mass of the meteor, we have  ˜ [(1 – 5mv sin /2MErEE)(1 – 5m/2ME)]E ˜ (1 – 5mv sin /2MErEE – 5m/2ME)E

= {1 – [5(7.0 1010 kg)(1.0 104 m/s) sin 45°/2(6.0 1024 kg)(6.4 106 m)(7.27 10–5 rad/s)] – [5(7.0 1010 kg)/2(6.0 1024 kg)]}E –13 = (1 – 4.7 10 )E.

38. (a) By walking to the edge, the moment of inertia of the person changes. Because the system of person and platform is isolated, angular momentum will be conserved: L = (Iplatform + Iperson1)1 = (Iplatform + Iperson2)2 ; [670 kg · m2 + (55 kg)(0)2](2.0 rad/s) = [670 kg · m2 + (55 kg)(2.5 m)2]2 , which gives 2 = 1.3 rad/s. (b) For the kinetic energies, we have K1 = !(Iplatform + Iperson1)12 = !(670 kg · m2 + 0)(2.0 rad/s)2 = 1.34 103 J; K2 = !(Iplatform + Iperson2)22 = 2 = Thus there is a loss of 4.6 102 J, a decrease of 34%.

39. (a) We find the speed of the center of mass from the conservation of linear momentum: Mv + 0 = (M + m)vCM; (200 kg)(18 m/s) = (200 kg + 50 kg)vCM , which gives vCM = 14 m/s. (b) During the collision, angular momentum about the center of mass will be conserved. We find the location of the center of mass relative to the center of the beam: d = m(!¬)/(M + m) = (50 kg)!(2.0 m)/(200 kg + 50 kg) = 0.20 m. When we use the parallel-axis theorem for the moment of inertia of the beam, angular momentum conservation gives us Li = Lf ;

8.8 102 J.

M

¬ v

d

m

Mvd + 0 = Itotal = [(M¬2/12) + Md2 + m(!¬ – d)2];

(200 kg)(18 m/s)(0.20 m) = ((200 kg){[(2.0 m)2/12] + (0.20 m)2} + (50 kg)(1.0 m – 0.20 m)2), which gives = 6.8 rad/s. 40. We find the speed of the center of mass from the conservation of linear momentum: mv + 0 = (M + m)vCM, which gives vCM = mv/(M + m). During the collision, angular momentum about the center of mass will be conserved. We find the location of the center of mass relative to the center of the beam: d = m(#¬)/(M + m) = m¬/4(M + m). When we use the parallel-axis theorem for the moment of inertia of the beam, angular momentum conservation gives us Li = Lf ; mv(#¬ – d) + 0 = Itotal = [(M¬2/12) + Md2 + m(#¬ – d)2];

v m

¬/4

mv{#¬ – [m¬/4(M + m)]} = ((M¬2 /12) + M[(m¬/4(M + m)]2 + m{#¬ – [m¬/4(M + m)]}2); mv[M/4(M + m)] = {(M/12) + M[(m/4(M + m)]2 + m[M/4(M + m)]2}¬; Page 10

d

¬

M

Chapter 11

mv[M/4(M + m)] = ({(M/12)[16(M + m)2] + Mm2 + mM2}/16(M + m)2)¬;

mv = ({(1/12)[4(M + m)]2 + m2 + mM}/4(M + m))¬; 12mv = [(16M2 + 32Mm + 16m2 + 12m2 + 12mM)/4(M + m)]¬; 12mv = 4[(4M2 + 11Mm + 7m2)/4(M + m)]¬ = (4M + 7m)¬, which gives  = 12mv/¬(4M + 7m). Thus the center of mass of the rod and ball moves with a speed mv/(M + m), and the system rotates about the center of mass with an angular speed 12mv/¬(4M + 7m). 41. We find the moment of inertia from the precession rate:  = Mgr/L = Mgr/I; (1 rev)(2p rad/rev)/(8.0 s) = (0.220 kg)(9.80 m/s2)(0.035 m)/I(15 rev/s)(2p rad/rev), 1.02 10–3 kg · m2. which gives I = 42. (a) We find the precession time from  = 2p/T = Mgr/L = Mgr/I = Mgr/(!MR2) = 2gr/R2; (2p rad/rev)/T = 2(9.80 m/s2)(0.085 m)/(0.055 m)2(70 rev/s)(2p rad/rev), which gives T = 5.0 s. (b) If we form the ratio of the two times, we get T2/T1 = (R2/R1)2(r1/r2) T2/(5.0 s) = (2)2(!), which gives T2 =

10 s.

43. We find the rate of precession from  = Mgr/L = Mgr/I = Mgr/(!MR2) = 2gr/R2 = 2(9.80 m/s2)(0.10 m)/(0.060 m)2(250 rad/s) =

2.2 rad/s (0.35 rev/s).

44. Because the mass is placed on the axis of rotation, the moment of inertia will not change. If the length of the axle is d, the distance from the support to the center of mass will be r = [M(!d) + (!M)d]/(M + !M) = %d. If we form the ratio of the two precession rates, we get 2/1 = M2r2/ M1r1 ;

2/(2.18 rad/s) = (*M/M)(%d/!d), which gives 2 =

4.4 rad/s (0.69 rev/s).

45. In the rotating frame of the platform, there will be an outward pseudoforce with a magnitude Fr = ma = mv2/r = mr2. Thus the effective gravity makes an angle from the vertical given by tan  = mr2/mg = r2/g. The grass grows opposite to gravity, therefore inward at an angle = tan–1 (r2/g).

z

 mr 2 mg

r



46. Newton’s third law is invalid for the pseudoforce in the rotating frame because it acts only on one body and thus is not one of a pair of forces.

Page 11

Chapter 11

47. A convenient dimensionless factor is g/rE2 = (9.80 m/s2)/(6.38106 m)[(2p rad)/(86,400 s)]2  r = 290.4. mg  (a) At the North Pole there is no radial acceleration, mr 2 mg so the effective acceleration of gravity is  g, along a radial line. x mg  (b) At a latitude  there will be a pseudoforce mr2 away from the axis, where r = rE cos . We use the coordinate mg mR 2 system shown on the diagram, with positive y down y R along the radial line. For the components of g we have gx = r2 sin  = rE2 sin  cos ; gy = g – r2 cos  = g – rE2 cos2 . We find the angle from tan  = gx/gy = rE2 sin  cos /(g – rE2 cos2 ) = sin  cos /[(g/rE2) – cos2 ]. At a latitude of 45°, we get tan  = sin 45° cos 45°/(290.4 – cos2 45°) = 1.72510–3, or  = 0.0988°. We can find the magnitude from g = gx/sin  = g sin  cos /(g/rE2) sin  = g sin 45° cos 45°/(290.4) sin 0.0988° = 0.998g. Thus the effective acceleration of gravity is 0.998g, 0.0988° south from a radial line. (c) At the equator  = 0, so we have tan  = 0; as expected the effective acceleration is along a radial line. We find the magnitude from g = gy = g – rE2 cos2  = g[1 – 1/(g/rE2)] = g[1 – 1/(290.4)] = 0.997g, along a radial line.

48. (a) In the inertial frame, the ball has the tangential speed of B, vB = rB, perpendicular to the radial line. This is greater than the tangential speed of the woman at A, vA = rA, so the ball passes in front of A, to the right of its motion. (b) In the inertial frame, the ball has a radial velocity v, so the time to reach A is found from rB – rA = vt. Because the ball has the tangential speed of B, it will travel a distance vBt = rBt perpendicular to the radial line. The woman at A will travel vAt = rAt. Thus the additional distance traveled by the ball is ?s = rBt – rAt = (rB – rA)t = vt2. The deflection is ?s = vt2. If we ascribe this to a constant acceleration, we have acor = 2v, as before. ?s = !acort2 = vt2, or 49. There is no Coriolis force when the velocity is parallel to the axis of rotation. Thus at the equator there will be none for velocities in the north or south direction. Note that for velocities parallel to the equator, the Coriolis force will be vertical. A ball thrown parallel to the equator will not deviate to the right; the Coriolis force would effectively change the acceleration of gravity. Page 12

Chapter 11

50. The deflection of the velocity will be to the right or east. The component of velocity perpendicular to the Earth’s rotation is v sin . The Coriolis force is Fcor = macor = m2v sin  = (1200 kg)2(2p/86,400 s)([500 km/h)/(3.6 ks/h)] sin 40° = 15.6 N east.

v

N



r



51. (a) In the northern hemisphere, the projectile will be deflected to the right, that is, south. (b) The initial velocity of the projectile is perpendicular to the rotation axis, so the Coriolis acceleration is acor = 2v0 . The direction of this acceleration is perpendicular to the axis and v (thus, parallel to the equator), so it has components acor,up = 2v0 cos , and acor,south = 2v0 sin . If we ignore the vertical acceleration, which is an effective g, the distance the projectile travels east in a time t is D = v0t. In this time, the projectile will have deflected south a distance ssouth = !(2v0 sin )t2. Thus we have ssouth = v0 sin (D/v0)2 = D2 sin /v0. (c) For the given data, we have ssouth = (2p/86,400 s)(3.0 103 m)2 sin 45°/(1000 m/s) =

52. The vertical forces are mg and the normal force from the spoke. The pseudoforces are the Coriolis force in the negative x-direction, and the radial force away from the center. Because the ant is moving at constant velocity, these forces must be balanced by a force from the spoke, which we treat as two forces, one perpendicular to the spoke and one, friction, along the spoke. Thus we have ?F = 0; (mr2 – Ffr)i + (Fspoke – 2mv)j + (FN – mg)k = 0.

53. (a) The angular momentum relative to the origin is

Page 13

0.46 m.

x FR

y

v

Fspoke Ffr

Fcor



Chapter 11

i j k L = r  mv = (0.50 kg) 0 2.0 3.0 (m 2 / s) 8.0 6.0 0 = (– 18)i – ( – 24)j + (– 16)k (0.50 kg m 2/ s) =

– 9.0 i + 12j – 8.0k kg m 2/ s.

(b) The torque is

i j k  = r  F= 0 2.0 3.0 (m N ) = 3.0 0 0

9.0j – 6.0k m N .

54. The initial angular speed of the Earth is E  (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 10–5 rad/s.

v m

We approximate the Earth as a uniform sphere, with IE = ^MErE2. We assume the asteroid attaches to the Earth. For the system of asteroid and Earth, angular momentum is conserved: mvrE + IEarthE = (IEarth + Iasteroid). This gives  = (IEarthE + mvrE)/(IEarth + Iasteroid)

Earth rE

= [(1 + mvrE/^MErE2E)/(1 + mrE2/^MErE2)]E = [(1 + 5mv/2MErEE)/(1 + 5m /2ME)]E. Because the mass of the Earth is much greater than the mass of the asteroid, we have  ˜ [(1 + 5mv/2MErEE)(1 – 5m /2ME)]E. ˜ (1 + 5mv/2MErEE – 5m/2ME)E

= {1 + [5(1 105 kg)(30 103 m/s)/2(6.0 1024 kg)(6.4 106 m)(7.27 10–5 rad/s)] – [5(1 105 kg)/2(6.0 1024 kg)]}E –18 = (1 + 2.6 10 )E. Thus the fractional change is ( – E)/E = [(1 + 2.6 10–18)E – E]/E = 3 10–18.

55. (a) The torque produced by the normal force about the center of mass causes a change ?L in the direction of the tire’s motion. When this is added to the angular momentum L, we see that the tire will turn in the direction of the lean. (b) There is no vertical acceleration, so we have FN = mg. The change in the angular momentum about the center of mass is ?L =  ?t = mgr sin  ?t = (9.0 kg)(9.80 m/s2)(0.32 m) sin 10° (0.20 s) = 0.98 kg · m2/s. The original angular momentum is L0 = I = (0.83 )[(2.1 m/s)/(0.32 m)] = 5.5 kg · m2/s. Thus we have ?L/L0 = (0.98 kg · m2/s)/(5.5 kg · m2/s) = 0.18, so ?L = 0.18L0.

 L

FN

mg

56. Because the electric force provides the necessary radial acceleration of the circular orbit, we have ke2/r2 = mv2/r, which gives r = ke2/mv2. The angular momentum of the electron is Page 14

Chapter 11

¬ = mvr = nh/2p, n = 1, 2, 3, …, which gives v = nh/2pmr. When we use this in the first result, we get r = ke2(4p2m2r2)/mn2h2, which gives r = n2h2/4p2kme2, n = 1, 2, 3, … . 57. (a) The angular momentum delivered to the waterwheel in a time ?t is that lost by the water: ?L/?t = ?m R(vi – vf)/?t = (?m/?t)R(vi – vf) = (150 kg/s)(3.0 m)(7.0 m/s – 3.0 m/s) = 1.8 103 kg · m2/s2. (b) The torque applied to the waterwheel increases the angular momentum:  = ?L/?t = 1.8 103 m · N. (c) The power delivered is P =  = (1.8 103 m · N)(2p/5.5 s) = 2.1 103 W. 58. (a) If we consider an axis through the center of mass parallel to the velocity vector (that is, parallel to the ground), there is no angular acceleration about this axis. If d is the distance from the center of mass to the contact point on the ground, we have   = I ; FNd sin  – Ffrd cos  = 0, which gives tan  = Ffr/FN. (b) The friction force is producing the necessary radial acceleration for the turn: Ffr = FN tan  = mg tan  = mv2/r, or tan  = v2/gr = (3.2 m/s)2/(9.80 m/s2)(2.8 m) = 0.373, so  = 20°. (c) From Ffr = mv2/r, we see that the minimum turning radius requires the maximum static friction force: smg = mv2/rmin , or rmin = v2/sg = (3.2 m/s)2/(0.65)(9.80 m/s2) = 1.6 m. 59. Because there is no friction, the center of mass must fall straight down. The vertical velocity of the right end of the stick must always be zero. If  is the angular velocity of the stick just before it hits the table, the velocity of the right end with respect to the center of mass will be (¬/2) up. Thus we have (¬/2) – vCM = 0, or  = 2vCM/¬. The kinetic energy will be the translational energy of the center of mass and the rotational energy about the center of mass. With the reference level for potential energy at the ground, we use energy conservation to find the speed of the center of mass just before the stick hits the ground: Ki + Ui = Kf + Uf;

¬

60. (a) Because the hoop is rolling down the string, the acceleration of the center of mass is related to the angular acceleration: a = R. For the linear motion of the center of mass we have ?F = Ma; Mg – FT = Ma = MR. For the angular motion about the center of mass we have ? = dL/dt = I; FTR = MR2 = MRa. When we combine these equations, we get  = a/R = !g/R. Thus we have L = !MRgt. dL/dt = (MR2)(!g/R) = !MRg, so Page 15

FN

mg

Ffr

A

CM

 y =0

0 + Mg!¬ = !MvCM2 + !(M¬2/12)2 + 0; Mg!¬ = !MvCM2 + !(M¬2/12)(2vCM/¬)2 = !(4MvCM2/3), which gives



CM

vCM = (3g¬/4)1/2 .

Chapter 11

(b) The tension in the string is FT = Ma = M(!g) = !Mg.

61. We assume the tensions in the other cables immediately go to zero. The torque created by the weight changes the angular momentum about the pivot point of the tower:  = dL/dt; mg(!¬ sin ) = @m¬2 d/dt. This equation contains three variables, , , t. We use the definition of angular velocity to eliminate t: mg(!¬ sin ) = @m¬2 (d/d)(d/dt) = @m¬2 d/d, or 3(g/¬) sin  d = 2 d. When we integrate this, we get   2 d  = (3g/ ) sin  d  ; 0

0

 2 = ( – 3 g/ ) cos 

 0

= (3g / )(1 – cos  ).

Thus the speed of the top of the tower is v = ¬ = [3g¬(1 – cos )]1/2 = [3(9.80 m/s2)(12 m)(1 – cos )]1/2 =

(19 m/s)(1 – cos )1/2.

62. We find the location of the center of mass: 0.84 m 0.84 m  x dm x 0.56 kg/ m + (3.5 kg/ m 3 )x 2 dx 0 0 d CM = = 0.84 m

0

=

dm

0.84 m

0

0.56 kg/ m + (3.5 kg / m 3 )x 2 dx

2 4 (0.56 kg/ m ) x + (3.5 kg / m 3 ) x 2 4 3

(0.56 kg / m)x + (3.5 kg / m 3 ) x 3

0.84 m 0 0.84 m 0

Page 16

=

0.632 kg m = 0.545 m . 1.162 kg

 ¬

mg

Chapter 11

Note that the mass of the bat is 1.162 kg.

x=0

xS dCM CM

d0

d

We integrate to find the moment of inertia about the preferred point on the handle: 0.84 m 0.84 m  2 2 I handle = x – d0 d m = (x – 0.050 m ) 0.56 kg / m + (3.5 kg / m 3)x 2 d x 0

0

3 2 = (0.56 kg / m) x – (0.100 m) x + (0.0025 m 2)x 3 2

+ (3.5 kg / m 3)

0.84 m 0

x5 x4 x3 – (0.100 m ) + (0.0025 m 2 ) 5 4 3

0.84 m 0

= 0.343 kg m 2.

If we have pure rotation about the handle, the acceleration of the center of mass is related to the angular acceleration about the handle by aCM = (dCM – d0). For the translational effect of the force we have F = maCM = m(dCM – d0). For the rotational effect about the handle we have Fd = Ihandle. When we combine the two equations, we get md(dCM – d0) = Ihandle; (1.162 kg)d(0.545 m – 0.050 m) = 0.343 kg · m2, which gives d = 0.60 m (0.65 m from end). 63. (a) If the thrown-off mass carries off no angular momentum, from conservation of angular momentum for the star we have Iii = Iff ;

^Mrs21 = ^(#M)rn22 , which gives

2 = 4(rs/rn)21 = 4(7.0 105 km/10 km)2(1 rev/10 days)/(86,400 s/day) = 2.3 104 rev/s. (b) If the thrown-off mass carries off & of the initial angular momentum, from conservation of angular momentum for the star we have #Iii = Iff ; #(^Mrs2)1 = ^(#M)rn22 , which gives 2 = (rs/rn)21 = (7.0 105 km/10 km)2(1 rev/10 days)/(86,400 s/day) = 5.7 103 rev/s.

Page 17

Chapter 12

CHAPTER 12 – Static Equilibrium; Elasticity and Fracture 1.

2.

From the force diagram for the sapling we can write ?Fx = F1 – F2 sin 20° – F3 cos  = 0; 380 N – (255 N) sin 20° – F3 cos  = 0, or F3 cos  = 293 N. ?Fy = F2 cos 20° – F3 sin  = 0; F3 sin  = (255 N) cos 20° = 240 N. Thus we have F3 = [(293 N)2 + (240 N)2]1/2 = 379 N. tan  = (240 N)/(293 N) = 0.818,  = 39.3°. 141°. So  = 180° –  =

y F2

20°

F1



x



F3

y

From the force diagram for the junction we can write ?Fx = F2 – F1 sin 45° = 0. This shows that F1 > F2 , so we take F1 to be the maximum. ?Fy = F1 cos 45° – Mg = 0; Mg = (1150 N) sin 45° = 813 N.

F1 45°

F2 x

Mg

3. 

We choose the coordinate system shown, with positive torques clockwise. For the torque from the person’s weight about the point B we have  B = MgL = (56 kg)(9.80 m/s2)(3.0 m) = 1.6103 m · N.

A

B L

d

4. 

5.

We choose the coordinate system shown, with positive torques clockwise. For the torque from the person’s weight about the point A we have  A = Mgx; 1000 m · N = (56 kg)(9.80 m/s2)x, which gives x = 1.82 m.

We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the leg: ?A = MgD – FTL = 0; (15.0 kg)(9.80 m/s2)(0.350 m) – FT(0.805 m), which gives FT = 63.9 N. Because there is no acceleration of the hanging mass, we have FT = mg, or m = FT/g = (63.9 N)/(9.80 m/s2) = Page 1

A

Mg

B x

Mg

FT

FT

L Hip Joint

A

CM

D

mg Mg

6.52

Chapter 12

kg. 6.

7.

8. 







9.

We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the support point A from the force diagram for the board and people: ?A = – m1g(L – d) + m2gd = 0; – (23.0 kg)(10.0 m – d) + (67.0 kg)d = 0, which gives d = 2.56 m from the adult. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the support point A from the force diagram for the board and people: ?A = – m1g(L – d) – Mg(!L – d) + m2gd = 0; – (23.0 kg)(10 m – d) – (12.0 kg)(5.0 m – d) + (67.0 kg)d = 0, which gives d = 2.84 m from the adult.

A

m 1g

(20.0 kg)!(1.20 m) – (66.0 kg)(0.50 m – x) = 0, which gives x = 0.32 m.

Page 2

m 2g

FN

d

L–d A

m 1g

m 2g

Mg FN

(a) We choose the coordinate system shown, with positive torques clockwise. For the torques about the point B we have F1   ?B = F1d + MgD = 0; A F1(1.0 m) + (56 kg)(9.80 m/s2)(3.0 m) = 0, which gives F1 = – 1.6103 N (down). For the torques about the point A we have  ?A = – F2d + Mg(D + d)= 0; F2(1.0 m) = (56 kg)(9.80 m/s2)(3.0 m + 1.0 m), which gives F2 = 2.2103 N (up). (b) For the torques about the point B we have   ?B = F1d + MgD + mg[!(D + d) – d] = 0; F1(1.0 m) = – (56 kg)(9.80 m/s2)(3.0 m) – (35 kg)(9.80 m/s2)(1.0 m), which gives F1 = – 2.0103 N (down). For the torques about the point A we have   ?A = – F2d + Mg(D + d) + mg!(D + d) = 0; F2(1.0 m) = (56 kg)(9.80 m/s2)(4.0 m) + (35 kg)(9.80 m/s2)(2.0 m), which gives F2 = 2.9103 N (up). We choose positive torques clockwise. We write ? = Iabout the point B from the force diagram for the table and person: ?B = Mg(d – x) + FAD – mg(!D) = 0. At the point of tipping, FA = 0, so we have

d

L–d

F2 B d

D

F1

F2

A

B

D Mg

d mg

L

D

d FA A

Mg

mg

x FB B

Mg

Chapter 12

10. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the two beams:   ?A = mg#L + Mg!L – FN2L = 0, which gives

y

FN2 = #mg + !Mg

B D mg

= (1100 kg)(9.80 m/s2) + !(1100 kg)(9.80 m/s2) – 6.74103 N =

Mg

9.43103 N.

y FT2 FT1



x

mg

12. From the force diagram for the hanging light and junction we can write ?Fx = FT1 cos 1 – FT2 cos 2 = 0; FT1 cos 37° = FT2 cos 53°; ?Fy = FT1 sin 1 + FT2 sin 2 – mg = 0; FT1 sin 37° + FT2 sin 53° = (30 kg)(9.80 m/s2). When we solve these two equations for the two unknowns, FT1 , and FT2 , we get FT1 = 1.8102 N, and FT2 = 2.3102 N.

y

FT2

FT1

2

1

mg

13. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the support point A from the force diagram for the cantilever: ?A = – F2d + Mg! (d + D) = 0;



F2

F1 d

A – F2(20.0 m) + (1200 kg)(9.80 m/s2) !(20.0 m + 30.0 m) = 0, B 4 which gives F2 = 1.4710 N. For the forces in the y-direction we have  ?Fy = F1 + F2 – Mg = 0, or F1 = Mg – F2 = (1200 kg)(9.80 m/s2) – (1.47104 N) = – 2.94103 N (down).

Page 3

FN 2

L

A

= #[!(1100 kg)](9.80 m/s2) + !(1100 kg)(9.80 m/s2) = 6.74103 N. We write ?Fy = may from the force diagram for the two beams: FN1 + FN2 – Mg – mg = 0, which gives FN1 = Mg + mg – FN2

11. From the force diagram for the mass we can write ?Fx = FT1 – FT2 cos = 0, or FT1 = FT2 cos 30°. ?Fy = FT2 sin – mg = 0, or FT2 sin 30° = mg = (200 kg)(9.80 m/s2), which gives FT2 = 3.9103 N. Thus we have FT1 = FT2 cos 30° = (3.9103 N) cos 30° = 3.4103 N.

x

FN 1

D

Mg

x

Chapter 12

14. From the force diagram for the sheet we can write ?Fx = FT2 cos  – FT1 cos  = 0, which gives FT2 = FT1. ?Fy = FT1 sin  + FT2 sin  – mg = 0; 2FT1 sin  = mg; 2FT1 sin 3.5°= (0.60 kg)(9.80 m/s2), which gives FT1 = 48 N. The tension is so much greater than the weight because only the vertical components balance the weight.

y FT1

FT2 x 



mg

15. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the lower hinge B from the force diagram for the door:   ?B = FAx(H – 2D) – Mg!w = 0;

w

FAy A D

FAx FAx[2.30 m – 2(0.40 m)] – (13.0 kg)(9.80 m/s2)!(1.30 m), y H FBy which gives FAx = 55.2 N. Mg We write ?F = ma from the force diagram for the door: FBx x B ?Fx = FAx + FBx = 0; D 55 N + FBx = 0, which gives FBx = – 55.2 N. The top hinge pulls away from the door, and the bottom hinge pushes on the door. ?Fy = FAy + FBy – Mg = 0. Because each hinge supports half the weight, we have FAy = FBy = !(13.0 kg)(9.80 m/s2) = 63.7 N. Thus we have top hinge: FAx = 55.2 N, FAy = 63.7 N; bottom hinge: FBx = – 55.2 N, FBy = 63.7 N.

16. Because FT = FB , from the symmetry we see that F1 = F2 . We choose one handle as the system and write ? = I  about the hinge P: ?P = FTL1 cos  – F2L2 cos  = 0, which gives F2 = FTL1/L2 = (11.0 N)(8.50 cm)/(2.70 cm) = 34.6 N.

L1 F1 FT

P



FB

L2 F2

17.



(b) We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the pin B from the force diagram for the trailer:   ?B = FNA(a + b) – Mgb = 0; FNA = Mgb /(a + b) = (2.2103 kg)(9.80 m/s2)(5.5 m)/(2.5 m + 5.5 m) Page 4

(a)

FNB

FNA

B A

a

b Mg

Chapter 12



= 1.5104 N. (c) For the forces in the y-direction we have   ?Fy = FNA + FNB – Mg = 0; 1.48104 N + FNB – (2.2103 kg)(9.80 m/s2) = 0, which gives FNB =

18. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the elbow joint from the force diagram for the lower arm:   ? = mgD + MgL – FMd = 0; (2.8 kg)(9.80 m/s2)(0.12 m) + (7.3 kg)(9.80 m/s2)(0.300 m) – FM(0.025 m) = 0, 2 which gives FM = 9.910 N.

6.7103 N.

FJ

L

D

d

Elbow joint

CG

Mg

mg FM

19. Because the person is standing on one foot, the normal force on the ball of the foot must support the weight: FN = Mg. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the bone in the foot:   ?A = FTd – FND = 0; FTd – FN(2d) = 0, which gives FT = 2FN = 2(70 kg)(9.80 m/s2) = 1.4103 N (up).  We write ?Fy = may from the force diagram: FT + FN – Fbone = 0, which gives Fbone = FT + FN = 3FN = 3(70 kg)(9.80 m/s2) = 2.1103 N (down).

Fbone FT

20. We choose the coordinate system shown, with positive torques FM clockwise. (a) We write ? = Iabout the shoulder joint A from  B the force diagram for the arm: A     ?A = mgD – (FM sin )d = 0; FJ (3.3 kg)(9.80 m/s2)(0.24 m) – (FM sin 15°)(0.12 m) = 0, d which gives FM = 2.5102 N. D (b) We write ? = Iabout the point B from the force diagram for the arm:    ?B = mg(D – d) – (FJ sin )d = 0; (3.3 kg)(9.80 m/s2)(0.24 m – 0.12 m) – (FJ sin )(0.12 m) = 0, which gives FJ sin  = 32.3 N. For the forces in the x-direction we have    ?Fy = FJ cos  – FM cos  = 0, which gives FJ cos  = FM cos  = (2.5102 N) cos 15° = 2.41102 N. The magnitude of the force FJ is FJ = [(32.3 N)2 + (2.41102 N)2]1/2 = 2.4102 N.

Page 5

FN A d

D

CG

mg

y x

Chapter 12

21. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the shoulder joint A from the force diagram for the arm:   ?A = mgD + MgL – (FM sin )d = 0; (3.3 kg)(9.80 m/s2)(0.24 m) + (15 kg)(9.80 m/s2)(0.52 m) – (FM sin 15°)(0.12 m) = 0, 3 which gives FM = 2.710 N. Note that this is more than 10 times the result for Problem 20.

A

x

CM

 FJ d

mg

Mg

D L

22. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the support point A from the force diagram for the seesaw and boys: ?A = + m2g!L + m3gx – m1g!L = 0;

L A

+ (35 kg)!(3.6 m) + (25 kg)x – (50 kg)!(3.6 m) = 0, which gives x = 1.1 m. The girl should be 1.1 m from pivot on side of lighter boy.

23. We choose the coordinate system shown, with positive torques clockwise. For the torques about the center of gravity we have   ?CG = FN1(L – x) – FN2x = 0; (35.1 kg)g(170 cm – x) – (31.6 kg)gx = 0, which gives x = 89.5 cm from the feet.

y

FM

m 1g

x

m 3g

m 2g

FN

FN 1

FN 2

CG

A

B

x

mg L 24. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the beam and piano:   ?A = Mg#L + mg!L – FN2L = 0, which gives FN2 = #Mg + !mg

FN 1 A

L/4 = #(285 kg)(9.80 m/s2) + !(76.0 kg)(9.80 m/s2) 3 = 1.0710 N. Mg We write ?Fy = may from the force diagram for the beam and piano: FN1 + FN2 – Mg – mg = 0, which gives FN1 = Mg + mg – FN2 = (285 kg)(9.80 m/s2) + (76.0 kg)(9.80 m/s2) – 1.07103 N = 2.47103 N. The forces on the supports are the reactions to these forces: 2.47103 N down, and 1.07103 N down.

Page 6

FN 2

L

B mg

y x

Chapter 12

25. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the support point B from the force diagram for the beam: ?B = + F1(d1 + d2 + d3 + d4) – F3(d2 + d3 + d4) – F4(d3 + d4) – F5d4 – Mg!(d1 + d2 + d3 + d4) = 0; F1(10.0 m) – (4000 N)(8.0 m) – (3000 N)(4.0 m) – (2000 N)(1.0 m) – (250 kg)(9.80 m/s2)(5.0 m) = 0, F1 = 5.8103 N. which gives We write ? = Iabout the support point A from the force diagram for the beam: ?A = – F2(d1 + d2 + d3 + d4) + F3d1 + F4(d1 + d2) +

F3 F1

F4

d1

A

d2

F5 F2 d3 d4 B

y Mg x

F5(d1 + d2 + d3) + Mg!(d1 + d2 + d3 + d4) = 0; – F2(10.0 m) + (4000 N)(2.0 m) + (3000 N)(6.0 m) + (2000 N)(9.0 m) + (250 kg)(9.80 m/s2)(5.0 m) = 0, F2 = 5.6103 N. which gives

26. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the beam:   ?A = – (FT sin )L + Mg!L = 0;

L FW

FT

– FT sin 40° + (30 kg)(9.80 m/s2)! = 0,   which gives FT = 2.3102 N. A Note that we find the torque produced by the tension by finding Mg the torques produced by the components. We write ?F = ma from the force diagram for the beam: ?Fx = FWx – FT cos  = 0; FWx – (2.29102 N) cos 40° = 0, which gives FWx = 175 N. ?Fy = FWy + FT sin  – Mg = 0; FWy + (2.29102 N) sin 40° – (30 kg)(9.80 m/s2) = 0, which gives FWy = 147 N. For the magnitude of FW we have FW = (FWx2 + FWy2)1/2 = [(175 N)2 + (147 N)2]1/2 = 2.3102 N. We find the direction from tan  = FWy/FWx = (147 N)/(175 N) = 0.84, which gives = 40°. Thus the force at the wall is FW = 2.3102 N, 40° above the horizontal.

Page 7

y x B

Chapter 12

27. Because the backpack is at the midpoint of the rope, the angles are equal. The force exerted by the backpacker is FT2 the tension in the rope. From the force diagram for the backpack and junction we can write  ?Fx = FT1 cos  – FT2 cos  = 0, or FT1 = FT2 = F; ?Fy = FT1 sin  + FT2 sin  – mg = 0, or 2F sin  = mg. (a) We find the angle from tan  = h/!L = (1.5 m)/!(7.6 m) = 0.395, or  = 21.5°. When we put this in the force equation, we get 2F sin 21.5° = (16 kg)(9.80 m/s2), which gives F = 2.1102 N. (b) We find the angle from tan  = h/!L = (0.15 m)/!(7.6 m) = 0.0395, or  = 2.26°. When we put this in the force equation, we get 2F sin 2.26° = (16 kg)(9.80 m/s2), which gives F = 2.0103 N.

y FT1 

mg

28. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the board:   ?A = mg(!L) cos  – FDoorL sin  = 0, which gives FDoor = !mg/tan  = !(80 kg)(9.80 m/s2)/tan 45° = 3.9102 N. The reaction to this force is the force exerted on the door by the board. To open the door a person must push with this force at the edge: 3.9102 N. 29. Because the board will slide up on the door, the friction force will be down. With the board on the verge of slipping, we write ? = Iabout the point A from the force diagram for the board:   ?A = mg(!L) cos  – FDoorL sin  + FDoorL cos  = 0;

y

FDoor

FDoor =

L

FGy



31. We choose the coordinate system shown, with positive torques Page 8

mg O

FGx A y

FDoor

B

x

 FGx

Ffr

L

FGy

mg O

A

30. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the beam and sign:   ?A = – (FT sin )D + MgL + mg!L = 0; – FT (sin 35.0°)(1.35 m) + (215 N)(1.70 m) + (135 N)!(1.70 m) = 0, which gives FT = 620 N. Note that we find the torque produced by the tension by finding the torques produced by the components. We write ?F = ma from the force diagram for the beam and sign: ?Fx = Fhingex – FT cos  = 0; Fhingex – (620 N) cos 35.0° = 0, which gives Fhingex = 508 N. ?Fy = Fhingey + FT sin  – Mg – mg = 0; Fhingey + (620 N) sin 35.0° – 215 N – 135 N = 0, which gives Fhingey = – 6 N (down).

B

x

!(80 kg)(9.80 m/s2) cos 45° – FDoor sin 45° +

(0.45)FDoor cos 45° = 0, which gives 7.1102 N.

x

L D y Fhinge x

FT

Fhinge y A

mg

x

 B

Mg

Chapter 12



clockwise. We write ? = Iabout the point A from the force diagram for the pole and light:  ?A = – FTH + MgL cos  + mg!L cos  = 0; – FT (3.80 m) + (12.0 kg)(9.80 m/s2)(7.5 m) cos 37° +

FT

B

 FAV

H !(7.5 m) cos 37° = 0, (8.0 kg)(9.80 mg FT = 2.5102 N. which gives y  We write ?F = ma from the force diagram for the pole and light: A FAH ?Fx = FAH – FT = 0; 2 2 FAH – 2.510 N = 0, which gives FAH = 2.510 N. ?Fy = FAV – Mg – mg = 0; FAV – (12.0 kg)(9.80 m/s2) – (8.0 kg)(9.80 m/s2) = 0, which gives FAV = 2.0102 N. m/s2)

32. We choose the coordinate system shown, with positive torques counterclockwise. We write ? = Iabout the point A from the force diagram for the ladder:   ?A = mg(!¬) cos  – FN2¬ sin  = 0, which gives FN2 = mg/2 tan . We write ?Fx = max from the force diagram for the ladder: Ffr – FN2 = 0, which gives Ffr = FN2 = mg/2 tan .   We write ?Fy = may from the force diagram for the ladder: FN1 – mg = 0, which gives FN1 = mg. For the bottom not to slip, we must have Ffr = sFN1 , or mg/2 tan  = smg, from which we get tan  = 1/2s. The minimum angle is min = tan–1(1/2s).

33. We choose the coordinate system shown, with positive torques clockwise. (a) For the torques about the point B we have    ?B = FT1(!L + D) – MgD = 0; FT1(0.500 m + 0.400 m) – (0.230 kg)(9.80 m/s2)(0.400 m) = 0, which gives FT1 = 1.00 N. (b) For the torques about the point A we have    ?A = – FT2(!L + D) + Mg!L = 0; – FT2(0.500 m + 0.400 m) + (0.230 kg)(9.80 m/s2)(0.500 m), which gives FT2 = 1.25 N.

Page 9

Mg

x

B y

FN 2 x ¬ mg

 Ffr

O

A FN 1

y FT1

x

FT2

D

A

B Mg L

Chapter 12

34.





(a) We choose the coordinate system shown, with positive torques clockwise. (b) We write ?Fx = max from the force diagram: L FAH + FT sin  = 0, which gives FAV FAH = (70 N) sin 37°  = – 42 N (– x-direction).  FAH  We write ?Fy = may from the force diagram: x FAV + FT cos  – W – Mg = 0, which gives A W FAV = 20 N + (3.0 kg)(9.80 m/s2) – (70 N) cos 37° = – 6.5 N (– y-direction). (c) For the torques about the point A we have   ?A = Wx sin  + Mg!L sin  – FT cos  (L sin ) + FT sin  (L cos ) = 0;

 B

y x

Mg

(20 N) x sin 53° + (3.0 kg)(9.80 m/s2)!(5.0 m) sin 53° – (70 N) cos 37° (5.0 m) sin 53° + (70 N) sin 37° (5.0 m) cos 53° = 0, which gives x = 2.4 m. Note that we find the torque of FT by finding the torques of the components. 35. We choose the coordinate system shown, with positive torques clockwise. We assume the normal force acts a distance x from the edge of the crate. If the crate does not tip, ? = 0, which we write about the edge of the crate: (mg cos )!h – (mg sin )d – FNx = 0  We write ?Fy = may from the force diagram: FN – mg cos  = 0, which gives FN = mg cos . When we use this in the torque equation, we get x = !h – d tan . If the crate is not to tip over, x = 0; !h = d tan . , or max = 40°. tan max = !h/d = !(2.0 m)/(1.2 m) = 0.833, If the crate were sliding, the friction force would be kinetic, but in the limiting case would still act at the corner of the base, so the angle would be the same.

Page 10

y x

FN d

x

h

mg

 Ffr

FT



Chapter 12

36. We choose the coordinate system shown, with positive torques clockwise. We assume the normal force acts a distance x from the center line of the crate. In the accelerating frame of the truck, we add a pseudoforce – ma at the center of mass.  We write ?Fx = max from the force diagram: Ffr – ma = 0, which gives Ffr = ma.  We write ?Fy = may from the force diagram: FN – mg = 0, which gives FN = mg. If the crate does not tip, ? = 0, which we write about the center of mass of the crate: FNx – Ffr!h = 0. When we combine the equations, we get a = 2gx/h. Because the maximum value of x is !w, we have amax = gw/h = (9.80 m/s2)(1.0 m)/(2.5 m) = 3.9 m/s2.

w y ma

mg

FN

a

x

CM

h

x Ffr

37. We choose the coordinate system shown, with FB positive torques clockwise.  B We find the angle from  B cos  = !D/L = !(1.8 m)/(2.5 m) = 0.360,  = 68.9°. (a) For the torques on the left side of the ladder FB about the top point B we have FT d    ?B = FN1!D –FT!L sin  – Mg(L – d) cos  = 0. y For the torques on the right side of the ladder FT Mg about the top point B we have FN 1 x    ?B = – FN2!D + FT!L sin  = 0.  If we subtract the two equations, we get A D/2 (FN1 + FN2)!D = Mg(L – d) cos  + FTL sin . We write ?Fy = may for the entire ladder: FN1 + FN2 – Mg = 0, or FN1 + FN2 = Mg. When we use this in the previous equation, we get FT = Mg[!D – (L – d) cos  ]/L sin  = (60 kg)(9.80 m/s2)[0.90 m – (2.5 m – 2.0 m) cos 68.9°]/(2.5 m) sin 68.9° = 182 N. (b) From the second torque equation we get FN2 = FTL sin  /D = (182 N)(2.5 m) sin 68.9°/(1.8 m) = 236 N. From the force equation we get FN1 = Mg – FN2 = (60 kg)(9.80 m/s2) – 235 N = 352 N. (c) We write ?Fx = max for the right side of the ladder: FB cos  – FT = 0, or FB cos  = FT = 182 N. We write ?Fy = may for the right side of the ladder: – FB sin  + FN2 = 0, or FB sin  = FN2 = 236 N. We find the angle from tan  = FB sin /FB cos  = (236 N)/(182 N) = 1.30, so  = 52.4°. We find the magnitude from FB = FB cos /cos  = (182 N)/cos 52.4° = 298 N.

Page 11

L FN 2 C

Chapter 12

38. We choose the coordinate system shown, with positive torques clockwise. We write ?Fx = Max from the force diagram for the lamp: FP – Ffr = 0, which gives FP = Ffr.  We write ?Fy = May from the force diagram for the lamp: FN – Mg = 0, which gives FN = Mg. If the lamp slides without acceleration, we have FP = Ffr = FN = Mg. We write ? = Iabout the center of the base from the force diagram for the lamp:   ? = FPy – FNd = Mgy – Mgd = 0, or y = d/. The maximum height without tipping will be when d is maximum, which is L: ymax = L/ = (0.10 m)/(0.20) = 0.50 m = 50 cm. 39. From the symmetry of the wires, we see that the angle between a horizontal line on the ground parallel to the net and the line from the base of the pole to the anchoring point is  = 30°. We find the angle between the pole and a wire from tan  = d/H = (2.0 m)/(2.6 m) = 0.769, which gives  = 37.6°. Thus the horizontal component of each tension is FT sin . We write ? = Iabout the horizontal axis through the base A perpendicular to the net from the force diagram for the pole:   ?A = FnetH – 2(FT sin  cos )H = 0, or Fnet = 2FT sin  cos  = 2(95 N) sin 37.6° cos 30° = 1.0102 N.

FP

y

x

L

Fnet FT FT 

H A axis

41. (a) We find the stress from 1.2105 N/m2. Stress = F/A = (25,000 kg)(9.80 m/s2)/(2.0 m2) = (b) We find the strain from 2.410–6. Strain = Stress/E = (1.2105 N/m2)/(50109 N/m2) = 42. We use the strain to find how much the column is shortened: Strain = ?L/L0; 2.410–6 = ?L/(12 m), which gives ?L = 2.910–5 m = 0.029 mm. 43. (a) We find the stress from 1.3105 N/m2. Stress = F/A = (2000 kg)(9.80 m/s2)/(0.15 m2) = (b) We find the strain from 6.510–7. Strain = Stress/E = (1.3105 N/m2)/(200109 N/m2) = (c) We use the strain to find how much the girder is lengthened: Strain = ?L/L0; 6.510–7 = ?L/(9.50 m), which gives ?L = 6.210–6 m = 0.0062 mm.

Page 12

FN d

Ffr

40. We find the increase in length from the elastic modulus: E = Stress/Strain = (F/A)/(?L/L0); 5109 N/m2 = [(250 N)/p(0.50010–3 m)2]/[?L/(30.0 cm)], which gives ?L =

44. We find the volume change from ?P = – B ?V/V0 ; (2.6106 N/m2 – 1.0105 N/m2) = – (1.0109 N/m2)?V/(1000 cm3),

y

Mg



d

1.91 cm.

Chapter 12

which gives ?V = – 2.5 cm3. The new volume is V0 + ?V = 1000 cm3 + (– 2.5 cm3) =

998 cm3.

45. We find the elastic modulus from E = Stress/Strain = (F/A)/(?L/L0) = [(13.4 N)/#p(8.510–3 m)2]/[(0.37 cm)/(15 cm)] =

9.6106 N/m2.

46. The tension in each wire produces the stress. We find the strain from Strain = Stress/E = FT/EA. For wire 1 we have Strain1 = (1.8102 N)/(200109 N/m2)p(0.5010–3 m)2 = 1.110–3 = For wire 2 we have Strain2 = (2.4102 N)/(200109 N/m2)p(0.5010–3 m)2 = 1.510–3 =

0.11%. 0.15%.

47. The pressure needed is determined by the bulk modulus: ?P = – B ?V/V0 = – (90109 N/m2)(– 0.1010–2) = 9.0107 N/m2. 7 2 5 2 This is (9.010 N/m )/(1.010 N/m · atm) = 9.0102 atm. 48. We will take the change in pressure to be 200 atm. We find the volume change from ?P = – B ?V/V0 ; (200 atm)(1.0105 N/m2 · atm) = – (90109 N/m2)?V/V0 , which gives ?V/V0 = – 2.210–4 = – 0.022%. 49. We find the shear force applied along one edge of the square plate from G = Stress/Strain = (F/A)/(?L/L0); 80109 N/m2 = [F/(0.50 m)(0.011 m)]/(0.050), which gives F = 2.2107 N. 50. If we treat the abductin as an elastic spring, we find the effective spring constant from k = F/?L = EA/L0 = (2.0106 N/m2)(0.5010–4 m2)/(3.010–3 m) = 3.33104 N/m. We find the elastic potential energy stored in the abductin from U = !kx2 = !(3.33104 N/m)(1.010–3 m)2 = 0.017 J. 51. (a) For the torque from the sign’s weight about the point A we have    A = Mgd = (5.1 kg)(9.80 m/s2)(2.2 m) = 1.1102 m · N. (b) The balancing torque must be exerted by the wall, which is the only other contact point. Because the torque from the sign is clockwise, this torque must be counterclockwise. (c) If we think of the torque from the wall as a pull on the top of the pole and a push on the bottom of the pole, there is tension along the top of the pole and compression along the bottom. There must also be a vertical force at the wall which, in combination with the weight of the sign, will create a shear stress in the pole. Thus all three play a part.

FA d A

A

52. We find the maximum compressive force from the compressive strength of bone: Fmax = (Compressive strength)A = (170106 N/m2)(3.010–4 m2) = 5.1104 N.

Page 13

Mg

Chapter 12

53. We find the maximum tension from the tensile strength of nylon: FTmax = (Tensile strength)A = (500106 N/m2)p(0.5010–3 m)2 = 3.9102 N. We can increase the maximum tension by increasing the area, so we use thicker strings. The impulse on the ball that changes its momentum must be provided by an increased tension, so that the maximum strength is exceeded.

FT

Fball

FT

54. (a) We determine if the compressive strength, 1.7108 N/m2, is exceeded: Stress = F/A = (3.6104 N)/(3.610–4 m2) = 1.0108 N/m2. Because this is less than the compressive strength, the bone will not break. (b) We find the change in length from Strain = ?L/L0 = Stress/E, or ?L = (Stress)L0/E = (1.0108 N/m2)(0.20 m)/(15109 N/m2) = 1.310–3 m =

1.3 mm.

55. (a) We want the maximum stress to be (1/7.0) of the tensile strength: Stressmax = F/Amin = (Tensile strength)/7.0; (320 kg)(9.80 m/s2)/Amin = (500106 N/m2)/7.0, which gives Amin = 4.410–5 m2. (b) We find the change in length from Strain = ?L/L0 = Stress/E, or ?L = (Stress)L0/E = [(500106 N/m2)/7.0](7.5 m)/(200109 N/m2) = 2.710–3 m = 2.7 mm. 56. We choose the coordinate system shown, with positive torques F1 F2 clockwise. We write ? = Iabout the support point A from the force diagram for the cantilever: ?A = – F2d + Mg!L = 0; d – F2(20.0 m) + (2600 kg)(9.80 m/s2)(25.0 m) = 0, A B which gives F2 = 3.19104 N. We assume the force is parallel to the grain. We want the maximum stress to be (1/8.5) of the compressive strength: Stressmax2 = F/Amin2 = (Compressive strength)/8.5; (3.19104 N)/Amin2 = (35106 N/m2)/8.5, which gives Amin2 = 7.710–3 m2. For the forces in the y-direction we have   ?Fy = F1 + F2 – Mg = 0, or F1 = Mg – F2 = (2600 kg)(9.80 m/s2) – (3.19104 N) = – 6.42103 N (tension). We assume the force is parallel to the grain. We want the maximum stress to be (1/8.5) of the tensile strength: Stressmax1 = F/Amin1 = (Tensile strength)/8.5; (6.42103 N)/Amin1 = (40106 N/m2)/8.5, which gives Amin1 = 1.410–3 m2.

D

Mg

57. We want the maximum shear stress to be (1/6.0) of the shear strength: Stressmax = F/Amin = (Shear strength)/6.0; (3200 N)/#pdmin2 = (170106 N/m2)/6.0, which gives dmin = 1.210–2 m =

Page 14

1.2 cm.

Chapter 12

58. We find the required tension from ?Fy = may: FT – mg = ma, or FT = m(a + g) = (3100 kg)(1.2 m/s2 + 9.80 m/s2) = 3.41104 N. We want the maximum stress to be (1/7.0) of the tensile strength: Stressmax = F/Amin = (Tensile strength)/7.0; N)/#pdmin = which gives dmin = 2.510–2 m = (3.41104

2

(500106

+y

FT

a

N/m2)/7.0,

2.5 cm.

mg 59. We find the acceleration, and then the force, required to stop: v2 = v02 + 2as; 0 = (60 m/s)2 +2a(1.0 m); which gives a = – 1.8103 m/s2. The required force is F = ma = (75 kg)(– 1.8103 m/s2) = – 1.35105 N. This will produce a stress of Stress = F/A = (1.35105 N)/(0.30 m2) = 4.5105 N/m2. Because this is less than the ultimate strength of 5105 N/m2, it is possible to escape serious injury. 60. We choose the coordinate system shown. (a) From the symmetry of the force diagram for the truss, we see that F1 = F2 = !W. If we take the pin A as the object, we have the forces shown in the diagram, where the directions have been chosen so the forces can sum to zero. We write ?F = ma from the diagram: ?Fy = F1 – FAB sin 60° = 0;

B

60°

A F1

Page 15

F2

W F1

!(1.25104 N) = FAB sin 60°, which gives FAB = 7.22103 N. ?Fx = FAC – FAB cos 60° = 0; FAC – (7.22103 N) cos 60° = 0, which gives FAC = 3.61103 N. If we take the pin B as the object, we have the forces shown in the diagram, where the directions have been chosen so the forces can sum to zero. We write ?F = ma from the diagram: ?Fx = FAB cos 60° – FBD cos 60° = 0, which gives FBD = FAB = 7.22103 N. ?Fy = FAB sin 60° + FBD sin 60° – FBC = 0; 2(7.22103 N) sin 60° – FBC = 0, which gives FBC = 1.25104 N. From the symmetry we see that FCD = FAC = 3.61103 N. This can also be seen from the force diagram for pin C.

D

60°

C

FAC

A

y

60°

x

FAB FBD 60°

FBC

FAB 60°

B

FBC FAC

FCD C W

Chapter 12

(b) We see that FAB pushes on pin A, so pin A pushes on AB, a compression. FAC pulls on pin A, so AC is under tension. FBC pulls on pin B, so BC is under tension. Thus we have AC, CD, BC under tension; AB, BD under compression.

61. (a) We choose the coordinate system shown, with positive torques counterclockwise. If we take the truss as the object, we can find the tension in the cable and the force at point A. d D We write ? = Iabout the point A from the force diagram for the truss: d    ?A = W(2d) – FT(d sin 60°) = 0; 60° which gives FT = 2.31W = 2.31(56.0 kN) = 129 kN. 60° E We write ?F = ma from the diagram: C ?Fx = FT – FA cos  = 0, which gives FA cos  = FT = 129 kN. y W ?Fy = FA sin  – W = 0, which gives FA sin  = W = 56.0 kN. When we divide these two equations, we get FDE tan  = 0.434,  = 23.5°; then FA = 141 kN. (b) If we take the pin E as the object, we have the forces shown FCE E 60° in the force diagram. We write ?F = ma from the diagram: ?Fy = FDE sin 60° – W = 0; FDE sin 60° – 56.0 kN = 0, which gives W FDE = 64.7 kN (tension) . ?Fx = FDE cos 60° – FCE = 0; (64.7 kN) cos 60° – FCE = 0, which gives FBC FCE = 32.3 kN (compression) .

From the forces shown on pin D as the object, we can see that FBD 60° the only way for three forces that form an equilateral triangle FAC C 60° FCE to have a zero resultant is for the forces to be equal, so we have FCD = FBD = FDE = 64.7 kN, FCD with CD under compression and BD under tension. From the forces shown on pin C as the object, we have ?Fy = FBC sin 60° – FCD sin 60° = 0, which gives FBC = FCD = 64.7 kN (tension). ?Fx = FBC cos 60° + FCD cos 60° + FCE – FAC = 0; (64.7 kN) cos 60° + (64.7 kN) cos 60° + 32.3 kN – FAC = 0, which gives FAC = 97.0 kN (compression). From the forces shown on pin B as the object, we have ?Fy = FAB sin 60° – FBC sin 60° = 0, which gives FAB = FBC = 64.7 kN (compression). ?Fx = FT – FAB cos 60° – FBC cos 60° – FBD = 0; 129 kN – (64.7 kN) cos 60° – (64.7 kN) cos 60° – 64.7 kN = 0, which gives a check on the calculations (within significant figures).

Page 16

FT

B FA 

A

x FCD D

60° 60°

FBD

FDE FAB 60°

B

60°

FBC

FT

Chapter 12

62. We choose the coordinate system shown, for one of the two trusses. From the symmetry of the force diagram for the B truss, we see that F1 = F2 = Mg/4. F1 This could be also be found by using ? = 0 about 60° one of the supports. If we take the pin A as the object, we have the forces A shown in the diagram, where the directions have been chosen so the forces can sum to zero. Mg /8 Mg /4 We write ?F = ma from the diagram: ?Fy = F1 – FAB sin 60° – Mg/8 = 0; (Mg/4) – FAB sin 60° – (Mg/8) = 0, F1 which gives FAB = Mg/4v3. FAC ?Fx = FAC – FAB cos 60° = 0; A 60° FAC – (Mg/4v3) cos 60° = 0, F AB which gives FAC = Mg/8v3. Mg /8 If we take the pin B as the object, we have the forces shown in the diagram, where the directions have been chosen so the forces can sum to zero. We write ?F = ma from the diagram: ?Fy = FAB sin 60° + FBC sin 60° = 0; which gives FBC = FAB = Mg/4v3. ?Fx = FAB cos 60° + FBC cos 60° – FBD = 0, 2(Mg/4v3) cos 60° – FBD = 0, which gives FBD = Mg/4v3. From the symmetry we see that FCD = FBC = FAB = FDE = FBD = Mg/4v3; FCE = FAC = Mg/8v3.

D F2 60°

C

E

L

Mg /8 y x B FBD

FAB 60° 60°

FBC

We see that each force is # of the force from Example 12-15.

63. (a) Because the tensile strength of steel is the same as the compressive strength, we choose a member with the greatest stress, FAB = Mg/v3. The safety condition is Stress = FAB /A = (strength)/6.0, or A = (6.0)FAB /strength = (6.0)(7.0105 kg)(9.80 m/s2)/(500106 N/m2)v3 = 4.810–2 m2. (b) If we simplify by putting the additional mass at the center of the bridge, and remember that each truss will support half of the load of the trucks, we have A = (6.0)FAB /strength = (6.0)[(7.0105 kg + !(50)(1.2104 kg)](9.80 m/s2)/(500106 N/m2)v3 =

Page 17

6.810–2 m2.

Chapter 12

64. We choose the coordinate system shown, with positive torques counterclockwise. If we take one of the trusses, which supports half of the B D weight, as the object, we can find the forces at the supports. We write ? = Iabout the point A from the force diagram F1 for the truss:   ?A = F2L – (Mg/2)(L/4) = 0; 60° 60° which gives F2 = Mg/8 = (44103 kg)(9.80 m/s2)/8 A C = 5.39104 N. L We write ? = Iabout the point E:   ?E = – F1L + (Mg/2)(3L/4) = 0; Mg /2 y which gives F1 = 3Mg/8 = 3(44103 kg)(9.80 m/s2)/8 FAC y FAC y = 1.62105 N. If we take the strut AC as the object, we can find the forces AC exerts on the pins at A and C by finding their reactions C FAC x A FAC x acting on the strut. The weight of the train at the center of the strut requires vertical forces at the pins. Because the Mg /2 weight of the train is at the midpoint of the strut, we can use symmetry to show the forces in the force diagram. F1 Thus we have 2FACy = Mg/2, or FAC x B A FACy = Mg/4 = (44103 kg)(9.80 m/s2)/4 = 1.04105 N. 60°

F2

E

x

FAB 60° 60°

If we take the pin A as the object, we have the forces shown FBD FAB FAC y FBC in the force diagram. We write ?F = ma from the diagram: ?Fy = F1 – FACy – FAB sin 60° = 0; (3Mg/8) – (Mg/4) – FAB sin 60° = 0, which gives FDE FBC FCD FAB = Mg/4v3 = 6.2104 N (compression). 60° 60° D 60° ?Fx = FACx – FAB cos 60° = 0; 60° FAC x C FCE FBD FACx – (Mg/4v3) cos 60° = 0, which gives FAC y 4 F FACx = Mg/8v3 = 3.110 N (tension). CD From the forces shown on pin B as the object, we have ?Fy = FAB sin 60° – FBC sin 60° = 0, which gives FBC = FAB = Mg/4v3 = 6.2104 N (tension). ?Fx = FAB cos 60° + FBC cos 60° – FBD = 0; (Mg/4v3) cos 60° + (Mg/4v3) cos 60° – FBD = 0, which gives FBD = Mg/4v3 = 6.2104 N (compression). From the forces shown on pin C as the object, we have ?Fy = FCD sin 60° + FBC sin 60° – FACy = 0; FCD sin 60° + (Mg/4v3) sin 60° – (Mg/4) = 0, which gives FCD = Mg/4v3 = 6.2104 N (tension). ?Fx = FCD cos 60° – FBC cos 60° + FCE – FACx = 0; (Mg/4v3) cos 60° – (Mg/4v3) cos 60° + FCE – (Mg/8v3) = 0, which gives FCE = Mg/8v3 = 3.2104 N (tension). From the forces shown on pin D as the object, we have ?Fy = FDE sin 60° – FCD sin 60° = 0, which gives FDE = FCD = Mg/4v3 = 6.2104 N (compression). Page 18

Chapter 12

?Fx = FBD – FDE cos 60° – FCD cos 60° = 0; (Mg/4v3) – (Mg/4v3) cos 60° – (Mg/4v3) cos 60° = 0, which gives a check on the calculations. Evaluating the forces at pin E will also confirm the calculations. Note that the vertical forces in strut AC create a shear stress.

65. We choose the coordinate system shown, with positive torques counterclockwise. If we take one of the trusses, which supports half of the B D weight, as the object, we can find the forces at the supports. We write ? = Iabout the point A from the force diagram F2 for the truss: F1   ?A = F2L – (Mg/2)d = 0, which gives 60° 60° F2 = Mgd/2L = (28103 kg)(9.80 m/s2)(22 m)/2(64 m) A E C d = 4.72104 N. L We write ? = Iabout the point E:   ?E = – F1L + (Mg/2)(L – d) = 0, which gives Mg /2 y F1 = Mg(L – d)/2L FAy FCy = (28103 kg)(9.80 m/s2)(64 m – 22 m)/2(64 m) x = 8.99104 N. If we take the strut AC as the object, we can find the forces C FAC x A FACx AC exerts on the pins at A and C by finding their reactions acting on the strut. The weight of the truck on Mg /2 the strut requires vertical forces at the pins. We write ? = Iabout the point A for the strut: F1 FAB   ?A = FCy(L /2) – (Mg/2)d = 0, which gives FAC x B 3 2 A 60° FCy = Mgd/L = (2810 kg)(9.80 m/s )(22 m)/(64 m) 60° 60° 4 = 9.4310 N. FBD FAB F FBC Ay We write ? = Iabout the point C:   ?C = – FAy(L /2) + (Mg/2)[(L /2) – d] = 0, which gives FAy = Mg[(L /2) – d] /L FDE FBC FCD = (28103 kg)(9.80 m/s2)(32 m – 22 m)/(64 m) 60° 60° D 60° = 4.28104 N. 60° FACx C If we take the pin A as the object, we have the forces shown FCE FBD FCy in the force diagram. We write ?F = ma from the diagram: FCD ?Fy = F1 – FAy – FAB sin 60° = 0; (8.99104 N) – (4.28104 N) – FAB sin 60° = 0, which gives FAB = 5.44104 N (compression). ?Fx = FACx – FAB cos 60° = 0; FACx – (5.44104 N) cos 60° = 0, which gives FACx = 2.72104 N (tension). From the forces shown on pin B as the object, we have ?Fy = FAB sin 60° – FBC sin 60° = 0, which gives FBC = FAB = 5.44104 N (tension). ?Fx = FAB cos 60° + FBC cos 60° – FBD = 0; (5.44104 N) cos 60° + (5.44104 N) cos 60° – FBD = 0, which gives FBD = 5.44104 N (compression). From the forces shown on pin C as the object, we have Page 19

Chapter 12

?Fy = FCD sin 60° + FBC sin 60° – FCy = 0; FCD sin 60° + (5.44104 N) sin 60° – (9.43104 N) = 0, which gives FCD = 5.44104 N (tension). ?Fx = FCD cos 60° – FBC cos 60° + FCE – FACx = 0; (5.44104 N) cos 60° – (5.44104 N) cos 60° + FCE – (2.72104 N) = 0, which gives FCE = 2.72104 N (tension). From the forces shown on pin D as the object, we have ?Fy = FDE sin 60° – FCD sin 60° = 0, which gives FDE = FCD = 5.44104 N (compression). ?Fx = FBD – FDE cos 60° – FCD cos 60° = 0; 5.44104 N – (5.44104 N) cos 60° – (5.44104 N) cos 60° = 0, which gives a check. Evaluating the forces at pin E will also confirm the calculations. 66. We choose the coordinate system shown. From the symmetry of the force diagram for the truss, we see that D G B F1 F2 F1 = F2 = 5F/2. This could be also be found by using ? = 0 about a J A 45° one of the supports. a E H C If we take the pin A as the object, we have the forces shown F F F F F in the force diagram. We write ?F = ma from the diagram: W ?Fy = F1 – F – FAB sin 45° = 0; (5F/2) – F – FAB sin 45° = 0, which gives F1 y FAB = 3F/v2 (compression). FAC A ?Fx = FAC – FAB cos 45° = 0; 45° x FAC – (3F/v2) cos 45° = 0, which gives FAB F FAC = 3F/2 (tension). FAB FBC From the forces shown on pin C as the object, we have FAC FCE ?Fy = FBC – F = 0, which gives FBC = F (tension). 45° B 45° ?Fx = FCE – FAC = 0, which gives FCE = FAC = 3F/2 FBD C FBE F (tension). FBC From the forces shown on pin B as the object, we have FDE FEG ?Fy = FAB sin 45° – FBE sin 45° – FBC = 0; FBE F F BD (3F/v2) sin 45° – FBE sin 45° – F = 0, DG D F 45° 45° EH which gives FBE = F/v2 (tension). FCE E FDE ?Fx = FAB cos 45° + FBE cos 45° – FBD = 0; (3F/v2) cos 45° + (F/v2) cos 45° – FBD = 0, F which gives FBD = 2F (compression). From the forces shown on pin D as the object, we have ?Fy = FDE = 0. ?Fx = FBD – FDG = 0, which gives FDG = FBD = 2F (compression) The members in the right half of the truss will mirror those on the left. This can be confirmed by evaluating forces at the other pins. Thus we have FGJ = FAB = 3F/v2 (compression), FGH = FBC = F (tension), FAC = FHJ = FCE = FEH = 3F/2 (tension), FEG = FBE = F/v2 (tension), FDG = FBD = 2F (compression), FDE = 0.

Page 20

Chapter 12

67. In each arch the horizontal force at the base must equal the horizontal force at the top. Because the two arches support the same load, we see from the force diagrams that the vertical forces will be the same and have the same moment arms. Thus the torque about the base of the horizontal force at the top must be the same for the two arches:  = Froundhround = FroundR = Fpointedhpointed; Fround(4.0 m) = @Froundhpointed , which gives hpointed = 12 m.

Fload

Fload Fround F2 hround

F4 A

R

F 2.5106 N.

hpointed F2

F1 A

68. We find the required tension from ?Fy = may: 2F sin  – Fload = 0; 2F sin 5° – 4.3105 N = 0, which gives F =

Fpointed

F



R

y 

x Fload

69. All elements are in equilibrium. For the C-D pair, L1 L2 we write ? = Iabout the point c from the force diagram: a L3 L4 ?c = MCgL6 – MDgL5 = 0; MC(5.00 cm) – MD(17.50 cm) = 0, or MC = 3.500MD. b L5 L6 M Ag The center of mass of C and D must be under the point c. c We write ? = Iabout the point b from the force diagram: M Bg ?b = MBgL4 – (MC + MD)gL3 = 0; MCg M Dg (0.735 kg)(5.00 cm) – (MC + MD)(15.00 cm) = 0, or MC + MD = 0.245 kg. When we combine these two results, we get MC = 0.191 kg, and MD = 0.0544 kg. The center of mass of B, C, and D must be under the point b. For the entire mobile, we write ? = Iabout the point a from the force diagram: ?a = (MB + MC + MD)gL2 – MAgL1 = 0; (0.735 kg + 0.245 kg)(7.50 cm) – MA(30.00 cm) = 0, which gives MA = 0.245 kg. 70. Because the walker is at the midpoint of the rope, the angles y are equal. We find the angle from tan  = h/!L = (3.4 m)/!(46 m) = 0.148, or  = 8.41°. FT2 FT1 From the force diagram for the walker we can write ?Fx = FT1 cos  – FT2 cos  = 0, or FT1 = FT2 = FT;   ?Fy = FT1 sin  + FT2 sin  – mg = 0, or 2FT sin 8.41° = (60 kg)(9.80 m/s2), which gives mg FT = 2.0103 N. No. There must always be an upward component of the tension to balance the weight.

Page 21

x

Chapter 12

71. (a) The cylinder will roll about the contact point A. We write  = I about the point A: Fa(2R – h) + FN1[R2 – (R – h)2]1/2 – Mg[R2 – (R – h)2]1/2 = IA. When the cylinder does roll over the curb, contact with the ground is lost and FN1 = 0. Thus we get Fa = {IA + – (R – }/(2R – h) = [IA/(2R – h)] + [Mg(2Rh – h2)1/2/(2R – h)]. The minimum force occurs when  = 0: Famin = Mg[h(2R – h)]1/2/(2R – h) = Mg[h/(2R – h)]1/2. (b) The cylinder will roll about the contact point A. We write  = I about the point A: Fb(R – h) + FN1[R2 – (R – h)2]1/2 – Mg[R2 – (R – h)2]1/2 = IA. When the cylinder does roll over the curb, contact with the ground is lost and FN1 = 0. Thus we get Mg[R2

Fa Fb

h)2]1/2

A

R

FN 1

h Mg

FN 2

Fb = {IA + Mg[R2 – (R – h)2]1/2}/(R – h) = [IA/(R – h)] + [Mg(2Rh – h2)1/2/(R – h)]. The minimum force occurs when  = 0: Fbmin = Mg[h(2R – h)]1/2/(R – h).

72. We choose the coordinate system shown, with positive torques clockwise. We write ? about the rear edge from the force diagram: ?edge = mg!L – FA!H = (1.8108 N)!(40 m) – (950 N/m2)(200 m)(70 m)!(200 m) = + 2.3109 m · N. Because the result is positive, the torque is clockwise, so the building will not topple. An alternative procedure is to find the location of the force FEy = mg. We write ? = 0 about the middle of the base from the force diagram: ?C = FEyx – FA!H = 0; (1.8108 N)x – (950 N/m2)(200 m)(70 m)!(200 m) = 0, which gives x = 7.4 m. Because this is less than 20 m, the building will not topple.

Page 22

L

y

H

FA FEy mg

x FEx B C x

Chapter 12

73. If the vertical line of the weight falls within the base of the truck, there will be a normal force FN1 required to make the net torque about the center of mass zero, and the truck will not tip over. The limiting case will be when the line passes through the corner of the base. Thus we find the limiting angle from max = 29°. tan max = !w/d = !(2.4 m)/(2.2 m) = 0.545, or

w

CG

d

mg FN1

h FN2 



74. (a) From Example 9–7, the force of the ground on one leg is Fleg = !(2.1105 N) = 1.05105 N. We find the stress in the tibia bone from Stress = Fleg/A = (1.05105 N)/(3.010–4 m) = 3.5108 N/m2. (b) The compressive strength of bone is 1.7108 N/m2. Thus the bone will break. (c) From Example 9–7, the force of the ground on one leg is Fleg = !(4.9103 N) = 2.45103 N. We find the stress in the tibia bone from Stress = Fleg/A = (2.45103 N)/(3.010–4 m) = 8.2106 N/m2. This is less than the compressive strength of bone, so the bone will not break. 75. The force is parallel to the grain. We want the maximum stress to be (1/12) of the compressive strength. For N studs we have Stressmax = (Mg/N)/A = (Compressive strength)/12; (12,600 kg)(9.80 m/s2)/N(0.040 m)(0.090 m) = (35106 N/m2)/12, which gives N = 11.8. Thus we need 6 studs on each side. There are five spaces between the studs, so they will be (10.0 m)/5 = 2.0 m apart.

76. From the force diagram for the section we can write ?Fx = FT1 cos 1 – FT2 sin 2 = 0, or FT1 cos 19° – FT2 sin 60° = 0; ?Fy = – FT1 sin 1 + FT2 cos 2 – mg = 0, or – FT1 sin 19° + FT2 cos 60° = mg. When we combine these equations, we get FT1 = 4.54mg, and FT2 = 4.96mg. We write ? = Iabout the point A from the force diagram: ?A = – (FT2 sin 2)h + (FT1 sin 1)d1 + mg!d1 = 0;

FT2



y

h

x

A

– (4.96mg sin 60°)h + (4.54mg sin19°)(343 m) + mg!(343 m) = 0, which gives h = 158 m. Page 23

d1

mg

FT1



Chapter 12

North span 77. Because there is no net horizontal force  Tower on the tower, from the force diagram for d1 FT2 FT3 the tower we can write FT2 ?Fx = FT2 sin 2 – FT3 sin 3 = 0, or   FT3 = FT2 (sin 2)/(sin 3). h From the force diagram for the north span FN we can write y A ?Fx = FT1 cos 1 – FT2 sin 2 = 0, or mg FT1 = FT2 (sin 2)/(cos 1). x ?Fy = + FT2 cos 2 – FT1 sin 1 – mg = 0, d2/2 or mg = FT2 cos 2 – FT1 sin 1. From the force diagram for one-half of the center span we can write FT4 ?Fy = + FT3 cos 3 – !Mg = 0, or Mg = 2FT3 cos 3. Center span Because the roadway is uniform, the Mg /2 length of each roadway is proportional to the mass: d2/d1 = M/m = (2FT3 cos 3)/(FT2 cos 2 – FT1 sin 1)

FT1 

FT3

= 2[FT2 (sin 2)/(sin 3)](cos 3)/{FT2 cos 2 – [FT2 (sin 2)/(cos 1)](sin 1)} = 2(cot 3)/(cot 2 – tan 1) = 2(cot 66°)/(cot 60° – tan 19°) = 3.8.

78. The minimum mass is placed symmetrically between two of the legs of the table, so the normal force on the opposite leg becomes zero, as shown in the top view of the table. We write ? = Iabout a horizontal axis that passes through the two legs where there is a normal force: ? = – MgR cos + mgR(1 – cos ) = 0; – (36 kg) cos 60° + m(1 – cos 60°) = 0, which gives m = 36 kg.

Fleg

Mg

(up)

(down)

mg (down)

 R

Fleg (up) axis 79. From the force diagram for the aircraft we can write ?Fx = FT – FD = 0, or h FD = FT = 5.0105 N. FD CM ?Fy = FL – W = 0, or FL = W = (77,000 kg)(9.80 m/s2) = 7.55105 N. We write ? = Iabout the center of mass: W ?CM = FDh – FLd + FTH = 0. When we use the previous results, we get FT(h + H) = Wd; (5.0105 N)(h + 1.6 m) = (7.55105 N)(3.2 m), which gives h = 3.2 m. Page 24

d

H

FL FT



Chapter 12

80. We select one-half of the cable for our system. From the force diagram for the section we can write ?Fx = FT1 – FT2 cos  = 0; FT1 – FT2 cos 60° = 0.

FT2

y



x

?Fy = + FT2 sin  – !mg = 0;

FT1

FT2 sin 60° – !mg = 0. When we combine these equations, we get mg /2 (a) FT1 = mg/(2 tan 60°) = 0.289mg; (b) FT2 = mg/(2 sin 60°) = 0.577mg. (c) The direction of the tension in each case is tangent to the cable: horizontal at the lowest point, and 60° above the horizontal at the attachment. 81. We choose the coordinate system shown, with positive torques clockwise. (a) The maximum weight will cause the force FA to be zero. FA We write ? = Iabout the support point B from the force diagram for the beam and person: ?B = – W(!L – d2) + wd2 + FAD = 0; A – (600 N)[!(20.0 m) – 5.0 m] + wmax(5.0 m) + 0 = 0, d1 which gives wmax = 600 N. (b) The maximum weight means the force FA = 0. We write ? = Iabout the support point A from the force diagram for the beam and person: ?A = + W(!L – d1) + w(D + d2) – FBD = 0;

FB L B x

D W

y

w d2

x

FB = 1200 N. + (600 N)[!(20.0 m) – 3.0 m] + (600 N)(17.0 m) – FB(12.0 m) = 0, which gives (c) We write ? = Iabout the support point B from the force diagram for the beam and person: ?B = – W(!L – d2) + wx + FAD = 0; FA = 150 N. – (600 N)[!(20.0 m) – 5.0 m] + (600 N)(2.0 m) + FA(12.0 m) = 0, which gives We write ? = Iabout the support point A from the force diagram for the beam and person: ?A = + W(!L – d1) + w(D + x) – FBD = 0; + (600 N)[!(20.0 m) – 3.0 m] + (600 N)(12.0 m + 2.0 m) – FB(12.0 m) = 0, which gives FB = 1050 N. (d) We write ? = Iabout the support point B from the force diagram for the beam and person: ?B = – W(!L – d2) + wx + FAD = 0; FA = 750 N. – (600 N)[!(20.0 m) – 5.0 m] + (600 N)(– 10.0 m) + FA(12.0 m) = 0, which gives We write ? = Iabout the support point A from the force diagram for the beam and person: ?A = + W(!L – d1) + w(D + x) – FBD = 0; N.

+ (600 N)[!(20.0 m) – 3.0 m] + (600 N)(12.0 m – 10.0 m) – FB(12.0 m) = 0, which gives

82. We choose the coordinate system shown, with positive torques clockwise.

Page 25

FB = 450

Chapter 12









(a) We write ?Fx = max from the force diagram: FGx – FW = 0, or FGx = FW.   We write ?Fy = may from the force diagram: FGy – mg = 0, which gives FGy = mg = (15.0 kg)(9.80 m/s2) = 147 N. We write ? = Iabout the point A from the force diagram for the ladder:   ?A = mg(!¬ sin ) – FW¬ cos  = 0;

FW B

y x

 ¬

Mg d

mg (15.0 kg)(9.80 m/s2)!(sin 20.0°) – FW(cos 20.0°) = 0, which gives FW = 26.8 N. FGx Thus the components of the force at the ground are O A FGx = 26.8 N, FGy = 147 N. FGy (b) We write ?Fx = max from the force diagram: FGx – FW = 0, or FGx = FW.  We write ?Fy = may from the force diagram: FGy – mg – Mg = 0, which gives FGy = (m + M)g = (15.0 kg + 70 kg)(9.80 m/s2) = 833 N. We write ? = Iabout the point A from the force diagram for the ladder and person:   ?A = mg(!¬ sin ) + Mg(d sin ) – FW¬ cos  = 0; (15.0 kg)(9.80 m/s2)!(7.0 m)(sin 20.0°) + (70 kg)(9.80 m/s2)&(7.0 m)(sin 20.0°) – FW(7.0 m)(cos 20.0°) = 0, which gives FW = 214 N = FGx. Because the ladder is on the verge of slipping, we must have FGx = FGy , or  = FGx/FGy = (214 N)/(833 N) = 0.257.

83. Because the backpack is at the midpoint of the rope, the angles are equal. From the force diagram for the backpack and junction we can write ?Fx = FT1 cos  – FT2 cos  = 0, or FT1 = FT2 = F; ?Fy = FT1 sin  + FT2 sin  – mg – Fbear = 0, or 2 F sin  = mg + Fbear. When the bear is not pulling, we have 2F1 sin 1 = mg; 2F1 sin 15° = (23.0 kg)(9.80 m/s2), which gives F1 = 435 N. When the bear is pulling, we have 2F2 sin 2 = mg + Fbear; 2(2)(435 N) sin 30° = (23.0 kg)(9.80 m/s2) + Fbear , which gives Fbear = 6.5102 N.

y FT2

FT1 



Fbear

x

mg

84. The maximum stress in a column will be at the bottom, caused by the weight of the material. If the column has density , height h, and area A, we have Stress = F/A = mg/A = Vg/A = Ahg/A = gh, which is independent of area. The column will buckle when this stress exceeds the compressive strength: hmax = (Compressive strength)/g. (a) For steel we have hmax = (500106 N/m2)/(7.8103 kg/m3)(9.80 m/s2) = 6.5103 m. (b) For granite we have hmax = (170106 N/m2)/(2.7103 kg/m3)(9.80 m/s2) = 6.4103 m.

Page 26

Chapter 12

85. We assume when the brick strikes the floor there is an average force which produces an average stress in the brick, creating an average strain: ?L/L0 = (F/A)/E. If we use this average strain for the distance the CM moves while the brick comes to rest, the work done by the average force is – F ?L. When we use the work-energy principle from the release point to the final resting point, we have – F ?L = ?K + ?U = 0 + (0 – mgh), or h = F ?L/mg = F(F/A)L0/Emg = A(F/A)2L0/Emg. If we assume that the stress varies linearly from zero at the top of the brick to maximum at the bottom, the brick will break when the average stress exceeds one-half the compressive strength, so we have hmin = (0.150 m)(0.060 m)[!(35106 N/m2)]2(0.040 m)/(14109 N/m2)(1.2 kg)(9.80 m/s2) = 0.67 m. Note that AL0 is the volume of the brick, so the answer is independent of which face strikes the ground. 86. We write ?Fx = max from the force diagram: F – Ffr = 0, or F = Ffr.  We write ?Fy = may from the force diagram: FN – Mg = 0, or FN = Mg. We find the location of the force FN when the static friction force reaches its maximum value: F = Ffr =sFN = sMg. We write ? = 0 about the edge A of the block from the force diagram: ?A = – Mg!¬ + FNx + Fh = 0, which gives

¬ F

y ¬

Mg

FN A

Ffr x

h

x

x = (!Mg¬ – Fh)/FN = (!Mg¬ – sMgh)/Mg = !¬ – sh. (a) For the block to slide, we must have x > 0, or !¬ > sh, which gives s < ¬/2h. (b) For the block to tip, we must have x < 0, or !¬ < sh, which gives s > ¬/2h. 87. The ropes can only provide a tension, so the scaffold will be stable D if FT1 and FT2 are greater than zero. The tension will be least in FT1 FT2 the rope farthest from the painter. To find how far the painter can walk from the right rope toward the right end, we set FT1 = 0. d d d d We write ? = 0 about B from the force diagram: A B   ?B = Mgxright + FT1(D + 2d) – mpailg(D + d) – mgD = 0; m pail g 2 2 (60 kg)(9.80 m/s )xright + 0 – (4.0 kg)(9.80 m/s )(2.0 m + 1.0 m) – Mg mg (25 kg)(9.80 m/s2)(2.0 m) = 0, L which gives xright = 1.03 m. Because this is greater than the distance to the end of the plank, 1.0 m, walking to the right end is safe. To find how far the painter can walk from the left rope toward the left end, we set FT2 = 0. We write ? = 0 about A from the force diagram:   ?A = – Mgxleft – FT2(D + 2d) + mpailgd + mg2d = 0; – (60 kg)(9.80 m/s2)xleft – 0 + (4.0 kg)(9.80 m/s2)(1.0 m) + (25 kg)(9.80 m/s2)2(1.0 m) = 0, which gives xleft = 0.90 m. Because this is less than the distance to the end of the plank, 1.0 m, walking to the left end is not safe. The painter can safely walk to within 0.10 m of the left end.

Page 27

Chapter 12

88. We choose positive torques clockwise. We write ? = 0 about the point B from the force diagram: ?B = – Mg(L – D) – mg!L + F1L = 0, which gives F1 = [Mg(L – D)/L] + !mg = ({M[1 – (vt/L)]} + !m)g

= {(95,000 kg)[1 – (80 km/h)t/(3.6 ks/h)(220 m)] + !(23,000 kg)}(9.80 m/s2) = 1.04106 N – (9.40104 N/s)t. We write ? = 0 about the point A from the force diagram: ?A = MgD + mg!L – F2L = 0, which gives

F1

F2

L

B

A D = vt

mg

Mg

y x

F2 = (MgD/L) + !mg = [(Mvt/L) + !m]g = [(95,000 kg)(80 km/h)t/(3.6 ks/h)(220 m) + !(23,000 kg)](9.80 m/s2) 5 = 1.1310 N + (9.40104 N/s)t. 89. (a) We choose positive torques clockwise. We write ? = 0 about the point A from the force diagram: ?A = mg!L – FLD = 0, which gives FL = mgL/2D = (10 kg)(9.80 m/s2)(2.0 m)/2(0.30 m) = 3.3102 N up. We write ? = 0 about the point B from the force diagram: ?B = mg(!L – D) – FRD = 0, which gives

FL L A D

B mg

FR

y x

FR = mg(!L – D)/D = (10 kg)(9.80 m/s2)[!(2.0 m) – 0.30 m]/(0.30 m) = 2.3102 N down. (b) For the force arrangement shown, FL = FR + mg, so FL > FR. We write ? = 0 about the point A from the force diagram, with FL = 150 N: ?A = mg!L – FLD = 0, which gives

!(10 kg)(9.80 m/s2)(2.0 m) = (150 N)D, which gives D = 0.65 m =

65 cm from right hand. (c) We write ? = 0 about the point A from the force diagram, with FL = 80 N: ?A = mg!L – FLD = 0;

!(10 kg)(9.80 m/s2)(2.0 m) = (80 N)D, which gives D = 1.23 m =

123 cm from right hand. To check on the magnitude of FR , we write ? = 0 about the new point B: ?B = – mg(D – !L) – FRD = 0, (10 kg)(9.80 m/s2)[1.23 m – !(2.0 m)] = – FR(1.23 m), which gives FR = – 18 N (up), which is less than 80 N.

Page 28

Chapter 12

90. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force diagram for the beam and mass:   ?A = – (FT sin )¬ + mgx + Mg!¬ = 0, FT = [(mx/¬) + !M]g/ sin  which gives We write ?F = ma from the force diagram for the beam and mass: ?Fx = FhingeH – FT cos  = 0;

¬ y x

Fhinge V

FT

Fhinge H A

FhingeH – {[(mx/¬) + !M]g/ sin } cos  = 0,

 B

x

FhingeH = [(mx/¬) + !M]g/ tan . which gives ?Fy = FhingeV + FT sin  – Mg – mg = 0;

Mg

mg

FhingeV – {[(mx/¬) + !M]g/ sin } sin  – (m + M)g= 0,

which gives

FhingeV = [(m(1 – x/¬) + !M]g.

91. From the symmetry we see that the force each beam exerts on the other must be horizontal. We write ?F = ma from the force diagram for the left beam: ?Fx = Ffr – F = 0, or F = Ffr. ?Fy = FN – Mg = 0, or FN = Mg. We write ? = Iabout the point B from the force diagram:   ?B = FNL cos  – Mg!L cos  – FfrL sin  = 0; MgL cos  – Mg!L cos  – FfrL sin  = 0; which gives Ffr = Mg/2 tan  For the friction to remain static, we have Ffr = Mg/2 tan  = µsMg, so tan  = 1/2µs = 1.2(0.60) = 0.833,  = 40°.

B F L Mg FN



y x

Ffr A

92. We find the angles of the truss from B tan A = h/d1 = (6.0 m)/(4.0 m) = 1.50, A = 56.3°. tan D = h/d2 = (6.0 m)/(6.0 m) = 1.00, D = 45.0°. We write ? = 0 about the point A from the force diagram: h A A D ?A = F2(d1 + d2) – Wd1 = 0, which gives D 4 F2 = (1.2010 N)(4.0 m)/(4.0 m + 6.0 m) C d2 d1 = 4.8103 N. F1 F2 W Thus F1 = W – F2 = 1.20104 N – 4.8103 N = 7.2103 N. If we take the pin A as the object, we have the F1 forces shown in the diagram, where the directions y FAC have been chosen so the forces sum to zero. A We write ?F = ma from the diagram: A x ?Fy = F1 – FAB sin A = 0; FAB 7.2103 N = FAB sin 56.3°, FAB FBD FBC which gives FAB = 8.65103 N compression. D A FAC FCD ?Fx = FAC – FAB cos A = 0; B FAC – (8.65103 N) cos 56.3° = 0, C FBC which gives FAC = 4.80103 N tension. W From the forces shown on pin C as the object, we have ?Fy = FBC – W = 0, which gives FBC = 1.20104 N tension. ?Fx = FCD – FAC = 0, which gives FCD = FAC = 4.80103 N tension. Page 29

Chapter 12

If we take the pin B as the object, we have the forces shown in the diagram, where the directions have been chosen so the forces sum to zero: ?Fx = FAB cos A – FBD cos D = 0; (8.65103 N) cos 56.3° – FBD cos 45.0° = 0, which gives FBD = 6.79103 N compression. ?Fy = FAB sin A + FBD sin D – FBC = 0; (8.65103 N) sin 56.3° + (6.79103 N) sin 45.0° – FBC = 0, which gives FBC = 1.20104 N tension, which agrees with our previous result. These results can be confirmed by summing the forces on pin D.

93. (a) The maximum distance for the top brick to L remain on the next brick will be reached when its center of mass is directly over the m edge of the next brick. Thus the top brick 2m x1 will overhang by x1 = L/2. 3m x2 The maximum distance for the top two bricks to x3 4m remain on the next brick will be reached when x4 the center of mass of the top two bricks is directly over the edge of the third brick. If we take the edge of the third brick as the origin, we have xCM = [m(x2 – L/2) + mx2]/2m = 0, which gives x2 = L/4. The maximum distance for the top three bricks to remain on the next brick will be reached when the center of mass of the top three bricks is directly over the edge of the fourth brick. If we take the edge of the fourth brick as the origin, we have xCM = [m(x3 – L/2) + 2mx3 ]/3m = 0, which gives x3 = L/6. The maximum distance for the four bricks to remain on the table will be reached when the center of mass of the four bricks is directly over the edge of the table. If we take the edge of the table as the origin, we have xCM = [m(x4 – L/2) + 3mx4 ]/4m = 0, which gives x4 = L/8. (b) With the origin at the table edge, we find the position of the left edge of the top brick from D = x1 + x2 + x3 + x4 – L = (L/2) + (L/4) + (L/6) + (L/8) – L = L/24, which is beyond the table. (c) We can generalize our results by recognizing that the ith brick is a distance L/2i beyond the edge of the brick below it. The total distance spanned by n bricks is n D =L 1. i = 1 2i (d) Each side of the arch must span 0.50 m. From our general result, we have  m = (0.30 m ) n 1 . 0.50 i = 1 2i If we evaluate this numerically, such as by using a spreadsheet, we find that 16 bricks will create a span of 0.507 m, so a minimum of 32 bricks is necessary, not counting the one on top. If the bottom brick is flush with the opening, as shown in Fig. 9–32, then two more bricks will be needed.

Page 30

Chapter 12

94. If we take the pin A as the object, we have the forces shown in the diagram, where the directions have been chosen so the forces can sum to zero. We write ?F = ma from the diagram: ?Fx = FAD cos 45° – FAB = 0; which gives FAD = v2 FAB = v2 F tension. ?Fy = FAC – FAD sin 45° = 0; which gives FAC = F compression. From the symmetry the forces in the diagonal members must be the same, so we have FBC = FAD = v2 F tension, FBD = FAC = F compression, FCD = FAB = F compression.

A

C

B

45°

45°

D FAC

y

A 45°

FAB

FAD

x

FBC C

45°

FCD FAC

95. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point C:   ?C = mgL sin A – (FTB sin B)L cos A + (FTB cos B)L sin A = 0, which gives FTB = mg /[(sin B / tan A) – cos B] = (20.0 kg)(9.80 m/s2)/[(sin 50°/tan 20°) – cos 50°] = 134 N. Note that we have found the torque produced by FTB by finding the torques of the components. We write ?F = ma from the force diagram: ?Fx = FTB sin B – FTA sin A = 0, which gives FTA = FTB sin B /sin A = (134 N) sin 50°/sin 20° = 300 N. 96. When the engine is raised a distance h, there will be that much more rope on the side that is pulled. Thus the angle the rope makes with the vertical is found from cos  = !¬/(!¬ + !h) = (20.0 m)/(20.0 m + 0.50 m) = 0.976, so  = 12.7°. From the forces on the engine, we see that the tension in the rope is FT = mg. From the symmetry of the force diagram for the point A, the horizontal force that must be exerted is F = 2FT sin  = 2(250 kg)(9.80 m/s2) sin 12.7° = 1.08103 N. 3 Note that the weight of the engine is 2.4510 N.

Page 31

C A

L

y FTA

x FTB

B

mg



y FT



F



FT

A

¬

FT h mg

x

Chapter 12

97. We choose the coordinate system shown, with positive torques clockwise. We write ? = Iabout the point A from the force d1 diagram for the torso: d2   ?A = w1(d1 + d2 + d3) cos  + w2(d2 + d3) cos  + d3 w3d3 cos  – (FM sin )(d2 + d3) = 0; w1 0.07w(24 cm + 12 cm + 36 cm) cos 30° +  FM  A 0.12w(12 cm + 36 cm) cos 30° + w2 0.46w(36 cm) cos 30° – y FV FM(sin 12°)(12 cm + 36 cm) = 0, V w3 which gives FM = 2.37w. x We write ?Fx = max from the force diagram: FVx – FM cos ( – ) = 0, which gives FVx = FM cos ( – ) = (2.37w) cos (30° – 12°) = 2.26w.  We write ?Fy = may from the force diagram: FVy – FM sin ( – ) – w1 – w2 – w3 = 0, which gives FVy = FM sin ( – ) + w1 + w2 + w3 = (2.37w) sin (30° – 12°) + 0.07w + 0.12w + 0.46w = 1.38w. For the magnitude we have FV = (FVx2 + FVy2)1/2 = [(2.26w)2 + (1.38w)2]1/2 = 2.6w. We find the direction from tan V = FVy/FVx = (1.38w)/(2.26w) = 0.61, so V = 31° above horizontal.

Page 32

Chapter 13

CHAPTER 13 – Fluids 1. 2. 3. 4.

When we use the density of granite, we have  m = V = (2.7103 kg/m3)(1108 m3) =

31011 kg.

When we use the density of air, we have  m = V = LWH = (1.29 kg/m3)(4.8 m)(3.8 m)(2.8 m) =

66 kg.

When we use the density of gold, we have  m = V = LWH = (19.3103 kg/m3)(0.60 m)(0.25 m)(0.15 m) = If we assume a mass of 65 kg with the density of water, we have m = V; 6.510–2 m3 65 kg = (1.0103 kg/m3)V, which gives V =

4.3102 kg

(˜ 950 lb!).

(˜ 65 L).

5.

From the masses we have mwater = 98.44 g – 35.00 g = 63.44 g; mfluid = 88.78 g – 35.00 g = 53.78 g. Because the water and the fluid occupy the same volume, we have SGfluid = fluid/water = mfluid/mwater = (53.78 g)/(63.44 g) = 0.8477.

6.

The definition of the specific gravity of the mixture is SGmixture = mixture/water. The density of the mixture is mixture = mmixture/V = (SGantifreezewaterVantifreeze + SGwaterwaterVwater)/V, so we get SGmixture = (SGantifreezeVantifreeze + SGwaterVwater)/V = [(0.80)(5.0 L) + (1.0)(4.0 L)]/(9.0 L) = 0.89.

7.

(a) The normal force on the four legs must equal the weight. The pressure of the reaction to the normal force, which is exerted on the floor, is P = FN/A = mg/A = (60 kg)(9.80 m/s2)/4(0.0510–4 m2) = 3107 N/m2. (b) For the elephant standing on one foot, we have P = FN/A = mg/A = (1500 kg)(9.80 m/s2)/(80010–4 m2) = 2105 N/m2. Note that this is a factor of ˜ 100 less than that of the loudspeaker!

8.

(a) The force of the air on the table top is F = PA = (1.013105 N/m2)(1.6 m)(2.9 m) = 4.7105 N (down). (b) Because the pressure is the same on the underside of the table, the upward force has the same This is why the table does not move! magnitude: 4.7105 N.

9.

The pressure difference on the lungs is the pressure change from the depth of water: ?P = g ?h; (80 mm-Hg)(133 N/m2 · mm-Hg) = (1.00103 kg/m3)(9.80 m/s2)?h, which gives ?h =

1.1 m.

10. There is atmospheric pressure outside the tire, so we find the net force from the gauge pressure. Because the reaction to the force from the pressure on the four footprints of the tires supports the automobile, we have 4PA = mg; 2.0103 kg. 4(240103 N/m2 )(20010–4 m2) = m(9.80 m/s2), which gives m = 11. Because the force from the pressure on the cylinder supports the automobile, we have PA = mg; 8.28103 kg. (17.0 atm)(1.013105 N/m2 · atm)#p(24.510–2 m)2 = m(9.80 m/s2), which gives m = Note that we use gauge pressure because there is atmospheric pressure on the outside of the cylinder. Page 1

Chapter 13

12. The pressure from the height of alcohol must balance the atmospheric pressure: P = gh; 13 m. 1.013105 N/m2 = (0.79103 kg/m3)(9.80 m/s2)h, which gives h = 13. The pressure at a depth h is P = P0 + gh = 1.013105 N/m2 + (1.00103 kg/m3)(9.80 m/s2)(2.0 m) = 1.2105 N/m2. The force on the bottom is F = PA = (1.2105 N/m2)(22.0 m)(8.5 m) = 2.3107 N (down). The pressure depends only on depth, so it will be the same: 1.2105 N/m2. 14. The pressure is produced by a column of air: P = gh; 1.013105 N/m2 = (1.29 kg/m3)(9.80 m/s2)h, which gives h = 8.0103 m = 15. Because points a and b are at the same elevation of water, the pressures must be the same. Each pressure is due to the atmospheric pressure at the top of the column and the height of the liquid, so we have Pa = Pb , or P0 + oilghoil = P0 + waterghwater; oilg(27.2 cm) = (1.00103 kg/m3)g(27.2 cm – 9.41 cm), which gives oil = 6.54102 kg/m3.

Oil

8.0 km.

hoil

hwater

a

b

Water

16. We find the gauge pressure from the water height: P = waterghwater = (1.00103 kg/m3)(9.80 m/s2)[5.0 m + (100 m) sin 60°] = 9.0105 N/m2 (gauge). If we neglect turbulence and frictional effects, we know from energy considerations that the water would rise to the elevation at which it started: h = 5.0 m + (100 m) sin 60° = 92 m. 17. When we take into account the change in density with pressure, we use the result from Example 13–4: P = P0 e – (  0 g/ P0) y 5 3 2 2 5 = (1.013  10 N / m 2)e – [(1.29 kg / m )(9.80 m/ s )/ (1.01310 N/ m )](8850 m) =

4

3.36  10 N/ m 2 (0.331 a t m).

18. The minimum gauge pressure would cause the water to come out of the faucet with very little speed. This means the gauge pressure must be enough to hold the water at this elevation: Pgauge = gh; = (1.00103 kg/m3)(9.80 m/s2)(36.5 m) = 3.58105 N/m2. 19. (a) When the height of the mercury in the open tube is greater than the height of the mercury on the tank side of the manometer, the pressure in the tank is Ptank = Patm + ggh = (1040 mbar)(100 Pa/mbar) + (28.0 cm Hg)(1330 Pa/cm Hg) = 1.41105 Pa. (b) When the height of the mercury in the open tube is less than the height of the mercury on the tank side of the manometer, we use a negative value for h: Ptank = Patm + ggh = (1040 mbar)(100 Pa/mbar) + (– 4.2 cm Hg)(1330 Pa/cm Hg) = 9.8104 Pa.

Page 2

Chapter 13

20. When we apply ? = I to the lever about the pivot point B, we have F2L – F1L = 0, or F1 = 2F. Because the pressure in the hydraulic fluid is constant, we have F2/A2 = F1/A1 , or F2 = F1(A2/A1) = 2F(D2/D1)2. From the forces on the large cylinder we have Fsample = F2. Thus the pressure on the sample is Psample = Fsample /Asample = 2F(D2/D1)2/Asample = 2(300 N)[(10.0 cm)/(2.0 cm)]2/ (4.010–4 m2) 7 = 3.810 N/m2.

Fsample F L D1

L

B

F2

21. (a) For the mass of water in the tube we have m = V = pr2H = (1.00103 kg/m3)p(0.0030 m)2(12 m) = 0.34 kg. (b) The net force on the lid of the barrel is due to the gauge pressure of the water at the top of the barrel. Because this gauge pressure is from the mass of water in the tube, we have F = PAB = (mg/A)AB = mg(R/r)2 = (0.34 kg)(9.80 m/s2)[(20 cm)/(0.30 cm)]2 = 1.5104 N (up).

D1 F1

r

H R

22. (a) In the accelerated frame of the beaker, there is a pseudoforce opposite to the direction of the acceleration. There is an a effective g at an angle from the vertical given by h a tan  = a/g.  The surface of the water will be perpendicular to this effective g, and thus will make an angle g   = tan–1 a/g with the horizontal. g (b) The left edge of the water surface, opposite to the direction of the acceleration, will be higher. (c) The magnitude of the effective acceleration of gravity is g = (a2 + g2)1/2. A point at a vertical distance h below the surface will be a distance h along the direction of g from the surface, where h = h cos . The pressure at this point is P = P0 + gh = P0 + (a2 + g2)1/2h[g/(a2 + g2)1/2] = P0 + gh.

Page 3

Chapter 13

23. (a) Because the pressure is a function of the depth, the force on the wall will also be a function of the depth. We choose a coordinate system with y = 0 at the bottom of the dam and the water level at height h. We find the force on a differential slice dy of width b at height y and then integrate to get the total force: h  F = p water d A =  g(h – y)b d y 0

= gb hy –

1 y2 2

h 0

2

=  gb h –

1h2 2

=

1  gbh 2 2

y t h dF

dy

F H

y y=0

(b) To find the height H of the effective point of action, the torque produced by this force must equal the sum (integral) of the torques produced by all of the differential elements: H  FH = g(h – y)yb d y 0

H =  gb h12 y 2 – 13 y 3 =  gb 12 h 3 – 13 h 3 = 16 gb h 3. 0

When we use the result for the force, we have 1  gbh 2 H = 16  gbh 3, which gives H = @h. 2  (c) To prevent overturning, the torque about the base of the dam from the weight of concrete must be greater than the torque from the water pressure: mconcretegt/2 = water gbh3/6; concrete(hbt)gt/2 = water gbh3/6, or t2 = (water/concrete)h2/3 = [(1.00103 kg/m3)/(2.3103 kg/m3)]h2/3, which gives t = 0.38h. It is not necessary to add in atmospheric pressure, because it acts on both sides of the dam. 24. We consider a mass of water m that occupies a volume V0 at the surface. The pressure increase at a depth h will decrease the volume to V, where m = 0V0 = V. We use the bulk modulus to find the volume change: ?V/V0 = – ?P/B = – 0gh/B = – (1.025103 kg/m3)(9.80 m/s2)(6.0103 m)/(2.0109 N/m2) = – 0.030. The fact that this is small justifies using the surface density for the pressure calculation. The volume at the depth h is V = V0 + ?V = (1 – 0.030)V0 = 0.970V0, so the density is  = 0(V0/V) = (1.025103 kg/m3)(1/0.970) = 1.057103 kg/m3. For the fractional change in density, we have ?/0 = (V0/V) – 1 = (1/0.970) – 1 = 1.030 – 1 = + 0.030 (3%).

Page 4

Chapter 13

25. In the rotating frame of the bucket, there will be a pseudoforce r2 dm acting on a differential mass of the water. We choose a ring of height ?z, radius r and thickness dr, with mass dm = (2pr ?z dr), so the magnitude of the pseudoforce is constant. The force from the differential pressure difference across the ring must balance the pseudoforce: 2pr ?z dP = r2(2pr ?z dr), or dP = r2 dr. We integrate to find the pressure at a radius r: P P0

dP =

r 0



r 2 d m

O r

dr

 2r dr;

P – P0 = 12 2 r 2, or P = P0 + 12  2r 2.

26. Because the mass of the displaced liquid is the mass of the hydrometer, we have m = waterhwaterA = liquidhliquidA; (1000 kg/m3)(22.5 cm) = liquid (22.9), which gives liquid = 983 kg/m3. 27. Because the mass of the displaced water is the apparent change in mass of the rock, we have ?m = waterV. For the density of the rock we have rock = mrock/V = (mrock/?m)water = [(7.85 kg)/(7.85 kg – 6.18 kg)](1.00103 kg/m3) = 4.70103 kg/m3. 28. When the aluminum floats, the net force is zero. If the fraction of the aluminum that is submerged is f, we have Fnet = 0 = Fbuoy – mAlg = Hgg f V – AlgV; (13.6103 kg/m3)g fV = (2.70103 kg/m3)gV, which gives f = 0.199.

Fbuoy m Alg

29. When the balloon and cargo float, the net force is zero, so we have Fnet = 0 = Fbuoy – mHeg – mballoong – mcargog; 0 = airgVballoon – HegVballoon – mballoong – mcargog; 0 = (1.29 kg/m3 – 0.179 kg/m3))p(7.35 m)3 – (1000 kg + mcargo), which gives mcargo = 8.5102 kg.

Fbuoy

m He g

m balloon g

m cargo g

30. Because the mass of the displaced water is the apparent change in mass of the person, we have ?m = waterVlegs. For the mass of one leg we have mleg = SGbodywater!Vlegs = SGbodywater! ?m/water = SGbody! ?m Page 5

Chapter 13

= (1.00)!(78 kg – 54 kg) =

12 kg.

31. Because the mass of the displaced water is the apparent change in mass of the sample, we have ?m = waterV. For the density of the sample we have  = m/V = (m/?m)water = [(63.5 g)/(63.5 g – 56.4 g)](1.00103 kg/m3) = 8.94103 kg/m3. From the table of densities, the most likely metal is copper. 32. Because the mass of the displaced air is the apparent change in mass of the aluminum, we have ?m = m – m = airV = air(m/Al), or [1 – (air/Al)]m = m;

{1 – [(1.29 kg/m3)/(2.70103 kg/m3)]}m = 2.0000 kg, which gives m =

33. (a) The buoyant force on the chamber is Fbuoy = watergV

2.0010 kg.

Fbuoy

= (1.025103 kg/m3)(9.80 m/s2))p(3.0 m)3 = 1.14106 N. (b) Because the net force is zero, we have Fnet = 0 = Fbuoy – mg – FT ; 1.14106 N – (75,000 kg)(9.80 m/s2) = FT = 4.0105 N.

mg

FT

34. (a) The buoyant force on the completely submerged diver is Fbuoy = watergV = (1.025103 kg/m3)(9.80 m/s2)(65.010–3 m3) = (b) With the positive direction upward, the net force is Fnet = Fbuoy – mg = 653 N – (63.0 kg)(9.80 m/s2) = + 36 N. The net force is up, so the diver will float.

653 N.

35. (a) The buoyant force is a measure of the net force on the partially submerged object due to the pressure in the fluid. In order to remove the object and have no effect on the fluid, we would have to fill the submerged volume with an equal volume of fluid. As expected, the buoyant force on this fluid is the weight of the fluid. To have no net torque on the fluid, the buoyant force and the weight of the fluid would have to act at the same point, the center of gravity. (b) From the diagram we see that, if the center of buoyancy Fbuoy Fbuoy is above the center of gravity, when the ship tilts, the net torque will tend to restore the ship’s position. From the CG CB diagram we see that, if the center of buoyancy is below the CG center of gravity, when the ship tilts, the net torque will tend CB to continue the tilt. If the center of buoyancy is at the center of mg mg gravity, there will be no net torque from these forces, so other torques (from the wind and waves) would determine the motion of the ship. Thus stability is achieved when the center of buoyancy is above the center of gravity. Page 6

Chapter 13

36. (a) Because the mass of the displaced liquid is the apparent change in mass of the ball, we have ?m = liquidV, or liquid = ?m/V = (?m/m)ball ; = [(3.40 kg – 2.10 kg)/(3.40 kg)](2.70103 kg/m3) = 1.03103 kg/m3. (b) For a submerged object, from part (a) we have liquid = [(m – mapparent)/m]object. 37. Because the mass of the displaced alcohol is the apparent change in mass of the wood, we have ?m = alcoholV. We find the specific gravity of the wood from SGwoodwater = mwood/V = (mwood/?m)alcohol ; SGwood(1.00103 kg/m3) = [(0.48 kg)/(0.48 kg – 0.047 kg)](0.79103 kg/m3), which gives SGwood = 0.88. As expected, the specific gravity is less than 1, so the wood floats in water. 38. When the ice floats, the net force is zero. If the fraction of the ice that is above water is f, we have Fnet = 0 = Fbuoy – miceg = swg(1 – f )V – icegV, or (1.025)wg(1 – f )V = (0.917)wgV, which gives f = 0.105 (10.5%). 39. From Problem 38, we know that the initial volume out of the water, Fice without the bear on the ice, is 3 3 V1 = 0.105V0 = 0.105(10 m ) = 1.05 m . Thus we find the submerged volume of ice with the bear on the ice from V2 = V0 – !V1 = 10 m3 – !(1.05 m3) = 9.48 m3. m ice g Because the net force is zero, we have Fnet = 0 = Fbear + Fice – mbearg – miceg, or sea waterg(0.30)Vbear + sea watergV2 = mbearg + icegV0 ; sea waterg(0.30)(mbear/bear) + sea watergV2 = mbearg + icegV0 ; (1.025)w(0.30)[mbear/(1.00)w] + (1.025103 kg/m3)(9.48 m3) = mbear + (0.917)(1.00103 kg/m3)(10 m3), which gives mbear = 40. The minimum mass of lead will suspend the wood under water. Because the net force is zero, we have Fnet = 0 = Flead + Fwood – mleadg – mwoodg, or watergVlead + watergVwood = mleadg + mwoodg; waterg(mlead/lead) + waterg(mwood/wood) = mleadg + mwoodg. We can rearrange this: mlead[1 – (1/SGlead)] = mwood[(1/SGwood) – 1]; mlead[1 – (1/11.3)] = (3.15 kg)[(1/0.50) – 1], which gives mlead = 3.46 kg.

Fwood

m wood g Flead

m lead g

Page 7

Fbear

m bear g

7.9102 kg.

Chapter 13

41. The apparent weight is the force required to hold the system, so the net force is zero. When only the sinker is submerged, we have Fnet = 0 = w1 + Fbuoy1 – w – mg, or w1 + watergVsinker = w + mg. When the sinker and object are submerged, we have Fnet = 0 = w2 + Fbuoy1 + Fbuoy2 – w – mg, or w2 + watergVsinker + watergVobject = w + mg. If we subtract the two results, we get w1 – w2 – watergVobject = 0, or waterg(mobject/object) = w/SGobject = w1 – w2 , so SGobject = w/(w1 – w2).

w1

w2

w

Fbuoy2

w Fbuoy1

mg

Fbuoy1

mg

42. The flow rate in the major arteries must be the flow rate in the aorta: v1A1 = v2A2; (30 cm/s)p(1.0 cm)2 = v2(2.0 cm2), which gives v2 = 47 cm/s. 43. From the equation of continuity we have Flow rate = Av; (9.2 m)(5.0 m)(4.5 m)/(12 min)(60 s/min) = p(0.15 m)2v, which gives v =

4.1 m/s.

44. If we use the condition of no flow, v1 = v2 = 0, in Bernoulli’s equation we have P1 + !v12 + gy1 = P2 + !v22 + gy2; P1 + 0 + gy1 = P2 + 0 + gy2 , or P2 = P1 + g(y1 – y2) = P1 + gh. 45. If we ignore viscosity, we can use Bernoulli’s equation. We choose the initial point at the top of the water, where the velocity is essentially zero, and the final point at the hole. Thus we have P1 + !v12 + gy1 = P2 + !v22 + gy2; Patm + 0 + gh = Patm + !v22 + 0, which gives v2 = (2gh)1/2 = [2(9.80 m/s2)(4.6 m)]1/2 = 9.5 m/s. 46. From the equation of continuity we have Flow rate = Apoolh/t = Ahosev;

#p(6.1 m)2(1.2 m)/t = #p[(0.625 in)(0.0254 m/in)]2(0.33 m/s), which gives t = 5.37105 s =

6.2 days.

47. If we choose the initial point at the water main where the water is not moving and the final point at the top of the spray, where the water also is not moving, from Bernoulli’s equation we have P1 + !v12 + gy1 = P2 + !v22 + gy2; P1 + 0 + 0 = Patm + 0 + (1.00103 kg/m3)(9.80 m/s2)(15 m), which gives P1 – Patm = Pgauge = 1.5105 N/m2 = 1.5 atm. 48. From the equation of continuity we have Flow rate = A1v1 = A2v2 , or v2 = (A1/A2)v1 = (D1/D2)2v1. From Bernoulli’s equation for the horizontal pipe we have P1 + !v12 + gy1 = P2 + !v22 + gy2; P1 + !v12 + 0 = P2 + !(D1/D2)4v12 + 0, which gives v12 = 2(P1 – P2)/[(D1/D2)4 – 1]; = 2[(32103 Pa) – (24103 Pa)]/(1.00103 kg/m3)[(6.0 cm/4.0 cm)4 – 1], which gives v1 = 1.98 m/s. Page 8

Chapter 13

Thus the volume rate of flow is ?V/?t = A1v1 = #pD12v1 = #p(6.010–2 m)2(1.98 m/s) =

5.610–3 m3/s.

49. If we choose the initial point at the pressure head where the water is not moving and the final point at the faucet, from Bernoulli’s equation we have P1 + !v12 + gh1 = P2 + !v22 + gh2; Patm + 0 + (1.00103 kg/m3)(9.80 m/s2)(12.0 m) = Patm + !(1.00103 kg/m3)v22, which gives v2 = 15.3 m/s. For the flow rate we have 4.1110–3 m3/s. Flow rate = Av = #p(0.0185 m)2 (15.3 m/s) = 50. The pressure under the roof will be atmospheric. If we choose the initial point where the air is not moving and the final point above the roof, from Bernoulli’s equation we have P1 + !v12 + gh1 = P2 + !v22 + gh2; Patm + 0 + 0 = P + !(1.29 kg/m3)(25 m/s)2, which gives Patm – P = 4.0102 N/m2. The net upward force on the roof is F = (Patm – P)A = (4.0102 N/m2)(240 m2) = 9.7104 N. 51. If we consider the points at the top and bottom surfaces of the wing compared to the air flow in front of the wing, from Bernoulli’s equation we have P0 + !v02 + gh0 = P1 + !v12 + gh1 = P2 + !v22 + gh2; P1 + !(1.29 kg/m3)(340 m/s)2 + 0 = P2 + !(1.29 kg/m3)(290 m/s)2 + 0, which gives P2 – P1 = 2.03104 N/m2. The net upward force on the wing is F = (P2 – P1)A = (2.03104 N/m2)(86 m2) = 1.7106 N.

52. If we consider the points far away from the center of the hurricane and at the center of the hurricane, from Bernoulli’s equation we have P1 + !v12 + gh1 = P2 + !v22 + gh2; 1.013105 N/m2 + 0 + 0 = P2 + !(1.29 kg/m3)[(300 km/h)/(3.6 ks/h)]2 + 0, which gives P2 = 9.7104 N/m2. 53. If we consider the volume of fluid in the pipe, at each end of the pipe there is a force toward the fluid of PA. If the area of the pipe is constant, the net force on the fluid is Fnet = (P1 – P2)A. The required power is Power = Fnetv = (P1 – P2)Av = (P1 – P2)Q. 54. The flow rate in the pipe at street level must be the flow rate at the top floor: v1A1 = v2A2; 2.2 m/s. (0.60 m/s)#p(5.0 cm)2 = v2 #p(2.6 cm)2, which gives v2 = 2.22 m/s = If we use Bernoulli’s equation between the street level and the top floor, we have P1 + Patm + !v12 + gh1 = P2 + Patm + !v22 + gh2; 3.8 atm + Patm + !(1.00103 kg/m3)(0.60 m/s)2/(1.013105 N/m2 · atm) + 0 = P2 + Patm + [!(1.00103 kg/m3)(2.22 m/s)2 + (1.00103 kg/m3)(9.80 m/s2)(20 m)]/ (1.013105 N/m2 · atm), which gives P2 = 1.8 atm (gauge). Page 9

Chapter 13

55. From the equation of continuity we have Flow rate = A1v1 = A2v2 , or v2 = (A1/A2)v1 . From Bernoulli’s equation we have P1 + !v12 + gy1 = P2 + !v22 + gy2; Patm + !v12 + 0 = Patm + !(A1/A2)2v12 + gh, which gives v1 = {2gh/[1 – (A1/A2)2]}1/2.

56. (a) From Bernoulli’s equation we have PA + !v12 + gy1 = P2 + PA + 0 + gy2 , which gives

v2

v1 = {2[(P2/) + g(y2 – y1)]}1/2. (b) For the given data we have

y2 – y1

v1 = (2{[(0.85 atm)(1.013105 N/m2 · atm)/(1.00103 kg/m3)] + =

(9.80 m/s2)(2.1 m)})1/2

15 m/s.

v1

57. (a) From the equation of continuity we have Flow rate = A0v = Avinside , or vinside = (A0/A)v .

Because A0 « A, we can assume that vinside ˜ 0. From Bernoulli’s equation for the assumed steady flow we have P + !vinside2 + gy = P0 + !v2 + gy;

P + 0 + 0 = P0 + !v2 + 0, which gives v = [2(P – P0)/]1/2. (b) The thrust force on the rocket from the emitted gases is F = v dm/dt = v dV/dt = vA0v = A0v2 = 2A0(P – P0).

58. (a) The flow rate through the venturi meter is constant: v1A1 = v2A2. If we use Bernoulli’s equation between the segments of the meter, we have P1 + !v12 + gh1 = P2 + !v22 + gh2; P1 + ! + 0 = P2 + ! + 0, or P1 – P2 = ! – When we substitute for v2 from the flow rate, we get

v12

v22

(v22

P1

v12).

P2 v2

v1 A1

P1 – P2 = ![(v1A1/A2)2 – v12] = !v12[(A12 – A22)/A22], which gives v1 = A2[2(P1 – P2)/(A12 – A22)]1/2. (b) With the given data, we have v1 = A2[2(P1 – P2)/(A12 – A22)]1/2 = D22[2(P1 – P2)/(D14 – D24)]1/2

A2

= (1.0 cm)2{2(18 mm-Hg)(133 N/m2 · mm-Hg)/(1.00103 kg/m3)[(3.0 cm)4 – (1.0 cm)4]}1/2 = 0.24 m/s.

Page 10

Chapter 13

v2

59. (a) We assume v2 = 0. From Bernoulli’s equation we have P1 + !

v12

+ gy1 = P2 + !

v22

h2

+ gy2;

Patm + !v12 + gh1 = Patm + 0 + gh2 , which gives v1 = [2g(h2 – h1)]1/2. After the fluid leaves the tank, its horizontal velocity is constant. We find the time to reach the ground from y = y0 + !(– g)t2;

v1

h1

h=0

which gives t = 0 = h1 – ! In this time the water will have traveled a horizontal distance given by x = v1t = [2g(h2 – h1)]1/2(2h1/g)1/2 = 2[h1(h2 – h1)]1/2. (b) To have the same range for a height h1, we must have h1(h2 – h1) = h1(h2 – h1). We could solve this quadratic, but we can see by inspection that h1 = h2 – h1. gt2,

x

(2h1/g)1/2.

60. (a) From the equation of continuity we have Flow rate = A1v1 = A2v2 , or v1 = (A2/A1)v2 . From Bernoulli’s equation we have P1 + !v12 + gy1 = P2 + !v22 + gy2; Patm + !(A2/A1)2v22 + 0 = Patm + !v22 + gh, which gives

v2 = {2gh/[(A2/A1)2 – 1]}1/2. We know that the height of the liquid decreases, so dh/dt = – v2 = – [2ghA12/(A22 – A12)]1/2. (b) We integrate to find the height as a function of time: h h0

dh = – h

t

2

2gA 1 2

0

dt;

2

A2– A1

2

2 h – h0 = –

2gA 1 t, or A 22 – A 21

2

h = h0 –

gA 1 t. 2(A 22 – A 21)

(c) The two areas are A1 = #pD12 = #p(0.5010–2 m)2 = 1.9610–5 m2; A2 = V/h0 = (1.010–3 m3)/(9.410–2 m) = 1.0610–2 m2. We find the time to empty from h1/2 = h01/2 – [gA12/2(A22 – A12)]1/2t;

0 = (9.410–2 m)1/2 – {(9.80 m/s2)(1.9610–5 m2)2/2[(1.0610–2 m2)2 – (1.9610–5 m2)2]}1/2t, which gives t = 75 s.

61. The torque corresponds to a tangential force on the inner cylinder: F = /R1. The layer of fluid has a thickness R2 – R1 , the fluid next to the outer cylinder is at rest, and the fluid next to the inner cylinder has a speed v = R1, where  = (62 rev/min)(2p rad/rev)/(60 s/min) = 6.49 rad/s. If we use the average radius for the area over which the torque acts in the definition of viscosity, we have  = FL/Av = (R2 – R1)/R1[2p!(R1 + R2)H]R1 = (R2 – R1)/pR12(R1 + R2)H = (0.024 m · N)(0.0530 m – 0.0510 m)/p(0.0510 m)2(0.0510 m + 0.0530 m)(0.120 m)(6.49 rad/s) = 0.072 Pa · s. 62. The same volume of water is used, so the time is inversely proportional to the flow rate. Taking into account the viscosity of the water, with the only change the diameter of the hose, we have Page 11

Chapter 13

t  1/Q  1/r4  1/d4. For the two hoses we have. t2/t1 = (d1/d2)4 = (3/5)4 =

1/7.7 = 0.13.

63. From Poiseuille’s equation we have Q = pr4(P1 – P2)/8L; (5.610–6 m3/min)/(60 s/min) = p(0.9010–3 m)4(P1 – P2)/8(20010–3 Pa · s)(5.510–2 m), which gives P1 – P2 = 4.0103 Pa. 64. From Poiseuille’s equation we have Q = pr4(P1 – P2)/8L; 45010–6 m3/s = p(0.145 m)4(P1 – P2)/8(0.20 Pa · s)(1.9103 m), which gives P1 – P2 = 9.9102 N/m2. 65. From Poiseuille’s equation we have Q = V/t = pr4(P1 – P2)/8L; (9.0 m)(12.0 m)(4.0 m)/(10 min)(60 s/min) = pr4(0.7110–3 atm)(1.013105 N/m2 · atm)/8(0.01810–3 Pa · s)(17.5 m), which gives r = 0.053 m. Thus the diameter needed is 0.106 m = 11 cm. 66. For the viscous flow of the blood, we have Q  r4. For the two flows we have. Q2/Q1 = (r2/r1)4; 0.25 = (r2/r1)4, which gives r2/r1 = 0.71 (reduced by 29%) . 67. (a) We find the Reynolds number for the blood flow: Re = 2ær/ = 2(0.30 m/s)(0.010 m)(1.05103 kg/m3)/(4.010–3 Pa · s) = 1600. Thus the flow is laminar but close to turbulent. (b) If the only change is in the average speed, we have Re2/Re1 = æ2/æ1 ; Re2/1600 = 2, or Re2 = 3200, so the flow is turbulent. 68. Normally there would be two contributions to the necessary pressure: the elevation change and the viscosity: ?P = gh + 8QL/pr4. To determine the pressure for a small flow, Q ˜ 0, we ignore the viscosity term to have 2.0105 N/m2. ?P = gh = (1.00103 kg/m3)(9.80 m/s2)(20 m) =

69. We find the pressure difference required across the needle from Poiseuille’s equation: Q = pr4(P1 – P2)/8L; (4.010–6 m3/min)/(60 s/min) = p(0.2010–3 m)4(P1 – P2)/8(4.010–3 Pa · s)(0.040 m), which gives P1 – P2 = 1.70104 N/m2. The pressure P2 is the pressure in the vein. The pressure P1 is produced from the elevation of the bottle: P1 = Patm + gh. Thus we have Patm + gh – P2 = 1.70104 N/m2, or gh = P2 – Patm + 1.70104 N/m2; (1.05103 kg/m3)(9.80 m/s2)h = (18 torr)(133 N/m2 · torr) + 1.70104 N/m2, which gives h = 1.9 m.

Page 12

Chapter 13

70. For the two surfaces, top and bottom, we have = F/2¬ = (5.110–3 N)/2(0.070 m) = 0.036 N/m. 71. For the two surfaces, top and bottom, we have F = 2¬ = (0.025 N/m)2(0.182 m) = 9.110–3 N. 72. From Example 13–15, the six upward surface tension forces must balance the weight. We find the required angle from 6(2pr cos ) ˜ w; 6(2)p(3.010–5 m)(0.072 N/m) cos  ˜ (0.01610–3 kg)(9.80 m/s2), which gives cos  ˜ 1.9. would not remain on the surface. Because the maximum possible value of cos  is 1, the insect 73. (a) We assume that the net force from the weight and the buoyant force is much smaller than the surface tension. For the two surfaces, inner and outer circumferences, we have F = 2L = 2(2pr), or  = F/4pr. (b) For the given data we have  = F/4pr = (8.4010–3 N)/4p(0.028 m) = 0.024 N/m. 74. We consider half of the soap bubble. The forces on the hemisphere will be the surface tensions on the two circles and the net force from the excess pressure between the inside and the outside of the bubble. This net force is the sum of all of the forces perpendicular to the surface of the hemisphere, but must be parallel to the surface tension. Therefore we can find it by finding the force on the circle that is the base of the hemisphere. The total force must be zero, so we have 2(2pr) = (pr2) ?P, which gives ?P = 4/r.

F F

A²P r F F

75. The liquid pressure is produced from the elevation of the bottle: ?P = gh. 0.88 m. (a) (65 mm-Hg)(133 N/m2 · mm-Hg) = (1.00103 kg/m3)(9.80 m/s2)h, which gives h = 0.55 m. (b) (550 mm-H2O)(9.81 N/m2 · mm-H2O) = (1.00103 kg/m3)(9.80 m/s2)h, which gives h = (c) If we neglect viscous effects, we must produce a pressure to balance the blood pressure: 0.24 m. (18 mm-Hg)(133 N/m2 · mm-Hg) = (1.0010–3 kg/m3)(9.80 m/s2)h, which gives h =

76. (a) If the fluid is incompressible, the pressure must be constant, so we have P = Fneedle/Aneedle = Fplunger/Aplunger; Fneedle/#p(0.020 cm)2 = (2.4 N)/#p(1.3 cm)2, which gives Fneedle = 5.710–4 N. (b) Just before the fluid starts to move, the pressure must be the gauge pressure in the vein: P = Fplunger/Aplunger; (18 mm-Hg)(133 N/m2 · mm-Hg) = Fplunger/#p(1.310–2 m)2, which gives Fplunger =

0.32 N.

77. If the motion is such that we can consider the air to be in equilibrium, we find the applied force from F = PA; Fi = PiA = (210103 N/m2)#p(0.030 m)2 = 1.5102 N; Page 13

Chapter 13

Ff = PfA = (310103 N/m2)#p(0.030 m)2 = 2.2102 N. Thus the range of the applied force is 1.5102 N = F = 2.2102 N. 78. The pressure is P = Patm + gh = 1 atm + (917 kg/m3)(9.80 m/s2)(4103 m)/(1.013105 N/m2 · atm) ˜ 79. The pressure difference is produced from the elevation: ?P = gh = (1.29 kg/m3)(9.80 m/s2)(410 m)/(1.013105 N/m2 · atm) =

4102 atm.

0.051 atm.

80. The pressure difference in the blood is produced by the elevation change: ?P = g ?h = (1.05103 kg/m3)(9.80 m/s2)(6 m)/(1.013105 N/m2 · atm) =

0.6 atm.

81. The pressure difference on the ear drum is the change produced by the elevation change: ?P = g ?h. The net force is F = A ?P = Ag ?h. 0.63 N. = (0.5010–4 m2)(1.29 kg/m3)(9.80 m/s2)(1000 m) = 82. The pressure at the bottom of the U must be the same on each side of the boundary. Each pressure is due to the atmospheric pressure at the top of the column and the height of the liquid, so we have Pa = Pb , or P0 + alcoholghalcohol = P0 + waterghwater; (0.79103 kg/m3)g(18.0 cm) = (1.00103 kg/m3)ghwater, which gives hwater = 14 cm.

83. The net force on the floating continent is zero: Fnet = 0 = Fbuoy – mcontg = rockg Ahrock – contgAhcont; (3300 kg/m3)gAhrock = (2800 kg/m3)gA(35 km), which gives hrock = 30 km. Thus the height of the continent above the surrounding rock is hcont – hrock =

halcohol

hwater

5 km.

84. The sum of the magnitudes of the forces on the ventricle wall is F = A ?P; 1.3102 N. = (8210–4 m2)(120 mm-Hg)(133 N/m2 · mm-Hg) = Note that the forces on the wall are not all parallel, so this is not the vector sum. 85. We assume that g is constant. The pressure on a small area of the Earth’s surface is produced by the weight of the air above it: P = mg/A = mtotalg/Atotal = mtotalg/4pREarth2; 1.013105 N/m2 = mtotal(9.80 m/s2)/4p(6.37106 m)2, which gives mtotal = 5.31018 kg. 86. The pressure difference on the water in the straw produces the elevation change: ?P = gh (80 mm-Hg)(133 N/m2 · mm-Hg) = (1.00103 kg/m3)(9.80 m/s2)h, which gives h =

Page 14

1.1 m.

Chapter 13

87. If we choose the initial point at the pressure head, where the water is not moving, and the final point at the faucet, from Bernoulli’s equation we have P1 + !v12 + gh1 = P2 + !v22 + gh2; Patm + 0 + (1.00103 kg/m3)(9.80 m/s2)h1 = Patm + !(1.00103 kg/m3)(7.2 m/s)2 + 0, which gives h1 = 2.6 m. 88. The mass of the unloaded water must equal the mass of the displaced sea water: m = sea waterAh = (1.025103 kg/m3)(2650 m2)(8.50 m) = 2.31 107 kg. Note that this is a volume of V = m/water = (2.31 107 kg)/(1.00103 kg/m3) = 2.31104 m3, which is greater than the volume change of the sea water, which is (2650 m2)(8.50 m) = 2.25104 m3. 89. We assume that for the maximum number of people, the logs are completely in the water and the people are not. Because the net force is zero, we have Fbuoy = mpeopleg + mlogsg; waterg(r2)NlogsL = Npeoplemg + Nlogslogsg(r2)L. Because the specific gravity is the ratio of the density to the density of water, this can be written (1 – SGlogs)Nlogswaterr2L = Npeoplem; (1 – 0.60)(10)(1.00103 kg/m3)(0.19 m)2(6.1 m) = Npeople(70 kg), which gives Npeople = 39.5. Thus the raft can hold a maximum of 39 people. 90. The work done in each heartbeat is W = Fd = PAd = PV. If n is the heart rate, the power output is Power = nW = nPV = [(70 beats/min)/(60 s/min)](105 mm-Hg)(133 N/m2 · mm-Hg)(7010–6 m3) = 91. We find the effective g by considering a volume of the water. Fbuoy1 The net force must produce the acceleration: Fbuoy1 – mg = ma; watergVwater – waterVwaterg = waterVwatera; g = g + a = g + 2.4g = 3.4g. a Note that this agrees with adding a pseudoforce – ma in the mg accelerating frame of the bucket. The buoyant force on the rock is Fbuoy2 = watergVrock = waterg(mrock/rock) = gmrock/SGrock = (3.4)(9.80 m/s2)(3.0 kg)/(2.7) = 37 N. To see if the rock will float, we find the net force acting on the rock in the accelerated frame: Fnet = Fbuoy2 – mrockg = 37 N – (3.0 kg)(3.4)(9.80 m/s2) = – 63 N. Because the result is negative, the rock will not float.

1.1 W.

Fbuoy2

m rock g

92. We choose the reference level at the nozzle. If we apply Bernoulli’s equation from the exit of the nozzle to the top of the spray, we have P1 + !v12 + gh1 = P2 + !v22 + gh2; Patm + !v12 + 0 = Patm + 0 + gh2 , which gives v12 = 2gh2. If we use the equation of continuity from the pump to the nozzle, we have Flow rate = A3v3 = A1v1 , or v3 = (A1/A3)v1 = (D1/D3)2v1. If we apply Bernoulli’s equation from the pump to the exit of the nozzle, we have Page 15

Chapter 13

P3 + !v32 + gh3 = P1 + !v12 + gh1; Ppump + ![(D1/D3)2v1]2 + gh3 = Patm + !v12 + 0, or Ppump – Patm = !v12[1 – (D1/D3)4] – gh3 = g{h2[1 – (D1/D3)4] – h3}

= (1.00103 kg/m3)(9.80 m/s2){(0.16 m)[1 – (0.60 cm/1.2 cm)4] – (– 1.1 m)} = 1.2104 N/m2.

93. We let d represent the diameter of the stream a distance y below the faucet. If we use the equation of continuity, we have Flow rate = A0v0 = A1v1 , or v1 = (A0/A1)v0 = (D/d)2v0. We choose the reference level at the faucet. If we apply Bernoulli’s equation to the stream, we have P0 + !v02 + gh0 = P1 + !v12 + gh1; Patm + !v02 + 0 = Patm + !v12 + g(– y); v12 = v02 + 2gy = (D/d)4v02, which gives

d = D[v02/(v02 + 2gy)]1/4.

94. (a) Because we assume zero air resistance, we can find the speed of the water from the nozzle from the horizontal range equation for projectile motion: R = (v02/g) sin 2; 8.0 m = (v02/9.80 m/s2) sin 2(30°), which gives v0 = 9.5 m/s. (b) The flow rate of the four nozzles is Q = 4A0v0 = 4[#p(3.010–3 m)2](9.5 m/s)(1.00103 L/m3) = 0.27 L/s. (c) From the equation of continuity, we have 4A0v0 = A1v1 ; 4[#pD02]v0 = [#pD12]v1; 4(3.010–3 m)2(9.5 m/s) = (1.910–2 m)2v1, which gives v1 =

0.95 m/s.

95. (a) We choose the reference level at the bottom of the sink. If we apply Bernoulli’s equation to the flow from the average depth of water in the sink to the pail, we have P0 + !v02 + gh0 = P1 + !v12 + gh1; Patm + 0 + gh0 = Patm + !v12 + gh1 , or

v1 = [2g(h0 – h1)]1/2 = {2(9.80 m/s2)[0.020 m – (– 0.50 m)]}1/2 = (b) We use the flow rate to find the time: Q = Av = V/t; p(0.010 m)2(3.19 m/s) = (0.48 m2)(0.040 m)/t, which gives t =

3.2 m/s.

19 s.

96. We choose the reference level at the bottom of the lower vessel. We assume the higher vessel is large enough that the water velocity in it can be neglected and the flow is turbulent. If we apply Bernoulli’s equation to the flow, we have P0 + !v02 + gh0 = P1 + !v12 + gh1; Patm + 0 + gh0 = Patm + !v12 + gh1 , or v1 = [2g(h0 – h1)]1/2. The flow rate is Q = A1v1 = A1[2g(h0 – h1)]1/2 = #p(1.210–2 m)2[2(9.80 m/s2)(0.64 m)]1/2 =

4.010–4 m3/s.

97. If we consider only the Bernoulli effect, the force from the difference in pressure on the wing must balance the weight of the airplane: (P2 – P1)A = mg, so P2 – P1 = mg/A. If we consider the points at the top and bottom surfaces of the wing compared to the air flow in front of the Page 16

Chapter 13

wing, from Bernoulli’s equation we have P0 + !v02 + gh0 = P1 + !v12 + gh1 = P2 + !v22 + gh2; P1 + !v12 + 0 = P2 + !v22 + 0, or P2 – P1 = !(v12 – v22) = mg/A;

!(1.29 kg/m3)[v12 – (100 m/s)2] = (2.0106 kg)(9.80 m/s2)/(1200 m2), which gives v1 =

1.9102 m/s.

98. (a) We let 1 refer to the input piston and 2 refer to the output piston. Because the pressure in the hydraulic fluid is constant, we have F2/A2 = F1/A1 , or F2 = F1(A2/A1) = F1(D2/D1)2. The force of the output piston must balance the weight of the car, so we have F2 = F1(A2/A1) = mg;

(b) (c)

(d) (e)

4.510–4 m2. (250 N)#p(1510–2 m)2/A1 = (1000 kg)(9.80 m/s2), which gives A1 = –4 2 1/2 The diameter of the piston is [4(4.510 m )/p] = 0.024 m = 2.4 cm. The work done by F2 is W = F2h = mgh = (1000 kg)(9.80 m/s2)(0.10 m) = 9.8102 J. Because the fluid is assumed to be incompressible, the volume is constant, so we have A1h1 = A2h2 , or h2 = h1(A1/A2) = h1(D1/D2)2 = (12 cm)[(2.4 cm)/(15 cm)]2 = 0.306 cm = 0.31 cm. We find the number of strokes from N = h/h2 = (10 cm)/(0.306 cm) = 32.7 = 33 strokes. The work done by F1 during the 33 strokes is W = F1Nh1 = (250 N)(32.7)(0.12 m) = 9.8102 J. This is the same as the work required to lift the car.

Page 17

Chapter 14

CHAPTER 14 – Oscillations 1.

In one period the particle will travel from one extreme position to the other (a distance of 2A) and back again. The total distance traveled is d = 4A = 4(0.15 m) = 0.60 m.

2.

(a) We find the spring constant from the elongation caused by the weight: k = mg/?x = (3.7 kg)(9.80 m/s2)/(0.028 m) = 1.30103 N/m. (b) Because the fish will oscillate about the equilibrium position, the amplitude will be the distance the fish was pulled down from equilibrium: A = 2.5 cm. The frequency of vibration will be f = (k/m)1/2/2p = [(1.30103 N/m)/(3.7 kg)]1/2/2p = 3.0 Hz.

3.

We find the spring constant from the compression caused by the increased weight: k = mg/x = (80 kg)(9.80 m/s2)/(0.0140 m) = 5.60104 N/m. The frequency of vibration will be f = (k/m)1/2/2p = [(5.60104 N/m)/(1080 kg)]1/2/2p = 1.15 Hz.

4.

(a) Because the motion starts at the maximum extension, we have x = A cos (t) = A cos (2pt/T) = (8.8 cm) cos [2pt/(0.75 s)]. (b) At t = 1.8 s we get x = (8.8 cm) cos [2p(1.8 s)/(0.75 s)] = – 7.1 cm.

5.

(a) We find the effective spring constant from the frequency: f1 = (k/m1)1/2/2p; 10 Hz = [k/(0.6010–3 kg)]1/2/2p, which gives k = 2.4 N/m. (b) The new frequency of vibration will be f2 = (k/m2)1/2/2p = [(2.37 N/m)/(0.4010–3 kg)]1/2/2p = 12 Hz.

6.

The general expression for the displacement is x = A cos (t + , so x(t = 0) = A cos .  = p (or –p). (a) – A = A cos , which gives cos  = – 1, so  = p/2 (or 3p/2). (b) 0 = A cos , which gives cos  = 0, so  = 0. (c) A = A cos , which gives cos  = + 1, so  = p/3 (or –p/3). (d) !A = A cos , which gives cos  = !, so  = 2p/3 (or 4p/3). (e) – A /2 = A cos , which gives cos  = – !, so  = p/4 (or –p/4). (f) A /v2 = A cos , which gives cos  = 1/v2, so

7.

Because the mass is released at the maximum displacement, we have x = xmax cos (t); v = – vmax sin (t); a = – amax cos (t). (a) We find t from v = – !vmax = – vmax sin (t), which gives t = 30°. Thus the distance is x = xmax cos (t) = xmax cos 30° = 0.866 xmax . (b) We find t from a = – !amax = – amax cos (t), which gives t = 60°. Thus the distance is x = xmax cos (t) = xmax cos 60° = 0.500 xmax .

Page 1

Chapter 14

8.

(a) We find the effective spring constant from the frequency: f1 = (k/m1)1/2/2p; 2.5 Hz = [k/(0.050 kg)]1/2/2p, which gives k = 12 N/m. (b) Because the size and shape are the same, the spring constant, which is determined by the buoyant force, will be the same. The new frequency of vibration will be f2 = (k/m2)1/2/2p = [(12 N/m)/(0.25 kg)]1/2/2p = 1.1 Hz.

9.

For a general displacement, we have x = A cos (t + ); v = – A sin (t + ). We find t +  from v = – !A = – A sin (t + ), which gives t +  = 30°. Thus the displacement is x = A cos (t + ) = A cos (30°) = 0.866 A.

10. The dependence of the frequency on the mass is f = (k/m)1/2/2p. Because the spring constant does not change, we have f2/f = (m/m2)1/2; (0.48 Hz)/(0.88 Hz) = [m/(m + 1.25 kg)]1/2, which gives m =

0.53 kg.

11. In the equilibrium position, the net force is zero. When the mass is pulled down a distance x, the net restoring force is the sum of the additional forces from the springs, so we have Fnet = ?F2 + ?F1 = – k2x – k1x = – (k1 + k2)x, which gives an effective force constant of k1 + k2 . We find the frequency of vibration from f = (keff/m)1/2/2p = [(k1 + k2)/m]1/2/2p.

F1 x

12. (a) From the graph we see that the period is x 0.69 s and the amplitude is 0.82 cm. We find the spring constant from the period: 0.82 cm T = 2p(m/k)1/2; 0.69 s = 2p[(14.310–3 kg)/k]1/2, 0.43 cm which gives k = 1.19 N/m. (b) For a general displacement, we have x = A cos (t + ). We find the phase constant from the condition at t = 0: 0.69 s 0.43 cm = (0.82 cm) cos , which gives cos = 0.524,  = – 58° = – 1.02 rad. We choose the negative angle to have a positive slope at t = 0. Thus the equation is x = A cos [2p(t /T) + ] (0.82 cm) cos [(9.1 s–1)t – 1.02]. = (0.82 cm) cos {2p[t/(0.69 s)] – 1.02} =

Page 2

F2

mg

t

Chapter 14

13. (a) When we compare x = (3.8 m) cos [(7p/4 s–1)t + p/6] to a general displacement, x = A cos (t + ), we see that  = 7p/4 s–1, so the period is T = 2p/ = 2p/(7p/4 s–1) = 8/7 s. The frequency is f = 1/T = 7/8 Hz = 0.875 Hz. (b) The velocity is v = dx/dt = – (3.8 m)(7p/4 s–1) sin [(7p/4 s–1)t + p/6] = – (20.9 m/s) sin [(7p/4 s–1)t + p/6]. At t = 0 , we have x0 = (3.8 m) cos(p/6) = 3.3 m; v0 = – (20.9 m/s) sin (p/6) = – 10.4 m/s. (c) The acceleration is a = dv/dt = – (3.8 m)(7p/4 s–1)2 cos [(7p/4 s–1)t + p/6] = – (115 m/s2) cos [(7p/4 s–1)t + p/6]. At t = 2.0 s , we have v = dx/dt = – (20.9 m/s) sin [(7p/4 s–1)(2.0 s) + p/6] = + 18 m/s; 2 –1 a = – (115 m/s ) cos [(7p/4 s )(2.0 s) + p/6] = – 57 m/s2. 14. The angular frequency of the motion is  = 2pf = 2p(264 Hz) = 1.66103 s–1. (a) The maximum speed is vmax = A = (1.66103 s–1)(1.510–3 m) = 2.5 m/s. (b) The maximum acceleration is amax = 2A = (1.66103 s–1)2(1.510–3 m) = 4.1103 m/s2. 15. The dependence of the frequency on the mass is f = (k/m)1/2/2p. Because the spring constant does not change, we have f2/f1 = (m1/m2)1/2; f2/(3.0 Hz) = [(0.50 kg)/(0.35 kg)]1/2, which gives f2 =

3.6 Hz.

16. (a) We find the derivatives of the given expression: x = a sin t + b cos t; dx/dt = a cos t – b sin t; d2x/dt2 = – 2a sin t – 2b cos t = – 2x. Thus Eq. 14–3 is satisfied if 2 = k/m. (b) We use a trigonometric identity to expand Eq. 14–4: x = A cos (t + ) = A(cos t cos  – sin t sin ). If we compare this to the given solution, we see that a = – A sin , and b = A cos . 17. We use a coordinate system with down positive. With x0 positive, at the equilibrium position we have ?F = –kx0 + mg = 0. If the spring is compressed a distance x (so x is negative) from the equilibrium position, we have ?F = – k(x + x0) + mg. When we use the equilibrium condition, we get ?F = F = – kx. Note that x is negative, so the restoring force is positive (down). If we stretch the spring a distance x (so x is positive), we still have Page 3

+x x0

x x=0

Chapter 14

?F = – k(x + x0) + mg. When we use the equilibrium condition, we get ?F = F = – kx. Note that x is positive, so the restoring force is negative (up). 18. We find the spring constant from the elongation caused by the mass: k = ?mg/?x = (1.62 kg)(9.80 m/s2)/(0.315 m) = 50.4 N/m. The period of the motion is independent of amplitude: T = 2p(m/k)1/2 = 2p[(1.62 kg)/(50.4 N/m)]1/2 = 1.13 s. The time to return to the equilibrium position is one-quarter of a period: t = #T = #(1.13 s) = 0.282 s. 19. (a) We find the frequency from f = (k/m)1/2/2p = [(345 N/m)/(0.250 kg)]1/2/2p = 5.91 Hz, so  = 2pf = 2p(5.91 Hz) = 37.1 s–1. Because the mass starts at the equilibrium position moving in the negative (downward) direction, we have a sine function: y = – A sin (t) = – (0.220 m) sin [(37.1 s–1)t]. (b) The period of the motion is T = 1/f = 1/(5.91 Hz) = 0.169 s. It will take one-quarter period to reach the maximum extension, so the spring will have maximum extensions at 0.0423 s, 0.211 s, 0.381 s,  . It will take three-quarters period to reach the minimum extension, so the spring will have minimum extensions at 0.127 s, 0.296 s, 0.465 s,  . 20. (a) We find the frequency from the period: f = 1/T = 1/(0.55 s) = 1.82 Hz, so  = 2pf = 2p(1.82 Hz) = 11.4 s–1. The amplitude is the compression: 0.10 m. Because the mass is released at the maximum displacement, we have a cosine function: y = A cos (t) = (0.10 m) cos [(11.4 s–1)t]. (b) The time to return to the equilibrium position is one-quarter of a period: t = #T = #(0.55 s) = 0.14 s. (c) The maximum speed is v0 = A = (11.4 s–1)(0.10 m) = 1.1 m/s. (d) The maximum acceleration is amax = 2A = (11.4 s–1)2(0.10 m) = 13 m/s2. The maximum magnitude of the acceleration occurs at the endpoints of the motion, so it will be attained first at the release point. 21. If the spring is stretched x0 when the stick is held in equilibrium in a horizontal position, for the torques about the point A, with clockwise positive, we have ?A = Mg(!¬) – kx0¬ = 0. If the stick is displaced through a small angle , the end will be displaced an additional distance x, so we have ?A = Mg(!¬) – k(x + x0)¬ = I = @M¬2 d2/dt2. When we use the result for equilibrium, we get – kx¬ = @M¬2 d2/dt2. For a small angle, x = L, so we have – k¬2 = @M¬2 d2/dt2, or d2/dt2 = – (3k/M). The motion will be simple harmonic, with Page 4

A

F



Mg

Chapter 14

2 = 3k/M, so the frequency is

f = (3k/M)1/2/2p.

22. The impulse, which acts for a very short time, will change the momentum of the mass, giving it an initial velocity. Because this occurs at the equilibrium position, the velocity will be the maximum velocity v0. Thus we have J = ?p = m ?v = mv0. The angular frequency is  = (k/m)1/2, so v0 = A = A(k/m)1/2. The SHM starts at the equilibrium position, x = 0 at t = 0, so we use a sine function: x = A sin t = v0(m/k)1/2 sin (k/m)1/2t = (J/m)(m/k)1/2 sin (k/m)1/2t = [J/(mk)1/2] sin (k/m)1/2t. 23. (a) If we apply a force F to stretch the springs, the total displacement ?x is the sum of the displacements of the two springs: ?x = ?x1 + ?x2. The effective spring constant is defined from F = – keff ?x. Because they are in series, the force must be the same in each spring: F1 = F2 = F = – k1 ?x1 = – k2 ?x2. Then ?x = ?x1 + ?x2 becomes – F/keff = – (F/k1) – (F/k2), or 1/keff = (1/k1) + (1/k2). For the period we have T = 2p(m/keff)1/2 = 2p{m[(1/k1) + (1/k2)]}1/2. (b) In the equilibrium position, we have Fnet = F20 – F10 = 0, or F10 = F20. When the object is moved to the right a distance x, we have Fnet = F20 – k2x – (F10 + k1x) = – (k1 + k2)x. The effective spring constant is keff = k1 + k2 , so the period is T = 2p(m/keff)1/2 = 2p[m/(k1 + k2)]1/2.

(a) k2 F

k1

(b)

k1

F1

F2 k2

x equilibrium

24. (a) At the equilibrium positions the net force on each mass is zero. F1 F12 F12 If we displace each mass to the right and assume that x2 > x1, the magnitudes of the additional forces are F1 = kx1, F12 = k(x2 – x1), F2 = kx2. x1 x2 The spring on the left and the middle spring will + have an additional tension, while the spring on equil equil the right will have an additional compression. The directions of the forces are indicated on the diagram, so for the two masses we have ?F = – kx1 + k(x2 – x1) = m1a1 = m1 d2x1/dt2; ?F = – k(x2 – x1) – kx2 = m2a2 = m2 d2x2/dt2. Because the masses are equal, we let k/m1 = 12 = k/m2 = 22 = 02 = k/m, and get d2x1/dt2 = – 02(2x1 – x2) ; and d2x2/dt2 = – 02(2x2 – x1) . (b) For the assumed solutions we have x1 = A1 cos t; d2x1/dt2 = – A12 cos t; x2 = A2 cos t; d2x2/dt2 = – A22 cos t. Page 5

F2

Chapter 14

When we substitute these in the previous equations, we get – 02(2A1 – A2) cos t = – A12 cos t, or A202 = A1(202 – 2); – 02(2A2 – A1) cos t = – A22 cos t, or A2(202 – 2) = A102. If we divide these equations, we get (202 – 2)2 = 04, or 202 – 2 = ± 02, which gives 2 = 02, 302, or  = (k/m)1/2, v3(k/m)1/2. Note that when  = 0 , A1 = A2 , a symmetric motion; and when  = v30 , A1 = – A2 , an antisymmetric motion. 25. The angular frequency is = 2pf = (k/m)1/2 = [(250 N/m)/(0.380 kg)]1/2 = 25.6 s–1. The period is T = 2p/= 2p/(25.6 s–1) = 0.245 s. (a) For a general displacement, we have x = A cos (t + ); v = – A sin (t + ). Because v = vmax = A, when t0 = 0.110 s, we have sin [(25.6 s–1)(0.110 s) + ] = – 1, or 2.82 +  = 3p/2, which gives  = 1.89 rad = 108°. Note that this also gives x = 0. Thus the equation for the motion is x = (12.0 cm) cos [(25.6 s–1)t + 1.89 rad]. (b) Because the mass passes through the equilibrium position toward positive x at t0 , it will reach the maximum length, x = – 12.0 cm, &T later: tmax1 = t0 + &T = 0.110 s + &(0.245 s) = 0.294 s. It will be at this position at intervals of T: tmax = 0.294 s, 0.539 s, 0.784 s, … . The mass will reach the minimum length, x = + 12.0 cm, #T after t0: tmin1 = t0 + #T = 0.110 s + #(0.245 s) = 0.171s. It will be at this position at intervals of T: tmin = 0.171 s, 0.416 s, 0.661 s, … . (c) The displacement at t = 0 is x = (12.0 cm) cos (0 + 1.89 rad) = – 3.77 cm. (d) Because the net force produces the SHM, the force in the spring at t = 0 is F = – kx + mg = – (250 N/m)(– 0.0377 cm) + (0.380 kg)(9.80 m/s2) = + 13.1 N (up). (e) The maximum speed is vmax = A= (0.120 m)( 25.6 s–1) = 3.07 m/s. Because the initial phase is less than 180°, the maximum speed is reached at t = t0 = 0.110 s. 26. (a) The total energy is the maximum potential energy, so we have U = !E = !Umax;

!kx2 = !(!kA2), which gives x =

(b) For the potential energy we have U/E = !kx2/!kA2 = (!A)2/A2 = #. Then we have K/E = 1 – U/E = &.

± 0.707A.

27. (a) The amplitude is the maximum value of x: 0.650 m. (b) We find the frequency from the coefficient of t: 1.34 Hz. 2pf = 8.40 s–1, which gives f = (c) The maximum speed is v0 = A = (8.40 s–1)(0.650 m) = 5.46 m/s. Page 6

Chapter 14

We find the total energy from the maximum kinetic energy: E = Kmax = !mv02 = !(2.00 kg)(5.46 m/s)2 = 29.8 J. (d) We find the velocity at the position from v = v0[1 – (x2/A2)]1/2

= (5.46 m/s){1 – [(0.260 m)2/(0.650 m)2]}1/2 = 5.00 m/s. The kinetic energy is K = !mv2 = !(2.00 kg)(5.00 m/s)2 = 25.0 J. The potential energy is U = E – K = 29.8 J– 25.0 J = 4.8 J.

28. The angular frequency is  = 2pf = 2p(3.0 Hz) = 6.0 p s–1. (a) The velocity at the equilibrium point is maximum: vmax = A = (6.0 p s–1)(0.15 m) = 2.8 m/s. (b) We find the velocity at the given position from v = v0[1 – (x2/A2)]1/2

= (2.8 m/s){1 – [(0.10 m)2/(0.15 m)2]}1/2 = 2.1 m/s. (c) We find the total energy from the maximum kinetic energy: E = Kmax = !mvmax2 = !(0.35 kg)(2.8 m/s)2 = 1.4 J. (d) Because the mass starts at the maximum position, we use a cosine function: x = A cos t = (0.15 m) cos (6.0 p s–1)t.

29. We find the spring constant from the compression: k = F/x = (95.0 N)/(0.185 m) = 514 N/m. The kinetic energy will be maximum when the ball leaves the gun, and thus equal to the initial potential energy: !mv2 = !kx2; 9.37 m/s. (0.200 kg)v2 = (514 N/m)(0.185 m)2, which gives v = 30. Immediately after the collision, the block-bullet system will have its maximum velocity at the equilibrium position. We find this velocity from energy conservation: Ki + Ui = Kf + Uf ;

!(M + m)v02 + 0 = 0 + !kA2; !(0.300 kg + 0.0125 kg)v02 = !(2.25103 N/m)(0.124 m)2, which gives v0 = 10.5 m/s.

We find the initial speed of the bullet from momentum conservation for the impact: mv + 0 = (M + m)v0; (0.0125 kg)v = (0.300 kg + 0.0125 kg)(10.5 m/s), which gives v = 263 m/s. 31. We can compare the two maximum potential energies: U1/U2 = !k1A12/!k2A22 = (k1/k2)(A1/A2)2; 10 = 2(A1/A2)2, or A1 = 2.24A2. 32. (a) We find the spring constant from 2pf = (k/m)1/2; 1.2102 N/m. 2p(3.5 Hz) = [k/(0.240 kg)]1/2, which gives k = (b) We find the total energy from the maximum potential energy: E = Umax = !kA2 = !(1.16102 N/m)(0.045 m)2 = 0.12 J.

33. (a) The work done in compressing the spring is stored as the maximum potential energy, so we have Page 7

Chapter 14

E = Umax = !kA2; 4.2102 N/m. 3.0 J = !k(0.12 m)2, which gives k = 4.17102 N/m = (b) We find the mass from Fmax = kA = mamax ; (4.17102 N/m)(0.12 m) = m(15 m/s2), which gives m = 3.3 kg. 34. (a) The total energy is the maximum potential energy: !mv2 + !kx2 = !kA2; (2.1 kg)(0.55 m/s)2 + (280 N/m)(0.020 m)2 = (280 N/m)A2, which gives A = 0.052 m = (b) We find the maximum velocity from the maximum kinetic energy: !mvmax2 = !kA2; (2.1 kg)vmax2 = (280 N/m)(0.052 m)2, which gives vmax = 0.60 m/s.

5.2 cm.

35. For the collision of the bullet and block momentum is conserved: mv = (m + M)V, so V = mv/(m + M). The kinetic energy of the bullet and block immediately after the collision is stored in the potential energy of the spring when the spring is fully compressed: !(m + M)V2 = !kA2; (m + M)[mv/(m + M)]2 = kA2; 352.6 m/s. [(0.007870 kg)v]2/(0.007870 kg + 6.023 kg) = (142.7 N/m)(0.09460 m)2, which gives v = 36. (a) The period of the motion is independent of amplitude: T = 2p(m/k)1/2 = 2p[(0.650 kg)/(184 N/m)]1/2 = 0.373 s. The frequency is f = 1/T = 1/(0.373 s) = 2.68 Hz. (b) Because the mass is struck at the equilibrium position, the initial speed is the maximum speed. We find the amplitude from v0 = A = 2pfA; 2.26 m/s = 2p(2.68 Hz)A, which gives A = 0.134 m. (c) The maximum acceleration is amax = 2A = (2pf)2A = [2p(2.68 Hz)]2(0.134 m) = 38.0 m/s2. (d) Because the mass starts at the equilibrium position, we have a sine function. If we take the positive x-direction in the direction of the initial velocity, we have x = A sin (t) = A sin (2pft) = (0.134 m) sin [2p(2.68 Hz)t] = (0.134 m) sin [(16.8 s–1)t]. (e) We find the total energy from the maximum kinetic energy: E = Kmax = !mv02 = !(0.650 kg)(2.26 m/s)2 = 1.66 J. (f) We find the kinetic energy from the total energy: E = K + !kx2; 1.40 J. 1.66 J = K + !(184 N/m)[(0.40)(0.134 m)]2; which gives K = 37. The total energy of the oscillator is: E = !kA2 = !mv2 + !kx2; which becomes v = dx/dt = ± [(k/m)(A2 – x2)]1/2. We separate the variables: dx/(A2 – x2)1/2 = ± (k/m)1/2 dt, and integrate, with x = x0 at t = 0:

Page 8

Chapter 14 x



x0

dx = A2 – x 2

sin – 1 x A

x x0

k m

t 0

d t;

x = sin – 1 x – sin – 1 0 =  A A

k m t.

If we let sin–1 (x0/A) = , we have k t + . x = A sin  m Of course we could have a cosine function by a change in phase. The choice of sign is made from the sign of the velocity at t = 0. 38. (a) We find the period from the time for N oscillations: T = t/N = (50 s)/42 = 1.2 s. (b) The frequency is f = 1/T = 1/(1.19 s) = 0.84 Hz. 39. We find the length from T = 2p(L/g)1/2; 2.000 s = 2p[L/(9.80 m/s2)]1/2, which gives L =

0.9929 m.

40. The period of a simple pendulum is given by T = 2p(L/g)1/2. If we form the ratio for the two values of g, we get TMars/TEarth = (gEarth/gMars)1/2; TMars/(0.80 s) = (1/0.37)1/2, which gives TMars =

1.3 s.

41. (a) We find the length from T = 2p(L/g)1/2; 1.00 s = 2p[L/(9.80 m/s2)]1/2, which gives L = (b) We find the period from T = 2p(L/g)1/2 = 2p[(1.00 m)/(9.80 m/s2)]1/2 =

0.248 m. 2.01 s.

42. (a) For the period on Earth we have T = 2p(L/g)1/2 = 2p[(0.73 m)/(9.80 m/s2)]1/2 = 1.7 s. (b) In a freely falling elevator, the effective g is zero, so the period would be

infinite (no swing).

43. We assume that 14° is small enough that we can consider this a simple pendulum, with a period T = 2p(L/g)1/2 = 2p[(0.30 m)/(9.80 m/s2)]1/2 = 1.10 s, and  = 2p/T = 2p/(1.10 s) = 5.71 s–1. Because the pendulum is released at the maximum angle, the angle will oscillate as a cosine function:  = 0 cos (t) = (14°) cos [(5.71 s–1)t]. (a)  = (14°) cos [(5.71 s–1)(0.65 s)] = – 12°. This is reasonable, because the time is slightly over half a period. + 1.9°. (b)  = (14°) cos [(5.71 s–1)(1.95 s)] = – 13°. (c)  = (14°) cos [(5.71 s–1)(5.00 s)] = This is reasonable; the time is 4 periods and 0.60 s, so it should be close to the answer for part (a).

Page 9

Chapter 14

44. We use energy conservation between the release point and the lowest point: Ki + Ui = Kf + Uf ; 0 + mgh = !mv02 + 0, or v02 = 2gh = 2gL(1 – cos ). When we use a trigonometric identity, we get v02 = 2gL(2 sin2 !).



For a simple pendulum  is small, so we have sin ! ˜ !. Thus we get v02 = 2gL2 (!)2, or v0 = (gL)1/2.

h

45. This is most easily found from the associated circular motion for SHM. We see from the diagram that, out of the 2p radians for a complete circular motion, the total angle corresponding to the pendulum being between + !A and – !A is 4. We find  from sin  = !A/A = !, so  = p/6. Thus the fraction of time spent between + !A and – !A is @. fraction = 4(p/6)/2p =

46. (a) For the frequency we have f = (g/L)1/2/2p = [(9.80 m/s2)/(0.68 m)]1/2/2p = 0.60 Hz. (b) We use energy conservation between the release point and the lowest point: Ki + Ui = Kf + Uf ; 0 + mgh = !mv02 + 0;



A

 

 A/2

 

0 L cos 0

(9.80 m/s2)(0.68 m)(1 – cos 12°) = !v02, which gives v0 = 0.54 m/s.

h

Page 10

L

L cos 

L

x

Chapter 14

47. We use the parallel-axis theorem to find the moment of inertia about the pin A: IA = ICM + Mh2 = !MR2 + Mh2 = M(!R2 + h2). The period for small oscillations is T = 2p(IA/Mgh)1/2 = 2p[(!R2 + h2)/gh]1/2

d A

= 2p{[!(0.200 m)2 + (0.180 m)2] /(9.80 m/s2)(0.180 m)}1/2 = 1.08 s.

h

R M

48. (a) For the moment of inertia of the system about the suspension point we have I = ML2 + @mL2 = (M + @m)L2. The distance of the center of mass of the system from the suspension point is h = [ML + m(!L)]/(M + m) = (M + !m)L/(M + m). The period for small oscillations is T = 2p[I/(M + m)gh]1/2 = 2p[(M + @m)L2/(M + !m)Lg]1/2 = 2p[(M + @m)L/(M + !m)g]1/2. (b) If we were to use the expression for a simple pendulum, the fractional error would be error = (T – Tsimple)/T

m L

M

= {2p[(M + @m)L/(M + !m)g]1/2 – 2p(L/g)1/2}/2p[(M + @m)L/(M + !m)g]1/2 = {[(M + @m)/(M + !m)]1/2 – 1}/[(M + @m)/(M + !m)]1/2 = 1 – [(M + !m)/(M + @m)]1/2 = 1 – [(1 + m/2M)/(1 + m/3M)]1/2.

49. We can find the effective torsional constant from = K; 1.110–5 m · N = K(60°)(p/180°), which gives K = 1.0510–5 m · N/rad. We use the expression for the period from Problem 50: T = 1/f = 2p(I/K)1/2 = 2p(mR2/K)1/2; 1/3.10 Hz = 2p[m(0.0095 m)2/(1.0510–5 m · N/rad)]1/2, which gives m = 3.110–4 kg = 50. (a) When we apply ?= I to the twisting motion of the torsional pendulum, we have – K = I = I d2/dt2. When we write this as d2/dt2 = – (K/I) = – 2, we see that the angular acceleration is proportional to the displacement, so we have SHM. We can write the equation for the motion as  = 0 sin (t + ), where  = (K/I)1/2. (b) The period of the motion is T = 2p/ = 2p(I/K)1/2.

Page 11

0.31 g.

Chapter 14

51. (a) We use the parallel-axis theorem for the lower mass to find the moment of inertia of the leg about the hip: I = m1L12/3 + m2L22/12 + m2(L1 + !L2)2 = (m1/3 + m2/12 + 9m2/4)L2 = [(7.0 kg)/3 + (4.0 kg)/12 + 9(4.0 kg)/4](0.50 m)2 = 2.92 kg · m2. The distance of the center of mass of the leg from the hip is h = [m1!L1 + m2(L1 + !L2)]/(m1 + m2) = (!m1 + *m2)L/(m1 + m2)

L1

m1

= [!(7.0 kg) + *(4.0 kg)](0.50 m)/(7.0 kg + 4.0 kg) = 0.432 m. The natural period is T = 2p[I/(m1 + m2)gh]1/2 = 2p[(2.92 kg · m2)/(7.0 kg + 4.0 kg)(9.80 m/s2)(0.432 m)]1/2 = 1.6 s.

L2 m2

52. We use the parallel-axis theorem to find the moment of inertia about the pin: I = mL2/12 + mh2 = m(h2 + L2/12). The period is T = 2p(I/mgh)1/2 = 2p[(h2 + L2/12)/gh]1/2. To find the minimum we set dT/dh = 0. Because T ? 0, this is simpler if we differentiate T2: dT2/dh = 2T dT/dh = 4p2{(2h/gh) – [(h2 + L2/12)/gh2]} = 4p2[(2h2 – h2 – L2/12)/gh2] = 0, which gives h = L/v12 = 0.289L. 0.211 m from the end. Thus the hole should be !L – h = The period of the motion is T = 2p[(h2 + L2/12)/gh]1/2 = 2p[(L2/12 + L2/12)/g(L/v12)]1/2 = 2p(L/gv3)1/2 = 2p[(1.00 m)/(9.80 m/s2)v3]1/2 = 1.53 s.

h m

53. From Problem 50 the period of the motion is T = 2p(I/K)1/2. Because the torsional constant does not change, the ratio of the two periods is T2/T1 = (I2/I1)1/2 = [(m2L22/12)/m1L12/12]1/2 T2/(6.0 s) = [(0.70 m1)(0.70L1)2/m1L12]1/2, which gives T2 = 3.5 s. 54. We find the torsional constant with the result from Problem 50: T = 1/f = 2p(I/K)1/2; 1/(0.331 Hz) = 2p[!(0.500 kg)(0.0625 m)2/K]1/2, which gives K =

4.2210–3 m · N/rad.

55. (a) The angular frequency for the damped motion is  = [(k/m) – (b2/4m2)]1/2 = {[(56.0 N/m)/(0.750 kg)] – [(0.162 N · s/m)2/4(0.750 kg)2]}1/2 = 8.64 s–1. The period is T = 2p/ = 2p/(8.64 s–1) = 0.727 s. (b) We evaluate the factor b/2m = (0.162 N · s/m)/2(0.750 kg) = 0.108 /s. The amplitude is proportional to e –bt/2m. The fractional decrease over one period is Page 12

Chapter 14

fractional decrease = [e –bt/2m – e –b(t + T)/2m]/e –bt/2m = 1 – e –bT/2m 0.0755. = 1 – e – (0.108 /s)(0.727 s) = (c) The general expression for the displacement is x = Ae –bt/2m cos (t + ). We use the given data to evaluate the constants: 0 = Ae –0 cos , which gives  = – p/2, or we can change to a sine function. 0.120 m = Ae – (0.108 /s)(1.00 s) sin [(8.64 s–1)(1.00 s)], which gives A = 0.189 m. Thus the displacement is x = (0.189 m)e – (0.108 /s)t sin [(8.64 s–1)t]. 56. (a) From the assumed expression for the angular displacement,  = Ae – t cos t, we see that the  is maximum at t = 0, so A = 15°. After a time t, the amplitude of the cosine function is A e – t, so at t = 8.0 s we have  = 0.125 s–1. (15°) e –(8.0 s) = 5.5°, which gives (b) The angular frequency of the undamped motion is 0 = (mgh/I)1/2 = [mg(L/2)/(mL2/3)]1/2 = (3g/2L)1/2 = [3(9.80 m/s2)/2(0.80 m)]1/2 = 4.29 s–1. For the damped motion we have  = [02 – 2]1/2 = [(4.29 s–1)2 – (0.125 s–1)2]1/2 ˜ 4.29 s–1. Thus the approximate period is T = 2p/ = 2p/(4.29 s–1) = 1.47 s. (c) To reduce the amplitude to !, we have e – t = !, or (0.125 s–1)t = ln 2 = 0.693, which gives t = 5.5 s. 57. (a) The energy of the oscillator is all potential energy when the cosine factor is 1: E = !kxmax2 = !kA2 e –bt/m. If we compare the energies for t = 0 and t = T, we have E/E0 = e –bT/m = 1 – 0.050, which gives bT/m = 0.0513. We assume light damping, so T ˜ 2p/0 , to get 2pb/0m = 0.0513, or b/2m = 0.004080 . The angular frequency is  = [(k/m) – (b2/4m2)]1/2 = [02 – (0.004080)2]1/2 = (1 – 1.6710–5)1/20 ˜ (1 - 8.310–6)0 . This justifies our assumption of light damping. The fractional difference is ( – 0)/0 = 8.310–6 = 8.310–4 %. (b) After n periods, the ratio of amplitudes is e –bnT/2m = e–1, or bnT/2m = 2pnb/2m0 = 1; 2pn(0.00408) = 1, which gives n = 39 periods.

58. (a) If we assume the oscillator starts at maximum displacement, we have x = Ae –bt/2m cos t. The velocity is v = dx/dt = A(–b/2m)e –bt/2m cos t – Ae –bt/2m  sin t. For a lightly damped oscillator,  » b/2m, so we ignore the cosine term and get Page 13

Chapter 14

v ˜ – Ae –bt/2m  sin t. For the total energy we have E = !mv2 + !kx2 = !mA2 e –bt/m 2 sin2 t + !kA2 e –bt/m cos2 t. For a lightly damped oscillator, 2 ˜ 02 = k/m, so we get = !mA2 e –bt/m (k/m) sin2 t + !kA2 e –bt/m cos2 t = !kA2 e –bt/m (sin2 t + cos2 t ) = !kA2 e –bt/m = E0e –bt/m . (b) The fractional energy lost in one period is – ?E/E = [E0e –bt/m – E0e –b(t + T)/m ]/E0e –bt/m = 1 – e –bT/m . For a lightly damped oscillator,  ˜ 0 , so the period is T = 2p/0. The exponent will be small, so we use the expansion ex ˜ 1 + x for the exponential: – ?E/E = 1 – e –bT/m ˜ 1 – (1 – 2pb/m0) = 2pb/m0 = 2p/Q, where Q = m0/b. E

59. (a) In the equilibrium position, we have k Fnet = F20 – F10 = 0, or F10 = F20. k F1 F2 When the object is moved to the right a distance x, we have Fnet = F20 – kx – (F10 + kx) = – 2kx. x The effective spring constant is keff = 2k, so the frequency is f = (keff/m)1/2/2p = (2k/m)1/2/2p equilibrium = [2(100 N/m)/(0.200 kg)]1/2/2p = 5.03 Hz. (b) From the expression for the displacement, x = Ae –t cos t, we see that the amplitude of the cosine function is Ae –t, so to reduce the amplitude to ! in n periods, we have e –n = !, or  nT = n/f = ln 2;  (55)/(5.03 Hz) = 0.693, which gives  = 0.0634 s–1. (c) To reduce the amplitude to # of the initial value, we have e –n = #, or  nT =  n/f = ln 4 = 2 ln 2; 110 oscillations. (0.0634 s–1)n/(5.03 Hz) = 2(0.693), which gives n = This is expected, because the exponential function has the property that the same time is required to reduce the value by !, no matter what the initial value is. Thus to reduce by #, we reduce by ! (55 oscillations) and then ! again (55 oscillations). 





60. (a) At resonance,  = 0 , so we have 0 = tan–1 (02 – 2)/(b/m) = tan–1 0 = 0°. (b) The driving force, Fext = F0 cos t, is maximum at t = 0. Thus the displacement is x = A0 sin (t + 0) = A0 sin (0 + 0) = 0. The driving force, Fext = F0 cos t, is zero at t = ± p/2. Thus the displacement is x = A0 sin (t + 0) = A0 sin (± p/2 + 0) = ± A0. (c) We see that the phase difference between the force and the displacement is 90°.

61.

Page 14

Chapter 14

 A 0=

F0/ k

 2/

2 0

–1

2

+  2/ 02Q 2

A 0 k/ F0 4

3

2

1

0

0

0.5

1.0

1.5

2.0

 / 0

62. At resonance,  = 0 , so the amplitude is A0 = (F0/m)/[(2 – 02) + (b22/m2)]1/2 = (F0/m)m/b0 = (Q/02)(F0/m); 28.6(F0/m) = {Q/[2p(382 Hz)]2}(F0/m), which gives Q =

1.65108.

63. We find the derivatives of the assumed solution: x = A0 sin (t + 0); dx/dt = A0 cos (t + 0); d2x/dt2 = – 2A0 sin (t + 0). We use these in the equation of motion: m d2x/dt2 + b dx/dt + kx = F0 cos t; m[– 2A0 sin (t + 0)] + b[A0 cos (t + 0)] + k[A0 sin (t + 0)] = F0 cos t. We expand the trigonometric functions: A0(k – m2)(sin t cos 0 + cos t sin 0) + A0b(cos t cos 0 – sin t sin 0) = F0 cos t; [A0(k – m2) cos 0 – A0b sin 0] sin t + [A0(k – m2) sin 0 + A0b cos 0 – F0] cos t = 0. For this to be true for any t, the coefficient of each variable function must equal zero. From the coefficient of sin t, we get 1/2 A0(k – m2) cos 0 – A0b sin 0 = 0; 2] ) m b/ tan 0 = sin 0/cos 0 = (k – m2)/b ( + 2 2) = [(k/m) – 2]/(b/m) 0 2 – 2  2 – 2 2 = (0 –  )/(b/m). [(  0 We can represent this in a triangle, as shown. 0 From the coefficient of cos t, we get b/ m A0(k – m2) sin 0 + A0b cos 0 – F0 = 0;

{A0(k – m2)(02 – 2)/[(02 – 2)2 + (b/m)2]1/2} + {A0b (b/m)/[(02 – 2)2 + (b/m)2]1/2} = F0; A0m[(02 – 2)2 + (b/m)2]/[(02 – 2)2 + (b/m)2]1/2 = F0 , or A0 = (F0/m)/[(2 – 02)2 + (b/m)2]1/2.

Page 15

Chapter 14

64. To find the maximum for A0 , we set the first derivative equal to zero: A0 = (F0/m)/[(2 – 02)2 + (b/m)2]1/2; = – !(F0/m)[2(2 – 02)(2) + 2b2/m2]/[(2 – 02)2 + (b/m)2]3/2 = – (F0/m)[2(2 – 02) + b2/m2]/[(2 – 02)2 + (b/m)2]3/2 = 0; 2(2 – 02) + b2/m2 = 0, or 2 = 02 – b2/2m2.

dA0/d

65. (a) For a lightly damped oscillator,  ˜ 0 = 2p/T = (g/L)1/2 = [(9.80 m/s2)/(0.50 m)]1/2 = 4.43 s–1. The amplitude of the cosine function is Ae –t = Ae –bt/2m , so to reduce the amplitude by %, we have e –bt/2m = @, or bt/2m = 0t/2Q = ln 3; (4.43 s–1)t/2(400) = 1.098, which gives t = 198 s. (b) From Problem 58, the energy is E = E0e –bt/m . For the rate of change we differentiate and express as a loss: – dE/dt = (b/m)E0e –bt/m = (b/m)!kA02 = !b02A02 = !m03A02/Q = (0.20 kg)(4.43 s–1)3(0.020 m)2/2(400) = 8.710–6 W. (c) We find the width of the peak at half-maximum from ?/0 = 1/Q; ?/(4.43 s–1) = 1/400, which gives ? = 1.110–2 s–1. The frequency difference from the natural frequency is 8.810–4 Hz on either side of f0. ?f = !?/2p = !(1.110–2 s–1)/2p =

Page 16

Chapter 14

66. (a) For the motion of a driven oscillator we have x = A0 sin (t + 0); v = A0 cos (t + 0). Thus the power input is P = Fextv = F0 cos t [A0 cos (t + 0)] = F0A0 cos t (cos t cos 0 – sin t sin 0). When we use the result for A0 we get 2

 P=

F0 

cos 2 t cos 0 – sin t cos t sin 0 2 2 m (  2 – 02) +  2b / m 2 F02 cos 2 t cos  0 – 12 sin 2t sin 0 = . 2 m ( 2 – w02) +  2b2/ m 2 (b) The average power input is P = F0A 0 cos 2 t av cos 0 – 12 sin 2t av sin 0 .

(c)

Over a cycle, cos2 t is always positive with an average of !; sin 2t is positive and negative with an average of 0. Thus we have P = 12 F0A 0 cos 0 = 12 F0v max cos 0 .

6

P/( F0 2 /2 m 0 ) 4

2

0

0

0.5

1.0

1.5

/  0 67. The amplitude of the forced oscillator is A0 = (F0/m)/[(2 – 02)2 + (b/m)2]1/2 = (F0/m)/[(2 – 02)2 + 202/Q2]1/2. At resonance,  = 0 , so A0,res = F0/(m02/Q); and at the half-width we have A02/A0,res2 = (04/Q2)/[(2 – 02)2 + 202/Q2] = !, or 204/Q2 = 4 – 2022 + 04 + 202/Q2. We can simplify the algebra if we let x = /0. After rearranging we have x4 – (2 – 1/Q2)x2 + (1 – 2/Q2) = 0. This is a quadratic equation for x2 with the solutions x2 = !{(2 – 1/Q2) ± [(2 – 1/Q2)2 – 4(1 – 2/Q2)]1/2} = 1 – 1/2Q2 ± !(4/Q2 + 1/Q4)1/2. We know that Q > 1 and assume that Q2 » 1, which gives Page 17

2.0

Chapter 14

x2 ˜ 1 – 1/2Q2 ± (1/Q2)1/2 ˜ 1 ± 1/Q, or x = (1 ± 1/Q)1/2 ˜ 1 ± 1/2Q. The width of the curve is ?/0 = ?x = x+ – x– = (1 + 1/2Q) – (1 – 1/2Q) = 1/Q. 68. We choose h = 0 at the unstretched position of the net and let the stretch of the net be x. We use energy conservation between the release point and the lowest point to find the spring constant: Ki + Ui = Kf + Uf ; 0 + mghi = 0 + mg(– x1) + !kx12, or mg(hi + x1) = !kx12; (52 kg)(9.80 m/s2)(20.0 m + 1.1 m) = !k(1.1 m)2, which gives k = 1.78104 N/m. When the person lies on the net, the weight causes the deflection: mg = kx2; 2.9 cm. (52 kg)(9.80 m/s2) = (1.78104 N/m)x2 , which gives x2 = 0.0287 m = We use energy conservation between the release point and the lowest point to find the stretch: Ki + Ui = Kf + Uf ; 0 + mghi = 0 + mg(– x3) + !kx32, or mg(hi + x3) = !kx32; (52 kg)(9.80 m/s2)(35 m + x3) = !(1.78104 N/m)x32. This is a quadratic equation for x3 , for which the positive result is

1.4 m.

69. (a) For the frequency we have f = (g/L)1/2(1/2p) = [(9.80 m/s2)/(0.63 m)]1/2(1/2p) = 0.63 Hz. (b) We use energy conservation between the release point and the lowest point: Ki + Ui = Kf + Uf ; 0 + mgh = !mv02 + 0; (9.80 m/s2)(0.63 m)(1 – cos 15°) = !v02, which gives v0 = 0.65 m/s. (c) The energy stored in the oscillation is the initial potential energy: Ui = mgh = (0.365 kg)(9.80 m/s2)(0.63 m)(1 – cos 15°) = 0.077 J. Note that this is also the maximum kinetic energy, !

mv02.

70. (a) The amplitude is the maximum value of x: 0.25 m. (b) We find the frequency from the coefficient of t: 0.875 Hz. 2pf = 5.50 s–1, which gives f = (c) The period is T = 1/f = 1/(0.875 Hz) = 1.14 s. (d) We find the total energy from the maximum kinetic energy: E = Kmax = !mv02 = !mA22 = !(0.650 kg)(0.25 m)2(5.50 s–1)2 = 0.61 J. (e) The potential energy is U = !kx2 = !m2x2 = !(0.650 kg)(5.50 s–1)2(0.10 m)2 = 0.098 J. The kinetic energy is K = E – U = 0.61 J– 0.098 J = 0.51 J. 71. We find the period from the time for N oscillations: T = t/N = (34.7 s)/8 = 4.34 s. From this we can get the spring constant: T = 2p(m/k)1/2; 151 N/m. 4.34 s = 2p[(72.0 kg)/k]1/2, which gives k = At the equilibrium position, we have mg = kx0; Page 18

0 L cos 0

h

L

Chapter 14

(72.0 kg)(9.80 m/s2) = (151 N/m)x0 , which gives x0 = 4.68 m. Because this is how much the cord has stretched, we have L = D – x0 = 25.0 m – 4.68 m = 20.3 m.

72. (a) The stress from the tension in the cable causes the strain. We find the effective spring constant from E = stress/strain = (FT/A)/(?L/L0), or k = FT/?L = EA/L0 = (200109 N/m2)p(3.210–3 m)2/(20.0 m) = 3.22105 N/m. We find the period from T = 2p(m/k)1/2 = 2p[(1200 kg)/(3.22105 N/m)]1/2 = 0.38 s. (b) The elongation at the equilibrium position is produced by the stress from the weight of the car: E = (F1/A)/(?L1/L0); 200109 N/m2 = [(1200 kg)(9.80 m/s2)/p(3.210–3 m)2]/(?L1/20.0 m), which gives ?L1 = 3.6610–2 m. The elongation that will break the cable is produced by a stress equal to the strength of the cable: E = (F2/A)/(?L2/L0); 200109 N/m2 = (500106 N/m2)/(?L2/20.0 m), which gives ?L2 = 5.0010–2 m. Thus the amplitude to reach the breaking point is A = ?L2 – ?L1 = 5.0010–2 m – 3.6610–2 m = 1.3410–2 m = 1.34 cm. 73. If we ignore losses due to friction, the initial kinetic energy becomes potential energy: !mv02 = !kx2; (1500 kg)(2.0 m/s)2 = (500103 N/m)x2, which gives x = 0.11 m. 74. We find the length from T = 2p(L/g)1/2. In Austin, Texas, we have 2.000 s = 2p[LAustin/(9.793 m/s2)]1/2, which gives LAustin = 0.9922 m. In Paris we have 2.000 s = 2p[LParis/(9.809 m/s2)]1/2, which gives LParis = 0.9939 m. Thus the additional length is LParis – LAustin = 0.9939 m – 0.9922 m = 0.0017 m = On the Moon we have 2.000 s = 2p[LMoon/(1.62 m/s2)]1/2, which gives LMoon = 0.164 m.

1.7 mm.

75. The shear stress and shear strain are related by the shear modulus: F ²x G = (F/A)/(?x/h), or F = (GA/h) ?x, where A is the area of the top of the block. CM Because the center of mass oscillates with half the displacement of the top, the effective force constant is keff = F/(?x/2) = 2GA/h. The frequency is f = (keff/m)1/2/2p = [(2GA/h)/Ah]1/2/2p = [2(520 N/m2)/(1300 kg/m3)(0.040 m)2]1/2/2p = 3.6 Hz. 76. The effective value of g is increased when the acceleration is upward and decreased when the acceleration is downward. Because the length does not change, for the ratio of frequencies we have f /f = (g/g)1/2. (a) For the upward acceleration we get f /f = (g/g)1/2 = [(g + !g)/g]1/2, which gives f  = 1.22f. Page 19

h

Chapter 14

(b) For the downward acceleration we get f /f = (g/g)1/2 = [(g – !g)/g]1/2, which gives f  =

0.71f.

77. (a) We find the effective spring constant from the displacement caused by the additional weight: k ?y = mg, or k = mg/?y = (75 kg)(9.80 m/s2)/(0.035 m) = 2.10104 N/m. We find the frequency of vibration from f = (k/m)1/2(1/2p) = [(2.10104 N/m)/(420 kg)]1/2(1/2p) = 1.1 Hz. (b) The total energy is the maximum potential energy, so we have E = Umax = !kA2 = !(2.10104 N/m)(0.035 m)2 = 13 J. Note that this is similar to the weight hanging on a spring. If we measure from the equilibrium position, we ignore changes in mgh. 78. (a) We find the speed from energy conservation: Ki + Ui = Kf + Uf ;

!Mv02 + 0 = 0 + !kA2; !(5.0 kg)v02 = !(310 N/m)(0.24 m)2, which gives v0 =

1.9 m/s.

(b) We find the period of the oscillation from T = 2p(m/k)1/2 = 2p[(5.0 kg)/(310 N/m)]1/2 = 0.798 s. The car will be in contact with the spring for half a cycle, so the time is t = !T = !(0.798 s) = 0.40 s. 79. (a) We find the effective spring constant from the displacement caused by the additional weight: k ?y = mg, or k = mg/?y = (0.55 kg)(9.80 m/s2)/(0.060 m) = 90 N/m. (b) We assume the collision takes place before the springs start to compress. We use momentum conservation to find the speed of the table and clay at the beginning of the oscillation: mv + 0 = (m + M)V; (0.55 kg)(1.65 m/s) = (0.55 kg + 1.60 kg)V, which gives V = 0.422 m/s. The table and clay are 6.0 cm above the equilibrium position when they have this speed. At the maximum amplitude, the speed is zero. From energy conservation, we have Ki + Ui = Kf + Uf ;

!(m + M)V2 + !kx2 = 0 + !kA2; !(0.55 kg + 1.60 kg)(0.422 m/s)2 + !(89.8 N/m)(0.060 m)2 = !(89.8 N/m)A2,

which gives A = 8.8710–2 m =

8.9 cm.

80. The object will leave the surface when the maximum acceleration of the SHM becomes greater than g, so the normal force becomes zero. For the pebble to remain on the board, we have amax = 2A = (2pf)2A < g; [2p(5.0 Hz)]2A < 9.80 m/s2, which gives A < 9.9310–3 m = 0.99 cm. 81. In the equilibrium position, the net force is zero, so we have

Page 20

Chapter 14

Fbuoy = mg. When the block is pushed into the water, there will be an additional buoyant force, equal to the weight of the additional water displaced, to bring the block back to the equilibrium position. When the block is pushed down a distance ?x, this net upward force is Fnet = – watergA ?x. Because the net restoring force is proportional to the displacement, the block will oscillate with SHM. We find the effective force constant from the coefficient of ?x: k = watergA.

Fbuoy mg

82. The distance the mass falls is the distance the spring is stretched. We use energy conservation between the initial point, where the spring is unstretched, and the lowest point, our reference level for the gravitational potential energy, to find the spring constant: Ki + Ui = Kf + Uf ; 0 + mgh = 0 + !kh2, which gives k = 2mg/h. We find the frequency from f = (k/m)1/2/2p = (2mg/hm)1/2/2p = (2g/h)1/2/2p = [2(9.80 m/s2)/(0.220 m)]1/2/2p = 83. When the water is displaced a distance ?x from equilibrium, the net restoring force is the unbalanced weight of water in the height 2 ?x: Fnet = – 2gA ?x. We see that the net restoring force is proportional to the displacement, so the water will oscillate with SHM. We find the effective spring constant from the coefficient of ?x: k = 2gA. From the formula for k, we see that the effective spring constant depends on the density and the cross section.

84.

(b) At equilibrium the force is zero, so we have F = – (C/r02) + (D/r03) = 0, which gives r0 = D/C. (c) For a small displacement from equilibrium, ?r = r – r0 , the net force is F = – [C/(r0 + ?r)2] + [D/(r0 + ?r)3] = [1/(r0 + ?r)3][– C(r0 + ?r) + D] = [1/(r0 + ?r)3][– Cr0 + D – C ?r]. When we use the equilibrium condition for r0, we get F = – C ?r/(r0 + ?r)3 ˜ – (C/r03) ?r. Thus the net force is a restoring force Page 21

(a)

²x

1.50 Hz.

²x

equil

F

0

1

2

r/( D/ C)

Chapter 14

proportional to the small displacement, so the motion will be simple harmonic. (d) We see that the effective force constant is keff = C/r03 = C4/D3. (e) We find the period of the motion from T = 2p(m/keff)1/2 = 2p(mr03/C)1/2 = 2p(mD3/C4)1/2.

85. In the equilibrium position, each spring is stretched ?a, F1 F2 so the unstretched length of a spring is a – ?a. When the k k mass is pulled aside a distance x, the length of each x spring will be (a2 + x2)1/2. From the symmetry we see  that the net force will be toward the equilibrium position: equilibrium a Fnet = – 2k[(a2 + x2)1/2 – (a – ?a)] sin  = – 2k[(a2 + x2)1/2 – (a – ?a)] [x/(a2 + x2)1/2] = – 2kx[1 – (a – ?a)/(a2 + x2)1/2] = – 2kx{1 – (1 – ?a/a)/[1 + (x/a)2]1/2} ˜ – 2kx{1 – (1 – ?a/a)[1 – !(x/a)2]} ˜ – 2kx[(?a/a) + !(x/a)2] ˜ – 2k(?a/a)x, where we have used the fact that x « a. Because the restoring force is proportional to the small displacement, the motion will be simple harmonic, with keff = 2k(?a/a). The period is T = 2p(m/keff)1/2 = 2p[m/2k(?a/a)]1/2 = 2p(ma/2k ?a)1/2. 86. We find the spring constant of each bond from f = (k/m)1/2/2p; 2.831013 Hz = [k/16(1.6710–27 kg)]1/2/2p, which gives k =

845 N/m.

87. (a) The period is T = 2p(I/mgh)1/2 = 2p[(m¬2/3)/mg(¬/2)]1/2 = 2p(2¬/3g)1/2 = 2p[2(1.00 m)/3(9.80 m/s2)]1/2 = 1.64 s. (b) For the simple pendulum to have the same period, we have T = 2p(2¬/3g)1/2 = 2p(¬0/g)1/2, or ¬0 = %¬ = %(1.00 m) = 0.67 m.

Page 22

Chapter 14

88. When a mass m is a distance x from the center of the Earth, there will be a gravitational force only from the mass within a sphere of radius x. This mass is M = (ME/)prE3))px3 = MEx3/rE3. The force will be toward the center, opposite to x: F = – GMm/x2 = – G(MEx3/rE3)m/x2 = – (GMEm/rE3)x. Thus we see that the restoring force is proportional to the displacement, so the motion will be simple harmonic, with keff = GMEm/rE3. The apple will take one period to return to the initial location: T = 2p(m/keff)1/2 = 2p[m/(GMEm/rE3)]1/2 = 2p(rE3/GME)1/2 = 2p[(6.38106 m)3/(6.6710–11 N · m2/kg2)(5.981024 kg)]1/2 = 5.07103 s = 84.5 min.

Page 23

m

x

rE

Chapter 15

CHAPTER 15 – Wave Motion 1.

The speed of the wave is v = f = /T = (9.0 m)/(4.0 s) =

2.3 m/s.

2.

For AM we find the wavelengths from AMhigher = v/fAMlower = (3.00108 m/s)/(550103 Hz) = 545 m; AMlower = v/fAMhigher = (3.00108 m/s)/(1600103 Hz) = 188 m. For FM we have FMhigher = v/fFMlower = (3.00108 m/s)/(88106 Hz) = 3.4 m; FMlower = v/fFMhigher = (3.00108 m/s)/(108106 Hz) = 2.78 m.

3.

We find the wavelength from v = f; 330 m/s = (262 Hz), which gives  =

1.26 m.

4.

We find the speed of the longitudinal (compression) wave from v = (B/)1/2 for fluids and v = (E/)1/2 for solids. (a) For water we have v = (B/)1/2 = [(2.0109 N/m2)/(1.00103 kg/m3)]1/2 = 1.4103 m/s. (b) For granite we have v = (E/)1/2 = [(45109 N/m2)/(2.7103 kg/m3)]1/2 = 4.1103 m/s.

5.

The speed of the longitudinal (compression) wave is v = (E/)1/2, so the wavelength is  = v/f = (E/)1/2/f = [(100109 N/m2)/(7.8103 kg/m3)]1/2/(5000 Hz) =

6.

We find the speed of the wave from v = [FT/(m/L)]1/2 = {(120 N)/[(0.65 kg)/(30 m)]}1/2 = 74.4 m/s. We find the time from t = L/v = (30 m)/(74.4 m/s) = 0.40 s.

7.

We find the tension from the speed of the wave: v = [FT/(m/L)]1/2;

(4.8 m)/(0.85 s) = {FT/[(0.40 kg)/(4.8 m)]}1/2 , which gives FT =

8.

9.

0.72 m.

2.7 N.

The speed of the longitudinal wave is v = (B/)1/2, so the distance down and back that the wave traveled is 2D = vt = (B/)1/2t; 2D = [(2.0109 N/m2)/(1.00103 kg/m3)]1/2(3.5 s), which gives D = 2.5103 m = (a) Because the pulse travels up and back, the speed is v = 2L/t = 2(600 m)/(16 s) = 75 m/s. (b) The mass density of the cable is  = m/L = AL/L = A. We find the tension from v = (FT/)1/2 = (FT/A)1/2; 75 m/s = [FT/(7.8103 kg/m3)p(0.7510–2 m)2]1/2 , which gives FT =

Page 1

2.5 km.

7.8103 N.

Chapter 15

10. (a) The shape is maintained and moves 1.80 m in 1.00 s.

down

up

down

up

down

2 cm 1 cm

A

0 – 1 cm

0

1m

2m

3m

4m

5m

– 2 cm

(b) At the instant shown, point A is moving down. We use the slope of the string to estimate the vertical speed. In the time to move vertically from 1 cm to – 1 cm, we estimate the string moves 30 cm, so the time is ?t = (0.30 m)/(1.80 m/s) = 0.17 s. The vertical velocity is – (2 cm)/(0.17 s) = – 0.1 m/s (down). 11. (a) Because both waves travel the same distance, we have ?t = (d/vS) – (d/vP) = d[(1/vS) – (1/vP)];

94 s = d{[1/(5.5 km/s)] – [1/(9.0 km/s)]}, which gives d = 1.3103 km. (b) The direction of the waves is not known, thus the position of the epicenter cannot be determined. Because two circles have two intersections, it would take at least two more stations.

12. Because the speed, frequency, and medium are the same for the two waves, the intensity depends on the amplitude only: I  DM2. For the ratio of intensities we have I2/I1 = (DM2/DM1)2 ; 2 = (DM2/DM1)2 , which gives DM2/DM1 = 1.41. 13. We assume that the wave spreads out uniformly in all directions. (a) The intensity will decrease as 1/r2, so the ratio of intensities is I2/I1 = (r1/r2)2 = [(10 km)/(20 km)]2 = 0.25. 2 (b) Because the intensity depends on DM , the amplitude will decrease as 1/r, so the ratio of amplitudes is DM2/DM1 = r1/r2 = (10 km)/(20 km) = 0.50. 14. We assume that the wave spreads out uniformly in all directions. (a) The intensity will decrease as 1/r2, so the ratio of intensities is I2/I1 = (r1/r2)2; I2/(2.2106 W/m2) = [(100 km)/(4.0 km)]2 , which gives I2 = 1.4109 W/m2. (b) We can take the intensity to be constant over the small area, so we have P2 = I2S = (1.38109 W/m2)(5.0 m2) = 6.9109 W. 15. If we consider two concentric circles around the spot where the waves are generated, the same energy must go past each circle in the same time. The intensity of a wave depends on DM2, so for the energy passing through a circle of radius r, we have E = I(2r) = kDM22r = a constant. Thus DM must vary with r; in particular, we have DM  1/vr.

Page 2

Chapter 15

16. (a) We consider a wave traveling through an area A at speed v. After a time t, the energy that will have passed through A is contained in a volume Avt, so we have E = IAt = Avt, or I = v. (b) If we assume all of the energy is radiated, we have I = P/A = v; 6.610–9 J/m3. (100 W)/4p(2.0 m)2 = (3.00108 m/s), which gives  = Actually some of the energy will be transferred by conduction and convection, so the energy density will be less than this. 17. (a) The speed of the wave is v = (FT/)1/2 = (FT/A)1/2 = [(4.5 N)/(7.8103 kg/m3)p(0.5010–3 m)2]1/2 = 27.1 m/s. The power output is P = 2p2Avf 2DM2 = 2p2(7.8103 kg/m3)p(0.5010–3 m)2(27.1 m/s)(60 Hz)2(0.5010–2 m)2 = (b) If we form the ratio for the two conditions, we have P 2/ P 1 = f22DM22/f12DM12; 0.25 cm. 1 = (2)2(DM2/0.50 cm)2, which gives DM2 = 18. (a) The linear mass density is  = m/L = AL/L = A. The average power is P = 2p2Avf 2DM2 = 2p2vf 2DM2. (b) The speed of the wave is v = (FT/)1/2 = (FT/A)1/2 = [(100 N)/(0.10 kg/m)]1/2 = 31.6 m/s. The power output is P = 2p2vf 2DM2 = 2p2(0.10 kg/m)(31.6 m/s)(120 Hz)2(2.010–2 m)2 =

0.30 W.

3.6102 W.

19. To represent a wave traveling to the left, we replace x by x + vt: D = DM sin (2px/ + ) = DM sin [(2p/)(x + vt) + ] = DM sin [2p(x/ + t/T) + ]. 20. The traveling wave is D = (0.48 m) sin [(5.6 m–1)x + (84 s–1)t]. (a) We find the wavelength from the coefficient of x: 1.12 m. (5.6 m–1)x = 2px/, which gives  = (b) We find the frequency from the coefficient of t: 13 Hz. (84 s–1)t = 2pft, which gives f = (c) From the positive sign between the x and t terms, the wave is traveling in the – x direction, with speed v = f = (13.4 Hz)(1.12 m) = 15 m/s (toward negative x). (d) The amplitude is the coefficient of the sine function: DM = 0.48 m. (e) The speed of a particle is u = ?D/?t = (0.48 m)(84 s–1) cos [(5.6 m–1)x + (84 s–1)t] = (40 m/s) cos [(5.6 m–1)x + (84 s–1)t]. Thus we have umax = (40 m/s)(1) = 40 m/s; umin = (40 m/s)(0) = 0.

Page 3

Chapter 15

21. The traveling wave is D = (0.026 m) sin [(45 m–1)x – (1570 s–1)t + 0.66], with  = 1570 s–1. (a) Each point has simple harmonic motion, so the maximum velocity is umax = DM = (0.026 m)(1570 s–1) = 41 m/s. (b) The maximum acceleration is amax = DM2 = (0.026 m)(1570 s–1)2 = 6.4104 m/s2. (c) For any point on the string we have u = ?D/?t = – DM cos [(45 m–1)x – (1570 s–1)t + 0.66]; a = ?2D/?t2 = – DM2 sin [(45 m–1)x – (1570 s–1)t + 0.66]. For the given point and time we get u = – (41 m/s) cos [(45 m–1)(1.00 m) – (1570 s–1)(2.0 s) + 0.66] = 41 m/s; a = – (6.4104 m/s2) sin [(45 m–1)(1.00 m) – (1570 s–1)(2.0 s) + 0.66] = 8.2103 m/s2. 22. If we represent the transverse wave by D = DM sin (kx – t), the transverse speed of a particle is u = ?D/?t = – DM cos (kx – t). The slope is slope = ?D/?x = DMk cos (kx – t) = – uk/ = – u/v. 23. (a) At t = 0 the traveling wave is D = (0.45 m) cos [(3.0 m–1)x + 1.2], with  = 2p/(3.0 m–1) = 2.09 m. (b) For a wave traveling to the right (+ x), we replace x with x – vt: D = (0.45 m) cos [(3.0 m–1)(x – vt) + 1.2] = (0.45 m) cos {(3.0 m–1)[x – (2.0 m/s)t] + 1.2} = (0.45 m) cos [(3.0 m–1)x – (6.0 s–1)t + 1.2]. (d) For a wave traveling to the left (– x), we replace x with x + vt: D = (0.45 m) cos [(3.0 m–1)(x + vt) + 1.2] = (0.45 m) cos {(3.0 m–1)[x + (2.0 m/s)t] + 1.2} = (0.45 m) cos [(3.0 m–1)x + (6.0 s–1)t + 1.2].

D 0.45 cm 0.27 cm 0.16 cm

1m

0.04 cm

0

24. (a) We find the wavelength from  = v/f = (345 m/s)/(440 Hz) = 0.784 m. (b) The phase change is produced by a time interval: ? = 2pf ?t; 5.6810–4 s. (90°)(p/180°) = 2p(440 Hz) ?t, which gives ?t = (c) The phase change for a change in x is ? = 2p ?x/; 20°. = [2p(4.410–2 m)/(0.784 m)](180°/p) = 25. For the general expression for a wave traveling to the right, we have D = DM sin [2p(x/ – t/T) + ]. We find the phase angle from the initial conditions: – 0.020 cm = (0.020 cm) sin (0 – 0 + ), which gives sin  = –1,  = – p/2. Thus we can write the expression as D = DM sin [2p(x/ – t/T) – p/2] = – DM cos (2px/ – 2pft). Using the data from Problem 24, we have Page 4

t=0 t = 1 s right t = 1 s left

2m

3m

x

Chapter 15

D = – DM cos [2p(x/ – ft)]

= – (0.020 cm) cos {2p[x/(0.784 m) – 2p(440 Hz)t]} = – (0.020 cm) cos [(8.01 m–1)x – (2.76103 s–1)t] .

26.

(a) (b) For the general expression for a wave traveling in the – x direction, we have D = DM sin [kx + t + ]. We find the phase angle from the initial conditions: 0.80 cm = (1.00 cm) sin (0 – 0 + ), which gives sin  = 0.80,  = 53°, 127°. The speed of a particle is u = ?D/?t = DM cos (kx + t + ). Because u(0, 0) is positive, we choose the phase with a positive value for the cosine function:  = 53° = 0.927 rad. Characteristics of the wave are k = 2p/ = 2p/(3.00 cm) = 2.09 cm–1;  = 2pf = 2p(200 Hz) = 1.26103 s–1. Thus the function is D = (1.00 cm) sin [(2.09 cm–1)x + (1.26103 s–1)t + 0.927].

D 1.00 cm

1 cm

0

– 1.00 cm

27. We find the various derivatives for the given function: D = DM sin kx cos t; ?D/?x = kDM cos kx cos t; ?2D/?x2 = – k2DM sin kx cos t; ?D/?t = – DM sin kx sin t; ?2D/?t2 = – 2DM sin kx cos t. From v = f = (/2p)(2p/k), we see that k = /v. If we use this in ?2D/?x2, we get ?2D/?x2 = – k2DM sin kx cos t = – (2/v2)DM sin kx cos t = (1/v2) ?2D/?t2, which is the wave equation. Thus the function is a solution. 28. (a) We find the various derivatives for the given function: D = DM ln (x + vt); ?D/?x = DM/(x + vt); ?2D/?x2 = – DM /(x + vt)2; ?D/?t = vDM/(x + vt); ?2D/?t2 = – v2DM /(x + vt)2. Thus we have ?2D/?x2 = (1/v2) ?2D/?t2, which is the wave equation. (b) We find the various derivatives for the given function: D = (x – vt)4; ?D/?x = 4(x – vt)3; ?2D/?x2 = 12(x – vt)2; ?D/?t = 4(– v)(x – vt)3; ?2D/?t2 = 12(– v)2(x – vt)3 = 12v2(x – vt)2. Thus we have ?2D/?x2 = (1/v2) ?2D/?t2, which is the wave equation.

Page 5

2 cm

3 cm

x

Chapter 15

29. We find the various derivatives for the function from Eq. 15–13: D = DM sin (kx + t); ?D/?x = kDM cos (kx + t); ?2D/?x2 = – k2DM sin (kx + t); ?D/?t = DM cos (kx + t); ?2D/?t2 = – 2DM sin (kx + t). From v = f = (/2p)(2p/k), we see that k = /v. If we use this in ?2D/?x2, we get ?2D/?x2 = – k2DM sin (kx + t) = – (2/v2) DM sin (kx + t) = (1/v2) ?2D/?t2, which is the wave equation. We find the various derivatives for the function from Eq. 15–15: D = D(x + vt); ?D/?x = [dD/d(x + vt)][?(x + vt)/?x] = dD/d(x + vt); ?2D/?x2 = [d2D/d2(x + vt)][?(x + vt)/?x] = d2D/d2(x + vt); ?D/?t = [dD/d(x + vt)][?(x + vt)/?t] = v dD/d(x + vt); ?2D/?t2 = v [d2D/d2(x + vt)][?(x + vt)/?t] = v2 d2D/d2(x + vt). Thus we have ?2D/?x2 = (1/v2) ?2D/?t2, which is the wave equation. 30. We find the various derivatives for the linear combination: D = C1D1 + C2D2 ; ?D/?x = C1 ?D1/?x + C2 ?D2/?x; ?2D/?x2 = C1 ?2D1/?x2 + C2 ?2D2/?x2; ?D/?t = C1 ?D1/?t + C2 ?D2/?t; ?2D/?t2 = C1 ?2D1/?t2 + C2 ?2D2/?t2; Because both D1 and D2 satisfy the wave equation, we have ?2D/?x2 = C1 ?2D1/?x2 + C2 ?2D2/?x2 = (1/v2)C1 ?2D1/?t2 + (1/v2)C2 ?2D2/?t2 = (1/v2)(C1 ?2D1/?t2 + C2 ?2D2/?t2 ) = (1/v2)?2D/?t2, which is the wave equation. 31. (a) The speed of the wave in a string is v = [FT/]1/2. Because the tensions must be the same anywhere along the string, for the ratio of velocities we have v2/v1 = (1/2)1/2. (b) Because the motion of one string is creating the motion of the other, the frequencies must be the same. For the ratio of wavelengths we have 2/1 = v2/v1 = (1/2)1/2. (c) From the result for part (b) we see that, if 2 > 1 , we have 2 < 1 , so the lighter cord will have the greater wavelength. 32. (a) For the traveling wave in the lighter cord, D1 = (0.050 m) sin [(6.0 m–1)x – (12.0 s–1)t], we find the wavelength from the coefficient of x: 1.05 m. (6.0 m–1)x = 2px/1 , which gives 1 = (b) We find the tension from the velocity: v1 = /k1 = (FT/1)1/2; Page 6

Chapter 15

(12.0 s–1)/(6.0 m–1) = [FT/(0.10 kg/m)]1/2, which gives FT = 0.40 N. (c) The tension and frequency do not change, so we have v2 = /k2 = (FT/2)1/2, or k1/k2 = 2/1 = (1/2)1/2; 2/(1.05 m) = [(0.10 kg/m)/(0.20 kg/m)]1/2, which gives 2 = 0.74 m.

33. The tension and frequency will not change across the boundary. The wave number and velocity will change: 1 = 2 , k1v1 = k2v2 . The equations for the waves are DI = A sin [k1(x – v1t)]; DR = AR sin [k1(x + v1t)]; DT = AT sin [k2(x – v2t)]. (a) Because the rope is continuous, at x = 0 we have DI + DR = DT ; A sin (– k1v1t) + AR sin (k1v1t) = AT sin (– k2v2t). If we use sin (– ) = – sin , and k1v1 = k2v2 , we get – A sin (k1v1t) + AR sin (k1v1t) = – AT sin (k1v1t), or A = AT + AR. (b) The slopes of the waves are DI/x = Ak1 cos [k1(x – v1t)]; DR/x = ARk1 cos [k1(x + v1t)]; DT/x = ATk2 cos [k2(x – v2t)]. For the slope to be the same on both sides of x = 0, we have DI/x + DR/x = DT/x; Ak1 cos (– k1v1t) + ARk1 cos (k1v1t) = ATk2 cos (– k2v2t); If we use cos (– ) = + cos , and k1v1 = k2v2 , we get Ak1 cos (k1v1t) + ARk1 cos (k1v1t) = ATk2 cos (k1v1t), or Ak1 + ARk1 = ATk2 . If we use the previous result, we get Ak1 + ARk1 = (A – AR)k2 , or AR = [(k2 – k1)/(k2 + k1)]A. In terms of v this is AR = {[k1(v1/v2) – k1]/[k1(v1/v2) + k1]}A = [(v1 – v2)/(v1 + v2)]A. (c) For AT we have AT = A – AR = A – [(k2 – k1)/(k2 + k1)]A = [2k1/(k2 + k1)]A = [2v2/(v1 + v2)]A. 34.

(a)

(b)

(c) Because all particles of the string are at equilibrium positions, there is no potential energy. Particles of the string will have transverse velocities, so they have kinetic energy. 35. (a) For the sum of the two waves we have D = D1 + D2 = DM sin (kx – t) + DM sin (kx – t + ) = 2DM sin !(kx – t + kx – t + ) cos !(kx – t – kx + t – ) = 2DM cos !(– ) sin !(2kx – 2t + ) = 2DM cos (!) sin (kx – t + !). (b) The amplitude is the coefficient of the sine function: The variation in x and t is purely sinusoidal. Page 7

2DM cos (!).

Chapter 15

(c) If  = 0, 2p, 4p, … ; ! = 0, p, 2p, … ; so cos (!) = ± 1. Thus the amplitude is maximum and we have complete constructive interference. If  = p, 3p, 5p, … ; ! = p/2, 3p/2, 5p/2, … ; so cos (!) = 0. Thus the amplitude is zero and we have destructive interference. (d) If  = p/2, we have D = 2DM cos (p/4) sin (kx – t + p/4) = v2DM sin (kx – t + p/4). The resultant wave has amplitude DMv2, travels toward + x, and at x = 0, t = 0, the displacement is + DM.

36. From the diagram the initial wavelength is 2L, and the final wavelength is 3L/2. The tension has not changed, so the velocity has not changed: v = f11 = f22; (294 Hz)(2L) = f2(3L/2), which gives f2 = 392 Hz.

L

Unfingered

Fingered

37. All harmonics are present in a vibrating string. Because the harmonic specifies the multiple of the fundamental, we have fn = nf1 , n = 1, 2, 3,  : f1 = 1f1 = (1)(440 Hz) = 440 Hz; f2 = 2f1 = (2)(440 Hz) = 880 Hz; f3 = 3f1 = (3)(440 Hz) = 1320 Hz; f4 = 4f1 = (4)(440 Hz) = 1760 Hz. 38. From the diagram the initial wavelength is L/2. We see that the other wavelengths are 1 = 2L, 2 = L and 3 = 2L/3. The tension has not changed, so the velocity has not changed: v = f = fnn; 66 Hz; (264 Hz)(L/2) = f1(2L), which gives f1 = (264 Hz)(L/2) = f2(L), which gives f2 = 132 Hz; (264 Hz)(L/2) = f3(2L/3), which gives f3 = 198 Hz.

39. The oscillation corresponds to the fundamental with a frequency: f1 = 1/T = 1/(2.0 s) = 0.50 Hz. This is similar to the vibrating string, so all harmonics are present: fn = nf1 = n(0.50 Hz), n = 1, 2, 3,  . We find the corresponding periods from Tn = 1/fn = 1/nf1 = T/n = (2.0 s)/n, n = 1, 2, 3,  . 40. We find the wavelength from v = f; 270 m/s = (131 Hz), which gives  = 2.06 m. The distance between adjacent nodes is !, so we have d = ! = !(2.06 m) = 1.03 m. Page 8

L

Chapter 15

41. All harmonics are present in a vibrating string: fn = nf1 , n = 1, 2, 3,  . The difference in frequencies for two successive harmonics is ?f = fn+1 – fn = (n + 1)f1 – nf1 = f1 , so we have f1 = 350 Hz – 280 Hz = 70 Hz. Note that the given harmonics correspond to n = 4 and 5. 42. We find the speed of the wave from v = [FT/(m/L0)]1/2 = {(520 N)/[(0.0036 kg)/(0.900 m)]}1/2 = 361 m/s. The wavelength of the fundamental for a string is 1 = 2L. We find the fundamental frequency from f1 = v/1 = (361 m/s)/2(0.60 m) = 300 Hz. All harmonics are present so the first overtone is the second harmonic: f2 = (2)f1 = (2)300 Hz = 600 Hz. The second overtone is the third harmonic: f3 = (3)f1 = (3)300 Hz = 900 Hz. 43. The speed of the wave depends on the tension and the mass density: v = (FT/)1/2. The wavelength of the fundamental for a string is 1 = 2L. We find the fundamental frequency from f1 = v/1 = (1/2L)(FT/)1/2. All harmonics are present in a vibrating string, so we have fn = nf1 = (n/2L)(FT/)1/2, n = 1, 2, 3,  . 44. The hanging weight creates the tension in the string: FT = mg. The speed of the wave depends on the tension and the mass density: v = (FT/)1/2 = (mg/)1/2. The frequency is fixed by the vibrator, so the wavelength is  = v/f = (1/f )(mg/)1/2. With a node at each end, each loop corresponds to /2. (a) For one loop, we have 1/2 = L, or 2L = v1/f = (1/f)(m1g/)1/2; 2(1.40 m) = (1/60 Hz)[m1(9.80 m/s2)/(4.810–4 kg/m)]1/2, which gives m1 = 1.4 kg. (b) For two loops, we have 2/2 = L/2, or L = v2/f = (1/f)(m2g/)1/2; 1.40 m = (1/60 Hz)[m2(9.80 m/s2)/(4.810–4 kg/m)]1/2, which gives m2 = 0.35 kg. (c) For five loops, we have 5/2 = L/5, or 2L/5 = v5/f = (1/f)(m5g/)1/2; 2(1.40 m)/5 = (1/60 Hz)[m5(9.80 m/s2)/(4.810–4 kg/m)]1/2, which gives m5 = 0.055 kg. The amplitude of the standing wave can be much greater than the vibrator amplitude because of the resonance built up from the reflected waves at the two ends of the string. 45. The hanging weight creates the tension in the string: FT = mg. The speed of the wave depends on the tension and the mass density: v = (FT/)1/2 = (mg/)1/2, and thus is constant. The frequency is fixed by the vibrator, so the constant wavelength is  = v/f = (1/f)(mg/)1/2 = (1/60 Hz)[(0.080 kg)(9.80 m/s2)/(4.810–4 kg/m)]1/2 = 0.674 m. The different standing waves correspond to different integral numbers of loops, starting at one loop. With a node at each end, each loop corresponds to /2. The lengths of the string for the possible standing wavelengths are Ln = n/2 = n(0.674 m)/2 = n(0.337 m), n = 1, 2, 3,  , or Ln = 0.337 m, 0.674 m, 1.010 m, 1.347 m, 1.684 m,  . Thus we see that there are 4 standing waves for lengths between 0.10 m and 1.5 m.

Page 9

Chapter 15

46. The standing wave is D = (8.6 cm) sin [(0.60 cm–1)x] cos [(58 s–1)t]. (a) The distance between nodes is d = ! = !2p/k = p/(0.60 cm–1) = 5.2 cm. (b) The component waves travel in opposite directions with the same amplitude, frequency, and speed. They are DM1 = !DM = !(8.6 cm) = 4.3 cm; f = /2p = (58 s–1)/2p = 9.2 Hz; v = /k = (58 s–1)/(0.60 cm–1) = 97 cm/s. (c) For any point on the string we have u = ?D/?t = – (8.6 cm)(58 s–1)sin [(0.60 cm–1)x] sin [(58 s–1)t] = – (8.6 cm)(58 s–1)sin [(0.60 cm–1)(3.20 cm)] sin [(58 s–1)(2.5 s)] 2.2102 cm/s. = – 2.2102 cm/s, so the speed is

47. The traveling wave is D = (4.2 cm) sin [(0.71 cm–1)x – (47 s–1)t + 2.1]. (a) We get a wave traveling in the opposite direction with the same amplitude, frequency and speed by changing the relative sign of the x and t terms: D2 = (4.2 cm) sin [(0.71 cm–1)x + (47 s–1)t + 2.1] . (b) When we add the two sine functions and use a trigonometric identity, we get Dresultant = D + D2 = (4.2 cm) sin [(0.71 cm–1)x – (47 s–1)t + 2.1] + (4.2 cm) sin [(0.71 cm–1)x + (47 s–1)t + 2.1] = 2(4.2 cm) sin [(0.71 cm–1)x + 2.1] cos [– (47 s–1)t] = (8.4 cm) sin [(0.71 cm–1)x + 2.1] cos [(47 s–1)t]. 48. With an antinode at each side of the tub, the width is !. Thus the speed of the water wave is v = f = (0.85 Hz)2(0.60 m) = 1.0 m/s. 49. The velocity depends on the tension and mass density, so the frequency is f = v/ = (1/)(FT/)1/2. The length of the string and thus the wavelength does not change. If we form the ratio, we get f2/f1 = (FT2/FT1)1/2; f2/(294 Hz) = (1.10FT1/FT1)1/2, which gives f2 = 308 Hz. 50. The two traveling waves are D1 = (0.15 m) sin [(3.5 m–1)x – (1.2 s–1)t]; D2 = (0.15 m) sin [(3.5 m–1)x + (1.2 s–1)t]. Their sum is DR = D1 + D2 = (0.30 m) sin [(3.5 m–1)x] cos [(1.2 s–1)t]. At t = 1.0 s, we have DR = (0.30 m) sin [(3.5 m–1)x] cos (1.2) = (0.109 m) sin [(3.5 m–1)x] . This agrees with the plot.

D, m 0.15 0.10

DR

0.05 0.00

node 0

node

-0.05 -0.10 -0.15

Page 10

antinode

D1

0.5

node

1.0

1.5

D2 antinode

2.0

x, m

Chapter 15

51. (a)

(b)

D, m

D, m

0.30

0.30

0.20 0.10 0.00

0.20

D2

0.10

DR 0

1

2

3

4

5

-0.10 -0.20

t, s 0.00

0

1

2

3

4

5

t, s

-0.10

D1

-0.20

-0.30

DR

-0.30

D 1 and D 2

From the plots we see that DR = 0 at x = 0 for any t, so it is a node. At x = /4, the displacement varies with the maximum amplitude, so it is an antinode. 52. (a) Because two loops are a wavelength, we have  = %L = %(1.80 m) = 1.20 m. The wave characteristics are k = 2p/ = 2p/(1.20 m) = 5.24 m–1;  = 2pf = 2p(120 Hz) = 240p s–1. The wave function is D = DM sin (kx) cos (t) = (6.0 cm) sin [(5.24 m–1)x] cos [(240p s–1)t]. (b) The two waves traveling in opposite directions will have half the amplitude of the standing wave, but the same values for k and : D1 = (3.0 cm) sin [(5.24 m–1)x – (240p s–1)t]; D2 = (3.0 cm) sin [(5.24 m–1)x + (240p s–1)t]. 53. For the refraction of the waves we have v2/v1 = (sin 2)/(sin 1); v2/(8.0 km/s) = (sin 31°)/(sin 50°), which gives v2 =

5.4 km/s.

54. For the refraction of the waves we have v2/v1 = (sin 2)/(sin 1); (2.5 km/s)/(2.8 km/s) = (sin 2)/(sin 40°), which gives 2 =

35°.

55. The speed of the longitudinal (compression) wave for the solid rock depends on the modulus and the density: v = (E/)1/2 . The modulus does not change, so we have v  (1/)1/2. For the refraction of the waves we have v2/v1 = (1/2)1/2 = (SG1/SG2)1/2 = (sin 2)/(sin 1); [(3.7)/(2.8)]1/2 = (sin 2)/(sin 25°), which gives 2 = 29°. 56. (a) For the refraction of the waves we have v2/v1 = (sin 2)/(sin i); Because v2 > v1, 2 > i. When we use the maximum value of 2 , we get v2/v1 = (sin 90°)/(sin iM) = 1/(sin iM), or iM = sin–1(v1/v2). (b) We have iM = sin–1(v1/v2) = sin–1[(7.5 km/s)/(9.3 km/s)] = 54°. Thus for angles > 54° there will be only reflection.

Page 11

Chapter 15

57. The approximate expression for the speed of sound at these temperatures is v = [331 + (0.60 /C°)T] m/s. For the refraction we have v2/v1 = (sin 2)/(sin i);

{[331 + (0.60 /C°)(– 10°C)] m/s}/{[331 + (0.60 /C°)(+ 10°C)] m/s} = (sin 2)/(sin 25°),

which gives 2 =

24°.

58. If we approximate the sloshing as a standing wave at the fundamental frequency with antinodes at each lip, we have  = 2D. We find the speed of the waves from v = f = (1.0 Hz)2(0.08 m) = 0.16 m/s. 59. The speed of the longitudinal wave in a solid is given by v = (B/)1/2. If we form the ratio for two rods with the same bulk modulus, we get v2/v1 = (1/2)1/2 = (1/2)1/2, or v1 = v2v2 . The speed will be greater in the less dense rod by a factor of v2.

60. Because the speed and frequency are the same for the two waves, the intensity (and thus the power) depends on the amplitude: P  I  A2. For the ratio of the powers we have P2/P1 = (A2/A1)2 ; 3 = (A2/A1)2 , which gives A2/A1 = 1.73. 61. (a) The amplitude will be half the total vertical distance: DM = !d = !(0.10 m) = 0.050 m. (b) The bug will undergo SHM, so the maximum K is also the maximum U, which occurs at the maximum displacement. For the ratio of energies we have K2/K1 = U2/U1 = (DM2/DM1)2 = [!(0.15 m)/!(0.10 m)]2 = 2.3. 62. We assume that the change in tension does not change the mass density, so the velocity variation depends only on the tension. Because the wavelength does not change, we have  = v1/f1 = v2/f2 , or FT2/FT1 = (f2/f1)2. For the fractional change we have (FT2 – FT1)/FT1 = (FT2/FT1) – 1 = (f2/f1)2 – 1 = [(200 Hz)/(205 Hz)]2 – 1 = – 0.048. Thus the tension should be decreased by 4.8%. 63. An object will leave the ground when the maximum acceleration of the ground during the SHM as the wave passes is greater than the acceleration due to gravity: amax = DM2 > g, or DM > g/2 = g/4p2f 2 = (9.80 m/s2)/4p2(0.50 Hz)2 = 0.99 m. 64. (a) The speed of the wave at a point h above the lower end depends on the tension at that point. From the equilibrium of the lower portion, we know that the tension equals the mass of the lower segment: FT = mg = (m/L)hg. Thus the speed is v = (FT/)1/2 = [(m/L)gh/(m/L)]1/2 = (gh)1/2. Page 12

Chapter 15

(b) We measure h from the bottom of the cord. The distance traveled by the pulse in a time dt at this location is dh = v dt = (gh)1/2 dt. We separate variables and integrate to find the total time to reach the top of the cord: L 0

dh = 1/ 2 h

which gives

65.

t 0

g1/ 2 dt ;

2h 1/ 2

L 0

= 2L 1/ 2 = g 1/ 2 t,

t = 2(L/g)1/2.

(a) (b) To represent a wave traveling to the right, we replace x by x – vt: D = (4.0 m3)/[(x – vt)2 – 2.0 m2] = (4.0 m3)/{[x – (3.0 m/s)t]2 – 2.0 m2}. Note that at t = 0 the function is discontinuous at x = ± (2.0 m2)1/2. (c) The wave will travel (3.0 m/s)(0.50 s) = 1.5 m. (d) To represent a wave traveling to the left, we replace x by x + vt: D = (4.0 m3)/[(x + vt)2 – 2.0 m2] = (4.0 m3)/{[x + (3.0 m/s)t]2 – 2.0 m2}. In 0.50 s the wave will travel (3.0 m/s)(0.50 s) = 1.5 m.

D, m 100

t=0

-5.0

-2.5

t = 0.50 s

50

0

0

2.5

x, m

5.0

–50

–100

D, m 100

t = 0.50 s

-5.0

-2.5

0

–50

–100

Page 13

t=0

50

0

2.5

5.0

x, m

Chapter 15

66. (a) The wavelength of the fundamental for a string is 2L, so the fundamental frequency is f = (1/2L)(FT/)1/2. When the tension is changed, the change in frequency is ?f = f – f = (1/2L)[(FT/)1/2 – (FT/)1/2] = (1/2L){(FT/)1/2[(FT/FT)1/2 – 1]} = f {[(FT + ?FT)/FT]1/2 – 1]}

= f {[1 + (?FT/FT)]1/2 – 1]}. If ?FT/FT is small, we have

[1 + (?FT/FT)]1/2 ˜ 1 + !(?FT/FT), so we get ?f ˜ f[1 + !(?FT/FT) – 1] = !(?FT/FT)f. (b) With the given data, we get ?f = !(?FT/FT)f; 1.8% (increase). 442 Hz – 438 Hz = !(?FT/FT)(438 Hz), which gives ?FT/FT = 0.018 = (c) For each overtone there will be a new wavelength, but the wavelength does not change when the tension changes, so the formula will apply to the overtones.

67. (a) All harmonics are present in a vibrating string: fn = nf1 , n = 1, 2, 3,  . The first overtone is f2 and the second overtone is f3. For G we have f2 = 2(392 Hz) = 784 Hz; f3 = 3(392 Hz) = 1176 Hz. For A we have f2 = 2(440 Hz) = 880 Hz; f3 = 3(440 Hz) = 1320 Hz. 1/2 (b) The speed of the wave in a string is v = [FT/(M/L)] . Because the lengths are the same, the wavelengths of the fundamentals must be the same. For the ratio of frequencies we have fA/fG = vA/vG = (MG/MA)1/2; (440 Hz)/(392 Hz) = (MG/MA)1/2, which gives MG/MA = 1.26. (c) Because the mass densities and the tensions are the same, the speeds must be the same. The wavelengths are proportional to the lengths, so for the ratio of frequencies we have fA/fG = G/A = LG/LA; (440 Hz)/(392 Hz) = LG/LA , which gives LG/LA = 1.12. (d) The speed of the wave in a string is v = [FT/(M/L)]1/2. Because the lengths are the same, the wavelengths of the fundamentals must be the same. For the ratio of frequencies we have fA/fG = vA/vG = (FTA/FTG)1/2; (440 Hz)/(392 Hz) = (FTA/FTG)1/2, which gives FTG/FTA = 0.794. 68. Without the support the bridge is like a string, so the wavelength is 1 = 2L. Adding the support creates a node at the middle, so the wavelength is 2 = L. The wave speed in the bridge has not changed. We find the new frequency from v = 1f1 = 2f2 ; 2L(4.0 Hz) = Lf2 , which gives f2 = 8.0 Hz. Yes, because this resonant frequency is higher than the expected earthquake frequencies, the modification Page 14

Chapter 15

did some good. 69. The fundamental standing wave will have a node at the fixed end and an antinode at the free end. The distance between an adjacent node-antinode pair is /4, so the wavelength of the fundamental is 4L. The first overtone will have an additional node-antinode pair, so the wavelength is 4L/3. The wavelength of the next overtone, with another node-antinode pair, is 4L/5. Thus the general form for the wavelengths is n = 4L/(2n – 1), n = 1, 2, 3, … .

A

N N

A

N

A

A

N N

A

N

A

70. The work done by the force F to pull the string into the triangular shape is stored in the string as elastic potential energy. If we assume the tension is constant as the string vibrates, this energy will oscillate between kinetic and potential. We ignore the small change in gravitational potential energy. If the string is slowly pulled by the force F, we see from the force diagram that F = 2FT sin  = 2FTy/[(!L)2 + y2]1/2. For the next differential distance dy, the work done will be dW = F dy. We integrate to find the total work pulling the center of the string a distance h:

W=

h 0

h

Fd y = 0

2FT y d y 1 2 4L

+y

2 1/ 2

= 2FT 14 L2 + y 2

1/ 2

h 0

=

FT L 2 + 4h 2

1/ 2

–L.

71. From the figure we see that DM = 3.5 cm, and  = 6.0 cm, so k = 2p/ = 2p/(6.0 cm) = 1.05 cm–1. The displacement y = 0 is at x = 1.5 cm at t = 0, and at x = 5.5 cm at t = 3.0 s. Thus the velocity of the wave is v = (5.5 cm – 1.5 cm)/(3.0 s) = 1.33 cm/s, and  = vk = (1.33 cm/s)(1.05 cm–1) = 1.39 s–1. The wave is moving to the right. We also see that y is maximum at x = 0 and t = 0, so we can use a cosine function without a phase angle: y = DM cos (kx – t) = (3.5 cm) cos[(1.05 cm–1)x – (1.39 s–1)t]. 72. From the given data DM = 0.50 m,  = 2.5 m, T = 5.0 s, we can find the wave speed: v = f = /T = (2.5 m)/(5.0 s) = 0.50 m/s. If we use the density of sea water and estimate the area of the chest as (0.25 m)2, we have P = 2p2Avf 2DM2 = 2p2(1.025103 kg/m3)(0.25 m)2(0.50 m/s)(1/5.0 s)2(0.50 m)2 = 6.3 W.

Page 15

Chapter 15

73. In 1.0 s, each pulse will travel (5.0 cm/s)(1.0 s) = 5.0 cm.

t=0 0

10

20

30

40

0

10

20

30

40

0

10

20

30

40

0

10

20

30

40

x, cm

t = 1.0 s x, cm

t = 2.0 s x, cm

t = 3.0 s x, cm

74. Because the radiation is uniform, the same energy must pass through every spherical surface, which has an area 4pr2. Thus the intensity, which is proportional to the square of the amplitude, must decrease as 1/r2. Thus the amplitude will decrease as 1/r. The radial motion will be sinusoidal, so we have D = (A/r) sin (kr – t).

Page 16

Chapter 16

CHAPTER 16 – Sound 1.

Because the sound travels both ways across the lake, we have L = !vt = !(343 m/s)(1.5 s) = 2.6102 m.

2.

(a) We find the extreme wavelengths from 1 = v/f1 = (343 m/s)/(20 Hz) = 17 m; 2 = v/f2 = (343 m/s)/(20,000 Hz) = 1.710–2 m = 1.7 cm. The range of wavelengths is 1.7 cm =  = 17 m. (b) We find the wavelength from  = v/f = (343 m/s)/(10106 Hz) = 3.410–5 m.

3.

The speed in the concrete is determined by the elastic modulus: vconcrete = (E/)1/2 = [(20109 N/m2)/(2.3103 kg/m3)]1/2 = 2.95103 m/s. For the time interval we have ?t = (d/vair) – (d/vconcrete);



1.4 s = d{[1/(343 m/s)] – [1/(2.95103 m/s)]}, which gives d =

5.4102 m.

4. (a) For the time interval in sea water we have 0.64 s. ?t = d/vwater = (1.0103 m)/(1560 m/s) = (b) For the time interval in the air we have 2.9 s. ?t = d/vair = (1.0103 m)/(343 m/s) = 5.

If we let L1 represent the thickness of the top layer, the total transit time is t = (L1/v1) + [(L – L1)/v2)]; 4.5 s = [L1/(331 m/s)] + [(1500 m – L1)/(343 m/s)], which gives L1 = 1200 m. Thus the bottom layer is 1500 m – 1200 m = 300 m.

6.

Because the distance is d = vt, the change in distance from the change in velocity is ?d = t ?v; so the percentage change is (?d/d)100 = (?v/v)100 = (343 m/s – 331 m/s)(100)/(343 m/s) = 3.5%.

7.

We find the displacement amplitude from ?PM = 2pvDMf. (a) For the frequency of 100 Hz, we have 1.110–8 m. 3.010–3 Pa = 2p(1.29 kg/m3)(331 m/s)DM(100 Hz), which gives DM = (b) For the frequency of 10 kHz, we have 1.110–10 m. 3.010–3 Pa = 2p(1.29 kg/m3)(331 m/s)DM(10103 Hz), which gives DM =

8.

We find the pressure amplitude from ?PM = 2pvDMf. (a) For the frequency of 50 Hz, we have 410–5 Pa. ?PM = 2p(1.29 kg/m3)(331 m/s)(310–10 m)(50 Hz) = (b) For the frequency of 5.0 kHz, we have 410–3 Pa. ?PM = 2p(1.29 kg/m3)(331 m/s)(310–10 m)(5.0103 Hz) =

Page 1

Chapter 16

9.

The pressure variation is ?P = ?PM sin (kx – t). (a) For the frequency of 50 Hz, we have  = 2pf = 2p(50 Hz) = 315 s–1; k = /v = (315 s–1)/(331 m/s) = 0.949 m–1. The pressure variation is ?P = (410–5 Pa) sin [(0.949 m–1)x – (315 s–1)t]. (b) For the frequency of 5.0 kHz, we have  = 2pf = 2p(5.0103 Hz) = 3.15104 s–1; k = /v = (3.15104 s–1)/(331 m/s) = 94.9 m–1. The pressure variation is ?P = (410–3 Pa) sin [(94.9 m–1)x – (3.15104 s–1)t].

10. The pressure variation is ?P = (0.0025 Pa) sin {[(p/3) m–1]x – (1700p s–1)t}. We compare this to the general expression: ?P = ?PM sin (kx – t). (a) The wavelength is  = 2p/k = 2p/[(p/3) m–1] = 6.0 m. (b) The frequency is f = /2p = (1700p s–1)/2p = 850 Hz. (c) The speed is v = f = (850 Hz)(6.0 m) = 5.1103 m/s. (d) We find the displacement amplitude from ?PM = 2pvDMf; 0.0025 Pa = 2p(2.7103 kg/m3)(5.1103 m/s)DM(850 Hz), which gives DM =

3.410–14 m.

11. (a) We find the intensity level from  = 10 log10(I/I0) = 10 log10[(8.510–8 W/m2)/(10–12 W/m2)] = 49 dB. (b) We find the intensity from  = 10 log10(I/I0); 25 dB = 10 log(I/10–12 W/m2), which gives I = 3.210–10 W/m2. 12. From Figure 16–5, we see that a 100-Hz tone with an intensity level of 50 dB has a loudness level of 20 phons. If we follow this loudness curve to 6000 Hz, we find that the intensity level must be 25 dB. 13. From Figure 16–5, we see that an intensity level of 30 dB intersects the loudness level of 0 phons at 150 Hz. At the high frequency end, the highest frequency is reached, so the range is 150 Hz to 20,000 Hz. 14. We find the ratio of intensities from  = 10 log10(Isignal/Inoise); 63 dB = 10 log10(Isignal/Inoise), which gives Isignal/Inoise =

2.0106.

15. (a) The intensity of the sound wave is given by I = 2p2f 2DM2v. Because the frequency, density, and velocity are the same, for the ratio we have I2/I1 = (DM2/DM1)2 = 32 = 9. (b) We find the change in intensity level from 9.5 dB. ? = 10 log10(I2/I1) = 10 log10(9) =

Page 2

Chapter 16

16. We find the ratio of intensities of the sounds from  = 10 log10(I2/I1); 2.0 dB = 10 log10(I2/I1), which gives I2/I1 = 1.58. Because the intensity is proportional to the square of the amplitude, we have I2/I1 = (A2/A1)2; 1.58 = (A2/A1)2, which gives A2/A1 = 1.3. 17. (a) The pressure amplitude is ?PM = 2pvDMf. In the same medium,  and v are fixed. When we form the ratio for the two waves, we get ?PM2/?PM1 = f2/f1 = 2. The pressure amplitude of the wave with the higher frequency is greater by a factor of 2. (b) The intensity of the sound wave is given by I = 2p2f 2DM2v. Because the displacement amplitude, density, and velocity are the same, for the ratio we have I2/I1 = (f2/f1)2 = 22 = 4. 18. When three of the four engines are shut down, the intensity is reduced by a factor of 4, so we have 2 – 1 = 10 log10(I2/I0) – 10 log10(I1/I0) = 10 log10(I2/I1);

2 – 120 dB = 10 log10(#I1/I1), which gives 2 =

114 dB.

19. (a) We find the intensity of the sound from  = 10 log10(I/I0); 40 dB = 10 log10(I/10–12 W/m2), which gives I = 1.010–8 W/m2. The rate at which energy is absorbed is the power of the sound wave: P = IA = (1.010–8 W/m2)(5.010–5 m2) = 5.010–13 W. (b) We find the time from t = E/P = (1.0 J)/(5.010–13 W) = 2.01012 s = 6.3104 yr. 20. (a) We find the intensity of the sound at a distance of 50 cm from  = 10 log10(I/I0); 65 dB = 10 log10(I/10–12 W/m2), which gives I = 3.210–6 W/m2. The rate at which energy passes through a hemisphere is P = IA = (3.210–6 W/m2)2p(0.50 m)2 = 5.010–6 W. (b) Because the intensities add, the number is N = Ptotal/P = (100 W)/(5.010–6 W) = 2.0107. Note that this is 20 million, all speaking from the same location! 21. The intensity with one firecracker is one-half that of two firecrackers. Thus the change in intensity level is 2 – 1 = 10 log10(I2/I1);

2 – 90 dB = 10 log10(!), which gives 2 =

87 dB.

22. The intensity of the sound wave is given by I = 2p2f 2DM2v = 2p2(1.29 kg/m3)(330 Hz)2(1.310–3 m)2(343 m/s) = 1.61103 W/m2. We find the intensity level from  = 10 log10(I/I0) = 10 log10[(1.61103 W/m2)/(10–12 W/m2)] = 1.5102 dB.

Page 3

Chapter 16

23. (a) We find the intensity of the sound from  = 10 log10(I/I0); 120 dB = 10 log(I/10–12 W/m2), which gives I = 1.00 W/m2. We find the maximum displacement from I = 2p2f 2DM2v 1.00 W/m2 = 2p2(1.29 kg/m3)(210 Hz)2DM2(343 m/s), which gives DM = 5.1010–5 m. (b) We find the pressure amplitude from I = (?PM)2/2v; 1.00 W/m2 = (?PM)2/2(1.29 kg/m3)(343 m/s), which gives ?PM = 29.8 Pa. 24. (a) If we assume that one channel is connected to the speaker, the rating is the power in the sound, and the sound spreads out uniformly, we have I = P/4pr2, so we get I1 = (250 W)/4p(2.5 m)2 = 3.18 W/m2; I2 = (40 W)/4p(2.5 m)2 = 0.509 W/m2. For the intensity levels we have 1 = 10 log10(I1/I0) = 10 log10[(3.18 W/m2)/(10–12 W/m2)] = 125 dB; 2 = 10 log10(I2/I0) = 10 log10[(0.509 W/m2)/(10–12 W/m2)] = 117 dB. (b) The difference in intensity levels is 8 dB. A change of 10 dB corresponds to a doubling of the loudness, so the expensive amp is almost twice as loud. 25. (a) We find the intensity of the sound from  = 10 log10(I/I0); 130 dB = 10 log(I/10–12 W/m2), which gives I = 10 W/m2. The power output of the speaker is P = IA = I4pr2 = (10 W/m2)4p(3.4 m)2 = 1.5103 W. (b) We find the intensity of the sound from  = 10 log10(I/I0); 90 dB = 10 log(I/10–12 W/m2), which gives I = 1.010–3 W/m2. We find the distance from P = IA; 3.4102 m. 1.45103 W = (1.010–3 W/m2)4pr2, which gives r = 26. (a) If we assume negligible absorption in 30 m, we find the intensity of the sound from P = IA = I4pr2; 5.0105 W = I4p(30 m)2 , which gives I = 44.2 W/m2. We find the intensity level from  = 10 log10(I/I0) = 10 log10[(44.2 W/m2)/(10–12 W/m2)] = 136 dB. (b) We find the intensity of the sound without air absorption from P = I1A1 = I14pr12; 5.0105 W = I14p(1000 m)2, which gives I1 = 3.9810–2 W/m2. We find the intensity level from 10 = 10 log10(I1/I0) = 10 log10[(3.9810–2 W/m2)/(10–12 W/m2)] = 106 dB. When we consider air absorption, we have 1 = 10 – (7.0 dB/km)r1 = 106 dB – (7.0 dB/km)(1.0 km) = 99 dB. (c) We find the intensity of the sound without air absorption from P = I2A2 = I24pr22; Page 4

Chapter 16

5.0105 W = I24p(5000 m)2, which gives I2 = 1.5910–3 W/m2. We find the intensity level from 20 = 10 log10(I2/I0) = 10 log10[(1.5910–3 W/m2)/(10–12 W/m2)] = 92 dB. When we consider air absorption, we have 2 = 20 – (7.0 dB/km)r2 = 92 dB – (7.0 dB/km)(5.0 km) = 57 dB. 27. (a) Because the sounds are in the same medium, the variation in intensity is due only to the change in pressure amplitude. Thus we have  = 10 log10(I/I0) = 10 log10[(?PM)2/(?PM0)2] = 10 log10(?PM/?PM0)2 = 20 log10(?PM/?PM0). (b) For the given data we have  = 20 log10(?PM/?PM0) = 20 log10[(1.013105 Pa)/(3.010–5 Pa)] = 190 dB. 28. The wavelength of the fundamental frequency for a string is  = 2L, so the speed of a wave on the string is v = f = 2(0.32 m)(196 Hz) = 125 m/s. We find the tension from v = [FT/(m/L)]1/2; 125 m/s = {FT/[(0.6810–3 kg)/(0.32 m)]}1/2 , which gives FT =

33 N.

29. (a) The empty soda bottle is approximately a closed pipe with a node at the bottom and an antinode at the top. The wavelength of the fundamental frequency is  = 4L. We find the frequency from v = f; 343 m/s = 4(0.15 m)f, which gives f = 570 Hz. (b) The length of the pipe is now % of the original length. We find the frequency from v = f; 860 Hz. 343 m/s = 4(%)(0.15 m)f, which gives f = 30. The fundamental wavelength for a flute has an antinode at each end, so the wavelength is  = 2L. Uncovering a hole shortens the length. We find the new length from v = f; 343 m/s = 2L(294 Hz), which gives L = 0.583 m. Thus the hole must be 0.655 m – 0.583 m = 0.072 m = 7.2 cm from the end. 31. For an open pipe the wavelength of the fundamental frequency is  = 2L. We find the required lengths from v = f = 2Lf; 343 m/s = 2Llowest(20 Hz), which gives Llowest = 8.6 m; 343 m/s = 2Lhighest(20,000 Hz), which gives Lhighest = 8.610–3 m = 8.6 mm. Thus the range of lengths is 8.6 mm < L < 8.6 m. 32. (a) For an open pipe the wavelength of the fundamental frequency is  = 2L. We find the fundamental frequencies from v = f = 2Lf, or f = v/2L; f1 = (343 m/s)/2(3.0 m) = 57 Hz; f2 = (343 m/s)/2(2.5 m) = 69 Hz; f3 = (343 m/s)/2(2.0 m) = 86 Hz; f4 = (343 m/s)/2(1.5 m) = 114 Hz; f5 = (343 m/s)/2(1.0 m) = 172 Hz. Note that for the shorter pipes, these results may be affected by the diameter being relatively large compared to the length of the pipe. We have neglected the presence of overtones. (b) During a noisy day, many frequencies will be generated in the room, so the presence of the fundamental frequencies that will resonate with all of the pipes is more likely.

Page 5

Chapter 16

33. (a) For a closed pipe the wavelength of the fundamental frequency is  = 4L. We find the fundamental frequency from v = 1f 1; 343 m/s = 4(0.780 m)f , which gives f = 110 Hz. Only the odd harmonics are present, so we have f3 = 3f1 = (3)(110 Hz) = 330 Hz; f5 = 5f1 = (5)(110 Hz) = 550 Hz; f7 = 7f1 = (7)(110 Hz) = 770 Hz. (b) For an open pipe the wavelength of the fundamental frequency is  = 2L. We find the fundamental frequency from v = 1f 1 ; 343 m/s = 2(0.780 m)f1 , which gives f1 = 220 Hz. All harmonics are present, so we have f2 = 2f1 = (2)(220 Hz) = 440 Hz; f3 = 3f1 = (3)(220 Hz) = 660 Hz; f4 = 4f1 = (4)(220 Hz) = 880 Hz. 

34. (a) The wavelength of the fundamental frequency for a string is  = 2L. Because the speed of a wave on the string does not change, we have v = 1f1 = 2f2 ; 2(0.73 m)(330 Hz) = 2L2(440 Hz), which gives L2 = 0.55 m. Thus the finger must be placed 0.73 m – 0.55 m = 0.18 m from the end. (b) The frequency must be the same: 440 Hz. The wave length will be  = v/f2 = (343 m/s)/(440 Hz) = 0.78 m. 35. (a) We find the speed of sound at 21°C: v = (331 + 0.60T) m/s = [331 + (0.60/C°)(21°C)] m/s = 344 m/s. The fundamental wavelength for an open pipe has an antinode at each end, so the wavelength is  = 2L. We find the length from v = f; 344 m/s = 2L(262 Hz), which gives L = 0.656 m. (b) The frequency must be the same: 262 Hz. The wavelength will be  = 2L = 2(0.656 m) = 1.31 m. (c) The wavelength and frequency in the outside air will be the same as in the air in the organ pipe: 1.31 m, 262 Hz. 36. If we consider the ear canal as a closed pipe, the wavelength of the fundamental frequency is 1 = 4L. We find the fundamental frequency from v = 1f1 = 4Lf1 ; 343 m/s = 4(0.025 m)f1 , which gives f1 = 3400 Hz. Because only odd harmonics are present in a closed pipe, the audible resonant frequencies are 3400 Hz, 10,200 Hz, 17,000 Hz. Page 6

Chapter 16

From Fig. 16–5, we see that the

fundamental frequency is the most sensitive frequency.

37. We assume that the length, and thus the wavelength, has not changed. The frequency change is due to the change in the speed of sound. We find the speed of sound at 5°C: v = (331 + 0.60T) m/s = [331 + (0.60/C°)(5°C)] m/s = 334 m/s. Because the frequency is proportional to the velocity, for the percent change in frequency we have (?f/f)100 = (?v/v)100 = [(334 m/s – 343 m/s)/(343 m/s)]100 = – 2.6%.

38. (a) For an open pipe all harmonics are present, the difference in frequencies is the fundamental frequency, and all frequencies will be integral multiples of the difference. For a closed pipe only odd harmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencies will not be integral multiples of the difference but odd multiples of half the difference. For this pipe we have ?f = 616 Hz – 440 Hz = 440 Hz – 264 Hz = 176 Hz. Because we have frequencies that are not integral multiples of this, the pipe is closed. (b) The fundamental frequency is f1 = !?f = !(176 Hz) = 88 Hz. Note that the given frequencies are the third, fifth, and seventh harmonics. 39. (a) We find the speed of sound at 15°C: v = (331 + 0.60T) m/s = [331 + (0.60/C°)(15°C)] m/s = 340 m/s. For an open pipe, the wavelength of the fundamental frequency is 1 = 2L. We find the length from v = 1f 1 = 2Lf1 ; 340 m/s = 2L(294 Hz), which gives L = 0.578 m. (b) For helium we have v = 1f 1 = 2Lf1 ; 1005 m/s = 2(0.578 m)f1 , which gives f1 = 869 Hz. Note that we have no correction for the 5 C° temperature change. 40. For an open pipe all harmonics are present, the difference in frequencies is the fundamental frequency, and all frequencies will be integral multiples of the difference. For a closed pipe only odd harmonics are present, the difference in frequencies is twice the fundamental frequency, and frequencies will not be integral multiples of the difference but odd multiples of half the difference. For this pipe we have ?f = 280 Hz – 240 Hz = 40 Hz. Because we have frequencies that are integral multiples of this, the pipe is open, with a fundamental frequency of 40 Hz. The wavelength of the fundamental frequency is  = 2L. We find the length from v = 1f 1 = 2Lf1 ; 343 m/s = 2L(40 Hz), which gives L = 4.3 m. 41. For an open pipe all harmonics are present, the difference in frequencies is the fundamental frequency, and all frequencies will be integral multiples of the difference. Thus we have f1 = ?f = 330 Hz – 275 Hz = 55 Hz. The wavelength of the fundamental frequency is 1 = 2L. We find the speed of sound from v = 1f 1 = 2Lf1 = 2(1.95 m)(55 Hz) = 215 m/s.

Page 7

Chapter 16

42. (a) The wavelength of the fundamental frequency is 1 = 2L. We find the fundamental frequency from v = 1f1 = 2Lf1 ; 343 m/s = 2(2.16 m)f1 , which gives f1 = 79.4 Hz. We find the highest harmonic from fn = nf1 ; 20,000 Hz = n(79.4 Hz), which gives n = 251.9. Because all harmonics are present in an open pipe and the fundamental frequency is within the audible range, 251 harmonics are present, which means 250 overtones. (b) The wavelength of the fundamental frequency is  = 4L. We find the fundamental frequency from v = 1f1 = 4Lf1 ; 343 m/s = 4(2.16 m)f1 , which gives f1 = 39.7 Hz. Because only the odd harmonics are present in a closed pipe, the frequencies are given by fn = (2n – 1)f1 , n = 1, 2, 3, … . We find the highest harmonic from fn = (2n – 1)f1 ; 20,000 Hz = (2n – 1)(39.7 Hz), which gives n = 252. The fundamental frequency is within the audible range, so there are 251 overtones. 

43. Because the intensity is proportional to the square of both the amplitude and the frequency, we have I2/I1 = (A2/A1)2(f2/f1)2 = (0.4)2(2)2 = 0.64; I3/I1 = (A3/A1)2(f3/f1)2 = (0.15)2(3)2 = 0.20. We find the relative intensity levels from 2 – 1 = 10 log10(I2/I1) = 10 log10(0.64) = – 2 dB; 3 – 1 = 10 log10(I3/I1) = 10 log10(0.20) = – 7 dB. 44. The beat frequency is the difference in frequencies, so we have ?f = fbeat = 1/(2.0 s) = 0.50 Hz. Note that the frequency of the second string could be higher or lower. 45. The beat frequency is the difference in frequencies, so we have fbeat = ?f = f2 – f1; ± 5.0 kHz = f2 – 23.5 kHz, which gives f2 = 18.5 kHz, 28.5 kHz. Because the second whistle cannot be heard by humans, its frequency is

28.5 kHz.

46. The beat frequency will be the difference in frequencies. We assume that the change in tension does not change the mass density, so the velocity variation depends only on the tension. Because the wavelength does not change, we have v = f = (FT/)1/2. If we treat the small changes in f and FT as differentials, we get

 df = !(1/FT)1/2 dFT. Page 8

Chapter 16

When we divide both sides by v, we get df/f = ! dFT/FT; df/(294 Hz) = !(0.020), which gives df =

2.9 Hz.

47. (a) The beat frequency will be the difference in frequencies. Because the second frequency could be higher or lower, we have fbeat = ?f = f2 – f1; ± (3 beats)/(2.0 s) = f2 – 132 Hz, which gives f2 = 130.5 Hz, or 133.5 Hz. (b) We assume that the change in tension does not change the mass density, so the velocity variation depends only on the tension. Because the wavelength does not change, we treat the small changes in f and FT as differentials to get

 df = !(1/FT)1/2 dFT. When we divide both sides by v, we get df/f = (± 1.5 Hz)/(132 Hz) = ! dFT/FT , which gives dFT/FT = ± 0.023 =

± 2.3%.

48. Because the flutes are identical, the fundamental wavelengths are the same. The temperature difference causes a change in velocity, and thus a change in frequency. The velocities are v1 = (331 + 0.60T1) m/s = [331 + (0.60/C°)(5.0°C)] m/s = 334 m/s; v2 = (331 + 0.60T2) m/s = [331 + (0.60/C°)(25.0°C)] m/s = 346 m/s. If we assume 262 Hz for the lower temperature, for the ratio of frequencies we have f2/f1 = v2/v1 , so f2 – f1 = [(v2/v1) – 1]f1, or ?f = (?v/v1)f1 = [(346 m/s – 334 m/s)/(334 m/s)](262 Hz) = 9.4 Hz. 49. For destructive interference, the path difference is an odd multiple of half the wavelength. (a) Because the path difference is fixed, the lowest frequency corresponds to the longest wavelength. We find this from L = 1/2, so we have f1 = v/1 = (343 m/s)/2(3.5 m – 3.0 m) = 343 Hz. (b) We find the wavelength for the next frequency from L = 32/2, so we have f2 = v/2 = (343 m/s)/[2(3.5 m – 3.0 m)/3] = 1030 Hz. We find the wavelength for the next frequency from L = 53/2, so we have f3 = v/3 = (343 m/s)/[2(3.5 m – 3.0 m)/5] = 1715 Hz. 50. The 180° phase difference of the speakers is equivalent to a path difference of !. We consider a point a distance x from one speaker on the line between the two speakers, which are a distance L apart. (a) Because of the 180° phase difference, constructive interference will occur when the path difference from the speakers to the point is !: (L – x) – x = !, or L = 2x + !. 0.69 m. If x = 0 (at a speaker), Lmin = ! = !(343 m/s)/(250 Hz) = (b) Because of the 180° phase difference, destructive interference will occur when the path difference from the speakers to the point is 0, which is the midpoint of the line: (L – x) – x = 0, or L = 2x. Page 9

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Thus we have destructive interference for any separation, with the minimum being

0.

51. The two sound waves travel in the same medium, so the same wavelength means the same frequency. We take the initial phase of each wave to be zero. If the sources radiate uniformly in all directions, the intensity decreases as 1/r2, so the amplitude will decrease as 1/r. If we add the displacements of the two waves, we have D = D1 + D2 = (DM/rA) sin (krA – t) + (DM/rB) sin (krB – t). Because rA ˜ rB , the two coefficients of the sine functions are approximately equal, so we have D ˜ (DM/rA)[sin (krA – t) + sin (krB – t)] = (DM/rA)2 sin [!k(rA + rB) – t] cos [!k(rA – rB)]. Because rA ˜ rB , we have rA + rB ˜ 2rA , and we get D ˜ (2DM/rA) sin (krA – t) cos [!k(rA – rB)], or D ˜ (2DM/rA) cos [(p/)(rA – rB)] sin (krA – t). This is a traveling wave with an amplitude of (2DM/rA) cos [(p/)(rA – rB)] .

52. The wavelength of the sounds is  = v/f = (343 m/s)/(440 Hz) = 0.780 m. (a) When the microphone is equidistant from the speakers, there is no path difference from the speakers. Because a maximum is recorded, the speakers are in phase. When the microphone is moved a distance x to the first minimum, the path difference must be !: s2 – s1 = !;

D

s2

L

s1

[(!D + x)2 + L2]1/2 – [(!D – x)2 + L2]1/2 = !, or x [(!D + x)2 + L2]1/2 = ! + [(!D – x)2 + L2]1/2. When we square and cancel terms, we get 2Dx – 2/4 = [(!D – x)2 + L2]1/2. If we square again and collect terms, we get (4D2 – 2)x2 = 2[L2 + (D2/4) – (2/16)]; [4(3.00 m)2 – (0.780 m)2]x2 = (0.780 m)2{(3.20 m)2 + [(3.00 m)2/4] – [(0.780 m)2/16]}, which gives x = ± 0.46 m. Thus the microphone must be moved 0.46 m to the right. (b) Because of the 180° phase difference, the maxima and minima will be interchanged, so the minimum will be at the midpoint, x = 0; maximum will be at x = 0.46 m. 53. Because the beat frequency increases for the fork with the higher frequency, the string frequency must be below 350 Hz. Thus we have fstring = f1 – fbeat1 = f2 – fbeat2 = 350 Hz – 4 Hz = 355 Hz – 9 Hz = 346 Hz. 54. When A and B are sounded, we have fA – fB = 3 Hz, so fA = fB ± 3 Hz = 440 Hz ± 3 Hz, so fA = 437 Hz, or 443 Hz. When C and B are sounded, we have fC – fB = 4 Hz, so fC = fB ± 4 Hz = 440 Hz ± 4 Hz, so fC = 436 Hz, or 444 Hz. The possible beat frequencies when A and C are sounded are fA – fC = 437 Hz – 436 Hz = 443 Hz – 444 Hz = 1 Hz, and fA – fC = 437 Hz – 444 Hz = 443 Hz – 436 Hz = 7 Hz.

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55. We see that there is no path difference for a listener on the bisector of the two speakers. The maximum path difference occurs for a listener on the line of the speakers. If this path difference is less than /2, there will be no location where destructive interference can occur. Thus we have path difference = dB – dA = d = /2.

Listener

A no path difference

d B

56. (a) We find the two frequencies from f1 = v/1 = (343 m/s)/(2.64 m) = 129.9 Hz; f2 = v/2 = (343 m/s)/(2.76 m) = 124.3 Hz. The beat frequency is fbeat = ?f = f2 – f1 = 129.9 Hz – 124.3 Hz = 5.6 Hz = 6 Hz. (b) The intensity maxima will travel with the speed of sound, so the separation of regions of maximum intensity is the “wavelength” of the beats: beat = v/fbeat = (343 m/s)/(5.6 Hz) = 61 m. 57. Because the police car is at rest, the wavelength traveling toward you is 1 = v/f0 = (343 m/s)/(1550 Hz) = 0.221 m. (a) This wavelength approaches you at a relative speed of v + vL. You hear a frequency f1 = (v + vL)/1 = (343 m/s + 30.0 m/s)/(0.221 m) = 1690 Hz. (b) The wavelength approaches you at a relative speed of v – vL. You hear a frequency f2 = (v – vL)/1 = (343 m/s – 30.0 m/s)/(0.221 m) = 1410 Hz. 58. Because the bat is at rest, the wavelength traveling toward the object is 1 = v/f0 = (343 m/s)/(50,000 Hz) = 6.8610–3 m. The wavelength approaches the object at a relative speed of v – vobject. The sound strikes and reflects from the object with a frequency f1 = (v – vobject)/1 = (343 m/s – 25.0 m/s)/(6.8610–3 m) = 46,360 Hz. This frequency can be considered emitted by the object, which is moving away from the bat. Because the wavelength behind a moving source increases, the wavelength approaching the bat is 2 = (v + vobject)/f1 = (343 m/s + 25.0 m/s)/(46,360 Hz) = 7.9410–3 m. This wavelength approaches the bat at a relative speed of v, so the frequency received by the bat is f2 = v/2 = (343 m/s)/(7.9410–3 m) = 43,200 Hz. 59. The wavelength in front of a moving object decreases, so the wavelength traveling toward the wall is 1 = (v – vbat)/f0 . Because the wall is stationary, this is also the wavelength of the reflected sound. This wavelength approaches the bat at a relative speed of v + vbat , so the frequency received by the bat is f = (v + vbat)/1 = [(v + vbat)/(v – vbat)]f0 Page 11

Chapter 16

= [(343 m/s + 5.0 m/s)/(343 m/s – 5.0 m/s)](30,000 Hz) =

30,890 Hz.

60. Because the wavelength in front of a moving source decreases, the wavelength from the approaching tuba is 1 = (v – v1)/f0 = (343 m/s – 10.0 m/s)/(75 Hz) = 4.44 m. This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f1 = v/1 = (343 m/s)/(4.44 m) = 77 Hz. Because the frequency from the stationary tube is unchanged, the beat frequency is fbeat = ?f = f1 – f0 = 77 Hz – 75 Hz = 2 Hz. 61. Because the wavelength in front of a moving source decreases, the wavelength from the approaching automobile is 1 = (v – v1)/f0 = (343 m/s – 15 m/s)/f0. This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f1 = v/1 = (343 m/s)/[(343 m/s – 15 m/s)/f0] = (343 m/s)f0/(328 m/s). Because the frequency from the stationary automobile is unchanged, the beat frequency is fbeat = ?f = f1 – f0 ; 5.5 Hz = [(343 m/s)f0/(328 m/s)] – f0 , which gives f0 = 120 Hz.

62. Because the wavelength in front of a moving source decreases, the wavelength from the approaching source is 1 = (v – v1)/f0 . This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f1 = v/1 = v/[(v – v1)/f0] = vf0/(v – v1) = (343 m/s)(2000 Hz)/(343 m/s – 15 m/s) = 2091 Hz. The wavelength from a stationary source is 2 = v/f0 .   This wavelength approaches the moving receiver at a relative speed of v + v1 , so the frequency heard is f2 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 15 m/s)(2000 Hz)/(343 m/s) = 2087 Hz. The two frequencies are not exactly the same, but close. For the other speeds we have, at 150 m/s: f3 = v/1 = v/[(v – v1)/f0] = vf0/(v – v1) = (343 m/s)(2000 Hz)/(343 m/s – 150 m/s) = 3554 Hz; f4 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 150 m/s)(2000 Hz)/(343 m/s) = 2875 Hz. at 300 m/s: f3 = v/1 = v/[(v – v1)/f0] = vf0/(v – v1) = (343 m/s)(2000 Hz)/(343 m/s – 300 m/s) = 15950 Hz; f4 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 300 m/s)(2000 Hz)/(343 m/s) = 3750 Hz. The Doppler formulas are not symmetric; the speed of the source creates a greater shift than the speed of the observer. We can write the expression for the frequency from an approaching source as f1 = v/1 = vf0/(v – v1) = f0/[1 – (v1/v)]. If v1 « v, we use 1/[1 – (v1/v)] ˜ 1 + (v1/v), so we have f1 ˜ f0[1 + (v1/v)], which is the expression f2 = (v + v1)f0/v for the frequency for a stationary source and a moving listener. Note that we have carried more significant figures than justified to show differences.

63. Because the source is at rest, the wavelength traveling toward the blood is 1 = v/f0 . This wavelength approaches the blood at a relative speed of v – vblood. The ultrasound strikes and reflects Page 12

Chapter 16

from the blood with a frequency f1 = (v – vblood)/1 = (v – vblood)/(v/f0) = (v – vblood)f0/v. This frequency can be considered emitted by the blood, which is moving away from the source. Because the wavelength behind the moving blood increases, the wavelength approaching the source is 2 = (v + vblood)/f1 = (v + vblood)/[(v – vblood)f0/v] = [1 + (vblood/v)]v/f0[1 – (vblood/v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is f2 = v/2 = v/{[1 + (vblood/v)]v/f0[1 – (vblood/v)]} = [1 – (vblood/v)]f0/[1 + (vblood/v)]. Because vblood « v, we use 1/[1 + (vblood/v)] ˜ 1 – (vblood/v), so we have f2 ˜ f0[1 – (vblood/v)]2 ˜ f0[1 – 2(vblood/v)]. For the beat frequency we have fbeat = f0 – f2 = 2(vblood/v)f0 = 2[(0.020 m/s)/(1540 m/s)](3.5106 Hz) =

91 Hz.

64. Because the source is at rest, the wavelength traveling toward the heart is 1 = v/f0 . If we assume that the heart is moving away, this wavelength approaches the heart at a relative speed of v – vheart. The ultrasound strikes and reflects from the heart with a frequency f1 = (v – vheart)/1 = (v – vheart)/(v/f0) = (v – vheart)f0/v. This frequency can be considered emitted by the heart, which is moving away from the source. Because the wavelength behind the moving heart increases, the wavelength approaching the source is 2 = (v + vheart)/f1 = (v + vheart)/[(v – vheart)f0/v] = [1 + (vheart/v)]v/f0[1 – (vheart/v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is f2 = v/2 = v/{[1 + (vheart/v)]v/f0[1 – (vheart/v)]} = [1 – (vheart/v)]f0/[1 + (vheart/v)]. Because vheart « v, we use 1/[1 + (vheart/v)] ˜ 1 – (vheart/v), so we have f2 ˜ f0[1 – (vheart/v)]2 ˜ f0[1 – 2(vheart/v)]. The maximum beat frequency occurs for the maximum heart velocity, so we have fbeat = f0 – f2 = 2(vheart/v)f0 ; 500 Hz = 2[vheart/(1.54103 m/s)](2.25106 Hz), which gives vheart = 0.171 m/s.

65. In Problem 64, we assumed that the heart is moving away from the source. Because the heart velocity is much less than the wave speed, the same beat frequency will occur when the heart is moving toward the source. Thus the maximum beat frequency will occur twice during each beat of the heart, so the heartbeat rate is !(180 maxima/min) = 90 beats/min. 66. (a) The frequency heard by a stationary observer from a source moving toward the observer is f  = f/[1 – (vs/v)] = f [1 – (vs/v)]–1. When we use the binomial expansion with vs « v, we get f  = f [1 + (– 1)(– vs/v) + (– 1)(– 2)(vs/v)2/2 + … ] ˜ f [1 + (vs/v)], which is the frequency an observer hears when moving toward a stationary source. Page 13

Chapter 16

(b) The percent error is error

= 100({f/[1 – (vs/v)]} – {f [1 + (vs/v)]})/{f/[1 – (vs/v)]}

= 100{1 – [1 + (vs/v)][1 – (vs/v)]} = 100(vs/v)2. For the given speed we get error = 100[(22 m/s)/(343 m/s)]2 = 0.41%.

67. Because the wind velocity is a movement of the medium, it adds or subtracts from the speed of sound in the medium. (a) Because the wind is blowing away from the observer, the effective speed of sound is v – vwind. Therefore the wavelength traveling toward the observer is a = (v – vwind)/f0 . This wavelength approaches the observer at a relative speed of v – vwind . The observer will hear a frequency fa = (v – vwind)/a = (v – vwind)/[(v – vwind)/f0] = f0 = 570 Hz. (b) Because the wind is blowing toward the observer, the effective speed of sound is v + vwind. From the analysis in part (a), we see that there will be no change in the frequency: 570 Hz. (c) Because the wind is blowing perpendicular to the line toward the observer, the effective speed of sound is v. Because there is no relative motion of the whistle and the observer, there will be no change in the frequency: 570 Hz. (d) Because the wind is blowing perpendicular to the line toward the observer, the effective speed of sound is v. Because there is no relative motion of the whistle and the observer, there will be no change in the frequency: 570 Hz. (e) Because the wind is blowing toward the cyclist, the effective speed of sound is v + vwind. Therefore the wavelength traveling toward the cyclist is e = (v + vwind)/f0 . This wavelength approaches the cyclist at a relative speed of v + vwind + vcycle . The cyclist will hear a frequency fe = (v + vwind + vcycle)/e = (v + vwind + vcycle)/[(v + vwind)/f0] = (v + vwind + vcycle)f0/(v + vwind) = (343 m/s +12.0 m/s + 15.0 m/s)(570 Hz)/(343 m/s + 12.0 m/s) = 594 Hz. (f) Because the wind is blowing perpendicular to the line toward the cyclist, the effective speed of sound is v. Therefore the wavelength traveling toward the cyclist is Page 14

Chapter 16

f = v/f0 . This wavelength approaches the cyclist at a relative speed of v + vcycle . The cyclist will hear a frequency ff = (v + vcycle)/f = (v + vcycle)/(v/f0) = (v + vcycle)f0/v = (343 m/s + 15.0 m/s)(570 Hz)/(343 m/s) = 595 Hz. 68. (a) From the definition of the Mach number, we have v = (Mach number)vsound = (0.33)(343 m/s) = 1.1102 m/s. (b) We find the speed of sound from v = (Mach number)vsound ; (3000 km/h)/(3.6 ks/h) = (3.2)vsound , which gives vsound = 2.6102 m/s. 69. In a time t the shock wave moves a distance vsoundt perpendicular to the wavefront. In the same time the object moves vobjectt. We see from the diagram that sin  = (vsoundt)/(vobjectt) = vsound/vobject .

70. (a) We find the angle of the shock wave from = vsound/vobject = vsound/(Mach number)vsound sin  = 1/(Mach number) = 1/2.3 = 0.435, so  = 26°. (b) From the diagram we see that, when the shock wave hits the ground, we have tan  = h/vairplanet = h/(Mach number)vsoundt; tan 26° = (7100 m)/(2.3)(310 m/s)t, which gives t = 21 s.

v sound t

 v object t

v airplane t



v sound t h

71. (a) From the definition of the Mach number, we have v = (Mach number)vsound ; (15,000 km/h)/(3.6 ks/h) = (Mach number)(35 m/s), which gives Mach number = (b) We find the angle of the shock wave from sin  = vsound/vobject = (35 m/s)/[(15,000 km/h)/(3.6 ks/h)] = 0.0084, so  = 0.48°. Thus the apex angle is 2 = 0.96°.

120.

72. (a) We find the angle of the shock wave in the air from 2.46°. sin  = vsound/vobject = (343 m/s)/(8000 m/s) = 0.0429, so  = (b) We find the angle of the shock wave in the ocean, before the meteorite slows down, from 11.2°. sin  = vsound/vobject = (1560 m/s)/(8000 m/s) = 0.193, so  = 73. (a) From the diagram we see that, when the shock wave hits the ground, we have 37°. tan  = h/dplane = (1.5 km)/(2.0 km) = 0.75, so  = (b) We find the speed of the plane from h sin  = vsound/vplane = vsound/(Mach number)vsound = 1/(Mach number); sin 37° = 1/(Mach number), which gives Mach number = 1.7.

Page 15

dplane = v plane t v sound t



Chapter 16

74. The angle of the shock wave is = vsound/vplane = vsound/(Mach number)vsound sin  = 1/(Mach number) = 1/1.8 = 0.556,  = 33.7°. From the diagram we see that, when the shock wave hits the ground, we have tan  = h/dplane ; tan 33.7° = (10 km)/dplane , which gives dplane = 15 km.

dplane = v plane t v sound t



h

75. Because the reflected pulse must be received before the emission of the next pulse, the minimum time between pulses is tmin = 2d/vsound = 2(200 m)/(1440 m/s) = 0.278 s. 76. Because the frequency doubles for each octave, we have fhigh/flow = 2x; (20,000 Hz)/(20 Hz) = 2x, or log10(1000) = x log10(2). which gives x ˜ 10. 77. We choose a coordinate system with origin at the top of the cliff, down positive, and t = 0 when the stone is dropped. If we call t1 the time of fall for the stone, we have y = y01+ v01t1 + !gt12; h = 0 + 0 + !gt12, or t1 = (2h/g)1/2. For the time t2 for the sound to reach the top of the cliff, we have t2 = h/vsound . Thus for the total time we have t = t1 + t2 = (2h/g)1/2 + h/vsound ; 3.5 s = [2h/(9.80 m/s2)]1/2 + [h/(343 m/s)]. This is a quadratic equation for h1/2, which has the positive result h1/2 = 7.39 m1/2, so the height of the cliff is 55 m. 78. The intensity for a 0 dB sound is I0 = 10–12 W/m2. For 1000 mosquitoes, we find the intensity level from  = 10 log10(I1/I0) = 10 log10(1000I0/I0) = 30 dB.

79. Because the wavelength in front of a moving source decreases, the wavelength from the approaching car is 1 = (v – vcar)/f0 . This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f1 = v/1 = v/[(v – vcar)/f0] = vf0/(v – vcar). Because the wavelength behind a moving source increases, the wavelength from the receding car is 2 = (v + vcar)/f0 . This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f2 = v/2 = v/[(v + vcar)/f0] = vf0/(v + vcar). If the frequency drops by one octave, we have f1/f2 = [vf0/(v – vcar)]/[vf0/(v + vcar)] = (v + vcar)/(v – vcar) = 2, or 2(v – vcar) = v + vcar , which gives vcar = @v = @(343 m/s) = 114 m/s =

410 km/h (255 mi/h).

80. The wavelength of the fundamental frequency for a string is  = 2L. Because the speed of a wave on the string does not change, we have v = 1f1 = 1f1; 2L[(540 Hz)/3] = 2(0.60L)f1, which gives f1 = 300 Hz. Page 16

Chapter 16

81. The tension and mass density of the string determines the velocity: v = (FT/)1/2. Because the strings have the same length, the wavelengths are the same, so for the ratio of frequencies we have fn+1/fn = vn+1/vn = (n/n+1)1/2 = 1.5, or n+1/n = (1/1.5)2 = 0.444. With respect to the lowest string, we have n+1/1 = (0.444)n. If we call the mass density of the lowest string 1, we have 1, 0.444, 0.198, 0.0878, 0.0389. 82. We find the intensity of each sound from  = 10 log10(I/I0); 80 dB = 10 log10[I1/(1.010–12 W/m2)], which gives I1 = 1.010–4 W/m2; 85 dB = 10 log10[I2/(1.010–12 W/m2)], which gives I2 = 3.1610–4 W/m2. The intensities add, so the resultant intensity level is  = 10 log10[(I1 + I2)/I0] = 10 log10[(1.010–4 W/m2 + 3.1610–4 W/m2)/(1.010–12 W/m2)] =

86 dB.

83. We find the intensity of the sound from  = 10 log10(I/I0); 100 dB = 10 log(I/10–12 W/m2), which gives I = 1.0010–2 W/m2. If the speaker radiates equally in all directions, the power output of the speaker is P = IA = I4pr2 = (1.0010–2 W/m2)4p(12.0 m)2 = 18.1 W.

84. (a) The wavelength of the fundamental frequency for a string is  = 2L, so the speed of a wave on the string is v = f = 2(0.32 m)(440 Hz) = 2.8102 m/s. We find the tension from v = (FT/)1/2; 282 m/s = [FT/(6.110–4 kg/m)]1/2 , which gives FT = 49 N. (b) For a closed pipe the wavelength of the fundamental frequency is  = 4L. We find the required length from v = f = 4Lf; 343 m/s = 4L(440 Hz), which gives L = 0.195 m = 19.5 cm. (c) All harmonics are present in the string, so the first overtone is the second harmonic: f2 = 2f1 = 2(440 Hz) = 880 Hz. Only the odd harmonics are present in a closed pipe, so the first overtone is the third harmonic: f3 = 3f1 = 3(440 Hz) = 1320 Hz. 85. We find the ratio of intensities from ? = 10 log10(I2/I1); Page 17

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– 10 dB = 10 log10(I2/I1), which gives I2/I1 = 0.10. If we assume uniform spreading of the sounds, the intensity is proportional to the power output, so we have P2/P1 = I2/I1 ; P2/(150 W) = 0.10, which gives P2 = 15 W. 86. At each resonant position the top of the water column is a node. Thus the distance between the two readings corresponds to the distance between adjacent nodes, or half a wavelength: ?L = /2, or  = 2 ?L. For the frequency we have f = v/ = (343 m/s)/2(0.395 m – 0.125 m) = 635 Hz.

L1

L2

87. Because the speakers are at rest, the wavelength traveling toward the person is  = v/f0. This wavelength from the speaker in front of the person approaches the person at a relative speed of v + vperson. The frequency heard is f1 = (v + vperson)/. This wavelength from the speaker behind the person approaches the person at a relative speed of v – vperson. The frequency heard is f2 = (v – vperson)/. Thus the beat frequency is ?f = f1 – f2 = [(v + vperson)/] – [(v – vperson)/] = 2vperson/ = 2vpersonf0/v = 2(1.4 m/s)(280 Hz)/(343 m/s) = 2.3 Hz.

88. We find the intensity of the sound at a distance of 30 m from  = 10 log10(I/I0); 140 dB = 10 log10(I/10–12 W/m2), which gives I = 1.0102 W/m2. The power intercepted by the ear is P = IA = Ipr2 = (1.0102 W/m2)p(0.020 m)2 = 0.13 W. 89. The intensity of the sound wave is given by I = 2p2f 2DM2v. Because the frequency, density, and velocity are the same, for the ratio we have I/I0 = (DM/DM0)2 = 1012, which gives DM/DM0 = 106. The pressure amplitude is proportional to the displacement amplitude, so we have ?PM/?PM0 = DM/DM0 = 106. 90. The smallest displacement maximum will occur for the lowest intensity, or intensity level. From Fig. 16–5, we estimate the intensity levels. For the four frequencies we have 35 Hz:  = 10 log10(I1/I0); Page 18

Chapter 16

60 dB = 10 log(I1/10–12 W/m2), which gives I1 = 1.010–6 W/m2. We find the displacement maximum from I1 = 2p2f 2DM12v 1.010–6 W/m2 = 2p2(1.29 kg/m3)(35 Hz)2DM12(343 m/s), which gives DM1 = 3.110–7 m. 1000 Hz:  = 10 log10(I2/I0); 0 dB = 10 log(I2/10–12 W/m2), which gives I2 = 1.010–12 W/m2. We find the displacement maximum from I2 = 2p2f 2DM22v 1.010–12 W/m2 = 2p2(1.29 kg/m3)(1000 Hz)2DM22(343 m/s), which gives DM2 = 1.110–11 m. 5000 Hz:  = 10 log10(I3/I0); 0 dB = 10 log(I3/10–12 W/m2), which gives I3 = 1.010–12 W/m2. We find the displacement maximum from I3 = 2p2f 2DM32v 1.010–12 W/m2 = 2p2(1.29 kg/m3)(5000 Hz)2DM32(343 m/s), which gives DM3 = 2.110–12 m. 15,000 Hz:  = 10 log10(I4/I0); 20 dB = 10 log(I4/10–12 W/m2), which gives I4 = 1.010–10 W/m2. We find the displacement maximum from I4 = 2p2f 2DM42v 1.010–10 W/m2 = 2p2(1.29 kg/m3)(15,000 Hz)2DM42(343 m/s), which gives DM4 = 7.110–12 m. The ear is most sensitive to displacement at a frequency of 5000 Hz. 91. We find the gain from  = 10 log10(P2/P1) = 10 log10[(100 W)/(110–3 W)] =

50 dB.

92. (a) Because both sources are moving toward the observer at the same speed, they will have the same Doppler shift, so the beat frequency will be 0. (b) Because the wavelength in front of a moving source decreases, the wavelength from the approaching loudspeaker is 1 = (v – vcar)/f0 = (343 m/s – 10.0 m/s)/(200 Hz) = 1.665 m. This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f1 = v/1 = (343 m/s)/(1.665 m) = 206 Hz. Because the wavelength behind a moving source decreases, the wavelength from the receding loudspeaker is 2 = (v + vcar)/f0 = (343 m/s + 10.0 m/s)/(200 Hz) = 1.765 m. This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is f2 = v/2 = (343 m/s)/(1.765 m) = 194 Hz. The beat frequency is fbeat = ?f = 206 Hz – 194 Hz = 12 Hz. Page 19

Chapter 16

Note that this frequency may be too high to be heard as beats. (c) Because both sources are moving away from the observer at the same speed, they will have the same Doppler shift, so the beat frequency will be 0. 93. Because the wavelength in front of a moving source decreases, the wavelength from the approaching train whistle is 1 = (v – vtrain)/f0 . Because you are stationary, this wavelength approaches you at a relative speed of v, so the frequency heard by you is f1 = v/1 = v/[(v – vtrain)/f0 ] = vf0 /(v – vtrain). Because the wavelength behind a moving source increases, the wavelength from the receding train whistle is 2 = (v + vtrain)/f0 . This wavelength approaches you at a relative speed of v, so the frequency heard by you is f2 = v/2 = v/[(v + vtrain)/f0] = vf0/(v + vtrain). For the ratio of frequencies, we get f2/f1 = (v – vtrain)/(v + vtrain); (486 Hz)/(538 Hz) = (343 m/s – vtrain)/(343 m/s + vtrain), which gives vtrain = 17.5 m/s. 94. For a closed pipe the wavelength of the fundamental frequency is 1pipe = 4L. Thus the frequency of the third harmonic is f3pipe = 3f1pipe = 3v/4L = 3(343 m/s)/4(0.75 m) = 343 Hz. This is the fundamental frequency for the string. The wavelength of the fundamental frequency for the string is 1string = 2L, so we have vstring = f1string2L = [FT/(m/L)]1/2; (343 Hz)2(0.75 m) = {FT/[(0.00210 kg)/(0.75 m)]}1/2, which gives FT =

7.4102 N.

95. Because the source is at rest, the wavelength traveling toward the blood is 1 = v/f0 . This wavelength approaches the blood at a relative speed of v – vblood. The ultrasound strikes and reflects from the blood with a frequency f1 = (v – vblood)/1 = (v – vblood)/(v/f0) = (v – vblood)f0/v. This frequency can be considered emitted by the blood, which is moving away from the source. Because the wavelength behind the moving blood increases, the wavelength approaching the source is 2 = (v + vblood)/f1 = (v + vblood)/[(v – vblood)f0/v] = [1 + (vblood/v)]v/f0[1 – (vblood/v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is f2 = v/2 = v/{[1 + (vblood/v)]v/f0[1 – (vblood/v)]} = [1 – (vblood/v)]f0/[1 + (vblood/v)]. Because vblood « v, we use 1/[1 + (vblood/v)] ˜ 1 – (vblood/v), so we have f2 ˜ f0[1 – (vblood/v)]2 ˜ f0[1 – 2(vblood/v)]. For the beat frequency we have fbeat = f0 – f2 = 2(vblood/v)f0 = 2[(0.32 m/s)/(1.54103 m/s)](5.50106 Hz) = 2.29103 Hz = Page 20

2.3 kHz.

Chapter 16

96. The initial path difference is ?D = 2[(¬/2)2 + d2]1/2 – ¬. When the obstacle is moved ?d, the new path difference is ?D = 2[(¬/2)2 + (d + ?d)2]1/2 – ¬. Because ?d « d, we get ?D = 2[(¬/2)2 + d2 + ?d(2d + ?d)]1/2 – ¬ = 2[(¬/2)2 + d2 + 2d ?d]1/2 – ¬

Detector

¬

= 2[(¬/2)2 + d2]1/2(1 + {2d ?d/[(¬/2)2 + d2]})1/2 – ¬. = 2[(¬/2)2 + d2]1/2(1 + !{2d ?d/[(¬/2)2 + d2]}) – ¬.

= 2[(¬/2)2 + d2]1/2(1 + {d ?d/[(¬/2)2 + d2]}) – ¬. To create destructive interference, the change in path difference must be /2, so we have ?D – ?D = /2;

S

d

2[(¬/2)2 + d2]1/2(1 + {d ?d/[(¬/2)2 + d2]}) – ¬ – {2[(¬/2)2 + d2]1/2 – ¬} = /2; 2[(¬/2)2 + d2]1/2d ?d/[(¬/2)2 + d2] = /2, which gives ?d = (/4d)[(¬/2)2 + d2]1/2.

97. When the person is equidistant from the sources, there is constructive interference. To move to a point where there is destructive interference, the path difference from the sources must be ?L = n/2, with n = 1, 3, 5, … . If we assume that n = 1, we find the frequency from f = v/ = v/2 ?L = (343 m/s)/2(0.31 m) = 550 Hz. Because larger values of n would give a frequency out of the given range, this is the result. 98. Because the wavelength in front of a moving source decreases, the wavelength approaching the moth is 1 = (v – vbat)/f0 . The wavelength approaches the moth at a relative speed of v + vmoth. The sound strikes and reflects from the object with a frequency f1 = (v + vmoth)/1 = (v + vmoth)/[(v – vbat)/f0 ] = (v + vmoth)f0/(v – vbat). This frequency can be considered emitted by the moth, which is moving toward the bat. Because the wavelength in front of a moving source decreases, the wavelength approaching the bat is 2 = (v – vmoth)/f1 = (v – vmoth)/[(v + vmoth)f0/(v – vbat)] = (v – vmoth)(v – vbat)/(v + vmoth)f0. This wavelength approaches the bat at a relative speed of v + vbat , so the frequency received by the bat is f2 = (v + vbat)/2 = (v + vbat)(v + vmoth)f0/(v – vmoth)(v – vbat) = (343.0 m/s + 6.5 m/s)(343.0 m/s + 5.0 m/s)(51.35 kHz)/(343.0 m/s – 5.0 m/s)(343.0 m/s – 6.5 m/s) = 54.91 kHz.

99. (a) The rod will have a node at the middle and antinodes at the ends. The wavelength for the fundamental frequency is 1 = 2L. Thus the fundamental frequency is f1 = v/1 = v/2L = (5100 m/s)/2(0.90 m) = 2.8103 Hz. (b) The wavelength in the rod is rod = 1 = 2L = 2(0.90 m) = 1.80 m. (c) The frequency in the air is the frequency in the rod, so the wavelength in the air is air = v/f1 = (343 m/s)/(2.8103 Hz) = 0.12 m. 100.We consider standing waves along each dimension, with nodes at the walls, so the wavelength is twice the dimension. Thus we have f1 = v/2L = (343 m/s)/2(5.0 m) = 34 Hz. Page 21

Chapter 16

f2 = v/2W = (343 m/s)/2(4.0 m) = 43 Hz. f3 = v/2H = (343 m/s)/2(2.8 m) = 61 Hz. Note that there could be more complicated standing waves from reflections in two or three directions. 101. We find the intensity of the sound from  = 10 log10(I/I0); 100 dB = 10 log10(I/10–12 W/m2), which gives I = 1.010–2 W/m2. (a) We find the maximum displacement from I = 2p2f 2DM2v 1.010–2 W/m2 = 2p2(1.29 kg/m3)(10103 Hz)2DM2(343 m/s), which gives DM = 1.110–7 m. The cone will vibrate with SHM, so the total motion is d = 2DM = 2(1.110–7 m) = 2.210–7 m. (b) For the given frequency we get I = 2p2f 2DM2v 1.010–2 W/m2 = 2p2(1.29 kg/m3)(40 Hz)2DM2(343 m/s), which gives DM = 2.710–5 m. The cone will vibrate with SHM, so the total motion is d = 2DM = 2(2.710–5 m) = 5.410–5 m. 102. The velocity component of the blood parallel to the sound transmission is vblood cos 45° = 0.707vblood. Because the source is at rest, the wavelength traveling toward the blood is 1 = v/f0 . This wavelength approaches the blood at a relative speed of v – 0.707vblood. The ultrasound strikes and reflects from the blood with a frequency f1 = (v – 0.707vblood)/1 = (v – 0.707vblood)/(v/f0) = (v – 0.707vblood)f0/v. This frequency can be considered emitted by the blood, which is moving away from the source. Because the wavelength behind the moving blood increases, the wavelength approaching the source is 2 = (v + 0.707vblood)/f1 = (v + 0.707vblood)/[(v – 0.707vblood)f0/v] = [1 + (0.707vblood/v)]v/f0[1 – (0.707vblood/v)]. This wavelength approaches the source at a relative speed of v, so the frequency received by the source is f2 = v/2 = v/{[1 + (0.707vblood/v)]v/f0[1 – (0.707vblood/v)]} = [1 – (0.707vblood/v)]f0/[1 + (0.707vblood/v)]. Because vblood « v, we use 1/[1 + (0.707vblood/v)] ˜ 1 – (0.707vblood/v), so we have f2 ˜ f0[1 – (0.707vblood/v)]2 ˜ f0[1 – 2(0.707vblood/v)]. For the beat frequency we have fbeat = f0 – f2 = 2(0.707vblood/v)f0 900 Hz = 2[(0.707vblood)/(1.54103 m/s)](5.0106 Hz), which gives vblood =

Page 22

0.20 m/s.

Chapter 17

CHAPTER 17 – Temperature, Thermal Expansion, and the Ideal Gas Law 1.

The number of atoms in a mass m is given by N = m/Mmatomic. Because the masses of the two rings are the same, for the ratio we have NAu/NAg = MAg/MAu = 108/197 = 0.548.

2.

The number of atoms in a mass m is given by N = m/Mmatomic = (3.410–3 kg)/(63.5 u)(1.6610–27 kg/u) =

3.

(a) T(°C) = (5/9)[T(°F) – 32] = (5/9)(68°F – 32) = 20°C. (b) T(°F) = (9/5)T(°C) + 32 = (9/5)(1800°C) + 32 = 3272°F

4.

(a) T(°F) = (9/5)T(°C) + 32 = (9/5)(– 15°C) + 32 = (b) T(°C) = (5/9)[T(°F) – 32] = (5/9)(– 15°F – 32) =

5.

T(°F) = (9/5)T(°C) + 32 = (9/5)(40.0°C) + 32.0 =

6.

Because the temperature and length are linearly related, we have ?T/?L = (100.0°C – 0.0°C)/(22.85 cm – 11.82 cm) = 9.067 C°/cm. (a) (T1 – 0.0°C)/(16.70 cm – 11.82 cm) = 9.067 C°/cm, which gives T1 = (b) (T2 – 0.0°C)/(20.50 cm – 11.82 cm) = 9.067 C°/cm, which gives T2 =

3.21022 atoms.

˜ 3300°F.

5°F. – 26°C. 104.0°F.

44.2°C. 78.7°C.

7.

We set T(°F) = T(°C) = T in the conversion between the temperature scales: T(°F) = (9/5)T(°C) + 32 T = (9/5)T + 32, which gives T = – 40°F = – 40°C.

8.

At any temperature below 20°C the expansion cracks will increase. Thus the expansion from 20°C to 50°C must eliminate the cracks. Any higher temperature will cause stress in the concrete. If the cracks have a width ?L, we have 0.43 cm. ?L = L0 ?T = [1210–6 (C°)–1](12 m)(50°C – 20°C) = 4.310–3 m =

9.

For the expansion ?L, we have 2.010–6 m. ?LInvar = InvarL0 ?T = [0.210–6 (C°)–1](2.0 m)(5.0 C°) = For the other materials we have ?Lsteel = steelL0 ?T = [1210–6 (C°)–1](2.0 m)(5.0 C°) = 1.210–4 m. ?Lmarble = marbleL0 ?T = [2.510–6 (C°)–1](2.0 m)(5.0 C°) = 2.510–5 m.

10. We find the height change from ?L = L0 ?T = [1210–6 (C°)–1](300 m)(25°C – 2°C) = 8.310–2 m =

8.3 cm.

11. We can treat the change in diameter as a simple change in length, so we have ?L = L0 ?T; 1.869 cm – 1.871 cm = [1210–6 (C°)–1](1.871 cm)(T – 20°C), which gives T =

– 69°C.

12. For the expanded dimensions, we have ¬ = ¬(1 +  ?T); w = w(1 +  ?T). Thus the change in area is ?A = A – A = ¬w – ¬w = ¬w(1 +  ?T)2 – ¬w = ¬w[2 ?T + ( ?T)2] = ¬w ?T(2 +  ?T). Because  ?T « 2, we have ?A = 2¬w ?T.

Page 1

Chapter 17

13. The contraction of the glass causes the enclosed volume to decrease as if it were glass. The volume of water that can be added is ?V = ?Vglass – ?Vwater = V0glass ?T – V0water ?T = V0(glass – water)?T = (350 mL)[2710–6 (C°)–1 – 21010–6 (C°)–1](20°C – 100°C) = 5.1 mL. 14. (a) The expansion of the container causes the enclosed volume to increase as if it were made of the same material as the container. The volume of water that was lost is ?V = ?Vwater – ?Vcontainer = V0water ?T – V0container ?T = V0(water – container)?T; (0.35 g)/(0.98324 g/mL) = (55.50 mL)[21010–6 (C°)–1 – container](60°C – 20°C), which gives container = 5010–6 (C°)–1. (b) From Table 17–1, copper is the most likely material. 15. We find the change in volume from ?V = V0 ?T = )pr3 ?T = )p(4.375 cm)3[110–6 (C°)–1](200°C – 30°C) =

0.06 cm3.

16. We can treat the change in diameter as a simple change in length, so we have D = D0(1 +  ?T). The two objects must reach the same diameter: D = D0brass(1 + brass ?T) = D0iron(1 + iron ?T);

(8.753 cm){1 + [1910–6 (C°)–1](T – 20°C)} = (8.743 cm){1 + [1210–6 (C°)–1](T – 20°C)}, which gives T= – 1.4102 °C.

17. We assume that we can ignore the change in cross sectional area of the tube. The volume change of the fluid is the increased volume in the column: ?V = AL ?T = A ?L, or ?L = L ?T. When we compare this to the expression for linear expansion, ?L = L ?T, we see that  = . 18. (a) We consider a fixed mass of the substance. The change in volume from the temperature change is ?V = V0 ?T. Because the density is mass/volume, for the fractional change in the density we have ?/ = [(1/V) – (1/V0)]/(1/V0) = (V0 – V)/V ˜ (V0 – V)/V0 = – ?V/V0 = –  ?T, which we can write ? = –  ?T. (b) For the lead sphere we have 0.0057 (0.57%). ?/ = – [8710–6 (C°)–1](– 40°C – 25°C) =

Page 2

Chapter 17

19. The increase in temperature will cause the length of the brass rod to increase. The period of the pendulum depends on the length, T = 2p(L/g)1/2, so the period will be greater. This means the pendulum will make fewer swings in a day, so the clock will be slow and the clock will lose time. We use TC for the temperature to distinguish it from the period. For the length of the brass rod, we have L = L0(1 +  ?TC). Thus the ratio of periods is T/T0 = (L/L0)1/2 = (1 +  ?TC)1/2. Because  ?TC is much less than 1, we have T/T0 ˜ 1 + ! ?TC , or ?T/T0 = ! ?TC. The number of swings in a time t is N = t/T. For the same time t, the change in period will cause a change in the number of swings: ?N = (t/T) – (t/T0) = t(T0 – T)/TT0 ˜ – t(?T/T0)/T0 , because T ˜ T0. The time difference in one year is ?t

= T0 ?N = – t(?T/T0) = – t(! ?TC) = – (1 yr)(3.16107 s/yr) = – 2.4103 s =

– 40 min.

20. (a) The radius will increase as if it were a length: r = r(1 +  ?T). The new surface area will be A = 4pr2 = 4pr2(1 +  ?T)2 = A(1 +  ?T)2. Thus the change in area is ?A = A – A = A[2 ?T + ( ?T)2] = 2A ?T (1 + ! ?T). Because ! ?T « 1, we have ?A = 2A ?T = 8pr2 ?T. (b) For the iron sphere we have 3.1102 cm2. ?A = 8p(60.0 cm)2[1210–6 (C°)–1](310°C – 20°C) = 21. We find the change in radius from ?R = R ?T. Because the bearings are frictionless, angular momentum will be conserved: I11 = I22 , with I = !mR2 for a solid cylinder.  For the fractional change in the angular velocity, we have ?/ = (2 – 1)/1 = [(I11/I2) – 1]/1 = (I1 – I2)/I2 . Because the mass is constant, we have ?/ = [R2 – (R + ?R)2]/(R + ?R)2 = – [2R ?R + (?R)2]/(R + ?R)2 ˜ – 2R ?R/R2 = – 2 ?R/R. From the temperature change, we have – 2.810–3 (0.28%). ?/ = – 2 ?T = – 2[2510–6 (C°)–1](75.0°C – 20.0°C) = 22. The compressive strain must compensate for the thermal expansion. From the relation between stress and strain, we have Stress = E(Strain) = E ?T. When we use the ultimate strength of concrete, we have 20106 N/m2 = (20109 N/m2)[1210–6 (C°)–1](T – 10.0°C), which gives T = 93°C. 23. The compressive strain must compensate for the thermal expansion. From the relation between stress and strain, we have Stress = E(Strain) = E ?T; Page 3

Chapter 17

F/A = (70109 N/m2)[2510–6 (C°)–1](35°C – 15°C) =

3.5107 N/m2.

24. (a) The tensile strain must compensate for the thermal contraction. From the relation between stress and strain, we have Stress = E(Strain) = E ?T; F/A = (200109 N/m2)[1210–6 (C°)–1](– 30°C – 30°C) = – 1.4108 N/m2 (tensile). 6 2 No, because the ultimate strength of steel is 50010 N/m = 5.0108 N/m2. (b) (c) For concrete we have F/A = (20109 N/m2)[1210–6 (C°)–1](– 30°C – 30°C) = – 1.4107 N/m2 (tensile). will fracture. Because the ultimate tensile strength of concrete is 2106 N/m2, it 25. (a) As the iron band expands, the inside diameter will increase as if it were iron. We can treat the change in inside diameter as a simple change in length, so we have D = D0(1 +  ?T), or D – D0 = D0 ?T; 134.122 cm – 134.110 cm = (134.110 cm)[1210–6 (C°)–1](T – 20°C), which gives T = 27°C. (b) We assume that the barrel is rigid. The tensile strain in the circumference of the band is p ?D/pD0 = ?D/D0 , which is the thermal strain. The tensile strain must compensate for the thermal contraction. From the relation between stress and strain, we have Stress = E(Strain) = E ?T = E ?D/D; F/(0.074 m)(0.0065 m) = (100109 N/m2)(134.122 cm – 134.110 cm)/(134.110 cm), which gives F = 4.3103 N. 26. (a) T(K) = T(°C) + 273 = 86°C + 273 = 359 K. (b) T(°C) = (5/9)[T(°F) – 32] = (5/9)(78°F – 32) = 26°C. T(K) = T(°C) + 273 = 26°C + 273 = 299 K. (c) T(K) = T(°C) + 273 = – 100°C + 273 = 173 K. (d) T(K) = T(°C) + 273 = 5500°C + 273 = 5773 K. 27. On the Celsius scale, absolute zero is T(°C) = T(K) – 273.15 = 0 K – 273.15 = – 273.15°C. On the Fahrenheit scale, we have T(°F) = (9/5)T(°C) + 32 = (9/5)(– 273.15°C) + 32 =

– 459.7°F.

28. (a) T1(K) = T1(°C) + 273 = 4000°C + 273 = 4273 K; T2(K) = T2(°C) + 273 = 15106 °C + 273 = 15106 K. (b) The difference in each case is 273, so we have Earth: (273)(100)/(4273) = 6.4%; Sun: (273)(100)/(15106) = 0.0018%. 29. For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; (3.20 atm/1.00 atm)(V2/3.00 m3) = (311 K/273 K), which gives V2 = 30. For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; (40 atm/1.00 atm)(1/9) = T2/(293 K), which gives T2 = 1300 K.

Page 4

1.07 m3.

Chapter 17

31. If we assume oxygen is an ideal gas, we have PV = nRT = (m/M)RT; (1.013105 Pa)V = [m/(32 g/mol)(10–3 kg/g)](8.315 J/mol · K)(273 K), which gives m/V = 1.43 kg/m3. 32. The volume, temperature, and mass are constant. For the two gases we can write P1V = n1RT = (m/M1)RT, and P2V = n2RT = (m/M2)RT, which can be combined to give P2/P1 = M1/M2 ; P2/(3.65 atm) = (28 g/mol)/(44 g/mol), which gives P2 = 2.32 atm. 33. (a) For the ideal gas we have PV = nRT = (m/M)RT; (1.013105 Pa)V = [(18.5 kg)(103 g/kg)/(28 g/mol)](8.315 J/mol · K)(273 K), 14.8 m3. which gives V = (b) With the additional mass in the same volume, we have PV = nRT = (m/M)RT; P(14.8 m3) = [(18.5 kg + 15.0 kg)(103 g/kg)/(28 g/mol)](8.315 J/mol · K)(273 K), 1.81 atm. which gives P = 1.83105 Pa = 34. (a) For the ideal gas we have PV = nRT; (1.000 atm + 0.350 atm)(1.013105 Pa/atm)V = (18.75 mol)(8.315 J/mol · K)(283 K), 0.323 m3. which gives V = (b) For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; [(1.00 atm + 1.00 atm)/(1.00 atm + 0.350 atm)](1/2) = (T2/283 K), which gives T2 = 210 K = – 63°C. 35. If we assume argon is an ideal gas, we have PV = nRT = (m/M)RT; P(35.010–3 m3) = [(105.0 kg)(103 g/kg)/(40 g/mol)](8.315 J/mol · K)(293 K), 1.80103 atm. which gives P = 1.83108 Pa = 36. For an ideal gas we have PV = nRT = (m/M)RT; For the two gases we can write P1V1 = (m1/M1)RT1 and P2V2 = (m2/M2)RT2 , which can be combined to give (P2/P1)(V2/V1) = (m2/m1)(M1/M2)(T2/T1); [(7.00 atm + 1.00 atm)/(8.70 atm + 1.00 atm)](1) = (m2/30.0 kg)(32 g/mol/4 g/mol)(1), which gives m2 = 3.09 kg He. 37. The density of an ideal gas is  = m/V = mP/nRT = MP/RT. The net lift is the buoyant force less the weight of gas in the balloon: Flift = (cold – hot)gV = (MP/R)[(1/Tcold) – (1/Thot)]gV; 2700 N = [(29 g/mol)(1.013105 Pa)/(8.315 J/mol · K)(103 g/kg)]  [(1/273 K) – (1/Thot)](9.80 m/s2)(1800 m3), which gives Thot = 310 K = 37°C. The maximum altitude is limited because the pressure decreases with altitude, which decreases the density. The buoyant force is due to the change in pressure over the height of the balloon, which is reduced by the decreased density. There is also a limit on the temperature inside the balloon.

Page 5

Chapter 17

38. The volume and pressure are constant. For the fractional change in the number of moles, we can write ?n/n = [(PV/RT2) – (PV/RT1)]/(PV/RT1) = (T1 – T2)/T2 = (288 K – 311 K)/(311 K) = – 0.074 (7.4%).

39. For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; (P2/2.45 atm)(48.8 L/61.5 L) = (323.2 K/291.2 K), which gives P2 =

3.43 atm.

40. For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; (0.70 atm/1.00 atm)(V2/V1) = (278.2 K/293.2 K), which gives V2/V1 =

1.4.

41. If we assume water vapor is an ideal gas, we have PV = nRT = (m/M)RT; (1.013105 Pa)V = [m/(18 g/mol)(10–3 kg/g)](8.315 J/mol · K)(373 K), which gives m/V = 0.588 kg/m3. This is less than the listed value of 0.598 kg/m3. We expect a difference because the tendency of steam to form droplets indicates an attractive force, so water vapor is not an ideal gas. 42. The pressure at the bottom of the lake is Pbottom = Ptop + gh = 1.013105 Pa + (1000 kg/m3)(9.80 m/s2)(37.0 m) = 4.64105 Pa. For the two states of the gas we can write PbottomVbottom = nRTbottom , and PtopVtop = nRTtop , which can be combined to give (Pbottom/Ptop)(Vbottom/Vtop) = Tbottom/Ttop ; (4.64105 Pa/1.013105 Pa)(1.00 cm3/Vtop) = (278.7 K/294.2 K), which gives Vtop = 43. If we write the ideal gas law as PV = NkT, we have N/V = P/kT = (1.013105 Pa)/(1.3810–23 J/K)(273 K) =

4.83 cm3.

2.691025 molecules/m3.

44. We find the number of moles from n = V/M = (1.000 g/cm3)(1.000103 cm3)/(18 g/mol) = 55.6 mol. For the number of molecules we have N = nNA = (55.6 mol)(6.021023 molecules/mol) = 3.341025 molecules. 45. We find the number of moles in one breath from PV = nRT; (1.013105 Pa)(2.010–3 m3) = n(8.315 J/mol · K)(300 K), which gives n = 8.1210–2 mol. For the number of molecules in one breath we have N = nNA = (8.1210–2 mol)(6.021023 molecules/mol) = 4.91022 molecules. 46. (a) We find the number of moles from n = &V/M = &4pR2d/M = (1000 kg/m3)3p(6.4106 m)2(3103 m)(103 g/kg)/(18 g/mol) = 61022 mol. (b) For the number of molecules we have N = nNA = (61022 mol)(6.021023 molecules/mol) = 41046 molecules.

Page 6

Chapter 17

47. The volume and mass are constant. For the two states of the gas we can write P1V1 = nRT1 , and P2V2 = nRT2 , which can be combined to give P2/P1 = T2/T1 ; (P2/1.00 atm) = (453 K/293 K), which gives P2 = 1.55 atm. We find the length of a side of the box from V = L3; 5.110–2 m3 = L3, which gives L = 0.371 m. The net force is the same on each side of the box. Because there is atmospheric pressure outside the box, the net force is F = A ?P = L2(P2 – P1) = (0.371 m)2(1.55 atm – 1.00 atm)(1.013105 Pa/atm) = 7.7103 N. Note that we have assumed no change in dimensions from the increased pressure. 48. We find the number of moles in one breath from PV = nRT; (1.013105 Pa)(2.010–3 m3) = n(8.315 J/mol · K)(300 K), which gives n = 8.1210–2 mol. For the number of molecules in one breath we have N = nNA = (8.1210–2 mol)(6.021023 molecules/mol) = 4.91022 molecules. We assume that all of the molecules from the last breath that Galileo took are uniformly spread throughout the atmosphere, so the fraction that are in one breath is given by V/Vatmosphere . We find the number now in one breath from N /N = V/Vatmosphere = V/4pR2h; N /(4.91022 molecules) = (2.010–3 m3)/4p(6.4106 m)2(10103 m), 20 molecules. which gives N  ˜ 49. (a) For the constant-volume gas thermometer, we have T = (273.16 K)(P/Ptp); (444.6 + 273.15) K = (273.16 K)(187 torr/Ptp), which gives Ptp = (b) We find the temperature read by the thermometer from T = (273.16 K)(P/Ptp) = (273.16 K)(112 torr/71.2 torr) = 430 K = 50. For the constant-volume gas thermometer, we have T = (273.16 K)(P/Ptp); (100.00 + 273.15) K = (273.16 K)(P/Ptp), which gives P/Ptp =

71.2 torr. 157°C.

1.3660.

51. (a) From Fig. 17–16, we read a temperature of 373.34 K from the oxygen curve at a pressure of 268 torr. Thus the inaccuracy is ?T = 373.34 K – (273.15 + 100.00) K = 0.19 K. (b) As a percentage, this is (?T/T)100 = (0.19 K/373.15)100 = 0.051%.

Page 7

Chapter 17

52. The two temperatures of the gas are T1 = (273.16 K)(P1/Ptp) = (273.16 K)(218 torr/286 torr) = 208.21 K; T2 = (273.16 K)(P2/Ptp) = (273.16 K)(128 torr/163 torr) = 214.51 K. For a constant-volume thermometer, the actual temperature is T = (273.16 K) lim P/ Ptp .

T T mp T2 T1 P2

0

Ptp  0

P1

Ptp

We do this limiting procedure by assuming a linear relation and extrapolating to Ptp = 0: (T – T1)/(P1 – 0) = (T – T2)/(P2 – 0); (T - 208.21 K)/286 torr = (T – 214.51 K)/163 torr, which gives T = 222.9 K. 53. (a) The tape will expand, so the numbers will be beyond the true length, so it will read low. (b) The percentage error will be 0.017%. (?L/L)(100) = (L ?T/L)(100) = ( ?T)(100) = [1210–6 (C°)–1](34°C – 20°C)(100) = 54. Because we neglect the glass expansion, when the 300 mL cools to room temperature, the change in volume of the water will be ?V = V0 ?T = [21010–6 (C°)–1](300 mL)(20°C – 80°C) = – 3.8 mL (– 1.3%). 55. For the two conditions of the gas in the cylinder, we can write P1V = n1RT, and P2V = n2RT, which can be combined to give P2/P1 = n2/n1 ; (5 atm + 1 atm)/(35 atm + 1 atm)) = n2/n1 , which gives n2/n1 =

1/6.

56. The ideal gas law is PV = nRT = (m/M)RT, where m is the mass and M is the molecular weight. We write this as P = (m/V)RT/M = RT/M. 57. We use the ideal gas law: PV = NkT; (1.013105 Pa)(8.0 m)(6.0 m)(4.2 m) = N(1.3810–23 J/K)(293 K), 5.11027 molecules. which gives N = We find the number of moles from n = N/NA = (5.051027 molecules)/(6.021023 molecules/mol) =

8.4103 mol.

58. We use the ideal gas law: PV = NkT, or N/V = P/kT = (110–12 N/m2)/(1.3810–23 J/K)(273 K)(106 cm3/m3) =

3102 molecules/cm3.

59. The pressure at a depth h is P = P0 + gh = 1.013105 Pa + (1000 kg/m3)(9.80 m/s2)(10 m) = 1.99105 Pa. For the two states of the gas we can write PV = nRT, and P0V0 = nRT, which can be combined to give P/P0 = V0/V; (1.99105 Pa/1.013105 Pa) = (V0/5.5 L), which gives V0 = 11 L. This doubling of the volume is definitely not advisable.

Page 8

Chapter 17

60. When the rod has a length L, a small (differential) change in temperature will cause a small (differential) change in length: dL = LdT, or dL/L =  dT. (a) If we integrate this for a constant , we get L2

d L = T2  d T ; L1 L T1 L2 =  T 2 – T 1 , or L1 (b) If  = (T), we have ln

 (T 2 – T1)

L2 = L1e

.

L2

dL = T 2  (T) d T ; L1 L T1 ln

T2 L2 =  (T) d T , or L1 T1

T2

L2 = L 1 e

T1

(T) d T

.

(c) If  = 0 + bT, we have L2

d L = T 2 ( + bT) d T ; 0 L1 L T1 ln

L2 2 2 =  0 T 2 – T 1 + 12 b T 2 – T 1 , or L1

2 2  0(T 2 – T1) + 12b T 2 – T1

L 2 = L 1e

.

61. (a) The ideal gas law is PV = nRT = (m/M)RT; (1.013105 Pa)(770 m3) = [m/(29 g/mol)](8.315 J/mol · K)(293 K), 9.3102 kg. which gives m = 9.3105 g = (b) At the lower temperature we have (1.013105 Pa)(770 m3) = [m/(29 g/mol)](8.315 J/mol · K)(263 K), which gives m = 1.03106 g = 10.3102 kg. Thus the mass that has entered the house is 10.3102 kg – 9.3102 kg =

1.0102 kg.

62. (a) The ideal gas law is PV = nRT. For a small change in volume at constant pressure we have P dV = nR dT, or dV/V = dT/T. The thermal expansion is dV/V =  dT, so we see that  = 1/T. At 293 K we have  = 1/293 K = 3.4110–3 (C°)–1, which agrees with the value of 340010–6 (C°)–1 in Table 17–1. (b) When the pressure and volume change at constant temperature, we can differentiate the Page 9

Chapter 17

ideal gas law: P dV + V dP = 0 , or dV/V = – dP/P. From the definition of the bulk modulus, we have B = – dP/(dV/V) = – dP/(– dP/P) = P. 63. The pressure on a small area of the surface can be considered to be due to the weight of the air column above the area: P = Mg/A. When we consider the total surface of the Earth, we have Mtotal = PAtotal/g = P4pR2/g = (1.013105 Pa)4p(6.37106 m)2/(9.80 m/s2) = 5.271018 kg. If we use the average mass of an air molecule, we find the number of molecules from N = Mtotal/m = (5.271018 kg)/(28.8 u)(1.6610–27 kg) = 1.11044 molecules. 64. (a) The volume of the bulb is so much greater than the volume of mercury in the tube that we can ignore any changes in the tube dimensions. The additional length of the mercury column in the tube will be due to the increased expansion of the mercury in the bulb compared to the expansion of the glass bulb. The volume of mercury that adds to the length in the tube is ?V = ?Vmercury – ?Vglass = V0mercury ?T – V0glass ?T = V0(mercury – glass)?T = (0.315 cm3)[18010–6 (C°)–1 – 910–6 (C°)–1](33.0°C – 11.5°C) = 1.1610–3 cm3. We find the additional length from L = ?V/Atube = ?V/#pd2 = 4(1.1610–3 cm3)/p(0.0140 cm)2 = 7.52 cm. (b) If we combine the two expressions from part (a), we get L#pd2 = ?V = V0(mercury – glass)?T, which gives L = 4V0(mercury – glass)?T/pd2. 65. We find the molecular density from the ideal gas law: PV = NkT, or N/V = P/kT = (1.013105 Pa)/(1.3810–23 J/K)(273 K)(106 cm3/m3) = 2.691019 molecules/cm3. If we assume that each molecule occupies a cube of side a, we can find a, which is the average distance between molecules, from the volume occupied by a molecule: V/N = a3; 3.310–7 cm. 1/(2.691019 molecules/cm3) = a3, which gives a = 66. (a) If V0Fe is the volume of the iron and V0Hg is the volume of mercury that is displaced, the fraction of the volume of the iron that is submerged is f = V0Hg/V0Fe . Each volume will increase as the temperature is raised, so the new fraction will be f = V0Hg(1 + Hg ?T)/V0Fe(1 + Fe ?T) = f(1 + Hg ?T)/(1 + Fe ?T).

Because  ?T « 1, we use the approximation 1/(1 + Fe ?T) ˜ 1 – Fe ?T, so we have f = f(1 + Hg ?T)(1 – Fe ?T) = f [1 + Hg ?T – Fe ?T – HgFe(?T)2] , or (f  – f)/f = Hg ?T – Fe ?T – HgFe(?T)2. The last term is the product of two small numbers, so it can be neglected, and we have ?f/f = Hg ?T – Fe ?T = (Hg – Fe)?T. Because Hg > Fe , the fraction that is submerged will increase, so the cube will float lower. (b) For the percent change in the fraction submerged, we have (?f/f )(100) = (Hg – Fe)(?T)(100) = [18010–6 (C°)–1 – 3510–6 (C°)–1](25°C – 0°C)(100) = 0.36%.

67. We treat the circumference of the band as a length, which will expand according to 2pR = 2pR0(1 +  ?T), or Page 10

Chapter 17

R – R0 = R0 ?T = (6.38106 m)[1210–6 (C°)–1](35°C – 20°C) =

1.1103 m.

68. We consider a fixed mass of iron. The change in volume from the temperature change is ?VT = V0 ?T. The change in volume from the pressure change depends on the bulk modulus: ?VP = – V0/B) ?P. Because the density is mass/volume, for the fractional change in the density we have ?/0 = [(1/V) – (1/V0)]/(1/V0) = (V0 – V)/V ˜ (V0 – V)/V0 = – ?V/V0 = –  ?T + ?P/B = – [3510–6 (C°)–1](2000°C – 20°C) + (5000 atm)(1.013105 N/m2 · atm)/(90109 N/m2) = – 0.064 = – 6.4%.

69. As the oxygen expands into the atmosphere, we can find the volume it would occupy at atmospheric pressure. For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; [(1.013105 Pa)/(1.38107 Pa + 1.013105 Pa)](V2/16 L) = 1, which gives V2 = 2.20103 L. If we ignore the 16 L left in the tank, we find the time from t = V2/rate = (2.20103 L)/(2.4 L/min) = 915 min = 15 h. 70. We find the number of moles in the balloon from PV = nRT; (1.05 atm)(1.013105 Pa/atm))p(0.180 m)3 = n(8.315 J/mol · K)(293 K), which gives n = The mass of the helium is m = nM = (1.07 mol)(4 g/mol) = 4.27 g.

1.07 mol.

71. We consider a fixed mass of the substance: m = 0V0 = V = V0(1 +  ?T); 0 = (1 +  ?T);

0.68103 kg/m3 = {1 + [95010–6 (C°)–1](32°C – 0°C)}, which gives  =

0.66103 kg/m3.

72. We can treat the change in radius as a simple change in length, so we have R = R0(1 +  ?T). The difference in the radii of the lid and jar is ?R = Rbrass – Rglass = R0(1 + brass ?T) – R0(1 + glass ?T) = R0(brass – glass) ?T = (4.0 cm)[1910–6 (C°)–1 – 910–6 (C°)–1](60°C – 20°C) = 1.610–3 cm. 73. We find the temperature change that would produce a change in length equal to the accuracy: ?L = L0 ?T; ± 1.010–6 m = [9.010–6 (C°)–1](1.00 m) ?T, which gives ?T = ± 0.11 C°. 74. Because brass has a larger coefficient of expansion, a compressive stress will be created in the brass and a tensile stress will be created in the concrete so they have the same length. Thus the change in length from the two causes for the brass will be equal to the change in length for the concrete: (?Lthermal + ?Lstress)brass = (?Lthermal + ?Lstress)concrete. We assume the stress has the same magnitude in the brass and the concrete: Page 11

Chapter 17

brassL0 ?T – (stress)L0/Ebrass = concreteL0 ?T + (stress)L0/Econcrete , or (stress)[(1/Econcrete) + (1/Ebrass)] = (brass – concrete) ?T; (stress)[(1/20109 N/m2) – (1/100109 N/m2)] = [1910–6 (C°)–1 – 1210–6 (C°)–1](20 C°), which gives stress = 3.5106 N/m2. not stay in one piece. The ultimate tensile strength of the concrete is 2106 N/m2, so it will 75. (a) If P refers to the pressure in the atmosphere, for the two states of the gas we can write 1.05P0V0 = nRT0 and 1.05PV = nRT1 , which can be combined to give (P/P0)(V/V0) = T1/T0 . When we use the dependence of the pressure in the atmosphere on the altitude, P = P0 e –cy, where c = 0g/P0 , we get (P0 e –cy/P0)(V/V0) = T1/T0 , or V = V0(T1/T0) e +cy. (b) The density of the air is  = m/V, so /0 = V0/V = (P/P0)(T0/T1). The buoyant force at an altitude y is Fbuoy = gV = 0(P/P0)(T0/T1)gV0(T1/T0)e +cy = 0 e –cygV0 e +cy = 0gV0 = Fbuoy0 . 76. (a) For the two states of the gas we can write P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give (P2/P1)(V2/V1) = T2/T1 ; (1.00 atm/200 atm)(V2/11.3 L) = 1, which gives V2 = 2.26103 L. (b) On the surface, because the air expands as the person breathes, we have t = V2/rate = (2.26103 L)/(2.0 L/breath)(12 breaths/min) = 94 min = 1.6 h. (c) The pressure on the lungs at the depth of 20.0 m is P3 = P1 + gh = 1.00 atm + (1.025103 kg/m3)(9.80 m/s2)(20.0 m)/(1.013105 Pa/atm) = 2.98 atm. We find the volume the air would occupy at this pressure and temperature: (P3/P1)(V3/V1) = T3/T1 ; (2.98 atm/200 atm)(V2/11.3 L) = (283 K/293 K), which gives V2 = 7.33102 L. Thus the time is t = V3/rate = (7.33102 L)/(2.0 L/breath)(12 breaths/min) = 30 min. 77. The arc of the composite strip subtends an angle , as shown in the diagram. We take the average length of each strip, of thickness d, as the length of its centerline: Lbrass = (R + !d) = L0(1 + brass ?T); Lsteel = (R – !d) = L0(1 + steel ?T). If we divide the two equations, we get (R + !d)/(R – !d) = (1 + brass ?T)/(1 + steel ?T); [1 + (d/2R)]/[1 – (d/2R)] = (1 + brass ?T)/(1 + steel ?T).

Because d/2R « 1 and  ?T « 1, we can use the approximation, 1/(1 ± x) ˜ 1 — x: [1 + (d/2R)][1 + (d/2R)] = (1 + brass ?T)(1 – steel ?T). When we expand each side and keep only first-order terms, we have 1 + (d/R) = 1 + (brass – steel) ?T, or R = d/(brass – steel) ?T



= (2.010–3 m)/{[1910–6 (C°)–1] – [1210–6 (C°)–1]}(100°C – 20°C) =

Page 12

Steel R Brass

3.6 m.

Chapter 18

CHAPTER 18 – Kinetic Theory of Gases 1.

(a) The average kinetic energy depends on the temperature: !mvrms2 = *kT = *(1.3810–23 J/K)(273 K) = 5.6510–21 J. (b) For the total translational kinetic energy we have K = N(!mvrms2) = *nNAkT = *(2.0 mol)(6.021023 molecules/mol)(1.3810–23 J/K)(293 K) =

2.

3.

The average kinetic energy depends on the temperature: !mvrms2 = *kT, which gives vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(6000 K)/(4 u)(1.6610–27 kg/u)]1/2 = The average kinetic energy depends on the temperature: !mvrms2 = *kT. If we form the ratio for the two temperatures, we have (vrms2/vrms1)2 = T2/T1 = 373 K/273 K, which gives vrms2/vrms1 =

7.3103 J.

6.1103 m/s.

1.17.

4.

The average kinetic energy depends on the temperature: !mvrms2 = *kT. We form the ratio at the two temperatures: (vrms2/vrms1)2 = T2/T1 ; (2)2 = T2/293 K, which gives T2 = 1172 K = 899°C.

5.

(a) We find the mean speed from vrms = (?v)/N = (6 + 2 + 4 + 6 + 0 + 4 + 1 + 8 + 5 + 3 + 7 + 8)/12 = 4.5. (b) We find the rms speed from vrms = [(?v2)/N]1/2 = [(62 + 22 + 42 + 62 + 02 + 42 + 12 + 82 + 52 + 32 + 72 + 82)/12]1/2 = Note that this is greater than the mean speed of 4.5.

6.

The average kinetic energy depends on the temperature: !mvrms2 = *kT. We treat the small changes as differentials. We find the relationship between the changes by differentiating: mvrms dvrms = *k dT, or, after dividing by !mvrms2 = *kT, 2 dvrms/vrms = dT/T; 2(0.010vrms)/vrms = dT/293.2 K, which gives dT = 5.9 K. Thus the new temperature is T + dT = 293.2 K + 5.9 K = 299.1 K = 25.9°C.

7.

The average kinetic energy depends on the temperature: !mvrms2 = *kT. We form the ratio at the two temperatures, and use the ideal gas law: (vrms2/vrms1)2 = T2/T1 = P2V2/P1V1 = P2/P1 ; (vrms2/vrms1)2 = 2, which gives vrms2/vrms1 = v2.

8.

The average kinetic energy depends on the temperature: !mvrms2 = *kT, or kT/m = @vrms2. With M the mass of the gas and m the mass of a molecule, we write the ideal gas law as PV = NkT = (M/m)kT, or P = (M/V)(kT/m) = (@vrms2) = @vrms2, or vrms = (3P/)1/2.

Page 1

5.2.

Chapter 18

9.

The average kinetic energy depends on the temperature: !mvrms2 = *kT. We form the ratio at the two temperatures, and use the ideal gas law: (m2/m1)(vrms2/vrms1)2 = T2/T1 = 1, so we have (vrms2/vrms1)2 = (m1/m2), or vrms2/vrms1 = (m1/m2)1/2.

10. We use the ideal gas law to find the temperature: PV = nRT; (2.1 atm)(1.013105 Pa/atm)(8.5 m3) = (1300 mol)(8.315 J/mol · K)T, which gives T = 167 K. The average kinetic energy depends on the temperature: !mvrms2 = *kT, which gives vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(167 K)/(28 u)(1.6610–27 kg/u)]1/2 =

3.9102 m/s.

11. (a) The average kinetic energy depends on the temperature: !mvrms2 = *kT, which gives vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(273 K)/(32 u)(1.6610–27 kg/u)]1/2 = 461 m/s. (b) The molecule, on the average, will have a component in one direction less than the average speed. If we take the rms speed as the average speed, from the analysis of the molecular motion, we know that (vx2)av = @vrms2. Thus the time to go back and forth is t = 2¬/(vx)av ˜ 2¬v3/vrms . The frequency of collisions with one wall is N = 1/t = vrms/2¬v3 = (461 m/s)/2(7.0 m)v3 = 19 s–1. 12. We find the molecular density from the ideal gas law: PV = NkT, or N/V = P/kT = (1.013105 Pa)/(1.3810–23 J/K)(273 K)(106 cm3/m3) = 2.691019 molecules/cm3. If we assume that each molecule occupies a cube of side a, we can find a, which is the average distance between molecules, from the volume occupied by a molecule: V/N = a3; 3.310–7 cm. 1/(2.691019 molecules/cm3) = a3, which gives a = 13. The average kinetic energy depends on the temperature: !mvrms2 = *kT. We form the ratio for the two masses: (m235/m238)(vrms235/vrms238)2 = T2/T1 = 1, so we have (vrms235/vrms238)2 = (m238/m235), or

vrms235/vrms238 = (m238/m235)1/2 = {[6(19 u) + 238 u]/[6(19 u) + 235 u]}1/2 =

1.00429.

14. (a) We find the average speed from vrms = (?v)/N = [(2)(10 m/s) + (7)(15 m/s) + (4)(20 m/s) + (3)(25 m/s) + (6)(30 m/s) + (1)(35 m/s) + (2)(40 m/s)]/25 = 23 m/s. (b) We find the rms speed from vrms = [(?v2)/N]1/2 = [(2)(10 m/s)2 + (7)(15 m/s)2 + (4)(20 m/s)2 + (3)(25 m/s)2 + (6)(30 m/s)2 + (1)(35 m/s)2 + (2)(40 m/s)2]/25]1/2 = 25 m/s. Note that this is greater than the average speed of 23 m/s. (b) The speed with the largest number of particles is the most probable speed: 15 m/s.

Page 2

Chapter 18

15. (a) The total number of molecules is represented by the area under the Maxwell distribution, which we find by integration:  3/2 2 – mv 2/2 kT  (v f ) dv = 4N m v e dv . 2kT 0 This can be simplified by a change in variable: 1/2 1/2 u= m v; d u = m d v. 2kT 2kT The integral becomes  3/ 2 3/ 2  2  1/ 2 = N . (v f ) dv = 4 N 1 u 2 e –u d u = 4N 1   4 0 where we have used the result from integral tables. (b) We make the same change in variable for the integral to find the rms speed:  3/ 2 4 – 2 v 2 (v) f d v/ N = 4 m v e m v / 2kT d v 2kT 0

= 4 1 

3/ 2

2kT m

 0

4

u e

–u 2

d u = 4 1 

3/ 2

2kT 3 1/ 2 = 3kT . 8 m m

16. (a) From Figure 18–7 we see that at atmospheric pressure CO2 can exist as a (b) From Figure 18–7 we see that CO2 may be a liquid when 5.11 atm < P < 73 atm, and – 56.6°C < T < 31°C. 17. From Figure 18–7 we see that CO2 is a

vapor

18. (a) From Figure 18–6 we see that the phase is (b) From Figure 18–6 we see that the phase is

solid or vapor.

at 30 atm and 30°C. vapor. solid.

19. (a) From Figure 18–6 we see that water is a gas at 220 atm and 100°C. As we lower the pressure, it becomes a liquid at 218 atm, and a vapor at 1.0 atm. (b) Water is a gas at 220 atm and 0.0°C. As we lower the pressure, it becomes a liquid at 218 atm, a solid at 1.0 atm, and a vapor at a pressure < 0.006 atm. 20. The saturated vapor pressure at 25°C is 23.8 torr. At 50% humidity the partial vapor pressure is P = 0.50Ps = 0.50(23.8 torr) = 11.9 torr. This corresponds to a saturated vapor pressure between 10°C and 15°C. We assume a linear change between values listed in Table 18–2 and use the values at 10°C and 15°C to find the dew point; T = 10°C + [(15°C – 10°C)(11.9 torr – 9.21 torr)/(12.8 torr – 9.21 torr)] = 14°C. 21. Water boils when the saturated vapor pressure equals the air pressure. From Table 18–2 we see that the saturated vapor pressure at 90°C is 7.01104 Pa = 0.69 atm. 22. Water boils when the saturated vapor pressure equals the air pressure. From Table 18–2 we see that 0.85 atm = 8.6104 Pa lies between 90°C and 100°C. We use the values at 90°C and 100°C to find the temperature; T = 90°C + [(100°C – 90°C)(8.6104 Pa – 7.0104 Pa)/(10.1104 Pa – 7.0104 Pa)] = 95°C. 23. We find the saturated vapor pressure from Page 3

Chapter 18

P = (RH)Ps ; 530 Pa = 0.40Ps , which gives Ps = 1325 Pa. This saturated vapor pressure corresponds to a temperature between 10°C and 15°C. We use the values at 10°C and 15°C to find the temperature; T = 10°C + [(15°C – 10°C)(1.325103 Pa – 1.23103 Pa)/(1.71103 Pa – 1.23103 Pa)] = 11°C. 24. From Table 18–2 we see that the saturated vapor pressure at 25°C is 3.17103 Pa. At 35% humidity the partial vapor pressure is P = 0.35Ps = 0.35(3.17103 Pa) = 1.11103 Pa. 25. Water boils when the saturated vapor pressure equals the air pressure. From Table 18–2 we see that the saturated vapor pressure at 120°C is 1.96 atm. 1.99105 Pa = 26. From Table 18–2 we see that the saturated vapor pressure at 25°C is 3.17103 Pa. Water can evaporate until the saturated vapor pressure is reached. The initial pressure is (relative humidity)(saturated vapor pressure). Because the volume and temperature are constant, we use the ideal gas law to find the number of moles that can evaporate: ?n = ?P(V/RT) = (1 – RH)PsV/RT = (1 – 0.80)(3.17103 Pa)(240 m3)/(8.315 J/mol · K)(298 K) = 61.4 mol. We find the mass from m = M ?n = (18 g/mol)(61.4 mol) = 1.1103 g = 1.1 kg. 27. Because there is only steam in the autoclave, the saturated vapor pressure is the gauge pressure plus atmospheric pressure: 1.0 atm + 1.0 atm = 2.0 atm = 2.03105 Pa. From Table 18–2 we see this saturated vapor pressure occurs at 120°C. 28. Because the outside air is at the dew point, its vapor pressure is the saturated vapor pressure at 5°C, which is 872 Pa. We consider a constant mass of gas, that is a fixed number of moles, that moves from outside to inside. Because the pressure is constant, we have V2/T2 = V1/T1 . The vapor pressure inside is P2 = nRT2/V2 = nRT1/V1 = P1 = 872 Pa. The saturated vapor pressure at 25°C is 3170 Pa, so the relative humidity is (872 Pa)/(3170 Pa) = 0.28 = 28%. 29. (a) The molar density is V/n = (0.40 L)(10–3 m3/L)/(1.0 mol) = 0.4010–3 m3/mol. We find the pressure from the Van der Waals equation: [P + a/(V/n)2][(V/n) – b] = RT; [P + (0.14 N · m4/mol2)/(0.4010–3 m3/mol)2](0.4010–3 m3/mol – 3.210–5 m3/mol) = (8.315 J/mol · K)(273 K), which gives P = 5.3106 Pa. (b) We find the pressure from the ideal gas equation: P(V/n) = RT; P(0.4010–3 m3/mol) = (8.315 J/mol · K)(273 K), which gives P = 5.7106 Pa. 30. If we consider one mol of oxygen, the volume occupied by the molecules is b, so the volume occupied by one molecule is b/NA. If we assume the molecule is a sphere of radius r, we have b/NA = )pr3; (3.210–5 m3/mol)/(6.021023 molecules/mol) = )pr3, 4.610–10 m. which gives r = 2.310–10 m, so d = Page 4

Chapter 18

Note: see Problem 32 for a refinement of this calculation.

31. (a) We write the Van der Waals equation as P = – (an2/V2) + nRT/(V – nb). When we differentiate, with the temperature constant, we get dP/dV = (2an2/V3) – nRT/(V – nb)2. At the critical point, we have dP/dV = 0 = (2an2/Vcr3) – nRTcr/(Vcr – nb)2 = [2an2(Vcr – nb)2 – nRTcrVcr3]/Vcr3(Vcr – nb)2, so 2an(Vcr – nb)2 = RTcrVcr3. For the second derivative we get d2P/dV2 = – (6an2/V4) + 2nRT/(V – nb)3. Because the critical point is an inflection point, we have d2P/dV2 = 0 = – (6an2/Vcr4) + 2nRTcr/(Vcr – nb)3 = – [6an2(Vcr – nb)3 – 2nRTcrVcr4]/Vcr4(Vcr – nb)3, so 3an(Vcr – nb)3 = RTcrVcr4. When we divide the two equations, we get 3(Vcr – nb) = 2Vcr , or Vcr = 3nb. If we use this result in one of the equations, we get 2an(3nb – nb)2 = RTcr(3nb)3, which gives Tcr = 8a/27bR. For the pressure at the critical point, we have Pcr = – (an2/Vcr2) + nRTcr/(Vcr – nb) = – [an2/(3nb)2] + nR(8a/27bR)/(3nb – nb) = a/27b2. (b) We find b from Tcr/Pcr = (8a/27bR)/(a/27b2) = 8b/R; (304 K)/(72.8 atm)(1.013105 N/m2 · atm) = 8b/(8.315 J/mol · K), b = 4.2810–5 m3/mol. which gives We find a from Tcr2/Pcr = (8a/27bR)2/(a/27b2) = 64a/27R2; (304 K)2/(72.8 atm)(1.013105 N/m2 · atm) = 64a/27(8.315 J/mol · K)2, a = 0.365 N · m4/mol2. which gives 32. (a) If we consider the relative motion of one molecule with respect to the other, we see that if the direction of motion of the center of one molecule lies on a line that is within 2r of the center of the other molecule, there will be a collision. Thus the center of one molecule is excluded from a sphere of radius 2r. (b) If we consider one mol (NA molecules) of gas, the effective volume occupied by the molecules is b. From part (a) we see that each pair of molecules occupies an effective volume equivalent to a sphere of radius 2r. Thus we have b = ![)p(2r)3NA] = 16pr3NA/3. (c) For CO2 we have 4.210–5 m3/mol. = 16pr3(6.021023 molecules/mol)/3, 3.210–10 m. which gives r = 1.610–10 m, so d =

r

33. (a) If we use the ideal gas law, PV = NkT, in the expression for the mean free path, we have ¬M = 1/4pv2r2(N/V) = kT/4pv2r2P; 1.0 m = (1.3810–23 J/K)(273 K)/4pv2(1.510–10 m)2P, 10–7 atm. which gives P = 9.410–3 Pa ˜ Page 5

r

Chapter 18

(b) For a mean free path equal to a diameter, we have ¬M = 1/4pv2r2(N/V) = kT/4pv2r2P; 310–10 m = (1.3810–23 J/K)(273 K)/4pv2(1.510–10 m)2P, 300 atm. which gives P = 3.1107 Pa ˜

34. (a) If we use the ideal gas law, PV = NkT, in the expression for the mean free path, we find the radius from ¬M = 1/4pv2r2(N/V) = kT/4pv2r2P; 5.610–8 m = (1.3810–23 J/K)(273 K)/4pv2r2(1.013105 Pa), 3.910–10 m. which gives r = 1.9310–10 m, so d = (b) For helium we have ¬M = 1/4pv2r2(N/V) = kT/4pv2r2P; 2510–8 m = (1.3810–23 J/K)(273 K)/4pv2r2(1.013105 Pa), 1.810–10 m. which gives r = 9.1510–11 m, so d = 35. The length of a side of the cube is L = V1/3 = (4.410–3 m3)1/3 = 0.164 m. (a) We assume that vigorous shaking will disperse the marbles throughout the box. Because the diameter of a marble ˜ 0.1L and there are on average about four marbles along a side, we use the approximation of an ideal gas: ¬M = 1/4pv2r2(N/V) = 1/4pv2(0.7510–2 m)2(70/4.410–3 m3) = 6.2910–2 m = 6.3 cm. (b) If the box is slightly shaken , all marbles will be on the floor of the box. There are on average about eight marbles along a side. Because 8(1.5 cm) = 12 cm < L, there will be space between marbles. The volume occupied by the marbles is Vb = A(2r) = 2L2r. We assume we can still use the expression for the mean free path: ¬Mb = 1/4pv2r2(N/Vb) = 2L2r/4pv2r2N = v2L2/4prN = v2(0.164 m)2/4p(0.7510–2 m)(70) = 5.810–3 m = 0.58 cm. 36. If we use the ideal gas law, PV = NkT, in the expression for the mean free path, we have ¬M = 1/4pv2r2(N/V) = kT/4pv2r2P; = (1.3810–23 J/K)(273 K)/4pv2(1.510–10 m)2(10–6 torr)(1.013105 Pa/760 torr) = 71 m. The time between collisions is proportional to the distance between collisions. If we take the length of a side of the box for the distance between collisions with a wall (it will actually be slightly greater than this because a molecule on the average will not be moving perpendicular to a wall), we have molecular collisions/wall collisions = twall/tmolecule = L/¬M = (1.20 m)/(71 m) = 1.710–2 ˜ 1/60. 37. We compare the rms speed of a hydrogen molecule with the rms speed of an air molecule: vrms,H/vrms,air = (3kT/MH)1/2/(3kT/Mair)1/2 = (Mair/MH)1/2 = [(29 g/mol)/(2 g/mol)]1/2 = 3.8. Because the hydrogen molecules are moving much faster than the air molecules, we assume stationary targets with an effective radius of rH + rair. Thus the mean free path is ¬M = 1/p(rH + rair)2(N/V) = kT/p(rH + rair)2P = (1.3810–23 J/K)(298 K)/p(1.010–10 m + 1.510–10 m)2(1.013105 Pa) = 210–7 m. 38. The electrons are moving much faster than the air molecules and are much smaller. Thus we assume stationary targets with an effective radius of rair. The probability of collision is proportional to the distance the electron travels. The mean free path is the distance in which the probability of collision is 1. If we want a collision probability of 2% in the length of the cathode ray tube, we have ¬M/1 = L/0.02. Thus we have Page 6

Chapter 18

¬M = L/0.02 = 1/prair2(N/V) = kT/prair2P; (0.38 m)/0.02 = (1.3810–23 J/K)(300 K)/p(1.510–10 m)2P, 3.110–3 Pa = 2.310–5 torr. which gives P = Note that we have assumed room temperature.

39. (a) If the average speed of a molecule is æ, the average time between collisions is ?t = ¬M/æ, so the frequency of collisions is f = 1/?t = æ/¬M. When we use the expression for the mean free path, we get f = æ/[1/4pv2r2(N/V)] = 4v2 pr2æN/V. (b) We make use of the ideal gas law and the Maxwell distribution to get f = 4v2 pr2æN/V = 4v2 pr2(8kT/pm)1/2P/kT = 16r2P(p/kTm)1/2 = 16(1.510–10 m)2(1.010–2 atm)(1.013105 Pa)  [p/(1.3810–23 J/K)(293 K)(28 u)(1.6610–27 kg/u)]1/2 = 4.7107 s–1. 40. From the Maxwell distribution, the average speed of an air molecule is æ = (8kT/pm)1/2 = [8(1.3810–23 J/K)(273 K)/p(29 u)(1.6610–27 kg/u)]1/2 = 446 m/s. From Problem 39, the collision frequency is f = æ/¬M = (446 m/s)/(910–8 m) = 5109 s–1. 41. For collisions of a type 1 molecule with type 1 molecules the volume of the cylinder along the path is p(2r1)2æ?t. For collisions of a type 1 molecule with type 2 molecules the volume of the cylinder along the path is p(r1 + r2)2æ ?t. The number of collisions of either type in a time ?t is # = n1p4r12æ ?t + n2p(r1 + r2)2æ ?t. Thus the mean free path is ¬M = æ ?t/# = 1/[4pr12 n1 + p(r1 + r2)2n2]. 42. The probability that a molecule will collide in a small distance dx is dx/¬M. At an instant when we have N molecules, the average number that will collide in a distance dx is N dx/¬M. Because every collision removes a molecule from the survivors, we have N dx/¬M = – dN, or dN/N = – dx/¬M. When we integrate, we get N

x

dN = – dx ; N0 N 0 ¬M –x/ ¬M ln N – ln N 0 = – x , or N = N 0 e . ¬M

43. From the result of Example 18–8, we have 2 2 (1.5m) t = C (x ) = 1 = 2.8  104 / s = 7.8 h. – 5 2 4.0  10 m 2 / s C D Our experience is that the odor is detected much sooner than this, which means that convection is much more important than diffusion. 44. From the result of Example 18–8, we have

Page 7

Chapter 18 2

–6

2

C (x ) = (1.00mol / m 3 + 0.40 mol / m 3 )/ 2 (15  10 m ) = 0.28 s. C D (1.00mol / m 3 – 0.40 mol / m 3 ) 95  10– 11 m 2/ s The diffusion speed is v = ?x/t = (1510–6 m)/(0.28 s) = 5.410–5 m/s. We find the rms speed from vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(293 K)/(75 u)(1.6610–27 kg/u)]1/2 = t=

3.1102 m/s.

45. (a) From the ideal gas law, we have C0 = n/V = P/RT = (0.21)(1.013105 Pa)/(8.315 J/mol · K)(293 K) = 8.7 mol/m3. (b) The concentration change is ?C = C0 – !C0 = !C0 = !(8.7 mol/m3) = 4.35 mol/m3. We find the diffusion rate from J = DA ?C/?x = (110–5 m2/s)(210–9 m2)(4.35 mol/m3)/(210–3 m) = 410–11 mol/s. (c) For the average concentration we have Cav = !(C0 + !C0) = &C0 = &(8.7 mol/m3) = 6.53 mol/m3. We find the average time from t = N/J = CavV/J = (6.53 mol/m3)(210–9 m2)(210–3 m)/(410–11 mol/s) = 0.6 s. 46. We use the ideal gas law to find the temperature: PV = nRT; (3.42 atm)(1.013105 Pa/atm)(12.8 m3) = (1800 mol)(8.315 J/mol · K)T, which gives T = 296 K. The average kinetic energy depends on the temperature: !mvrms2 = *kT, which gives vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(296 K)/(28 u)(1.6610–27 kg/u)]1/2 = 47. The average kinetic energy depends on the temperature: !mvrms2 = *kT, which gives vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(2.7 K)/(1 u)(1.6610–27 kg/u)]1/2 = We find the pressure from PV = NkT, or P = (N/V)kT = (1 atom/cm3)(106 cm3/m3)(1.3810–23 J/K)(2.7 K) = 410–17 N/m2 ˜ 410–22 atm.

514 m/s.

2.6102 m/s.

48. We find the total number of molecules from the total number of moles of water plus others: N = {[(0.70)/(18 g/mol)](6.021023 molecules/mol) + [(0.30)/(105 u)(1.6610–24 g/u)]}(2.010–12 g) = 4.71010 molecules. Because each molecule has an average kinetic energy of *kT, the total translational kinetic energy is K = *NkT = *(4.71010 molecules)(1.3810–23 J/K)(310 K) = 3.010–10 J. 49. (a) We find the rms speed from vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(310 K)/(89 u)(1.6610–27 kg/u)]1/2 = 2.9102 m/s. (b) For the protein we have vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(310 K)/(50,000 u)(1.6610–27 kg/u)]1/2 = 12 m/s. 50. (a) We find the temperature from æ = (8kT/pm)1/2 ; Page 8

Chapter 18

1.12104 m/s = [8(1.3810–23 J/K)T/p(32 u)(1.6610–27 kg/u)]1/2, which gives T= 1.90105 K. (b) For helium atoms we have æ = (8kT/pm)1/2 ; 1.12104 m/s = [8(1.3810–23 J/K)T/p(4 u)(1.6610–27 kg/u)]1/2, which gives T= 2.37104 K.

51. We find the average volume occupied by a molecule from the ideal gas law: PV = NkT, or V/N = kT/P = (1.3810–23 J/K)(273 K)/(1.013105 Pa) = 3.710–26 m3/molecule. The volume of a molecule is Vmolecule = )pr3 = )p(0.110–9 m)3 = 410–30 m3 ˜ 0.01% of average volume occupied by a molecule. Thus the assumption is reasonable. If we consider the average volume occupied by a molecule to be a cube, the side of the cube (which we can use as the average distance between molecules) is ¬ = (3.710–26 m3)1/3 = 3.310–9 m. If we scale the diameter of a molecule to the size of a ping-pong ball, for the average distance between molecules we have D = [(4 cm)/(0.210–9 m)](3.310–9 m) ˜ 70 cm. 52. If we consider one molecule moving in the container, the time between collisions with a particular wall, say one perpendicular to the x-axis, is ?t = 2L/vx. If we approximate the average component from vx2 = @vrms2 = @(3kT/m) = kT/m, for the frequency of collision we have f = 1/?t = (kT/m)1/2/2L. Because this will be the average frequency for all the molecules, we have ftotal = nNAf = nNA(kT/m)1/2/2L = (2.0 mol)(6.021023 molecules/mol)  [(1.3810–23 J/K)(293 K)/(32 u)(1.6610–27 kg/u)]1/2/2(1.0 m) =

1.71026 s–1.

53. The gravitational potential energy of a molecule at the top of the box is mgh = (32 u)(1.6610–27 kg/u)(9.80 m/s2)(0.50 m) = 2.610–25 J. The average kinetic energy of a molecule is !mvrms2 = *kT = *(1.3810–23 J/K)(293 K) = 6.110–21 J. Thus the ratio is mgh/!mvrms2 = (2.610–25 J)/(6.110–21 J) = 4.310–5, or mgh = 4.310–5(!mvrms2) . It is reasonable to neglect the gravitational potential energy. 54. From Table 18–2 we see that the saturated vapor pressure at 20°C is 2.33103 Pa. The vapor pressure is (relative humidity)(saturated vapor pressure). Because the volume and temperature are constant, we use the ideal gas law to find the number of moles that must be removed: ?n = ?P(V/RT) = (RH2 – RH1)PsV/RT = (0.30 – 0.95)(2.33103 Pa)(85 m2)(2.8 m)/(8.315 J/mol · K)(293 K) = – 148 mol. We find the mass from m = M ?n = (18 g/mol)(148 mol) = 2.7103 g = 2.7 kg.

Page 9

Chapter 18

55. For the two conditions of the gas in the cylinder, we can write P1V = nRT1 , and P2V = nRT2 , which can be combined to give P2/P1 = T2/T1 ; P2/P1 = (563 K/393 K) = 1.43. The average kinetic energy depends on the temperature: !mvrms2 = *kT. We form the ratio for the two temperatures: (vrms2/vrms1)2 = T2/T1 = (563 K/393 K), which gives vrms2/vrms1 =

1.20.

56. (a) The volume of each gas is !Vtank. We use the ideal gas law: PVO = NOkT; (10 atm + 1 atm)(1.013105 Pa/atm)!(2800 cm3)/(106 cm3/m3) = NO(1.3810–23 J/K)(293 K), which gives NO = 3.91023 molecules = NHe , because the molecular mass is not in the ideal gas law. (b) The average kinetic energy depends on the temperature: !mvrms2 = *kT. Because the gases are at the same temperature, the ratio of average kinetic energies is 1. (c) We see that 2.8. (vrmsHe/vrmsO)2 = mO/mHe = (32 u)/(4 u), which gives vrmsHe/vrmsO = 57. The rms speed is the speed of the nitrogen molecules, so we have !mvrms2 = *kT; (28 u)(1.6610–27 kg/u)[(40,000 km/h)/(3.6 ks/h)]2 = 3(1.3810–23 J/K)T, 1.4105 K. which gives T = 58. If we neglect the thermal energy of the water molecules in the liquid, the energy required to evaporate the water becomes the kinetic energy of the molecules in the vapor. If mevap is the evaporated mass and m is the mass of a molecule, we have E = N(!mv2) = (mevap/m)(!mv2) = !mevapv2; 2.45103 J = !(1.0010–3 kg)v2, which gives v = 2.2103 m/s. The rms speed is vrms = (3kT/m)1/2 = [3(1.3810–23 J/K)(293 K)/(18 u)(1.6610–27 kg/u)]1/2 = 6.4102 m/s. Thus we have v/vrms = (2.2103 m/s)/(6.4102 m/s) = 3.5. 59. (a) From Table 18–2 the saturated vapor pressure at 30°C is 4.24103 Pa. At 40% humidity the water vapor pressure is P = 0.40Ps = 0.40(4.24103 Pa) = 1.7103 Pa. (b) From Table 18–2 the saturated vapor pressure at 5°C is 8.72102 Pa. At 80% humidity the water vapor pressure is P = 0.80Ps = 0.80(8.72102 Pa) = 7.0102 Pa. The ratio of summer to winter is 2.4. 60. We find the pressure from the ideal gas equation: PidealV = nRT; Pideal(0.200 m3) = (8.50 mol)(8.315 J/mol · K)(300 K), which gives We find the pressure from the Van der Waals equation: [PVW + a/(V/n)2][(V/n) – b] = RT; Page 10

Pideal = 1.060105 Pa.

Chapter 18

{PVW + (3.610–3 N · m4/mol2)/[(0.200 m3)/(8.50 mol)]2}[(0.200 m3)/(8.50 mol) – 4.210–5 m3/mol] =

(8.315 J/mol · K)(300 K), which gives PVW = 1.062105 Pa. The error is [(Pideal – PVW)/PVW]100 = [(1.060105 Pa. – 1.062105 Pa)/(1.062105 Pa)]100 = – 0.2%. 61. The mean free path is ¬M = 1/4pv2r2(N/V) = 1/4pv2[!(1.010–10 m)]2[1 atom/(110–6 m3)] =

21013 m.

62. If we use the ideal gas law, PV = NkT, in the expression for the mean free path, we have ¬M = 1/4pv2r2(N/V) = kT/4pv2r2P. For the given data we have ¬M = (1.3810–23 J/K)(300 K)/4pv2(1.510–10 m)2(10 atm)(1.013105 Pa/atm) = 1.010–8 m.

Page 11

Chapter 19

CHAPTER 19 – Heat and the First Law of Thermodynamics 1.

The required heat flow is ?Q = mc ?T = (30.0 kg)(4186 J/kg · C°)(95°C – 15°C) =

2.

We find the temperature from ?Q = mc ?T; 7700 J = (3.0 kg)(4186 J/kg · C°)(T – 10.0°C), which gives T =

3.

1.0107 J.

10.6°C.

The heat flow generated must equal the kinetic energy loss: ?Q = – (!mvf2 – !mvi2) = !m(vi2 – vf2) = !(3.010–3 kg)[(400 m/s)2 – (200 m/s)2] =

1.8102 J.

4.

We convert the units: ?Q = mc ?T; 1 Btu = (1 lb)(1.00 kcal/kg · C°)(1 F°)(0.454 kg/lb)(5 C°/9 F°) = 0.252 kcal; 1 Btu = (0.252 kcal)(4186 J/kcal) = 1055 J.

5.

We find the mass per hour from ?Q/t = (m/t)c ?T; 7200 kcal/h = (m/t)(1.00 kcal/kg · C°)(50°C – 15°C), which gives m/t =

6.

7.

8.

9.

2.1102 kg/h.

We find the time from ?Q = mc ?T; (350 W)t = (250 mL)(1.00 g/mL)(10–3 kg/g)(4186 J/kg · C°)(60°C – 20°C), which gives t = The heat flow generated must equal the kinetic energy loss: ?Q = !mv2 = !(1000 kg)[(95 km/h)/(3.6 ks/h)]2(1 kcal/4186 J) = We find the specific heat from ?Q = mc ?T; 135103 J = (5.1 kg)c (30°C – 20°C), which gives c =

120 s.

83 kcal.

2.6103 J/kg · C°.

The required heat flow is ?Q = mc ?T = (16 L)(1.00 kg/L)(4186 J/kg · C°)(90°C – 20°C) =

4.7106 J.

10. We find the temperature from heat lost = heat gained; mwatercwater ?Twater = mglasscglass ?Tglass; (135 mL)(1.00 g/mL)(10–3 kg/g)(4186 J/kg · C°)(T – 39.2°C) = (0.035 kg)(840 J/kg · C°)(39.2°C – 21.6°C), which gives T = 40.1°C. 11. If all the kinetic energy in the hammer blows is absorbed by the nail, we have K = 10(!mv2) = mc ?T; 40 C°. 10[!(1.20 kg)(6.5 m/s)2] = (0.014 kg)(450 J/kg · C°) ?T, which gives ?T = 12. We find the temperature from heat lost = heat gained; mCucCu ?TCu = (mAlcAl + mwatercwater) ?TAl ; (0.245 kg)(390 J/kg · C°)(300°C – T) = [(0.150 kg)(900 J/kg · C°) + (0.820 kg)(4186 J/kg · C°)](T – 12.0°C), which gives T =

Page 1

19.5°C.

Chapter 19

13. We find the temperature from heat lost = heat gained; mshoecshoe ?Tshoe = (mpotcpot + mwatercwater) ?Tpot ; (0.40 kg)(450 J/kg · C°)(T – 25°C) = [(0.30 kg)(450 J/kg · C°) + (1.35 L)(1.00 kg/L)(4186 J/kg · C°)](25°C – 20°C), which gives T = 186°C. 14. We find the specific heat from heat lost = heat gained; mxcx ?Tx = (mAlcAl + mwatercwater + mglasscglass) ?Twater ; (0.215 kg)cx (330°C – 35.0°C) = [(0.100 kg)(900 J/kg · C°) + (0.150 kg)(4186 J/kg · C°) + (0.017 kg)(840 J/kg · C°)](35.0°C – 12.5°C), 260 J/kg · C° . which gives cx = 15. The water must be heated to the boiling temperature, 100°C. We find the time from t = (heat gained)/P = [(mAlcAl + mwatercwater) ?Twater]/P = [(0.360 kg)(900 J/kg · C°) + (0.75 L)(1.00 kg/L)(4186 J/kg · C°)](100°C – 8.0C)/(750 W) = 425 s = 7.1 min. 16. The Calorie content of 100 g of brownie will be 10 times the thermal energy released when 10 g is ignited: Q = 10(heat gained) = 10(mbombcAl + mwatercwater + mcupcAl) ?T = 10[(0.615 kg)(0.22 kcal/kg · C°) + (2.00 kg)(1.00 kcal/kg · C°) + (0.524 kg)(0.22 kcal/kg · C°)](36.0°C – 15.0°C) = 470 kcal. 17. (a) The heat required at a temperature T to raise the temperature by a small amount dT is dQ = mc(T) dT. We add the total heat required to raise the temperature from T1 to T2 by integrating:

Q=

Q 0

dQ =

T2 T1

mc(T) d T .

(b) For the given relationship, we get

Q=

T2 T1

mc0(1+ aT) d T =

mc0 (T2 – T 1) + a(T22 – T12)/ 2 .

(c) For the mean value of c, we have Q = mcmean (T2 – T1) = mc0[(T2 – T1) + a(T22 – T12)/2] = mc0(T2 – T1)[1 + a(T2 – T1)(T2 + T1)/2(T2 – T1)] , so we get cmean = c0[1 + !a(T2 + T1)]. 18. The silver must be heated to the melting temperature, 961°C, and then melted. We find the heat required from Q = msilvercsilver ?Tsilver + msilverLsilver = (15.50 kg)(230 J/kg · C°)(961°C – 20° C) + (15.50 kg)(0.88105 J/kg) = 4.7106 J. 19. If we assume that the heat is required just to evaporate the water, we have Q = mwaterLwater ; 180 kcal = mwater(539 kcal/kg), which gives mwater = 0.334 kg (0.334 L). 20. Because the liquid oxygen is at the boiling point, there will be only a change in phase: Q = moxygenLoxygen; 2.80105 J = moxygen(2.1105 J/kg), which gives moxygen = 1.3 kg.

Page 2

Chapter 19

21. We assume all the ice melts and all the steam condenses, so we have only water. With msteam = mice = m, we have heat lost = heat gained; mLs + mcwater ?Tsw = mLi + mci ?Tiw; 22.6105 J/kg + (4186 J/kg · C°)(100°C – T) = 3.33105 J/kg + (4186 J/kg · C°)(T – 0°C), which gives T = 280°C. Because we can not have all water at this temperature, our initial assumption is wrong. We must have some water and some steam at 100°C. If the amount of steam that has condensed is xm, we have heat lost = heat gained; xmLs = mLi + mci ?Tiw; x(22.6105 J/kg) = 3.33105 J/kg + (4186 J/kg · C°)(100°C – 0°C), which gives x = 0.333. %m steam and )m water at 100°C. Thus we have 22. We assume there is enough liquid nitrogen that the temperature of the liquid nitrogen will not change, so the ice will cool to 77 K. We find the amount of nitrogen that has evaporated from heat lost = heat gained; micecice ?Tice = mnitrogenLnitrogen ; (0.040 kg)(2100 J/kg · C°)(273 K – 77 K) = mnitrogen(200103 J/kg), which gives mnitrogen = 0.082 kg. Note that a change of 1 K is equal to a change of 1 C°. 23. The temperature of the ice will rise to 0°C, at which point melting will occur, and then the resulting water will rise to the final temperature. We find the mass of the ice cube from heat lost = heat gained; (mAlcAl + mwatercwater) ?TAl = mice(cice ?Tice + Lice + cwater ?Twater); [(0.075 kg)(900 J/kg · C°) + (0.300 kg)(4186 J/kg · C°)](20°C – 17°C) = mice{(2100 J/kg · C°)[0°C – (– 8.5°C)] + (3.33105 J/kg) + (4186 J/kg · C°)(17°C – 0°C)}, which gives mice = 9.410–3 kg = 9.4 g. 24. (a) We find the heat required to reach the boiling point from Q1 = (mFecFe + mwatercwater)?T = [(230 kg)(450 J/kg · C°) + (760 kg)(4186 J/kg · C°)](100°C – 20° C) = 2.63108 J. We find the time from t1 = Q1/P = (2.63108 J)/(52,000103 J/h) = 5.1 h. (b) As the water changes to steam, there is no change in the temperature, so the additional heat required to change the water into steam is Q2 = mwaterLsteam = (760 kg)(22.6105 J/kg) = 1.72109 J. We find the additional time from t2 = Q2/P = (1.72109 J)/(52,000103 J/h) = 33.0 h. Thus the total time required is t = t1 + t2 = 5.1 h + 33.0 h = 38.1 h. 25. We use the heat of vaporization at body temperature: 585 kcal/kg. If all of the energy supplied by the bicyclist evaporates the water, we have Q = mwaterLwater = (8.0 L)(1.00 kg/L)(585 kcal/kg) = 4.7103 kcal.

Page 3

Chapter 19

26. The steam will condense at 100°C, and then the resulting water will cool to the final temperature. The ice will melt at 0°C, and then the resulting water will rise to the final temperature. We find the mass of the steam required from heat lost = heat gained; msteam(Lsteam + cwater ?T1) = mice(Lice + cwater ?T2); msteam[(22.6105 J/kg) + (4186 J/kg · C°) (100°C – 20°C)] = (1.00 kg)[(3.33105 J/kg) + (4186 J/kg · C°)(20°C – 0°C)], which gives msteam = 0.16 kg. 27. We find the latent heat of fusion from heat lost = heat gained; (mAlcAl + mwatercwater) ?Twater = mHg(LHg + cHg ?THg) [(0.620 kg)(900 J/kg · C°) + (0.430 kg)(4186 J/kg · C°)](12.80°C – 5.06°C) = (1.00 kg){LHg + (138 J/kg · C°)[5.06°C – (– 39.0°C)]}, 4 which gives LHg = 1.2210 J/kg. 28. We assume that the water created by the melting of the ice stays at 0°C. Because the work done by friction, which decreases the kinetic energy, generates the heat flow, we have Q = !(?K); miceLice = !(!mv2); mice(3.33105 J/kg) = #(54.0 kg)(4.8 m/s)2, which gives mice = 9.310–4 kg =

0.93 g.

29. The work done by friction, which decreases the kinetic energy, generates the heat flow. In general a fraction of the heat flow is used to raise the temperature of the lead bullet and then melt the lead bullet. The larger this fraction, the smaller the bullet velocity needed. We determine the minimum muzzle velocity by assuming that this fraction is 1: Q = ?K; mlead(clead ?T + Llead) = !mleadvmin2; [(130 J/kg · C°)(327°C – 20°C) + 0.25105 J/kg] = !vmin2, which gives vmin =

360 m/s.

30. We use the first law of thermodynamics to find the change in internal energy: ?U = Q – W = – 2.78103 J – (– 1.6103 J) = – 1.2103 J. 31. (a) The internal energy of an ideal gas depends only on the temperature, U = *nRT, so we have 0. ?U = *nR ?T = (b) We use the first law of thermodynamics to find the heat absorbed: ?U = Q – W; 5.00103 J. 0 = Q – 5.00103 J , which gives Q = 32. For the isothermal process, we have PAVA = PBVB ; (6.0 atm)(1.0 L) = (1.0 atm)VB , which gives VB = 6.0 L.

P (atm) A

6.0

1.0

C

0 1.0

Page 4

B 6.0

V (L)

Chapter 19

33. For the isothermal process, we have PBVB = PCVC ; (1.0 atm)(1.0 L) = PC (2.0 L), which gives PC = 0.5 atm.

P (atm) B

1.0 0.5

A

C 0

1.0

2.0

V (L)

34. (a) Because the pressure is constant, we find the work from W = P(V2 – V1) = (1.013105 N/m2)(18.2 m3 – 12.0 m3) = 6.3105 J. (b) We use the first law of thermodynamics to find the change in internal energy: ?U = Q – W = + (1400 kcal)(4186 J/kcal) – 6.3105 J = 5.23106 J. 35. (a) Because there is no change in volume, we have W= 0. (b) We use the first law of thermodynamics to find the change in internal energy: ?U = Q – W = – 1300 kJ – 0 = – 1300 kJ. 36. (a) In an adiabatic process there is no heat flow: Q= 0. (b) We use the first law of thermodynamics to find the change in internal energy: ?U = Q – W = 0 – (– 2350 J) – 0 = + 2350 J. (c) The internal energy of an ideal gas depends only on the temperature, U = *nRT, so we see that an increase in the internal energy means that the temperature must rise. 37. (a) Work is done only in the constant pressure process: W = Pa(Vb – Va) = (5.0 atm)(1.013105 N/m2 · atm)[(0.710 – 0.400)10–3 m3] = 1.6102 J. (b) Because the initial and final temperatures are the same, we know that ?U = 0. We use the first law of thermodynamics to find the heat flow for the entire process: ?U = Q – W; + 1.6102 J. 0 = Q – (+ 1.6102 J), which gives Q =

Page 5

P (atm) 5.0

a

b

c

0

400

710

V (mL)

Chapter 19

38. (a) Work is done only in the constant pressure process: W = Pb(Vc – Vb) = (1.5 atm)(1.013105 N/m2 · atm)[(10.0 – 6.8)10–3 m3] = 4.9102 J. (b) Because the initial and final temperatures are the same, we know that ?U = 0. (c) We use the first law of thermodynamics to find the heat flow for the entire process: ?U = Q – W; 0 = Q – (+ 4.9102 J), which gives Q = + 4.9102 J (into the gas).

P (atm) a

2.2

c

1.5

b

0

39. The work done during an isothermal process is P W = nRT ln(V2/V1) = (2.00 mol)(8.315 J/mol · K)(300 K) ln(7.00 m3/3.50 m3) = 3.46103 J. The internal energy of an ideal gas depends only on the temperature, so ?U = 0. We use the first law of thermodynamics to find the heat flow for the process: 0 ?U = Q – W; + 3.46103 J (into the gas). 0 = Q – (+ 3.46103 J), which gives Q =

6.8

10.0

A

B 3.50

7.00

V (m 3 )

40. (a) The initial volume of the water is V1 = (1.00 kg)/(1.00103 kg/m3) = 1.0010–3 m3. We find the final volume from the ideal gas law: PV2 = nRT; (1.013105 N/m2)V2 = [(1.00103 g)/(18 g/mol)](8.315 J/mol · K)(373 K), which gives V2 = 1.70 m3, so the initial volume is negligible. The work done in the constant pressure process is W = P(V2 – V1) = (1.013105 N/m2)(1.70 m3) = 1.72105 J. (b) The heat added to the system is Q = mL = (1.00 kg)(22.6105 J/kg) = 2.26106 J. We use the first law of thermodynamics to find the internal energy change: ?U = Q – W = 2.26106 J – 1.72105 J = 2.09106 J. 41. The work done during an isothermal process is W = nRT ln(V2/V1) = P1V1 ln(V2/V1) = (1.013105 N/m2 )(2.5010–3 m3) ln(1.50 L/2.50 L) = – 129 J. Thus the work done on the gas is + 129 J. 42. (b) Work done in the constant pressure process is W = P1(V2 – V1) = (450 N/m2)(8.00 m3 – 2.00 m3) = 2.70103 J. The internal energy of an ideal gas depends only on the temperature, so we have ?U = *nR ?T = *(nRT2 – nRT1) = *(P2V2 – P1V1) 4.05103 J. = *P1(V2 – V1) = *W = *(2.70103 J) = (d) Because both paths have the same initial and final states, the change in internal energy is the same: ?U = 4.05103 J.

Page 6

P (N/m 2) 500 400 300 200 100

0

V (L)

1

(A)

2

(B) 2.0 4.0 6.0 8.0 10.0

V (m 3)

Chapter 19

43. We consider a differential area dA on the surface of the arbitrary volume. Because the force from the outside pressure is perpendicular to the surface, the work done when the surface expands a distance d¬ is dW = P dA d¬. The work done in a finite expansion is W = ? dW = ??P dA d¬. The pressure is the same over the surface of the volume, so we can integrate over the area: W = ? P [? dA] d¬ = ? PA d¬ = ? P dV. 44. (a) We can find the internal energy change Uc – Ua from the information for the curved path ac: Uc – Ua = Qac – Wac = – 63 J – (– 35 J) = – 28 J. For the path abc, we have Uc – Ua = Qabc – Wabc = Qabc – Wab ; – 28 J = Qabc – (– 48 J), which gives Qabc = – 76 J. (b) For the path cda, work is done only during the constant pressure process, cd, so we have Wcda = Pc(Vd – Vc) = !Pb(Va – Vb)

d¬ dA

P b

a

c

d



V

0

+ 24 J. = !Wba = – !Wab = – !(– 48 J) = (c) We use the first law of thermodynamics for the path cda to find Qcda : Ua – Uc = – (Uc – Ua) = Qcda – Wcda ; – (– 28 J) = Qcda – (24 J), which gives Qcda = + 52 J. (d) Ua – Uc = – (Uc – Ua) = – (– 28 J) = + 28 J. (e) Because there is no work done for the path da, we have Ua – Ud = (Ua – Uc) + (Uc – Ud) = (Ua – Uc) – (Ud – Uc) = Qda – Wda ; + 28 J – (+ 5 J) = Qda – 0, which gives Qda = + 23 J. 45. (a) We can find the internal energy change Ua – Uc from the information P for the curved path ac: (Uc – Ua) = – (Ua – Uc) = Qac – Wac = – 80 J – (– 55 J) = – 25 J, b so Ua – Uc = + 25 J. (b) We use the first law of thermodynamics for the path cda to find Qcda : Ua – Uc = Qcda – Wcda ; c + 25 J = Qcda – (+ 38 J), which gives Qcda = + 63 J. (c) For the path abc, work is done only during the constant 0 pressure process, ab, so we have Wabc = Pa(Vb – Va) = (2.5)Pd(Vc – Vd) = (2.5)Wdc = – (2.5)Wcd = – (2.5)(+ 38 J) = – 95 J. (d) We use the first law of thermodynamics for the path abc to find Qabc : Uc – Ua = Qabc – Wabc ; – 25 J = Qabc – (– 95 J), which gives Qabc = – 120 J. (e) Because there is no work done for the path bc, we have Uc – Ub = (Uc – Ua) + (Ua – Ub) = Qbc – Wbc ; – 25 J + (+ 10 J) = Qbc – 0, which gives Qbc = – 15 J.

Page 7

a

d V

Chapter 19

46. Because the directions along the path are opposite to the directions P in Problem 45, all terms for Q and W will have the opposite sign. (a) Leg b  a is an isobaric expansion, W > 0; b a Leg a  d is an isochoric contraction, W = 0; Leg d  c is an isobaric contraction, W < 0; Leg c  b is an isochoric expansion, W = 0; (b) For the work done around the cycle, we have d c Wcycle = Wba + Wdc = Wba – Wcd = Pa(Va – Vb) – Pd(Vd – Vc) V 0 = 2.5Pd(Vd – Vc) – Pd(Vd – Vc) = 1.5Wcd = 1.5(+ 38 J) = + 57 J. (c) For the cycle, because the gas returns to its initial state, there is no change in internal energy, so Qcycle = Wcycle = + 57 J. We can confirm this using the results of Problem 45: For the heat flow during the cycle, we have Qcycle = Qcba + Qadc = – Qabc – Qcda = – (– 120 J) – (+ 63 J) = + 57 J. (d) For the internal energy change of the cycle, we have ?Ucycle = 0. Because the system returns to the initial state, this must always be true for a cycle. (e) The intake heat is Qcba . For the efficiency, we have e = Wcycle/Qcba = (57 J)/(120 J) = 0.475 = 48%. 47. The pressure is a function of volume for a van der Waals gas: [P + a/(V/n)2][(V/n) – b) = RT. With n = 1, we can write this as P = [RT/(V – b)] – a/V2. We find the work by integrating:

W = =

V2 V1

P dV =

V2 V1

RTl n

RT – a d V = RT l n V – b + a (V – b) V 2 V

V2 V1

V2 – b +a 1 – 1 . V1 – b V2 V1

48. If both rotational and vibrational degrees of freedom are active, CV = 7R/2, so we have U = nCVT = (4.50 mol)[7(8.315 J/mol · K)/2](600 K) = 7.86104 J. 49. If there are no heat losses, we have Q = mcV ?T = VcV ?T, so the rate is Q/?t = VcV ?T/?t; (1.8106 J/h)/(4186 J/kcal) = (1.29 kg/m3)(6.5 m)(4.6 m)(3.0 m)(0.17 kcal/kg · C°) ?T/?t, which gives ?T/?t = 22 C°/h. 50. For one mole of a gas, each degree of freedom has an average energy of !RT, so the internal energy of a mole of the gas is U = n(!RT) = CVT, which gives CV = (n/2)R. For the molar specific heat at constant pressure we have CP = CV + R = [(n/2) + 1]R = [(n + 2)/2]R. 51. Because hydrogen is diatomic, at room temperature there are 5 degrees of freedom. Thus we have CV = 5R/2 = 5(1.99 cal/mol · K)/2 = 4.98 cal/mol · K; Page 8

Chapter 19

cV = CV /M = (4.98 cal/mol · K)/(2 g/mol) = 2.49 cal/g · K = 2.49 kcal/kg · K; CP = CV + R = 4.98 cal/mol · K + 1.99 cal/mol · K = 6.97 cal/mol · K; cP = CP /M = (6.97 cal/mol · K)/(2 g/mol) = 3.48 cal/g · K = 3.48 kcal/kg · K. 52. When we apply the first law of thermodynamics for an adiabatic process, we get ?U = Q – W = 0 – W = – W. Thus the work done by the gas is W = – ?U = – nCV ?T = – nCV(T2 – T1) = nCV(T1 – T2). 53. If the specific heat does not change with temperature, the gas must be monatomic, with CV = McV = *R; M(0.0356 kcal/kg · C°)(4186 J/kcal) = *(8.315 J/mol · K)(103 g/kg), which gives M = From the table of atomic masses, we see that the gas is krypton.

83.7 g/mol.

54. (a) The molar specific heat at constant volume is CV = McV = (34 g/mol)(0.182 kcal/kg · K) = 6.19 cal/mol · K. The molar specific heat at constant pressure is CP = CV + R = 6.19 cal/mol · K + 1.99 cal/mol · K = 8.18 cal/mol · K. Thus the specific heat is cP = CP /M = (8.18 cal/mol · K)/(34 g/mol) = 0.241 cal/g · K = 0.24 kcal/kg · K. triatomic. (b) From Table 19–3, it is likely that the gas is 55. If there are no heat losses, we have Q = mcV ?T = VcV ?T, so the rate is (2500)(70 W)(2.0 h)(3600 s/h)/ (4186 J/kcal) = (1.29 kg/m3)(30,000 m3)(0.17 kcal/kg · C°) ?T, which gives ?T = 46 C°. 56. (a) The heat added to the gas is Q = nCP ?T = (770 mol)(6.95 cal/mol · C°)(180°C – 40°C)(4.186 J/cal) = 3.14106 J. (b) The work done at constant pressure is W = P ?V = nR ?T = (770 mol)(8.315 J/mol · K)(180°C – 40°C) = 8.96105 J. (c) We use the first law of thermodynamics to find the change in internal energy: 2.24106 J. ?U = Q – W = 3.14106 J – 8.96105 J = Note that we could also have used ?U = nCV ?T . 57. (a) The change in internal energy is 2.08103 J. ?U = nCV ?T = (1.00 mol)(4.96 cal/mol · C°)(100°C – 0°C)(4.186 J/cal) = (b) The work done at constant pressure is W = P ?V = nR ?T = (1.00 mol)(8.315 J/mol · K)(100°C – 0°C) = 8.32102 J. (c) We use the first law of thermodynamics to find the heat added to the gas: ?U = Q – W; 2.91103 J. 2.08103 J = Q – 8.32102 J, which gives Q =

Page 9

Chapter 19

58. (a) The change in internal energy is 4.78103 J. ?U = nCV ?T = n(R ?T = (1.00 mol)((8.315 J/mol · K)(720°C – 490°C) = (b) The pressure is a linear function of T: P = P0 + aT. We find the constants from P = P0 + aT; 1.00 atm = P0 + a(490 K); 1.60 atm = P0 + a(720 K). When we combine these two equations, we get P0 = – 0.279 atm, a = 2.6110–3 atm/K. The work done by the gas is W = ? P dV = ? (P0 + aT) dV. We use the ideal gas law to express V as a function of P: V = nRT/P = nR[(P – P0)/a]/P = (nR/a) – (nRP0/aP); dV = nRP0 dP/aP2. We integrate to find the work done by the gas:

W=

P dV =

P2 P1

P

nRP0 aP

2

dP =

nRP0 P2 d P nRP0 P = ln 2 . P1 a P1 P a

W = [(1.00 mol)(8.315 J/mol · K)(– 0.279 atm)/(2.6110–3 atm/K)] ln(1.60 atm/1.00 atm) = – 418 J. (c) We use the first law of thermodynamics to find the heat added to the gas: ?U = Q – W; 4.36103 J. 4.78103 J = Q – (– 418 J), which gives Q = 59. For an adiabatic process we have P2V2 = P1V1, or P2/P1 = (V1/V2); P2/1.00 atm = (!)7/5, which gives P2 = 0.379 atm. We find the temperature from the ideal gas equation: P2V2/P1V1 = T2/T1 ; (0.379 atm/1.00 atm)(2) = T2/293 K, which gives T2 = 222 K =

– 51°C.

60. The pressure is a function of volume, PV = (constant), so we find the work by integrating: 

W =

P dV = (constant )

V2 V1

1 –

dV = (constant ) V V 1–  P1 V1 V11 – 

P2V 2 V21 – 

(constant ) 1 –  – V1 – V 21 –  =  –1 – 1 where we have used (constant) = P2V2 = P1V1. =

V2 V1

=

1 –

= (constant )

1 –

V2 V – 1 1–  1– 

P1V1 – P2 V2 ,  –1

61. We find the initial and final volumes of the gas: P1V1 = nRT1; (3.0 atm)(1.013105 N/m2 · atm)V1 = (1.0 mol)(8.315 J/mol · K)(300 K), which gives V1 = 8.2110–3 m3. P2V2 = P1V1; (1.0 atm)(1.013105 N/m2 · atm)V25/3= (3.0 atm)(1.013105 N/m2 · atm)(8.2110–3 m3)5/3, which gives V2 = 1.5910–2 m3. If we use the result from Problem 60, we get W = (P1V1 – P2V2)/( – 1) Page 10

Chapter 19

= [(3.0 atm)(1.013105 N/m2 · atm)(8.2110–3 m3) – (1.0 atm)(1.013105 N/m2 · atm)(1.5910–2 m3)]/[(5/3) – 1] =

1.33103 J.

62. (a) For an adiabatic process we have P2V2 = P1V1, or P2/P1 = (V1/V2) = (1/4.2)5/3 = 9.1510–2. We find the temperature from the ideal gas equation: P2V2/P1V1 = T2/T1 ; (9.1510–2)(4.2) = T2/400 K, which gives T2 = 154 K. (b) For a diatomic gas with no vibrations,  = 7/5, so we have P2V2 = P1V1, or P2/P1 = (V1/V2) = (1/4.2)7/5 = 0.134. We find the temperature from the ideal gas equation: P2V2/P1V1 = T2/T1 ; (0.134)(4.2) = T2/400 K, which gives T2 = 225 K. (c) For a diatomic gas with vibrations,  = 9/7, so we have P2V2 = P1V1, or P2/P1 = (V1/V2) = (1/4.2)9/7 = 0.158. We find the temperature from the ideal gas equation: P2V2/P1V1 = T2/T1 ; (0.158)(4.2) = T2/400 K, which gives T2 = 265 K. 63. (a) We find the initial temperature from the ideal gas equation: P1V1 = nRT1 ; (1.013105 N/m2)(0.1210 m3) = (4.65 mol)(8.315 J/mol · K)T1 , which gives T1 = 317 K. For an adiabatic process we have P2V2 = P1V1, or P2/P1 = (V1/V2) = (0.1210 m3/0.750 m3)7/5 = 7.7810–2. We find the final temperature from the ideal gas equation: P2V2/P1V1 = T2/T1 ; (7.7810–2)(0.750 m3/0.1210 m3) = T2/317 K, which gives T2 = 153 K. (b) The change in internal energy of the ideal gas is – 1.59104 J. ?U = nCV ?T = n(R ?T = (4.65 mol)((8.315 J/mol · K)(153 K – 317 K) = (c) If we use the result from Problem 60, for the work done by the gas we have W = (P1V1 – P2V2)/( – 1) = [(1.00 atm)(0.1210 m3) – (7.7810–2 atm)(0.750 m3)](1.013105 N/m2 · atm)/[(7/5) – 1] = 1.59104 J. Thus the work done on the gas is Won = – W = – 1.59104 J. (d) For an adiabatic process, there is no heat flow, so Q = 0. 64. From the ideal gas equation we have P2V2/P1V1 = T2/T1 ; (P2/P1)(V2/V1) = 205 K/298 K = 0.688, or P2/P1 = 0.688(V1/V2). For the adiabatic process we have P2V2 = P1V1, or P2/P1 = (V1/V2); 0.688(V1/V2) = (V1/V2)5/3 , or (0.084 m3/V2)2/3 = 0.688, which gives V2 =

Page 11

0.15 m3.

Chapter 19

65.

(a) P (b) For a diatomic gas with no vibrations,  = 5/3, so we have P2V2 = P1V1, or P2/P1 = (V1/V2). 1 T1 P1 When we use the ideal gas equation for the adiabatic process, we have P2V2/P1V1 = T2/T1 ; T2 T3 P2/P1 = (T2/T1)(V1/V2) = (V1/V2) , or V1/V2 = (T2/T1)1/( – 1). 2 3 When we use the ideal gas equation for the constant pressure V process, we have 0 P3V3/P2V2 = T3/T2 . Because P3 = P2 , and V3 = V1 , this becomes V1/V2 = T3/T2 . When we combine this with the previous result, we get T3 = T2 /( – 1)/T11/( – 1) = (389 K)(5/3)/(2/3)/(550 K)1/(2/3) = 231 K. (c) For the adiabatic process, 1 2, we have no heat flow: Q12 = 0; internal energy change: – 2.01103 J; ?U12 = nCV ?T = n*R ?T = (1.00 mol)*(8.315 J/mol · K)(389 K – 550 K) = work done by the gas, from the first law of thermodynamics: ?U12 = Q12– W12; – 2.01103 J = 0 – W12, which gives W12 = + 2.01103 J. For the constant pressure process, 2 3, we have work done by the gas: W23 = P ?V = nR ?T = (1.00 mol)(8.315 J/mol · K)(231 K – 389 K) = – 1.31103 J; internal energy change: – 1.97103 J; ?U23 = nCV ?T = n*R ?T = (1.00 mol)*(8.315 J/mol · K)(231 K – 389 K) = heat flow added to the gas, from the first law of thermodynamics: ?U23 = Q23 – W23; – 1.97103 J = Q23 – (– 1.31103 J), which gives Q23 = – 3.28103 J. For the constant volume process, 3 1, we have work done by the gas: W31 = P ?V = 0; internal energy change: + 3.98103 J; ?U31 = nCV ?T = n*R ?T = (1.00 mol)*(8.315 J/mol · K)(550 K – 231 K) = heat flow added to the gas, from the first law of thermodynamics: ?U31 = Q31– W31; + 3.98103 J = Q31 – 0, which gives Q31 = + 3.98103 J. (d) For the complete cycle, we have Wcycle = W12 + W23 + W31 = + 2.01103 J + (– 1.31103 J) + 0 = + 0.70103 J. 3 3 Qcycle = Q12 + Q23 + Q31 = 0 + (– 3.2810 J) + 3.9810 J = + 0.70103 J. ?Ucycle = ?U12 + ?U23 + ?U31 = – 2.01103 J + (– 1.97103 J) + 3.98103 J = 0.

66. The strong gusty winds will provide heat convection away from the outside of the window, so the Page 12

Chapter 19

temperature at the outside of the window will be the external air temperature. We find the rate of heat flow from ?Q/?t = kA(?T/L) 1.6104 W. = (0.84 J/s · m · C°)(3.0 m2)[15°C – (– 5°C)]/(3.210–3 m) =

67. (a) We find the radiated power from ?Q/?t = eAT4 = (0.35)(5.6710–8 W/m2 · K4)4(0.180 m)2(298 K)4 = 64 W. (b) We find the net flow rate from ?Q/?t = eA(T24 – T14) = (0.35)(5.6710–8 W/m2 · K4)4(0.180 m)2[(298 K)4 – (268 K)4] = 68. We find the distance for the conduction from ?Q/?t = kA(?T/L) 200 W = (0.2 J/s · m · C°)(1.5 m2)(0.50 C°)/L, which gives L = 7.510–4 m =

22 W.

0.75 mm.

69. The cross-sectional area of the beam that falls on an area A is A cos . Thus the rate at which energy is absorbed is P = IeA cos  = (1000 W/m2)(0.70)(0.80 m2) cos 30° = 4.8102 W.

I

 A



70. We find the rate of heat flow through the wall from ?Q/?t = kA(?T/L) = (0.84 J/s · m · C°)(4.0 m)(4.0 m)(30°C – 10°C)/(0.15 m) = 1.79 103 W. We find the number of bulbs required to provide this heat flow from N =(?Q/?t)/P = (1.79103 W)/(100 W) = 17.9 = 18 bulbs. 71. The cross-sectional area of the beam that falls on an area A is A cos . Thus the rate at which energy is absorbed is P = IeA cos . There is no change in temperature of the ice or melted water. We find the time to provide the energy to melt the ice from Q = mL = AhL = Pt = (IeA cos )t; (917 kg/m3)(0.016 m)(3.33105 J/kg) = (1000 W/m2)(0.050)(cos 30°)t, which gives t = 1.13105 s = 31 h. Note that the result is independent of the area. We have used the value of I given in Problem 69. 72. In the steady state, the intermediate temperature does not change, so the heat flow must be the same through the two rods: ?Q/?t = kCuACu(Thot – T)/LCu = kAlAAl(T – Tcold)/LAl . The rods have the same area and length, so we have kCu(Thot – T) = kAl(T – Tcold); (380 J/s · m · C°)(250°C – T) = (200 J/s · m · C°)(T – 0.0°C), which gives T = 164°C. 73. (a) We use the cross-sectional area of the Earth to find the rate at which solar energy is received: Page 13

Chapter 19

P = (1350 W/m2)p(6.38106 m)2 = 1.731017 W = 1.71017 W. (b) We estimate the average temperature from the power radiated into space: P = eAT4; 1.731017 W = (1.0)(5.6710–8 W/m2 · K4)4p(6.38106 m)2T4, which gives T =

278K (5°C).

74. The rate of thermal energy flow is the same for the brick and T int the insulation: ?Q/?t = A(T1 – T2)/Reff = A(T1 – Tint)/R2 = A(Tint – T2)/R1 , T2 T1 where Tint is the temperature at the brick-insulation interface. By equating the first term to each of the others, we have Reff(T1 – Tint) = R2(T1 – T2), and Reff(Tint – T2) = R1(T1 – T2). ²Q ²t Ta – Tb = [(?Q/?t)/A]¬2/k2 ; Tb – T1 = [(?Q/?t)/A]¬3/k3 . k1 k2 k3 If we add these equations, we get T2 – T1 = [(?Q/?t)/A][(¬1/k1) + (¬2/k2) + (¬3/k3)], which gives ?Q/?t = A(T2 – T1)/[(¬1/k1) + (¬2/k2) + (¬3/k3)]. ¬1 ¬3 ¬2 (b) We can generalize this by recognizing that more layers will mean more equations, similar to the three that we had. When we eliminate the intermediate temperatures by adding all the equations, we get ?Q/?t = A(T2 – T1)/?(¬i/ki). 76. The total area of the six sides of the icebox is A = 2[(0.25 m)(0.35 m) + (0.25 m)(0.50 m) + (0.35 m)(0.50 m)] = 0.775 m2. As the ice melts, the inside temperature of the icebox remains at 0°C. The rate at which heat flows through the walls of the icebox is ?Q/?t = miceLice/?t = kA(?T/L); (8.50 kg)(3.33105 J/kg)/?t = 2(0.023 J/s · m · C°)(0.775 m2)(30°C – 0°C)/(0.015 m), 11 h. which gives ?t = 3.97104 s = Page 14

T1

Chapter 19

77. We assume that the rate of heat loss is proportional to the temperature difference, so the heat flow in a time t is Q = K(?T)t. We assume that the constant K takes into account all forms of heat loss, but does not depend on the temperature difference, and thus is the same day and night. When the thermostat is turned down, we have Q1 = K(?Tday)tday + K(?T1night)tnight = K[(22°C – 8°C)(17.0 h) + (12°C – 0°C)(7.0 h)] = (322 h · C°)K. When the thermostat is not turned down, we have Q2 = K(?Tday)tday + K(?T2night)tnight = K[(22°C – 8°C)(17.0 h) + (22°C – 0°C)(7.0 h)] = (392 h · C°)K. For the percentage increase we have 22%. (?Q/Q1)(100) = {[(392 h · C°)K – (322 h · C°)K]/[(322 h · C°)K]}(100) = 78. (a) We choose a cylindrical shell of length L, radius R and thickness dR. The heat flow through any cylindrical shell is the same, so we have dQ/dt = constant = kA dT/dR = k2pRL dT/dR. We separate variables and integrate: T R2 d R = 2 2kL d T ; ln R 2 = 2kL T – T . 2 1 R1 d Q/ d t R1 R T1 d Q/ d t

R Wac. The net work is the sum of the works done in each leg. For a b positive net work, Wbc > 0, so we have an expansion from b to c, Tb = Tc and the path must be clockwise. (b) We find the ratio of volumes from the ideal gas equation: a (Pc/Pa)/(Vc/Va ) = Tc/Ta ; (1)(Vc/Va ) = Tc/Ta , so Vc/Va = Vc/Vb = Tc/Ta = Tb/Ta . 0 The work done during the isothermal expansion is Wbc = nRTb ln(Vc/Vb) = nRTb ln(Tb/Ta) = (1.0 mol)(8.315 J/mol · K)(423 K) ln(423 K/273 K) = 1.54103 J. The work done during the constant pressure compression is Wca = Pa(Va – Vc) = PaVa[1 – (Vc/Va)] = nRTa [1 – (Vc/Va)] = (1.0 mol)(8.315 J/mol · K)(273 K)[1 – (423 K/273 K)] = – 1.25103 J. For the isothermal expansion ?Ubc = 0, so Qbc = Wbc = 1.54103 J. For the constant volume compression Wab = 0, so Qab = ?Uab = nCV(Tb – Ta) = (1.0 mol)*(8.315 J/mol · K)(423 K – 273 K) = 1.87103 J. The efficiency is e = Wnet/Qadded = (Wbc + Wca)/(Qbc + Qab) = (1.54103 J – 1.25103 J)/(1.54103 J + 1.87103 J) = 0.085 = 8.5%.

Page 1

c V

Chapter 20

7.

We find the heat input rate from e = W/QH ; 0.38 = (810 MW)/(QH/t), which gives QH/t = 2132 MW. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 2132 MW – 810 MW = 1322 MW. If this heat flow warms the air, we have QL/t = (n/t)cP ?T; (1322106 W)(3600 s/h)(24 h/day) = (n/t)(7.0 cal/mol · C°)(7.5 C°)(4.186 J/cal), which gives n/t = 5.201011 mol/day. To find the volume rate, we use the ideal gas law: P(V/t) = (n/t)RT; (1.013105 Pa)(V/t) = (5.201011 mol/day)(8.315 J/mol · K)(293 K), 13 km3/day. which gives V/t = 1.251010 m3/day = Depending on the dispersal by the winds, the local climate could be heated significantly. We find the area from A = (V/t)t/h = (12.5 km3/day)(1 day)/(0.200 km) = 63 km2.

8.

(a) We find the heat flow and work done for each leg. p Adiabatic compression (a  b): Qab = 0; b c ?Uab = nCV(Tb – Ta); Wab = – ?Uab = – nCV(Tb – Ta). Isobaric expansion (b  c): Qbc = nCP(Tc – Tb) absorbed; ?Ubc = nCV(Tc – Tb); Wbc = Qbc – ?Ubc = n(CP – CV)(Tc – Tb); Adiabatic compression Adiabatic expansion (c  d): Qcd = 0; ?Ucd = nCV(Td – Tc); Wcd = – ?Ucd = – nCV(Td – Tc). O Constant volume (d  a): Wda = 0; ?Uda = nCV(Ta – Td); Qda = ?Uda + Wda = nCV(Ta – Td) liberated.  The net work is W = Wbc + Wcd + Wda + Wab = n(CP – CV)(Tc – Tb) – nCV(Td – Tc) + 0 – nCV(Tb – Ta) = nCP(Tc – Tb) – nCV(Td – Ta). The efficiency is e W/Qabs = W/Qbc = [nCP(Tc – Tb) – nCV(Td – Ta)]/nCP(Tc – Tb) = 1 – [(Td – Ta)/(Tc – Tb)]. For an adiabatic process, PV = nRTV – 1 = constant. For the adiabatic processes of the cycle we have Td = Tc(Vc/Vd) – 1, and Ta = Tb(Vb/Va) – 1. When we substitute these, we get



Adiabatic expansion d

a

e  1 – {[Tc(Vc/Vd) – 1 – Tb(Vb/Va) – 1]/(Tc – Tb)}. From the other processes we have Vd = Va , and Tb = TcVb/Vc . When we use these we get e  1 – {[(Vc/Va) – 1 – (Vb/Vc)(Vb/Va) – 1]/[1 – (Vb/Vc)]} = 1 – {[(Vc/Va) – (Vb/Va)]/[(Vc/Va) – (Vb/Va)]}

= 1 – {[(Va/Vc)–  – (Va/Vb)– ]/[(Va/Vc)– 1 – (Va/Vb)– 1]}. (b) For an ideal diatomic gas,  = 1.4, so we get Page 2

V

Chapter 20

e = 1 – {[(5.0)– 1.4 – (15)– 1.4]/(1.4)[(5.0)– 1 – (15)– 1]} = 0.56 = 9. 

56%.

The maximum efficiency is the efficiency of the Carnot cycle:  e = 1 – (TL/TH) = 1 – [(578 K)/(803 K)] = 0.280 = 28.0%.

10. We find the temperature from the Carnot efficiency: e = 1 – (TL/TH); 0.36 = 1 – [(493 K)/TH], which gives TH = 770 K =

500°C.

11. (a) The net work is W = W ab + W bc + W cd + W da

= =

b a b a

P dV + P dV +

c b c b

P dV + P dV –

p

d c

P dV +

c d

P dV +

a d

a

P dV

d a

P dV .

The sum of the first two terms is the area under the abc path, and the sum of the last two terms is the area under the adc path. Thus the net work done is the area enclosed by the cycle. (b) For any reversible cycle, we can consider it a large number of paths, with W = ?Wi. We select the two points with the maximum and minimum volumes. We could then apply the above reasoning to the upper and lower paths to arrive at the same result.

b d

12. We find the high temperature from the Carnot efficiency: e1 = 1 – (TL1/TH); 0.35 = 1 – [(633 K)/TH], which gives TH = 974 K. Because the high temperature does not change, for the new efficiency we have e2 = 1 – (TL2/TH); 0.50 = 1 – [TL2/(974 K)], which gives TL2 = 487 K = 214°C. 13. The efficiency of the plant is e = 0.75eCarnot = 0.75[1 – (TL/TH)] = 0.75{1 – [(633 K)/(933 K)]} = 0.241. We find the intake heat flow rate from e = W/QH ; 0.241 = (1.1109 J/s)(3600 s/h)/(QH/t), which gives QH/t = 1.641013 J/h. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 1.641013 J/h – (1.1109 J/s)(3600 s/h) = 1.21013 J/h. 14. The efficiency of the engine is e1 = ![1 – (TL/TH)] = ![1 – (553 K/798 K)] = 0.154. We find the rate at which heat is exhausted from e = W/QH = W/(W + QL) = (W/t)/[(W/t) + (QL/t)]; 0.154 = (850 kW)/[850 kW + (QL/t)], which gives QL/t = (4.67103 kW)(103 W/kW)(3600 s/h) =

1.681010 J/h.

15. If we assume the efficiency of a reversible engine, we have e1 = 1 – (TL/TH) = 1 – (293 K/310 K) = 0.055. We find the rate at which work can be done from e = W/QH ; 0.055 = W/(4000 kcal/day), which gives W = 219 kcal/day. We can estimate the maximum height by assuming all this work increases the potential energy: Page 3

c

V

Chapter 20

W = mghmax ; (219 kcal/day)(4186 J/kcal) = (65 kg)(9.80 m/s2)hmax , which gives hmax =

1.4103 m/day.

16. We find the efficiency from e = W/QH = (W/t)/(QH/t) = (570103 J/s)/(1350 kcal)(4186 J/kcal) = 0.101. We find the temperature from the Carnot efficiency: e = 1 – (TL/TH); 0.101 = 1 – [TL/(853 K)], which gives TL = 767 K = 494°C. 17. We find the low temperature from the Carnot efficiency: e1 = 1 – (TL/TH1); 0.29 = 1 – [TL/(853 K)], which gives TL = 606 K. Because the low temperature does not change, for the new efficiency we have e2 = 1 – (TL/TH2); 0.35 = 1 – [(606 K)/TH2], which gives TH2 = 932 K = 660°C. 18. For the efficiencies of the engines, we have e1 = 0.65eCarnot = 0.60[1 – (TL1/TH1)] e2

T L2

= 0.65{1 – [(703 K)/(953 K)]} = 0.171; = 0.65eCarnot = 0.60[1 – (TL2/TH2)]

= 0.65{1 – [(553 K)/(688 K)]} = 0.128. Because coal is burned to produce the input heat to the first engine, we need to find QH1. We relate W1 to QH1 from the efficiency: e1 = W1/QH1 , or W1 = 0.171QH1 . Because the exhaust heat from the first engine is the input heat to the second engine, we have e2 = W2/QH2 = W2/QL1 , or W2 = 0.128QL1 . For the first engine we know that QL1 = QH1 – W1 , so we get W2 = 0.128QL1 = 0.128(QH1 – 0.171QH1) = 0.106QH1 . For the total work, we have W = W1 + W2 = 0.171QH1 + 0.106QH1 = 0.277QH1 . When we use the rate at which this work is done, we get 900106 W = 0.277(QH1/t), which gives QH1/t = 3.25109 J/s. We find the rate at which coal must be burned from m/t = (3.25109 J/s)/(2.80107 J/kg) = 116 kg/s.

Q L2 Engine W2 T L1

Engine W1 Q H1

19. We find the discharge heat flow for the plant from QL2/t = (QH1/t) – (W/t) = 3.29109 J/s – 900106 J/s = 2.39109 J/s. We find the rate at which water must pass through the plant from QL2/t = (m/t )c ?T; (2.39109 J/s)(3600 s/h) = (m/t )(4186 J/kg · C°)(5.5 C°), which gives m/t =

Page 4

T H2

Q L1 = Q H2

T H1

3.7108 kg/h.

Chapter 20

20. For an ideal gas, ?U = 0 for the isotherm, so we have QH = WT absorbed. Because there are no other heat flows, this absorbed heat must equal the net work done. Thus we have a cycle whose sole effect is to transfer heat into work, which violates the Kelvin-Planck statement of the second law of thermodynamics. Thus adiabatic lines cannot cross.

P

Q=0

T = constant

Q=0

V

21. (a) We find the pressures from the ideal gas equation: p PV = nRT; a Pa(6.010–3 m3) = (0.50)(8.315 J/mol · K)(743 K), which gives Pa = 5.15105 Pa. Pb(15.010–3 m3) = (0.50)(8.315 J/mol · K)(743 K), which gives Pb = 2.06105 Pa. (b) For an adiabatic process, PV = nRTV – 1 = constant; THVb – 1 = TLVc – 1; (743 K)(15.0 L)1.4 – 1 = (563 K)Vc1.4 – 1, which gives Vc = 30.0 L.  – 1 THVa = TLVd – 1; (743 K)(6.0 L)1.4 – 1 = (563 K)Vd1.4 – 1, which gives Vd = 12.0 L. (c) The work done during the isothermal process is Wab = nRTH ln(Vb/Va) = (0.50 mol)(8.315 J/mol · K)(743 K) ln(15.0 L/6.0 L) = Note that this is also QH. (d) The heat transfer during the isothermal process is Qcd = QL = Wcd = nRTL ln(Vd/Vc) = (0.50 mol)(8.315 J/mol · K)(563 K) ln(12.0 L/30.0 L) = (e) For the adiabatic processes we have W = – ?U = – nCV ?T; Wbc = – nCV (Tc – Tb) = – nCV (TL – TH); Wda = – nCV (Ta – Td) = – nCV (TH – TL) = – Wbc . Thus the net work done for the cycle is W = Wab + Wcd = 2.83103 J – 2.14103 J = 0.69103 J. (f) The efficiency is e = W/QH = (0.69103 J)/(2.83103 J) = 0.24 = 24%. From the temperatures we have e = 1 – (TL/TH) = 1 – (563 k/743 K) = 0.24.

Page 5

TH b d

TL

c

V

2.83103 J.

– 2.14103 J.

Chapter 20

p 22. (a) We find the initial volume from the ideal gas equation: a PaVa = nRTa ; (10 atm)(1.013105 Pa/atm)Va = (1.00)(8.315 J/mol · K)(623 K), which gives Va = 5.1110–3 m3. TH For point b we have –3 3 Vb = 2Va = 10.210 m , and Pb = !Pa = 5.0 atm. b For an adiabatic process, PV = nRTV – 1 = constant;  – 1  – 1 d THVb = TLVc ; (623 K)(10.210–3 m3)(5/3) – 1 = (483 K)Vc(5/3) – 1, TL which gives Vc = 14.910–3 m3. THVa – 1 = TLVd – 1; (623 K)(5.1110–3 m3)(5/3) – 1 = (483 K)Vd(5/3) – 1, which gives Vd = 7.4910–3 m3. We find the pressures from the ideal gas equation: PV = nRT; Pc(14.910–3 m3) = (1.00)(8.315 J/mol · K)(483 K), which gives Pc = 2.70105 Pa = 2.7 atm; Pd(7.4910–3 m3) = (1.00)(8.315 J/mol · K)(483 K), which gives Pd = 5.39105 Pa = 5.3 atm. The pressures and volumes are Pa = 10 atm, Va = 5.1110–3 m3; Pb = 5.0 atm, Vb = 10.210–3 m3; Pc = 2.7 atm, Vc = 14.910–3 m3; Pd = 5.3 atm, Vd = 7.4910–3 m3. (b) For each segment we have Isothermal expansion (a  b): ?Uab = 0; Qab = Wab = nRTH ln(Vb/Va) = (1.00 mol)(8.315 J/mol · K)(623 K) ln(2) = 3.59103 J. Adiabatic expansion (b  c): Qbc = 0; Wbc = – ?Ubc = – nCV(Tc – Tb) = – (1.00 mol)*(8.315 J/mol · K)(483 K – 623 K) = 1.75103 J. Isothermal compression (c  d): ?Ucd = 0; Qcd = Wcd = nRTL ln(Vd/Vc) = (1.00 mol)(8.315 J/mol · K)(483 K) ln(!) = – 2.78103 J. Adiabatic compression (d  a): Qda = 0; Wda = – ?Uda = – nCV(Ta – Td) = – (1.00 mol)*(8.315 J/mol · K)(623 K – 483 K) = – 1.75103 J. Page 6

c

V

Chapter 20

Thus we have Qab = 3.59103 J, Wab = 3.59103 J, ?Uab = 0; Qbc = 0, Wbc = 1.75103 J, ?Ubc = – 1.75103 J; Qcd = – 2.78103 J, Wcd = – 2.78103 J, ?Ucd = 0; Qda = 0, Wda = – 1.75103 J, ?Uda = 1.75103 J. (c) The net work is W = Wbc + Wcd + Wda + Wab = Wab + Wcd . The efficiency is e = W/Qabs = (Wab + Wcd)/Qab = (3.59103 J – 2.78103 J)/(3.59103 J) = 0.226 = Note that this agrees with e = (TH – TL)/TH = (623 K – 483 K)/(623 K) = 0.225.

22.6% .

23. The maximum coefficient of performance for the cooling coil is CP = QL/W = TL/(TH – TL) = (258 K)/(303 K – 258 K) = 5.7. 24. The coefficient of performance for the refrigerator is CP = QL/W = TL/(TH – TL) = (258 K)/(295 K – 258 K) =

7.0.

25. We find the low temperature from the coefficient of performance: CP = QL/W = TL/(TH – TL); 5.0 = TL/(302 K – TL), which gives TL = 252 K = – 21°C. 26. If we compare the definition of the coefficient of performance for a heat pump with that for a refrigerator, we see that CP = TH/(TH – TL). (a) The coefficient of performance for the heat pump is CP1 = TH/(TH – TL1) = (295 K)/(295 K – 273 K) = 13.4. We find the work from CP1 = QH/W1 ; 13.4 = (2800 J)/W1 , which gives W1 = 2.1102 J. (b) The coefficient of performance for the heat pump is now CP2 = TH/(TH – TL2) = (295 K)/(295 K – 258 K) = 7.97. We find the work from CP2 = QH/W2 ; 7.97 = (2800 J)/W2 , which gives W2 = 3.5102 J. 27. The efficiency of the engine is e = 1 – (TL/TH). For the coefficient of performance, we have CP = TH/(TH – TL) = 1/[1 – (TL/TH)] = 1/e = 1/0.35 =

2.9.

28. The efficiency of the engine is e = 1 – (TL/TH). We find the coefficient of performance of the refrigerator from the efficiency of the engine: CP = QL/W = TL/(TH – TL) = (TL/TH)/[1 – (TL/TH)] = (1 – e)/e = (1 – 0.30)/(0.30) = 2.33. The heat flow required to cool and freeze the water is QL = m(c ?T + L). We find the time from t = W/P = (QL/CP)/P = m(c ?T + L)/(CP)P = (12)(0.040 kg){[(4186 J/kg · C°)(20°C – 0°C)] + 3.33105 J/kg}/(2.33)(450 W) = 190 s = 3.2 min.

Page 7

Chapter 20

29. (a) The coefficient of performance for the refrigerator is CP = QL/W = TL/(TH – TL) = (256 K)/(298 K – 256 K) = The heat that is to be removed is QL = m(cwater ?Twater + L + cice ?Tice)

6.09.

= (0.50 kg){(4186 J/kg · C°)(25°C – 0°C) + 3.33105 J/kg + (2100 J/kg · C°)[0°C – (– 17°C)]} = 2.37105 J. We find the work from CP = QL/W; 6.09 = (2.37105 J)/W, which gives W = 3.9104 J. (b) The power output is P = W/t = QL/(CP)t = m(cwater ?Twater + L )/(CP)t 200 W = (0.50 kg)[(4186 J/kg · C°)(25°C – 0°C) + 3.33105 J/kg ]/(6.09)t, 3.0 min. which gives t = 1.8102 s =

30. (a) For a Carnot refrigerator, we have QH = nRTH ln(Vb/Va) = nRTH ln(Vb/Va); QL = nRTL ln(Vd/Vc) = nRTL ln(Vc/Vd) = nRTL ln(Vb/Va), because Vc/Vd = Vb/Va; W = QH – QL. The coefficient of performance is CP = QL/W = QL/(QH – QL) = nRTL ln(Vb/Va)/[nRTH ln(Vb/Va) – nRTL ln(Vb/Va)] = TL/(TH – TL). (b) The efficiency of the reversible heat engine is e = 1 – (TL/TH), so TL/TH = 1 – e. Thus we have CP = TL/(TH – TL) = (TL/TH)/[1 – (TL/TH)] = (1 – e)/e. (c) The coefficient of performance is CP = TL/(TH – TL) = (257 K)/(295 K – 257 K) = 6.8. 31. The power input is P = W/t = QL/(CP)t = mL /(CP)t = VL/(CP)t; 1000 W = (1.00103 kg/m3)V(3.33105 J/kg)/(7.0)(3600 s), 76 L. which gives V = 7.610–2 m3 = 32. (a) The coefficient of performance is CP = 0.24CPideal = (0.24)TL/(TH – TL) = (0.24)(297 K)/(311 K – 297 K) = 5.1. (b) The power required is P = W/t = QL/t(CP) = (36103 Btu/h)(1.054103 J/Btu )/(5.09)(3600 s/h) = 2.07103 W = 2.1 kW. 2.8 hp. (c) P = (2.07103 W)/(746 W/hp) = 33. We assume that the loss in kinetic energy is transferred to the environment as a heat flow. For the entropy change we have ?S = Q/T = !mv2/T = !(10 kg)(3.0 m/s)2/(293 K) = 0.15 J/K. 34. The freezing occurs at constant temperature. Because there is a heat flow out of the water, we have ?S = Q/T = – mL/T = – (1.00 m3)(1000 kg/m3)(79.7 kcal/kg)/(273 K) = – 292 kcal/K. 35. We assume when the ice is formed at 0°C it is removed, so its entropy change will be the same as in Page 8

Chapter 20

Problem 34. The heat flow from the water went into the great deal of ice at – 10°C. Because there is a great deal of ice, its temperature will not change. We find the entropy change of the block of ice from ?Sice = Q/Tice = + mL/T = (1.00 m3)(1000 kg/m3)(79.7 kcal/kg)/(263 K) = + 303 kcal/K. Thus the total entropy change is ?Stotal = ?Sice + ?S = + 303 kcal/K + (– 292 kcal/K) = + 11 kcal/K. If the new ice were not removed, we would include an additional heat term which would be negative for cooling the new ice and positive for the great deal of ice. The net additional entropy change would be small. 36. (a) The change in entropy of the water is ?Swater = Q/T = mL/T = – (1.00 kg)(539 kcal/kg)/(373 K) = + 1.45 kcal/K. (b) Because the heat to vaporize the water comes from the surroundings, we have ?Ssurr = – Q/T = – 1.45 kcal/K. (c) For the universe we have ?Suniverse = ?Swater + ?Ssurr = + 1.45 kcal/K + (– 1.45 kcal/K) = 0. This must be so for a reversible process. (d) If the process were irreversible, Tsurr > 100°C, so ?Ssurr would be less negative. Thus ?Suniverse > 0. 37. The total rate of the entropy change is ?Stotal/t = ?Ssource/t + ?Swater/t = (– Q/t)/Tsource + (+ Q/t)/Twater = (– 7.50 cal/s)/(513 K) + (+ 7.50 cal/s)/(300 K) = + 0.0104 cal/K · s. 38. The aluminum and water are isolated, so we find the final temperature from heat lost = heat gained; mAlcAl ?TAl = mwatercwater ?Twater ; (3.8 kg)(0.22 kcal/kg · C°)(30°C – T) = (1.0 kg)(1.00 kcal/kg · C°)(T – 20°C), which gives T = 24.55°C. The heat flow from the aluminum to the water is Q = mAlcAl ?TAl = (3.8 kg)(0.22 kcal/kg · C°)(30°C – 24.55°C) = 4.55 kcal. The heating and cooling do not occur at constant temperature. We add (integrate) the differential changes in entropy: = ? (mAlcAl dT)/T = mAlcAl ln(T/Thot) ?SAl = (3.8 kg)(0.22 kcal/kg · C°) ln(297.8 K/303.2 K) = – 1.5010–2 kcal/K; ?Swater = ? (mwatercwater dT)/T = mwatercwater ln(T/Tcold) = (1.0 kg)(1.00 kcal/kg · C°) ln(297.8 K/293.2 K) = + 1.5610–2 kcal/K. The total entropy change is ?S = ?SAl + ?Swater = – 1.5010–2 kcal/K + 1.5610–2 kcal/K = 610–4 kcal/K = + 0.6 cal/K. 39. The heat flow to freeze the water is Q = mwaterL = (2.5 kg)(3.33105 J/kg ) = 8.33105 J. We find the final temperature of the ice from Q = micecice ?Tice ; 8.33105 J = (450 kg)(2100 J/kg · C°)[T – (– 15°C)], which gives T = – 14.12°C. The heating of the ice does not occur at constant temperature. Because the temperature change is small, to estimate the entropy change, we will use the average temperature: Tice,av = !(Tice + T) = ![(– 15°C) + (– 14.12°C)] = – 14.56°C; The total entropy change is ?S = ?Swater + ?Sice = (– Q/Twater) + (+ Q/Tice,av) = [(– 8.33105 J)/(273.2 K)] + [(+ 8.33105 J)/(258.6 K)] = 172 J/K = + 1.7102 J/K. 40. We find the final temperature from heat lost = heat gained; Page 9

Chapter 20

mhotcwater ?Thot = mcoldcwater ?Tcold ; (3.0 kg)(1000 cal/kg · C°)(80°C – T) = (2.0 kg)(1000 cal/kg · C°)(T – 20°C), which gives T = 56°C. The heating and cooling do not occur at constant temperature. We add (integrate) the differential changes in entropy: = ? (mhotcwater dT)/T = mhotcwater ln(T/Thot) ?Shot = (3.0 kg)(1000 cal/kg · C°) ln(329 K/353 K) = – 211 cal/K; ?Scold = ? (mcoldcwater dT)/T = mcoldcwater ln(T/Tcold) = (2.0 kg)(1000 cal/kg · C°) ln(329 K/293 K) = + 232 cal/K. The total entropy change is ?S = ?Shot + ?Scold = – 211 cal/K + 232 cal/K = + 21 cal/K. 41. Because entropy is a state function, if we have a system that is a reversible engine, there is no entropy change of the system in a complete cycle: ?Ssystem = 0. We assume that we can run this engine and do work while absorbing heat QH but without a heat flow to the lower temperature. This means a heat flow from the surroundings, so we have – QH/TH . Thus the total entropy change is ?S = ?Ssystem + ?Ssurr = 0 – QH/TH < 0, which violates the principle of entropy increase. To have ?S > 0, there must be a heat flow QL back to the surroundings. 42. We add (integrate) the differential changes in entropy: ?S = ? (nCV dT)/T = nCV ln(Tf/Ti) = (2.0 mol)((8.315 J/mol · K) ln(318 K/298 K) =

2.7 J/K.

43. (a) The heating does not occur at constant temperature. We add (integrate) the differential changes in entropy: ?Swater = ? (mwatercwater dT)/T = mwatercwater ln(Tf/Ti) = (1.00 kg)(1.00 kcal/kg · C°) ln(373 K/273 K) = 0.312 kcal/K. (b) If the process were reversible, the energy change of the universe would be zero, so ?Ssurr = – ?Swater = – 0.312 kcal/K. Because the process is not reversible, ?Ssurr will be greater, that is, less negative: ?Ssurr > – 0.312 kcal/K. 44. (a) The aluminum and water are isolated, so we find the final temperature from heat gained = heat lost; mAlcAl ?TAl = mwatercwater ?Twater ; (150 g)(0.22 cal/g · C°)(T – 20°C) = (240 g)(1.00 cal/g · C°)(100°C – T), which gives T = 90.3°C = 90°C. (b) We add (integrate) the differential changes in entropy: ?S = ?Scold + ?Shot = ? (mAlcAl dT)/T + ? (mwatercwater dT)/T = mAlcAl ln(T/TAl) + mwatercwater ln(T/Twater) = (150 g)(0.22 cal/g · C°) ln(363.5 K/293.2 K) + (240 g)(1.00 cal/g · C°) ln(363.5 K/373.2 K) = + 0.77 cal/K (3.2 J/K). 45. (a) For the isothermal process we find the ratio of pressures from the ideal gas equation: (P2iV2/P1V1) = T2/T1= 1, so P2i/P1 = V1/V2 = 2. For the adiabatic process, P1V1 = P2aV2, so P2a/P1 = (V1/V2) = 2. Because  > 1, P2a/P1 > 2, so the final pressure is greater for the adiabatic process. (b) For the isothermal process we have ?U = 0; Qi = Wi. Page 10

Chapter 20

Thus ?Si = Wi/T = [nRT ln(V2/V1)]/T = nR ln(1/2) = – nR ln 2. For the adiabatic process we have Qa = 0. Thus ?Sa = 0. (c) Because each process is reversible, the energy change of the universe is zero. For the isothermal process we have ?Ssurr,i = – ?Si = nR ln 2. For the adiabatic process we have ?Ssurr,a = – ?Sa = 0. 46. (a) Ideal gases do not interact, so each gas expands to twice the volume at constant temperature. Thus we have ?SN = ?SA = nR ln(V2/V1) = nR ln(2) = nR ln 2. The total change of the system is ?Ssys = ?SN + ?SA = 2nR ln 2 = 2(1 mol)(8.315 J/mol · K) ln 2 = 11.5 J/K. (b) Because there is no interaction with the environment, ?Senv = 0. (c) For one gas, say nitrogen, the volume increases by a factor of 3 and for the other the volume increases by a factor of 1.5. Thus we have ?SN = nR ln(V2N/V1) = nR ln(3) = nR ln 3. ?SA = nR ln(V2A/V1) = nR ln(1.5) = nR ln 1.5 . The total change of the system is ?Ssys = ?SN + ?SA = nR ln 3 + nR ln 1.5 = (1 mol)(8.315 J/mol · K)(ln 3 + ln 1.5) = 12.5 J/K. 47. (a) Because entropy is a state function and the system returns to the initial state in a cycle, the change in entropy for the system is zero. Because all processes are reversible, the change in entropy for the universe is zero, so the change in entropy for the environment is zero. (b) For the adiabatic processes ?Q = 0, so ?Sad = ? dQ/T = 0. We let V2 represent the higher volume for the isothermal processes, so the entropy change is ?S = ?Shot + ?Scold = nR ln(V2H/V1H) + nR ln(V1L/V2L). For a Carnot cycle, V2H/V1H = V2L/V1L , so we get ?S = nR [ln(V2H/V1H) – ln(V2L/V1L)] = nR [ln(V2H/V1H) – ln(V2H/V1H)] = 0. 48. The cooling does not occur at constant temperature. We add (integrate) the differential changes in entropy: T2  S = d Q/ T = nC V dT / T = n a + bT 2 dT T1

T2 3 3 3 = n aT + 13bT = n a T 2 – T 1 + 13b T 2 – T1 . T 1

For the given data we have ?S = (0.25 mol){(2.08 mJ/mol · K2)(1.0 K – 3.0 K) + @(2.57 mJ/mol · K4)[(1.0 K)3 – (3.0 K)3]} = – 6.6 mJ/K. 49. (a) The kinetic energy the rock loses when it hits the ground becomes a heat flow to the ground. We assume the temperature of the ground, TL , does not change appreciably. The entropy change is ?S = Q/TL , or TL ?S = Q = !mv2. This energy is no longer available to do useful work. (b) We find the entropy change for the free expansion of an ideal gas by considering a reversible isothermal process at TL . Because ?U = 0, Q = W = nRTL ln(V2/V1). Thus we have TL ?S = Q = nRTL ln(V2/V1) = W. Although this work could have been done in a reversible process, no work was done in the Page 11

Chapter 20

irreversible free adiabatic expansion, so this energy is no longer available for useful work. (c) The entropy change when QH is conducted from TH to TL is ?S = QH/TL – QH/TH . Thus we have TL ?S = QH(1 – TL/TH) = QHeCarnot . For a Carnot engine QHeCarnot = W, with less heat discharged at the lower temperature. Thus this work, which was not done during the conduction, is no longer available for useful work. 50. We find the total energy stored in the copper block from the heat flow that raised its temperature from 290 K to 420 K: Q = mCucCu ?TCu = (5.0 kg)(390 J/kg · C°)(420 K – 290 K) = 2.54105 J. If we extract this energy in a reversible process, the entropy changes are = ? (mCucCu dT)/T = mCucCu ln(Tcold/Thot) ?SCu = (5.0 kg)(390 J/kg · C°) ln(290 K/420 K) = – 7.22102 J/K; ?Ssurr = Q/Tcold = (2.54105 J)/(290 K) = + 8.76102 J/K. The total entropy change is ?S = ?SCu + ?Ssurr = – 7.22102 J/K + 8.76102 J/K = 1.54102 J/K. From the result for Problem 49, the unavailable energy is Tcold ?S = (290 K)(1.54102 J/K) = 4.5104 J. Thus the maximum work available is 2.54105 J – 4.5104 J = 2.1105 J. 51. We use H for head, and T for tail. For the microstates we construct the following table: Macrostate Microstates 6 heads HHHHHH 5 heads, 1 tail H H H H H T, H H H H T H, H H H T H H, H H T H H H, H T H H H H, T H H H H H 4 heads, 2 tails H H H H T T, H H H T H T, H H H T T H, H H T H H T, H H T H T H, H H T T H H,

Number 1 6 15

H T H H H T, H T H H T H, H T H T H H, H T T H H H, T H H H H T, T H H H T H, T H H T H H, T H T H H H,T T H H H H 3 heads, 3 tails H H H T T T, H H T H T T, H H T T H T, H H T T T H, H T H H T T, H T H T H T, H T H T T H, 20 H T T H H T, H T T H T H, H T T T H H, T H H H T T, T H H T H T, T H H T T H, T H T H H T, T H T H T H, T H T T H H, T T H H H T, T T H H T H, T T H T H H, T T TH H H 2 heads, 4 tails H H T T T T, H T H T T T, H T T H T T, H T T T H T, H T T T T H, T H H T T T, 15 T H T H T T, T H T T H T, T H T T T H, T T H H T T, T T H T H T, T T H T T H, T T T H H T, T T T H T H, T T T T H H 1 head, 5 tails H T T T T T, T H T T T T, T T H T T T, T T T H T T, T T T T H T, T T T T T H 6 6 tails TTTTTT 1

There are a total of 64 microstates. (a) The probability of obtaining three heads and three tails is P33 = (20 microstates)/(64 microstates) = 5/16. (b) The probability of obtaining six heads is P60 = (1 microstate)/(64 microstates) = 1/64.

52. Because each die has six possible results, there are (6)(6) = 36 possible microstates. (a) We find the number of microstates that give a 7 by listing all the possibilities: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1), so there are 6 microstates. Thus the probability of obtaining a 7 is P7 = (6 microstates)/(36 microstates) = 1/6. (b) We find the number of microstates that give an 11 by listing all the possibilities: (5, 6), and (6, 5), so there are 2 microstates. Thus the probability of obtaining an 11 is P11 = (2 microstates)/(36 microstates) = 1/18. Page 12

Chapter 20

(c) We find the number of microstates that give a 5 by listing all the possibilities: (1, 4), (2, 3), (3, 2), and (4, 1), so there are 4 microstates. Thus the probability of obtaining a 5 is P5 = (4 microstates)/(36 microstates) = 1/9.

53. We are not concerned with the order that the cards are placed in the hand. (a) One of the possible microstates is the four aces and one of the four kings. Because the suit of the king is not specified, there are four different possibilities, and thus the macrostate of four aces and one king has 4 microstates. (b) Because the deck contains only one of each card specified, there is only 1 microstate for the macrostate of six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades. (c) If we call the four jacks J1, J2, J3, J4, without regard to order we have the following possible pairs: J1J2, J1J3, J1J4, J2J3, J2J4, J3J4, so there are 6 combinations for the jacks. Similarly, there will be 6 combinations for the queens, but only 4 combinations for the ace. Thus the total number of microstates for the macrostate of two jacks, two queens, and an ace is (6)(6)(4) = 144. (d) We construct the hand by considering the number of ways we can draw each card. Because we are not concerned with any specific values, there will be 52 possibilities for the first card. For the second draw, because we cannot have any of the cards equal in value to the first one, there will be only 48 possibilities. Similarly, there will be 44 possibilities for the third draw, 40 possibilities for the fourth draw, and 36 possibilities for the fifth draw. If we were concerned with order, the total number of possibilities would be the product of these. The number of microstates must be reduced by dividing by the number of ways of arranging five cards, which is (5)(4)(3)(2)(1) = 120. Thus the number of microstates for a hand with no two equal-value cards is (52)(48)(44)(40)(36)/120 = 1.32106. The probability will increase with the number of microstates, so the order is (b), (a), (c), (d). 54. (a) There is only one microstate for 4 tails: T T T T. For two heads and two tails, we have H H T T, H T H T, H T T H, T H H T, T H T H, T T H H. Thus there are six microstates for two heads and two tails. The change in entropy is Page 13

Chapter 20

?S = k(ln W2 – ln W1) = k ln(W2/W1) = (1.3810–23 J/K) ln(6/1) = 2.4710–23 J/K. (b) There is only one microstate for 100 heads: W2 = 1. For 100 coins there are 100! ways of arranging them. There are 50! ways of arranging 50 heads, and 50! ways of arranging 50 tails. Thus the number of microstates for any arrangement of 50 heads and 50 tails is W1 = 100!/50! 50! = (9.3310157)/(3.041064)(3.041064) = 1.01029. The change in entropy is ?S = k ln(W2/W1) = (1.3810–23 J/K) ln(1/1.01029) = – 9.210–22 J/K. (c) These changes are much smaller than those for ordinary thermodynamic systems, which have much larger numbers of microstates, and thus much greater entropy changes. 55. The maximum possible efficiency is the efficiency of the Carnot cycle:   e = 1 – (TL/TH) = 1 – [(90 K)/(293 K)] = 0.69 = 69%. 56. The maximum possible efficiency is the efficiency of the Carnot cycle:   e = 1 – (TL/TH) = 1 – [(277 K)/(300 K)] = 0.077 = 7.7%. The engine might be feasible because the great amount of water in the ocean could allow a large flow rate through the engine. Possible adverse environmental effects would be that mixing the waters on a large scale would change the environment for those creatures that live in the cooler water, and change in the surface temperature of the ocean over a large area could cause atmospheric changes. 57. We assume that the loss in kinetic energy is transferred to the environment as a heat flow. For the entropy change we have ?S = Q/T = 2(!mv2)/T = mv2/T = (1100 kg)[(95 km/h)/(3.6 ks/h)]2/(293 K) = 2.6103 J/K. 58. (a) We find the final temperature from heat lost = heat gained; mwatercwater ?Twater = mAlcAl ?TAl ; (0.210 kg)(4186 J/kg · C°)(50°C – T) = (0.120 kg)(900 J/kg · C°)(T – 15°C), which gives T = 46.2°C. (b) The heating and cooling do not occur at constant temperature. We add (integrate) the differential changes in entropy: ?Swater = ? (mwatercwater dT)/T = mwatercwater ln(T/Twater) = (0.210 kg)(4186 J/kg · C°) ln(319.4 K/323.2 K) = – 10.4 J/K; = ? (mAlcAl dT)/T = mAlcAl ln(T/TAl) ?SAl = (0.120 kg)(900 J/kg · C°) ln(319.4 K/288.2 K) = + 11.1 J/K. The total entropy change is ?S = ?Swater + ?SAl = – 10.4 J/K + 11.1 J/K = + 0.70 J/K. 59. (a) The coefficient of performance for the heat pump is CP = TH/(TH – TL) = (297 K)/(297 K – 279 K) = 16.5 = 17. (b) We find the heat delivered at the high temperature from CP = QH/W; 16.5 = QH/(1000 J/s)(3600 s/h) , which gives QH = 5.9107 J/h. 60. First we check to see if energy is conserved: QL + W = 1.50 MW + 1.50 MW = 3.00 MW = QH , so energy is conserved. The efficiency of the engine is e = W/QH = (1.50 MW)/(3.00 MW) = 0.500 = 50.0%. The maximum possible efficiency is the efficiency of the Carnot cycle: Page 14

Chapter 20



 e = 1 – (TL/TH) = 1 – [(215 K)/(425 K)] = 0.494 = 49.4%. Yes, there is something fishy, because his claimed efficiency is not possible.

61. (a) The efficiency of the plant is e = 1 – (TL/TH) = 1 – [(285 K)/(600 K)] = 0.525 = 52.5%. We find the heat input rate from e = W/QH ; 0.525 = (900 MW)/(QH/t), which gives QH/t = 1714 MW. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 1714 MW – 900 MW = 814 MW. If this heat flow warms some river water, which mixes with the rest of the river water, we have QL/t = (n/t)c ?T; (814106 W) = (37 m3/s)(1000 kg/m3)(4186 J/kg · C°) ?T, which gives ?T = 5.3 C°. (b) The heat flow does not occur at constant temperature. We add (integrate) the differential change in entropy: ?S = ? (mc dT)/T = mc ln(Tf/Ti). Thus we have ?S/m = c ln(Tf/Ti) = (4186 J/kg · K) ln(290.3 K/285 K) = + 77 J/kg · K.

62. (a) The efficiency of the Carnot cycle is    eCarnot = 1 – (TL/TH) = 1 – [(358 K)/(773 K)] = 0.537 = 53.7%. Thus we have e/eCarnot = 15%/53.7% = 0.28, so e = (28%)eCarnot . (b) The rate at which heat flows into the engine is QH/t = (100 hp)(746 W/hp) = 7.46104 W. The useful power, which moves the car, is W/t = eQH/t = (0.15)(7.46104 W) = 1.12104 W. The rate at which heat is exhausted is QL/t = QH/t – W/t = [7.46104 W – 1.12104 W](3600 s/h) = 2.28108 J/h = 5.45104 kcal/h. 63. The kinetic energy the rock loses when it hits the ground becomes a heat flow to the ground. We assume the temperature of the ground, T, does not change appreciably. The entropy change is ?S = Q/T = K/T. 64. We find the final temperature from heat lost = heat gained; mhotcwater ?Thot = mcoldcwater ?Tcold ; (0.500 kg)(4186 J/kg · C°)(50°C – T) = (0.500 kg)(4186 J/kg · C°)(T – 0°C), which gives T = 25°C. The heating and cooling do not occur at constant temperature. We add (integrate) the differential changes in entropy: ?S = ?Shot + ?Scold = ? (mhotcwater dT)/T + ? (mcoldcwater dT)/T = mcwater [ln(T/Thot) + ln(T/Tcold)] = (0.500 kg)(4186 J/kg · C°)[ln(298 K/323 K) + ln(298 K/273 K)] = + 15 J/K. Page 15

Chapter 20

65. (a)

(b)

T TH

TL

a

b

d

c

The area within the curve is area = ? T dS = ? dQ = Qnet . Because there is no internal energy change, we have area = Qnet = Wnet .

S

66. (a) The actual efficiency of the engine is eactual = W/QH = (600 J)/(1600 J) = 0.375 = 37.5%. The ideal efficiency is eideal = 1 – (TL/TH) = 1 – [(400 K)/(850 K)] = 0.529 = 52.9%. Thus the engine is running at eactual /eideal = (0.375 )/(0.529 ) = 0.708 = 70.8% of ideal. (b) We find the heat exhausted in one cycle from QL = QH – W = 1600 J – 600 J = 1000 J. In one cycle of the engine there is no entropy change for the engine. The input heat is taken from the universe at TH and the exhaust heat is added to the universe at TL. The total entropy change is ?Stotal = (– QH/TH) + (QL/TL) = [– (1600 J)/(850 K)] + [(1000 J)/(400 K)] = + 0.618 J/K. (c) For a Carnot engine we have W = eQH = 0.529QH ; QL = QH – W = (1 – e)QH = (1 – 0.529 )QH = 0.471 QH . The total entropy change is ?Stotal = (– QH/TH) + (QL/TL) = [– QH/(850 K)] + [+ 0.471 QH/(400 K)] = 0, as expected for an ideal engine. (d) For a Carnot engine we have WCarnot = eQH = 0.529QH = 0.529(1600 J) = 846 J. For the real engine Wreal = 600 J, so the difference is WCarnot – Wreal = 846 J – 600 J = 246 J. For TL ?S we get Page 16

Chapter 20

TL ?S = (400 K)(0.618) = 247 J, which agrees within significant figures. 67. Because process ab is isothermal, ?Uab = 0, and Qab = Wab = nRTH ln(Vb/Va). p Because process bc is constant volume, we have a TH Qbc = ?Ubc = nCV(TL – TH) = n*R(TL – TH); Wbc = 0. Because process cd is isothermal, ?Ucd = 0, and Qcd = Wcd = nRTL ln(Vd/Vc) b d = nRTL ln(Va/Vb) = – nRTL ln(Vb/Va). Because process da is constant volume, we have Qda = ?Uda = nCV(TH – TL) = n*R(TH – TL); c TL Wda = 0. The net work done by the cycle is Va Vb W = Wab + Wcd = nRTH ln(Vb/Va) – nRTL ln(Vb/Va) = nR(TH – TL)ln(Vb/Va). The heat added to the system is Qin = Qab + Qda = nRTH ln(Vb/Va) + n*R(TH – TL). The efficiency is eStirling = W/Qin = nR(TH – TL)ln(Vb/Va)/[nRTH ln(Vb/Va) + n*R(TH – TL)]

V

= (TH – TL)ln(Vb/Va)/[TH ln(Vb/Va) + *(TH – TL)]. The efficiency of a Carnot cycle is eCarnot = (TH – TL)/TH . We rearrange the expression for the Stirling cycle: eStirling = [(TH – TL)/TH]{1/[1 + *(TH – TL)/TH ln(Vb/Va)]}. Because the denominator of the second factor is greater than 1, we see that eStirling < eCarnot .

P 68. For a monatomic gas CP = 5/2, CV = 3/2,  = CP/CV = 5/3. (a) We find the pressure at b from the ideal gas equation: a (Pb/Pa)/(Vb/Va) = Tb/Ta ; (Pb/1 atm)(2.5) = 1, so Pb = 0.40 atm. For the adiabatic compression, we have Isothermal PaVa = PcVc, or Pc/Pa = (Va/Vc); b Pc/1 atm = (1/2.5)5/3, so Pc = 0.217 atm. Adiabatic (b) We find the temperature at c from the ideal gas equation: (Pc/Pa)/(Vc/Va) = Tc/Ta ; c (0.217 atm/1 atm)(2.5) = Tc/273 K, so Tc = 148 K. Va Vb (c) For process ab, the isothermal expansion, we have ?Uab = 0; Qab = Wab = nRTa ln(Vb/Va) = (1 mol)(8.315 J/mol · K)(273 K) ln(2.5) = 2.08103 J. 3 ?Sab = Qab/Ta = (2.0810 J)/(273 K) = 7.62 J/K. For process bc, at constant volume, we have Wbc = 0; Qbc = ?Ubc = nCV(Tc – Tb) = (1 mol)*(8.315 J/mol · K)(148 K – 273 K) = ?Sbc = nCV ln(Tc/Tb) = (1 mol)*(8.315 J/mol · K) ln(148 K/273 K) = For process ca, the adiabatic compression, we have Qca = 0; Page 17

– 1.56103 J. – 7.64 J/K.

V

Chapter 20

Wca = – ?Uca = – nCV(Ta – Tc) = – (1 mol)*(8.315 J/mol · K)(273 K – 148 K) = – 1.56103 J. ?Sca = 0. (d) The efficiency of the cycle is e = W/QH = (Wab + Wca)/Qab = (2.08103 J – 1.56103 J)/(2.08103 J) = 0.25 = 25%. 69. The rate at which heat is conducted through the walls is dQ/dt = kA ?T/?L = (0.050 W/m · K)(6.0 m2)[20°C – (– 10°C)]/(0.10 m) = 90 W. In a time t, the amount of heat entering the freezer is (dQ/dt)t, which, at a minimum, must be removed by the unit in 15% of the time: (dQ/dt)t < 0.15t(QL/t), so (QL/t) > (dQ/dt)/0.15 = (90 W)/0.15 = 600 W. We find the power requirement from the coefficient of performance: CP = QL/W = (QL/t)/(W/t) = TL/(TH – TL); (600 W)/(W/t) < (263 K)/(293 K – 263 K), which gives (W/t) > 68 W = 0.091 hp. 70. We find the heat transfers, which occur for the constant pressure processes: P Qbc = nCP(Tc – Tb), (into the gas); Qda = nCP(Ta – Td), (from the gas). b The net heat flow in the cycle is also the net work done: W = Q = Qbc + Qda = nCP(Tc – Tb + Ta – Td). The efficiency of the cycle is eBrayton = W/Qbc = nCP(Tc – Tb + Ta – Td)/nCP(Tc – Tb) Adiabatic = 1 – (Td – Ta)/(Tc – Tb). compression When we use the ideal gas equation, we get eBrayton = 1 – (PdVd – PaVa)/(PcVc – PbVb) = 1 – Pa(Vd – Va)/Pb(Vc – Vb), because Pd = Pa , and Pc = Pb . For the adiabatic processes, we have PaVa = PbVb, or Va = (Pb/Pa)1/Vb ; PdVd = PcVc, or Vd = (Pc/Pd)1/Vc = (Pb/Pa)1/Vc . When we use these in the expression for the efficiency, we get e = 1 – (Pa/Pb)[(Pb/Pa)1/Vc – (Pb/Pa)1/Vb]/(Vc – Vb), = 1 – (Pb/Pa)–1(Pb/Pa)1/(Vc – Vb)/(Vc – Vb) = 1 – (Pb/Pa)(1 – )/

c Adiabatic expansion

a

d

V

71. (a) We assume that our hand is the first dealt and we receive the aces in the first four cards. The probability of the first card being an ace is 4/52. For the next three cards dealt to the other players, there must be no aces, so the probabilities are 48/51, 47/50, 46/49. The next card is ours; the probability of it being an ace is 3/48. For the next three cards, there must be no aces, so the probabilities are 45/47, 44/46, 43/45. For the next cards, we have 2/44, 42/43, 41/42, 40/41, 1/40, 39/39, 38/38,. For the product of all these, we have P1 = (4/52)(48/51)(47/50)(46/49)(3/48)(45/47)(44/46)(43/45)(2/44)(42/43)(41/42)(40/41)(1/40) = (4)(3)(2)(1)/(52)(51)(50)(49). Because we do not have to receive the aces as the first four cards, we multiply this probability by the number of ways 4 cards can be drawn from a total of 13, which is 13!/4!9! = (13)(12)(11)(10)/(4)(3)(2)(1). Thus the probability of being dealt four aces is P = [(4)(3)(2)(1)/(52)(51)(50)(49)][(13)(12)(11)(10)/(4)(3)(2)(1)] = (13)(12)(11)(10)/(52)(51)(50)(49) = 0.00264 = 1/379. [Note that this is (48! 4!/52!)(13!/4! 9!).] (b) We assume that our hand is the first dealt. Because the suit is not specified, any first card is acceptable, so the probability is (52/52) = 1. For the next three cards dealt to the other players, there must be no cards in the suit we were dealt, so the probabilities are Page 18

Chapter 20

39/51, 38/50, 37/49. The next card is ours; the probability of it being in the same suit is 12/48. For the next three cards, there must be no cards in the suit we were dealt, so the probabilities are 36/47, 35/46, 34/45. For the next cards, we have 11/44, 33/43, 32/42, 31/41, 10/40, . For the product of all these, we have P1 = (1)(39/51)(38/50)(37/49)(12/48)(36/47)(35/46)(34/45)(11/44)(33/43)(32/42)(31/41)(10/40) = 12! 39!/51! = 6.310–12 = 1/1.591011.

Page 19

Chapter 21 p. 1

CHAPTER 21 – Electric Charge and Electric Field 1.

The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0109 N · m2/C2)(2.50 C)(2.50 C)/(3.0 m)2 =

6.3109 N.

2.

The number of electrons is N = Q/(– e) = (– 30.010–6 C)/(– 1.6010–19 C/electrons) =

3.

The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0109 N · m2/C2)(26)(1.6010–19 C)(1.6010–19 C)/(1.510–12 m)2 =

4.

5.

The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0109 N · m2/C2)(1.6010–19 C)(1.6010–19 C)/(5.010–15 m)2 = The magnitude of the Coulomb force is F = kQ1Q2/r2 = (9.0109 N · m2/C2)(2510–6 C)(3.010–3 C)/(0.35 m)2 =

6.

The magnitude of the Coulomb force is F = kQ1Q2/r2. If we divide the expressions for the two forces, we have F2/F1 = (r1/r2)2; F2/(4.210–2 N) = (8)2, which gives F2 = 2.7 N.

7.

The magnitude of the Coulomb force is F = kQ1Q2/r2. If we divide the expressions for the two forces, we have F2/F1 = (r1/r2)2; 3 = [(15.0 cm)/r2]2, which gives r2 = 8.66 cm.

8.

The number of excess electrons is N = Q/(– e) = (– 4010–6 C)/(– 1.6010–19 C/electrons) = The mass increase is ?m = Nme = (2.51014 electrons)(9.1110–31 kg/electron) =

9.

1.881014 electrons.

2.710–3 N.

9.2 N.

5.5103 N.

2.51014 electrons. 2.310–16 kg.

The number of molecules in 1.0 kg H2O is N = [(1.0 kg)(103 g/kg)/(18 g/mol)](6.021023 molecules/mol) = 3.341025 molecules. Each molecule of H2O contains 2(1) + 8 = 10 electrons. The charge of the electrons in 1.0 kg is q = (3.341025 molecules)(10 electrons/molecule)(– 1.6010–19 C/electron) = – 5.4107 C.

10. Using the symbols in the figure, we find the magnitudes of the three individual forces: F12 = F21 = kQ1Q2/r122 = kQ1Q2/L2 Q2 Q1 Q3 = (9.0109 N · m2/C2)(7010–6 C)(4810–6 C)/(0.35 m)2 F21 F31 F12 F13 2 = 2.4710 N. + + – F13 = F31 = kQ1Q3/r132 = kQ1Q3/(2L)2 F23 F32 = (9.0109 N · m2/C2)(7010–6 C)(8010–6 C)/[2(0.35 m)]2 L L

Chapter 21 p. 2

= 1.03102 N. F23 = F32 = kQ2Q3/r232 = kQ2Q3/L2 = (9.0109 N · m2/C2)(4810–6 C)(8010–6 C)/(0.35 m)2 = 2.82102 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the net forces, we get F1 = F13 – F12 = 1.03102 N – 2.47102 N = – 1.4 102 N (left). 2 2 F2 = F21 + F23 = 2.4710 N + 2.8210 N = + 5.3 102 N (right). F3 = – F31 – F32 = – 1.03102 N – 2.82102 N = – 3.9 102 N (left). Note that the sum for the three charges is zero. 11. Because all the charges and their separations are equal, y we find the magnitude of the individual forces: F1 F1 F1 = kQQ/L2 = kQ2/L2 + 9 2 2 –6 2 2 = (9.010 N · m /C )(11.010 C) /(0.150 m) Q = 48.4 N. The directions of the forces are determined from the signs L of the charges and are indicated on the diagram. F 60° Q Q F1 For the forces on the top charge, we see that the horizontal 1 + + x components will cancel. For the net force, we have F1 F1 F = F1 cos 30° + F1 cos 30° = 2F1 cos 30° = 2(48.4 N) cos 30° = 83.8 N up, or away from the center of the triangle. From the symmetry each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is 83.8 N away from the center of the triangle. Note that the sum for the three charges is zero. 12. We find the magnitudes of the individual forces on the y F1 charge at the upper right corner: F3 2 2 2 L F1 = F2 = kQQ/L = kQ /L + + = (9.0109 N · m2/C2)(6.0010–3 C)2/(0.100 m)2 Q F2 Q = 3.24107 N. F3 = kQQ/(Lv2)2 = kQ2/2L2 L = (9.0109 N · m2/C2)(6.0010–3 C)2/2(0.100 m)2 x 7 = 1.6210 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the Q Q + forces on the upper-right charge, we see that the net force + will be along the diagonal. For the net force, we have F = F1 cos 45° + F2 cos 45° + F3 = 2(3.24107 N) cos 45° + 1.62107 N = 6.20107 N along the diagonal, or away from the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is 6.20107 N away from the center of the square. Note that the sum for the three charges is zero. 13. Because the magnitudes of the charges and the distances y have not changed, we have the same magnitudes of the F3 individual forces on the charge at the upper right corner: L F1 – + Q F1 = F2 = kQQ/L2 = 3.24107 N. –Q 2 2 7 F3 = kQ /2L = 1.6210 N. F2 The directions of the forces are determined from the signs L of the charges and are indicated on the diagram. For the x forces on the upper-right charge, we see that the net force will be along the diagonal. For the net force, we have F = – F1 cos 45° – F2 cos 45° + F3 Q –Q – = – 2(3.24107 N) cos 45° + 1.62107 N + = – 2.96107 N along the diagonal, or toward the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction toward the

Chapter 21 p. 3

center: The net force on each charge is 2.96107 N toward the center of the square. Note that the sum for the four charges is zero. 14. The magnitudes of the individual forces on the charges are F14 F13 F12 = kQ2Q/¬2 = 2kQ2/¬2; F13 = kQ3Q/(¬v2)2 = 3kQ2/2¬2; + F14 = kQ4Q/¬2 = 4kQ2/¬2; F12 Q F23 = k2Q3Q/¬2 = 6kQ2/¬2; F24 = k2Q4Q/(¬v2)2 = 4kQ2/¬2; F34 = k3Q4Q/¬2 = 12kQ2/¬2. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. F34 For charge Q we have 4Q + F1 = (– F12 – F13 cos 45°)i + (F13 sin 45° + F14)j F24 = [– (2kQ2/¬2) – (3kQ2/2¬2)(v2/2)]i + F14 [(3kQ2/2¬2)(v2/2) + (4kQ2/¬2)]j = (kQ2/¬2)[(– 2 – 3v2/4)i + (4 + 3v2/4)j]. For charge 2Q we have F2 = (F12 + F24 cos 45°)i + (F24 sin 45° + F23)j = [(2kQ2/¬2) + (4kQ2/¬2)(v2/2)]i + [(4kQ2/¬2)(v2/2) + (6kQ2/¬2)]j = (kQ2/¬2)[(2 + 2v2)i + (6 + 2v2)j]. For charge 3Q we have F3 = (F34 + F13 cos 45°)i + (– F13 sin 45° – F23)j = [(12kQ2/¬2) + (3kQ2/2¬2)(v2/2)]i + [– (3kQ2/2¬2)(v2/2) – (6kQ2/¬2)]j = (kQ2/¬2)[(12 + 3v2/4)i + (– 6 – 3v2/4)j]. For charge 4Q we have F4 = (– F34 – F24 cos 45°)i + (– F24 sin 45° – F14)j = [– (12kQ2/¬2) – (4kQ2/¬2)(v2/2)]i + [– (4kQ2/¬2)(v2/2) – (4kQ2/¬2)]j = (kQ2/¬2)[(– 12 – 2v2)i + (– 4 – 2v2)j].

15. The magnitudes of the individual forces on the charges are F14 y F12 = kQ2Q/¬2 = 2kQ2/¬2; F12 F13 = kQ3Q/(¬v2)2 = 3kQ2/2¬2; + 2 2 2 F14 = kQ4Q/¬ = 4kQ /¬ ; F13 ¬ Q F23 = k2Q3Q/¬2 = 6kQ2/¬2; F24 = k2Q4Q/(¬v2)2 = 4kQ2/¬2; F34 = k3Q4Q/¬2 = 12kQ2/¬2. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For charge Q we have 4Q + F1 = (– F12 + F13 cos 45°)i + (– F13 sin 45° + F14)j F34 F24 = [– (2kQ2/¬2) + (3kQ2/2¬2)(v2/2)]i + F 14 [– (3kQ2/2¬2)(v2/2) + (4kQ2/¬2)]j 2 2 = (kQ /¬ )[(– 2 + 3v2/4)i + (4 – 3v2/4)j]. For charge 2Q we have F2 = (F12 + F24 cos 45°)i + (F24 sin 45° – F23)j = [(2kQ2/¬2) + (4kQ2/¬2)(v2/2)]i + [(4kQ2/¬2)(v2/2) – (6kQ2/¬2)]j = (kQ2/¬2)[(2 + 2v2)i + (– 6 + 2v2)j]. For charge – 3Q we have F3 = (– F34 – F13 cos 45°)i + (F13 sin 45° + F23)j

F23

y ¬

F24

2Q

+

F12

¬

3Q

x F34

+

F13

F23

F24

2Q

+

F12 F23

¬

x

F13

F34

F23

–

– 3Q

Chapter 21 p. 4

= [(– 12kQ2/¬2) – (3kQ2/2¬2)(v2/2)]i + [(3kQ2/2¬2)(v2/2) + (6kQ2/¬2)]j = (kQ2/¬2)[(– 12 – 3v2/4)i + (6 + 3v2/4)j]. For charge 4Q we have F4 = (F34 – F24 cos 45°)i + (– F24 sin 45° – F14)j = [(12kQ2/¬2) – (4kQ2/¬2)(v2/2)]i + [– (4kQ2/¬2)(v2/2) – (4kQ2/¬2)]j = (kQ2/¬2)[(12 – 2v2)i + (– 4 – 2v2)j]. 16. For the two forces, we have Felectric = kq1q2/r122 = ke2/r2 = (9.0109 N · m2/C2)(1.610–19 C)2/(0.5310–10 m)2 = 8.210–8 N. Fgravitational = Gm1m2/r2 = (6.6710–11 N · m2/kg2)(9.1110–31 kg)(1.6710–27 kg)/(0.5310–10 m)2 = 3.610–47 N. The ratio of the forces is Felectric /Fgravitational = (8.210–8 N)/(3.610–47 N) = 2.31039. 17. If the separation is r and one charge is Q1 , the other charge will be Q2 = QT – Q1 . For the repulsive force, we have F F = kQ1Q2/r2 = kQ1(QT – Q1)/r2. (a) If we plot the force as a function of Q1 , we see that the maximum occurs when Q 1 = Q 2 = !Q T , which we would expect from symmetry, since we could interchange the two charges without changing the force. (b) We see from the plot that the minimum occurs 0 when either charge is zero: Q1 (or Q2) = 0.

QT

Q1

18. If one charge is Q1 , the other charge will be Q2 = Q – Q1 . For the force to be repulsive, the two charges must have the same sign. Because the total charge is positive, each charge will be positive. We account for this by considering the force to be positive: F = kQ1Q2/r2 = kQ1(Q – Q1)/r2; 22.8 N = (9.0109 N · m2/C2)Q1(56010–6 C – Q1)/(1.10 m)2, which is a quadratic equation: Q12 – (56010–6 C)Q1 + 3.0710–9 C2 = 0, which gives Q1 = 5.5410–4 C, 5.5310–6 C. Note that, because the labels are arbitrary, we get the value of both charges. 19. Because the charges have the same sign, they repel each other. –Q 0 x F21 F23 The force from the third charge must balance the repulsive force – + – x for each charge, so the third charge must be positive and Q F12 F13 –3Q 0 ¬ between the two negative charges. For each of the negative charges, we have Q0 : kQ0Q/x2 = kQ0(3Q0)/¬2, or ¬2Q = 3x2Q0 ; 3Q0 : k3Q0Q/(¬– x)2 = kQ0(3Q0)/¬2, or ¬2Q = (¬ – x)2Q0 . Thus we have 3x2 = (¬ – x)2, which gives x = –1.37¬, + 0.366¬. Because the positive charge must be between the charges, it must be 0.366¬ from Q0. When we use this value in one of the force equations, we get Q = 3(0.366¬)2Q0/¬2 = 0.402Q0. Thus we place a charge of 0.402Q0 , 0.366¬ from Q0. Note that the force on the middle charge is also zero. 20. If we place a positive charge, it will be repelled by the positive charge and attracted by the negative charge. Thus the third charge must be placed along the line of the

L

x

+

–

+Q 1

–Q 2

P

+

F +Q F

x

Chapter 21 p. 5

charges, but not between them. For the net force to be zero, the magnitudes of the individual forces must be equal: F = kQ1Q/r12 = kQ2Q/r22 , or Q1/(L + x)2 = Q2/x2; (7.7 C)/(0.185 m + x)2 = (3.5 C)/x2, which gives x = 0.38 m, – 0.074 m. The negative result corresponds to the position between the charges where the magnitudes and the directions are the same. Thus the third charge should be placed 0.38 m beyond the negative charge. Note that we would have the same analysis if we placed a negative charge.

21. If one charge is Q1 , the other charge will be Q2 = Q – Q1 . For the force to be repulsive, the two charges must have the same sign. Because the total charge is positive, each charge will be positive. We account for this by considering the force to be positive: F = kQ1Q2/r2 = kQ1(Q – Q1)/r2; 12.0 N = (9.0109 N · m2/C2)Q1(90.010–6 C – Q1)/(1.16 m)2, which is a quadratic equation: Q12 – (90.010–6 C)Q1 + 1.7910–9 C2 = 0, which gives Q1 = 60.210–6 C, 29.810–6 C. Note that, because the labels are arbitrary, we get the value of both charges. For an attractive force, the charges must have opposite signs, so their product will be negative. We account for this by considering the force to be negative: F = kQ1Q2/r2 = kQ1(Q – Q1)/r2; – 12.0 N = (9.0109 N · m2/C2)Q1(90.010–6 C – Q1)/(1.16 m)2, which is a quadratic equation: Q12 – (90.010–6 C)Q1 – 1.7910–9 C2 = 0, which gives Q1 = – 16.810–6 C, 106.810–6 C.

22. (a) We assume the angles are small enough that the Coulomb force FT1 FT2 between the charges can be treated as being horizontal. From the force diagram, we apply ?F = 0 on each charge: F F 1 Q1: FT1 sin 1 = F = kQ1Q2/r2 = 2kQ2/r2;  2 Q Q1 2 FT1 cos 1 = m1g, or tan 1 = 2kQ2/m1gr2. m g 1 Q2: FT2 sin 2 = F = kQ1Q2/r2 = 2kQ2/r2; m2g 2 2 FT2 cos 2 = m2g, or tan 2 = 2kQ /m2gr . If we divide the two results and use tan  ˜  for small angles, we get 1. tan 1/tan 2 = 1/2 = m2/m1 = (b) The analysis of forces is the same, so we have 2. tan 1/tan 2 = 1/2 = m2/m1 = 2m/m = (c) The distance between the charges is r = ¬ sin 1 + ¬ sin 2 ˜ ¬1 + ¬2. For the conditions in part (a), we have ra = 2¬1 = 2¬(2kQ2/mgra2), which gives ra = (4k¬Q2/mg)1/3. For the conditions in part (b), we have rb = ¬1 + ¬2 = 3¬2 = 3¬(2kQ2/m2grb2)= 3¬(2kQ2/2mgrb2), which gives rb = (3k¬Q2/mg)1/3.

Chapter 21 p. 6

23. From the symmetry, we see that there are only three magnitudes for the seven forces from the other charges: 3 adjacent corners: F1 = kQQ/¬2 = kQ2/¬2; 3 diagonal corners: F2 = kQQ/(¬v2)2 = kQ2/2¬2; 1 opposite corner: F3 = kQQ/(¬v3)2 = kQ2/3¬2. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. We could add the seven forces, but we can use the symmetry to reduce the process. Each of the components will have the same magnitude and will be in the corresponding negative direction. Thus we find one of them: Fx = – F1 – F2/v2 – F2/v2 – F3/v3 = – [(kQ2/¬2) + 2(kQ2/2¬2)/v2 + (kQ2/3¬2)/v3] = – (kQ2/¬2)[1 + (1/v2) + (1/3v3)] = – 1.90kQ2/¬2. Thus the resultant force is F = – (1.90kQ2/¬2)(i + j + k).

z ¬

Q+

+

Q

¬ Q +

+

F2 F1 F3 F2

Q

¬

F1

+ Q

+

Q

y

F2 F1

+– Q

+

x

24. The acceleration is produced by the force from the electric field: F = qE = ma; 1.051014 m/s2. (1.6010–19 C)(600 N/C) = (9.1110–31 kg)a, which gives a = Because the charge on the electron is negative, the direction of force, and thus the acceleration, is opposite to the direction of the electric field. 25. If we take the positive direction to the east, we have F = qE = (– 1.6010–19 C)(+ 1360 N/C) = – 2.1810–16 N, or 26. If we take the positive direction to the south, we have F = qE ; 2.7510–14 N = (+ 1.6010–19 C)E, which gives E =

2.1810–16 N (west).

+ 1.72105 N/C (south).

27. The electric field above a positive charge will be away from the charge, or up. We find the magnitude from E = kQ/r2 = (9.0109 N · m2/C2)(33.010–6 C)/(0.200 m)2 = 7.43106 N/C (up). 28. The directions of the fields are determined from the signs of the charges and are indicated on the diagram. The net electric field will be to the left. We find its magnitude from E = kQ1/L2 + kQ2/L2 = k(Q1 + Q2)/L2 = (9.0109 N · m2/C2)(8.010–6 C + 7.010–6 C)/(0.040 m)2 = 8.4107 N/C. Thus the electric field is 8.4107 N/C toward the negative charge. 29. From the definition of the electric field we have E = F/Q = (5.8510–4 N)j/(4.2010–6 C) =

–

–Q 1

E2 E1

L +

+Q 2

(1.39102 N/C)j.

30. From the definition of the electric field we have E = F/Q = [(3.010–3 N)i – (5.010–3 N)j]/(1.2510–6 C) =

(2.4103 N/C)i – (4.0103 N/C)j.

x

Chapter 21 p. 7

31. We know that the field lines from a point charge are radial. If we choose two spherical surfaces centered at the point charge, the field lines will be perpendicular to the surfaces and must have the same number passing through each sphere. If the field varies as 1/r2 + x, and we assume that the number per unit area is proportional to the electric field magnitude, we have N1 = b(1/r12 + x)4pr12, and N2 = b(1/r22 + x)4pr22. For the ratio we get N1/N2 = (r22 + x/r12 + x)(r12/r22) = (r2/r1)x ? 1. Thus we could not draw the field lines proportional to the magnitude of the field. 32.

r2 Q + r1

33.

+Q +

–

– 3Q

+

+

34. The acceleration is produced by the force from the electric field: F = qE = ma; (1.6010–19 C)E = (1.6710–27 kg)(1106)(9.80 m/s2), which gives E =

+

+

+

0.10 N/C.

35. The acceleration is produced by the force from the electric field: F = qE = ma; (– 1.6010–19 C)E = (9.1110–31 kg)(145 m/s2), which gives E = – 8.2610–10 N/C. Because the charge on the electron is negative, the direction of the force, and thus the acceleration, is opposite to the direction of the electric field, so the electric field is 8.2610–10 N/C (south). 36. The directions of the fields are determined from the signs of the charges and are in the same direction, as indicated on the diagram. The net electric field will be to the left. We find its magnitude from E = kQ1/L2 + kQ2/L2 = k(Q + Q)/L2 = 2kQ/L2 845 N/C = 2(9.0109 N · m2/C2)Q/(0.080 m)2 , which gives Q = 3.010–10 C. 37. At point A, from the diagram, we see that the electric fields produced

–

–Q

E2 E1

L +

+Q

x

Chapter 21 p. 8

+

+

+

+

y by the charges will have the same magnitude, and the resultant field will be up. We find the angle  from EA tan  = (0.050 m)/(0.100 m) = 0.500, or  = 26.6°. E2A E1A For the magnitudes of the individual fields we have  2 E1A = E2A = kQ/rA A (0, 5) +Q = (9.0109 N · m2/C2)(7.010–6 C)/[(0.100 m)2 + (0.050 m)2] +Q  x = 5.04106 N/C. (–10, 0) (10, 0) From the symmetry, the resultant electric field is EA = 2E1A sin  = 2(5.04106 N/C) sin 26.6° = 4.5106 N/C up. y For point B we find the angles for the directions of the fields from EB tan 1 = (0.050 m)/(0.050 m) = 1.00, or 1 = 45.0°. tan 2 = (0.050 m)/(0.150 m) = 0.333, or 2 = 18.4°. E1B E2B For the magnitudes of the individual fields we have B(–5, 5) E1B = kQ/r1B2 +Q  +Q 2 1 9 2 2 –6 2 2 = (9.010 N · m /C )(7.010 C)/[(0.050 m) + (0.050 m) ] x (–10, 0) (10, 0) = 1.26107 N/C. E2B = kQ/r2B2 = (9.0109 N · m2/C2)(7.010–6 C)/[(0.150 m)2 + (0.050 m)2] = 2.52106 N/C. For the components of the resultant field we have EBx = E1B cos 1 – E2B cos 2 = (1.26107 N/C) cos 45.0° – (2.52106 N/C) cos 18.4° = 6.52106 N/C; EBy = E1B sin 1 + E2B sin 2 = (1.26107 N/C) sin 45.0° + (2.52106 N/C) sin 18.4° = 9.71106 N/C. We find the direction from tan B = EBy/EBx = (9.71106 N/C)/(6.52106 N/C) = 1.49, or 1 = 56.1°. We find the magnitude from EB = EBx/cos B = (6.52106 N/C)/cos 56.1° = 1.17107 N/C. Thus the field at point B is 1.2107 N/C, 56° above the horizontal. These results are consistent with Fig. 21–33b.

x

y

38. The directions of the individual fields will be along the diagonals of the square, as shown. We find the magnitudes of the individual fields: E1 = kQ1/(L/v2)2 = 2kQ1/L2 = 2(9.0109 N · m2/C2)(45.010–6 C)/(0.525 m)2 L – – –Q 2 –Q 2 = 2.94106 N/C. E3 E4 E2 = E3 = E4 = kQ2/(L/v2)2 = 2kQ2/L2 9 2 2 –6 2 E1 = 2(9.010 N · m /C )(27.010 C)/(0.525 m) E = 1.76106 N/C. E2 From the symmetry, we see that the resultant field will be along the diagonal shown as the x-axis. For the net field, we have +Q1 E = E1 + E3 = 2.94106 N/C + 1.76106 N/C = 4.70106 N/C. –Q 2 – + Thus the field at the center is 4.70106 N/C away from the positive charge.

Chapter 21 p. 9

y

x

39. The directions of the individual fields are shown in the figure. We find the magnitudes of the individual fields: E3 E1 = E3 = kQ/L2 E2 = (9.0109 N · m2/C2)(3.2510–6 C)/(1.00 m)2 E1 + L = 2.93104 N/C. +Q E2 = kQ/(Lv2)2 = !kQ2/L2 = !(9.0109 N · m2/C2)(3.2510–6 C)/(1.00 m)2 = 1.46104 N/C. From the symmetry, we see that the resultant field will be along the diagonal shown as the x-axis. For the net field, we have E = 2E1 cos 45° +E2 = 2(2.93104 N/C) cos 45° + 1.46104 N/C = 5.61104 N/C. Thus the field at the unoccupied corner is 5.61104 N/C away from the opposite corner.

+Q

+

+

+

40. (a) The directions of the individual fields are shown in the figure. y We find the magnitudes of the individual fields: +Q E1 = E2 = kQ/L2. For the components of the resultant field we have L Ex = – E2 sin 60° = – 0.866kQ/L2; L Ey = – E1 – E2 cos 60° = – kQ/L2 – 0.500kQ/L2 = – 1.500kQ/L2. We find the direction from 60° tan  = Ey/Ex = (– 1.50kQ/L2)/(– 0.866kQ/L2) = 1.73, or  = 60°. L O We find the magnitude from E = Ex/cos  = (0.866kQ/L2)/cos 60° = 1.73kQ/L2. E1 E2 2 Thus the field is 1.73kQ/L , 60° below the – x-axis. (b) The directions of the individual fields are shown in the figure. y The magnitudes of the individual fields will be the same: +Q E1 = E2 = kQ/L2. For the components of the resultant field we have L Ex = + E2 sin 60° = + 0.866kQ/L2; Ey = – E1 + E2 cos 60° = – kQ/L2 + 0.500kQ/L2 = – 0.500kQ/L2. L We find the direction from 60° tan  = Ey/Ex = (– 0.500kQ/L2)/(+ 0.866kQ/L2) = – 0.577, or  = – 30°. L E2 O We find the magnitude from E = Ex/cos  = (0.866kQ/L2)/cos 30° = kQ/L2. E1 kQ/L2, 30° below the + x-axis. Thus the field is 41. For the electric field to be zero, the individual fields must have ¬/3 opposite directions, so the two charges must have the same sign. + For the net field to be zero, the magnitudes of the individual E2 E1 Q1 fields must be equal: ¬ E = kQ1/r12 = kQ2/r22 , or Q1/(@¬)2 = Q2/(%¬)2,

+Q

+ +Q

x

+

which gives

–

–Q

x

+

x Q2

Q1/Q2 = #.

42. In Example 21–9, the field produced on the axis of a single ring is given in terms of the distance from the center of the ring. We use two expressions with the origin shifted to the position between the two rings: E = (Q/4pÅ0)({(x + !¬)/[(x + !¬)2 + R2]3/2} +

{(x – !¬)/[(x – !¬)2 + R2]3/2})i.

y

R

R

!¬

!¬ O

E2

x E1

Chapter 21 p. 10

43. (a) From the symmetry of the charges, we see that the electric field points along the y-axis. Thus we have E = 2(Q/4pÅ0) sin /(y2 + ¬2) j = 2Qy/4pÅ0(y2 + ¬2)3/2 j. (b) To find the position where the magnitude is maximum, we differentiate and set the first derivative equal to zero: 2 2 2 (3/ 2)y(2 y) 2Q y + – 3 y d E = 2Q 1 – = = 0, 3/ 2 5 / 2 5/ 2 d y 4Å 2 2 4Å0 y 2 + 2 0 y2 + y2+ which gives y = ± ¬/v2.

y E E1

Q

+

E2

 r  ¬

y +

¬

Q

x

44. The field along the x-axis is E = (Q/4pÅ0)[x/(x2 + a2)3/2] i. To find the position where the magnitude is maximum, we differentiate and set the first derivative equal to zero: Q x 2 + a2 – 3x 2 (3/ 2)x(2x) dE = Q 1 – = = 0, 3/ 2 5/ 2 d y 4Å x 2 + a2 4Å0 x 2 + a2 5/ 2 x 2 + a2 0 which gives xM = ± a/v2. Note that E = 0 at x = 0, and x = 8.

45. The linear charge density of the half-ring is  = Q/pa. We select a differential element of the ring which makes an angle  with the – z-axis of length a d, thus charge dQ = a d. The magnitude of the differential field produced by this element is dE = dQ/4pÅ0r2. We get the components of the field from the diagram: dE = (a d/4pÅ0r2)(cos  i – sin  sin  j – sin  cos  k) = (a d/4pÅ0r2)[(x/r) i – (a/r) sin  j – (a/r) cos  k] z 2 2 3/2 = [a/4pÅ0(x + a ) ][(x d) i – (a sin  d) j – (a cos  d) k]. We see from the symmetry that the total field will have no z-component. We integrate to find the other components:   a E= x d  i – a si n  d  j 0 4Å0 x 2 + a2 3/ 2 0  Q x i – (2a/ ) j a = x i + a cos  j = . 2 2 3/ 2 0 4Å0 x 2 + a2 3/ 2 4Å0 x + a

y ad a

r



x

P  dE

46. Because 2.8 cm « 2.0 m, we can use the expression for the electric field of a long wire: E = (1/4pÅ0)2/x = (9.0109 N · m2/C2)2[(4.7510–6 C)/(2.0 m)]/(0.028 m) = 1.5106 N/C away from the wire.

x

Chapter 21 p. 11

47. We choose a differential element of the rod dy a distance y from the center of the rod, as shown in the diagram. The charge of this element is dq = (Q/L) dy. We find the field, which has both x- and y-components, by integrating along the rod:  y = + L/ 2 dq E= 1 + cos  i – sin  j 4Å0 y = – L/ 2 r 2

=  4Å0

y = + L/ 2 y = – L/ 2

y dq

dy

r

y



x

x

L

dy + cos  i – sin  j . r2

P dE

From the symmetry we see that there will be no y-component. To perform the integration, we must eliminate variables until we have one, for which we choose . From the diagram we see that r = x/cos , and y = x tan . This gives dy = x sec2 d = (x d)/cos2 . The limits for  are ± 0 = ± sin–1 {!L/[x2 + (!L)2]1/2} = ± sin–1 [L/(4x2 + L2)1/2]. When we make these substitutions, we have

 E(x, 0) =

 4Å0

0 –

(x d  )/ cos 2  x/ cos 

0

2

cos  i =

0  cos  d  i 4Å0x –  0

 sin   0 i =  2 sin  i. 0 – 0 4Å0 x 4Å0 x Thus for the magnitude we have E = L/2pÅ0x(L2 + 4x2)1/2. =

y 48. We choose a differential element of the dE dq rod dx a distance x from the origin of O x the coordinate system, as shown in the x x diagram. Because the positive direction d x of x is to the left, the limits for x are 0 to ¬. The charge of the element is dq = (Q/¬) dx. We find the field by integrating along the rod: E= 1 4Å0 =

dq 0

x + x

Q –1 4Å0 x + x

2

i= i=

0

Q 4Å0

0

d x x + x

2

i

–Q 1 – 1 i= x 4Å0 x +

49. We select a differential element of the arc which makes an angle  with the x-axis of length R d, thus charge dQ = R d. The magnitude of the differential field produced by this element is dE = dQ/4pÅ0R2. We see from the symmetry that the total field will have only an x-component. We integrate to find the total field:

Q i. 4Å0x(x + )

R d R

 dE

0

x

Chapter 21 p. 12

E = R 4Å0

0

d  – cos  i = –  2 – 0 R 4Å0 R

0 –

0

cos  d  i



0 = –  si n  i = –  2 si n  0 i = – 0 4Å0R 4Å0R

– 2 si n  0 i. 4Å0 R

50. (a) The differential element of the arc which makes an angle  with the x-axis of length R d will now have a charge dQ = R d = R cos  d. The magnitude of the differential field produced by this element is dE = dQ/4pÅ0R2 = R cos  d/4pÅ0R2, so the field is dE = ( cos  d/4pÅ0R)(– cos  i – sin  j). We integrate to find the total field: 0 0 – 0  0 E= cos  d  – cos  i – si n  j = cos 2  d  i + si n  cos  d  j –  4Å0R 4Å0R –  0 0 0 – 0 1 – 0 2 1 1  0 + 12 si n 2 0 i. 2  + 4 si n 2 i + 2 si n  j –  = 4Å0R 4Å0 R 0 (b) The differential element of the arc will now have a charge dQ = R d = R sin  d. The magnitude of the differential field produced by this element is dE = dQ/4pÅ0R2 = R sin  d/4pÅ0R2, so the field is dE = ( sin  d/4pÅ0R)(– cos  i – sin  j). We integrate to find the total field: 0 0 – 0  0 E= si n  d  – cos  i – si n  j = si n  cos  d  i + si n2  d  j 4Å0R –  0 4Å0 R –  0

=

=

0 – 0 1 2 si n  i + 12  – 14 si n 2 j = 2 – 4Å0R 0

51. (a) We choose a differential element of the rod dy a distance y from the end of the rod, as shown in the diagram. The charge of this element is dq =  dy. We find the field, which has both x- and y-components, by integrating along the rod:  y=L dq E= 1 + cos  i – sin  j 4Å0 y = 0 r 2

=  4Å0

y=L y =0

– 0  – 1 si n 2 0 j. 4Å0R 0 2

y

L dy

dy + cos  i – sin  j . r2

To perform the integration, we must eliminate variables until we have one, for which we choose . From the diagram we see that r = x/cos , and y = x tan . This gives dy = x sec2 d = (x d)/cos2 .

dq y

r 

x

P dE

x

Chapter 21 p. 13

The limits for  are 0 and 0 = sin–1 [L/(x2 + L2)1/2]. When we make these substitutions, we have

E=  4Å0

0

(x d  )/ cos 2 

0

x/ cos 

2

+ cos  i – si n  j =

 4Å0x

0 0

cos  d  i – si n  d  j



 si n  i + cos  j 0 =  si n  i + cos  – 1 j 0 0 4Å0x 0 4Å0x L x  2 2 =  i+ 1/ 2 – 1 j = 1/ 2 L i + x – x + L 2 2 2 2 4Å0x x 2 + L 2 1/ 2 x +L 4Å0x x + L (b) The angle the field makes with the x-axis is found from tan  = Ey/Ex = [x – (x2 + L2)1/2]/L. When L  8, we have tan   (x – L)/L = – 1, so the angle is 45° below the x-axis, independent of x. =

52. (a) We choose a differential element of the rod dy a distance y from the end of the rod, as shown in the diagram. The charge of this element is dq =  dy. We find the field, which has both x- and y-components, by integrating along the rod:  y = L1 d q E= 1 + cos  i – sin  j 4Å0 y = – L 2 r 2

1/ 2

y

dy

L1 dq

y

r 

x dE

dy =  + cos  i – sin  j . 4Å0 y = – L 2 r 2 To perform the integration, we must eliminate variables until we have one, for which we choose . From the diagram we see that r = x/cos , and y = x tan . This gives dy = x sec2 d = (x d)/cos2 . The limits for  are 1 = tan–1 (L1/x) = tan–1 (4.0 m/0.250 m) = tan–1 16.0 = 86.4°, and 2 = tan–1 (L2/x) = tan–1 (2.0 m/0.250 m) = tan–1 8.00 = 82.9°. When we make the substitutions, we have

P

x

y = L1

L2

 2 2 (x d )/ cos 2  E=  + cos  i – sin  j =  cos  d  i – sin  d  j 2 4Å0 –  4Å0x –  1 x/ cos  1 

 sin  i + cos  j 2 =  sin  2 – sin (– 1 ) i + cos  2 – cos (– 1) j 4Å0x – 1 4Å0x =  sin  2 + sin  1 i + cos  2 – cos  1 j . 4Å0x For the given data we have E = (9.0109 N · m2/C2)[(3.1510–6 C)/(6.0 m)(0.250 m)]  [(sin 82.9° + sin 86.4°) i + (cos 82.9° – cos 86.4°) j] = (3.76104 N/C) i + (1.17103 N/C) j. (b) If we use the result from Example 21–10, we get E = 2/4pÅ0x = 2(9.0109 N · m2/C2)[(3.1510–6 C)/(6.0 m)(0.250 m)] = 3.78104 N/C. The errors are errorx = (Ex – E)/E = (3.76104 N/C – 3.78104 N/C)/(3.78104 N/C) = – 0.005 = – 0.5%. 3 4 – 3.1%. errory = Ey/E = (1.1710 N/C)/(3.7810 N/C) = – 0.031 = =

j .

Chapter 21 p. 14

53. For a differential element of the plate we choose a strip of width dy a distance y from the x-axis, as shown in the diagram. The charge on a length L is dq = L dy, so the linear charge density is d =  dy. For the field, we can use the result for an infinite rod: dE = 2 d/4pÅ0r. From the symmetry, we see that the resultant field will be in the z-direction. We find the field by integrating over the surface:  y = + dy E = 2 + cos  k . 4Å y = –  r

z dE

P z

– / 2

(z d  )/ cos 2  cos  k = 2 z / cos  4Å0

/ 2 – / 2

r

y

0

/ 2



L

To perform the integration, we must eliminate variables until we have one, for which we choose . From the diagram we see that r = z/cos , and y = z tan . This gives dy = z sec2 d = (z d)/cos2 . The limits for  are – p/2 to p/2. Thus we have

E = 2 4Å0

L

dy

x

d  k = 2  k = 4Å0

 k. 2Å0

54. (a) The force is opposite to the direction of the electron. We find the acceleration produced by the electric field: – qE = ma; – (1.6010–19 C)(11.4103 N/C) = (9.1110–31 kg)a, which gives a = – 2.001015 m/s2. Because the field is constant, the acceleration is constant, so we find the distance from v2 = v02 + 2ax; 0 = (21.5106 m/s)2 + 2(– 2.001015 m/s2)x, which gives x = 1.1510–1 m = 11.5 cm. (b) We find the time from x = v0t + !at2; 0 = (21.5106 m/s)t + !(– 2.001015 m/s2)t2, which gives t = 0 (the starting time), and 2.1510–8 s =

21.5 ns.

55. (a) We find the acceleration produced by the electric field: qE = ma; (– 1.6010–19 C)[(2.0104 N/C) i + (8.0104 N/C) j] = (9.1110–31 kg)a, a = – (3.51015 m/s2) i – (1.411016 m/s2) j. which gives Because the field is constant, the acceleration is constant. (b) We find the velocity from v = v0 + at = (8.0104 m/s) i + [– (3.51015 m/s2) i – (1.411016 m/s2) j](1.010–9 s) = (– 3.43106 m/s) i – (1.41107 m/s) j. The direction of the electron is the direction of its velocity:  = – 104°. tan  = vy/vx = (– 1.41107 m/s) /(– 3.43106 m/s) = 4.11, or 56. The weight must be balanced by the force from the electric field: mg = )pr3g = NeE ; (1000 kg/m3))p(2.010–5 m)3(9.80 m/s2) = N(1.6010–19 C)(150 N/C), 1.4107 electrons. which gives N =

y

Chapter 21 p. 15

57. We find the vertical acceleration produced by the electric field: ay = – qE/m = – (1.6010–19 C)(5.0103 N/C)/(9.1110–31 kg) = – 8.781014 m/s2. The horizontal velocity is constant, so we find the time to pass through the plates from x = v0t; 6.010–2 m = (1.0107 m/s)t, which gives t = 6.010–9 s. As the electron leaves the plates, its vertical velocity is vy = ayt = (– 8.781014 m/s2)(6.010–9 s) = – 5.27106 m/s. We find the angle from  = – 28°. tan  = vy/vx = (– 5.27106 m/s) /(1.0107 m/s) = – 0.527, or 58. The vertical acceleration produced by the electric field is ay = – qE/m = – (1.6010–19 C)(5.0103 N/C)/(9.1110–31 kg) = – 8.781014 m/s2. The vertical velocity at a height y is given by vy2 = v0y2 + 2ayy. To avoid striking the upper plate there is no real solution for vy , so we have vy2 = v0y2 + 2ay(!H) = 0, or v0 sin 45° = (– ayH)1/2 = [– (– 8.781014 m/s2)(0.010 m)]1/2, which gives v0 = 4.2106 m/s. Note that the analysis of the time from the vertical motion at the limiting condition will show that the electron will strike the bottom plate before exiting. 59. (a) The field along the axis of the ring is E = (– Q/4pÅ0)[x/[(x2 + R2)3/2] i, so the force on the charge is F = qE = (– qQ/4pÅ0)[x/[(x2 + R2)3/2] i = (– qQx/4pÅ0R3)/[1 + (x/R)2]3/2 i.

If x « R, we can use the approximation (1 + u)–n ˜ 1 – nu: F ˜ (– qQx/4pÅ0R3)[1 – *(x/R)2] ˜ – qQx/4pÅ0R3. We see that the force is a restoring force proportional to the displacement, so the motion will be simple harmonic. (b) The effective spring constant is k = qQ/4pÅ0R3, so the period is T = 2p(m/k)1/2 = 2p(4pÅ0mR3/qQ)1/2.

60. (a) The dipole moment is p = Q¬ = (1.6010–19 C)(0.6810–9 m) = 1.110–28 C · m. (b) The torque on the dipole is  = pE sin  = (1.110–28 C · m)(2.7104 N/C) sin 90° = 2.910–24 m · N. Note that this is the maximum torque. (c) The torque on the dipole is  = pE sin  = (1.110–28 C · m)(2.7104 N/C) sin 45° = 2.110–24 m · N. (d) The work done on the dipole changes its potential energy: W = ?U = (– p · E)f – (– p · E)i = [– (– pE)] – (– pE) = 2pE = 2(1.110–28 C · m)(2.7104 N/C) = 5.910–24 J.

61. (a) We find the net charge from p = Q¬ ; 3.410–20 C. 3.410–30 C · m = Q(1.010–10 m), which gives Q = (b) No, this is not an integral multiple of e. The covalent bonding means the electron is shared between the H and Cl atoms, so the effective net charge is less than e. (c) The maximum torque on the dipole is

Chapter 21 p. 16

 = pE sin  = (3.410–30 C · m)(2.5104 N/C) sin 90° = 8.510–26 m · N. (d) The lowest potential energy is when the dipole and electric field are parallel. Thus the energy needed to change the potential energy is W = ?U = (– p · E)f – (– p · E)i = (– pE cos 45°) – (– pE) = pE(1 – cos 45°) 2.510–26 J. = (3.410–30 C · m)(2.5104 N/C)(1 – 0.707) =

62. (a) From the symmetry we see that the resultant field will be in the y-direction: E = 2E1 sin  = 2{Q/4pÅ0[r2 + (!¬)2]}{r/[r2 + (!¬)2]1/2}

y E1

E1

= 2Qr/4pÅ0[r2 + (!¬)2]3/2.

If r » ¬, this becomes E ˜ 2Q/4pÅ0r2. (b) For like charges, the field far away is the sum of the individual fields and therefore is the field of a point charge 2Q. For unlike charges, the field far away is the small difference of the individual fields and therefore decreases more rapidly.

r 

Q +

¬/2

Q +

¬/2

x

63. (a) The torque on the dipole, which is in a direction to decrease , produces an angular acceleration:  = I; – pE sin  = I d2/dt2.  « 1, sin  ˜ , so we get If – pE = I d2/dt2, which produces simple harmonic motion, with the effective force constant k = pE. (b) The period is T = 2p(I/k)1/2 = 2p(I/pE)1/2, so the frequency is f = 1/T = (pE/I)1/2/2p. 64. If the charges of the dipole are separated by dx, the dipole moment is p = Q dx i. If the negative charge is at x, where the electric field is E(x), the electric field at the positive charge is E(x + dx) = E(x) + (dE/dx) dx. The net force on the dipole is F = F+ – F– = [QE(x + dx) – QE(x)] i = Q (dE/dx) dx i = p(dE/dx) i = [p · (dE/dx)]i. 65. (a) Along the x-axis the fields from the two charges are parallel, so the magnitude of the net field is E = E+ – E– = (Q/4pÅ0)[1/(r – !¬)2 – 1/(r + !¬)2] = (Q/4pÅ0){[(r + !

¬)2

– (r – !

= (Q/4pÅ0)2r¬/[r2 – (!¬)2]2.

¬)2]/(r

+!

¬)2(r

–!

}

¬)2

E(x) dx –

x

+

–Q

+Q

y –Q

+Q

–

E(x) x

+

¬

If r » ¬, we have E ˜ 2Qr¬/4pÅ0r4 = 2p/4pÅ0r3. (b) The electric field points in the direction of the dipole. 66. When we equate the two forces, we have mg = ke2/r2; (9.1110–31 kg)(9.80 m/s2) = (9.0109 N · m2/C2)(1.6010–19 C)2/r2, which gives r =

5.08 m.

67. Because the charge on the Earth can be considered to be at the center, we can use the expression for the force between two point charges. For the Coulomb force to be equal to the weight, we have kQ2/R2 = mg; 6.8103 C. (9.0109 N · m2/C2)Q2/(6.38106 m)2 = (1050 kg)(9.80 m/s2), which gives Q = 68. Because a copper atom has 29 electrons, we find the number of electrons in the penny from N = [(3.0 g)/(63.5 g/mol)](6.021023 atoms/mol)(29 electrons/atom) = 8.241023 electrons. We find the fractional loss from ?q/q = (5.510–6 C)/(8.241023 electrons)(1.6010–19 C/electron) = 4.210–11.

Chapter 21 p. 17

69. For the electrical attraction to replace gravity, we equate the two forces: GmMmE/rME2 = kQ2/rME2; (6.6710–11 N · m2/kg2)(7.351022 kg)(5.971024 kg) = (9.0109 N · m2/C2)Q2, 5.71013 C. which gives Q = 70. From the symmetry we see that the resultant field will be in the y-direction: E = 2E+ sin  – E– = 2[Q/4pÅ0(r2 + ¬2)][r/(r2 + ¬2)1/2] – 2Q/4pÅ0r2

y E+

E+

= (2Q/4pÅ0){[r/(r2 + ¬2)3/2] – (1/r2)}

=(2Q/4pÅ0){[r3 – (r2 + ¬2)3/2]/r2(r2 + ¬2)3/2}

E–

= (2Q/4pÅ0){[1 – (1 + ¬2/r2)3/2]/r2(1 + ¬2/r2)3/2}.

If r » ¬, we can use the approximation (1 + ¬2/r2)3/2 ˜ 1 + *¬2/r2: E ˜ (2Q/4pÅ0){[1 – (1 + *¬2/r2)]/r2(1 + *¬2/r2)} ˜ – 3Q¬2/4pÅ0r4.

r Q +

– 2Q – ¬



¬

Q +

71. We find the magnitude of the forces between the pairs: y F12 = kQ1Q2/L2 = (9.0109 N · m2/C2)(4.010–6 C)(8.010–6 C)/(1.20 m)2 Q + 1 = 0.20 N; F12 F13 F13 = kQ1Q3/L2 = (9.0109 N · m2/C2)(4.010–6 C)(6.010–6 C)/(1.20 m)2 L = 0.15 N; F12 F13 2 60° F23 = kQ2Q3/L – – = (9.0109 N · m2/C2)(8.010–6 C)(6.010–6 C)/(1.20 m)2 F23 F23 Q3 Q2 = 0.30 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the resultant forces, we have F1 = (F13 sin 30° – F12 sin 30°) i – (F13 cos 30° + F12 cos 30°) j = (0.15 N sin 30° – 0.20 N sin 30°) i – (0.15 N cos 30° + 0.20 N cos 30°) j = (– 0.025 N) i – (0.30 N) j , with magnitude F1 = [(0.025 N)2 + (0.30 N)2]1/2 = 0.30 N. For the direction we have tan 1 = (– 0.30 N)/(– 0.025 N) = 12.1, 1 = 265°, so F1 = 0.30 N, 265° from x-axis. F2 = (F12 cos 60° – F23) i + (F12 sin 60°) j

x

x

Chapter 21 p. 18

= (0.20 N cos 60° – 0.30 N) i + (0.20 N sin 60°) j = (– 0.20 N) i + (0.17 N) j , with magnitude F2 = [(0.20 N)2 + (0.17 N)2]1/2 = 0.26 N. For the direction we have tan 2 = (0.17 N)/(– 0.20 N) = – 0.866, 2 = 139°, so F2 = 0.26 N, 139° from x-axis. F3 = (F23 – F13 cos 60°) i + (F13 sin 60°) j = (0.30 N – 0.15 N cos 60°) i + (0.15 N sin 60°) j = (0.225 N) i + (0.13 N) j , with magnitude F1 = [(0.225 N)2 + (0.13 N)2]1/2 = 0.26 N. For the direction we have tan 3 = (0.13 N)/(0.225 N) = 0.577, 3 = 30°, so F3 = 0.26 N, 30° from x-axis.

1.0210–7 N/C (up).

y +

Q1 E3 E2

+

Q2 +

L

x

73. The directions of the individual fields will be along the diagonals of the square, as shown. All distances are the same. We find the magnitudes of the individual fields: E1 = kQ1/(L/v2)2 = 2kQ1/L2 = 2(9.0109 N · m2/C2)(1.010–6 C)/(0.35 m)2 = 1.47105 N/C. E2 = kQ2/(L/v2)2 = 2E1 = 2(1.47105 N/C) = 2.94105 N/C. E3 = kQ3/(L/v2)2 = 3E1 = 3(1.47105 N/C) = 4.41105 N/C. E4 = kQ4/(L/v2)2 = 4E1 = 4(1.47105 N/C) = 5.88105 N/C. We simplify the vector addition by using the xy-coordinate system shown. For the components of the resultant field we have Ex = E4 – E2 = 5.88105 N/C – 2.94105 N/C = 2.94105 N/C; Ey = E3 – E1 = 4.41105 N/C – 1.47105 N/C = 2.94105 N/C. Thus we see that the resultant will be in the y-direction: E = 2Ex cos 45° = 2(2.94105 N/C) cos 45° = 4.2105 N/C up.

y

72. The weight must be balanced by the force from the electric field: mg = qE; (1.6710–27 kg)(9.80 m/s2) = (1.6010–19 C)E, which gives E =

E4

x E1

Q4

Q3

+

74. The attractive Coulomb force provides the centripetal acceleration of the electron: ke2/r2 = mv2/r, or r = ke2/mv2; r = (9.0109 N · m2/C2)(1.6010–19 C)2/(9.1110–31 kg)(1.1106 m/s)2 = 2.110–10 m. 75. Because the charges have the same sign, they repel each other. –Q 0 x F13 F12 Q The force from the third charge must balance the repulsive force – + – x F23 F12 –4Q 0 for each charge, so the third charge must be positive and between ¬ the two negative charges. For each of the negative charges, we have Q0 : kQ0Q/x2 = kQ0(4Q0)/¬2, or ¬2Q = 4x2Q0 ; 4Q0 : k4Q0Q/(¬– x)2 = kQ0(4Q0)/¬2, or ¬2Q = (¬ – x)2Q0 . Thus we have 4x2 = (¬ – x)2, which gives x = –¬, + 0.333¬. Because the positive charge must be between the charges, it must be 0.333¬ from Q0. When we use this value in one of the force equations, we get Q = 4(0.333¬)2Q0/¬2 = 0.444Q0. Thus we place a charge of 0.444Q0 , 0.333¬ from Q0. Note that the force on the middle charge is also zero.

Chapter 21 p. 19

76. Because the charge moves in the direction of the electric field, it must be positive. We find the angle of the string from the dimensions: cos  = (0.43 m)/(0.55 m) = 0.782, or  = 38.6°. Because the charge is in equilibrium, the resultant force is zero. We see from the force diagram that tan  = QE/mg; tan 38.6° = Q(10,000 N/C)/(1.010–3 kg)(9.80 m/s2), 7.810–7 C. which gives Q =

L

FT FT



mg  Q

E

QE QE

mg

77. Because the charges have opposite signs, the location where the electric + field is zero must be outside the two charges, as shown. Q1 The fields from the two charges must balance: kQ1/x2 = kQ2/(x – L)2; (1.8510–5 C)/x2 = (7.6510–6 C)/(x – 2.00 m)2, which gives x = 1.22 m, 5.60 m. Because 1.22 m is between the charges, the location is 5.60 m from the positive charge, and 3.60 m from the negative charge.

78. We find the force between the groups by finding the force on the CO group from the HN group. A convenient numerical factor will C+ be d1 ke2/(10–9 m/nm)2 9 2 2 –19 2 –9 2 = (9.010 N · m /C )(1.6010 C) /(10 m/nm) = 2.3010–10 N · nm2. For the forces on the atoms, we have FO = kQO{[QH/(L – d2)2] – (QN/L2)} = ke2fO fH{[1/(L – d2)2] – (1/L2)} FC

x

P

–

Q2

L

E2 E1

x

+ O–

H+

N– d2

L

= (2.3010–10 N · nm2)(0.40)(0.20){[1/(0.28 nm – 0.10 nm)2] – [1/(0.28 nm)2]} = 3.3310–10 N. = kQC{[QN/(L + d1)2] – QH/(L + d1 – d2)2]} = ke2fC fN{[1/(L + d1)2] – [1/(L + d1 – d2)2]}

= (2.3010–10 N · nm2)(0.40)(0.20){[1/(0.28 nm + 0.12 nm)2] – [1/(0.28 nm + 0.12 nm – 0.10 nm)2]} = – 8.9410–11 N. Thus the net force is F = FO + FC = 3.3310–10 N – 8.9410–11 N = 2.410–10 N (attraction). 79. (a) The acceleration of the electron, and thus the force produced by the electric field, must be opposite its velocity. Because the electron has a negative charge, the direction of the electric field will be opposite that of the force, so the direction of the electric field is in the direction of the velocity, to the right. (b) Because the field is constant, the acceleration is constant, so we find the required acceleration from v2 = v02 + 2ax; 0 = (2.0106 m/s)2 + 2a(0.054 m), which gives a = – 3.701013 m/s2. We find the electric field from F = qE = ma;

Chapter 21 p. 20

(– 1.6010–19 C)E = (9.1110–31 kg)(– 3.701013 m/s2), which gives E =

80. (a) To estimate the force between a thymine and an adenine, we assume that only the atoms with an indicated charge make a contribution. Because all charges are fractions ofthe electronic charge, we let QH = QN = f1e, and QO = QC = f2e. A convenient numerical factor will be ke2/(10–10 m/Å)2 = (9.0109 N · m2/C2)(1.6010–19 C)2/ (10–10 m/Å)2 = 2.3010–8 N · Å2. For the first contribution we find the force for the bond of the oxygen on the thymine with the H-N pair on the adenine. From Newton's third law, we know that the force on one must equal the force on the other. We find the attractive force on the oxygen: FO = kQO{[QH/(L1 – a)2] – (QN/L12)}

2.1102 N/C.

T 120°

a

O –

H +

+C a

a

C

N –

H

L1 a

N –

C

L2 a

a

H +

–

a O

a

N a C

H

= ke2f2 f1{[1/(L1 – a)2] – (1/L12)}

= (2.3010–8 N · Å2)(0.4)(0.2){[1/(2.80 Å – 1.00 Å)2] – [1/(2.80 Å)2]} = 3.3310–10 N. For the force for the lower bond of the H-N pair on the thymine with the nitrogen on the adenine, we find the attractive force on the nitrogen: FN = kQN{[QH/(L2 – a)2] – (QN/L22)} = ke2f1 f1{[1/(L2 – a)2] – (1/L22)}

= (2.3010–8 N · Å2)(0.2)(0.2){[1/(3.00 Å – 1.00 Å)2] – [1/(3.00 Å)2]} = 1.2810–10 N. There will be a repulsive force between the oxygen of the first bond and the nitrogen of the second bond. To find the separation of the two, we note that the distance between the two nitrogens of the adenine, which is approximately perpendicular to L1, is 2a cos 30° = 1.73a. We find the magnitude of this force from FO-N = kQO{QN/[L12 + (1.73a)2]} = ke2f2 f1{1/[L12 + (1.73a)2]}

= (2.3010–8 N · Å2)(0.4)(0.2){1/[(2.80 Å)2 + (1.73 Å)2]} = 1.710–10 N. We find the angle that this force makes with the line of the other bonds from tan  = 1.73a/L1 = 1.73 Å/2.80 Å= 0.62, or  = 32°.

a

A

Chapter 21 p. 21

Thus the component that contributes to the bond is (1.710–10 N) cos 32° = 1.410–10 N. The other contribution will be from the carbon atom on the thymine. Because the distance is slightly greater and there will be attraction to the nitrogens and repulsion from the hydrogen, we neglect this contribution. Thus the estimated net bond is 3.3310–10 N + 1.2810–10 N – 1.410–10 N ˜ 310–10 N. (b) To estimate the net force between a cytosine and a guanine, we note that there are two oxygen bonds, one nitrogen bond, and one repulsive O-N force. We neglect the other forces because they involve cancellation from the involvement of both hydrogen and nitrogen. If we ignore the small change in distances, we have 2(3.3310–10 N) + 1.2810–10 N – 1.410–10 N ˜ 710–10 N. (c) The total force for the DNA molecule is ˜ 10–4 N. (310–10 N + 710–10 N)(105 pairs)

81. The vertical acceleration produced by the electric field is ay = – qE/m. The horizontal velocity is constant, so we find the time for the electron to arrive at the top plate from x = v0xt; or t = !L/v0 cos 0. The vertical velocity of the electron must be zero at this time, so we have vy = v0y + ayt; 0 = v0 sin 0 + ay(!L/v0 cos 0), or ay = – 2(v02/L) sin 0 cos 0. For the vertical motion we have y = v0yt + !ayt2;

!H = (v0 sin 0)(!L/v0 cos 0) + 2; H = L tan 0 – !(L tan 0), or tan 0 = 2H/L = 2(0.010 m)/(0.060 m) = 0.333,

0 = 18°.

82. We find the electric field at the location of Q1 due to the plates and Q2. For the field of Q2 we have E2 = kQ2/x2 = (9.0109 N · m2/C2)(1.310–6 C)/(0.34 m)2 = 1.01105 N/C (left). The field from the plates is to the right, so we have Enet = Eplates – E2 = 73,000 N/C – 1.01105 N/C = – 2.8104 N/C (left). For the force on Q1 , we have F1 = Q1Enet = (– 6.710–6 C)(– 2.8104 N/C) = + 0.19 N (right).

–

+ + + E2 +

– Q1

x Eplates

– +

Q2

–

+

–

+

–

83. The angular frequency of the SHM is  = (k/m)1/2 = [(126 N/m)/(0.800 kg)]1/2 = 12.5 s–1. If we take down as positive, with respect to the equilibrium position, the ball will start at maximum displacement, so the position as a function of time is x = A cos(t) = (0.0500 m) cos [(12.5 s–1)t]. Because the charge is negative, the electric field at the table will be up and the distance from the table is r = H – x = 0.150 m – (0.0500 m) cos [(12.5 s–1)t]. The electric field is E = kQ/r2 = (9.0109 N · m2/C2)(3.0010–6 C)/{0.150 m – (0.0500 m) cos [(12.5 s–1)t]}2 = (1.08107 N/C)/{3.00 – cos [(12.5 s–1)t]}2 up. 84. We consider the forces on one ball. (The other will be the same except

–

Chapter 21 p. 22

for the reversal.) The separation of the charges is r = 2L sin 30° = 2(0.75 m) sin 30° = 0.75 m. From the equilibrium force diagram, we have tan  = F/mg = [k(!Q)(!Q)/r2]/mg; tan 30° = #(9.0109 N · m2/C2)Q2/ (0.75 m)2(2210–3 kg)(9.80 m/s2), –6 5.6 C. which gives Q = 5.610 C =

FT

L mg 

FT

 Q/2

F

Q/2 F mg

85. The magnitudes of the electric fields of the – – + three plates are E1 = 1/2Å0 A B C D – – + = (0.5010–6 C/m2)/2(8.8510–12 C2/N · m2) E E2 E2 E1 E 1 E2 – – + E1 2 = 2.82104 N/C; E2 = 2/2Å0 – – + E1 E3 E3 E3 E3 = (0.1010–6 C/m2)/2(8.8510–12 C2/N · m2) 3 – – + = 5.6510 N/C; E3 = 3/2Å0 – – + = (0.3510–6 C/m2)/2(8.8510–12 C2/N · m2) 1 2 3 = 1.98104 N/C. The directions of the fields at each of the points are indicated on the diagram. The total fields are EA = E1 – E2 + E3 = 2.82104 N/C – 5.65103 N/C + 1.98104 N/C = 4.2104 N/C (right); EB = – E1 – E2 + E3 = – 2.82104 N/C – 5.65103 N/C + 1.98104 N/C = – 1.4104 N/C (left); 4 3 4 EC = – E1 + E2 + E3 = – 2.8210 N/C + 5.6510 N/C + 1.9810 N/C = – 2.8103 N/C (left); ED = – E1 + E2 – E3 = – 2.82104 N/C + 5.65103 N/C – 1.98104 N/C = – 4.2104 N/C (left). 86. The charge of all the electrons is Qelectrons = [(15 kg)(103 g/kg)/(27 g/mol)](6.021023 atoms/mol)  (13 electrons/atom)(– 1.6010–19 C/electron) = Because there are an equal number of protons, the net charge of the bar is 87. Because the charges have opposite signs, the location where the electric field is zero must be outside the two charges, as shown. The fields from the two charges must balance: + Q1 kQ1/(d + x)2 = kQ2/x2;

– 7.0108 C. 0.

x –

d

Q2

Q/(d + x)2 = Q/2x2, or d + x = ± xv2. which gives x = d/(v2 – 1), – d/(v2 + 1), or x = d(1 + v2), d(1 – v2), Because d(1 – v2) is between the charges, the location is d(1 + v2) from the negative charge, and d(2 + v2) from the positive charge. Other than at infinity, there is no place, not on the x-axis, where the vectors sum to zero.

P E2 E1

88. (a) The electric field from the long wire is E = 2/4pÅ0r = 2(9.0109 N · m2/C2)(0.1410–6 C/m)/r = (2.5103 N · m/C)/r, radially away from the wire. (b) This field produces a force on the electron toward the wire that provides the centripetal acceleration of the electron: eE = 2e/4pÅ0r = mv2/r, or v = (eEr/m)1/2 = [(1.6010–19 C)(2.5103 N · m/C)/(9.1110–31 kg)]1/2 = 2.1107 m/s. Note that this is independent of r.

x

Chapter 22 p. 1

CHAPTER 22 – Gauss’s Law 1.

Because the electric field is uniform, the flux through the circle is  = ? E · dA = E · A = EA cos . (a) When the circle is perpendicular to the field lines, the flux is  = EA cos  = EA = (5.8102 N/C)p(0.15 m)2 = 41 N · m2/C. (b) When the circle is at 45° to the field lines, the flux is  = EA cos  = EA = (5.8102 N/C)p(0.15 m)2 cos 45° = 29 N · m2/C. (c) When the circle is parallel to the field lines, the flux is  = EA cos  = EA = (5.8102 N/C)p(0.15 m)2 cos 90° = 0.

2.

Because the electric field is radial, it is perpendicular to the spherical surface just beyond the Earth’s surface. The field is also constant, so the flux through the sphere is  = ? E · dA = – EA = – (150 N/C)4p(6.38106 m)2 = – 7.71016 N · m2/C. Note that E and dA are in opposite directions.

3.

All field lines enter and leave the cube, so the net flux is net = 0. We find the flux through a face from  = ? E · dA. There are two faces with the field lines perpendicular to the face, say one at x = 0 and one at x = ¬. Thus for these two faces we have x = 0 = – EA = – (6.50103 N/C)¬2 = – (6.50103 N/C)¬2; x = ¬ = + EA = + (6.50103 N/C)¬2 = + (6.50103 N/C)¬2. For all other faces, the field is parallel to the face, so we have all others = 0.

4.

(a) Because the angle between the electric field and the area varies over the surface of the hemisphere, it would appear that we find the flux by integration. We see that the same flux must pass through the circular base of the hemisphere, where the field is constant and perpendicular to the surface. Thus we have  = EA = EpR2. (b) If E is perpendicular to the axis, every field line must enter and leave the surface, so we have  = 0.

5.

axis

 E

R dA

The total flux is depends only on the enclosed charge:  = Q/Å0 , or Q = Å0 = (8.8510–12 C2/N · m2)(1.45103 N · m2/C) = 1.2810–8 C =

12.8 nC.

Chapter 22 p. 2

6.

The net flux through each closed surface is determined by the net charge inside. Thus we have 1 = (Q – 3Q)/Å0 = – 2Q/Å0. 2 = (Q – 3Q + 2Q)/Å0 = 0. 3 = (– 3Q + 2Q)/Å0 = – Q/Å0. 4 = (0)/Å0 = 0. 5 = (+ 2Q)/Å0 = + 2Q/Å0.

S4 S5 + 2Q

+Q S1 – 3Q S2

7.

The total electric flux through the surface depends only on the enclosed charge:  = ı E · dA = Q/Å0 . The only contributions to the integral are from the faces perpendicular to the electric field. Over each of these two surfaces, the magnitude of the field is constant, so we have  = E¬A – E0A = (E¬ – E0)A = Q/Å0 ; (410 N/C – 560 N/C)(30 m)2 = Q/(8.8510–12 C2/N · m2), – 1.2 C. which gives Q = – 1.210–6 C =

S3

x=0 E0

E¬

¬

8.

The total electric flux through the surface depends only on the enclosed charge:  = ı E · dA = Q/Å0 . Because the charge is at the center of the cube, we know from symmetry that each of the six faces has the same flux through it: face = (1/6)total = Q/6Å0 .

9.

If we construct a spherical Gaussian surface just outside the ball, we have  = ı E · dA = Q/Å0 . Because the field over the surface has a constant magnitude and is perpendicular to the surface, we have ı E · dA = E4pr2 = Q/Å0 ; (– 2.75102 N/C)4p(3.5010–2 m)2 = Q/(8.8510–12 C2/N · m2), – 3.7510–11 C. which gives Q =

10. The charge on the spherical conductor will be uniformly distributed over the surface: Q = 4pr2. From the result of Example 22–3 we get E = Q/4pÅ0r2 = 4pr2/4pÅ0r2 = /Å0. 11. The field from a long thin wire is radial with a magnitude given by E = /2pÅ0r. (a) At a distance of 5.0 m the field is E = (– 2.810–6 C/m)/2p(8.8510–12 C2/N · m2)(5.0 m) = – 1.0104 N/C (toward the wire).

Chapter 22 p. 3

(b) At a distance of 2.0 m the field is E = (– 2.810–6 C/m)/2p(8.8510–12 C2/N · m2)(2.0 m) = – 2.5104 N/C (toward the wire). 12. (a) The charge on a conducting sphere must be on the surface. If we construct a spherical Gaussian surface inside the metal sphere, there will be no enclosed charge and thus Ea = 0. (b) Because the point is still inside the metal sphere, we have Eb = 0. (c) Because the point is outside the metal sphere, the field is radial, given by Ec = Q/4pÅ0rc2 = (– 3.5010–6 C)/4p(8.8510–12 C2/N · m2)(3.10 m)2 = – 3.28103 N/C (toward the sphere). (d) Because the point is outside the metal sphere, the field is radial, given by Ed = Q/4pÅ0rd2 = (– 3.5010–6 C)/4p(8.8510–12 C2/N · m2)(6.00 m)2 = – 8.75102 N/C (toward the sphere). (e) Because the charge on the shell is the same as the charge on the surface of the metal sphere, all of the fields will be the same. (f) For points inside the nonconducting sphere, only the charge inside a spherical surface with a radius to the point will provide the field: E(r = r0) = [(Q/)pr03)()pr3)]/4pÅ0r2 = Qr/4pÅ0r03. Thus we have Ea = (– 3.5010–6 C)(0.15 m)/4p(8.8510–12 C2/N · m2)(3.00 m)3 = – 1.75102 N/C (toward the center). Eb = (– 3.5010–6 C)(2.90 m)/4p(8.8510–12 C2/N · m2)(3.00 m)3 = – 3.38103 N/C (toward the center). At points outside the nonconducting sphere, the field will be the same as before; Ec = – 3.28103 N/C (toward the sphere); Ed = – 8.75102 N/C (toward the sphere). 13. For points inside the nonconducting sphere, only the charge inside a spherical surface with a radius to the point will provide the field: E(r = r0) = [(Q/)pr03)()pr3)]/4pÅ0r2 = Qr/4pÅ0r03. At points outside the nonconducting sphere, the field will be that of a point charge: E(r = r0) = Q/4pÅ0r2. The magnitude at the surface is E = Q/4pÅ0rc2

E (10 7 N/C) 1.9

0

7.5

= (12.010–6 C)/4p(8.8510–12 C2/N · m2)[!(0.150 m)]2 = 1.92107 N/C. 14. (a) Because 1.0 cm « 25 cm, we can approximate the sheet as an infinite sheet, with E = /2Å0 = Q/2AÅ0 = (3510–9 C)/2(0.25 m)2(8.8510–12 C2/N · m2) = 3.2104 N/C (away from sheet). (b) Because 20 m » 25 cm, we can approximate the sheet as a point charge, with E = (1/4pÅ0)Q/r2 = (9.0109 N · m2/C2)(3510–9 C)/(20 m)2 = 0.79 N/C (away from sheet).

15

r (cm)

Chapter 22 p. 4

15. From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent r2 of the direction. (a) For a spherical Gaussian surface within the spherical r cavity, we have 2 ı E · dA = E4pr = Qenclosed/Å0 = Q/Å0 , so we have Q r1 E = (1/4pÅ0)Q/r2 = (9.0109 N · m2/C2)(5.5010–6 C)/(0.030 m)2 = 5.5107 N/C (away from center). (b) The point 6.0 cm from the center is inside the conductor, thus the electric field is 0. Note that there must be a negative charge of – 5.50 C on the surface of the cavity and a positive charge of + 5.50 C on the outer surface of the sphere. 16. (a) From symmetry, for points inside the shell of radius r0 , the only field will be that of the point charge: Ea = (1/4pÅ0)Q/r2 (r < r0). (b) Inside the conducting shell the electric field is Eb = 0. (c) From symmetry, for points outside the shell, the field will be radial. If we were to construct a spherical Gaussian surface, the net enclosed charge is the point charge. Thus the electric field is Ec = (1/4pÅ0)Q/r2 (r > r0). (d) The shell does not affect the field due to Q alone (except, of course, inside the shell). The charge does affect the shell by inducing a charge – Q on the inner surface of the shell and a charge + Q on the outer surface of the shell. 17. (a) For a Gaussian surface within the cube just outside the spherical cavity, we have ı E · dA = Qenclosed/Å0. Because the field must be zero inside a conductor, the integral is zero, so the enclosed charge must be zero. With a charge + 8.00 C at the center, there must be on the surface of the cavity. a charge of – 8.00 C (b) There can be no free static charge inside the conducting cube. Because the net charge on the cube is – 7.00 C, if – 8.00 C is on the cavity surface, there must be on the outer surface. + 1.00 C

Q

Chapter 22 p. 5

18. From symmetry we know that any electric field will be perpendicular to the plates. (a) To find the field between the plates, for a Gaussian surface we choose a cylinder perpendicular to the plates with one end of area A inside a plate (where the field must be zero) and the other end between the plates. Thus we have ı E · dA = Qenclosed/Å0.

Ebetween

Eoutside

?end1 E · dA + ?end2 E · dA + ?sides E · dA = A/Å0 ;

0 + EbetweenA + 0 = A/Å0 , which gives Ebetween = /Å0 . (b) To find the field outside the plates, for a Gaussian surface we choose a cylinder perpendicular to the plates with one end of area A between the plates (where we know the field) and the other end outside the plates. Thus we have ı E · dA = Qenclosed/Å0.

–

+

?end1 E · dA + ?end2 E · dA + ?sides E · dA = – A/Å0 ;

– EbetweenA + EoutsideA + 0 = – A/Å0 ; – A/Å0 + EoutsideA + 0 = – A/Å0 , which gives Eoutside = 0. (c) If the plates were nonconductors, the fields from each plate would be the same, so the results will be unaffected. Note that the field inside the plates would change. 19. Each positive plate produces an electric field directed away from the plate with a magnitude E+ = /2Å0 . (a) Between the plates, the two fields are in opposite directions, so we have Ebetween = E+ – E+ = (/2Å0) – (/2Å0) = 0. (b) Outside the plates, the two fields are in the same direction, so we have Eoutside = E+ + E+ = (/2Å0) + (/2Å0) = /Å0 . (c) If the plates were nonconductors, the fields from each plate would be the same, so the results will be unaffected. Note that the field inside the plates would change.

E+

E+

E+

E+

E+

E+

+

20. Because 3.0 cm « 1.0 m, we can consider the plates to be infinite in size, with no edge effects. From Problem 18, the electric field between the plates depends only on the charge density: E = /Å0 = Q/AÅ0; 100 N/C = Q/(1.0 m)2(8.8510–12 C2/N · m2), which gives Q = 8.8510–10 C.

+

Chapter 22 p. 6

21. (a) In the region where r < r1 , we are inside both spherical r2 shells, so there will be no charge inside a Gaussian surface and we must have r < r1 ; E = 0. r1 (b) In the region where r1 < r < r2 , we are outside the inner shell, so it looks like a point charge; we are inside the outer shell, so it contributes no field: 1 r1 < r < r2 ; E = q1/4pÅ0r2 = 14pr12/4pÅ0r2 = 1r12/Å0r2. (c) In the region where r > r2 , we are outside both shells, so each one looks like a point charge: r > r2 ; E = (q1 + q2)/4pÅ0r2 = (14pr12 + 24pr22)/4pÅ0r2 = (1r12 + 2r22)/Å0r2. (d) To have a zero field in the region where r > r2 , we have 1r12 + 2r22 = 0, or 2/1 = – (r1/r2)2. (e) To have a zero field in the region where r1 < r < r2 , we have 1 = 0. 22. From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent of the direction. For a Gaussian surface we choose a sphere of radius r. On this surface, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. The charge density is  = Q/[)p(r03 – r13)]. (a) For the region where r < r1 , there is no charge inside the Gaussian surface, so we have    E = 0; r < r1. (b) For the region where r1 < r < r0 , we apply Gauss’s law:

r0 r r1

Q

ı E · dA = EA = Qenclosed/0; E4r2 = )p(r3 – r13)/Å0 , which gives E = )p(r3 – r13)/4Å0r2 = Q(r3 – r13)/4Å0(r03 – r13)r2; r1 < r < r0 .

(c) For the region where r > r0 , the electric field is that of a point charge; E = Q/4Å0r2; r > r0 . 23. From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent of the direction. The charge density of the sphere is  = Q/[)p(r03 – r13)]. We can add the field of the point charge at the center to the fields found in Problem 22: (a) For the region where r < r1 , the electric field is that of the point charge at the center:    E = q/4Å0r2; r < r1. (b) For the region where r1 < r < r0 , we add the two fields: E = Q(r3 – r13)/4Å0(r03 – r13)r2 + q/4Å0r2 = (1/4Å0)[Q(r3 – r13) + q(r03 – r13)]/(r03 – r13)r2; r1 < r < r0 .

r0 r q

r1

Q

r

2

Chapter 22 p. 7

(c) For the region where r > r0 , the electric field is that of a point charge equal to the total charge: E = (q + Q)/4Å0r2; r > r0 .

24. (a) For the region where r1 < r < r0 , which is inside the conductor, r0 the electric field must be zero. Thus the net charge inside a Gaussian surface must be zero, so the charge on the inner surface of the shell is – q. r Q (b) Because the net charge on the shell is Q, we have q r1 Qouter + Qinner = Q, or Qouter = Q – (– q) = Q + q. (c) For the region where r < r1 , the electric field is that of the point charge at the center:    E = q/4Å0r2; r < r1. (d) For the region where r1 < r < r0 , E = 0. (e) For the region where r > r0 , the electric field is that of a point charge equal to the total charge: E = (q + Q)/4Å0r2; r > r0 . 25. (a) For points inside the shell, the field will be due to the point charge only: E(r < r0) = q/4pÅ0r2. (b) At points outside the shell, the field will be that of an equivalent point charge equal to the E(r > r0) = (q + Q)/4pÅ0r2. total charge: (c) If q = Q, we have E(r < r0) = Q/4pÅ0r2; E(r > r0) = 2Q/4pÅ0r2. (d) If q = – Q, there is no change inside, so we have E(r < r0) = – Q/4pÅ0r2; E(r > r0) = 0. Thus there will be a field only inside the cavity. 26. We find the radius as a function of time from r = r0 + (?r/?t)t = r0 + (r0/T)t = r0(1 + t/T). From the symmetry of the charge distribution, we know that the electric field must be radial, away from the center of the balloon, with a magnitude independent of the direction. (a) Just outside the charged surface, the electric field is E = /Å0 = Q/4pr2Å0 = Q/4pÅ0r02(1 + t/T)2. (b) Because the radius of the balloon never becomes 4r0 , in the region outside the balloon the electric field is that of a point charge; E = Q/4pÅ0r2 = Q/4pÅ0(4r0)2 = Q/64pÅ0r02. 27. From the symmetry of the charge distribution, for points far from the ends and not too far from the shell, we know that the electric field must be radial, away from the axis of the cylinder, with a magnitude independent of the direction. For a Gaussian surface we choose a cylinder of length ¬ and radius r, centered on the axis. On the ends of this surface, the electric field is not constant but E and dA are perpendicular, so we have E · dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) For the region where r > R0 , the charge inside the Gaussian surface is Q =  2pR0¬. 

¬ r R0

Chapter 22 p. 8

For Gauss’s law we have ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ; 0 + E2pr¬ = 2pR0¬/Å0 , or E = R0/Å0r; r > R0 . (b) For the region where r < R0 , the charge inside the Gaussian surface is Q = 0, so we have    E = 0 for r < R0. (c) We find the equivalent linear charge density from Q = 2pR0L = eqL, which gives eq = 2pR0. If we treat the cylinder as a line of charge, the field is E = 2eq/4Å0r = 2(2pR0)/4Å0r = R0/Å0r; which is the same as the result for part (a). 28. From the symmetry of the charge distribution, for points far ¬ from the ends and not too far from the shell, we know that R the electric field must be radial, away from the axis of the r cylinder, with a magnitude independent of the direction. For a Gaussian surface we choose a cylinder of length ¬ and radius r, centered on the axis. On the ends of this surface, the electric field is not constant but E and dA are perpendicular, so we have E · dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) For the region where r > R0 , the charge inside the Gaussian surface is Q = pR02¬. For Gauss’s law we have ı E · dA = ?ends E · dA + ?side · dA = Qenclosed/Å0 ; 0 + E2pr¬ = pR02¬/Å0 , or E = R02/2Å0r; r > R0 . (b) For the region where r < R0 , the charge inside the Gaussian surface is Q = pr2¬, so we have

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ; 0 + E2pr¬ = pr2¬/Å0 , or

E = r/2Å0; r < R0 .

29. From the symmetry of the charge distribution, for points far –Q from the ends and not too far from the outer shell, we know ¬ R2 that the electric field must be radial, away from the axis of +Q the cylinders, with a magnitude independent of the direction. r For a Gaussian surface we choose a cylinder of length ¬ and R1 radius r, centered on the axis. On the ends of this surface, the electric field is not constant but E and dA are perpendicular, so we have E · dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) For the region where r < R1 , there is no charge inside the Gaussian surface, so we have    E = 0; r < R1. (b) For the region where R1 < r < R2 , the charge inside the Gaussian surface is Qenclosed = (Q/L)¬. For Gauss’s law we have ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ; 0 + E2pr¬ = (Q/L)¬/Å0 , or E = Q/2pÅ0Lr; R1 < r < R2 . (c) For the region where r > R2 , the charge inside the Gaussian surface is Qenclosed = (Q/L)¬ – (Q/L)¬. Thus there is no net charge inside the Gaussian surface, so we have    E = 0; r > R2 . (d) The velocity of the electron moving in a circular orbit is perpendicular to the electric field so the force is toward the axis and provides the centripetal acceleration:

Chapter 22 p. 9

eE = mv2/r. The kinetic energy is K = !mv2 = !eEr = !e(Q/2pÅ0Lr)r = eQ/4pÅ0L. Note that this is the kinetic energy for any orbit between the shells. 30. (a) From the result for part (c) of Problem 29 we see that E = 0 for any Q as long as there are equal and opposite charges on the shells. (b) From the result for part (b) of Problem 29 we see that in the region between the shells we will have E = 0 if Q = 0. Note that E = 0 also if L  8, but this means essentially zero charge density.

31. From the symmetry of the charge distribution, for points far Q2 from the ends and not too far from the outer shell, we know ¬ R that the electric field must be radial, away from the axis of 2 Q1 the cylinders, with a magnitude independent of the direction. r For a Gaussian surface we choose a cylinder of length ¬ and R1 radius r, centered on the axis. On the ends of this surface, the electric field is not constant but E and dA are perpendicular, so we have E · dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) A point 3.0 cm from the axis is inside the inner shell. For the region where r < R1 , there is no charge inside the Gaussian surface, so we have    E = 0; r = 3.0 cm. (b) A point 6.0 cm from the axis is between the shells. For the region where R1 < r < R2 , the charge inside the Gaussian surface is Qenclosed = (Q1/L)¬. For Gauss’s law we have

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

0 + E2pr¬ = (Q1/L)¬/Å0 , so E = Q1/2pÅ0Lr = (– 3.810–6 C)/2p(8.8510–12 C2/N · m2)(5.0 m)(0.060 m) = – 2.3105 N/C (toward the axis), r = 6.0 cm. (c) A point 12.0 cm from the axis is outside the shells. For the region where r > R2 , the charge inside the Gaussian surface is Qenclosed = [(Q1/L) + (Q2/L)]¬. For Gauss’s law we have

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

0 + E2pr¬ = [(Q1/L) + (Q2/L)]¬/Å0 , so E = (Q1 + Q2)/2pÅ0Lr = [(– 3.810–6 C) + (+ 3.210–6 C)]/2p(8.8510–12 C2/N · m2)(5.0 m)(0.120 m) = – 1.8104 N/C (toward the axis), r = 12.0 cm. 32. (a) The inward electric field will produce a force on the electron out along a radial line. The work done by this variable force will increase the kinetic energy of the electron: !mv2 – 0 = ?F dr = ? – eE dr. When we use the result from part (b) of Problem 31 and integrate, we get  R2 Q1 – eQ 1 R2 1 2 –e dr = ln ; 2 mv = R1 2Å Lr 2Å L R1

1 2

0

9.11  10 – 31 kg v 2 =

0

– 1.60 10– 19 C – 3 .8  10 – 6 C 2 8.85  10

– 12

2

C / N · m 2 5.0 m

ln 9.0 cm , which gives 5.0 cm

Chapter 22 p. 10

v = 5.3107 m/s. (b) The velocity of the proton moving in a circular orbit is perpendicular to the electric field so the force is toward the axis and provides the centripetal acceleration: + eE = mv2/r. We use the result for E from part (b) of Problem 31: 1.1106 m/s. (1.6010–19 C)(2.3105 N/C) = (1.6710–27 kg)v2/(0.060 m), which gives v =

33. From the symmetry of the charge distribution, for points far from the ends and not too far from the shell, we know that the electric field must be radial, away from the axis of the R2 R3 cylinders, with a magnitude independent of the direction. For a Gaussian surface we choose a cylinder of length ¬ and radius r, centered on the axis. On the ends of this surface, the R1 electric field is not constant but E and dA are perpendicular, so we have E · dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) For the region where r < R1 , the charge inside the Gaussian surface is Q = pr2¬, so we have

¬ r





ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

0 + E2pr¬ = pr2¬/Å0 , or E = r/2Å0; r < R1 . (b) For the region where R1 < r < R2 , the charge inside the Gaussian surface is Qenclosed = pR12¬, so we have

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

0 + E2pr¬ = pR12¬/Å0 , or E = R12/2Å0r; R1 < r < R2 . (c) For the region where R2 < r < R3 , the charge inside the Gaussian surface is Qenclosed = pR12¬ + p(r2 – R22)¬, so we have

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

0 + E2pr¬ = [pR12¬ + p(r2 – R22)¬]/Å0 , or E = (r2 + R12 – R22)/2Å0r; R2 < r < R3 . (d) For the region where r > R3 , the charge inside the Gaussian surface is Qenclosed = pR12¬ + p(R32 – R22)¬, so we have

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

0 + E2pr¬ = [pR12¬ + p(R32 – R22)¬]/Å0 , or E = (R32 + R12 – R22)/2Å0r; r > R3 . (e) For the given data we have E = (1510–6 C/m3)r/2(8.8510–12 C2/N · m2) = (8.5105 N/C · m)r, r < R1 ; E = (8.5105 N/C · m)(0.050 m)2/r = (2.1103 N · m/C)/r, R1 < r < R2 ; E = (8.5105 N/C · m)[r2 + (0.050 m)2 – (0.100 m)2]/r = (8.5105 N/C · m)[r2 – 0.0075 m2]/r, R2 < r < R3 ; E = (8.5105 N/C · m)[(0.150 m)2 + (0.050 m)2 – (0.100 m)2]/r = (1.3104 N · m/C)/r, r > R3 .

Chapter 22 p. 11

E (104 N/C) 10 8 6 4 2 0

0

5

10

15

20

r (cm)

34. From the symmetry of the charge distribution, we know that the electric field must be radial, with a magnitude independent of the direction. For a Gaussian surface we choose a sphere of radius r. On this surface, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) For the region where r < r1 , there is no charge inside the Gaussian surface, so we have    E = 0; r < r1. (b) For the region where r1 < r < r0 , we find the enclosed charge by integrating: r r  r Q enclosed =  E4r 2 dr = 4  0 1 r 2 dr = 4 0r 1 12 r 2 – r 1 2 . r r1 r1

r0 r

We apply Gauss’s law: ı E · dA = EA = Qenclosed/0; E4r2 = 2r1(r2 – r12)/Å0, which gives E = r1(r2 – r12)/2Å0r2; r1 < r < r0 . (c) For the region where r > r0 , the enclosed charge is the total charge on the sphere: r0 r0  r Q enclosed =  E4r 2 dr = 4   0 1 r 2 dr = 4  0 r1 12 r20 – r1 2 . r r1 r1

(d)

We apply Gauss’s law: ı E · dA = EA = Qenclosed/0 ; E4r2 = 2r1(r02 – r12)/Å0 , which gives

E = r1(r02 – r12)/2Å0r2; r > r0 .

r1



Chapter 22 p. 12

E

0

r1

r0

2r0

35. On the ends of the cylinder the electric field will vary in magnitude and direction. Thus we must integrate to find the flux through the ends. We choose a circular ring of radius y and thickness dy. From the diagram we see that R0 = r cos , y = R0 tan , dy = R0 sec2  d = (R0/cos2 ) d. The flux through one end is  Q  end = E  d A = E cos  d A = cos  2 y d y 4Å0r 2

= Q 2Å0

/ 4 0

cos  R 0 tan  R 0 / cos 2  d  R 0/ cos 

2

= Q 2Å0

/ 4 0

E dA

r

dy y

R0

r 

Q

R0

sin  d  = Q – cos  2Å0

/ 4 0

The total flux through the closed surface is Q/Å0 , so the flux through the curved sides is sides = total – 2end = (Q/Å0) – 2(Q/2Å0)[1 – 1/v2) = Q/Å0v2. 36. Because the slab is very large, we know from symmetry that the y field must be perpendicular to the slab, with a constant magnitude for a constant distance from the center. If  is positive, the field will be away from the center. For a Gaussian surface we choose a cylinder of length 2x and area A, centered on the axis. On the curved side of this surface, the electric field is not constant but x E and dA are perpendicular, so we have E · dA = 0. On the ends, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. (a) To find the field inside the slab, we use the fact that the field will be away from the center. If we place our Gaussian cylinder so that one end is at x, with x < !d, and the other end z is at – x, the fields at each end will be directed out of the Gaussian surface and have the same magnitude. When we apply Gauss’s law, we have ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ;

= Q 1– 1 . 2 2Å0

d

x A

E = x/Å0; x< !d . 2EA + 0 = 2xA/Å0 , or (b) To find the field outside the slab, we use the fact that the field will be away from the slab. If we place our Gaussian cylinder so that one end is at x, with x > !d, and the other end is at – x, the fields at each end will be directed out of the Gaussian surface and have the same magnitude. When we apply Gauss’s law, we have

x

Chapter 22 p. 13

ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ; E = d/2Å0; x > !d (away from slab). 2EA + 0 = Ad/Å0 , or 37. The gravitational field is F = – GM r. g=m r2 If we compare this to the electric field, E =  1 Q2 r, 4Å0 r we see that E  g, Q  – M, and Å0  1/4pG. If we make these substitutions in Gauss’s law for the electric field, we have  = ı E · dA = Q/Å0  = ı g · dA = – M/(1/4G) = – 4GM. Thus

ı g · dA = – 4GM

is Gauss’s law for the gravitational field, where M is the enclosed mass.

38. (a) For a conducting spherical surface, the radial electric field just outside the surface is E = /Å0 = Q/4prE2Å0 ; – 150 N/C = Q/4p(6.38106 m)2(8.8510–12 C2/N · m2), which gives Q = – 6.8105 C. (b) The surface density of electrons is n = Q/e4prE2 = Å0E/e = (8.8510–12 C2/N · m2)(150 N/C)/(1.6010–19 C/electron) = 8.3109 electrons/m2. 39. The flux through a Gaussian surface depends on the enclosed charge. Because the field is parallel to the y-axis, the only faces that will have flux through them are the ones perpendicular to the y-axis. Thus we have ı E · dA = ?y = ¬ E · dA + ?y = 0 E · dA = Qenclosed/Å0 ; (a + b¬)¬2 – (a + 0)¬2 = Qenclosed/Å0 , which gives Qenclosed = Å0b¬3.

z ¬ ¬

y ¬

x 40. (a) We find the value of b by integrating to get the total charge, with a spherical shell as the differential element: r0  Q = E d V = br4r 2 d r = br 40 , which gives b = Q/pr04. 0

(b) To find the field inside the sphere, we choose a Gaussian surface of radius r < r0. We find the charge within this surface by integrating: r  Q enclosed =  E d V = br4r 2 d r = br 4 = Qr4 / r04. 0

When we apply Gauss’s law, we have ı E · dA = EA = Qenclosed/Å0 ; E4r2 = Q(r4/r04)/Å0, which gives E = Qr2/4pÅ0r04; r < r0 . (c) To find the field outside the sphere, we choose a Gaussian surface of radius r > r0. The charge within this surface is the total charge, so we have ı E · dA = EA = Qenclosed/Å0 ; E4r2 = Q/Å0, which gives E = Q/4pÅ0r2; r > r0 .

E

r0 r



Chapter 22 p. 14

41. The flux through a Gaussian surface depends on the enclosed charge. For the sphere with a radius of 1.00 m, the only charge inside is the one at the origin, so we have  = Qenclosed/Å0 = Q1/Å0 = (+ 3.5010–9 C)/(8.8510–12 C2/N · m2) = 3.95102 N · m2/C. For the sphere with a radius of 2.00 m, both charges are inside the surface, so we have  = Qenclosed/Å0 = (Q1+ Q2)Å0 = (+ 3.5010–9 C – 5.0010–9 C)/(8.8510–12 C2/N · m2) = – 1.69102 N · m2/C.

y

ra

Q1

+

– Q2 rb

42. (a) The flux through any closed surface containing the charge must be the same, so the flux through the larger sphere is + 500 N · m2/C. (b) The flux through a Gaussian surface depends on the enclosed charge:  = Qenclosed/Å0 = Q/Å0 ; + 500 N · m2/C = Q/(8.8510–12 C2/N · m2), which gives Q = + 4.4310–9 C = + 4.43 nC.

43. (a) Because there is no charge within the sphere, every field line Q will enter and leave the sphere, so the net flux through r0/2 the sphere is 0. (b) The maximum electric field will be at the point on the sphere closest to the point charge, the top of the sphere: r0 Emax = Q/4Å0(!r0)2 = Q/Å0r02. The minimum electric field will be at the point on the sphere farthest from the point charge, the bottom of the sphere: Emin = Q/4Å0((r0)2 = Q/25Å0r02. E Thus the range of values is Q/25Å0r02 = E = Q/Å0r02. E (c) The electric field is not perpendicular at all points. E It is perpendicular only at the top and bottom of the sphere. At some points it is almost parallel to the surface of the sphere. (d) The electric field is not perpendicular or constant over the surface of the sphere. Gauss’s law is not useful for obtaining E because a Gaussian surface cannot be chosen that simplifies the integral for the flux.

x

Chapter 22 p. 15

44. The uniform fields from each of the three sheets are indicated on the diagram. We take the positive direction upward. The force on each sheet is produced by the net electric field from the other two sheets: FI/A = I(– EII + EIII) = I(– II + III)/2Å0 = (9.010–9 C/m2)(– 2.010–9 C/m2 + 5.010–9 C/m2)/ 2(8.8510–12 C2/N · m2) –6 2 = 1.510 N/m (up). FII/A = II(– EI + EIII) = II(– I + III)/2Å0 = (– 2.010–9 C/m2)(– 9.010–9 C/m2 + 5.010–9 C/m2)/ 2(8.8510–12 C2/N · m2) –7 2 = 4.510 N/m (up). FIII/A = III(– EI + EII) = III(– I + II)/2Å0 = (5.010–9 C/m2)(– 9.010–9 C/m2 + 2.010–9 C/m2)/ 2(8.8510–12 C2/N · m2) –6 2 = – 2.010 N/m (down).

I + EII

II – EI

45. (a) We determine the constant A by finding the total negative charge:   – e = E d V = – A exp – 2r/ a0 4r 2 d r. 0

If we change variable to x = r/a0 , we have   e = 4Aa30 exp – 2x x 2 d x = 4A a03 14 , which gives 0

A = e/pa03. We do a similar integration to find the charge inside a sphere with radius a0 :

 Q=

 E dV =

a0

0 1

– A exp – 2r/ a0 4r 2 dr

e a 3 exp – 2x x 2 dx = – 4e – 5 1 2 + 4 = a30 0 0 4 exp 1 Thus the negative charge is Q = {[5/(2.718)2] – 1}e = – 0.323e. Thus the net charge is = – 4

EIII

5 2 – 1 e. exp 1

EII

III +

Chapter 22 p. 16

Qnet = – 0.323e + e = 0.677e = + 1.0810–19 C. Note that we have used exp for the exponential to avoid confusion with the charge e. (b) The electric field at r = a0 will be due to the positive charge + e and the negative charge within a0 : Eb = Qnet/4pÅ0a02 = (+ 1.0810–19 C)(9.0109 N · m2/C2)/(0.5310–10 m)2 = 3.51011 N/C. 46. (a) The field from a large plate, not near the edge, is  perpendicular to the plate and uniform: E = /2Å0 . For regions outside the slab, it can be considered an E infinite number of plates. We can find the equivalent Eb Ea surface density by considering the charge in the slab with an area A: x Qslab = slabA = EAd, or slab = Ed. The electric field to the left of the plate is Ea = Eplate + Eslab A Ec = /2Å0 + slab/2Å0 = ( + Ed)/2Å0 (left). (b) The electric field to the right of the plate is d Eb = Eplate + Eslab = /2Å0 + slab/2Å0 = ( + Ed)/2Å0 (right). (c) To find the field inside the slab, we choose a cylinder for the Gaussian surface with one end of area A inside the slab parallel to the plate and the other end of area A to the left of the plate. The cylinder is a distance x inside the slab. When we apply Gauss’s law, we have ı E · dA = ?ends E · dA + ?sideE · dA = Qenclosed/Å0 ; EaA + EcA + 0 = (A + ExA)/Å0 ; ( + Ed)/2Å0 + Ec = ( + Ex)/Å0 , so Ec = [ + E(2x – d)]/2Å0 (right).

47. From symmetry, we know that the field inside a uniformly charged sphere must be radial and depends only on the distance from the center. At a distance r from the center, r = r0 , only the charge inside a spherical surface with a radius r will provide the field: E B E(r = r0) = Qenclosed/4pÅ0r2 = )pr3E/4pÅ0r2 = Er/3Å0 . A C We create the cavity by adding to the original sphere, with charge r0 /2 density E , a sphere with charge density – E and radius !r0 , centered r 0 at C. At any point the total field will be the sum of the fields from the two spheres, which we label E+ and E– . (a) At the point A we have EA = E+ + E– = 0 – [– E(!r0)/3Å0] = Er0/6Å0 (right). (b) At the point B the cavity can be treated as a point charge, so we have EB = E+ + E– = – Er0/3Å0 – [– E)p(!r03)/4pÅ0(*r0)2] = – 17Er0/54Å0 (left). By considering other points in the cavity, it can be shown that the field inside the cavity is uniform.

Chapter 22 p. 17

48. For a charged spherical surface, the radial electric field just outside the surface is E = /Å0 = Q/4pÅ0r2; 3106 N/C = Q/4p(8.8510–12 C2/N · m2)(0.37510–2 m)2, which gives Q = 510–9 C =

5 nC.

1 3 Q2 = 0 49. The positive sheet produces an electric field directed + – + – away from the plate with a magnitude E1 = 1/2Å0 . + The negative sheet produces an electric field directed – + – E1 E1 toward the plate with a magnitude b c E3 = 3/2Å0 . a + E3 – + (a) Because charges are free to move in a conductor, – E3 the field inside the middle sheet is 0. (b) Between the left and middle sheets, the two fields + – + – are in the same direction, so we have Eb = E1 + E3 = (1/2Å0) + (3/2Å0) = (1 + 3)/2Å0 = (5.0010–6 C/m2 + 5.0010–6 C/m2)/2(8.8510–12 C2/N · m2) = 5.65105 N/C (right). (c) Between the middle and right sheets, the two fields are in the same direction, so we have Eb = E1 + E3 = (1/2Å0) + (3/2Å0) = (1 + 3)/2Å0 = (5.0010–6 C/m2 + 5.0010–6 C/m2)/2(8.8510–12 C2/N · m2) = 5.65105 N/C (right). (d) To find the charge density on the surface of the left side of the middle sheet, we choose a cylinder for the Gaussian surface with one end of area A inside the sheet and the other end of area A between the left and middle sheets. When we apply Gauss’s law, we have ı E · dA = ?ends E · dA + ?side E · dA = Qenclosed/Å0 ; – EbA + 0 + 0 = leftA/Å0 ; – (1 + 3)/2Å0 = left/Å0 , so left = – (5.0010–6 C/m2 + 5.0010–6 C/m2)/2 = – 5.0010–6 2 C/m . (e) Because the middle sheet has no net charge, the charge density on the right side must be right = – left = + 5.0010–6 C/m2.

50. We find the net charge inside the cube by finding the flux through each face of the cube. Because the electric field has only x- and y-components, we know that z = 0 = ? E · dA = 0; z = a = ? E · dA = 0. For the other sides we choose a horizontal strip of height dz for a differential element and integrate to find the flux: a a  E0 1 + za i + E 0 za j  a d z ( – i) = – E0 a 1 + za d z x = 0 = E  d A = 0 0

= – E0 a a – 0 +

a2 – 0 = – 32 E0 a2; 2a

z a dz x

a a E y

Chapter 22 p. 18

 x = a =

E  dA =

 y = 0 =

E  dA =

y = a =

E  dA =

a 0 a 0 a 0

a

E0 1 + za i + E0 za j

 adz (i ) = E0 a

E0 1 + za i + E 0 za j

 a d z ( – j) = – E0

E 0 1 + za i + E0 za j

 a d z (j) = E0

For the total flux through the surface of the cube we have total = x = 0 + x = a + y = 0 + y = a + z = 0 + z = a = – *E0a2 + *E0a2 – !E0a2 + !E0a2 + 0 + 0 = 0. Thus there is no net charge inside the cube.

0

a 0

1 + za dz = 32 E 0a2 ; a 0

z d z = – 12 E0 a2;

z d z = 12 E0 a2 .

CHAPTER 23 – Electric Potential 1.

We find the work done by an external agent from the work-energy principle: Wab = ?K + ?U = 0 + q(Vb – Va) = (– 7.010–6 C)(+ 6.00 V – 0) = – 4.210–5 J (done by the field).

2.

We find the work done by an external agent from the work-energy principle: W = ?K + ?U = 0 + q(Vb – Va) = (1.6010–19 C)[(– 50 V) – (+ 100 V)] = – 2.4010–17 J (done by the field).

3.

Because the total energy of the electron is conserved, we have ?K + ?U = 0, or 3.410–15 J. ?K = – q(Vb – Va) = – (– 1.6010–19 C)(21,000 V) =

4.

Because the total energy of the electron is conserved, we have ?K + ?U = 0; ?K + q(VB – VA) = 0; 16.410–16 J + (– 1.6010–19 C)(VB – VA) = 0, which gives VB – VA = Plate B is at the higher potential.

1.03104 V.

5.

We find the potential difference from the work-energy principle: Wab = ?K + ?U = ?K + q(Vb – Va) 8.0010–4 J = 2.1010–4 J + (– 8.1010–6 C)(Vb – Va), which gives Vb – Va = – 72.8 V, or Va – Vb = + 72.8 V.

6.

For the uniform electric field between two large, parallel plates, we have E = ?V/d; 3.0 cm. 1500 V/m = (45 V)/d, which gives d = 3.010–2 m =

7.

For the uniform electric field between two large, parallel plates, we have E = ?V/d; 7.04 V. 640 V/m = ?V/(11.010–3 m), which gives ?V =

8.

For the uniform electric field between two large, parallel plates, we have E = ?V/d = (110 V)/(5.010–3 m) = 2.2104 V/m.

9.

The maximum charge will produce the electric field that causes breakdown in the air: E = Q/4pÅ0r2; 3106 V/m = (9.0109 N · m2/C2)Q/(0.050 m)2, which gives Q = 810–7 C = 0.8 C.

10. The electric field at the spherical surface is E = Q/4pÅ0r2,

while the potential of a sphere, with V = 0 at 8, is V = Q/4pÅ0r. Thus we have r = V/E, so rmin = V/Emax = (30,000 V)/(3106 V/m) = 110–2 m = 1 cm. At this radius the charge is Q = 4pÅ0rminV = (110–2 m)(30,000 V)/(9.0109 N · m2/C2) = 310–8 C.

11. The potential difference between two points in an electric field is found from ?V = – ? E · d¬. (a) For VBA we have B

V BA = –

A

E·d = –

B A

– 300 N/ C i · d yj =

0.

(b) For VCB we have C

V CB = –

B

E·d = –

C B

– 300 N/ C i · dx i =

= 300 N/ C – 3 m – 4 m = (c) For VCA we have C

V CA = –

A

E·d = –

C A

–3m 4m

– 2100 V.

– 300 N/ C i · dxi + d yj =

= 300 N/ C – 3 m – 4 m = Note that VCA = VCB + VBA .

300 N/ C dx

–3m 4m

300 N/ C dx

–2100 V.

12. The electric field produced by a large plate is uniform with magnitude /2Å0. If we take the potential of the plate to be V0 , we find the potential a distance x from the plate by integrating: x x  V – V0 = – E·d = – E d x = – ( / 2Å0)x, or 0

V = V0 – (/2Å0)x.

0

13. (a) The electric field at the spherical surface is E = Q/4pÅ0rE2,

while the potential of a sphere, with V = 0 at 8, is V = Q/4pÅ0rE . Thus we have V = ErE = (– 150 V/m)(6.38106 m) =

– 9.6108 V.

(b) The difference in potential between the surface and 8 must be the same, so we add the same constant to both locations: V(8) = + 9.6108 V. If there is an approximately equal positive charge in the ionosphere, the potential at the surface will be increased by the potential of the positive spherical charge. Because the radius of the ionospheric charge is greater than rE , its potential will have a positive value with a magnitude less than that found in part (a). Thus the potential at the surface of the Earth will be negative but with a much smaller magnitude. 14. (a) The potential at the surface of a charged sphere in terms of the charge density is V = Q/4pÅ0r = 4pr 2/4pÅ0r =  r/Å0; 500 V = (0.16 m)/(8.8510–12 C2/N · m2), which gives  = 2.810–8 C/m2. (b) If we form the ratio for the two distances, we have V1/V0 = r0/r1 ; (10 V)/(500 V) = (0.16 m)/r1 , which gives r1 = 8.0 m. 15. (a) After the connection, if the two spheres were at different potentials, there would be a flow of charge in the wire. Thus the potentials must be the same. (b) We assume the spheres are so far apart that the potential of one sphere at the other sphere is essentially zero. The initial potentials are V01 = Q/4pÅ0r1 , V02 = 0. After the connection, Q2 is transferred to the second sphere, so we have V1 = (Q – Q2)/4pÅ0r1 = V2 = Q2/4pÅ0r2 , or r2(Q – Q2) = r1Q2 , which gives Q2 = r2Q/(r1 + r2).

16. The radial electric field of the long wire is E = /2pÅ0r. We find the potential difference from

Vb – Va = –

Rb Ra

E·d = –

Rb Ra

E dr = –

Rb Ra

 dr = –  ln R b = Ra 2Å0r 2Å0

17. (a) The reference level for the potential is V = 0 at r = 8. At points outside the sphere, the electric field is that of a point charge: E = Q/4pÅ0r2 , r > r0 . Thus the potential when r > r0 is V = Q/4pÅ0r, r > r0 . (b) The charge density inside the sphere is  = Q/)pr03. The electric field at a distance r < r0 is due to the charge inside the sphere with radius r: E = ( )pr3)/4pÅ0r2 = Qr/4pÅ0r03, r < r0 . We find the potential by integrating along a radial line from r to r0 : r 0 r0 r0 dV = – E· d = – E d r; r

r

(c)

 l n Ra . Rb 2Å0 V

0

r0

r

E

r

r0

Q –Q –Q –V= rd r = r20 – r 2 , which gives 4Å0r 0 4Å0r03 r 8Å0r 30 V = (Q/8pÅ0r0)[3 – (r2/r02)], r < r0 .

0

r0

r

18. (a) The reference level for the potential is V = 0 at r = 8. (c) At points outside the sphere, the electric field is that of a point charge: E = Q/4pÅ0r2 , r > r0 . Thus the potential when r > r0 is V = Q/4pÅ0r, r > r0 . (b) If we let the charge density inside the sphere be E = br2, we relate the constant b to the total charge:  r0  4r2 d r = r 0 4br4 d r = 4br5 / 5. Q= E 0 0

r

0

The electric field at a distance r < r0 is due to the charge inside the sphere with radius r: r  4 4br d r 3 3 0 E= 1 = 1 4br = Qr 5 , r < r0 . 2 5 4Å0 4Å0 r 4Å0 r0 We find the potential by integrating along a radial line from r to r0 : r 0 r0 r0 dV = – E· d = – E dr; r

V

r

0

r0

E

r

r0 Q Q Q –V = r 3 dr = r 40 – r 4 , which gives 5 r 4Å0r 0 4Å0 r0 16Å0r 50 V = (Q/16pÅ0r0)[5 – (r4/r04)], r < r0 .

0

r

r0

19. The field outside the cylinder is the same as that of a long wire. We find the equivalent linear charge density from the charge on the length L: Q =  2pR0L = L, which gives  =  2pR0 . (a) The radial electric field outside the cylinder is E = /2pÅ0r =  2pR0 /2pÅ0r =  R0 /Å0r . We find the potential difference from r r  R0 dr R0 V – V0 = – E·d = – =– ln r , or Å r Å0 R0 R0 0 R 0

V = V0 + (R0/Å0) ln(R0/r), r > R0 . (b) The electric field inside the cylinder is zero, so the potential inside is constant and equal to the potential at the surface: V = V0 , r < R0 . (c) From the result in part (a) we see that the potential at r = 8 is undefined. would be charge at infinity for an infinite cylinder.

V?0

because there

20. (a) In each region the electric field is the same as that of a point charge equal to the net charge within a spherical surface. Thus we have E = (+ Q + !Q)/4pÅ0r2 = 3Q/8pÅ0r2, r > r2 . E = 0, r1 < r < r2

(inside a conductor), which means there is a negative charge – !Q on the

inner surface; thus a positive charge *Q on the outer surface; E = !Q/4pÅ0r2 = Q/8pÅ0r2, 0 < r < r1 .

(b) The reference level for the potential is V = 0 at r = 8. At points outside the spherical conducting shell, the potential is that of a point charge with net charge Q + !Q. Thus the potential when r > r2 is

(e)

V

V = (+ Q + !Q)/4pÅ0r = 3Q/8pÅ0r, r > r2 . (c) The electric field within the spherical conductor is zero, so its potential must be constant and therefore equal to the potential on the surface: V = 3Q/8pÅ0r2 = 3Q/16pÅ0r1 , r1 < r < r2 . (d) The potential inside the hollow region is due to the point charge at the center with an additive constant to make the potential at the inner surface of the conductor the constant potential of the conductor: V = !Q/4pÅ0r + constant; 3Q/16pÅ0r1 = Q/8pÅ0r1 + constant, which gives constant = Q/16pÅ0r1 . Thus the potential is V = (Q/8pÅ0)[(1/r) + (1/2r1)], 0 < r < r1 .

0

r1

r2

2r2

r1

r2

2r2

E

0

21. (a) The potential from the proton is V = Q/4pÅ0r = (9.0109 N · m2/C2)(1.6010–19 C)/(0.5010–10 m) = (b) We find the potential energy of the electron from U = qV = (– 1 e)(29 V) = – 29 eV (– 4.610–18 J). 22. We find the charge from V = Q/4pÅ0r; 125 V = (9.0109 N · m2/C2)Q/(1510–2 m), which gives Q = 2.110–9 C = 23. We find the electric potentials of the stationary charges at the initial and final points: Va = (1/4pÅ0)[(Q1/r1a) + (Q2/r2a)]

r

Q1 +

29 V.

2.1 nC.

a L

r

b d

Q2 +

= (9.0109 N · m2/C2){[(2510–6 C)/(0.030 m)] + [(2510–6 C)/(0.030 m)]} = 1.50107 V. Vb = (1/4pÅ0)[(Q1/r1b) + (Q2/r2b)]

x

= (9.0109 N · m2/C2){[(2510–6 C)/(0.040 m)] + [(2510–6 C)/(0.020 m)]} = 1.69107 V. Because there is no change in kinetic energy, we have Wa b = ?K + ?U = 0 + q(Vb – Va) = (0.1010–6 C)(1.69107 V – 1.50107 V) = + 0.19 J. 24. (a) The electric field will be zero when the fields Q1 E1 Q2 L from the two charges have the same magnitude and + – are in opposite directions, so the point must be outside b E2 a the two charges: x=0 E1 = E2 ; (1/4pÅ0)(Q1/r12) = (1/4pÅ0)(Q2/r22); (3.0 C)/x2 = (2.0 C)/(x – 4.0 cm)2, which gives x = 2.2 cm, 21.8 cm Because the point must be outside, the point is 22 cm from the positive charge and 18 cm from the negative charge. (b) The potential is a scalar that depends only on the distance. The potential for two charges is V = (1/4pÅ0)[(Q1/r1) + (Q2/r2)]. If the potential is 0 at a point x, we have 0 = (1/4pÅ0){[(3.0 C)/x] + [(– 2.0 C)/(x – 4.0 cm)]}, which gives 2.0x= 3.0(x – 4.0 cm). For a point between the two charges, we have 2.0x = 3.0(4.0 cm – x1), which gives x1 = 2.4 cm from positive charge. For a point outside the two charges, we have 2.0x = 3.0(x2 – 4.0 cm), which gives x2 = 12.0 cm from the positive charge and 8.0 cm from the negative charge.

x

25. When the proton is accelerated by a potential difference, it acquires a kinetic energy: K = QpVaccel . If it is far from the silicon nucleus, the potential is zero. The proton will slow as it approaches the positive charge of the nucleus, because the potential produced by the silicon nucleus is increasing. At the proton’s closest point the kinetic energy will be zero. We find the required accelerating potential from ?K + ?U = 0; 0 – K + Qp(VSi – 0) = 0, or QpVaccel = QpQSi/4pÅ0(rp + rSi); Vaccel = (9.0109 N · m2/C2)(14)(1.6010–19 C)/(1.210–15 m + 3.610–15 m) = 4.2106 V = 4.2 MV. 26. (a) We find the electric potentials at the two points: Va = Q/4pÅ0ra = (9.0109 N · m2/C2)(– 3.810–6 C)/(0.70 m) = – 4.89104 V. Vb = Q/4pÅ0rb = (9.0109 N · m2/C2)(– 3.810–6 C)/(0.80 m) = – 4.28104 V. Thus the difference is Vba = Vb – Va = – 4.28104 V – (– 4.89104 V) = + 6.1103 V. (b) We find the electric fields at the two points: Ea = Q/4pÅ0ra2 = (9.0109 N · m2/C2)(– 3.810–6 C)/(0.70 m)2 = 6.98104 N/C toward Q (down). Eb = Q/4pÅ0rb2 = (9.0109 N · m2/C2)(– 3.810–6 C)/(0.80 m)2

y a

Ea ra Eb

rb

+–

b

Q

Eb – E a 

Eb

–E a

x

= 5.34104 N/C toward Q (right). As shown on the vector diagram, we find the direction of Eb – Ea from tan  = Ea/Eb = (6.98104 N/C)/(5.34104 N/C) = 1.307, or  = 53° N of E. We find the magnitude from Eb – Ea = Eb/cos  = (5.34104 N/C)/cos 53° = 8.8104 N/C. 27. When the electron is far away, the potential from the fixed charge is zero. Because energy is conserved, we have ?K + ?U = 0; !mv2 – 0 + (– e)(0 – V) = 0, or !mv2 = – e(kQ/r) !(9.1110–31 kg)v2 = – (1.6010–19 C)(9.0109 N · m2/C2)(– 0.12510–6 C)/(0.725 m), 2.33107 m/s. which gives v = 28. We find the electric potential energy of the system by considering one of the charges to be at the potential created by the other charge. This will be zero when they are far away. Because the masses are equal, the speeds will be equal. From energy conservation we have ?K + ?U = 0; !mv2 + !mv2 – 0 + Q(0 – V) = 0, or 2(!mv2) = mv2 = Q(kQ/r) = kQ2/r; (1.010–6 kg)v2 = (9.0109 N · m2/C2)(7.510–6 C)2/(0.055 m), 3.0103 m/s. which gives v = 29. We find the electric potentials from the charges at the two points: VA = (1/4pÅ0){(+ q/b) + [– q/(d – b)]}

d +

= (1/4pÅ0)q{(1/b) – [1/(d – b)]} = (1/4pÅ0)q(d – 2b)/b(d – b).

VB

b

+q

= (1/4pÅ0){[+ q/(d – b)] + (– q/b)}

A

= (1/4pÅ0)q{[1/(d – b)] – (1/b)} = (1/4pÅ0)q(2b – d)/b(d – b). Thus we have VBA = VB – VA = [(1/4pÅ0)q(2b – d)/b(d – b)] – [(1/4pÅ0)q(d – 2b)/b(d – b)] = (1/2pÅ0)q(2b – d)/b(d – b). Note that VBA is negative and, as expected, VBA = 0 when b = !d. 30. For the potential at point A we have VA = (1/4pÅ0)(Q1/L + Q2/Lv2 + Q3/L ) = (Q/4pÅ0)(3/L + 1/Lv2 – 2/L) = (1 + 1/v2)Q/4pÅ0L.

Q2 = + Q +

L

Q 3 = – 2Q –

L

+ Q1 = + 3 Q

A

B

b

–

–q

31. We choose a ring of radius r and width dr for a differential element, with charge dq = 2pr dr. The potential of this element on the axis a distance x from the ring is dV = dq/4pÅ0(x2 + r2)1/2 = 2pr dr/4pÅ0(x2 + r2)1/2 =  r dr/2Å0(x2 + r2)1/2. We integrate to get the potential: R2

V= R1

=

 r dr 2Å0 x 2 + r 2

 2Å0

x 2 + R 22

1/ 2

R2 r

x

R2 =  x 2 + r 2 1/ 2 2Å0 R1

1/ 2

– x 2 + R 21

1/ 2

R1

.

y

32. We choose a differential element of the rod at position x, length dx, and charge dq =  dx = (Q/2L) dx. From the diagram, we see that r2 = x2 + y2. The potential on the y-axis from the differential element is dV = (1/4Å0) dq/r = (Q/2L) dx/4Å0r . The potential from the rod is

V=

=

Q 8Å0 L

x = L x = – L

Q ln 8Å0 L

d x = Q r 8Å0L 2

x + y 2 + x

L

=

Q ln 8Å0L

L L 2 + y2 + L 2

L + y2 – L Note that we have used a length of 2L for the rod. –L

d x

x

d x x 2 + y 2

–L L

r

33. We choose a differential element of the rod at position x, length dx, and charge dq =  dx = (Q/2L) dx. From the diagram, we see that r = x – x. The potential on the x-axis from the differential element is dV = (1/4Å0) dq/r = (Q/2L) dx/4Å0r . The potential from the rod is

x

L

.

y d x

x L

L

r x

V=

x = L L Q d x = Q dx 8Å0 L x = – L r 8Å0L – L x – x

L –Q –Q Q l n x – x = ln x – L = l n x + L , x > L. 8Å0 L 8Å0L x+ L 8Å0 L x –L –L Note that we have used a length of 2L for the rod.

=

y

34. We choose a differential element of the rod at position x, length dx, and charge dq =  dx = ax dx. (a) From the diagram, we see that r2 = x2 + y2. The potential on the y-axis from the differential element is dV = (1/4Å0) dq/r = ax dx/4Å0r . The potential from the rod is

V=

L

x = L

a x dx = a 4Å0 x = – L r 4Å0 –L

r

x dx x 2 + y 2

x

r

dx

x

L L L a x 2 + y 2 = a L2 + y 2 – L2 + y 2 = 0. 4Å0 4Å0 –L This is expected because the potential from the negative charge on the left half of the rod is balanced by the potential from the positive charge on the right half of the rod. (b) From the diagram, we see that r = x – x. The potential on the x-axis from the differential element is dV = (1/4Å0) dq/r = ax dx/4Å0r . The potential from the rod is

=

V=

L a x = L x d x = a x  d x 4Å0 x = – L r 4Å0 – L x – x L

a x – x – x l n x – x = a – L – L – x ln x – L 4Å0 4Å0 x+L –L Note that we have used a lengh of 2L for the rod. =

35. We choose a ring of radius r and width dr for a differential element, with charge dq = 2pr dr. The potential of this element on the axis a distance x from the ring is dV = dq/4pÅ0(x2 + r2)1/2 = 2pr dr/4pÅ0(x2 + r2)1/2 = ar3 dr/2Å0(x2 + r2)1/2. We integrate to get the potential:

=

a x l n x + L – 2L , x > L. 4Å0 x –L

R r

x

V= a 2Å0

r= R

r 3 dr 2 1/ 2 r= 0 x + r 2

= a r2 x 2 + r2 2Å0 = a 2Å0 = a 2Å0 =

1 3 1 3

x 2 + r2 2

x +R

1/ 2

1/ 2

2 1/ 2

a x 2 + R2 6Å0

– 23 x 2 + r2

R – 2x

1/ 2

R 0

R

r2 – 2x 2 2

3/ 2

0 2

– 13 x – 2x 2

R 2 – 2x 2 + 2x 3 .

36.

– –

– –

– –

– –

– –

– –

37. The field from the plate is uniform, with a magnitude given by E = /2Å0. The equipotential surfaces will be flat surfaces equally spaced. We find their separation from E = ?V/d; 3.2 mm. (0.5510–6 C/m2)/2(8.8510–12 C2/N · m2) = (100 V)/d, which gives d = 3.210–3 m = 38. The potential outside a charged sphere is V = Q/4pÅ0r, with the potential at the surface being V0 = Q/4pÅ0r0 . Thus the equipotential surfaces outside a charged sphere are spherical surfaces, with the higher potentials being closer to the charged sphere. (a) For the first equipotential surface we have V0 – V1 = Q/4pÅ0r0 – Q/4pÅ0r1 = 100 V; (0.5010–6 C)/4p(8.8510–12 C2/N · m2)[(1/0.30 m) – (1/r1)] = 100 V, which gives r1 = 0.302 m.

(b) For the tenth equipotential surface we have V0 – V10 = Q/4pÅ0r0 – Q/4pÅ0r10 = 10(100 V); (0.5010–6 C)/4p(8.8510–12 C2/N · m2)[(1/0.30 m) – (1/r10)] = 10(100 V), which gives r10 = 0.32 m. (c) For the 100th equipotential surface we have V0 – V100 = Q/4pÅ0r0 – Q/4pÅ0r100 = 100(100 V); (0.5010–6 C)/4p(8.8510–12 C2/N · m2)[(1/0.30 m) – (1/r100)] = 100(100 V), which gives r100 = 0.90 m. Note that the equipotential surfaces get farther apart at greater distances. 39. (a) We find the dipole moment from p = eL = (1.6010–19 C)(0.5310–10 m) = 8.510–30 C · m. (b) The dipole moment will point from the electron toward the proton. As the electron revolves about the proton, the dipole moment will spend equal times pointing in any direction. Thus the average over time will be zero.

40. With the dipole pointing along the axis, the potential at a point a distance r far from the dipole which makes an angle  with the axis is V = (p cos )/4pÅ0r2 = (9.0109 N · m2/C2)(4.810–30 C · m)(cos /(1.110–9 m)2 = (0.0357 V) cos . (a) Along the axis,  = 0, so we have V = (p cos )/4pÅ0r2 = (0.0357 V) cos 0° = 0.036 V. (b) Above the axis near the positive charge,  = 45°, so we have V = (p cos )/4pÅ0r2 = (0.0357 V) cos 45° = 0.025 V. (c) Above the axis near the negative charge,  = 135°, so we have V = (p cos )/4pÅ0r2 = (0.0357 V) cos 135° = – 0.025 V.

b

c

–

+



a

p

41. (a) With the distance measured from the center of the dipole, we find the potential from each charge: VO = QO/4pÅ0rO = (9.0109 N · m2/C2)(– 6.610–20 C)/(9.010–10 m – 0.610–10 m) = – 0.707 V. VC = QC/4pÅ0rC = (9.0109 N · m2/C2)(+ 6.610–20 C)/(9.010–10 m + 0.610–10 m) = + 0.619 V. Thus the total potential is V = VO + VC = – 0.707 V + 0.619 V = – 0.088 V. (b) The percent error introduced by the dipole approximation is % error = (100)(0.089 V – 0.088 V)/(0.088 V) = 1%.

42. Because the field is uniform, the magnitudes of the forces on the charges of the dipole will be equal: F+ = F– = QE. If the separation of the charges is ¬, the dipole moment will be p = Q¬. If we choose the center of the dipole for the axis of rotation, both forces create a CCW torque with a net torque of  = F+(!¬) sin  + F–(!¬) sin = 2QE(!¬) sin  = pE sin . Because the forces are in opposite directions, the net force is zero. If the field is nonuniform, there would be a torque produced by the average field. The magnitudes of the forces would not be the same, so there would be a resultant force that would cause a translation of the dipole.

y F+

E



p –

F–

43. (a) Because p1 = p2 , from the vector addition we have p = 2p1 cos (!) = 2qL cos (!);

r

6.110–30 C · m = 2q(0.9610–10 m) cos [!(104°)], 5.210–20 C. which gives q = (b) We find the potential by adding the potentials from the two dipoles: V = V1 + V2 = p1 cos( – !)/4pÅ0r2 + p2 cos( + !)/4pÅ0r2

p1

44. The potential gradient is the negative of the electric field: dV/dr = – E = – Q/4pÅ0r2 = – (9.0109 N · m2/C2)(92)(1.6010–19 C)/(7.510–15 m)2 =



/2 p

= (p1/4pÅ0r2)[cos( – !) + cos( + !)]. When we use a trigonometric identity and the above result for p we get V = (p1/4pÅ0r2)(2 cos  cos !) = p cos  /4pÅ0r2.

p2

– 2.41021 V/m.

45. The electric field is the negative of the potential gradient: E = – dV/dr = – (Q/4pÅ0)[d(1/r)/dr] = – (Q/4pÅ0)(– 1/r2) = Q/4pÅ0r2. 46. From the spatial dependence of the electric potential, V(x, y, z) = ay/(b2 + y2), we find the components of the electric field from the partial derivatives of V: Ex = – V/x = 0; Ey = – V/y = – a/(b2 + y2) – ay(– 2y)/(b2 + y2)2 = a(y2 – b2)/(b2 + y2)2 . Ez = – V/z = 0. E = a(y2 – b2)/(b2 + y2)2j . We can write the electric field: 47. From the spatial dependence of the electric potential, V(x, y, z) = y2 + 2xy – 4xyz, we find the components of the electric field from the partial derivatives of V: Ex = – V/x = – (2y – 4yz); Ey = – V/y = – (2y + 2x – 4xz); Ez = – V/z = – (– 4xy). E = 2y(2z – 1)i – 2(y + x – 2xz)j + (4xy)k. We can write the electric field: 48. (a) From Problem 32 the potential along the y-axis is 2  Q L + y2 + L Q V= ln = ln L 2 + y 2 + L – ln 2 8Å0L 2 8Å L 0 L + y –L

L2 + y2 – L .

+

We find the y-component of E from  y/ L 2 + y 2 y/ L 2 + y 2 Q E y = – V = – – 2 2 y 8Å0 L L + y2 + L L + y2 – L 2

2

Q y L + y2 – L – L + y2 + L Q = . 2 2 2 2 2 2 8Å0 L L + y L +y +L L + y –L 4Å0 y L 2 + y 2 We know from the symmetry that the electric field will be along the y-axis, so we have E = Q/4pÅ0y(L2 + y2)1/2j. =–

Note that we get the expression for a point charge when y » L. (b) From Problem 33 the potential along the x-axis is V = Q ln x + L = Q ln x + L – ln x – L . 8Å0L 8Å0 L x –L We find the x-component of E from Q 1 – 1 – 2L Ex = – V = – Q =– Q = . x 8Å0 L x + L x – L 8Å0 L x + L x – L 4Å0 x 2 – L 2 We know from the symmetry that the electric field will be along the x-axis, so we have E = Q/4pÅ0(x2 – L2)i. Note that we get the expression for a point charge when x » L.

49. (a) We find the potential energy of the system of two protons from the energy of one proton in the potential produced by the other proton: U = qV = q(q/4pÅ0d) = (+ 1 e)(9.0109 N · m2/C2)(1.6010–19 C)/(1510–15 m) = 9.6104 eV. (b) For the new separation we have U = qV = q(q/4pÅ0r) = (+ 1 e)(9.0109 N · m2/C2)(1.6010–19 C)/(7.510–15 m) = 1.9105 eV. 50. We find the work required to bring the three electrons in from infinity by bringing them in successively. Because there is no potential before the electrons are brought in, for the first electron we have W1 = (– e)V0 = 0. When we bring in the second electron, there will be a potential from the first: W2 = (– e)V1 = (– e)(– e)/4pÅ0r12 = e2/4pÅ0d. When we bring in the third electron, there will be a potential from the first two: W3 = (– e)V2 = (– e){[(– e)/4pÅ0r13] + [(– e)/4pÅ0r23]} = 2e2/4pÅ0d. The total work required is W = W1 + W2 + W3 = (e2/4pÅ0d) + (2e2/4pÅ0d) = 3e2/4pÅ0d = 3(9.0109 N · m2/C2)(1.6010–19 C)2/(1.010–10 m) = 6.910–18 J = 43 eV. 51. Because the total energy of the helium nucleus is conserved, we have ?K + ?U = 0; ?K + q?V = 0; – 2.4104 V. 48103 eV + [(3.210–19 C)/(1.6010–19 C/e)]?V; which gives ?V = The negative sign means the helium nucleus gains kinetic energy by going to a lower potential. 52. The data given are the kinetic energies, so we find the speed from (a) Ke = !mve2; (3.5103 eV)(1.6010–19 J/eV) = !(9.1110–31 kg)ve2, which gives ve = (b) Kp = !mvp

3.5107 m/s.

2;

(3.5103 eV)(1.6010–19 J/eV) = !(1.6710–27 kg)vp2, which gives vp =

8.1105 m/s.

53. (a) We find the potential energy from the work required to Q2 Q1 r12 bring the four charges in from infinity by bringing them in successively. Because there is no potential before the r 23 charges are brought in, for the first charge we have r13 r 15 r W1 = Q1V0 = 0. 25 Q3 When we bring in the second charge, there will be a r r 35 potential from the first: r14 24 r34 W2 = Q2V1 = Q2Q1/4pÅ0r12 = Q1Q2/4pÅ0r12 . r45 Q5 When we bring in the third charge, there will be a Q4 potential from the first two: W3 = Q3V2 = Q3[(Q1/4pÅ0r13) + (Q2/4pÅ0r23)] = Q1Q3/4pÅ0r13 + Q2Q3/4pÅ0r23 . When we bring in the fourth charge, there will be a potential from the first three: W4 = Q4V3 = Q4[(Q1/4pÅ0r14) + (Q2/4pÅ0r24) + (Q3/4pÅ0r34)] = Q1Q4/4pÅ0r14 + Q2Q4/4pÅ0r24 + Q3Q4/4pÅ0r34 . The electrostatic potential energy is the total work: U = W1 + W2 + W3 + W4 = (1/4pÅ0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14 + Q2Q3/r23 + Q2Q4/r24 + Q3Q4/r34). (b) From the result for part (a) we see that we have a term for each pair of charges. Adding a fifth charge will result in four more terms: U = (1/4pÅ0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14+ Q1Q5/r15 + Q2Q3/r23 + Q2Q4/r24 + Q2Q5/r25 + Q3Q4/r34 + Q3Q5/r35 + Q4Q5/r45). 54. We find the speed from K = !mv2; (5.53106 eV)(1.6010–19 J/eV) = !(6.6410–27 kg)v2, which gives v =

1.63107 m/s.

55. (a) The kinetic energy of the electron (q = – e) is Ke = – qVBA = – (– e)VBA = eVBA . The kinetic energy of the proton (q = + e) is Kp = – qVAB = – (+ e)(– VBA) = eVBA = 2.0 keV. (b) We find the ratio of their speeds, starting from rest, from !meve2 = !mpvp2, or ve/vp = (mp/me)1/2 = [(1.6710–27 kg)/(9.1110–31 kg)]1/2 = 56. (a) For the potential energy of the four charges we have Q + U = (1/4pÅ0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14 + Q2Q3/r23 + Q2Q4/r24 + Q3Q4/r34) = (Q2/4pÅ0)(1/b + 1/bv2 + 1/b + 1/b + 1/bv2 + 1/b) = (4 + v2)Q2/4pÅ0b . b (b) We find the potential energy with a charge at the center from the potential of the four charges at the center: UC = (+ Q)4Q/4pÅ0(b/v2) = v2Q2/pÅ0b. (c) To test for stability, we find the potential energy change Q + when the charge at the center is displaced slightly toward one of the corners. If the displacement is  « b, we have U = (+ Q/4pÅ0){2Q/[(b/v2)2 + 2]1/2 + Q/[(b/v2) + ] + Q/[(b/v2) – ]}

42.8.

b

+ Q

 C

+ Q

= (v2Q2/4pÅ0b){2/[1 + (v2/b)2]1/2 + 1/[1 + (v2/b)] + 1/[1 – (v2/b)]}

= (v2Q2/4pÅ0b){2/[1 + (v2/b)2]1/2 + 2/[1 – (v2/b)2]} If we use the approximation (1 ± x)–n ˜ 1 — nx, we get U ˜ (v2Q2/2pÅ0b){[1 – !(v2/b)2] + [1 + (v2/b)2]}

= (v2Q2/2pÅ0b)[1 – (/b)2 + 1 + 2(/b)2] = (v2Q2/pÅ0b)[1 + !(/b)2]. Thus we see that U > UC , so work would have to be done to move the fifth charge away from the center. The fifth charge is in stable equilibrium (d) Because the potential at the center from the charges at the four corners does not change, the potential energy of the fifth charge will be UC = (– Q)4Q/4pÅ0(b/v2) = – v2Q2/pÅ0b. If we again consider a small displacement from the center, the new potential energy will be U = – (v2Q2/pÅ0b)[1 + !(/b)2]. Thus U < UC , so the charge will acquire kinetic energy if it moves away from the center. The fifth charge is in unstable equilibrium. The maximum kinetic energy will be acquired when the fifth charge reaches one of the corners. 8. Because the potential goes to 8 there, the maximum kinetic energy would be This is due to the idealization of point charges. 57. (a) For the potential energy of the four charges we have b Q – + Q U = (1/4pÅ0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14 + Q2Q3/r23 + Q2Q4/r24 + Q3Q4/r34) 2 = (Q /4pÅ0)[– 1/b + 1/bv2 – 1/b – 1/b + 1/bv2 – 1/b] = (– 4 + v2)Q2/4pÅ0b . b (b) We find the potential energy with a charge at the center C from the potential of the four charges at the center: UC = (+ Q)(Q/4pÅ0)[– 1/(b/v2) + 1/(b/v2) – 1/(b/v2) + 1/(b/v2)] – Q Q + = 0. Note that the stability questions are much more complex. Along the lines through the center perpendicular to the sides the potential is zero, so displacement in these directions corresponds to neutral equilibrium. Motion along a diagonal toward a charge of the same sign will be stable, while motion along a diagonal toward the opposite sign will be unstable. 58. We consider adding an infinite number of differential charges dq to the sphere. When we add dq to the sphere when it has a charge q, the potential energy of the system increases by dU = V dq = (q/4pÅ0r) dq. We add (integrate) to find the total potential energy: Q

U=

q dq = 0 4Å0r

2

Q . 8Å0r

59. The charge density of the sphere is E = Q/)pr03. To find the total potential energy of the sphere, we consider it to be made up of differential shells and add (integrate) the work required to bring each shell in from infinity. If a sphere of radius r < r0 with charge q has been formed, the potential at the surface is V = (1/4pÅ0)(q/r) = (1/4pÅ0)(E)pr 3/r) = Er 2/3Å0 . The work to bring the charge of the next shell, dq = E4pr 2 dr, in from infinity is dW = dq V = (E4r 2 dr)(Er 2/3Å0) = E24r 4 dr/3Å0. The total work and thus the total potential energy stored is

W=

r0 0

 2E4r 4 d r = 4 2E 3Å0 3Å0

r0 0

r4 d r =

4  E2r05 = 15Å0

2

3Q . 20Å0r0

60. We find the rms speed from K = !mvrms2 = *kT; (9.1110–31 kg)v3002 = 3(1.3810–23 J/K)(300 K), which gives v300 = 1.17105 m/s. –31 –23 2 (9.1110 kg)v2500 = 3(1.3810 J/K)(2500 K), which gives v2500 = 3.37105 m/s.

61. We find the horizontal velocity of the electron as it enters the electric field from the accelerating voltage: !mv02 = eV;

!(9.1110–31 kg)v02 = (1.6010–19 C)(15103 V),

y

d h which gives v0 = 7.26107 m/s. V Because the force from the electric field is vertical, the x horizontal velocity is constant. The time to pass through E the field is L t1 = d/v0 = (0.028 m)/(7.26107 m/s) = 3.8610–10 s. The time for the electron to go from the field to the screen is t2 = L/v0 = (0.22 m)/(7.26107 m/s) = 3.0310–9 s. If we neglect the small deflection during the passage through the field, we find the vertical velocity when the electron leaves the field from the vertical displacement: vy = h/t2 = (0.11 m)/(3.0310–9 s) = 3.63107 m/s. This velocity was produced by the acceleration in the electric field: F = eE = may , or ay = eE/m. From the vertical motion in the field, we have vy = v0y + ayt1 ; 3.63107 m/s = 0 + [(1.6010–19 C)E/(9.1110–31 kg)](3.8610–10 s), 5.4105 V/m. which gives E =

62.

+

+

63. The potential at the surface of a charged sphere is V = Q/4pÅ0r = (9.0109 N · m2/C2)(10–8 C)/(0.10 m) =

9102 V.

64. There are 10 electrons in each water molecule, so the number of electrons in the drop is N = [(1000 kg/m3))p(1.510–3 m)3(103 g/kg)/(18 g/mol)]  (6.021023 molecules/mol)(10 electrons/molecule) = 4.71021 electrons. The change in the potential is ?V = ?Q/4pÅ0rE = (9.0109 N · m2/C2)(4.71021)(1.6010–19 C)/(6.38106 m) = 1.1106 V. 65. (a) We find the potential difference from U = Q ?V; 4.2 MJ = (4.0 C) ?V, which gives ?V = 1.1 MV. (b) We find the amount of water that can have its temperature raised to the boiling point from U = mc ?T; 13 kg. 4.2106 J = m(4186 J/kg · C°)(100°C – 20°C), which gives m = 66. (a) All eight charges are the same distance from the z center. For the potential at the center we have VC = 8(Q/4pÅ0)[1/!(Lv3)] = 16Q/4pÅ0Lv3. (b) For the seven charges that produce the potential at a corner, three are a distance L away, three are a Q + distance Lv2 away and one is a distance Lv3 away. The potential is Vcorner = 3(Q/4pÅ0L) + 3(Q/4pÅ0Lv2) + (Q/4pÅ0Lv3) = (Q/4pÅ0L)(3 + 3/v2 + 1/v3) = 5.70Q/4pÅ0L. L (c) We can find the energy for the charge at each corner, if we recognize that this would be counting the contribution from each pair twice. Thus we have U = !(8 Q)Vcorner = 4(5.70Q2/4pÅ0L) = 22.8Q2/4pÅ0L. +

+

Q

Q

+

Q +

y L Q

+ L

+ Q Q

+

x

67. When the proton is accelerated by a potential difference, it acquires a kinetic energy: K = QpVaccel . If it is far from the iron nucleus, the potential is zero. The proton will slow as it approaches the positive charge of the nucleus, because the potential produced by the iron nucleus is increasing. At the proton’s closest point the kinetic energy will be zero. We find the required accelerating potential from ?K + ?U = 0; 0 – K + Qp(VSi – 0) = 0, or QpVaccel = QpQSi/4pÅ0(rp + rSi); Vaccel = (9.0109 N · m2/C2)(26)(1.6010–19 C)/(1.210–15 m + 4.010–15 m) = 7.2106 V = 7.2 MV. 68. We find the velocity of the electron as it enters the electric field from the accelerating voltage: !mv02 = eV;

y

!(9.1110–31 kg)v02 = (1.6010–19 C)(14103 V),

d h V which gives v0 = 7.01107 m/s. x Because the force from the electric field is perpendicular, E this velocity component is constant. The time to pass through the field is L t1 = d/v0 = (0.026 m)/(7.01107 m/s) = 3.7110–10 s. The time for the electron to go from the field to the screen is t2 = L/v0 = (0.34 m)/(7.01107 m/s) = 4.8510–9 s. The electron will sweep both ways across the screen. If we neglect the small deflection during the passage through the deflecting plates, when the electron leaves the plates the horizontal velocity required to reach the edge of the screen is vymax = h/t2 = (0.15 m)/(4.8510–9 s) = 3.10107 m/s. This velocity was produced by the acceleration in the electric field: F = eEmax = maymax , or aymax = eEmax/m. From the horizontal motion in the field, we have vymax = v0y + aymaxt1 ; 3.10107 m/s = 0 + [(1.6010–19 C)Emax/(9.1110–31 kg)](3.7110–10 s), which gives Emax = 4.8105 V/m. Thus the range for the electric field is – 4.8105 V/m < E < 4.8105 V/m. 69. The electrons form a spherical shell. The electric field at the surface of the shell is E = Q/4pÅ0rE2 = Ne/4pÅ0rE2. For the magnitude of the electric force on an electron at the surface of the shell (which is up) to balance the force from the Earth’s gravity, we have eNe/4pÅ0rE2 = mg; (9.0109 N · m2/C2)N(1.6010–19 C)2/(6.38106 m)2 = (9.1110–31 kg)(9.80 m/s2), 1.581012 electrons. which gives N =

70. For the potential energy of the four charges we have U = (1/4pÅ0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14 + Q2Q3/r23 + Q2Q4/r24 + Q3Q4/r34) 2 = (Q /4pÅ0)[(1)(2)/b + (1)(– 3)/bv2 + (1)(2)/b + (2)(–3)/b + (2)(2)/bv2 + (– 3)(2)/b] 9 2 2 –6 2 = (9.010 N · m /C )(4.810 C)  [2 – 3/v2 + 2 – 6 + 4/v2 – 6]/(0.080 m) = – 19 J.

Q +

b

+

2Q

b – – 3Q

+ 2Q

71. The acceleration produced by a potential difference of 1000 V over a distance of 1 cm is a = eE/m = eV/md = (1.6010–19 C)(1000 V)/(9.1110–31 kg)(0.01 m) = 21016 m/s2. Because this is so much greater than g, yes, the electron can easily move upward. To find the potential difference to hold the electron stationary, we have mg = eE = eV/d; 1.710–12 V. (9.1110–31 kg)(9.80 m/s2) = (1.6010–19 C)V/(0.030 m), which gives V = 72. (a) The kinetic energy of the proton (q = + e) is Kp = – qVQP = – (+ e)VQP = – eVQP = 5.2 keV. The kinetic energy of the electron (q = – e) is Ke = – qVPQ = – (– e)(– VQP) = – eVQP = 5.2 keV. (b) We find the ratio of their speeds, starting from rest, from !meve2 = !mpvp2, or ve/vp = (mp/me)1/2 = [(1.6710–27 kg)/(9.1110–31 kg)]1/2 =

42.8.

73. For the motion of the electron from emission to the plate, the energy of the electron is conserved, so we have ?K + ?U = 0, or 0 – !mv2 + (– e) ?V = 0; 1.03106 m/s. – !(9.1110–31 kg)v2 + (– 1.6010–19 C)(– 3.02 V – 0) = 0, which gives v = 74. Because the electric field points downward, the potential is greater at the higher elevation. For the potential difference, we have ?V = – (150 V/m)(2.00 m) = – 300 V. For the motion of the falling charged balls, the energy is conserved: ?K + ?U = 0, or !mv2 – 0 + q ?V + mg(0 – h) = 0, which gives v2 = 2gh – 2(q/m) ?V. For the positive charge, we have v12 = 2gh – 2(q1/m) ?V = 2(9.80 m/s2)(2.00 m) – 2[(55010–6 C)/(0.540 kg)](– 300 V), which gives v1 = 6.31 m/s. For the negative charge, we have v22 = 2gh – 2(q2/m) ?V = 2(9.80 m/s2)(2.00 m) – 2[(– 55010–6 C)/(0.540 kg)](– 300 V), which gives v2 = 6.21 m/s. Thus the difference in speeds is 6.31 m/s – 6.21 m/s = 0.10 m/s.

75. The distances from the midpoint of a side to the three charges are L/2, L/2, and L cos 30°. At point a, we have Va = (1/4pÅ0){[(– Q)/(L/2)] + [(+ Q)/(L/2)] + [(– 3Q)/(L cos 30°)]} = (Q/4pÅ0L)[(– 2) + (+ 2) + (– 3/cos 30°)] = – 3.5 Q/4pÅ0L. At point b, we have Vb = (1/4pÅ0){[(+ Q)/(L/2)] + [(– 3Q)/(L/2)] + [(– Q)/(L cos 30°)]} = (Q/4pÅ0L)[(+ 2) + (– 6) + (– 1/cos 30°)] = – 5.2 Q/4pÅ0L. At point c, we have Vc = (1/4pÅ0){[(– 3Q)/(L/2)] + [(– Q)/(L/2)] + [(+ Q)/(L cos 30°)]} = (Q/4pÅ0L)[(– 6) + (– 2) + (+ 1/cos 30°)] = – 6.8 Q/4pÅ0L.

y –Q

L

+Q +

a

–

60°

L

y 76. We assume that r2 » ¬2, so we can use the potential produced by a dipole. We choose the coordinate system shown, so that x = r cos  and y = r sin . r Thus the potential produced by p is 2 2 2 3/2 V = p cos /4pÅ0r = (p/4pÅ0)x/(x + y ) . We find the components of E from – + Ex = – ?V/?x = – (p/4pÅ0) ?[x/(x2 + y2)3/2]/?x p = – (p/4pÅ0)[1/(x2 + y2)3/2 – 3x2/(x2 + y2)5/2] = (p/4pÅ0)[(2x2 – y2)/(x2 + y2)5/2] = (p/4pÅ0)(2 cos2  – sin2 )/r3; Ey = – ?V/?y = – (p/4pÅ0) ?[x/(x2 + y2)3/2]/?y = – (p/4pÅ0)[– 3xy/(x2 + y2)5/2] = (p/4pÅ0)[3xy/(x2 + y2)5/2] = (3p sin  cos )/4pÅ0r3;

c

b

–3Q

L

–

x

 x

77. (a) We find the electric potential of the proton from V = q/4pÅ0r = (9.0109 N · m2/C2)(1.6010–19 C)/(2.510–15 m) = 5.76105 V = 5.8105 V. (b) We find the electric potential energy of the system by considering one of the charges to be at the potential created by the other charge: U = qV = (1.6010–19 C)(5.76105 V) = 9.210–14 J = 0.58 MeV.

78. The charge density of the ring is  = Q/p[R2 –(!R)2] = 4Q/3pR2. We choose a ring of radius r and width dr for a differential element, with charge dq = 2pr dr. The potential of this element on the axis a distance x from the ring is dV = dq/4pÅ0(x2 + r2)1/2 = 2pr dr/4pÅ0(x2 + r2)1/2 = r dr/2Å0(x2 + r2)1/2. We integrate to get the potential: R

V= = =

R r

x R/2

R

 r dr =  x 2 + r2 1 / 2 2 2 1 / 2 2Å0 R/ 2 2Å x + r 0 R/ 2 2Q 3Å0R 2

x 2 + R2

2Q 3Å0R

1/ 2

– x 2 + R 2/ 4

1 + (x/ R) 2

1/ 2

1/ 2

1/ 2 – 1 1 + (2x/ R) 2 . 2

79. In the region between the wire and cylinder the radial electric field will be produced by the central wire: E = /2pÅ0r, Ra < r < Rb . We find the potential difference from

 Vb – Va = –

b a

Rb

E·d = – Ra

 dr = –  ln R b , or 2Å0r 2Å0 Ra

Va – Vb = (/2pÅ0) ln(Rb/Ra).

80. (a) The electric field at the spherical surface is E = Q/4pÅ0r2,

while the potential of a sphere, with V = 0 at 8, is V = Q/4pÅ0r. Thus we have r = V/E; 4.5105 V. 0.15 m = V/(3.0106 V/m), which gives V = (b) We find the charge on the sphere from V = Q/4pÅ0r; 4.5105 V = (9.0109 N · m2/C2)Q/(0.15 m), which gives Q = 7.510–6 C =

81. We assume that r2 » ¬2, so we can use the potential produced by a dipole. Thus the potential produced by the left dipole along the horizontal line is V1 = p1/4pÅ0r2. We find the interaction energy by finding the energy of each of the charges of the right dipole: U = Q[p1/4pÅ0(r + !¬)2] – Q[p1/4pÅ0(r – !¬)2]

7.5 C.

¬

p1 –

–

+ r

= (Qp1/4pÅ0r2)[1/(1 + !¬/r)2 – 1/(1 – !¬/r)2]. If we use the approximation (1 ± x)–n ˜ 1 — nx, we get U ˜ (Qp1/4pÅ0r2)[(1 – ¬/r) – (1 + ¬/r)] = (Qp1/4pÅ0r2)(– 2¬/r) = – p1p2/2pÅ0r3.

+ p2

82. We assume that r2 » ¬2, so we can use the potential produced by a dipole. We choose the coordinate system based on the left dipole as shown, so that x = r cos 1 and y = r sin 1 . Thus the potential produced by p1 at the center of p2 is V1 = p1 cos 1/4pÅ0r2 = (p1/4pÅ0)x/(x2 + y2)3/2. To find the potentials at the locations of the charges of p2 , we use the fact that they are a differential distance from the center, so we have V+ = V1 + !¬ cos (1 – 2) ?V1/?x + !¬ sin(1 – 2) ?V1/?y;

x p1 +

1

p2 r –

–

+

¬

y

V– = V1 – !¬ cos (1 – 2) ?V1/?x – !¬ sin(1 – 2) ?V1/?y. The partial derivatives are (suppressing the constant p1/4pÅ0) ?V1/?x = ?[x/(x2 + y2)3/2]/?x = 1/(x2 + y2)3/2 – 3x2/(x2 + y2)5/2 = (r2 – 3x2)/r5 = (1 – 3 cos2 1)/r3; ?V1/?y = ?[x/(x2 + y2)3/2]/?y = – 3xy/(x2 + y2)5/2 = – 3xy/r5 = – 3 cos 1 sin 1/r3; We find the interaction energy by finding the energy of each of the charges of p2 : U = QV+ + (– Q)V– = Q¬[cos (1 – 2) ?V1/?x + sin(1 – 2) ?V1/?y] = (p1p2/4pÅ0r3)[cos (1 – 2) (1 – 3 cos2 1) + sin(1 – 2) (– 3 cos 1 sin 1)] = (p1p2/4pÅ0r3){cos (1 – 2) – 3 cos 1 [cos (1 – 2) cos 1 + sin(1 – 2) sin 1]} = (p1p2/4pÅ0r3)[cos (1 – 2) – 3 cos 1 cos 2].

83. (a) The reference level for the potential is V = 0 at r = 8. At points outside the spherical shell, it is equivalent to a point charge. Thus the potential when r > r2 is V = Q/4pÅ0r = E)p(r23 – r13)/4pÅ0r = E(r23 – r13)/3Å0r , r > r2 . (b) The electric field within the spherical conductor, r1 < r < r2 , is due to the charge within a radius r: E = E)p(r3 – r13)/4pÅ0r2 = (E/3Å0)[r – (r13/r2)]. We find the potential by integrating along a radial line from r to r2 : r 2 r2 r2 dV = – E· d = – E d r; r

r

 E r32 – r31  –V =– E 3Å0 r2 3Å0

r

r2 r

r–

r13  E r 22 – r 2 r13 r31 d r = – + r – r , which gives 3Å0 2 2 r2

V = (E/6Å0)[3r22 – r2 – (2r13/r)], r1 < r < r2 . (c) Inside the cavity the electric field is zero, so the potential is constant and equal to the potential at the inner surface of the shell: V = V(r1) = (E/6Å0)[3r22 – r12 – (2r13/r1)] = (E/2Å0)(r22 – r12), r < r1 . The potential is continuous at r1 and r2 .

2

Chapter 24 p. 1

CHAPTER 24 – Capacitance, Dielectrics, Electric Energy Storage 1.

From Q = CV, we have 2500 C = C(950 V), which gives C =

2.

From Q = CV, we have 28.010–8 C = (12,00010–12 F)V, which gives V =

2.6 F. 23.3 V.

3.

From Q = CV, we have 75 pC = C(12.0 V), which gives C =

4.

The final potential on the capacitor will be the voltage of the battery. Positive charge will move from one plate to the other, so the charge that flows through the battery is Q = CV = (15.610–6 F)(12 V) = 1.910–4 C.

5.

From Q = CV, we see that ?Q = C ?V; 16 C = C(48 V – 28 V), which gives C =

6.

6.3 pF.

0.80 F.

When the capacitors are connected, some charge will flow from C1 to C2 until the potential difference across the two capacitors is the same: V1 = V2 = V. Because charge is conserved, we have Q0 = Q1 + Q2 . For the two capacitors we have Q1 = C1V, and Q2 = C2V. When we form the ratio, we get Q2/Q1 = (Q0 – Q1)/Q1 = C2/C1 , which gives Q1 = Q0C1/(C1 + C2). For Q2 we have Q2 = Q0 – Q1 = Q0{1 – [C1/(C1 + C2)]}, thus Q2 = Q0C2/(C1 + C2). We find the potential difference from Q1 = C1V; Q0C1/(C1 + C2) = C1V, which gives V = Q0/(C1 + C2).

7.

We assume that the charge transferred is small compared to the initial charge on the plates so the potential difference between the plates is constant. The energy required to move the charge is W = qV. Thus the charge on each plate is Q = CV = C(W/q) = (1610–6 F)(25 J)/(0.2010–3 C) = 2.0 C. Because this is much greater than the charge moved, our assumption is justified.

Chapter 24 p. 2

8.

We find the initial charges on the capacitors: Q1 = C1V1 = (2.40 F)(880 V) = 2112 C; Q2 = C2V2 = (4.00 F)(560 V) = 2240 C. (a) When the capacitors are connected with positive plates together, some charge will flow from C2 to C1 until the potential difference across the two capacitors is the same: V1 = V2 = V. Because charge is conserved, we have Q = Q1 + Q2 = Q1 + Q2 = 2112 C + 2240 C = 4352 C. For the two capacitors we have Q1 = C1V, and Q2 = C2V. When we add these, we get Q1 + Q2 = Q = (C1 + C2)V; 4352 C = (2.40 F + 4.00 F)V, which gives V = 680 V. The charge on C1 is Q1 = C1V = (2.40 F)(680 V) = 1.63103 C = 1.6310–3 C. The charge on C2 is Q2 = C2V = (4.00 F)(680 V) = 2.72103 C = 2.7210–3 C. (b) When the capacitors are connected with opposite plates together and charge flows from C2 to C1 , the combination of positive and negative charges will result in the cancellation of some charge until the potential difference across the two capacitors is the same: V1 = V2 = V. Because charge is conserved, we have Q = Q1 + Q2 = Q2 – Q1 = 2240 C – 2112 C = 128 C. For the two capacitors we have Q1 = C1V, and Q2 = C2V. When we add these, we get Q1 + Q2 = Q = (C1 + C2)V; 128 C = (2.40 F + 4.00 F)V, which gives V = 20 V. The charge on C1 is Q1 = C1V = (2.40 F)(20 V) = 48 C. The charge on C2 is Q2 = C2V = (4.00 F)(20 V) = 80 C.

9.

For a parallel-plate capacitor, we find the area from C = Å0A/d; 0.4010–6 F = (8.8510–12 C2/N · m2)A/(4.010–3 m), which gives A = If the area were a square, it would be ˜ 13 m on a side.

10. For a coaxial cable, we have C = L2Å0/ln(R2/R1) = L2Å0/ln(D2/D1), so C/L = 2(8.8510–12 F/m)/ln[(5.0 mm)/(1.0 mm)] = 11. The potential at the surface of a spherical conductor is V = Q/4pÅ0rE , so we have C = Q/V = 4pÅ0rE = 4p(8.8510–12 F/m)(6.38106 m) =

3.510–11 F/m.

7.110–4 F.

1.8102 m2.

Chapter 24 p. 3

12. From the symmetry of the charge distribution, we know that any electric field must be radial, away from the axis of R2 the cylinders, with a magnitude independent of the direction. ¬ For a Gaussian surface we choose a cylinder of length ¬ and r radius r, centered on the axis. On the ends of this surface, the R1 electric field is not constant but E and dA are perpendicular, so we have E · dA = 0. On the curved side, the field has a constant magnitude and E and dA are parallel, so we have E · dA = E dA. For the region where r < R1 , there is no charge inside the Gaussian surface, so we have 



–Q +Q

ı E · dA = ?ends E · dA + ?sideE · dA = Qenclosed/Å0 ;

0 + E2pr¬ = 0, or E = 0; r < R1. For the region where r > R2 , the charge inside the Gaussian surface is Qenclosed = (Q/L)¬ – (Q/L)¬ = 0. Thus there is no net charge inside the Gaussian surface, so we have E = 0; r > R2 . 13. The uniform electric field between the plates is related to the potential difference across the plates: E = V/d. For a parallel-plate capacitor, we have Qmax = CVmax = (Å0A/d)(Emaxd) = Å0AEmax = (8.8510–12 C2/N · m2)(8.510–4 m2)(3.0106 V/m) = 2.310–8 C = 23 nC. 14. The uniform electric field between the plates is related to the potential difference across the plates: E = V/d. For a parallel-plate capacitor, we have Q = CV = (Å0A/d)(Ed) = Å0AE = (8.8510–12 C2/N · m2)(21.010–4 m2)(2.80105 V/m) 5.20 nC. = 5.2010–9 C = 15. When the two cylinders are separated by d, we have Ra = Rb + d. For a cylindrical capacitor, we have C = L2Å0/ln(Ra/Rb) = L2Å0/ln[(Rb + d)/Rb] = L2Å0/ln[1 + (d/Rb)]. If d « Rb , we have C ˜ L2Å0/(d/Rb) = L2RbÅ0/d = Å0A/d, which is the expression for a parallel-plate capacitor.

16. The uniform electric field between the plates is related to the potential difference across the plates: E = V/d. For a parallel-plate capacitor, we have Q = CV = (Å0A/d)(Ed) = Å0AE; 4.210–6 C = (8.8510–12 C2/N · m2)A(2.0103 V/mm)(103 mm/m), 0.24 m2. which gives A = 17. We find the potential difference across the plates from Q = CV; 72 C = (0.80 F)V, which gives V = 90 V. We find the uniform electric field between the plates from E = V/d = (90 V)/(2.010–3 m) = 4.5104 V/m.

Chapter 24 p. 4

18. When the two spheres are separated by d, we have ra = rb + d. For a spherical capacitor, we have C = 4Å0rbra/(ra – rb) = 4Å0rb(rb + d)/d = 4Å0rb2[1 + (d/rb)]/d. If d « rb , we have C ˜ 4Å0rb2/d = Å0A/d, which is the expression for a parallel-plate capacitor.

19. (a) When the uncharged plate is placed between the two charged plates, charges will separate so that there is a charge + Q on the side facing the negative plate and a charge – Q on the side facing the positive plate. Thus we have the same uniform electric field in each gap: E = /Å0 = Q/AÅ0 . If x is the separation on one side of the sheet, the potentials across the gaps are V1 = Ex, V2 = E(d – ¬ – x). Thus the potential across the capacitor is V1 + V2 = Ex + E(d – ¬ – x) = (Q/AÅ0)(d – ¬). The capacitance is C = Q/(V1 + V2) = Å0A/(d – ¬). (b) If ¬ = %d, we have C/C0 = d/(d – ¬) = d/(d – %d) =

3.

20. We find the equivalent capacitance for a parallel connection from Cparallel = ?Ci = 6C1 = 6(1.8 F) = 10.8 F. When the capacitors are connected in series, we find the equivalent capacitance from 1/Cseries = ?(1/Ci ) = 6/C1 = 6/(1.8 F), which gives Cseries = 0.30 F. 21. We can decrease the capacitance by adding a series capacitor. We find the necessary capacitor from 1/C = (1/C1) + (1/C2); 1/(1600 pF) = [1/(3600 pF)] + (1/C2), which gives C2 = 2880 pF. Yes, it is necessary to break a connection to add a series component. 22. For the parallel network the potential difference is the same for all capacitors, and the total charge is the sum of the individual charges. We find the charge on each from Q1 = C1V = (Å0A1/d1)V; Q2 = C2V = (Å0A2/d2)V; Q3 = C3V = (Å0A3/d3)V. Thus the sum of the charges is Q = Q1 + Q2 + Q3 = (Å0A1/d1)V + (Å0A2/d2)V + (Å0A3/d3)V. The definition of the equivalent capacitance is Ceq = Q/V = (Å0A1/d1) + (Å0A2/d2) + (Å0A3/d3) = C1 + C2 + C3 .

Chapter 24 p. 5

23. (a) From the circuit, we see that C2 and C3 are in series and find their equivalent capacitance from 1/C4 = (1/C2) + (1/C3), which gives C4 = C2C3/(C2 + C3). From the new circuit, we see that C1 and C4 are in parallel, with an equivalent capacitance Ceq = C1 + C4 = C1 + [C2C3/(C2 + C3)] = (C1C2 + C1C3 + C2C3)/(C2 + C3). (b) Because V is across C1 , we have Q1 = C1V = (14.0 F)(25.0 V) = 350 C. Because C2 and C3 are in series, the charge on each is the charge on their equivalent capacitance: Q2 = Q3 = C4V = [C2C3/(C2 + C3)]V = [(14.0 F)(7.00 F)/(14.0 F + 7.00 F)](25.0 V) = 117 C.

C1 C2

c

a

C3 b

V C1

C4

a

b

V

24. Because C2 and C3 are in series, they must have the same charge: Q3 = Q2 = 24.0 C. Thus we can find the potential across each of these: V2 = Q2/C2 = (24.0 C)/(16.0 F) = 1.50 V; V3 = Q3/C3 = (24.0 C)/(16.0 F) = 1.50 V, so V2 = V3 = 1.50 V. C2 and C3 are in parallel with C1 , so we have V1 = V2 + V3 = 1.50 V + 1.50 V = 3.00 V. Thus we have Q1 = C1V1 = (16.0 F)(3.00 V) = 48.0 C; Vab = V1 = 3.00 V. 25. (a) From the circuit, we see that C2 and C3 are in series and find their equivalent capacitance from 1/C4 = (1/C2) + (1/C3); 1/C4 = (1/3.00 F)+ (1/4.00 F), which gives C4 = 1.71 F. From the new circuit, we see that C1 and C4 are in parallel, with an equivalent capacitance Ceq = C1 + C4 = 2.00 F + 1.71 F = 3.71 F. (b) Because Vab is across C1 , we have Vab = V1 = 26.0 V. For the series combination we have Q2 = Q3 = Q4 = C4Vab = (1.71 F)(26.0 V) = 44.6 C. We find the other voltages from V2 = Q2/C2 = (44.6 C)/(3.00 F) = 14.9 V; V3 = Q3/C3 = (44.6 C)/(4.00 F) = 11.1 V.

C1 C2 a

c

C3 b

V

C2

C3

c

a

C1

b

V ab C4

a

C1

V ab

b

Chapter 24 p. 6

As a check we see that V2 + V3 = Vab .

26. (a) Because the potential is the same across the top and bottom pairs, they are in parallel. d1 (b) Because d1 + d2 is much less than the dimensions of the plates, we can treat them as parallel-plate capacitors. d2 The equivalent capacitance is C = C1 + C2 = (Å0A/d1) + (Å0A/d2) = Å0A(d1 + d2)/d1d2 . (c) We see that the capacitance goes to infinity if d1 (or d2) = 0, for C maximum: Cmax = 8. This corresponds to a short circuit. If we let D = d1 + d2, we have C = C1 + C2 = Å0AD/d1(D – d1). We find the minimum by setting the first derivative equal to zero: dC/dd1 = Å0AD[– 1/d12(D – d1) + 1/d1(D – d1)2] = Å0AD(– D +2d1)/d12(D – d1)2 = 0. Thus we get the minimum capacitance when d1 = !D: Cmin = Å0A(!D + !D)/!D!D = 4Å0A/(d1 + d2 ) =

2Å0A/d1 , d1 = d2 .

27. The capacitance increases with a parallel connection, so the maximum capacitance is Cmax = C1 + C2 + C3 = 3000 pF + 5000 pF + 0.010 F = 3.0 nF + 5.0 nF + 10 nF = 18 nF (parallel). The capacitance decreases with a series connection, so we find the minimum capacitance from 1/Cmin = (1/C1) + (1/C2) + (1/C3) = [1/(3000 pF)] + [1/(5000 pF)] + [1/(0.010 F)] = [1/(3.0 nF)] + [1/(5.0 nF)] + [1/(10 nF)], which gives Cmin = 1.6 nF (series).

V

Chapter 24 p. 7

28. (a) We find the equivalent capacitance from 1/Cseries = (1/C1) + (1/C2) = [1/(0.20 F)] + [1/(0.30 F)], which gives Cseries = 0.12 F. The charge on the equivalent capacitor is the charge on each capacitor: Q1 = Q2 = Qseries = CseriesV = (0.12 F)(9.0 V) = 1.08 C. We find the potential differences from Q1 = C1V1 ; 1.08 C = (0.20 F)V1 , which gives V1 = 5.4 V. Q2 = C2V2 ; 1.08 C = (0.30 F)V2 , which gives V2 = 3.6 V. (b) As we found above Q1 = Q2 = 1.08 C. (c) For the parallel network, we have V1 = V2 = 9.0 V. We find the two charges from Q1 = C1V1 = (0.20 F)(9.0 V) = 1.8 C; Q2 = C2V2 = (0.30 F)(9.0 V) = 2.7 C.

29. (a) From the circuit, we see that C1 and C2 are in series and find their equivalent capacitance from 1/C5 = (1/C1) + (1/C2) = (1/C) + (1/C), which gives C5 = C/2. From the new circuit, we see that C3 and C5 are in parallel, with an equivalent capacitance C6 = C3 + C5 = C + C/2 = 3C/2. From the new circuit, we see that C4 and C6 are in series and find their equivalent capacitance from 1/Ceq = (1/C4) + (1/C6) = (1/C) + [1/(3C/2)], which gives Ceq = 3C/5. (b) The charge on the equivalent capacitor is also the charge on C4 and C6 : Qeq = Q4 = Q6 = CeqVab = (3C/5)V = 3CV/5. We find the potential difference between c and b from Vcb = Q6/C6 = (3CV/5)/(3C/2) = 2V/5. The charge on C5 is also the charge on C1 and C2 : Q5 = Q1 = Q2 = C5Vcb = (C/2)(2V/5) = CV/5. We find the potential differences from Vcd = Q1/C1 = (CV/5)/(C) = V/5; Vdb = Q2/C2 = (CV/5)/(C) = V/5; Vac = Q4/C4 = (3CV/5)/(C) = 3V/5. The charge on C3 is Q3 = C3Vcb = (C)(2V/5) = 2CV/5. Thus we have Q1 = Q2 = CV/5, Q3 = 2CV/5, Q4 = 3CV/5;

C1

C2

a

b

c V

C1

C2

a

b V

C1 C3

c

a

C2

d

C4

b

C5

c a

C3 C4 C6

c

a

b

C4

b

Chapter 24 p. 8

V1 = V2 = V/5, V3 = 2V/5, V4 = 3V/5. 30. Because C1 and C2 are in series, we have Q1 = Q2 = 12.4 C. Thus we have V1 = Vcd = Q1/C1 = 12.4 C/16.0 F = 0.775 V; V2 = Vdb = Q2/C2 = 12.4 C/16.0 F = 0.775 V. From the diagram we see that V3 = Vcd + Vdb = 0.775 V + 0.775 V = 1.55 V, so Q3 = C3V3 = (16.0 F)(1.55 V) = 24.8 C. For Q4 we have Q4 = Q1 + Q3 = 12.4 C + 24.8 C = 37.2 C, so V4 = Vac = Q4/C4 = 37.2 C/36.0 F = 1.03 V. From the diagram we see that Vab = Vac + Vcb = 1.03 V + 1.55 V = 2.58 V. Thus we have Q1 = Q2 = 12.4 C, Q3 = 24.8 C, Q4 = 37.2 C; V1 = V2 = 0.775 V, V3 = 1.55 V, V4 = 1.03 V, Vab = 2.58 V.

C1

c

a

d

C2

C3 C4

31. When the switch is down, the initial charge on C2 is Q2 = C2V0. When the switch is connected upward, some charge will flow from C2 to C1 until the potential difference across the two capacitors is the same: V1 = V2 = V. Because charge is conserved, we have Q = Q1 + Q2 = Q2 , or C1V + C2V = C2V0 , which gives V = C2V0/(C1 + C2). For the charges we have Q1 = C1V = C1C2V0/(C1 + C2); Q2 = C2V = C22V0/(C1 + C2).

b

C1 C2

S V0

Chapter 24 p. 9

32. (a) From the circuit, we see that C1 and C2 are in parallel, C3 C1 and C3 and C4 are in parallel: C5 = C1 + C2 ; C6 = C3 + C4 . We now have two capacitors in series, so the equivalent c C b a C2 4 capacitance is 1/Ceq = (1/C5) + (1/C6) = [1/(C1 + C2)] + [1/(C3 + C4)], which gives C5 Ceq = (C1 + C2)(C3 + C4)/(C1 + C2 + C3 + C4). C6 (b) For the series combination we have a b c Qeq = Q5 = Q6 = CeqVab = (C1 + C2)(C3 + C4)Vab/(C1 + C2 + C3 + C4). C eq We can now find Vac = V1 = V2 = Q5/C5 b a = (C1 + C2)(C3 + C4)Vab/(C1 + C2 + C3 + C4)(C1 + C2) = (C3 + C4)Vab/(C1 + C2 + C3 + C4). Vcb = V3 = V4 = Q6/C6 = (C1 + C2)(C3 + C4)Vab/(C1 + C2 + C3 + C4)(C3 + C4) = (C1 + C2)Vab/(C1 + C2 + C3 + C4). We find the charges on each capacitor from Q1 = C1V1 = C1(C3 + C4)Vab/(C1 + C2 + C3 + C4); Q2 = C2V2 = C2(C3 + C4)Vab/(C1 + C2 + C3 + C4); Q3 = C3V3 = C3(C1 + C2)Vab/(C1 + C2 + C3 + C4); Q4 = C4V4 = C4(C1 + C2)Vab/(C1 + C2 + C3 + C4). 33. Because C3 and C4 are in parallel, we have Vcb = V3 = V4 = Q3 /C3 = 30 C/8.0 F = 3.75 V. Thus we have Q4 = C4V4 = (16.0 F)(3.75 V) = 60 C. From the diagram we see that Q5 = Q6 = Q3 + Q4 = 30 C + 60 C = 90 C. We see that C1 and C2 are in parallel: C5 = C1 + C2 = 8.0 F + 16.0 F = 24.0 F, so we have V1 = V2 = V5 = Vac = Q5/C5 = 90 C/24.0 F = 3.75 V. We can now find the charges on C1 and C2 : Q1 = C1V1 = (8.0 F)(3.75 V) = 30 C; Q2 = C2V2 = (16.0 F)(3.75 V) = 60 C. As expected from the symmetry we have (a) Q1 = Q3 = 30 C; Q2 = Q4 = 60 C. (b) V1 = V2 = V3 = V4 = 3.75 V. (c) Vab = Vac + Vcb = 3.75 V + 3.75 V = 7.5 V.

C3

C1 a

C2

c

C5 a

C4

b

C6 c

b

34. For the two combinations we have Cp = C1 + C2 ; 35.0 F = C1 + C2 , and 1/Cs = (1/C1) + (1/C2) or Cs = C1C2/(C1 + C2) = C1C2/Cp ; 4.0 F = C1C2/(35.0 F). When we combine these two equations, we get a quadratic equation for C1: C12 – (35.0 F)C1 + (140 F2) = 0. The two solutions are C1 = 4.6 F and 30.4 F, which give C2 = 30.4 F and 4.6 F. Thus the two capacitors are 4.6 F, 30.4 F.

Chapter 24 p. 10

35. When there is no reading on the voltmeter, we have Vab = 0, so V1 = V2 , or Q1/C1 = Q2/ C2 ; and Vx = V3 , or Qx/Cx = Q3/ C3 . If we divide the two equations, we get (Q1/Qx)(Cx/C1) = (Q2/Q3)(C3/C2). Because Vab = 0, we could remove the connection between a and b without affecting the circuit. This means that Q1 = Qx , and Q2 = Q3 , so we have Cx/C1 = C3/C2 , or Cx = (C3/C2)C1 = [(6.0 F)/(18.0 F)](8.9 F) = 3.0 F.

a C1 c

Cx d

V

C2

b

C3

V0 36. We find the equivalent capacitance for the series combination from 1/Cseries = (1/C1) + (1/C2) = [1/(3200 pF)] + [1/(2200 pF)], which gives Cseries = 1300 pF. The charge on the equivalent capacitor is the initial charge on each capacitor: Q1 = Q2 = Qseries = CseriesV0 = (1300 pF)(12.0 V) = 1.56104 pC. When the capacitors are connected with positive plates together, some charge will flow until the potential difference across the two capacitors is the same: V1 = V2 = V. Because charge is conserved, we have Q = Q1 + Q2 = Q1 + Q2 = 1.56104 pC + 1.56104 pC = 3.12104 pC. For the two capacitors we have V = Q1/C1 = Q2/C2 ; Q1/3200 pF = (3.12104 pC – Q1)/2200 pF, which gives Q1 = 1.85104 pC. Thus Q2 = 3.12104 pC – 1.85104 pC = 1.27104 pC. 37. For a small angle , we have tan  ˜ . If we consider a differential element a distance y from the small end, the capacitance of the element is dC = Å0 dA/(d + y tan ) ˜ Å0¬ dy/(d + y), where ¬ is the length of a plate. The infinite number of elements are in parallel, so we find the total capacitance by integrating:

 C = Å0



dy Å d+  Å  = 0 ln = 0 ln 1 + . + y   d  d d 0

We use the expansion ln(1 + x) ˜ x –! for small x: C ˜ (Å0¬/)[(¬/d) – !(¬/d)2] = (Å0¬2/d)[1 – !(¬/d)] =

dy

y

x2,

38. (a) From the diagram we see that

(Å0A/d)[1 – !(vA/d)].

d

Chapter 24 p. 11

Vab = Vac + Vcb , or V = V1 + V2 ; (1) a Vcb = Vcd + Vdb , or V1 = V3 + V4 ; (2) C2 Vad = Vac + Vcd , or V5 = V2 + V3 . (3) c C5 At point a we see that the total charge is C 3 Q = Q2 + Q5 , or CV = C2V2 + C5V5 ; (4) C1 A similar reasoning at the other junctions gives d Q = Q1 + Q4 , or CV = C1V1 + C4V4 ; (5) C4 Q2 = Q1 + Q3 , or C2V2 = C1V1 + C3V3 . (6) We eliminate potentials by combining equations: b (5) & (2): CV = (C1 + C4)V1 – C4V3 ; (7) (4) & (3): CV = C2V2 + C5V2 + C5V3 ; (8) (8) & (6): CV = C1V1 + C3V3 + C5(C1/C2)V1 + C5(C3/C2)V3 + C5V3 ; (9) (6) & (1): V = V1 + (C1/C2)V1 + (C3/C2)V3 ; (10) (7) & (9): (C2C4 – C1C5)V1 = (C2C3 + C3C5 + C2C5 + C2C4)V3 ; (11) (7) & (11): CV = [C1 + C4 – C4(C2C4 – C1C5)/(C2C3 + C3C5 + C2C5 + C2C4)]V1 ; (12) (10) & (11): V = [1 + (C1/C2) + C3(C2C4 – C1C5)/C2(C2C3 + C3C5 + C2C5 + C2C4)]V1 ; (13) and finally dividing (12) by (13): C C C + C 1C2 C 4 + C 1 C2C 5 + C1C3 C5 + C 1 C4C5 + C 2C3 C 4 + C 2 C4C 5 + C3C4 C5 C= 1 2 3 C1 C3 + C1 C 4 + C 1 C5 + C2 C3 + C 2 C4 + C2 C 5 + C 3 C4 + C3C 5 C1C 2 C3 + C4 + C 5 + C 1 C5 C 3 + C4 + C 3 C4 C2 + C 5 + C 2 C4C 5 = . C1 C 3 + C 4 + C 5 + C2 C 3 + C4 + C5 + C3 C 4 + C5 (b) For the given data we get C = [(6.0 F)(8.0 F)(6.0 F + 8.0 F + 6.0 F) + (6.0 F)(6.0 F)(6.0 F + 8.0 F) + (6.0 F)(8.0 F)(8.0 F + 6.0 F) + (8.0 F)(8.0 F)(6.0 F)]/ [(6.0 F)(6.0 F + 8.0 F + 6.0 F) + (8.0 F)(6.0 F + 8.0 F + 6.0 F) + 6.9 F. (6.0 F)(8.0 F + 6.0 F)] = 39. The energy stored in the capacitor is U = !CV 2 = !(280010–12 F)(1200 V)2 =

2.010–3 J.

40. The energy density in the field is u = !Å0E2 = !(8.8510–12 C2/N · m2)(150 V/m)2 =

1.010–7 J/m3.

41. Because the separation of the plates is small compared to the dimensions of the plates, the capacitance is Å0A/d. We find the stored energy from U = !CV 2 = !Q 2/C = !Q 2d/Å0A = !(42010–6 C)2(1.510–3 m)/(8.8510–12 C2/N · m2)(8.010–2 m)2 = From E = V/d = Q/Cd = Q/Å0A, we see that this is also U = !Å0E2Ad.

2.3103 J.

42. (a) From C = Å0A/d, we see that separating the plates will change C. For the stored energy we have U = !CV 2 = !Q 2/C. Because the charge is constant, for the two conditions we have U2/U1 = C1/C2 = d2/d1 = 2. (b) The work changes the energy stored in the capacitor: W = U2 – U1 = !Q 2(1/C2 – 1/C1) = (!Q 2/Å0A)(d2 – d1) = Q 2d/2Å0A.

Chapter 24 p. 12

43. From Problem 23 we know that the equivalent capacitance is Ceq = (C1C2 + C1C3 + C2C3)/(C2 + C3) = 3C1/2 = 3(2200 pF)/2 = 3300 pF. Because this capacitance is equivalent to the three capacitors, the energy stored in it is the energy stored in the network: U = !CeqV2 = !(330010–12 F)(10.0 V)2 = 1.6510–7 J. 44. From Problem 29 we know that the equivalent capacitance is Ceq = 3C/5. Because this capacitance is equivalent to the three capacitors, the energy stored in it is the energy stored in the network: U = !CeqV2 = !(3C/5)V2 = 3CV2/10. 45. (a) When the capacitors are connected in parallel, we find the equivalent capacitance from Cparallel = C1 + C2 = 0.15 F + 0.20 F = 0.35 F. The stored energy is Uparallel = !CparallelV2 = !(0.3510–6 F)(12 V)2 = 2.510–5 J. (b) When the capacitors are connected in series, we find the equivalent capacitance from 1/Cseries = (1/C1) + (1/C2) = [1/(0.15 F)] + [1/(0.20 F)], which gives Cseries = 0.0857 F. The stored energy is Useries = !CseriesV2 = !(0.085710–6 F)(12 V)2 = 6.210–6 J. (c) We find the charges from Q = CeqV; Qparallel = CparallelV = (0.35 F)(12 V) = 4.2 C. Qseries = CseriesV = (0.0857 F)(12 V) = 1.0 C. 46. (a) The capacitance of the cylindrical capacitor is C = L2Å0/ln(Ra/Rb). Because the charge remains constant, we express the energy as U = !CV 2 = !Q 2/C. If we form the ratio for the two conditions, we have U2/U1 = C1/C2 = ln(2Ra/Rb)/ln(Ra/Rb) = 1 + [ln 2/ln(Ra/Rb)]. The increase in stored energy comes from the work required to separate the opposite charges on the two cylinders. (b) Because the voltage remains constant, we express the energy as U = !CV 2. If we form the ratio for the two conditions, we have U2/U1 = C2/C1 = ln(Ra/Rb)/ln(2Ra/Rb) = 1/{1 + [ln 2/ln(Ra/Rb)]}. The decrease in stored energy occurs because charge must leave the capacitor, returning to the voltage source.

Chapter 24 p. 13

47. (a) Because there is no stored energy on the uncharged 5.0-F capacitor, the total stored energy is Ua = !C1V02 = !(3.010–6 F)(12 V)2 = 2.210–4 J. (b) We find the initial charge on the 3.0-F capacitor when it is connected to the battery; Q = C1V0 = (3.0 F)(12 V) = 36 C. When the capacitors are connected, some charge will flow from C1 to C2 until the potential difference across the two capacitors is the same; so the two capacitors are connected in parallel: Ceq = C1 + C2 = 3.0 F + 5.0 F = 8.0 F. For the stored energy we have Ub = !CeqV2 = !Q 2 /Ceq = !(36 C)2/(8.0 F) = 81 J = 8.110–5 J. (c) The change in stored energy is – 1.410–4 J. ?U = Ub – Ua = 8.110–5 J – 2.210–4 J = During the flow of charge before the final (d) The stored potential energy is not conserved. steady state, some of the stored energy is dissipated as thermal and radiant energy. 48. (a) Because the voltage remains constant, we express the energy as U = !CV 2, so the work required is W = ?U = Uf – Ui = !(Å0A/d)V 2 – ![Å0A/(d – ¬)]V 2 = – Å0AV 2¬/2d(d – ¬). Note that, although positive work must be done by an outside agent to remove the sheet, the decrease in capacitance means charge (and energy) is returned to the voltage source, so the net energy change is negative. (b) Because the charge remains constant, we express the energy as U = !CV 2 = !Q 2/C, so the work required is W = ?U = Uf – Ui = !Q 2/(Å0A/d) – !Q 2/[Å0A/(d – ¬)] = Q 2¬/2Å0A. The charge on the capacitor is Q = [Å0A/(d – ¬)]Vi , so we have W = Å0AVi 2¬/2(d – ¬)2. 49. (a) Because the charge remains constant, we express the energy as U = !CV 2 = !Q 2/C = !Q 2x/Å0A, so the work required of an external force to increase the separation by dx is dW = dU; F dx = !Q 2 dx/Å0A. Thus the force required is F = !Q 2/Å0A. With no increase in kinetic energy, the force that each plate exerts on the other must have this magnitude. (b) We cannot use F = QE, where E is the electric field between the plates, because the electric field is produced by both plates. The electric field at one plate, due to the other plate, is !E.

Chapter 24 p. 14

50. (a) The electric field outside the spherical conductor is E = Q/4pÅ0r2. We choose a differential spherical shell of radius r and thickness dr and add (integrate) the energies in the shells:  2   Q Q 2  dr Q2 Q2 1Å U= u 4r2 d r = 4r 2 d r = =– 0–1 = . 2 2 2 0 R R 4Å0 r 8Å0 R r 8Å0 8Å0 R R (b) The energy stored in the spherical capacitor is U = !CV 2 = !Q 2/C = !Q 2/4pÅ0R = Q 2/8pÅ0R. (c) When there is a charge q on the sphere, its potential is V = q/4pÅ0R. The work required to bring a differential charge dq to the sphere is dW = V dq. We find the total work required by adding (integrating) the works required to bring all of the dq’s to the sphere: Q 2  Q q Q W= V dq = dq = . 0 8Å0 R 0 4Å0R 51. We find the capacitance from C = KÅ0A/d = KÅ0pr2/d = (7)(8.8510–12 C2/N · m2)p(0.050 m)2/(3.210–3 m) =

1.510–10 F.

52. We find the capacitance from C = KÅ0A/d = KÅ0¬2/d = (2.2)(8.8510–12 C2/N · m2)(0.055 m)2/(1.810–3 m) = 3.310–11 F =

33 pF.

53. The mica will change the capacitance. The potential difference is constant, so we have ?Q = Q2 – Q1 = (C2 – C1)V = (K – 1)C1V = (7 – 1)(350010–12 F)(22 V) = 4.610–7 C = 0.46 C. 54. Because the voltage remains constant, we express the energy as U = !CV 2, so the energy change of the capacitor is ?U = !(1 – K)C0V 2. The charge entering the battery is ?Qbattery = – ?Q, so the energy change in the battery is ?Ubattery = – V ?Q = – V2 ?C = – (1 – K)C0V 2. Thus the total work required is W = ?Ubattery + ?U = – (1 – K)C0V 2 + !(1 – K)C0V 2 = !(K – 1)C0V 2 = !(3.4 – 1)(8.8510–9 F)(100 V)2 = + 1.110–4 J. Note that the positive work done by an outside agent to remove the sheet means charge (and energy) is returned to the voltage source. 55. Because the charge remains constant, we express the energy as U = !CV 2 = !Q 2/C. If we form the ratio for the two conditions, we have U/U0 = C0/C = 1/K, so U = U0/K = (2.33103 J)/7 = 3.3102 J.

Chapter 24 p. 15

56. The potential difference must be the same on each half of the capacitor, so we can treat the system as two capacitors in parallel: C = C1 + C2 = [K1Å0(!A)/d] + [K2Å0(!A)/d]

d

K1

K2

= (Å0!A/d)(K1 + K2) = !(K1 + K2)(Å0A/d) = Å0A(K1 + K2)/2d . 57. If we think of a layer of equal and opposite charges on the interface between the two dielectrics, we see that they are in series. For the equivalent capacitance, we have 1/C = (1/C1) + (1/C2) = (!d/K1Å0A) + (!d/K2Å0A) = (d/2Å0A)[(1/K1) + (1/K2)] = (d/2Å0A)[(K1 + K2)/K1K2], which gives C = 2Å0AK1K2/d(K1 + K2). 58. If we think of a layer of equal and opposite charges on the interface between the two dielectrics, we see that they are in series. For the equivalent capacitance, we have 1/C = (1/C1) + (1/C2) = (d1/K1Å0A) + (d2/K2Å0A) = (1/Å0A)[(d1/K1) + (d2/K2)], which gives C = Å0AK1K2/(d1K1 + d2K2).

d

K1 K2

d1

K1

d2

K2

59. (a) Because each capacitor acquires Q0 , the (equal) initial capacitance of each is C = Q0/V0 . When the dielectric is inserted, the capacitors are still in parallel and charge will flow to make the potential the same. Charge is conserved so, if we call Q1 the charge on the capacitor without the dielectric, we have Q1/C = Q2/KC = (2Q0 – Q1)/KC, which gives Q1 = 2Q0/(1 + K) = 2Q0/(1 + 4.0) = 0.40Q0. For Q2 we have Q2 = 2Q0 – Q1 = 1.60Q0. (b) We find the common voltage from V = Q1/C = 0.40Q0/C = 0.40V0 .

Chapter 24 p. 16

60. (a) We can treat the system as a capacitor of length ¬ – x in ¬ parallel with a capacitor of length x: C = Cair + Cdielectric d K = [Å0¬(¬ – x)/d] + (KÅ0¬x/d) x = (Å0¬2/d)[1 – (x/¬) + K(x/¬)] = (Å0¬2/d)[1 + (K – x=0 + 1)(x/¬)]. (b) The energy stored is U = !CV02 = (Å0¬2V02/2d)[1 + (K – 1)(x/¬)]. (c) When the slab moves a small distance dx, the capacitance change is dC = (Å0¬2/d)(K – 1) dx/¬ = Å0¬(K – 1) dx/d. Because the voltage is constant, this increase in capacitance means a decrease in the charge on the capacitor plates, which means some charge is returned to the voltage source. The magnitude of the change in charge is dQ = V0 dC = Å0¬V0(K – 1) dx/d. We see from part (b) that there is an increase in the energy stored in the capacitor: dUcapacitor = Å0¬V02(K – 1) dx/2d. For the entire system we must include the decrease in energy in the voltage source: dUsource = – V0 dQ = – Å0¬V02(K – 1) dx/d. We find the external force required to produce the total energy change from dW = dUcapacitor + dUsource F dx = Å0¬V02(K – 1) dx/2d – Å0¬V02(K – 1) dx/d = – Å0¬V02(K – 1) dx/2d. Thus the external force is Fexternal = – Å0¬V02(K – 1)/2d, that is, to the right to oppose the attraction between the charges on the plates and the induced charges on the dielectric, and thus keep the slab from accelerating. Thus the force of attraction on the slab is Fslab = + Å0¬V02(K – 1)/2d, to the left, drawing the slab between the plates. 61. (a) The initial capacitance will be the same: C0 = Å0A/d = 111 pF. (b) The initial charge will be the same: Q0 = C0V0 = 1.6610–8 C. (c) Because the capacitance depends only on the physical characteristics, from Example 24–9 the new capacitance is C = 172 pF. Thus the charge on the each plate is now Q = CV0 = (17210–12 F)(150 V) = 2.5810–8 C. The induced charge is Qind = Q[1 – (1/K)] = (2.5810–8 C)[1 – (1/3.50)] = 1.8410–8 C. (d) We find the electric field in the gap from the charge density: Egap = Q/Å0A = (2.5810–8 C)/(8.8510–12 C2/N · m2)(2.5010–2 m2) = 1.17105 V/m. (e) The electric field in the dielectric is Edielectric = Egap/K = (1.17105 V/m)/3.50 = 3.34104 V/m.

Chapter 24 p. 17

(f) The potential difference is constant: V = V0 = 150 V. (g) Because the capacitance depends only on the physical characteristics, it will be the same: C= 172 pF. (h) The charge on each plate is now Q = CV0 = (17210–12 F)(150 V) = 2.5810–8 C.

62. If we think of a layer of equal and opposite charges at each side of the dielectric, we see that there are three capacitors in series. If one air gap has thickness d1 , for the equivalent capacitance, we have 1/C = (1/C1) + (1/C2) + (1/C3) = (d1/Å0A) + (¬/KÅ0A) + [(d – d1 – ¬)/Å0A] = (1/Å0A)[d1 + (¬/K) + d – d1 – ¬] = (1/KÅ0A)[Kd + ¬(1 – K)], so C = KÅ0A/[K(d – ¬) + ¬] = (3.50)(8.8510–12 C2/N · m2)(2.5010–2 m2)/[(3.50)(2.00 – 1.00)10–3 m + (1.0010–3 m)], which gives C = 1.7210–10 F = 172 pF. 63. We find the energy in each region from the energy density and the volume: Udielectric = udielectricVdielectric = !KÅ0Edielectric2A¬; Ugap = ugapVgap = !Å0Egap2A(d – ¬). When we use Egap = KEdielectric and cancel common factors, we have Udielectric/(Udielectric + Ugap) = KEdielectric2¬/[KEdielectric2¬ + Egap2(d – ¬)] 2 = KEdielectric ¬/[KEdielectric2¬ + K2Edielectric2(d – ¬)] = ¬/[¬ + K(d – ¬)] = (1.00 mm)/[1.00 mm + (3.50)(2.00 mm – 1.00 mm)] = 0.22 = 22%. 64. The capacitance is given by C = Q/V. If we use the results from Example 24–9, we have C = Å0AE0/E0[d – ¬ + (¬/K)] = KÅ0A/[K(d – ¬) + ¬]. 65. We can treat the capacitor as an air capacitor and a glass capacitor in series. We find the equivalent capacitance from 1/C = 1/Cair + 1/Cglass = (a/Å0A) + (b/KÅ0A), or C = Å0A/[a + (b/K)] = (8.8510–12 C2/N · m2)(1.85 m2)/[3.0010–3 m + (2.0010–3 m)/(5.80)] = 4.8910–9 F. The free charge on the capacitor plates is Q = CV = (4.8910–9 F)(90.0 V) = 4.4110–7 C. The induced charge on the glass is + + + + + + + + Qind = Q(1 – 1/K) = (4.4110–7 C)(1 – 1/5.80) = 3.6510–7 C. The electric field in the air is Eair Eair = /Å0 = Q/Å0A – – – – – + + + + Eglass + = (4.4110–7 C)/(8.8510–12 C2/N · m2)(1.85 m2) – – – – – – – – 4 = 2.6910 V/m. The electric field in the glass is Eglass = Eair/K = (2.69104 V/m)/(5.80) = 4.64103 V/m.

Chapter 24 p. 18

66. Fortunately the required capacitance is greater. We can increase the capacitance by adding a parallel capacitor, which does not require breaking the circuit. We find the necessary capacitor from C = C1 + C2 ; 16 F = 5.0 F + C2 , which gives C2 = 11 F in parallel. 67. We find the capacitance from U = !CV 2; 200 J = !C(6000 V)2, which gives C = 1.110–5 F =

11 F.

68. (a) The radius of the pie plate is r = !(9.0 in)(2.5410–2 m/in) = 0.114 m. If we assume that it approximates a parallel-plate capacitor, we have C = Å0A/d = Å0pr2/d = (8.8510–12 C2/N · m2)p(0.114 m)2/(0.10 m) = 3.610–12 F = 3.6 pF. (b) We find the charge on each plate from Q = CV = (3.6 pF)(9.0 V) = 33 pC. (c) We assume that the electric field is uniform, so we have E = V/d = (9.0 V)/(0.10 m) = 90 V/m. (d) The work done by the battery is the energy stored in the capacitor: W = U = !CV 2 = !(3.610–12 F)(9.0 V)2 = 1.510–10 J. (e) Because the battery is still connected, the electric field will not change. Insertion of the dielectric will change capacitance, charge, and work done by the battery. 69. (a) For the stored energy we have U = !CV 2. Because the capacitance does not change, we have U2/U1 = (V2/V1)2 = (2)2 = 4. (b) For the stored energy we have U = !CV 2 = !Q 2/C. The capacitance does not change, so we have U2/U1 = (Q2/Q1)2 = (2)2 = 4. (c) Because the battery is still connected, the potential difference will not change. From C = Å0A/d, we see that separating the plates will change C. For the stored energy we have U = !CV 2, so we get U2/U1 = C2/C1 = d1/d2 = !. 70. If we equate the heat flow to the stored energy, we have U = !CV 2 = mc ?T; !(7.0 F)V 2 = (2.5 kg)(4186 J/kg · C°)(95°C – 20°C), which gives V =

4.7102 V.

71. Because the charged capacitor is disconnected from the plates, the charge must be constant. The paraffin will change the capacitance, so we have Q = C1V1 = C2V2 = KC1V2 ; 24.0 V = (2.2)V2 , which gives V2 = 10.9 V. 72. If the plates initially have a charge Q on each plate, the energy to move a charge ?Q will increase the stored energy: ?U = U2 – U1 = (!Q2 2/C) – (!Q1 2/C) = [(Q + ?Q)2 – Q 2]/2C = [(2Q ?Q + (?Q)2]/2C = (2Q + ?Q) ?Q/2C; 10.6 mC. 18.5 J = (2Q + 13.010–3 C)(13.010–3 C)/2(12.010–6 F), which gives Q = 0.0106 C =

Chapter 24 p. 19

73. (a) For a coaxial cable with a dielectric, we have C/L = KC0/L = 2KÅ0/ln(Ra/Rb). (b) For the given data we get C/L = 2KÅ0/ln(Ra/Rb) = 2(2.6)(8.8510–12 F/m)/ln[(9.0 mm)/(3.5 mm)] =

1.510–10 F/m.

74. The uniform electric field between the plates is related to the potential difference across the plates: E = V/d. For a parallel-plate capacitor, we have Q = CV = CEd;   0.47510–6 C = C(9.21104 V/m)(1.9510–3 m), which gives C = 2.6410–9 F. We find the area of the plates from C = KÅ0A/d; 2.6410–9 F = (3.75)(8.8510–12 C2/N · m2)A/(1.9510–3 m), which gives A = 0.155 m2. 75. Because the capacitor is isolated, the charge will not change. The initial stored energy is U1 = !C1V12 = !Q 2/C1 , with C1 = Å0A/d1 . The changes will change the capacitance: C2 = KÅ0A/d2 . For the ratio of stored energies, we have U2/U1 = C1/C2 = (Å0A/d1)/(KÅ0A/d2) = d2/Kd1 = !/K = 1/2K. The stored energy decreases from two factors. Because the plates attract each other, when the separation is halved, work is done by the field, so the energy decreases. When the dielectric is inserted, the induced charges on the dielectric are attracted to the plates; again work is done by the field and the energy decreases. The uniform electric field between the plates is related to the potential difference across the plates: E = V/d. For a parallel-plate capacitor, we have Q = C1V1 = C1E1d1 = C2E2d2 , or E2/E1 = C1d1/C2d2 = Å0A/KÅ0A = 1/K. 76. We find the initial charge on the 3.5-F capacitor when it is connected to the battery: Q = C1V = (3.5 F)(12.4 V) = 43.4 C. When C1 is disconnected from the battery and then connected to C2 , some charge will flow from C1 to C2 . The flow will stop when the voltage across the two capacitors is the same: V1 = V2 = 5.2 V. Because charge is conserved, we have Q = Q1 + Q2 . We find the charge remaining on C1 from Q1 = C1V1 = (3.5 F)(5.2 V) = 18.2 C. The charge on C2 is

Chapter 24 p. 20

Q2 = Q – Q1 = 43.4 C – 18.2 C = 25.2 C. We find the value of C2 from Q2 = C2V2 ; 25.2 C = C2(5.2 V), which gives C2 = 4.8 F. 77. (a) The energy stored in the capacitor is U = !CV 2 = !(0.06010–6 F)(25103 V)2 = 19 J. (b) We find the power of the pulse from P = 0.10 U/t = (0.10)(19 J)/(1010–6 s) = 1.9105 W =

0.19 MW.

78. (a) We see from the diagram that all positive plates are connected d to the positive side of the battery, and that all negative plates are connected to the negative side of the battery, so the capacitors are connected in parallel. (b) For parallel capacitors, the total capacitance is the sum, so we have Cmin = 7(Å0Amin/d) = 7(8.8510–12 C2/N · m2)(2.010–4 m2)/(1.010–3 m) = 1.210–11 F = 12 pF. Cmax = 7(Å0Amin/d) = 7(8.8510–12 C2/N · m2)(9.010–4 m2)/(1.010–3 m) = 5.610–11 F = 56 pF. Thus the range is 12 pF = C = 56 pF.

+ V –

79. (a) Because the capacitor is disconnected from the power supply, the charge is constant. We find the new voltage from Q = C1V1 = C2V2 ; (10 pF)(10,000 V) = (1 pF)V2 , which gives V2 = 1.0105 V = 0.10 MV. (b) A major disadvantage is that, when the stored energy is used, the voltage will decrease exponentially, so it can be used for only short bursts. 80. For a series network, we have Q1 = Q2 = Qseries = 125 pC. We find the equivalent capacitance from Qseries = CseriesV; 125 pC = Cseries(25.0 V), which gives Cseries = 5.00 pF. We find the unknown capacitance from 1/Cseries = (1/C1) + (1/C2); 1/(5.00 pF) = [1/(150 pF)] + (1/C2), which gives C2 =

5.17 pF.

81. If we express the stored energy as U = !CV 2, we see that, because the voltage is constant, we must increase the capacitance. Thus we add a capacitor in parallel. If we form the ratio of energies, we have U2/U1 = (C1 + C2)/C1 = 3, which gives

Chapter 24 p. 21

C2 = 2C1 = 2(330 pF) =

660 pF in parallel.

82. If we have n identical capacitors in parallel, the equivalent capacitance is Cparallel = nC. If we have n identical capacitors in series, we have 1/Cseries = n/C, or Cseries = C/n. We use these results to reduce the following nine combinations.

C

C

C

C1 = 4C.

C

C6 =

C

C

C

C C

C

C

4C/3.

C/3

C

C C2 = C/4.

C

C

1/C7 = 1/C + 2/3C; C7 = 3C/5.

C/2

C

C C

C/2

C

C

C

C3 = C.

C

C/2

C

C

C

C

C C

C

C

3C

C

C

1/C4 = 1/C + 1/3C; C4 = 3C/4.

C

C

C/2

C8 = 5C/2.

Chapter 24 p. 22

C

C9 = 5C/3.

C C

C

C

C

C

C C

C

2C

2C/3

C 1/C5 = 1/C + 1/C + 1/2C; C5 = 2C/5.

83. When the switch is connected left, the initial charge on C1 is Q0 = C1Vab = (1.0 F)(24 V) = 24 C. a When the switch is connected right, some charge will flow from C1 to C2 and C3 until the potential difference across C1 is the potential difference across the series combination of C2 and C3: V1 = V23 = V. Because C2 and C3 are in series, we know that Q2 = Q3 , and we find b their equivalent capacitance from 1/C23 = 1/C2 + 1/C3 = 1/2.0 F + 1/3.0 F, which gives C23 = 1.2 F. Because charge is conserved, we have Q0 = Q1 + Q2 . Q0 = C1V + C23V, which gives V = Q0/(C1 + C23) = (24 C)/(1.0 F + 1.2 F) = 11 V = V1 . For the charges we have Q1 = C1V = (1.0 F)(11 V) = 11 C; Q2 = Q3 = C23V = (1.2 F)(11 V) = 13 C. We find the potential differences from V2 = Q2/C2 = (13 C)/(2.0 F) = 6.5 V; V3 = Q3/C3 = (13 C)/(3.0 F) = 4.4 V.

S C2 C1

C3

Chapter 24 p. 23

84. We put a linear charge density of –  on the innermost cylinder and +  on the outermost cylinder. Because the two cylinders between these are connected, they will acquire equal and opposite charge densities so that they are at the same potential and have no electric field in the region between them. A charged cylinder has no electric field within it and an electric field outside it that is the same as that of a line charge: E = /2pÅ0r. We find the potential difference between c and d by integration: Rc  Rc –  dr = +  ln R c . Vc – Vd = – E d = – Rd 2Å 2Å0 Rd R 0r

+

–

– +

Rc

Rd

Ra Rb

d

The electric field between a and b will be due to the net line charge. We integrate to find the potential difference between a and b: Ra  Ra –  dr = +  ln R a . Va – Vb = – E d = – Rb 2Å0 Rb R b 2Å0 r Because Vb = Vc , we can use these results to get the potential difference across the capacitor: Va – Vd = (Va – Vb) + (Vc – Vd) = + [(/2pÅ0) ln(Ra/Rb)] + [+ (/2pÅ0) ln(Rc/Rd)] = (/2pÅ0) ln (RaRc/RbRd). Thus the capacitance per unit length is C/L = Q/LVad = 2pÅ0/ln(RaRc/RbRd). Note that this is also the result if the arrangement is treated as two cylindrical capacitors is series. 85. To pull the plates apart requires a force to balance the attractive force between the charges. The electric field at a plate is due to the other plate and thus has magnitude E = !Q/AÅ0 . Thus the work done by this force is W = F ?x = Q(Q/2AÅ0)(2x – x) = Q2x/2AÅ0 . Because the charge remains constant, we find the energy change from ?U = ?(Q2/2C) = Q2/2(Å0A/2x) – Q2/2(Å0A/x) = Q2x/2AÅ0 , which is the same as the work done. 86. (a) We find the number of electrons from the charge on the capacitor: N = Q/e = CV/e = (3010–15 F)(3.0 V)/(1.6010–19 C/electron) = 5.6105 electrons. (b) The electric field in the dielectric is E = V/d. Thus we see that the minimum thickness is determined by the maximum field, which is the dielectric strength: Emax = V/dmin ; 5.0107 V/m = (3.0 V)/dmin , which gives dmin = 6.010–8 m = 60 nm. (c) We find the required area from C = KÅ0A/dmin ; 3010–15 F = (1.00104)(8.8510–12 C2/N · m2)A/(6.010–8 m), 2.010–14 m2. which gives A = 87. (a) The capacitance is C0 = Å0A/d = (8.8510–12 C2/N · m2)(2.5 m2)/(3.010–3 m) = 7.3810–9 F = The charge on the capacitor is Q0 = C0V = (7.3810–9 F)(45 V) = 3.3210–7 C = 0.33 C. The electric field is

7.4 nF.

Chapter 24 p. 24

E0 = V/d = (45 V)/(3.010–3 m) = 1.5104 V/m. The stored energy is U0 = !C0V2 = !(7.3810–9 F)(45 V)2 = 7.510–6 J. (b) With the addition of the dielectric we have C = KC0 = (3.6)(7.3810–9 F) = 2.6610–8 F = 27 nF; –8 –7 Q = CV = (2.6610 F)(45 V) =1 .2010 C = 1.2 C; E = V/d = (45 V)/(3.010–3 m) = 1.5104 V/m; U = !CV2 = !KC0V2 = KU0 = (3.6)(7.510–6 J) = 2.710–5 J. 88. Because the sphere is conducting, there is no field inside. The field outside is E = Q/4Å0r2, so the energy density is u = !Å0E2 = Q2/322Å0r4. If we let the radius of the sphere that contains half the energy be r1 , we use a spherical shell of radius r and thickness dr as a differential element and two integrals to determine the ratio of energies: r1 r1 r1  Q 2/ 32 2Å0r 4 4r 2 dr dr/ r2 – 1/ r r0 r0 1/ r 1 – 1/ r 0 r 1 – r0 U = r0 =  = = r .   = U total 0 – 1/ r0 1 2 2 4 2 2 Q / 32 Å0 r 4r dr dr/ r – 1/ r r0

r0

r0

If we want this ratio to be !, we have r1 = 2r0 . (r1 – r0)/r1 = !, which gives

89. (a) The capacitance is C0 = KÅ0A/d = (3.7)(8.8510–12 C2/N · m2)(8.5 in)(11 in)(2.5410–2 m/in)2/(0.03010–3 m) 66 pF. = 6.5810–8 F = (b) The electric field is E = V/d, so the charge on the capacitor is Qmax = C0Vmax = C0Emaxd = (6.5810–8 F)(15106 V/m)(0.03010–3 m) 30 C. = 2.9610–5 C = (c) From the diagram we see that we need the same number of sheets of paper as we have individual capacitors, but we need one more sheet .. . of aluminum than this. Thus the thickness is t = (n + 1)tAl + nd = (101)(0.04010–3 m) + (100)(0.03010–3 m) = 7.0410–3 m = 7.0 mm. (d) Because the capacitors are connected in parallel, we find the maximum voltage from the voltage across a single capacitor that will cause breakdown: Vmax = Emaxd = (15106 V/m)(0.03010–3 m) = 450 V.

Chapter 24 p. 25

90. (a) The capacitance is C = Å0A/d = (8.8510–12 C2/N · m2)(110106 m2)/(1500 m) = 6.510–7 F = (b) The charge on the capacitor is Q = CV = (6.510–7 F)(35106 V) = 23 C. (c) We find the stored energy from U = !CV2 = !(6.510–7 F)(35106 V)2 = 4.0108 J.

0.65 F.

Chapter 25 p. 1

CHAPTER 25 – Electric Currents and Resistance 1.

The rate at which electrons pass any point in the wire is the current: I = 1.50 A = (1.50 C/s)/(1.6010–19 C/electron) = 9.381018 electron/s.

2.

The charge that passes through the battery is ?Q = I ?t = (5.7 A)(7.0 h)(3600 s/h) = 1.4105 C.

3.

We find the current from I = ?Q/?t = (1000 ions)(1.6010–19 C/ion)/(7.510–6 s) =

4.

We find the resistance from V = IR; 110 V = (4.2 A)R, which gives R =

5.

We find the voltage from V = IR = (0.25 A)(3000 ) =

2.110–11 A.

26 .

7.5102 V.

6.

For the device we have V = IR. (a) If we assume constant resistance and divide expressions for the two conditions, we get V2/V1 = I2/I1; 0.90 = I2/(5.50 A), which gives I2 = 4.95 A. (b) With the voltage constant, if we divide expressions for the two conditions, we get I2/I1 = R1/R2; I2/(5.50 A) = 1/0.90, which gives I2 = 6.11 A.

7.

The rate at which electrons leave the battery is the current: I = V/R = [(9.0 V)/(1.6 )](60 s/min)/(1.6010–19 C/electron) =

8.

9.

We find the resistance from V = IR; 24 . 12 V = (0.50 A)R, which gives R = The energy taken out of the battery is Energy = Pt = IVt = (0.50 A)(12 V)(1 min)(60 s/min) =

3.6102 J.

(a) We find the resistance from V = IR; 16 . 120 V = (7.5 A)R, which gives R = (b) The charge that passes through the hair dryer is 6.8103 C. ?Q = I ?t = (7.5 A)(15 min)(60 s/min) =

10. We find the potential difference across the bird’s feet from V = IR = (2500 A)(2.510–5 /m)(4.010–2 m) = 2.510–3 V. 11. We find the radius from R = L/A = L/pr2; 0.22  = (5.610–8  · m)(1.00 m)/pr2, which gives r = 2.8510–4 m, 0.57 mm. so the diameter is 5.710–4 m = 12. We find the resistance from R = L/A = L/#pd2

2.11021 electron/min.

Chapter 25 p. 2

= (1.6810–8  · m)(3.5 m)/#(1.510–3 m)2 =

0.033 .

13. From the expression for the resistance, R = L/A, we form the ratio RAl/RCu= (Al/Cu)(LAl/LCu)(ACu/AAl) = (Al/Cu)(LAl/LCu)(dCu/dAl)2 = [(2.6510–8  · m)/(1.6810–8  · m)][(10.0 m)/(20.0 m)][(2.5 mm)/(2.0 mm)]2 = 1.2, or RAl = 1.2RCu. 14.

Yes, if we select the appropriate diameter. From the expression for the resistance, R = L/A, we form the ratio RT/RCu = (T/Cu)(LT/LCu)(ACu/AT) = (T/Cu)(dCu/dT)2; 1 = [(5.610–8  · m)/(1.6810–8  · m)][(2.5 mm)/dT]2, which gives dT = 4.6 mm.

15. Because the material and area of the two pieces are the same, from the expression for the resistance, R = L/A, we see that the resistance is proportional to the length: R1/R2 = L1/L2 = 5.0. Because L1 + L2 = L, we have 5.0L2 + L2 = L, or L2 = L/6.0, and L1 = 5.0L/6.0, so the wire should be cut at 1/6 the length. We find the resistance of each piece from R1 = (L1/L)R = (5.0/6.0)(10.0 ) = 8.3 ; R2 = (L2/L)R = (1/6.0)(10.0 ) = 1.7 . 16. We find the temperature change from R = R0(1 +  T), or ?R = R0T; 0.20R0 = R0[0.0068 (C°)–1] ?T, which gives ?T =

29 C°.

17. For the wire we have R = V/I. We find the temperature from R = R 0(1 +  T); (V/I) = (V/I0)(1 +  T);

(10.00 V/0.3618 A) = (10.00 V/0.4212 A){1 + [0.00429 (C°)–1](T – 20.0°C)}, which gives T =

58.3°C.

18. We find the temperature from 0,T = Cu = 0,Cu(1 + Cu T);

5.610–8  · m = (1.6810–8  · m){1 + [0.0068 (C°)–1](T – 20.0°C)}, which gives T =

363°C.

19. We find the temperature from R = R0(1 + Cu T);

140  = (12 ){1 + [0.0060 (C°)–1](T – 20.0°C)}, which gives T =

20. For each direction through the solid, the length and area will be constant, so we have R = L/A. (a) In the x-direction, we have Rx = Lx/LyLz = (3.010–5  · m)(0.010 m)/(0.020 m)(0.040 m) = 3.810–4 . (b) In the y-direction, we have Ry = Ly/LxLz = (3.010–5  · m)(0.020 m)/(0.010 m)(0.040 m) = 1.510–3 . (c) In the z-direction, we have

1.8103 °C.

y Lx Ly x Lz z

Chapter 25 p. 3

Rz = Lz/LxLy = (3.010–5  · m)(0.040 m)/(0.010 m)(0.020 m) = 6.010–3 . 21. The dependence of the resistance on the dimensions is R = L/A. When we form the ratio for the two conditions, we get R2/R1 = (L2/L1)(A1/A2) = (!)(!) = #, so R2 = #R1 . 22. We use the temperature coefficients at 20°C. For the total resistance we have R = RC + RN = RC0(1 +  C T) + RN0(1 +  N T) = RC0 + RN0 + RC0 C T + RN0 N T. If the total resistance does not change, we have RC + RN = RC0 + RN0 , or RC0 C T = – RN0 N T; RC0[– 0.0005 (C°)–1] = – RN0[0.0004 (C°)–1], which gives RC0 = 0.8RN0 . For the total resistance we have 4.70 k = RC0 + RN0 = 0.8RN0 + RN0 , which gives RN0 = 2.61 k, and RC0 = 2.09 k. 











23. (a) For a length x of the uniform wire we have V = IR = Ix/A. If we find the potential gradient by differentiating, we get dV/dx = I/A. Because the current is defined to flow from higher to lower potential, that is, opposite to the potential gradient, we must introduce a negative sign: I = dq/dt = – A dV/dx. (b) The expression for heat conduction through an area A is dQ/dt = – kA dT/dx, where the negative sign indicates heat flow from higher to lower temperature. We expect  and k to be related, because the free electrons can easily acquire thermal energy and transmit it along the wire. 24. (a) We choose a cylindrical shell of radius r and thickness dr as a differential element. The area of the element is 2pr¬. We add (integrate) the resistances of all the shells:  r2  r2  R= dr = ln r . 2 r 2 1 r1 (b) For the given data we have R = [(3.010–5  · m)/2p(0.030 m)] ln(1.8 mm/1.0 mm) = 9.410–5 . (c) For the resistance along the axis we have R = ¬/A = (3.010–5  · m)(0.030 m)/p[(1.810–3 m)2 – (1.010–3 m)2] = 0.13 .

dr

r r1

r2

25. We choose a spherical shell of radius r and thickness dr as a differential element. The area of the element is 4pr2. We add (integrate) the resistances of all the shells:

Chapter 25 p. 4

R=

r2 r1

1 1 – 1 = dr = – 1  4 r 2 r 1  4r 2

r2 – r1 . 4 r 1r 2

26. (a) When we take into account the change in resistivity and the change in dimensions, we get R = L/A = 0(1 +  ?T)L0(1 +  ?T)/A0(1 +  ?T)2, or R/R0 = (1 +  ?T)/(1 +  ?T); (140 )/(12 ) = [1 + (4.510–3 /C°) ?T]/[1 + (510–6 /C°)?T], which gives ?T = 2.4103 C°. Thus the operating temperature is T = T0 + ?T = 20°C + 2.4103 C° = 2.4103 °C. (b) The thermal expansion has the effect of lowering the resistance because the area increase is greater than the length increase. For the temperature change, the new resistance from thermal expansion is Rthermal = R0/(1 +  ?T) = (12 )/[1 + (510–6 /C°)(2.4103 C°)] = 11.9 . The percentage of the actual change is %thermal = (11.9  – 12 )/(140  – 12 ) = – 0.1%. The change in  increases the resistance. For the temperature change, the new resistance from the increase in  is Rresistivity = R0(1 +  ?T) = (12 )[1 + (4.510–3 /C°)(2.4103 C°)] = 141.6 . The percentage of the actual change is %resistivity = (141.6  – 12 )/(140  – 12 ) = + 101%. 27. We find the power from P = IV = (0.350 A)(9.0 V) =

3.2 W.

28. For an ohmic resistor, we have P = IV = V 2/R, or R = V 2/P = (240 V)2/(3.1103 W) =

19 .

29. From P = V 2/R, we see that the maximum voltage will produce the maximum power, so we have # W = Vmax2/(5.4103 ), which gives Vmax = 37 V. 30. (a) From P = V 2/R, we see that the lower power setting, resistance. (b) At the lower setting, we have P1 = V 2/R1 ; 600 W = (120 V)2/R1 , which gives R1 = 24 . (c) At the higher setting, we have P2 = V 2/R2 ; 1200 W = (120 V)2/R2 , which gives R2 = 12 . 31. (a) We find the resistance from P1 = V 2/R1 ; 60 W = (120 V)2/R1 , which gives R1 = 240 . The current is I1 = V/R1 = (120 V)/(240 ) = 0.50 A. (b) We find the resistance from

600 W,

must have the higher

Chapter 25 p. 5

P2 = V 2/R2 ; 150 W = (120 V)2/R2 , which gives R2 = The current is I2 = V/R2 = (120 V)/(96 ) = 1.25 A.

96 .

32. We find the operating resistance from P = V 2/R; 60 W = (240 V) 2/R, which gives R = 9.6102 . If we assume that the resistance stays the same, for the lower voltage we have P = V 2/R = (120 V)2/(9.6102 ) = 15 W. At one-quarter the power, the bulb will be much dimmer. 33. We find the energy used by the toaster from Energy = Pt = (0.550 kW)(10 min)/(60 min/h) = 0.092 kWh. The cost for a month would be Cost = Energy(rate) = (0.092 kWh/day)(5 days/wk)(4 wk/month)(12¢/kWh) = 34. The cost for a year would be Cost = Energy(rate) = Pt(rate) = (6010–3 kW)(1 yr)(365 days/yr)(24 h/day)($0.110/kWh) = Note that power companies may ignore significant figures.

22¢/month.

$57.82.

35. 90 A · h is the total charge that passed through the battery when it was charged. We find the energy from Energy = Pt = VIt = VQ = (12 V)(90 A · h)(10–3 kW/W) = 1.1 kWh = 3.9106 J. 36. (a) We find the maximum power output from Pmax = ImaxV = (0.028 A)(9.0 V) = 0.25 W. (b) The power output eventually becomes thermal energy. The circuit is designed to allow the dissipation of the maximum power, which we assume is the same. Thus we have Pmax = ImaxV = ImaxV; 0.25 W = Imax(7.0 V), which gives Imax = 0.036 A = 36 mA. 37. The total power will be the sum, so we have Ptotal = ItotalV; N(100 W) = (2.5 A)(120 V), which gives N =

3.

38. The power rating is the mechanical power output, so we have efficiency = output/input = Pmechanical/Pelectrical = (0.50 hp)(746 W/hp)/(4.6 A)(120 V) = 0.676 =

68%.

39. The required current to deliver the power is I = P/V, and the wasted power (thermal losses in the wires) is Ploss = I 2R. For the two conditions we have I1 = (520 kW)/(12 kV) = 43.3 A; Ploss1 = (43.3 A)2(3.0 )(10–3 kW/W) = 5.63 kW; I2 = (520 kW)/(50 kV) = 10.4 A; Ploss2 = (10.4 A)2(3.0 )(10–3 kW/W) = 0.324 kW. Thus the decrease in power loss is ?Ploss = Ploss1 – Ploss2 = 5.63 kW – 0.324 kW = 5.3 kW.

Chapter 25 p. 6

40. (a) We find the resistance from P = V 2/R; 20.6 . 2800 W = (240 V) 2/R, which gives R = (b) If 80% of the electrical energy is used to heat the water to the boiling point, we have 0.80Energyelec = 0.80 Pelect = mc ?T; (0.80)(2800 W)t = (100 mL)(1 g/mL)(10–3 kg/g)(4186 J/kg · C°)(100°C – 20°C), which gives t = 15 s. (c) We find the cost from Cost = Energy(rate) = Pelect (rate) = [(2.80 kW)(15 s)/(3600 s/h)](10¢/kWh) = 0.12¢. 41. For the water to remove the thermal energy produced, we have P = IV = (m/t)c ?T; (14.5 A)(240 V) = (m/t)(4186 J/kg · C°)(6.50 C°), which gives m/t =

0.128 kg/s.

42. If 60% of the electrical energy is used to heat the water, we have 0.60Energyelec = 0.60 IVt = mc ?T; (0.60)I(12 V)(6.0 min)(60 s/min) = (150 mL)(1 g/mL)(10–3 kg/g)(4186 J/kg · C°)(95°C – 5°C), which gives I = 22 A. The resistance of the heating coil is R = V/I = (12 V)/(22 A) = 0.55 . 43. We find the peak current from the peak voltage: V0 = v2 Vrms = I0R; v2(120 V) = I0(1.8103 ), which gives I0 = 0.094 A. 44. We find the peak current from the peak voltage: V0 = I0R; 180 V = I0(330 ), which gives I0 = 0.545 A. The rms current is Irms = I0/v2 = (0.545 A)/v2 = 0.386 A. 45. (a) Because the total resistance is Rtotal = V/I, when I = 0, the resistance is infinite. (b) With one light bulb on, we have P = IrmsVrms = Vrms2/R; 75 W = (120 V)2/R, which gives R = 1.9102 . 46. We find the rms voltage from P = IrmsVrms ;

Chapter 25 p. 7

1500 W = [(6.0 A)/v2]Vrms , which gives Vrms = 47. The peak voltage is V0 = v2 Vrms = v2(450 V) = 636 V. We find the peak current from P = IrmsVrms = (I0/v2)Vrms ; 1800 W = (I0/v2)(450 V), which gives I0 =

3.5102 V.

5.66 A.

48. The maximum instantaneous power is P0 = I0V0 =(v2 Irms)(v2 Vrms) = 2P = 2(3.0 hp) = 6.0 hp (4.5 kW). For the maximum current, we have P = IrmsVrms = (I0/v2)Vrms ; (3.0 hp)(746 W/hp) = (I0/v2)(240 V), which gives I0 = 13 A. 49. For the average power, we have P = IrmsVrms = Vrms2/R = (240 V)2/(38 W) = 1.5103 W = 1.5 kW. The maximum power is P0 = I0V0 =(v2 Irms)(v2 Vrms) = 2P = 2(1.5 kW) = 3.0 kW. Because the power is always positive, the minimum power is zero. 50. (a) We find the frequency from the coefficient of t: 33.4 Hz. 2pf = 210 s–1, which gives f = (b) The maximum current is 1.80 A, so the rms current is Irms = I0/v2 = (1.80 A)/v2 = 1.27 A. (c) For the voltage we have V = IR = (1.80 A)(42.0 ) sin (210 s–1)t = (75.6 V) sin (210 s–1)t. 51. From Example 25-12 we know that the density of free electrons in copper is n = 8.41028 m–3. (a) We find the drift speed from I = neAvd = ne(#pD2)vd ; 2.510–6 A = (8.41028 m–3)(1.6010–19 C)[#p(0.5510–3 m)2]vd , which gives vd = 7.810–10 m/s. (b) The current density is j = I/A = (2.510–6 A)/#p(0.5510–3 m)2 = 10.5 A/m2 along the wire. (c) We find the electric field from j = E/; 1.810–7 V/m. 10.5 A/m2 = E/(1.6810–8  · m), which gives E = 52. (a) We find the resistance from R = V/I = (22.0 mV)/(750 mA) = 0.0293 . (b) We find the resistivity from R = L/A; 1.810–8  · m. 0.0293  = (5.00 m)/p(1.010–3 m)2, which gives  = (c) The current density is j = I/A = (75010–3 A)/p(1.010–3 m)2 = 2.4105 A/m2 along the wire. (d) We find the electric field from

Chapter 25 p. 8

E = V/L = (22.010–3 V)/(5.00 m) = 4.4010–3 V/m. (e) We find the density of free electrons from the drift speed: I = neAvd = ne(pr2)vd ; 75010–3 A = n(1.6010–19 C)p(1.010–3 m)2(1.710–5 m/s), which gives n =

8.81028 m–3.

53. If we take north as the positive direction, for the current density we have I /A = n+(+ 2e)vd+ + n–(– e)vd– = (2.81012 m–3)(2)(1.6010–19 C)(2.0106 m/s) + 2.7 A/m2 north. (8.01011 m–3)(– 1.6010–19 C)(– 7.2106 m/s) = 54. The charge is ?Q = I ?t = (1.00 A · h)(3600 s/h) =

3.60103 C.

55. We find the current when the lights are on from P = IV; 92 W = I(12 V), which gives I = 7.67 A. 90 A · h is the total charge that passes through the battery when it is completely discharged. Thus the time for complete discharge is t = Q/I = (90 A · h)/(7.67 A) = 12 h. 56. We find the current from P = IV; (1.5 hp)(746 W/hp) = I(120 V), which gives I = 57. We find the conductance from G = 1/R = I/V = (0.800 A)/(12.0 V) =

9.3 A.

6.6710–2 S.

58. (a) For the resistance of each wire we have RCu = CuLCu/A = (1.6810–8  · m)(5.0 m)/p(0.5010–3 m)2 = 0.107 . RAl = AlLAl/A = (2.6510–8  · m)(5.0 m)/p(0.5010–3 m)2 = 0.169 . Thus the total resistance is R = RCu + RAl = 0.107  + 0.169  = 0.28 . (b) We find the current from V = IR; 89 A. 25 V = I(0.28 ), which gives I = (c) The current must be the same for the two wires, so we have VCu = IRCu = (89 A)(0.107 ) = 10 V. VAl = IRAl = (89 A)(0.169 ) = 15 V. 59. (a) We find the resistance from V = IR; 8.6 . 2(1.5 V) = (0.350 A)R, which gives R = The power dissipated is P = IV = (0.350 A)(3.0 V) = 1.1 W. (b) We assume that the resistance does not change, so we have P2/P1 = (V2/V1)2 = (6.0 V/3.0 V)2 = 4 . The increased power would last for a short time, until the increased temperature of the filament would burn out the bulb. 60. We find the resistance of the heating element from P = IV = V 2/R;

Chapter 25 p. 9

900 W = (110 V)2/R, which gives R = 13.4 . We find the diameter from R = L/A = L/#pD2 = 4L/pD2; 13.4  = 4(9.7110–8  · m)(5.4 m)/pD2, which gives D = 2.210–4 m =

0.22 mm.

61. (a) The daily energy use is Energy = (1.8 kW)(3.0 h/day) + 4(0.10 kW)(6.0 h/day) + (3.0 kW)(1.4 h/day) + 2.0 kWh/day = 14 kWh/day. The cost for a month is Cost = Energy(rate) = (14 kWh/day)(30 days/month)($0.105/kWh) = $44. (b) For a 35-percent efficient power plant, we find the amount of coal from 0.35m(7000 kcal/kg)(4186 J/kcal) = (14 kWh/day)(365 days/yr)(103 W/kW)(3600 s/h), 1.8103 kg/yr. which gives m = 62. The current required to deliver the power is I = P/V. Thus the loss in a length L of the two wires is Ploss = I2R = (P/V)22L/A, so the loss per unit length is Ploss/L = [(10106 W)/(120 V)]2(1.6810–8  · m)(2)/p(0.2510–2 m)2 = 1.19107 W/m. The cost of this loss is Cost/t = (Ploss/L)(rate) = (1.19104 kW/m)($0.10 /kWh) = $1.2103 /h · m. Note that the loss in one meter of line is greater than the power delivered, so the input power would have to be much greater! 63. (a) The dependence of the power output on the voltage is P = V 2/R. When we form the ratio for the two conditions, we get P2/P1 = (V2/V1)2 . For the percentage change we have [(P2 – P1)/P1](100) = [(V2/V1)2 – 1](100) = [(105 V/117 V)2 – 1](100) = – 19.5%. (b) The decreased power output would cause a decrease in the temperature, so the resistance would decrease. This means for the reduced voltage, the percentage decrease in the power output would be less than calculated. 64. The maximum current will produce the maximum rate of heating. We can find the resistance per meter from P/L = I 2R/L; 1.6 W/m = (30 A) 2(R/L ), which gives R/L = 1.7810–3 /m. From the dependence of the resistance on the dimensions, R = L/A, we get R/L = /#pD2 = 4/pD2; 3.5 mm. 1.7810–3 /m = 4(1.6810–8  · m)/pD2, which gives D = 3.510–3 m = 65. (a) If we assume the values are rms values, we find the power from P = IV = (12 A)(120 V) = 1.44103 W. (b) We find the resistance from R1 = L/A1 = (1.6810–8  · m)(15 m)/#p(1.62810–3 m)2 = 0.12 . The power dissipated in the wiring is Ploss1 = I2R1 = (12 A)2(0.12 ) = 17 W. (c) We find the resistance of the #12 wire from R2 = L/A2 = (1.6810–8  · m)(15 m)/#p(2.05310–3 m)2 = 0.076 . The power dissipated in the wiring is

Chapter 25 p. 10

Ploss2 = I2R2 = (12 A)2(0.076 ) = 11 W. (d) The difference in cost is cost1 – cost2 = (Ploss1 – Ploss2)t(rate) = (17 W – 11 W)(12 h/day)(10–3 kW/W)($0.10 /kWh) 0.8¢/day. = $7.810–3 /day ˜ 66. The dependence of the power output on the voltage is P = V 2/R. If the change in voltage is small, we can approximate it by a differential, so we have dP = (2V/R) dV, or dP/P = 2 dV/V, so the percentage change in power is twice the percentage change in voltage. If we assume that the drop from 60 W to 50 W is small, we have (dV/V)(100) =  = – 8.3%. Thus the required voltage drop is 8.3%.

67. (a) We find the input power from Poutput = (efficiency)Pinput ; 900 W = (0.60)Pinput , which gives Pinput = 1500 W = (b) We find the current from P = IrmsVrms ; 1500 W = Irms(120 V), which gives Irms = 12.5 A.

1.5 kW.

68. Because the volume is constant, we have AL = AL, or A/A = L/L = 1/3.00. The dependence of the resistance on the dimensions is R = L/A. When we form the ratio for the two wires, we get R/R = (L/L)(A/A) R/(1.00 ) = (3.00)(3.00), which gives R = 9.00 . 69. The dependence of the resistance on the dimensions is R = L/A. When we form the ratio for the two wires, we get R1/R2 = (L1/L2)(A2/A1) = (L1/L2)(D2/D1)2 = (2)(!)2 = !. For a fixed voltage, the power dissipation is P = V 2/R. When we form the ratio for the two wires, we get P1/P2 = R2/R1 = 1/! = 2. 70. Heat must be provided to replace the heat loss through the walls and to raise the temperature of the air brought in: P = Ploss + mc ?T = 850 kcal/h + (2 /h)(1.29 kg/m3)(68 m3)(0.17 kcal/kg · C°)(20°C – 5°C) = 1.30103 kcal/h = (1.30103 kcal/h)(4186 J/kcal)/(3600 s/h) = 1.51103 W = 1.5 kW. 71. For the resistance, we have R = L/A = L/#pd2; 6.50  = 4(1.6810–8  · m)L/d2. The mass of the wire is m = mAL; 0.0180 kg = (8.9103 kg/m3)#pd2L. This gives us two equations with two unknowns, L and d. When we solve them, we get

Chapter 25 p. 11

d = 3.0310–4 m =

0.303 mm,

and L =

28.0 m.

72. (a) We find the average power required to provide the force to balance the average retarding force from P = Fv = (240 N)(40 km/h)/(3.6 ks/h)(746 W/hp) = 3.6 hp. (b) We find the average current from P = IV ; (3.57 hp)(746 W/hp) = I(12 V), which gives I = 222 A. 52 A · h is the total charge that passes through the battery when it is completely discharged. Thus the time for complete discharge is t = Q/I = (26)(52 A · h)/(222 A) = 6.08 h. In this time the car can travel x = vt = (40 km/h)(6.08 h) = 2.4102 km.

73. (a) We find the initial power consumption from P = V 2/R0 = (120 V)2/(12 ) = 1.2103 W = (b) The power consumption when the bulb is hot is P = V 2/R = (120 V)2/(140 ) = 100 W. The designated power is the operating power.

1.2 kW.

74. The stored energy in the capacitor must provide the energy used during the lapse: U = !CV2 = Pt; !C(120 V)2 = (150 W)(0.10 s), which gives C = 2.110–3 F. 75. The time for a proton to travel completely around the accelerator is t = L/v. In this time all the protons stored in the beam will pass a point, so the current is I = Ne/t = Nev/L; 1.41012 protons. 1110–3 A = N(1.6010–19 C)(3.0108 m/s)/(6300 m), which gives N = 76. From Example 25–12 we know that the density of free electrons in copper is n = 8.41028 m–3. With the alternating current an electron will oscillate with SHM. We find the rms current from P = IrmsVrms ; 500 W = Irms(120 V), which gives Irms = 4.17 A, so the peak current is I0 = v2 Irms = v2(4.17 A) = 5.89 A. The peak current corresponds to the maximum drift speed, which we find from I0 = neAvdmax = ne(#pD2)vdmax ; 5.89 A = (8.41028 m–3)(1.6010–19 C)[#p(1.810–3 m)2]vdmax , which gives vdmax = 1.7210–4 m/s. For SHM the maximum speed is related to the maximum displacement: vdmax = A = A(2pf); 1.7210–4 m/s = A[2p(60 Hz)], which gives A = 4.5710–7 m. For SHM the electron will move from one extreme to the other, so the total distance covered is 2A: 9.110–7 m. 77. In the steady state, there is no buildup of charge in the conductor, so the current must be the same at each

Chapter 25 p. 12

end. Thus the current densities are ja = I/Aa = (2.0 A)/#p(3.010–3 m)2 = jb = I/Ab = (2.0 A)/#

p(4.010–3

m)2

=

2.8105 A/m2; 1.6105 A/m2.

78. To find the resistance of the cylinder, we choose a vertical slice at a distance x from the origin, with diameter d = a + x, where  = (b – a)/¬, and thickness dx. We find the resistance by integrating over these slices:

y

a/2 R= =–

 dL = A

0

 d x 4 = 2 1  4 d

0 dx 0 a + x

2

4 1 4 4 1 – 1 = =– =  a + x 0  a +  a a a + 

4 . a b

b/2 = a/2 + ¬ ¬

x

Chapter 26 p. 1

CHAPTER 26 – DC Circuits 1.

(a) For the current in the single loop, we have Ia = V/(Ra + r) = (8.50 V)/(68.0  + 0.900 ) = 0.123 A. For the terminal voltage of the battery, we have Va = å – Iar = 8.50 V – (0.123 A)(0.900 ) = 8.39 V. (b) For the current in the single loop, we have Ib = V/(Rb + r) = (8.50 V)/(680  + 0.900 ) = 0.0125 A. For the terminal voltage of the battery, we have Vb = å – Ibr = 8.50 V – (0.0125 A)(0.900 ) = 8.49 V.

2.

The voltage across the bulb is the terminal voltage of the four cells: V = IRbulb = 4(å – Ir); (0.62 A)(12 ) = 4[2.0 V – (0.62 A)r], which gives r = 0.23 .

3.

If we can ignore the resistance of the ammeter, for the single loop we have I = å/r; 0.060 . 25 A = (1.5 V)/r, which gives r =

4.

We find the internal resistance from V = å – Ir; 0.037 . 9.8 V = [12.0 V – (60 A)r], which gives r = Because the terminal voltage is the voltage across the starter, we have V = IR; 0.16 . 9.8 V = (60 A)R, which gives R =

5.

When the bulbs are connected in series, the equivalent resistance is Rseries = ?Ri = 4Rbulb = 4(90 ) = 360 . When the bulbs are connected in parallel, we find the equivalent resistance from 1/Rparallel = ?(1/Ri) = 4/Rbulb = 4/(90 ), which gives Rparallel = 23 .

6.

(a) When the bulbs are connected in series, the equivalent resistance is Rseries = ?Ri = 3R1 + 3R2 = 3(40 ) + 3(80 ) = 360 . (b) When the bulbs are connected in parallel, we find the equivalent resistance from 1/Rparallel = ?(1/Ri) = (3/R1) + (3/R2) = [3/(40 )] + [3/(80 )], which gives Rparallel =

7.

8.

If we use them as single resistors, we have R1 = 25 ; R2 = 70 . When the resistors are connected in series, the equivalent resistance is Rseries = ?Ri = R1 + R2 = 25  + 70  = 95 . When the resistors are connected in parallel, we find the equivalent resistance from 1/Rparallel = ?(1/Ri) = (1/R1) + (1/R2) = [1/(25 )] + [1/(70 )], which gives Rparallel =

8.9 .

18 .

Because resistance increases when resistors are connected in series, the maximum resistance is Rseries = R1 + R2 + R3 = 500  + 900  + 1400  = 2800  = 2.80 k. Because resistance decreases when resistors are connected in parallel, we find the minimum resistance from 1/Rparallel = (1/R1) + (1/R2) + (1/R3) = [1/(500 )] + [1/(900 )] + [1/(1400 )], which gives Rparallel = 261 .

Chapter 26 p. 2

9.

The voltage is the same across resistors in parallel, but is less across a resistor in a series connection. We connect three 1.0- resistors in series as shown in the diagram. Each resistor has the same current and thus the same voltage: Vi = @V = @(6.0 V) = 2.0 V. Thus we can get a 4.0-V output between a and c.

a

R

b

R

c

R

d

I

10. When the resistors are connected in series, as shown in A, we have A RA = ?Ri = 3R = 3(1.20 k) = 3.60 k. R R When the resistors are connected in parallel, as shown in B, we have 1/RB = ?(1/Ri) = 3/R = 3/(1.20 k), so RB = 0.40 k. In circuit C, we find the equivalent resistance of the two resistors in parallel: C R 1/R1 = ?(1/Ri) = 2/R = 2/(1.20 k), so R1 = 0.60 k. R This resistance is in series with the third resistor, R so we have RC = R1 + R = 0.60 k + 1.20 k = 1.80 k. In circuit D, we find the equivalent resistance of the two resistors in series: R2 = R + R = 1.20 k + 1.20 k = 2.40 k. This resistance is in parallel with the third resistor, so we have 1/RD = (1/R2) + (1/R) = (1/2.40 k) + (1/1.20 k), so RD = 0.80 k.

V

R

B R

R R

D

R

R R

Chapter 26 p. 3

11. We can reduce the circuit to a single loop by successively combining parallel and series combinations. We combine R1 and R2 , which are in series: R7 = R1 + R2 = 2.8 k + 2.8 k = 5.6 k. We combine R3 and R7 , which are in parallel: 1/R8 = (1/R3) + (1/R7) = [1/(2.8 k)] + [1/(5.6 k)], which gives R8 = 1.87 k. We combine R4 and R8 , which are in series: R9 = R4 + R8 = 2.8 k + 1.87 k = 4.67 k. We combine R5 and R9 , which are in parallel: 1/R10 = (1/R5) + (1/R9) = [1/(2.8 k)] + [1/(4.67 k)], which gives R10 = 1.75 k. We combine R10 and R6 , which are in series: Req = R10 + R6 = 1.75 k + 2.8 k = 4.6 k.

12. (a) In series the current must be the same for all bulbs. If all bulbs have the same resistance, they will have the same voltage: Vbulb = V/N = (110 V)/8 = 13.8 V. (b) We find the resistance of each bulb from Rbulb = Vbulb/I = (13.8 V)/(0.60 A) = 23 . The power dissipated in each bulb is Pbulb = IVbulb = (0.60 A)(13.8 V) = 8.3 W.

Chapter 26 p. 4

13. For the parallel combination, the total current from the source is I = NIbulb = 8(0.340 A) = 2.72 A. The voltage across the leads is Vleads = IRleads = (2.72 A)(1.5 ) = 4.1 V. The voltage across each of the bulbs is Vbulb = V – Vleads = 110 V – 4.1 V = 106 V. We find the resistance of a bulb from Rbulb = Vbulb/Ibulb = (106 V)/(0.340 A) = 310 . The power dissipated in the leads is IVleads and the total power used is IV, so the fraction wasted is IVleads/IV = Vleads/V = (4.1 V)/(110 V) = 0.037 = 3.7%.

R bulb

R leads

I V

14. In series the current must be the same for all bulbs. If all bulbs have the same resistance, they will have the same voltage: Vbulb = V/N = (110 V)/8 = 13.8 V. We find the resistance of each bulb from Pbulb = Vbulb2/Rbulb ; 7.0 W = (13.8 V)2/Rbulb , which gives Rbulb = 27 . 15. Fortunately the required resistance is less. We can reduce the resistance by adding a parallel resistor, which does not require breaking the circuit. We find the necessary resistance from 1/R = (1/R1) + (1/R2); 1/(320 ) = [1/(480 )] + (1/R2), which gives R2 = 960  in parallel. 16. The equivalent resistance of the two resistors connected in series is Rs = R1 + R2 . We find the equivalent resistance of the two resistors connected in parallel from 1/Rp = (1/R1) + (1/R2), or Rp = R1R2/(R1 + R2). The power dissipated in a resistor is P = V2/R, so the ratio of the two powers is Pp/Ps = Rs/Rp = (R1 + R2)2/R1R2 = 4. When we expand the square, we get R12 + 2R1R2 + R22 = 4R1R2 , or R12 – 2R1R2 + R22 = (R1 – R2)2 = 0, which gives R2 = R1 =

1.6 k.

17. With the two bulbs connected in parallel, there will be 110 V across each bulb, so the total power will be 75 W + 40 W = 115 W. We find the net resistance of the bulbs from P = V 2/R; 105 . 115 W = (110 V)2/R, which gives R =

Chapter 26 p. 5

18. We can reduce the circuit to a single loop by successively combining parallel and series combinations. We combine R1 and R2 , which are in series: R7 = R1 + R2 = 2.20 k + 2.20 k = 4.40 k. We combine R3 and R7 , which are in parallel: 1/R8 = (1/R3) + (1/R7) = [1/(2.20 k)] + [1/(4.40 k)], which gives R8 = 1.47 k. We combine R4 and R8 , which are in series: R9 = R4 + R8 = 2.20 k + 1.47 k = 3.67 k. We combine R5 and R9 , which are in parallel: 1/R10 = (1/R5) + (1/R9) = [1/(2.20 k)] + [1/(3.67 k)], which gives R10 = 1.38 k. We combine R10 and R6 , which are in series: Req = R10 + R6 = 1.38 k + 2.20 k = 3.58 k. The current in the single loop is the current through R6 : I6 = I = å/Req = (12 V)/(3.58 k) = 3.36 mA. For VAC we have VAC = IR10 = (3.36 mA)(1.38 k) = 4.63 V. This allows us to find I5 and I4 ; I5 = VAC/R5 = (4.63 V)/(2.20 k) = 2.11 mA; I4 = VAC/R9 = (4.63 V)/(3.67 k) = 1.26 mA. For VAB we have VAB = I4R8 = (1.26 mA)(1.47 k) = 1.85 V. This allows us to find I3 , I2 , and I1 ; I3 = VAB/R3 = (1.85 V)/(2.20 k) = 0.84 mA; I1 = I2 = VAB/R7 = (1.85 V)/(4.40 k) = 0.42 mA. From above, we have VAB = 1.85 V.

Chapter 26 p. 6

19. (a) When the switch is closed the addition of R2 to the parallel set will decrease the equivalent resistance, so the current from the battery will increase. This causes an increase in the voltage across R1 , and a corresponding decrease across R3 and R4. The voltage across R2 increases from zero. Thus we have V1 and V2 increase; V3 and V4 decrease. (b) The current through R1 has increased. This current is now split into three, so currents through R3 and R4 decrease. Thus we have I1 (= I) and I2 increase; I3 and I4 decrease. (c) The current through the battery has increased, so the power output of the battery increases. (d) Before the switch is closed, I2 = 0. We find the R1 a resistance for R3 and R4 in parallel from I 1/RA = ?(1/Ri) = 2/R3 = 2/(100 ), I S + which gives RA = 50 . R 2 I3 V R 3 I4 R4 For the single loop, we have – I = I1 = V/(R1 + RA) = (45.0 V)/(100  + 50 ) = 0.300 A. b This current will split evenly through R3 and R4 : R1 I3 = I4 = !I = !(0.300 A) = 0.150 A. a After the switch is closed, we find the I resistance for R2 , R3 , and R4 in parallel from + R 2 I3 1/RB = ?(1/Ri) = 3/R3 = 3/(100 ), V R 3 I4 R4 I2 – which gives RB = 33.3 . For the single loop, we have b I = I1 = V/(R1 + RB) = (45.0 V)/(100  + 33.3 ) = 0.338 A. This current will split evenly through R2 , R3 , and R4 : I2 = I3 = I4 =

@I = @(0.338 A) =

0.113 A.

20. (a) When the switch is opened, the removal of a resistor from the parallel set will increase the equivalent resistance, so the current from the battery will decrease. This causes a decrease in the voltage across R1 , and a corresponding increase across R2 . The voltage across R3 decreases to zero. Thus we have V1 and V3 decrease; V2 increases. (b) The current through R1 has decreased. The current through R2 has increased. The current through R3 has decreased to zero. Thus we have I1 (= I) and I3 decrease; I2 increases. (c) Because the current through the battery decreases, the Ir term decreases, so the terminal voltage of

Chapter 26 p. 7

the battery will increase. (d) When the switch is closed, we find the resistance for R2 and R3 in parallel from 1/RA = ?(1/Ri) = 2/R = 2/(5.50 ), which gives RA = 2.75 . For the single loop, we have I = V/(R1 + RA + r) = (12.0 V)/(5.50  + 2.75  + 0.50 ) = 1.37 A. For the terminal voltage of the battery, we have Vab = å – Ir = 12.0 V – (1.37 A)(0.50 ) = 11.3 V. (e) When the switch is open, for the single loop, we have I = V/(R1 + R2 + r) = (12.0 V)/(5.50  + 5.50  + 0.50 ) = 1.04 A. For the terminal voltage of the battery, we have Vab = å – Ir = 12.0 V – (1.04 A)(0.50 ) = 11.5 V. 21. We find the resistance of a bulb from the nominal rating: Rbulb = Vnominal2/Pnominal = (12.0 V)2/(3.0 W) = 48 . We find the current through each bulb when connected to the battery from: Ibulb = V/Rbulb = (11.8 V)/(48 ) = 0.246 A. Because the bulbs are in parallel, the current through the battery is I = 2Ibulb = 2(0.246 A) = 0.492 A. We find the internal resistance from V = å – Ir; 0.4 . 11.8 V = [12.0 V – (0.492 A)r], which gives r = 22. We find the resistance for R1 and R2 in parallel from 1/Rp = (1/R1) + (1/R2) = [1/(3.8 k)] + [1/(2.1 k)], which gives Rp = 1.35 k. Because the same current passes through Rp and R3 , the higher resistor will have the higher power dissipation, so the limiting resistor is R3 , which will have a power

R1 c

R3

b I3

R2 I2

dissipation of ! W. We find the current from P3max = I3max2R3 ; 0.50 W = I3max2(1.8103 ), which gives I3max = 0.0167 A. The maximum voltage for the network is Vmax = I3max(Rp + R3) = (0.0167 A)(1.35103  + 1.8103 ) = 53 V. 23. For the current in the single loop, we have I = V/(R1 + R2 + r) = (9.0 V)/(8.0  + 12.0  + 2.0 ) = 0.41 A. For the terminal voltage of the battery, we have Vab = å – Ir = 9.0 V – (0.41 A)(2.0 ) = 8.18 V. The current in a resistor goes from high to low potential. For the voltage changes across the resistors, we have Vbc = – IR2 = – (0.41 A)(12.0 ) = – 4.91 V; Vca = – IR1 = – (0.41 A)(8.0 ) = – 3.27 V. For the sum of the voltage changes, we have

I1

a

R4

a

R3

c I3

å +

a

– r

I

R1 R2 c

b

b

Chapter 26 p. 8

Vab + Vbc + Vca = 8.18 V – 4.91 V – 3.27 V = 0. 24. For the loop, we start at point a: – IR – å2 – Ir2 + å1 – Ir1 = 0; – I(6.6 ) – 12 V – I(2 ) + 18 V – I(1 ) = 0, which gives I = 0.625 A. The top battery is discharging, so we have V1 = å1 – Ir1 = 18 V – (0.625 A)(1 ) = 17.4 V. The bottom battery is charging, so we have V2 = å2 + Ir2 = 12 V + (0.625 A)(2 ) = 13.3 V.

25. For the conservation of current at point c, we have Iin = Iout ; I1 = I2 + I3 . For the two loops indicated on the diagram, we have loop 1: V1 – I2R2 – I1R1 = 0; + 9.0 V – I2(15 ) – I1(22 ) = 0; loop 2: V3 + I2R2 = 0; + 6.0 V + I2(15 ) = 0. When we solve these equations, we get I1 = 0.68 A, I2 = – 0.40 A, I3 = 1.08 A. Note that I2 is opposite to the direction shown. 26. For the conservation of current at point c, we have Iin = Iout ; I1 = I2 + I3 . When we add the internal resistance terms for the two loops indicated on the diagram, we have loop 1: V1 – I2R2 – I1R1 – I1r1 = 0; + 9.0 V – I2(15 ) – I1(22 ) – I1(1.2 ) = 0; loop 2: V3 + I2R2 – I3r3 = 0; + 6.0 V + I2(15 ) – I3(1.2 ) = 0. When we solve these equations, we get I1 = 0.60 A, I2 = – 0.33 A, I3 = 0.93 A.

r1 –

b

å1

+

a

I

R r2 –

d

V1 +

å2

+

R1

– b 1

I1 c

I2

R2

I3

– d V 3

V1 + r1

a 2

+

R1

– b 1

I1 c

c

I2 I3

– d V 3

R2 + r3

a 2

Chapter 26 p. 9

27. For the conservation of current at point b, we have Iin = Iout ; I1 + I3 = I2 . For the two loops indicated on the diagram, we have loop 1: å1 – I1R1 – I2R2 = 0; + 9.0 V – I1(15 ) – I2(20 ) = 0; loop 2: – å2 + I2R2 + I3R3 = 0; – 12.0 V + I2(20 ) + I3(40 ) = 0. When we solve these equations, we get I1 = 0.18 A right, I2 = 0.32 A left, I3 = 0.14 A up.

R1

c

å1 + å2

I1

1

R2

– a –

b I2

2

+

R3 I3

d

28. For the conservation of current at point b, we have Iin = Iout ; I1 + I3 = I2 . When we add the internal resistance terms for the two loops indicated on the diagram, we have loop 1: å1 – I1r1 – I1R1 – I2R2 = 0; + 9.0 V – I1(1.0 ) – I1(15 ) – I2(20 ) = 0; loop 2: – å2 + I3r2 + I2R2 + I3R3 = 0; – 12.0 V + I3(1.0 ) + I2(20 ) + I3(40 ) = 0. When we solve these equations, we get I1 = 0.17 A right, I2 = 0.31 A left, I3 = 0.14 A up. 29. When we include the current through the battery, we have six unknowns. For the conservation of current, we have junction a: I = I1 + I2 ; junction b: I1 = I3 + I5 ; junction d: I2 + I5 = I4 . For the three loops indicated on the diagram, we have loop 1: – I1R1 – I5R5 + I2R2 = 0; – I1(20 ) – I5(10 ) + I2(25 ) = 0; loop 2: – I3R3 + I4R4 + I5R5 = 0; – I3(2 ) + I4(2 ) + I5(10 ) = 0; loop 3: + å – I2R2 – I4R4 = 0. + 6.0 V – I2(25 ) – I4(2 ) = 0.

c

å1 å2

+ r1 – a – + r 2 d

R1 1

I1

R2 b 2

I2 R3 I3

b R1

Chapter 26 p. 10

a

I5

1

When we solve these six equations, we get I1 = 0.274 A, I2 = 0.222 A, I3 = 0.266 A, I4 = 0.229 A, I5 = 0.007 A, I = 0.496 A. We have carried an extra decimal place to show the agreement with the junction equations. 30. When the 20- resistor is shorted, points a and b become the same point and we lose I1 . For the conservation of current, we have junction a:

R3

I1

I4

I2

R4 d

3

å

b

+

I = I2 + I3 + I5 ;

a

I5

1

R2

R5 I2

c

2

R5

R2

I

I3

–junction d: R3

I3 2

c

I4

R4 + 6.0 V – I3(2 ) = 0; d

) = 0; loop 3 (around outside): + å – I3R3 = 0. 3 right, we have Vcd = I4R4 = (3.50 mA)(4.0 k) = 14.0 V. We can now find the current in R8 : I8 = Vcd/R8 = (14.0 I å V)/(8.0 k) = 1.75 mA. From conservation of current at the + – junction c, we have I = I4 + I8 = 3.50 mA + 1.75 mA = 5.25 mA. If we go clockwise around the outer loop, starting at a, we have Vba – IR5 – I4R4 – å – Ir = 0, or Vba = (5.25 mA)(5.0 k) + (3.50 mA)(4.0 k) + 12.0 V + (5.25 mA)(1.010–3 k) = 52 V. If we assume the current in R4 is to the left, all currents are reversed, so we have Vdc = 14.0 V; I8 = 1.75 mA, and I = 5.25 mA. If we go counterclockwise around the outer loop, starting at a, we have – Ir + å – I4R4 – IR5 + Vab = 0, or Vba = – Vab = – Ir + å – I4R4 – IR5 ; Vba = – (5.25 mA)(1.010–3 k) + 12.0 V – (3.50 mA)(4.0 k) – (5.25 mA)(5.0 k) = – 28 V. The negative value means the battery is facing the other direction. 32. The given current is I2 = + 0.90 A. For the conservation of current at point a, we have I2 + I3 = I1 , or 0.90 A + I3 = I1 . We assume r1 = r2 = r3 = r4 = 1.0 . For a loop CCW around the outside, we have å1 – I1r1 – I1R3 – I3R4 + å3 – I3r3 – I3R5 – I1R1 = 0; 12.0 V – I1(1.0 ) – I1(8.0 ) – (I1 – 0.90 A)(15 ) + 6.0 V – (I1 – 0.90 A)(1.0 ) – (I1 – 0.90 A)(18 ) – I1(12 ) = 0, which gives I1 = 0.884 A. For the top loop indicated on the diagram, we have loop 1: å1 – I1r1 – I1R3 + å2 – I2r2 – I2R – I1R1 = 0; + 12.0 V – (0.884 A)(1.0 ) – (0.884 A)(8.0 ) + 12.0 V – (0.90 A)(1.0 ) – (0.90 A)R – (0.884 A)(12 ) = 0, 5.0 . which gives R = 33. The given current is I2 = – 0.30 A. For the conservation of current at point a, we have

I2 + I5 = I

When we solve t

Chapter 26 p. 11

I2 + I3 = I1 , or – 0.30 A + I3 = I1 . For the top loop indicated on the diagram, we have loop 1: å1 – I1r1 – I1R3 + å2 – I2r2 – I2R2 – I1R1 = 0; + 12.0 V – I1(1.0 ) – I1(8.0 ) + 12.0 V – (– 0.30 A)(1.0 ) – (– 0.30 A)(10 ) – I1(12 ) = 0, which gives I1 = 1.30 A. Thus we have I3 = I1 – I2 = 1.30 A – (– 0.30 A) = 1.60 A. For the bottom loop indicated on the diagram, we have loop 2: å – I3r – I3R5 + I2R2 – å2 + I2r2 – I3R4 = 0; å – (1.60 A)(1.0 ) – (1.60 A)(18 ) + (– 0.30 A)(10 ) – 12.0 V + (– 0.30 A)(1.0 ) – (1.60 A)(15 ) = 0, which gives å = 70 V.

c

I1

å1

r1

b

1

R3 d R4

å 2 r2 I2

e

R1

R2

g

a

2

å

I3

34. For the conservation of current at point a, we have I2 + I3 = I1 . For the two loops indicated on the diagram, we have loop 1: å1 – I1r1 – I1R3 + å2 – I2r2 – I2R2 – I1R1 = 0; + 12.0 V – I1(1.0 ) – I1(8.0 ) + 12.0 V – I2(1.0 ) – I2(10 ) – I1(12 ) = 0; loop 2: å3 – I3r3 – I3R5 + I2R2 – å2 + I2r2 – I3R4 = 0; + 6.0 V – I3(1.0 ) – I3(18 ) + I2(10 ) – 12.0 V + I2(1.0 ) – I3(15 ) = 0. When we solve these equations, we get I1 = 0.77 A, I2 = 0.71 A, I3 = 0.055 A. For the terminal voltage of the 6.0-V battery, we have Vfe = å3 – I3r3 = 6.0 V – (0.055 A)(1.0 ) = 5.95 V.

35. The lower loop equation becomes loop 2: å3 – I3r3 + I2R2 – å2 + I2r2 – I3R4 = 0; + 6.0 V – I3(1.0 ) + I2(10 ) – 12.0 V + I2(1.0 ) – I3(15 ) = 0. The other equations are the same: I2 + I3 = I1 . loop 1: å1 – I1r1 – I1R3 + å2 – I2r2 – I2R2 – I1R1 = 0; + 12.0 V – I1(1.0 ) – I1(8.0 ) + 12.0 V – I2(1.0 ) – I2(10 ) – I1(12 ) = 0. When we solve these equations, we get I1 = 0.783 A, I2 = 0.686 A, I3 = 0.097 A. We have carried an extra decimal place to show the agreement with the junction equations.

R5

r f

Chapter 26 p. 12

d

36. (a) On the diagram, we show the potential difference applied between points a and c. Because all of the resistors are the same, symmetry means that the three currents leaving point a must be the same three currents entering point c. This means that there is no current in the resistor between points b and d, which can be removed without changing the currents. When we redraw the circuit, we see that we have three parallel branches between points a and c. We find the equivalent resistance from 1/Rac = (1/R) + (1/2R) + (1/2R), which gives Rac = R/2. (b) If we apply a potential difference between points a and b, the same symmetry exists, so all currents leaving a must enter b. Consequently there is no current in R, which can be removed. When we redraw the circuit, we get one identical to that in part (a), so the equivalent resistance is R/2.

R

R I3

R I3

b R

R I2

a

I2

I1

R

R

d

R

b

c R I3 R I2

a

I1

R

R

d

R

c

R I3 R

c

I2 a

b

R

I1

37. We find the equivalent resistance for the parallel connection from 1/Rparallel = (1/R) + (1/R) + … = n/R, or Rparallel = R/n. We find the equivalent resistance for the series connection from Rseries = R + R + … = nR. The power transformation is P = IV = V 2/R. If we form the ratio at constant voltage for the two connections, we get Pseries/Pparallel = Rparallel/Rseries = (R/n)/nR = 1/n2.

38. (a) For the conservation of current at point d, we have I4 = I1 + I3 . For the conservation of current at point e, we have I2 = I3 + I5 . For the three loops indicated on the diagram, we have loop 1: å1 – I1R1 – I4R4 = 0; + 14 V – (I4 – I3)(10 k) – I4(12 k) = 0; loop 2: å2 – I2R2 – I5R3 = 0;

R1

a

å1 + I1

1

– d

R2

c

2

R4 I3

b

I4 –

3

I5 +

å3

R3

å2

+ –

I2 e

Chapter 26 p. 13

+ 18 V – (I3 + I5)(15 k) – I5(20 k) = 0; loop 3: å3 + I5R3 – I4R4 = 0; + 12 V + I5(20 k) – I4(12 k) = 0. We now have three equations for three currents. When we solve these and use the junction equations, we get I1 = 0.0665 mA, I2 = 1.111 mA, I3 = 1.045 mA, I4 = 1.111 mA, I5 = 0.0663 mA. (b) If we start at point b and add the potential drops, we have Vb – I2R2 + I1R1 = Va , or Vab = – (1.111 mA)(15 k) + (0.0665 mA)(10 k) = – 16 V.

Chapter 26 p. 14

39. (a) On the diagram, we show the potential difference applied between points a and c, and the six currents. For the conservation of current we have at point a: I = I1 + I2 + I3 ; (1) at point b: I3 = I4 + I5 ; at point d: I2 + I4 = I6 . For a loop CW around the right triangle, we have – I4R – I6R + I5R = 0; – I4R – (I2 + I4)R + (I3 – I4)R = 0, or (2R + R)I4 = I3R – I2R. (2) For a loop CW around the left triangle, we have – I2R + I4R + I3R = 0, or I4 = I2 – I3 . (3) For a loop CW around the bottom triangle, we have – I3R – I5R + I1R = 0; – I3R – (I3 – I4)R + I1R = 0, or I4 = 2I3 – I1 . (4) When we combine (3) and (4), we get 3I3 = I1 + I2 . (5) When we combine (2) and (3), we get (2R + R)I2 – (2R + R)I3 = I3R – I2R, or I3 = 2(R + R)I2/(3R + R). (6) When we combine (5) and (6), we get I2 = (3R + R)I1/(3R + 5R). (7) When we combine (6) and (7), we get I3 = 2(R + R)I1/(3R + 5R). (8) When we use (7) and (8) in (1), we get I = 8(R + R)I1/(3R + 5R). Because I1 = Vac/R, for the equivalent resistance we have Req = Vac/I = I1R/I = R(3R + 5R)/8(R + R). (b) If we apply a potential difference between points a and b, all currents leaving a must enter b. From the symmetry I1 = I2 and I4 = I5 , so I2 = – I4 and I1 = – I5 . Consequently there is no current in R, which can be removed. When we redraw the circuit, we see that we have three parallel branches between points a and b. We find the equivalent resistance from 1/Rab = (1/R) + (1/2R) + (1/2R), which gives Rab = R/2.

d R

R I2

b

R

I4

I6

R

R I3

a I

I1

I5

R

I

R

d

R

R

c

I1 R I1

a

I3

R

b

c

Chapter 26 p. 15

40. (a) On the diagram, we have used the symmetry to reduce the number of currents to five, as labeled. For the conservation of current we have I3 at point a: I = 2I1 + I2 ; (1) g f at point c: I1 = I3 + I4 ; (2) I1 I1 at point d: I5 = 2I4 . I I a For a loop CW around the bottom of the cube, we have b I4 I2 I1 I4 – I4R – I5R– I4R + I3R = 0; I5 I 1 – 2I4R – (2I4)R + I3R = 0, or I3 = 4I4 . (3) e d For a loop CW around the top of the cube, we have I4 – I1R – I3R– I1R + I2R = 0, or I2 = 2I1 + I3 . (4) I4 h c When we combine (2) and (3), we get I3 I3 = 4I1/5. (5) When we combine (4) and (5), we get I2 = 14I1/5. (6) When we combine (1) and (6), we get I = 24I1/5. (7) Because I2 = Vab/R, for the equivalent resistance we have Req = Vab/I = I2R/I = (14I1/5)R/(24I1/5) = 7R/12. (b) On the diagram, we have used the symmetry to reduce the number of currents to three, as labeled. Note that there are two resistors with zero current. I3 For the conservation of current we have g f I2 at point a: I = 2I1 + I2 ; (1) I=0 I at point d: I2 = 2I3 . (2) a b For a loop CW around the top of the cube, we have I3 I3 I1 I 1 – I2R – I3R+ 0 + I1R = 0, or I1 = I2 + I3 . (3) I3 I1 e When we combine (2) and (3), we get d I2 = 2I1/3. (4) I=0 I2 When we combine (1) and (4), we get h c I I = 8I1/3. (5) I1 Because I1 = Vac/2R, for the equivalent resistance we have Req = Vac/I = I1(2R)/I = 2I1R/(8I1/3) = 3R/4. (c) On the diagram, we have used the symmetry to reduce the number of currents to two, as labeled. For the conservation of current we have I2 g f at point a: I = 3I1 ; I1 at point c: 2I2 = I1 . I2 I For a path from a to d we have a I1 b I2 Vad = I1R + I2R + I1R = (2I1 + I1/2)R = 5I1R/2. I1 I1 I1 I2 For the equivalent resistance we have d e Req = Vad/I = (5I1R/2)/3I1 = 5R/6. I I2 I 1 h c

I2

Chapter 26 p. 16

R 41. (a) We find the capacitance from  = RC; 5510–6 s = (15103 )C, – 3.7 nF. which gives C = 3.710–9 F = å C + (b) The voltage across the capacitance will increase to the final steady state value. The voltage across the resistor will start at the battery voltage and decrease exponentially: S VR = å e – t/ ; 16.0 V = (24.0 V)e – t/(55 s), or t/(55 s) = ln(24.0 V/16.0 V) = 0.405, which gives t = 42. The time constant of the circuit is  = RC = (6.7103 )(6.010–6 F) = 0.0402 s. The capacitor voltage will decrease exponentially: VC = V0 e – t/ ; 0.01V0 = V0 e – t/(0.0402 s), or t/(0.0402 s) = ln(100) = 4.61, which gives t = 0.19 s.

– V0

+

C

22 s.

R

S

43. The charge on the capacitor increases with time to a final charge Q0 : Q = Q0 (1 – e – t/ ). When we express the stored energy in terms of charge we have U = !CV2 = !Q2/C = !(Q02/C)(1 – e – t/ )2 = Umax(1 – e – t/ )2. We find the time to reach half the maximum from ! = (1 – e – t/ )2, or e – t/ = 1 – 1/v2, which gives t = 1.23. 44. We find the equivalent capacitance from R2 C R1 C1 1/Ceq = (1/C1) + (1/C2) = [1/(6.0 F)] + [1/(6.0 F)], 2 – + – + which gives Ceq = 3.0 F. We find the equivalent resistance from I Req = R1 + R2 =2.2 k + 2.2 k = 4.4 k. – + The time constant of the circuit is S å  = ReqCeq = (4.4 k)(3.0 F) = 13.2 ms. Initially there is no charge on the capacitors, so the applied voltage is across the equivalent resistance: å = I0Req ; 12.0 V = I0(4.4 k), which gives I0 = 2.73 mA. The current will decrease exponentially: I = I0 e – t/ ; 1.50 mA = (2.73 mA)e – t/ , which gives t = 0.60 = (0.60)(13.2 ms) = 7.9 ms.

Chapter 26 p. 17

45. (a) For the conservation of current at point a, we have I1 I3 R1 I1 = I2 + I3 . a For the loop on the right, we have I2 + I2R2 – Q/C = 0, or I2R2 = Q/C. R2 For the outside loop, we have å 1 å – I1R1 – Q/C = 0, or I1R1 = å – Q/C. C The current I3 is charging the capacitor: I3 = dQ/dt. b When we use these results in the junction equation, we get (å – Q/C)/R1 = (Q/R2C) + dQ/dt, which becomes å = R1 dQ/dt + (R1 + R2)Q/R2C. This has the same form as the simple RC circuit: å = R dQ/dt + Q/C, if we replace R with R1 , and C with R2C/(R1 + R2). Thus the time constant is  = R1R2C/(R1 + R2). (b) After a long time, the current through the capacitor will be zero and it will have its maximum charge. The current through the resistors will be I = å/(R1 + R2). The voltage across the capacitor will be the voltage across R2: V = IR2 . Thus the charge on the capacitor will be Qmax = IR2C = åR2C/(R1 + R2).

46. In the steady state there is no current through the capacitors. V c = 24 V c (a) The current through the resistors is I = Vcd/(R1 + R2) = (24 V)/(8.8  + 4.4 ) = 1.82 A. C1 R1 The potential at point a is Va = Vad = IR2 = (1.82 A)(4.4 ) = 8.0 V. b I a S (b) We find the equivalent capacitance of the two series capacitors: C2 R2 1/C = (1/C1) + (1/C2) = [1/(0.48 F)] + [1/(0.24 F)], which gives C = 0.16 F. d Vd = 0 We find the charge on each of the two in series: Q1 = Q2 = Q = CVcd = (0.16 F)(24 V) = 3.84 C. The potential at point b is Vb = Vbd = Q2/C2 = (3.84 C)/(0.24 F) = 16 V. (c) With the switch closed, the current is the same. Point b must have the same potential as point a: Vb = Va = 8.0 V. (d) We find the charge on each of the two capacitors, which are no longer in series: Q1 = C1Vcb = (0.48 F)(24 V – 8.0 V) = 7.68 C; Q2 = C2Vbd = (0.24 F)(8.0 V) = 1.92 C. When the switch was open, the net charge at point b was zero, because the charge on the negative plate of C1 had the same magnitude as the charge on the positive plate of C2. With the switch closed, these charges are not equal. The net charge at point b is Qb = – Q1 + Q2 = – 7.68 C + 1.92 C = – 5.8 C, which flowed through the switch.

Chapter 26 p. 18

47. Because we have no simple series or parallel connections, we analyze the circuit. On the diagram, we show the potential difference applied between points c and d, and the four currents. For the conservation of current at points c and d, we have Vc I = I1 + I3 = I2 + I4 . (1) c The current I3 is charging the capacitor C1 : I3 = dQ1/dt. I I3 I1 C1 The current I4 is charging the capacitor C2 : I4 = dQ2/dt. R For a CW loop acba, we have I1R1 – Q1/C1 = 0, or Q1/C1 = (I – I3)R1 = (I – dQ1/dt)R1 . For a CW loop dabd, we have I2R2 – Q2/C2 = 0, or Q2/C2 = (I – I4)R2 = (I – dQ2/dt)R2 . For the path cbd, we have Vcd = (Q1/C1) + (Q2/C2). (4) If we differentiate this, we get dVcd/dt = 0 = (1/C1)(dQ1/dt) + (1/C2)(dQ2/dt), or dQ2/dt = – (C2/C1) dQ1/dt. When we combine (3) and (5), we get Q2/C2 = R2[I + (C2/C1) dQ1/dt]. When we combine (2) and (6), we get Q2/C2 = (R2/R1C1)Q1 + [R2(C1 + C2)/C1] dQ1/dt. (7) When we combine (4) and (7), we get Vcd = [(R1 + R2)/R1C1]Q1 + [R2(C1 + C2)/C1] dQ1/dt. This has the same form as the simple RC circuit: å = R dQ/dt + Q/C, with R = R2(C1 + C2)/C1 , and C = R1C1/(R1 + R2). Thus the time constant is  = R1R2(C1 + C2)/(R1 + R2) = (8.8 )(4.4 )(0.48 F + 0.24 F)/(8.8  + 4.4 ) =

1

b

a (2)

(3)

Vd

d

(5) (6)

2.1 s.

48. We find the resistance of the voltmeter from R = (sensitivity)(scale) = (50,000 /V)(250 V) = 1.25107  =

I2 C 2

R2

12.5 M.

49. We find the current for full-scale deflection of the ammeter from I = Vmax/R = Vmax/(sensitivity)Vmax = 1/(20,000 /V) = 5.010–5 A =

50 A.

I4

Chapter 26 p. 19

50. (a) We make an ammeter by putting a resistor in parallel with the galvanometer. For full-scale deflection, we have Vmeter = IGr = IsRs ; (5010–6 A)(30 ) = (30 A – 5010–6 A)Rs , 5010–6  in parallel. which gives Rs = (b) We make a voltmeter by putting a resistor in series with the galvanometer. For full-scale deflection, we have Vmeter = I(Rx + r) = IG(Rx + r); 1000 V = (5010–6 A)(Rx + 30 ), 20 M in series. which gives Rx = 20106  =

Is

IG

I

Rs

G

r

Rx

r IG

I 51. (a) The current for full-scale deflection of the galvanometer is I = 1/(sensitivity) = 1/(35 k/V) = 2.8610–2 mA. We make an ammeter by putting a resistor in parallel with the galvanometer. For full-scale deflection, we have Vmeter = IGr = IsRs ; (2.8610–5 A)(20.0 ) = (2.0 A – 2.8610–5 A)Rs , 2.910–4  in parallel. which gives Rs = (b) We make a voltmeter by putting a resistor in series with the galvanometer. For full-scale deflection, we have Vmeter = I(Rx + r) = IG(Rx + r); 1.00 V = (2.8610–5 A)(Rx + 20 ), 35 k in series. which gives Rx = 3.5104  =

52. We can treat the milliammeter as a resistor and find its resistance from 1/RA = (1/Rs) + (1/r) = (1/0.20 ) + (1/30 ), which gives RA = 0.199 . We make a voltmeter by putting a resistor in series with the galvanometer. For full-scale deflection, we have Vmeter = I(Rx + RA); 10 V = (2010–3 A)(Rx + 0.199 ), 0.50 k in series. which gives Rx = 5.0102  = The sensitivity of the voltmeter is 50 /V. Sensitivity = (500 )/(10 V) = 53. Before connecting the voltmeter, the current in the series circuit is I0 = å/(R1 + R2) = (45 V)/(37 k + 28 k) = 0.692 mA. The voltages across the resistors are V01 = I0R1 = (0.692 mA)(37 k) = 25.6 V;

Is

IG

I

Rs

G

r

Rx

r

G IG

I

Is

IG I

G

Rx

Rs

r

G

Chapter 26 p. 20

V02 = I0R2 = (0.692 mA)(28 k) = 19.4 V. When the voltmeter is across R1 , we find the equivalent resistance of the pair in parallel: 1/RA = (1/R1) + (1/RV) = (1/37 k) + (1/100 k), which gives RA = 27.0 k. The current in the circuit is I1 = å/(RA + R2) = (45 V)/(27 k + 28 k) = 0.818 mA. The reading on the voltmeter is VV1 = I1RA = (0.818 mA)(27.0 k) = 22.1 V = 22 V. When the voltmeter is across R2 , we find the equivalent resistance of the pair in parallel: 1/RB = (1/R2) + (1/RV) = (1/28 k) + (1/100 k), which gives RB = 21.9 k. The current in the circuit is I2 = å/(R1 + RB) = (45 V)/(37 k + 22 k) = 0.764 mA. The reading on the voltmeter is VV2 = I2RB = (0.764 mA)(21.9 k) = 16.7 V = 17 V. We find the percent inaccuracies introduced by the meter: (25.6 V – 22.1 V)(100)/(25.6 V) = 14% low; (19.4 V – 16.7 V)(100)/(19.4 V) = 14% low.

54. We find the voltage of the battery from the series circuit with the ammeter in it: å = IA(RA + R1 + R2) = (4.2510–3 A)(60  + 700  + 400 ) = 4.93 V. Without the meter in the circuit, we have å = I0(R1 + R2); 4.93 V = I0(700  + 400 ), which gives I0 = 4.4810–3 A = 4.48 mA. 55. We find the equivalent resistance of the voltmeter in parallel with one of the resistors: 1/R = (1/R1) + (1/RV) = (1/9.0 k) + (1/11.5 k), which gives R = 5.05 k. The current in the circuit, which is read by the ammeter, is I = å/(r + RA + R + R2) = (12.0 V)/(1.0  + 0.50  + 5.05 k + 9.0 k) = 0.85 mA. The reading on the voltmeter is Vab = IR = (0.85 mA)(5.05 k) = 4.3 V.

IA

å

+ –

a

a

d RA

R1 b R2 c

I0

R1

+

å –

b R2 c

Chapter 26 p. 21

56. When the voltmeter is across R1 , for the junction b, we have I1A + I1V = I1 ; [(5.5 V)/R1] + [(5.5 V)/(15.0 k)] = (12.0 V – 5.5 V)/R2 ; [(5.5 V)/R1] + 0.367 mA = (6.5 V)/R2 . When the voltmeter is across R2 , for the junction e, we have I2 = I2A + I2V ; (12.0 V – 4.0 V)/R1 = [(4.0 V)/R2] + [(4.0 V)/(15.0 k)]; [(8.0 V)/R1] = [(4.0 V)/R2] + 0.267 mA. We have two equations for two unknowns, with the results: R1 = 9.4 k, and R2 = 6.8 k.

57. The resistance of the voltmeter is RV = (sensitivity)(scale) = (1000 /V)(3.0 V) = 3.0103  = 3.0 k. We find the equivalent resistance of the resistor and the voltmeter from 1/Req = (1/R) + (1/RV), or Req = RRV/(R + RV) = (8.4 k)(3.0 k)/(8.4 k + 3.0 k) = 2.21 k. The voltmeter measures the voltage across this equivalent resistance, so the current in the circuit is I = Vab/Req = (2.0 V)/(2.21 k) = 0.905 mA. For the series circuit, we have å = I(R + Req) = (0.905 mA)(8.4 k + 2.21 k) = 9.6 V.

Chapter 26 p. 22

58. We find the resistances of the voltmeter scales: RV100 = (sensitivity)(scale) = (20,000 /V)(100 V) = 2.0106  = 2000 k; RV30 = (sensitivity)(scale) = (20,000 /V)(30 V) = 6.0105  = 600 k. The current in the circuit is I = (Vab/RV) + (Vab/R1). For the series circuit, we have å = Vab + IR2 . When the 100-volt scale is used, we have I = [(25 V)/(2000 k)] + [(25 V)/(120 k)] = 0.221 mA. å = 25 V + (0.221 mA)R2 . When the 30-volt scale is used, we have I = [(23 V)/(600 k)] + [(23 V)/(120 k)] = 0.230 mA. å = 23 V + (0.230 mA)R2 . We have two equations for two unknowns, with the results: å = 74.1 V, and R2 = 222 k. Without the voltmeter in the circuit, we find the current: å = I (R1 + R2); 74.1 V = I (120 k + 222 k), which gives I = 0.217 mA. Thus the voltage across R1 is Vab = I R1 = (0.217 mA)(120 k) = 26 V. 59. (a) In circuit 1 the voltmeter is placed in parallel with the resistor, so we find their equivalent resistance from 1/Req1 = (1/R) + (1/RV). The ammeter measures the current through this equivalent resistance and the voltmeter measures the voltage across this equivalent resistance, so we have Req1 = V/I. Thus we have I/V = (1/R) + (1/RV), or 1/R = (I/V) – (1/RV). (b) In circuit 2 the ammeter is placed in series with the resistor, so we find their equivalent resistance from Req2 = R + RA . The ammeter measures the current through this equivalent resistance and the voltmeter measures the voltage across this equivalent resistance, so we have Req2 = V/I = R + RA , or R = (V/I) – RA .

60. We find the temperature of the junction from å = k(T – T0); 1.7210–3 V = (4010–6 V/C°)(T – 25°C), which gives T =

68°C.

61. We assume the lowest emf represents the uncertainty in a reading. We find the uncertainty in the temperature from ?å = k ?T;

Chapter 26 p. 23

0.50 V = (14 V/C°) ?T, which gives ?T =

3.610–2 C°.

62. (a) We let m be the mass density. Because the mass is constant, when a wire is stretched, its area must change. If we represent the small changes as differentials, we have A = m/mL, and dA = – (m/mL2) dL , or dA/A = – dL/L. The resistance will change from the change in length and the change in area: R = L/A, and dR = (/A) dL – (L/A2) dA, or dR/R = dL/L – dA/A = 2 dL/L. Thus the fractional change in resistance is proportional to the fractional change in length. (b) We find the values of the resistances from Rx0 = (R2/R1)R30 = (1.4800)(40.700 ) = 60.236 ; Rx = (R2/R1)R3 = (1.4800)(40.736 ) = 60.289 . We find the change in width from K= 1.8 = [(60.289  – 60.236 )/(60.236 )]/(?L/4.5 mm), which gives ?L = (?R/Rx0)/(?L/L);

a I

å

R1 + –

R body

b R2

c 63. The voltage is the same across resistors in parallel, 2.210–3 mm. but is less across a resistor in a series connection. We connect two resistors in series as shown in the diagram. Each resistor has the same current: I = V/(R1 + R2) = (6.0 V)/(R1 + R2). If the desired voltage is across R1 , we have Vab = IR1 = (6.0 V)R1/(R1 + R2); 0.25 V = (6.0 V)R1/(R1 + R2) = (6.0 V)/[1 + (R2/R1)], which gives R2/R1 = 23. When the body is connected across ab, we want very negligible current through the body, so the potential difference does not change. This requires Rbody = 2000  » R1 . If we also do not want a large current from the battery, a possible combination is R1 = 4 , R2 = 92 . 64. Because the voltage is constant and the power is additive, we can use two resistors in parallel. For the lower ratings, we use the resistors separately; for the highest rating, we use them in parallel. The rotary switch shown allows the B contact to successively connect to C and D. The A contact connects to C and D for the parallel connection. We find the resistances for the three settings from P = V 2/R; 50 W = (120 V)2/R1 , which gives R1 = 288 ; 100 W = (120 V)2/R2 , which gives R2 = 144 ; 150 W = (120 V)2/R3 , which gives R3 = 96 . As expected, for the parallel arrangement we have 1/Req = (1/R1) + (1/R2); 1/Req = [1/(288 )] + [1/(144 )], which gives Req = 96  = R3 . Thus the two required resistors are 288 , 144 . 65. The voltage drop across one of the wires is Vdrop = IR = (3.0 A)(0.0065 /m)(115 m) = The applied voltage at the apparatus is

2.2 V.

R2

A D C

R1 B

+V

Chapter 26 p. 24

V = V0 – 2Vdrop = 120 V – 2(2.2 V) =

116 V.

66. We find the current through the patient (and nurse) from the series circuit: I = V/(Rmotor + Rbed + Rnurse + Rpatient) = (220 V)/(104  + 0 + 104  + 104 ) = 7.310–3 A = 7.3 mA. 67. The time between firings is t = (60 s)/(72 beats) = 0.833 s. In this time the capacitor reaches 45% of maximum, so we have V = V0(1 – e – t/) = 0.45V0 , which gives e – t/ = 0.55, or  = t/ln(1.82) = (0.833 s)/ln(1.82) = 1.39 s. We find the required resistance from  = RC; 0.19 M. 1.39 s = R(7.5 F), which gives R =

68. We find the required current for the hearing aid from P = IV; 2 W = I(4.0 V), which gives I= 0.50 A. With this current the terminal voltage of the three mercury cells would be Vmercury = 3(åmercury – Irmercury) = 3[1.35 V – (0.50 A)(0.030 )] = 4.01 V. With this current the terminal voltage of the three dry cells would be Vdry = 3(ådry – Irdry) = 3[1.5 V – (0.50 A)(0.35 )] = 3.98 V. Thus the mercury cells would have a higher terminal voltage. 69. (a) We find the current from I1 = V/Rbody = (110V)/(1100 ) = 0.10 A. (b) Because the alternative path is in parallel, the current is the same: 0.10 A. (c) The current restriction means that the voltage will change. Because the voltage will be the same across both resistances, we have I2R2 = IbodyRbody ; I2(40 ) = Ibody(1100 ), or I2 = 27.5Ibody . For the sum of the currents, we have I2 + Ibody = 28.5Ibody = 1.5 A, which gives Ibody = 5.310–2 A = 53 mA. 70. The resistance of the platinum wire is Rx = (R2/R1)R3 = (46.0 /38.0 )(3.48 ) = 4.21 . We find the length from R = L/A = (10.610–8 · m)L/p(0.46010–3 m)2, which gives L =

26.4 m.

Chapter 26 p. 25

71. For the conservation of current at point b, we have I = I1 + I2 . For the two loops indicated on the diagram, we have loop 1: å1 – I1r1 – IR = 0; I1 + 2.0 V – I1(0.10 ) – I(4.0 ) = 0; loop 2: å2 – I2r2 – IR = 0; b + 3.0 V – I2(0.10 ) – I(4.0 ) = 0. When we solve these equations, we get I2 I1 = – 4.69 A, I2 = 5.31 A, I = 0.62 A. For the voltage across R we have Vab = IR = (0.62 A)(4.0 ) = 2.5 V. Note that one battery is charging the other with a significant current.

R

I

å1

–

+ c

1

r1 2

å2 r 2

–

a

+ d

72. Because the two sides of the circuit are identical, we find the resistance from the time constant:  = RC; 1.0 M. 3.0 s = R(3.0 F), which gives R = 73. The terminal voltage of a discharging battery is V = å – Ir. For the two conditions, we have 40.8 V = å – (7.40 A)r; 44.5 V = å – (2.20 A)r. We have two equations for two unknowns, with the solutions: å =

46.1 V,

0.71 .

and r =

74. One arrangement is to connect N resistors in series. Each resistor will have the same power, so we need N = Ptotal/P = 5 W/! W = 10 resistors. We find the required value of resistance from Rtotal = NRseries ; 1.2 k = 10Rseries , which gives Rseries = 0.12 k. Thus we have 10 0.12-k resistors in series. Another arrangement is to connect N resistors in parallel. Each resistor will again have the same power, so we need the same number of resistors: 10. We find the required value of resistance from 1/Rtotal = ?(1/Ri) = N/Rparallel ; 1/1.2 k = 10/Rparallel , which gives Rparallel = 12 k. Thus we have 10 12-k resistors in parallel.

a R pot

I + –

75. The resistance along the potentiometer is proportional to the length, so we find the equivalent resistance between points b and c: 1/Req = (1/xRpot) + (1/Rbulb), or Req = xRpotRbulb/(xRpot + Rbulb). We find the current in the loop from I = V/[(1 – x)Rpot + Req]. The potential difference across the bulb is Vbc = IReq , so the power expended in the bulb is P = Vbc2/Rbulb .

V

b

x c

R bulb

Chapter 26 p. 26

(a) For x = 1.00 we have Req = (1.00)(100 )(200 )/[(1.00)(100 ) + 200 ] = 66.7 . I = (120 V)/[(1 – 1.00)(100 ) + 66.7 ] = 1.80 A. Vbc = (1.80 A)(66.7 ) = 120 V. P = (120 V)2/(200 ) = 72.0 W. (b) For x = 0.50 we have Req = (0.50)(100 )(200 )/[(0.50)(100 ) + 200 ] = 40.0 . I = (120 V)/[(1 – 0.50)(100 ) + 40.0 ] = 1.33 A. Vbc = (1.33 A)(40.0 ) = 53.3 V. P = (53.3 V)2/(200 ) = 14.2 W. (c) For x = 0.25 we have Req = (0.25)(100 )(200 )/[(0.25)(100 ) + 200 ] = 22.2 . I = (120 V)/[(1 – 0.25)(100 ) + 22.2 ] = 1.23 A. Vbc = (1.23 A)(22.2 ) = 27.4 V. P = (27.4 V)2/(200 ) = 3.76 W.

–

+

76. (a) When there is no current through the galvanometer, the current I must pass through the long resistor R, so å RV I the potential difference between A and C is VAC = IR. R‘ Because there is no current through the measured emf, A B for the bottom loop we have C R å = IR. When different emfs are balanced, the current I is the same, S so we have ås = IRs , and åx = IRx . G When we divide the two equations, we get å x ås/åx = Rs /Rx , or åx = (Rx/Rs)ås . or (b) Because the resistance is proportional to the length, we have åS åx = (Rx/Rs)ås = (45.8 cm/25.4 cm)(1.0182 V) = 1.836 V. (c) If we assume that the current in the slide wire is much greater than the galvanometer current, the uncertainty in the voltage is ?V = ± IGRG = ± (0.015 mA)(30 ) = ± 0.45 mV. Because this can occur for each setting and there will be uncertainties in measuring the distances, the minimum uncertainty is ± 0.90 mV. (d) The advantage of this method is that there is no effect of the internal resistance, because there is no current through the cell.

–

+

Chapter 26 p. 27

77. (a) Normally there is no dc current in the circuit, so the voltage of the battery is across the capacitor. When there is an interruption, the capacitor voltage will decrease exponentially: VC = V0 e – t/ . We find the time constant from the need to maintain 70% of the voltage for 0.20 s: 0.70V0 = V0 e – (0.20 s)/ , or (0.20 s)/ = ln(1.43) = 3.57, which gives  = 0.56 s. We find the required resistance from  = RC; 40 k. 0.56 s = R(1410–6 F), which gives R = 4.0104  = (b) In normal operation, there will be no voltage across the resistor, so the device should be connected between b and c. 78. The charge on the capacitor and the current in the resistor decrease exponentially: Q = Q0 e – t/ , I = I0 e – t/ , with I0 = Q0 /. When we express the stored energy in terms of charge we have U = !CV2 = !Q2/C = !(Q02/C)(e – t/ )2. Thus the change in stored energy for an elapsed time equal to one time constant is ?Ustored = !(Q02/C)(e – 2/ ) – !(Q02/C) = !(Q02/C)(e – 2 – 1). Because the current varies, we find the thermal loss by integrating:   Q 0 2 RC – 2  R e – 1 = – U stored . U thermal = I 2 R d t = I 20 R e – 2t/  d t = – I 20R  e – 2t/  = – 2 RC 2 0 0

79. To get an output voltage of 100 V, it is necessary to put solar cells in series. The number of series cells required is Nseries = (100 V)/(0.80 V/cell) = 125 cells. To get an output current of 1.0 A, it is necessary to put solar cells in parallel. The number of parallel cells required is Nparallel = (1.0 A)/(0.350 A) = 2.8 = 3 cells. We connect 125 cells in series and then 3 of these units in parallel. Thus the total number of cells is N = NseriesNparallel = (125)(3) = 375 cells. The size of the panel is (125)(0.030 m)  (3)(0.030 m) = 3.8 m  0.090 m. Of course it is possible to adjust the dimensions by changing the wiring of the cells. To optimize the output it is necessary to have the panel move so that it is always perpendicular to the sunlight. 80. We have labeled the resistors and the currents through the resistors with the value of the resistance. For the conservation of current at point c, we have

Chapter 26 p. 28

I4 + I5 = I6 + I12 . å8 R5 c a For the three loops indicated on the diagram, we have + – e loop 1: å10 – I5R5 – I6R6 + å5 = 0; I6 3 å 10 + 1 + å4 + 10.00 V – I5(5.00 ) – I6(6.00 ) + 5.00 V = 0; 2 R6 – 15.00 V = I5(5.00 ) + I6(6.00 ); R 12 – I12 I4 I5 + – loop 2: å4 + å8 – I12R12 – I4R4 = 0; d b R4 f + 4.00 V + 8.00 V – I2(12.00 ) – I4(4.00 ) = 0; å5 12.00 V = I4(4.00 ) + I12(12.00 ); loop 3: I12R12 – I6R6 = 0; I12(12.00 ) – I6(6.00 ) = 0. When we solve these equations, we get I4 = 0.857 A, I5 = 1.296 A, I6 = 1.428 A, I12 = 0.714 A. We have carried an extra decimal place to show the agreement with the junction equations.

81. (a) Immediately after the switch is closed, there is no charge a on the capacitor, so the voltage across C, and thus across R2 , S is zero. Thus the current is R1 I å I0 = å/R1 = (10.0 V)/(20 ) = 0.50 A. + (b) After a long time the charge on the capacitor will be maximum, b – so there will be no current through the capacitor. C R2 Thus the current is I = å/(R1 + R2) = (10.0 V)/(20  + 40 ) = 0.17 A. c (c) For the potential difference Vbc we have Vbc = IR2 = Qmax/C; (0.167 A)(40 ) = Qmax/(0.50 F), which gives Qmax = 3.3 C. (d) When the switch is opened, the capacitor will discharge through R2 . The time constant for this circuit is  = R2C = (40 )(0.50 F) = 20 s. The capacitor charge will decrease exponentially: Q = Qmax e – t/ ; 0.20Qmax = Qmax e – t/(20 s), or t/(20 s) = ln(5.0) = 1.61, which gives t = 32 s.

82. (a) The current in the series circuit is I = V0/(R1 + R2). The output voltage is the voltage across R2 : VT = IR2 = V0R2/(R1 + R2); 3.0 V = (12.0 V)R2/(10.0  + R2), which gives R2 = 3.3 . (b) We find the equivalent resistance between b and c: 1/Req = (1/R2) + (1/Rload); 1/Req = (1/3.33 ) + (1/7.0 ), which gives Req = 2.26 . The current in the series circuit is I = V0/(R1 + Req). The output voltage is the voltage across Req : VT = IReq = V0Req/(R1 + Req) = (12.0 V)(2.26 )/(10.0  + 2.26 ) =

a

S V0

+ –

R1

I b VT

R2 c

2.2 V.

Chapter 26 p. 29

I1

+

83. (a) After a long time there will be a steady state; there a S will be no current in the capacitor branch: I5 = 0; I1 = I3 , and I2 = I4. I2 R 2 I5 I For the two resistor branches we have Va – Vb = å = I2(R2 + R4); c C 12.0 V = I2(1.0  + 9.0 ), which gives I2 = 1.20 A; å – I Va – Vb = å = I1(R1 + R3); R4 4 12.0 V = I1(10.0  + 5.0 ), which gives I1 = 0.80 A. We can find the potential difference across the capacitor from b Vc – Vd = (Vc – Vb) – (Vd – Vb) = I4R4 – I3R3 = (1.20 A)(9.0 ) – (0.80 A)(5.0 ) = + 6.8 V. The charge on the capacitor is Q = C(Vc – Vd) = (1.5 F)(6.8 V) = 10.2 C. (b) When the switch is opened, we find the equivalent resistance between c and d: 1/Req = 1/(R1 + R2) + 1/(R3 + R4); 1/Req = [1/(10.0  + 1.0 ) + 1/(5.0  + 9.0 )], which gives Req = 6.16 . The time constant for this circuit is  = ReqC = (6.16 )(1.5 F) = 9.24 s. The capacitor charge will decrease exponentially: Q = Qmax e – t/ ; 0.050Qmax = Qmax e – t/(9.24 s), or t/(9.24 s) = ln(20.0) = 3.00, which gives t = 28 s.

d I3

R S +

84. (a) The time constant of the RC circuit is  = RC = (33.0 k)(4.00 F) = 132 ms. During the charging cycle, the charge and the voltage on the capacitor increases exponentially: VC = V0 (1 – e – t/ ), so we find the time to reach 90.0 V from

V0 –

R1

C

90.0 V = (100.0 V) 1 – e – t 1/ 132 ms , or t1/(132 ms) = ln(10.0) = 2.30, which gives t1 = 304 ms. (b) When the neon bulb starts conducting, the voltage on the capacitor drops quickly to 70.0 V and

R3

Chapter 26 p. 30

then starts charging. We can find the recharging time by first finding the time for the capacitor to reach 70.0 V:

70.0 V = (100.0 V) 1 – e – t 3/ 132 ms , or

(c)

t3/(132 ms) = ln(3.33) = 1.20,

which gives t3 = 159 ms. Thus the time to charge from 70.0 V to 90.0 V is t4 = t1 – t3 = 304 ms – 159 ms = 145 ms. The time from when the switch is closed is t2 = t1 + t4 = 304 ms + 145 ms = 449 ms.

V 90.0 V 70.0 V

t4 0

t t3

t1

t2

Chapter 27 p. 1

CHAPTER 27 – Magnetism 1.

(a) The maximum force will be produced when the wire and the magnetic field are perpendicular, so we have Fmax = ILB, or Fmax/L = IB = (7.40 A)(0.90 T) = 6.7 N/m. (b) We find the force per unit length from F/L = IB sin 45.0° = (Fmax/L) sin 45.0° = (6.7 N/m) sin 45.0° = 4.7 N/m.

2.

The force on the wire is produced by the component of the magnetic field perpendicular to the wire: F = ILB sin  = (150 A)(240 m)(5.010–5 T) sin 60° = 1.6 N perpendicular to the wire and to B.

3.

For the maximum force the wire is perpendicular to the field, so we find the current from F = ILB; 0.900 N = I(4.20 m)(0.0800 T), which gives I = 2.68 A.

4.

The force on the wire is produced by the component of the magnetic field perpendicular to the wire: F = ILB sin 40° 2.410–4 N perpendicular to the wire and to B. = (4.5 A)(1.5 m)(5.510–5 T) sin 40° =

5.

The maximum force will be produced when the wire and the magnetic field are perpendicular, so we have Fmax = ILB; 1.18 N = (8.75 A)(0.555 m)B, which gives B = 0.243 T.

6.

The force is maximum when the current and field are perpendicular: Fmax = ILB. When the current makes an angle  with the field, the force is F = ILB sin . Thus we have F/Fmax = sin  = 0.45, or  = 27°.

7.

8.

I I B



Fmax (into page)

F (into page)

(a) We see from the diagram that the magnetic field is up, so the top pole face is a south pole. (b) We find the current from the length of wire in the field: F = ILB; 3.5102 A. 5.30 N = I(0.10 m)(0.15 T), which gives I = (c) The new force is F = ILB sin  = F sin  = (5.30 N) sin 80° = 5.22 N. Note that the wire could be tipped either way. We have the same upward force on the semicircular wire. The force on each of the horizontal wires will also be upward. Thus the total force is Ftotal = Fsemi + 2Fwire = 2IB0R + 2ILB0 = 2IB0(R + L) upward.

B

B I F (out of page)

Fsemi Fwire I

Fwire B0

I

Chapter 27 p. 2

9.

If we consider a length L of the wire, for the balanced forces, we have mg = pr2Lg = ILB; (8.9103 kg/m3)p(1.010–3 m)2(9.80 m/s2) = I(5.010–5 T), 5.5103 A. which gives I =

10. We find the force per unit length from F/L = I(i  B) = (3.0 A)[i  (0.20i – 0.30 j + 0.25k)T] = (3.0 A)(– 0.30k – 0.25j)T = (– 0.75j – 0.90k) N/m = 11. We choose the coordinate system shown in the diagram. We select a differential element of the curved wire, d¬ = dx i + dy j, on which the force is dF = I d¬  B = I(dx i + dy j)  (– Bk) = IB(dx j – dy i). We find the resultant force by integration: F = ? dF = IB ? (dx j – dy i) = IB(?x j – ?y i), where ?x = xb – xa , and ?y = yb – ya. If we have the same current in the straight wire, the resultant force is F = I(?x i + ?y j)  (– Bk) = IB(?x j – ?y i), which is the same result.

(– 7.510–3j – 9.010–3k) N/cm.

y

I

b

dF d¬ a

B x

12. If we select a differential length d¬ of the z wire, we see that the force on this element has a magnitude B   B dF = IB d¬ d¬ d¬ r and will have a direction perpendicular to   the wire at an angle  below the horizontal. dF dF For each element there will be one diametrically opposite with an opposite horizontal force component, so the net force will be vertical. We find the net force by adding the vertical components from all of the differential elements: F = ? dFz = ? IB d¬ (– sin ) = – IB sin  ? d¬ = – IB2pr sin  = – 2pIBr2/(r2 + d2)1/2 (downward). 13. To find the direction of the force on the electron, we point our fingers east and curl them upward into the magnetic field. Our thumb points south, which would be the direction of the force on a positive charge. Thus the force on the electron is north. F = qvB = (1.6010–19 C)(7.75105 m/s)(0.85 T) = 1.0510–13 N north. 14. To find the direction of the force on the negative charge, we point our fingers in the direction of v and curl them into the magnetic field B. Our thumb points in the direction of the force on a positive charge. Thus the force on the negative charge is opposite to our thumb. (a) Fingers out, curl down, thumb right, force left. (b) Fingers down, curl back, thumb right, force left. (c) Fingers in, curl right, thumb down, force up. (d) Fingers right, curl up, thumb out, force in. (e) Fingers left, but cannot curl into B, so force is zero. (f) Fingers left, curl out, thumb up, force down.

I

Chapter 27 p. 3

15. We assume that we want the direction of B that produces the maximum force, i. e., perpendicular to v. Because the charge is positive, we point our thumb in the direction of F and our fingers in the direction of v. To find the direction of B, we note which way we should curl our fingers, which will be the direction of the magnetic field B. (a) Thumb out, fingers left, curl down. (b) Thumb up, fingers right, curl in. (c) Thumb down, fingers in, curl right. 16. To find the direction of the force on the electron, we point our fingers upward and curl them forward into the magnetic field. Our thumb points left, which would be the direction of the force on a positive charge. Thus the force on the electron is to the right. As the electron deflects to the right, the force will always be perpendicular, so the electron will travel in a clockwise vertical circle. The magnetic force provides the radial acceleration, so we have F = qvB = mv2/r, so the radius of the path is r = mv/qB; 4.1010–5 m. = (9.1110–31 kg)(1.80106 m/s)/(1.6010–19 C)(0.250 T) = 17. The magnetic force provides the centripetal acceleration: qvB = mv2/r, or mv = p = qBr. 18. For the net force on the electron to be zero, the velocity is perpendicular to the crossed electric and magnetic fields. The magnitudes of the two forces will be equal: eE = evB, or v = E/B = (8.8103 V/m)/(3.510–3 T) = 2.5106 m/s. When the electric field is turned off, the magnetic force will always be perpendicular to the velocity. The magnetic force provides the radial acceleration, so we have F = evB = mv2/r, so the radius of the path is r = mv/eB; 0.41 cm. = (9.1110–31 kg)(2.5106 m/s)/(1.6010–19 C)(3.510–3 T) = 4.1010–3 m = 19. (a) The magnetic force provides the centripetal acceleration: qvB = mv2/r, or mv = qBr. The kinetic energy of the electron is K = !mv2 = !(qBr)2/m = (q2B2/2m)r2. (b) The magnetic force provides the centripetal acceleration: qvB = mv2/r, or mv = p = qBr. The angular momentum is L = mvr = qBr2. 20. The force on the electron is F =–evB = (– 1.6010–19 C)(4.0i – 6.0 j )  104 m/s  (– 0.80i + 0.60 j)T + (3.810–15 N)k. = – (1.6010–15 N)[(4.0)(0.60) – (– 6.0)(– 0.80)]k = 21. The magnetic force provides the radial acceleration, so we have F = evB = mv2/r, so r = mv/eB = (2mK)1/2/eB = [2(1.6710–27 kg)(5.0106 eV)(1.6010–19 J/eV)]1/2/(1.6010–19 C)(0.20 T) =

1.6 m.

Chapter 27 p. 4

22. The greatest force will be produced when the velocity and the magnetic field are perpendicular. We point our thumb down (a negative charge!), and our fingers north. We must curl our fingers to the east, which will be the direction of the magnetic field. We find the magnitude from F = qvB; 1.6 T east. 7.210–13 N = (1.6010–19 C)(2.9106 m/s)B, which gives B = 23. (a) We find the speed acquired from the accelerating voltage from energy conservation: 0 = ?K + ?U; 0 = !mv2 – 0 + q(– V), which gives v = (2qV/m)1/2 = [2(2)(1.6010–19 C)(2100 V)/(6.610–27 kg)]1/2 = 4.51105 m/s. For the radius of the path, we have r = mv/qB = (6.610–27 kg)(4.51105 m/s)/(2)(1.6010–19 C)(0.340 T) 2.7 cm. = 2.710–2 m = (b) The period of revolution is T = 2pr/v = 2pm/qB = 2p(6.610–27 kg)/(2)(1.6010–19 C)(0.340 T) = 3.810–7 s. 24. The magnetic force produces an acceleration perpendicular to the original motion: a = qvB/m = (13.510–9 C)(160 m/s)(5.0010–5 T)/(3.4010–3 kg) = 3.1810–8 m/s2. The time the bullet takes to travel 1.00 km is t = L/v = (1.00103 m)/(160 m/s) = 6.25 s. The small acceleration will produce a small deflection, so we assume the perpendicular acceleration is constant in magnitude and direction. We find the deflection of the electron from y = v0yt + !at2 = 0 + !(3.1810–8 m/s2)(6.25 s)2 = 6.2010–7 m. This justifies our assumption of constant acceleration. 25. If we assume the positively-charged ion is traveling west around the equator, the magnetic force will be down. This force and mg will produce the radial acceleration: mg + evB = mv2/r, or v2 – (eBr/m)v – gr = 0. We evaluate the constants: eBr/m = (1.6010–19 C)(0.4010–4 T)(6.38106 m + 5.0103 m)/(238 u)(1.6610–27 kg/u) = 1.034108 m/s; gr = (9.80 m/s2)(6.38106 m + 5.0103 m) = 6.25107 m2/s2. Thus we have v2 – (1.034108 m/s)v – 6.25107 m2/s2 = 0. The solutions to this quadratic equation are + 1.034108 m/s, – 0.605 m/s. We see that for the large positive velocity (to the west) the mg term is negligible, so the magnetic force provides the radial acceleration. For the small negative velocity (to the east), the radial acceleration is negligible because the radius is so large; the upward magnetic force essentially balances the mg force. Thus the appropriate answer is 1.034108 m/s (west) and gravity can be ignored. Note that this speed is close to c, so a relativistic correction is necessary. 26. We find the velocity acquired from the accelerating potential from energy conservation: K + U = 0; !mv2 + Q(– V) = 0, or mv = (2mQV)1/2. The magnetic force provides the radial acceleration, so we have F = QvB = mv2/r, so r = mv/QB = (2mQV)1/2/QB = (2mV/Q)1/2/B. When we form the ratio for two particles, we have

Chapter 27 p. 5

rd/rp = [(Qp/Qd)(md/mp)]1/2 = [(e/e)(2mp/mp)]1/2 = v2; r/rp = [(Qp/Q)(m/mp)]1/2 = [(e/2e)(4mp/mp)]1/2 = v2. 27. The total force on the proton is F = e(E + v  B) = e{(3.0i – 4.2 j)  103 V/m + [(6.0i + 3.0j – 5.0k)  103 m/s  (0.45i + 0.20 j)T]} = (1.6010–19 C)[(3.0i – 4.2 j) + (1.0i – 2.25j – 0.15k)]  103 V/m = (6.4i – 10.3j – 0.24k)]  10–16 N. 28. For the magnetic force we have F = – ev  B; (3.8i – 2.7j )  10–13 N = – (1.6010–19 C)(vxi + vyj + vzk)  (0.35 T)k; (3.8i – 2.7j )  10–13 N = – (1.6010–19 C)(0.35 T)(– vxj + vyi). When we equate each component, we get vx = (– 2.7  10–13 N)/(1.6010–19 C)(0.35 T) = – 4.8106 m/s; vy = – (3.8  10–13 N)/(1.6010–19 C)(0.35 T) = – 6.8106 m/s. Thus the velocity is v = (– 4.8106 m/s)i + (– 6.8106 m/s)j. 29. The component of the velocity parallel to the field does not change. The component perpendicular to the field produces a force which causes the circular motion. We find the radius of the circular motion from r = mv/qB = (9.1110–31 kg)(3.0106 m/s)(sin 45°)/(1.6010–19 C)(0.23 T) = 5.310–5 m. We find the time for one revolution: T = 2pr/v = 2p(5.2510–5 m)/(3.0106 m/s) sin 45° = 1.5610–10 s. In this time, the distance the electron travels along the field is p = vT = (3.0106 m/s)(cos 45°)(1.5610–10 s) = 3.310–4 m.

r

v

 B

30. The magnetic force provides the radial acceleration, so we have F = evB = mv2/r, so r = mv/eB = (2mK)1/2/eB. Because the protons are bent 90° their path will extend a distance into the field equal to the radius. Thus we have r = (2mK)1/2/eB = ¬, or B = (2mK/e2¬2)1/2.

Chapter 27 p. 6

31. (a) Because the velocity is perpendicular to the magnetic field, the proton will travel in a circular arc. From the symmetry of the motion we see that the upper half is a mirror image of the lower half, so the exit angle is the same as the incident angle: 45°. (b) The magnetic force provides the radial acceleration, so we have F = evB = mv2/r, so r = mv/eB = (1.6710–27 kg)(2.0105 m/s)/(1.6010–19 C)(0.850 T) = 2.4610–3 m. Thus the distance x is x = rv2 = 3.510–3 m.

 v

r B x r v 45°

32. When the loop is parallel to the magnetic field, the torque is maximum, so we have  = NIAB; 1.96 T. 0.185 m · N = (1)(7.10 A)p(0.0650)2B, which gives B = 33. We find the work required to rotate the coil from W = ?U = (–  · B)f – (–  · B)i = B( – cos f + cos i). (a) The work required to rotate from 0° to 180° is W = B( – cos 180° + cos 0°) = 2B. (b) The work required to rotate from 90° to – 90° is W = B[ – cos (– 90°) + cos 90°] = 0. 34. The angular momentum of the electron for the circular orbit is L = mvr. The time for the electron to go once around the orbit is T = 2pr/v, so the effective current is I = e/T = ev/2pr. The magnetic dipole moment is M = IA = (ev/2pr)pr2 = evr/2 = (e/2m)L. 35. (a) The angle between the normal to the coil and the field is 24.0°, so the torque is  = NIAB sin  = (12)(7.10 A)p(0.0850 m)2(5.5010–5 T) sin 24.0° = 4.3310–5 m · N. (b) From the directions of the forces shown on the diagram, the north edge of the coil will rise.

up F

I (west)

north I (east) BEarth

F



36. (a) The magnetic dipole moment of the coil is  = NIA = (20)(7.6 A)p(0.10 m)2 = 4.8 A · m2 perpendicular to the loop. The clockwise direction of the current produces a magnetic moment in the – z direction, so we have  = – (4.8 A · m2)k.

Chapter 27 p. 7

(b) The torque on the coil is  =   B = (– 4.8 A · m2)k (0.80i + 0.60j – 0.65k)T = (– 4.8 m · N)(0.80j – 0.60i) = (+ 2.9i – 3.8j) m · N. (c) If we take the reference level of U = 0 when the magnetic moment and field are perpendicular, the potential energy of the coil is U = –  ·· B = – (– 4.8 A · m2)k·· (0.80i + 0.60j – 0.65k)T = (4.8 A · m2)(– 0.65 T) = – 3.1 J.

37. The rotating charge is equivalent to a circular current. We choose a differential element of length dy a distance y from the axis of rotation. The charge on this element is dq = (Q/¬) dy. Because the time for one revolution is T = 2p/, the effective current of the element is dI = dq/T = (/2p) dq. Thus the magnetic moment of the element is d = A dI = (py2)(/2p) dq = (Q/2¬)y2 dy. We find the total magnetic moment by adding (integrating) the magnetic moments of the differential elements:

=



d  = (Q/ 2 )

3

0

+

dy

+



+ + +

y

+ + +

2

y 2 dy = (Q/ 2 )( / 3) = Q / 6.

38. When the coil comes to rest, the magnetic torque is balanced by the restoring torque: NIAB = k. Because the full-scale deflection is the same, we have I1B1 = I2B2 ; (63.0 A)B1 = I2(0.860B1), which gives I2 = 73.3 A. 39. When the coil comes to rest, the magnetic torque is balanced by the restoring torque: NIAB = k. Because the full-scale deflection is the same, we have I1/k1 = I2/k2 ; (36 A)/k1 = I2/(0.80k1), which gives I2 = 29 A. 40. If we assume that the magnetic field is constant, we have 2/1 = I2/I1 = 0.82, so the torque will decrease by 18%. If we assume that the magnetic field is produced by the current, it will be proportional to the current and will also decrease by 18%, so we have 2/1 = (I2/I1)(B2/B1) = (0.82)(0.82) = 0.67, so the torque will decrease by 33%.

Chapter 27 p. 8

41. For the net force on the particle to be zero when the two fields are present, the velocity is perpendicular to the crossed electric and magnetic fields. The magnitudes of the two forces will be equal: qE = qvB, or v = E/B. Without the electric field, the magnetic force will always be perpendicular to the velocity. The magnetic force provides the radial acceleration, so we have F = qvB = mv2/r, so q/m = v/Br = E/B2r = (200 V/m)/(0.46 T)2(8.010–3 m) = 1.2105 C/kg. 42. When the drop is held at rest, the electric field will produce an upward force to balance the force of gravity: mg = qE = NeV/d; (3.310–15 kg)(9.80 m/s2) = N(1.6010–19 C)(340 V)/(1.010–2 m), which gives N = 5.9 = 6 electrons.

43. (a) The Hall emf is across the width of the sample, so the Hall field is EH = åH/w = (6.510–6 V)/(0.030 m) = 2.210–4 V/m. (b) The forces from the electric field and the magnetic field balance. We find the drift speed from EH = vdB; 2.210–4 V/m = vd(0.80 T), which gives vd = 2.710–4 m/s. (c) We find the density from I = neAvd ; 42 A = n(1.6010–19 C)(0.030 m)(50010–6 m)(2.710–4 m/s), 6.41028 electrons/m3. which gives n = 44. The Hall field is EH = åH/w = (2.4210–6 V)/(0.015 m) = 1.61310–4 V/m. To determine the drift speed, we first find the density of free electrons: n = [(0.971)(1000 kg/m3)(103 g/kg)/(23 g/mol)](6.021023 free electrons/mol) = 2.5431028 electrons/m3. We find the drift speed from I = neAvd ; 12.0 A = (2.5431028 electrons/m3)(1.6010–19 C)(0.015 m)(1.3010–3 m)vd , which gives vd = 1.51310–4 m/s. The forces from the electric field and the magnetic field balance, so we have EH = vdB; 1.61310–4 V/m = (1.51310–4 m/s)B, which gives B = 1.07 T. 45. (a) The sign of the ions will not change the magnitude of the Hall emf, but will determine the polarity of the emf. (b) The forces from the electric field and the magnetic field balance. We find the flow velocity from åH = vdB/w; 0.1010–3 V = vd(0.070 T)(3.310–3 m), which gives vd = 0.43 m/s. 46. For the circular motion, the magnetic force provides the centripetal acceleration: qvB = mv2/r, or r = mv/qB. The velocities are the same because of the velocity selector, so we have m/m0 = r/r0 ;

Chapter 27 p. 9

m1/(76 u) = (21.0 cm)/(22.8 cm) = m2/(76 u) = (21.6 cm)/(22.8 cm) = m3/(76 u) = (21.9 cm)/(22.8 cm) = m4/(76 u) = (22.2 cm)/(22.8 cm) =

70 u; 72 u; 73 u; 74 u.

47. We find the velocity of the velocity selector from v = E/B = (2.48104 V/m)/(0.58 T) = 4.28104 m/s. For the radius of the path, we have r = mv/qB = [(4.28104 m/s)/(1.6010–19 C)(0.58 T)]m = (4.611023 m/kg)m. If we let A represent the mass number, we can write this as r = (4.611023 m/kg)(1.6610–27 kg)A = (7.6510–4 m)A = (0.765 mm)A. The separation of the lines is the difference in the diameter, or ?D = 2 ?r = 2(0.765 mm) ?A = (1.53 mm)(1) = 1.53 mm. If the ions were doubly charged, all radii would be reduced by one-half, so the separation would be 0.76 mm.

48. The velocity of the velocity selector is v = E/B. The separation on the film will be the difference in the diameters of the path in the magnetic field. For the radius of the path, we have r = mv/qB = mE/qBB, so r = km, and ?r = k ?m. If we form the ratio, we get ?m/m = ?r/r; 2.5 m. (28.0134 u – 28.0106 u)/(28.0134 u) = !(0.5010–3 m)/r, which gives r = 49. We find the speed acquired from the accelerating voltage from energy conservation: 0 = ?K + ?U; 0 = !mv2 – 0 + q(– V), which gives v = (2qV/m)1/2. We combine this with the expression for the radius of the path: R = mv/qB = m(2qV/m)1/2/qB, or m = qB2R2/2V. 50. For the circular motion, the magnetic force provides the radial acceleration: qvB = mv2/r, or v = qBr/m. To make the path straight, the forces from the electric field and the magnetic field balance: qE = qvB = q(qBr/m)B, or E = qB2r/m = (1.6010–19 C)(0.725 T)2(0.0510 m)/(1.6710–27 kg) = 2.57106 V/m perpendicular to B. 51. The magnetic force must be toward the center of the circular path, so the magnetic field must be up. The magnetic force provides the centripetal acceleration: qvB = mv2/r, or mv = qBr; 4.810–16 kg · m/s = (1.6010–19 C)B(1.0103 m), which gives B = 3.0 T up.

+ F

v

Chapter 27 p. 10

52. The radius of the path in the magnetic field is r = mv/eB, or mv = eBr. The kinetic energy is K = !mv2 = !(eBr)2/m. If we form the ratio for the two particles, we have Kp/Ke = (rp/re)2(me/mp); 1 = (rp/re)2[(9.1110–31 kg)/(1.6710–27 kg)], which gives rp/re =

42.8.

53. The magnetic force on the electron must be up, so the velocity must be toward the west. For the balanced forces, we have mg = evB; (9.1110–31 kg)(9.80 m/s2) = (1.6010–19 C)v(0.5010–4 T), 1.110–6 m/s west. which gives v = 54. The force on the airplane is F = qvB = (155010–6 C)(120 m/s)(5.010–5 T) =

9.310–6 N.

55. Even though the Earth’s field dips, the current and the field are perpendicular. The direction of the force will be perpendicular to both the cable and the Earth’s field, so it will be 68° above the horizontal toward the north. For the magnitude, we have F = ILB 0.17 N, 68° above the horizontal toward the north. = (330 A)(10 m)(5.010–5 T) = 56. (a) The force from the magnetic field will accelerate North the rod: F = I¬B = ma, which gives a = I¬B/m. Because the rod starts from rest and the acceleration is constant, we have I v = v0 + at = 0 + (I¬B/m)t = I¬Bt/m. F (b) The total normal force on the rod is mg, so there is a friction force of kmg. For the acceleration, we have B (out) ?F = I¬B – kmg = ma, which gives a = (I¬B/m) – kg. For the speed we have v = at = [(I¬B/m) – kg]t. east. (c) For a current toward the north in an upward field, the force will be to the

¬ East

57. We assume the magnetic field makes an angle  with the vertical. z From a view along the rod, we have the forces shown in the diagram. FN FB For the z-direction we have  FB sin  + FN – mg = 0, or FN = mg – ILB sin . B I  x To start motion in the x-direction, we have Ffr FB cos  > Ffr,max = sFN. When we combine the two equations, we get mg B > smg/IL(cos  + s sin ). To find the angle for the minimum field, we find the angle that will make the denominator maximum by differentiating: d(cos  + s sin )/d = – sin  + s cos  = 0, or tan  = s = 0.50,  = 26.6°.

Chapter 27 p. 11

Thus B > (0.50)(0.40 kg)(9.80 m/s2)/(40 A)(0.22 m)(cos 26.6° + 0.50 sin 26.6°) = The minimum magnetic field is 26.6° from the vertical.

0.20 T.

58. We find the speed acquired from the accelerating voltage from energy conservation: 0 = ?K + ?U; 0 = !mv2 – 0 + (– e)(V), or v = (2eV/m)1/2. If we assume that the deflection is small, the time the electron takes to reach the screen is t = L/v = L(m/2eV)1/2. The magnetic force produces an acceleration perpendicular to the original motion: a = evB/m. For a small deflection, we can take the force to be constant, so the deflection of the electron is d = !at2 = !(evB/m)(L/v)2 = !eBL2/mv = !BL2(e/2mV)1/2 = !(5.010–5 T)(0.20 m)2[(1.6010–19 C)/2(9.1110–31 kg)V]1/2 = (0.296 m · V1/2)/V1/2. (a) For a voltage of 2.0 kV, we have d = (0.296 m · V1/2)/V1/2 = (0.296 m · V1/2)/(2.0103 V)1/2 = 6.610–3 m = 6.6 mm. (b) For a voltage of 30 kV, we have d = (0.296 m · V1/2)/V1/2 = (0.296 m · V1/2)/(30103 V)1/2 = 1.710–3 m = 1.7 mm. These results justify our assumption of small deflection. 59. (a) The radius of the circular orbit is r = mv/qB. The time to complete a circle is T = 2pr/v = 2pm/qB, so the frequency is f = 1/T = qB/2pm. Note that this is independent of r. Because we want the ac voltage to be maximum when the proton reaches the gap and minimum (reversed) when the proton has made half a circle, the frequency of the ac voltage must be the same: f = 1/T = qB/2pm. (b) In a full circle, the proton crosses the gap twice. If the gap is small, the ac voltage will not change significantly from its maximum magnitude while the proton is in the gap. The energy gain from the two crossings is ?K = 2qV0 . (c) From r = mv/qB, we see that the maximum speed, and thus the maximum kinetic energy occurs at the maximum radius of the path. The maximum kinetic energy is Kmax = !mvmax2 = !m(qBrmax/m)2 = (qBrmax)2/2m = [(1.6010–19 C)(0.50 T)(2.0 m)]2/2(1.6710–27 kg) 48 MeV. = 7.6610–12 J = (7.6610–12 J)/(1.6010–13 J/MeV) = 60. With the direction of  as the positive direction, we see that the torque on the loop will be negative. If we let IM be the moment of inertia of the loop, for the angular motion we have – NIAB sin  = IM d2/dt2. If  « 1, sin ˜ , so we have – NIAB = IM d2/dt2, or d2/dt2 = – 2, where 2 = NIAB/IM . Thus we see that we have angular simple harmonic motion. The loop has two wires of length a and mass ma/(2a + 2b) and two wires of length b and mass mb/(2a + 2b). The moment of inertia of the loop is IM = 2[ma/(2a + 2b)](b/2)2 + 2[mb/(2a + 2b)](b2/12) = mb2(3a + b)/12(a + b). The period of the motion is

Chapter 27 p. 12

T = 2p/ = 2p(IM/NIAB)1/2 = 2p[mb2(3a + b)/12(a + b)abNIB]1/2 =

p[mb(3a + b)/3NIBa(a + b)]1/2.

61. If the beam is perpendicular to the magnetic field, the force from the magnetic field is always perpendicular to the velocity, so it will change the direction of the velocity, but not its magnitude. The radius of the path in the magnetic field is R = mv/qB. Protons with different speeds will have paths of different radii. Thus slower protons will deflect more, and faster protons will deflect less, than those with the design speed. We find the radius of the path from R = mv/qB = (1.6710–27 kg)(0.75107 m/s)/(1.6010–19 C)(0.33 T) = 0.237 m. Because the exit velocity is perpendicular to the radial line from the center of curvature, the exit angle is also the angle the radial line makes with the boundary of the field: sin  = L/R = (0.050 m)/(0.237 m) = 0.211, so  = 12°.

v d



R

v

B L

62. (a) The force on one side of the loop is F = ILB = (25.0 A)(0.200 m)(1.65 T) = 8.25 N. When the loop is perpendicular to the magnetic field, the forces at top and bottom will create a tension in each of the other two I 2 sides of !F. This produces a tensile stress of !F/A = F/2pr . When the loop is parallel to the magnetic field, the forces on right and left will create a shear in each of the other two sides. This produces a shear stress of !F/A = F/2pr2. F Because the magnitudes are the same and the tensile strength of aluminum is equal to the shear strength, we can use either condition B to determine the minimum diameter. With a safety factor of 10, we have F 10(F/2pr2) < Strength; 10(8.25 N)/2pr2 < 200106 N/m2, which gives r > 2.5610–4 m = 0.256 mm. Thus the minimum diameter is 0.512 mm. (b) The resistance of a single loop is R = 4L/A = 4L/pr2 = (2.6510–8  · m)(4)(0.200 m)/p(2.5610–4 m)2 = 0.103 .

F F

L

63. We find the required acceleration from v2 = v02 + 2ax; (30 m/s)2 = 0 + 2a(1.0 m), which gives a = 450 m/s2. This acceleration is provided by the force from the magnetic field: F = ILB = ma; I(0.20 m)(1.7 T) = (1.510–3 kg)(450 m/s2), which gives I = 2.0 A. The force is away from the battery, so fingers in the direction of I would have to curl down; thus the field points down.

Chapter 27 p. 13

64. (a) To make the path straight, the forces from the electric field and the magnetic field balance: eE = evB ; 2.110–3 T. 10,000 V/m = (4.8106 m/s)B, which gives B = (b) Because the electric force is down, the magnetic force must be up, so the magnetic field is out of the page. (c) If there is only the magnetic field, the radius of the circular orbit is r = mv/qB. The time to complete a circle is T = 2pr/v = 2pm/qB, so the frequency is f = 1/T = qB/2pm = (1.6010–19 C)(2.0810–3 T)/2p(9.1110–31 kg) =

FB

E

–e –

v B

FE

5.8107 Hz.

65. We find the speed acquired from the accelerating voltage from energy conservation: 0 = ?K + ?U; 0 = !mv2 – 0 + (– e)(V), or v = (2eV/m)1/2.  r In the uniform magnetic field, the magnetic force provides the radial acceleration for the circular arc:  evB = mv2/r, so B = mv/er = (2mV/e)1/2/r. V Thus we need to determine the radius from the h0 geometry. Because the velocity is perpendicular to v B the radial line from the center of curvature, the angle leaving the magnetic field is also the angle the arc d subtends: sin  = d/r. L We also see that h0 = r(1 – cos ), where h0 is the deflection caused by the magnetic field. From the linear motion after leaving the magnetic field we have tan  = (h – h0)/(L – d). The trigonometric functions make the solution for r rather messy. The equations can be combined to eliminate h0 and  to give a cubic equation for r. Instead we will use a trial approach and assume a value of 1.0 cm for the deflection h0. We can then find the angle from tan  = (h – h0)/(L – d) = (11 cm – 1.0 cm)/(22 cm – 3.5 cm) = 0.541, so  = 28.4°. We find the radius from sin  = d/r; sin 28.4° = (3.5 cm)/r, which gives r = 7.36 cm. To test our assumption we find h0 from

h

Chapter 27 p. 14

h0 = r(1 – cos ) = (7.36 cm)(1 – cos 28.4°) = 0.89 cm. Thus we refine our assumption and choose h0 = 0.9 cm. We now get tan  = (h – h0)/(L – d) = (11 cm – 0.9 cm)/(22 cm – 3.5 cm) = 0.546, so  = 28.6°. sin 28.6° = (3.5 cm)/r, which gives r = 7.30 cm. h0 = r(1 – cos ) = (7.30 cm)(1 – cos 28.6°) = 0.89 cm. Finally we find the magnetic field from B = (2mV/e)1/2/r = [2(9.1110–31 kg)(25103 V)/(1.6010–19 C)]1/2/(7.310–2 m) = 7.310–3 T. 66. (a) For a particle moving at speed v in a circle of radius r the frequency is f = 1/T = v/2pr, or v = 2prf. When there is no magnetic field, the radial acceleration is provided by the electrostatic attraction: ke2/r2 = mv02/r = 4p2mrf02. When there is a magnetic field perpendicular to the plane of rotation, we have ke2/r2 ± evB = mv2/r, or f02 ± (eB/2pm)f – f 2 = 0. From this quadratic equation we get f = !{± (eB/2pm) ± [(eB/2pm)2 + 4f02]1/2} = ± (eB/4pm) ± [f02 + (eB/4pm)2]1/2. If the force from the magnetic field is much less than that due to electrostatic attraction, (eB/4pm) « f0 , so we have f ˜ ± (eB/4pm) + f0 , where we have chosen the positive sign for the square root to get a positive value for f. Thus we have f – f0 = ?f = ± eB/4pm. (b) The ± sign indicates that the circulation of the electron may be clockwise or counterclockwise with respect to the magnetic field. 67. The magnetic force provides the centripetal acceleration of the circular motion: evB = mv2/r, or mv = eBr. The kinetic energy of the proton is K = !mv2 = !(eBr)2/m = (e2B2/2m)r2. Thus the change in the kinetic energy is ?K = (e2B2/2m)(rQ2 – rP2) = [(1.6010–19 C)2(0.010 T)2/2(1.6710–27 kg)][(8.510–3 m)2 – (10.010–3 m)2] = – 2.110–20 J.

Chapter 27 p. 15

68. For the two horizontal sides of the loop, the currents are in opposite directions so their forces will be in opposite directions. Although the magnetic field varies along the side, the current will be in the same average field and the magnitudes of the forces will be equal, so F3 + F4 = 0. For the sum of the two forces on the vertical sides of the loop, we have Fnet = F1 + F2 = IL1  B1 + IL2  B2 = Iaj  B0[1 – (a + b)/d]k + Ia(– j)  B0[1 – b/d]k = IaB0{[1 – (a + b)/d] – (1 – b/d)}i = – (Ia2B0/d)i (left).

y F3

B

F2

I

I

F1

a

a

b

F4 x

Chapter 28 p.1

CHAPTER 28 – Sources of Magnetic Field 1.

The magnetic field of a long wire depends on the distance from the wire: B = (0/4p)2I/r = (10–7 T · m/A)2(65 A)/(0.075 m) = 1.7  10–4 T. When we compare this to the Earth’s field, we get B/BEarth = (1.7  10–4 T)/(5.5  10–5 T) = 3.1.

2.

We find the current from B = (0/4p)2I/r; 5.5  10–5 T = (10–7 T · m/A)2I/(0.25 m), which gives I =

69 A.

3.

The two currents in the same direction will be attracted with a force of F = I1(0I2/2pd)L = 0I1I2L/2pd = (4p10–7 T · m/A)(35 A)(35 A)(45 m)/2p(0.060 m) = 0.18 N attraction.

4.

Because the force is attractive, the second current must be in the same direction as the first. We find the magnitude from F/L = 0I1I2/2pd 8.810–4 N/m = (4p10–7 T · m/A)(22 A)I2/2p(0.070 m), which gives I2 = 14 A upward.

5.

The magnetic field will be tangent to the circle with the current at the center. BC

C BD

I

D E

6.

BE

The magnetic field produced by the wire must be less than 1.0% of the magnetic field of the Earth. We find the current from B = (0/4p)2I/r < 0.01BEarth ; (10–7 T · m/A)2I/(1.00 m) < 0.010(5.5  10–5 T), which gives I < 2.8 A.

Chapter 28 p.2

7.

8.

9.

We find the direction of the field for each wire from y the tangent to the circle around the wire, as shown. For their magnitudes, we have r2 I2 B2 B1 = (0/4p)2I/r1 2 B1 = (10–7 T · m/A)2(25 A)/(0.120 m) = 4.1710–5 T. 1 2 B2 = (0/4p)2IB/r2 d = (10–7 T · m/A)2(25 A)/(0.050 m) = 1.0010–4 T. 1 r 1 We use the property of the triangle to find the angles shown: 2 2 2 r2 = r1 + d – 2r1d cos 1 ; (5.0 cm)2 = (12.0 cm)2 + (15.0 cm)2 – 2(12.0 cm)(15.0 cm) cos 1 , I1 which gives cos 1 = 0.956, 1 = 17.1°; r12 = r22 + d2 – 2r2d cos 2 ; (12.0 cm)2 = (5.0 cm)2 + (15.0 cm)2 – 2(5.0 cm)(15.0 cm) cos 2 , which gives cos 2 = 0.707, 2 = 45.0°; From the vector diagram, we have B = B1(– cos 1 i + sin 1 j) + B2 (cos 2 i + sin 2 j) = (4.1710–5 T)(– cos 17.1° i + sin 17.1° j) + (1.0010–4 T)(cos 45.0° i + sin 45.0° j) = (3.110–5 T) i + (8.310–5 T) j. We find the direction from tan  = By/Bx = (8.3010–5 T)/(3.0910–5 T) = 2.68,  = 70.1°. We find the magnitude from B = Bx/cos  = (3.0910–5 T)/cos 70.1° = 8.910–5 T, 70° above horizontal. The magnetic field from the wire at a point south of a downward current will be to the west, with a magnitude: Bwire = (0/4p)2I/r = (10–7 T · m/A)2(40 A)/(0.20 m) = 4.0  10–5 T. Because this is perpendicular to the Earth’s field, we find the direction of the resultant field, and thus of the compass needle, from tan  = Bwire/BEarth = (4.0  10–5 T)/(0.45  10–4 T) = 0.889, or  = 42° W of N.

x

I (down) B BEarth

 West

Bwire

The magnetic field to the west of a wire with a current to the north will be up, with a magnitude: Bwire = (0/4p)2I/r = (10–7 T · m/A)2(22.0 A)/(0.200 m) = 2.20  10–5 T. The net downward field is Bdown = BEarth sin 40° – Bwire = (5.0  10–5 T) sin 40° – 2.20  10–5 T = 1.01  10–5 T. The northern component is Bnorth = BEarth cos 40° = 3.83  10–5 T. We find the magnitude from B = [(Bdown)2 + (Bnorth)2]1/2 = [(1.01  10–5 T)2 + (3.83  10–5 T)2]1/2 = 4.0  10–5 T. We find the direction from tan  = Bdown/Bnorth = (1.01  10–5 T)/(3.83  10–5 T) = 0.264, or  = 15° below the horizontal.

10. Because a current represents the amount of charge that passes a given point, the effective current of the proton beam is I = ?q/?t = (1.5109 protons/s)(1.6010–19 C/proton) = 2.410–10 A. The magnetic field from this current will be B = (0/4p)2I/r

Chapter 28 p.3

= (10–7 T · m/A)2(2.410–10 A)/(2.0 m) =

2.4  10–17 T.

11. (a) When the currents are in the same direction, the fields between the currents will be in opposite directions, so at the midpoint we have Ba = B2 – B1 = [(0/4p)2I2/r] – [(0/4p)2I/r] = [(0/4p)2/r](I2 – I) = (10–7 T · m/A)2/(0.010 m)(15 A – I) = (2.010–5 T/A)(15 A – I) up, with the currents as shown. (b) When the currents are in opposite directions, the fields between the currents will be in the same direction, so at the midpoint we have Bb = B2 + B1 = [(0/4p)2I2/r] + [(0/4p)2I/r] = [(0/4p)2/r](I2 + I) = (10–7 T · m/A)2/(0.010 m)(15 A + I) = (2.010–5 T/A)(15 A + I) down, with the currents as shown.

B2

I



 B1

I



I2 B1

12. Because the currents are in opposite directions, the fields will be in opposite directions. For the net field we have B = B1 – B2 = [(0/4p)2I1/r1] – [(0/4p)2I2/r2] = [(0/4p)2I]{[1/(L – !d)] – [1/(L + !d)]}

= [(0/4p)2I/L]{[1/(1 – !d/L)] – [1/(1 + !d/L)]}.

Because d « L, we can use the approximation 1/(1 ± x) ˜ 1 — x: B = [(0/4p)2I/L][(1 + !d/L) – (1 – !d/L)] = [(0/4p)2I/L](d/L) = (0/4p)2Id/L2 = (10–7 T · m/A)2(25 A)(2.810–3 m)/(0.100 m)2 = 1.410–6 T up, with the currents as shown. This is (1.410–6 T)/(5.010–5 T) = 0.028 = 2.8% of the Earth’s field.

I2

B2

B1 I2

L

 d

I1

13. The magnetic field of the Earth points in the original direction of the compass needle. The field of the wire will be tangent to a circle centered at the wire. We see from the diagram that the field of the wire must be to the south to produce a greater angle for the resultant field. Thus the current in the wire must be down. From the vector diagram, we have B sin 2 = BEarth sin 1 ; I (down) B cos 2 = BEarth cos 1 – Bwire . When we divide the two equations, we get tan 2 = (BEarth sin 1)/(BEarth cos 1 – Bwire); tan 55° = [(0.5010–4 T) sin 20°]/[(0.5010–4 T) cos 20° – Bwire], which gives Bwire = 3.5010–5 T. We find the current from Bwire = (0/4p)2I/r; 3.5010–5 T = (10–7 T · m/A)2I/(0.120 m), which gives I = 21 A down.

B2

North BEarth

1

B wire

2 B East

Bwire

Chapter 28 p.4

14. The magnetic field at the loop from the wire will be into the page, and will depend only on the distance from the wire r: B = (0/4p)2I1/r, For the two vertical sides of the loop, the currents are in opposite directions so their forces will be in opposite directions. Because the current will be in the same average field, the magnitudes of the forces will be equal, so Fc + Fd = 0. For the sum of the two forces on the top and bottom of the loop, we have Fnet = Fa – Fb = I2BaL – I2BbL = I2[(0/4p)2I1/a]L – I2[(0/4p)2I1/(a + b)]L

I1 Fa Fd

a Fc

I2

b

B wire I2

Fb L

= (0/4p)(2I1I2L){(1/a) – [1/(a + b)]} = (10–7 T · m/A)2(2.5 A)(2.5 A)(0.100 m)[(1/0.030 m) – (1/0.080 m)] = 2.6  10–6 N toward the straight wire.

15. Because the currents are in the same direction, between the wires the fields will be in opposite directions. For the net field we have B = B1 – B2 = [(0/4p)2I1/x]j – [(0/4p)2I2/(d – x)]j

y B1 x

I1

I2

= (0/4p)2I{[(d – x) – x]/x(d – x)}j = [(0/4p)2I(d – 2x)/x(d – x)]j .

x

B2 d

16. Because the currents are in opposite directions, between the wires the fields will be in the same direction. With I1 = 2I2 , for the net field we have B = B1 + B2 = – [(0/4p)2I1/x]j – [(0/4p)2I2/(d – x)]j = – (0/4p)2I{[2(d – x) + x]/x(d – x)}j = – [(0/4p)2I(2d – x)/x(d – x)]j .

y d I1



17. We find the direction of the field for each wire from the tangent to the circle around the wire, as shown. For their magnitudes, we have BT = (0/4p)2IT/L = (10–7 T · m/A)2(20.0 A)/(0.100 m) = 4.0010–5 T. BB = (0/4p)2IB/L = (10–7 T · m/A)2(5.0 A)/(0.100 m) = 1.0010–5 T. Because the fields are perpendicular, we find the magnitude from B = (BT2 + BB2)1/2 = [(4.0010–5 T)2 + (1.0010–5 T)2]1/2 = 4.1210–5 T.

x

I2 B2

B1

IT L BT L

B B (out)

IB

x

Chapter 28 p.5

18. We find the direction of the field for each wire from y the tangent to the circle around the wire, as shown. B1 1 For their magnitudes, we have P B1 = (0/4p)2I1/L1 = (10–7 T · m/A)2(16.5 A)/(0.120 m) = 2.7510–5 T. 2 B B2 B2 = (0/4p)2I2/L2  = (10–7 T · m/A)2(16.5 A)/(0.130 m) = 2.5410–5 T. We use the property of the triangle to find the angles shown: r22 = r12 + d2 – 2r1d cos 1 ; r1 (13.0 cm)2 = (12.0 cm)2 + (7.0 cm)2 – 2(12.0 cm)(7.0 cm) cos 1 , which gives cos 1 = 0.143, 1 = 81.8°; r12 = r22 + d2 – 2r2d cos 2 ; 1 (12.0 cm)2 = (13.0 cm)2 + (7.0 cm)2 – 2(7.0 cm)(13.0 cm) cos 2 , which gives cos 2 = 0.407, 2 = 66.0°; d I1 From the vector diagram, we have B = B1(– sin 1 i + cos 1 j) + B2(– sin 2 i – cos 2 j) = (2.7510–5 T)(– sin 81.8° i + cos 81.8° j) + (2.5410–5 T)(– sin 66.0° i – cos 66.0° j) = (– 5.0410–5 T) i + (– 6.4110–6 T) j. For the direction of the field, we have tan  = By/Bx = (– 6.4110–6 T)/(– 5.0410–5 T) = 0.127,  = 187.2°. We find the magnitude from Bx = B cos ; – 5.0410–6 T = B cos 187.2°, 5.0810–5 T 187.2° from the x-axis. which gives B = 19. (a) The figure shows a view looking directly at the current. The sheet may be thought of as an infinite number of parallel wires. We choose a differential element of width dx a distance x from the center of the strip. This element has a current dI = (I/L) dx which produces a magnetic field dB = (0/4p)2 dI/r = (0I/2pL) dx/r, in the direction shown. Because a differential element at – x will produce a field of the same magnitude but below the horizontal, the symmetry means the resultant field will be parallel to the strip in the x-direction. We find the total field by adding (integrating) the x-components of the differential fields:

 B=

sin  d B =

L/ 2 – L/ 2

 0I y  Iy dx = 0 2L x 2 + y 2 2L

1 tan – 1 x y y

y

2 x

I2

dB



y

r2

r

B

 x

L

L/ 2 – L/ 2

=

x

dx

 0I tan – 1 L . L 2y

(b) If y » L, or L/2y « 1, the angle is small and equal to the tangent, so we have tan–1(L/2y) ˜ L/2y, and B ˜ (0I/pL))(L/2y) = (0/4p)(2I/y), which is the magnetic field produced by a long wire. This is what the strip will appear to be when

Chapter 28 p.6

we are far from the strip.

20. Because the electron is moving in a plane perpendicular y to the magnetic field of the wire, the force will be perpendicular to the velocity. Thus the magnitude of  F the velocity will not change. B (out) For the motion in the y-direction, we have evB cos  = ev(0I/2py) cos  = m dvy/dt.  This expression contains four variables, y, , vy , and t; v but we can use the chain rule and vy = v sin ; dvy = v cos  d, to reduce the number to two, which we choose to be y and : I ev(0I/2py) cos  = m dvy/dt = m (dvy/dy)(dy/dt ); 2 ev(0I/2py) cos  = m (v cos  d/dy)vy = mv cos  sin  d/dy. We separate variables and integrate from the initial position to the point closest to the wire: y2 2 dy = 2mv sin  d  ; y  Ie  y 0 1

1

ln (y2/y1) = – (2pmv/0Ie)(cos 2 – cos 1), or (2pmv/0Ie)(cos 2 – cos 1) = ln (y1/y2); [2p(9.11  10–31 kg)(3.4  106 m/s)/(4p  10–7 T · m/A)(1.6010–19 C)I](cos 0° – cos 45°) = ln (50 cm/1.0 cm), which gives I = 7.2 A. 21. We find the current in the solenoid from B = 0nI = 0NI/L; 0.38510–3 T = (4p10–7 T · m/A)[(1000 turns)/(0.400 m)]I, which gives I = 22. We find the number of turns in the solenoid from B = 0nI = 0NI/L; 0.30 T = (4p10–7 T · m/A)[N/(0.32 m)](5.7 A), which gives N =

0.123 A.

1.3104 turns.

23. We use the results from Example 28–4. (a) The magnetic field at the surface of the wire is Bsurface = 0I/2pR = (4p10–7 T · m/A)(40 A)/2p(1.2510–3 m) = 6.410–3 T. (b) Inside the wire we have Binside = (0I/2p)(r/R2) = (4p10–7 T · m/A)(40 A)(0.7510–3 m)/2p(1.2510–3 m)2 = 3.810–3 T. (c) Outside the wire we have Boutside = 0I/2pr = (4p10–7 T · m/A)(40 A)/2p(3.7510–3 m) = 2.110–3 T. 24. We use the result from Example 28–4 to find the maximum and minimum fields: B1 = (0/4p)2NI/r1 = (10–7 T · m/A)2(500)(25.0 A)/(0.250 m) = 1.00  10–2 T; B2 = (0/4p)2NI/r2 = (10–7 T · m/A)2(500)(25.0 A)/(0.270 m) = 0.926  10–2 T. Thus the range for B is 0.926  10–2 T = B = 1.00  10–2 T. 25. (a) If D is the diameter of the solenoid, the length of a coil is pD. Thus the number of turns is N = Lwire/Lcoil = Lwire/pD.

x

Chapter 28 p.7

Because the coils are tightly wrapped, the length of the solenoid is Lsolenoid = Nd = Lwired/pD = (20.0 m)(2.0010–3 m)/p(2.5010–2 m) = 0.509 m = 50.9 cm. (b) The magnetic field at the center of the solenoid is B = 0NI/Lsolenoid = 0I/d = (4p10–7 T · m/A)(20.0 A)/(2.0010–3 m) = 1.2610–2 T. 26. (a) The magnetic field at a point is B = B1 + B2 , where the magnitudes are B1 = 0I/2pr1 , B2 = 0I/2pr2. The components are B = B1(– sin 1 i + cos 1 j) + B2(– sin 2 i + cos 2 j). The plot of this field is Figure 28–8. (b)

y B

B1 B2 r1

I1

1

+ N N

r2

2

I2

x

+

(c) The diagrams are similar, but not identical. The similarity is because the electric potential also depends on 1/r. However, the magnetic field diagram is a vector diagram, while the potential is a scalar. At the midpoint N the magnetic field is zero because of the vector addition, but the electrostatic potential of two positive charges is not zero. 27. The current densities in the wires are Jinner = I0/pR12 and Jouter = I0/p(R32 – R22). Because of the cylindrical symmetry, we know that the magnetic fields will be circular. In each case we apply Ampere’s law to a circular path of radius r. (a) Inside the inner wire, r < R1:

ı B · ds = 0Ienclosed;

B2pr = 0Jinnerpr2 = 0I0pr2/pR12, which gives B = (0I0/2pR12)r circular CCW, r < R1. (b) Between the wires, R1 < r < R2:

ı B · ds = 0Ienclosed;

B2pr = 0I0 , which gives B = 0I0/2pr circular CCW, R1 < r < R2. (c) Inside the outer wire, R2 < r < R3:

ı B · ds = 0Ienclosed;

B2pr = 0[I0 – Jouter(pr2 – pR22)] = 0[I0 – I0(pr2 – pR22)/p(R32 – R22)] = 0I0[1 – (r2 – R22)/(R32 – R22)],

R1

r

R2

I0 (out) R 3 I0 (in)

Chapter 28 p.8

which gives B = (0I0/2pr)(R32 – r2)/(R32 – R22) circular CCW, R2 < r < R3. (d) Outside the outer wire, R3 < r:

ı B · ds = 0Ienclosed;

B2pr = 0(I0 – I0 ), which gives

B = 0, R3 < r.

28. We find the constants in the current densities in the wires by requiring that the total current in each wire be I0. For the inner wire we have R1  R1 I0 = J1 2r d r = 2C 1 r2 d r = 2C1 R 31 / 3, 0

0

which gives C1 = 3I0/2pR13. For the outer wire we have R3  R3 I0 = J2 2r dr = 2C2 r2 dr = 2C2 R 33 – R 23 / 3, R2

R2

which gives C2 = (3I0/2p)/(R33 – R23). Because of the cylindrical symmetry, we know that the magnetic fields will be circular. In each case we apply Ampere’s law to a circular path of radius r. (a) Inside the inner wire, r < R1:

ı B · ds = 0Ienclosed;

r  B2r = 0 C 1r2r d r = C 12 0 r3/ 3, which gives 0

B = 0C1r2/3 = (0r2/3)(3I0/2pR13) = (b) Between the wires, R1 < r < R2:

(0I0/2pR13)r2 circular CCW, r < R1.

ı B · ds = 0Ienclosed;

B2pr = 0I0 , which gives B = 0I0/2pr circular CCW, R1 < r < R2. (c) Inside the outer wire, R2 < r < R3:

ı B · ds = 0Ienclosed;  B2r = 0 I 0 –

r

R2

C2 r2r d r =  0 I 0 – C2 2/ 3 r3 – R 23 , which gives

B = (0I0/2pr)[1 – (r3 – R23)/(R33 – R23)] = (0I0/2pr)(R33 – r3)/(R33 – R23) circular CCW, R2 < r < R3. (d) Outside the outer wire, r > R3 .

ı B · ds = 0Ienclosed;

B2pr = 0(I0 – I0 ), which gives

B = 0, r > R3 .

29. The magnetic field along the axis of a dipole far from the dipole is B = (0/2p)/x3. If we form the ratio for the two distances, we have B2/B1 = (x1/x2)3; B2/(1.010–4 T) = [6.4103 km/(6.4103 km + 13103 km)]3, which gives B2 =

3.610–6 T.

Chapter 28 p.9

30. Because the point C is along the line of the two straight segments of the wire, there is no magnetic field from these segments. The magnetic field at the point C is the sum of two fields: B = Binner arc + Bouter arc. Each field is a portion of a circular loop, with the field of the inner arc out the page and that of the outer arc into the page, so we subtract the two magnitudes: B = (/2p)(0I/2R1) – (/2p)(0I/2R2) = (0I/4p)[(1/R1) – (1/R2)] = 0I(R2 – R1)/4pR1R2 out of the page.

31. Because the point C is along the line of the two straight segments of the wire, there is no magnetic field from these segments. The magnetic field at the point C is the sum of two fields: B = Blower semicircle + Bupper semicircle. Each field is half that of a circular loop, with the field of the lower semicircle out the page and that of the upper semicircle into the page, so we subtract the two magnitudes: B = !(0&I/2R) – !(0#I/2R) = 0I/8R out of the page.

I R2 I 

C

I/4 R C

I

I

3I/4

32. We assume that the inner loop is small enough that we can say it will be in the magnetic field at the center of the large loop, which is B1 = 0I1/2R1. The magnetic moment of the inner loop is 2 = I2A = I2pR22, and is perpendicular to the magnetic field. Thus the torque is  = 2B1 = (I2pR22)(0I1/2R1) = 0I1I2pR22/2R1 = (4p10–7 T · m/A)(7.0 A)(7.0 A)p(0.018 m)2/2(0.250 m) = 1.310–7 m · N, in a direction that will align the two loops.

33. (a) Because the point C is along the line of the two straight segments of the wire, there is no magnetic field from these segments. The magnetic field at the point C is the sum of two fields: B = Blower semicircle + Bupper semicircle. Each field is half that of a circular loop, with both fields into the page, so we add the two magnitudes: B = !(0I/2R2) + !(0I/2R1) = 0I(R1 + R2)/4R1R2 into the page. (b) The magnetic moment of the circuit is  = IA = I(!pR12 + !pR22) = !pI(R12 + R22) into the page.

R1

I1

2 B1

I2 R1

R1 C I R2 I

Chapter 28 p.10

34. The distance the charge moves in a time dt is d¬ = v dt. The current at a point is I = dq/dt. Thus we have I d¬ = (dq/dt)v dt = v dq. If we use this and r = r/r, in the Biot-Savart law we have

P

r

q

v

B = (0/4p)? (I d¬  r )/r2 = (0/4p)? (dq v  r)/r3 = (0/4p)(q v  r)/r3.

dr

35. (a) We can treat the disk as an infinite number of rings. We choose a differential element of radius r and thickness dr. The charge density on the disk is  = Q/pR2, so the current in the ring is dI =  dA/T = (Q/pR2)2pr dr/(2p/) = Q r dr/pR2. We find the magnetic dipole moment by integration:

r x Q

R R 2  Q  Ri. 3d =ir2dI=Q ir r = 2 4 0 R0

R



(b) The field from each of the rings is along the x-axis, so we integrate the magnitudes:

 B =

r=R

dB = r =0 R

= 0

0 dI r2 2 r2 + x 2

3/ 2

0Q  r 3 dr 0 Q  2 r2 + x 2 2 3/ 2 = 2 2 2R r + x 2R 2

0 Q  1/ 2 2 R2 2 R + x2 – 2x – 2 2 2R R + x2 Thus the magnetic field is =

1/ 2

R

r2 – 1/ 2 2 r + x2 0

0 Q  R 2 + 2x 2 – 2x R 2 + x 2 1/ 2 = . 1/ 2 1/ 2 2 2R 2 R + x2

1/2

2+2 2–2 2+ 2  Q  R x x R x B =02 1/2 i. 22 2R R +x (c) When x » R, we have B = (0Q /2pR2)x2{(R/x)2 + 2 – 2[1 + (R/x)2]1/2}/x[1 + (R/x)2]1/2

Chapter 28 p.11

˜ (0Q /2pR2)x{[2 + (R/x)2 – 2[1 + (R/x)2/2 – (R/x)4/8]}[1 – (R/x)2/2] ˜ (0Q /2pR2)x(R/x)4/4 = 0Q R2/8px3 = 0/2px3. This is Eq. 28–7b, so yes it does apply.

36. (a) We choose the y-axis along the wire. We choose the differential element dy and use the angle  indicated on the figure to specify the location of the point where we want to find the magnetic field. From the figure, we see that tan  = y/R, cos  = R/r, and sin  = y/r, and  =  + p/2. The angle  will vary from – 0 to 0 = tan–1 (¬/2R). We relate the change in angle to the change in y from y = R tan   dy = R sec2 d = (R/cos2) d. The field from each of the differential elements will be circular about the wire, that is, into the page, so the integration of dB becomes a scalar integration of the magnitude:  y= /2  I d y  r 0 I y = / 2 d y r sin  + / 2 0 B = dB = = r3 4 y = – / 2 r3 y = – / 2 4

=

0 I 4

0 – 0

dy

 r

y

 R

¬ I

0  I R/ cos 2  d  cos  = 0 sin  , which gives 2 4 R – 0 R/ cos 

B = (0I/4pR)(2 sin 0). With sin 0 = (¬/2)/[(¬/2)2 + R2]1/2 = ¬/(¬2 + 4R2)1/2, we get B = 0I¬/2pR(¬2 + 4R2)1/2 circular.

(b) We approximate an infinite wire by having ¬ » R: B = (0I/2pR)/[1 + (2R/¬)2]1/2 ˜ 0I/2pR, which is the field for an infinite wire.

B

Chapter 28 p.12

37. (a) With the wire along the x-axis, we choose a differential element d¬ that is parallel to the wire, so we have d¬ = dx i. The displacement from the element to the point Q is r = ri. We find the magnetic field of the I differential element from the Biot–Savart law: dB = (0/4p)I(d¬  r)/r3 = (0/4p)I [(dx i)  (ri)]/r3 = 0. (b) With the same differential element d¬ and the angle  indicated on the figure to specify the location of the point P where we want to find the magnetic field, from the figure, we see that tan  = – y/x, cos  = – x/r, and sin  = y/r. As x varies from – ¬ to 0, the angle  will vary from 0 = tan–1 (y/¬) to p/2. We relate the change in angle to the change in x from x = – y/tan   dx = y (sec2 /tan2d = (y/sin2) d. The field from all the differential elements will be circular about the wire, that is, out of the page, so the integration of dB becomes a scalar integration of the magnitude:

 B=

x=0

dB = x=– / 2

 I = 0 4  0

 0I dx  r 0I = 4 4 r3

x=0 x=–

y P B r

y

 dx

Q

x ¬

dx r sin  r3

2 / 2  I y/ sin  d  sin  = 0 – cos  , which gives 2 4 y 0 y/ sin 

B = – (0I/4py)(0 – cos 0). With cos 0 = ¬/(¬2 + y2)1/2, we get B = 0I¬/4py(¬2 + y2)1/2 circular.

I1 38. Although the current is the same, we label them so we can designate a side of the square. We choose ¬ the direction out of the page as positive. We find the magnetic field from each of the sides of the square: B1 = 0, I4 ¬ I2 because the point P is along the line of the side. 2 2 1/2 2 2 1/2 B2 = 0I¬/4py(¬ + y ) = 0I¬/4p¬(¬ + ¬ ) = 0I/4p¬v2. For B3 we consider the field from two currents: a current I3 in a wire of length 2¬ and a current – I3 in a I3 wire of length ¬. Thus we have B3 = – 0I2¬/4p¬(4¬2 + ¬2)1/2 + 0I¬/4p¬(¬2 + ¬2)1/2 = (0I/4p¬)[– (2/v5) + (1/v2)]. For B4 we have B4 = – 0I¬/4p2¬(¬2 + 4¬2)1/2 = – 0I/4p¬(2v5). Thus the net field is B = B1 + B2 + B3 + B4 = (0I/4p¬)[0 + (1/v2) – (2/v5) + (1/v2) – (1/2v5)] =

(v2 – !v5)0I/4p¬ out of the page.

¬

± I3

P

x

Chapter 28 p.13

39. (a) The angle subtended by a side of the polygon at the center I0 point P is  = 2p/n. The length of a side is P L = 2R sin (/2) = 2R sin (p/n). The perpendicular distance from the center to the side is  R D = R cos (/2) = R cos (p/n). D We use the result from Problem 36 to find the magnetic field of one side: L B1 = 0I0L/2pD(L2 + 4D2)1/2 = 0I0[2R sin (p/n)]/2p[R cos (p/n)][4R2 sin2 (p/n) + 4R2 cos2 (p/n)]1/2 = (0I0/2pR) tan (p/n). Thus the field from the n sides is B = nB1 = (0I0/2pR)n tan (p/n) into the page. (b) If we let n  8, the angle p/n  0, so tan (p/n)  p/n. Thus we have B  (0I0/2pR)n(p/n) = 0I0/2R, which is the expression for a circular loop. 40. With the solenoid along the x-axis, we choose a differential dx ring of width dx a distance x from the center which contains n dx turns. From the result of Example 28–10, the magnetic  R field of this ring is along the axis with a magnitude x dB = [(0nI/2) dx]R2/(R2 + x2)3/2. We can change variable to the angle , where x = R tan , and dx = R sec2 d. The differential field is then dB = [(0nI/2) (R sec2 d)]R2/(R2 + R2 tan2)3/2 = (0nI/2)(sec2 d)/sec3 = (0nI/2) cos  d. For a long solenoid, the angle  will vary from – p/2 to p/2. All fields will be along the axis, so we can integrate the magnitudes to find the field: / 2 / 2  0nI 0nI B = dB = cos  d  = sin  =  0nI. 2 2 – / 2

– / 2

41. The magnetic field from each side of the loop will be out of the page. We could find the field from each side by selecting a differential element and using the Biot-Savart law. We will use the result from Problem 37 by considering each side to be made of two pieces and adding the fields at P. We have labeled the currents to distinguish the side.

y b Chapter 28 p.14

I3

Thus we have

P (x, y) a

I2

I4

I1  I I x b– x B 1= 0 + 0 ; 4 y x 2 + y 2 1/ 2 4 y (b – x) 2 + y 2 1/ 2 x y a– y  0I 0I B2 = + ; 4 (b – x ) y 2 + (b – x) 2 1/ 2 4 (b – x) (a– y) 2 + (b – x) 2 1/ 2  I I x b–x B3 = 0 + 0 ; 4 (a – y) x 2 + (a– y) 2 1 / 2 4 (a– y) (b – x) 2 + (a – y) 2 1/ 2 y a– y  0I 0I B4 = + . 4 x y 2 + x 2 1 / 2 4 x (a – y) 2 + x 2 1/ 2 We can simplify the algebra by putting the constants with the field. Thus the net field is y a– y x b– x  0I = 4B/ + + 2 2 1/ 2 + 2 1/ 2 + 2 2 1/ 2 2 1/ 2 2 (b – x) y + (b – x ) (b – x) (a – y) 2 + (b – x) yx + y y (b – x) + y y a– y x b–x + 2 1/ 2 + 2 2 1/ 2 + 2 2 2 2 1/ 2 2 1/ 2 x y +x x (a – y) + x (a – y) x + (a – y) (a – y) (b – x) + (a – y) (x/ y) + (y/ x) [(b – x)/ y] + [y/( b – x)] [(a – y)/( b – x)] + [(b – x)/( a – y)] = + + + 1/ 2 1/ 2 2 2 1/ 2 x2 + y2 (b – x) + y 2 (a – y) 2 + (b – x) [x/( a – y)] + [(a – y)/ x] , so we get 2 1/ 2 x 2 + (a – y)

221/2221/2221/2221/2  +y(b–x)+y(a–y)+ (b–x)x+ (a–y) 0Ix

B =xy++ +),outofthepage. 4 y(b–x)(a–y)(b–x)x(a–y Note that, if a = b = L, at the center of the square, where x = y = L/2, we get B = (0I/4p)4(L/v2)/(L2/4) = 40I/pLv2. We can compare this to the result for Problem 39, where R = L/v2 and n = 4: B = (0I/2pR)n tan (p/n) = (0Iv2/2pL)4 tan (p/4) = 40I/pLv2. 42. (a) We use the result from Problem 36 to find the magnetic field from one side of the square; B1 = (0I/2pr)¬/(¬2 + 4r2)1/2, where r2 = (!¬)2 + x2. From the symmetry, we see that the components perpendicular to the x-axis of the fields from the four sides will cancel, so we add the four equal x-components: B = 4B1 cos  = 4[(0I/2pr)¬/(¬2 + 4r2)1/2](¬/2r) = (0I/p)¬2/r2(¬2 + 4r2)1/2 = (0I/p)¬2/[(!¬)2 + x2](¬2 + ¬2 + 4x2)1/2 = 40I¬2/p(¬2 + 4x2)(2¬2 + 4x2)1/2, along the x-axis.

I



r

B1

 x ¬ ¬

(b) If x » ¬, we get B ˜ 40I¬2/p(4x2)(4x2)1/2 = 0I¬2/2px3. This is the expression for the field far from a dipole, with  = I¬2. 43. (a) If the iron bar is completely magnetized, the dipoles of all of the atoms are aligned. Thus the dipole moment will be  = N1 = (V/M)NA1

x

Chapter 28 p.15

= (7.8103 kg/m3)(103 g/kg)(0.12 m)(1.210–2 m)2  (6.021023 atoms/mol)(1.810–23 A · m2)/(55.8 g/mol) = 26 A · m2. (b) Because the field is perpendicular to the dipole moment, the torque is maximal: = B = (26 A · m2)(1.2 T) = 31 m · N. 44. For the two magnetic fields we have B0 = 0nI; B = nI. Thus the magnetic permeability is µ (10–4 T · m/A )  = (B/B0)0. 70 B0 (10–4 T) B (T)  (10–4 T · m/A) 0 0 60 0.13 0.0042 4.1 50 0.25 0.010 5.0 0.50 0.028 7.0 40 0.63 0.043 8.6 0.78 0.095 15.3 30 1.0 0.45 56.6 20 1.3 0.67 64.8 1.9 1.01 66.8 10 2.5 1.18 59.3 6.3 1.44 28.7 0 13.0 1.58 15.3 5 10 15 B0 (10–4 T) 130 1.72 1.7 1300 2.26 0.22 10,000 3.15 0.04 Note that we have not plotted the last three points in order to show the changes at small fields. 45. The magnetic field in the toroid is B = nI = (3000)(4p10–7 T · m/A)(400 turns/m)(20 A) = 46. We find the permeability from B = nI; 2.2 T = [(600 turns)/(0.38 m)](48 A), which gives  =

30 T.

2.910–5 T · m/A.

47. Because the currents and the separations are the same, we find the force per unit length between any two wires from F/L = I1(0I2/2pd) = 0I2/2pd = (4p10–7 T · m/A)(8.00 A)2/2p(0.380 m) = 3.3710–5 N/m. The directions of the forces are shown on the diagram. The symmetry of the force diagrams simplifies the vector addition, so we have FM/L = 2(F/L) cos 30° = 2(3.3710–5 N/m) cos 30° = 5.8410–5 N/m up. FN/L = F/L = 3.3710–5 N/m 60° below the line toward P. FP/L = F/L = 3.3710–5 N/m 60° below the line toward N. 48. Because the currents and the separations are the same, we find the force on a length L of the top wire from

y FM F

F

IM d IN



F

60°

F FN

F FP

IP



x F

Chapter 28 p.16

y

F either of the two bottom wires from F F = I1(0I2/2pd)L = 0I1I2L/2pd I1 = (4p10–7 T · m/A)(20.0 A)I1L/2p(0.380 m) L = (1.0510–5 N/A · m)I1L. The directions of the forces are shown on the diagram. mg 60° The symmetry of the force diagram simplifies the I2 I3 vector addition, so for the net force to be zero, we have   2F cos 30° = mg = pr2Lg; 2(1.0510–5 N/A · m)I1L cos 30° = (8.9103 kg/m3)p(1.0010–3 m)2(9.80 m/s2)L, which gives I1 = 1.50104 A.

49. The component of the velocity parallel to the field does not change. The component perpendicular to the field produces a force which causes the circular motion. We find the radius of the circular motion from R = mv/qB = (9.1110–31 kg)(1.3107 m/s)(sin 7.0°)/(1.6010–19 C)(3.310–2 T) 0.27 mm. = 2.7210–4 m = We find the time for one revolution: T = 2pR/v = 2p(2.7210–4 m)/(1.3107 m/s) sin 7.0° = 1.0810–9 s. In this time, the distance the electron travels along the field, which is the pitch, is p = vT = (1.3107 m/s)(cos 7.0°)(1.0810–9 s) = 1.410–2 m = 1.4 cm. 50. (a) The magnetic field at the loop from the wire will be out of the page, and will depend only on the distance from the wire r: B = (0/4p)2I1/r, For the two horizontal sides of the loop, the currents are in opposite directions so their forces will be in opposite directions. Because the current will be in the same average field, the magnitudes of the forces will be equal, so Fc + Fd = 0. For the sum of the two forces on the vertical sides of the loop, we have Fnet = Fa – Fb = I2BaL – I2BbL = I2[(0/4p)2I1/a]L – I2[(0/4p)2I1/(a + b)]L

R

x

v

 B

Fc Bwire

I2

Fb

= (0/4p)(2I1I2L){(1/a) – [1/(a + b)]} = (10–7 T · m/A)2(10.0 A)(2.0 A)(0.26 m)[(1/0.10 m) – (1/0.22 m)] = 5.7  10–6 N toward the wire. (b) Because all forces on the loop lie in the same plane, the net torque is 0.

Fa L

I2

Fd b

a

I1

Chapter 28 p.17

51. The sheet may be thought of as an infinite number of parallel wires. The figure shows a view looking directly at the current. If we consider a point above the sheet, the wire directly underneath produces a magnetic field parallel to the sheet. By considering a pair of wires symmetrically placed about the first one, we see that the net field will be parallel to the sheet. Below the sheet, the field will be in the opposite direction. We apply Ampere’s law to the rectangular path shown in the diagram. For the sides perpendicular to the sheet, B is perpendicular to ds. For the sides parallel to the sheet, B is parallel to ds and constant in magnitude, because the upper and lower paths are equidistant from the sheet. We have B · d s =  I ;

D B y j

t

y B

0 enclosed

side s

B · ds +

lengths

B · ds = 0 + B

lengths

d s = B2D = 0 jtD.

This gives B = 0jt/2 parallel to the sheet, perpendicular to the current (opposite directions on the two sides). 52. (a) For the force produced by the magnetic field of the upper wire to balance the weight, it must be up, i. e., an attractive force. Thus IT the currents must be in the same direction. When we equate the d magnitudes of the two forces for a length L, we get IBBTL = mg; FT IB (0/4p)2IBITL/d = (pr2L)g; (10–7 T · m/A)2IB(48 A)/(0.15 m) = BT (8.9103 kg/m3)p(1.2510–3 m)2(9.80 m/s2), mg 6.7103 A to the right. which gives IB = (b) The magnetic force will decrease with increasing separation. If the wire is moved a small distance above or below the equilibrium position, there will be a net force, away from equilibrium, and the wire will be unstable. (c) If the second wire is above the first, there must be a repulsive magnetic force between the two wires to balance the weight, which means the currents must be opposite. Because the separation is the same, the magnitude of the current is the same: I2 = 6.7103 A to the left. The magnetic force will decrease with increasing separation. If the wire is moved a small distance above or below the equilibrium position, there will be a net force back toward equilibrium, and the wire will be stable for vertical displacements.

Chapter 28 p.18

53. The mass, and thus the volume, of the wire is fixed, so we have pr2L = k; a smaller radius will give a greater length. If we assume a given current (a variable voltage supply), the magnetic field of the solenoid will be determined by the density of turns: B = 0nI. The greatest density will be when the wires are closely wound. In this case, the separation of turns is 2r, so the density of turns is 1/2r, which would indicate that the radius should be very small. If D is the diameter of the solenoid, the number of turns is N = L/pD, so the length of the solenoid is N2r = 2Lr/pD = 2k/p2rD. The length of the solenoid must be much greater than the diameter, which will be true for small r, as long as the diameter is not large, which is the only restriction on the diameter. These considerations indicate that a long and thin wire should be used. However, we must be concerned with the resistance of the wire, because the thermal power generation, I 2R, must be dissipated in the solenoid. The resistance is R = L/pr2 = k/p2r4. Thus a very thin wire will create thermal dissipation problems, which means that the insulation and/or wire could melt. Thus the wire should be something between long, thin and short, fat. 54. The total field is the vector sum of the fields from the two currents. We can write the path integral on the circular path for the total field as the sum of two such integrals: ı B · d¬ = ı B1 · d¬ + ı B2 · d¬. To evaluate the integral for B2 we consider a differential angle d based at I2 which crosses the path at two locations at distances ra and rb from I2 . If we integrate clockwise around the path, the components of d¬ parallel to the field will be + ra d and – rb d. Thus the contribution to the integral from these two segments will be ? (B2ara d – B2brb d) = (0I2/2pra)ra d – (0I2/2prb)rb d = 0. The total integral will be composed of similar pairs and thus ı B2 · d¬ = 0. Because the circular path is centered on I1 , d¬ will be parallel to B1 , and B1 will have a constant magnitude, so we have

B2b

I1 rb B2a

ra d I2

ı B · d¬ = ı B1 · d¬ = B1 ı d¬ = (0I1/2pr)(2pr) = 0I1 . 55. We use the expression for the magnetic field on the axis of a circular loop: B = (0I/2)R2/(R2 + x2)3/2 = (0I/2)RE2/(RE2 + RE2)3/2 = 0I/4REv2; 110–4 T = (4p10–7 T · m/A)I/4v2(6.4106 m), which gives I = 3109 A. 56. The magnetic field from each of the sides will be directed out of the page. We can use the result of Problem 36 for the magnitude of the field from one side: Bside = 0IL/2pR(L2 + 4R2)1/2. At the center of the square, we have R = !L. The field will be the same for each side, so the total field is Bsquare = 40IL/2p!L[L2 + 4(!L)2]1/2 = 2v2 0I/pL. 57. For the circular loop to have the same length, 2pR = 4L, so the radius will be R = 2L/p. The magnetic field at the center is Bcircle = 0I/2R = 0I/2(2L/p) = 0Ip/4L. If we form the ratio for the two shapes, we get Bcircle/Bsquare = (0Ip/4L)/(2v2 0I/pL) = p2/8v2 = 0.87.

Chapter 28 p.19

Thus

B will decrease.

58. (a) We choose x = 0 at the left coil. The magnetic fields on the axis from the coils are in the same direction, so we find the magnitude of the total field from 2 2  NI R R B(x) = 0 + 3/ 2 3/ 2 2 R 2 + x2 R2 + R – x 2

(b)

B Bmax

1 1 0 N I + 2 3/ 2 2 3/ 2 . x x 2 1 + – 2 + x 2R R R R (c) At x = R/2, we have B(R/2) = (0NI/2R)({1/[1 + (1/2)2]3/2} + 0.5 Bmax x/ R 0 0.2 0.4 0.6 0.8 1.0 {1/[2 – 2(1/2) + (1/2)2]3/2}) = 0.7160NI/R = 0.716(4p10–7 T · m/A)(350 turns)(35 A)/(0.200 m) = 5.510–2 T. (d) When we differentiate the expression for B, we get  –3 2 1 x –3 2 –1 + 1 x dB = 0 NI 2 R R + 2 R R R 2 5/ 2 2 5/ 2 dx 2R x x x 1+ 2– 2 + R R R x x –1 3 NI R R =– 02 + ; 2 5/ 2 2 5/ 2 2R 2 –2 x + x 1+ x R R R =

d B = – 3 0NI dx 2 2R 2 2

=–

3 0NI 2R 3

2 –5 2 1 x R R 2 + 2 5/ 2 2 7/ 2 1+ x R 1 x –1 –5 2 –1 + 1 x R R R R R 2 + 2 5/ 2 2 7/ 2 x x x x 2– 2 + 2 –2 + R R R R 2 2 x 5R 5 xR – 1 1 – + – 2 5/ 2 x 2 7/ 2 x + x 2 5/ 2 2 – 2 x + x 2 1+ R 2 –2 R R R R

1 R 1+ x R + 1 x 1+ R

7/ 2

.

At x = R/2, we have



d B = – 3 0NI dx 2R 2 2 d B = – 3 0NI 3 dx 2 2R

=–

3 0NI 2R

3

1 1 –1 –1 30NI 2 2 2 = – + 2 5/ 2 5/ 2 = 0; 2 2 5/ 2 5 5 2R 1 1 + 2 –2 4 4 2 2 2 2 1 5 5 1 –1 1 1 2 2 – + – . 2 5/ 2 2 7/ 2 2 5/ 2 2 7/ 2 1 1 1 1 1 1 1+ 2– 2 1+ + 2 –2 + 2 2 2 2 2 2 5 5 1 1 4 4 – = 0. 5/ 2 – 7/ 2 + 5 5 5/ 2 5 5 7/ 2 4 4 4 4 1 2 1+ 1 2

5/ 2 +

59. Because the airplane is flying parallel to the wire, the circular magnetic field of the wire will be

Chapter 28 p.20

perpendicular to the velocity. Thus we have qvB = ma, which gives a = qvB/m = qv[(0/4p)2I/r]/m = (18.010–3 C)(2.8 m/s)(10–7 T · m/A)2(30 A)/(0.086 m)(0.175 kg)(9.80 m/s2/g) = 2.110–6 g. 60. (a) We let Rw represent the resistance of the wire, to distinguish it from the radius of the coil R. We find the required length of wire from the voltage at the maximum power: P = IV = V2/Rw = V2/(L/A), which gives L = V2A/P. Thus the number of turns is N = L/2pR = V2A/2pPR = (50 V)2(2.010–3 m)2/2p(1.6810–8  · m)(1.0103 W)(1.0 m) = 95 turns. (b) The magnetic field strength at the center of the coil is B = 0NI/2R = 0NP/V2R = (4p10–7 T · m/A)(95 turns)(1.0103 W)/(50 V)2(1.0 m) = 1.210–3 T. (c) We assume the same diameter for the coil, so a greater number of turns means a greater length of wire. The power supply produces a constant voltage, so the increased resistance of the wire means a lower power output. Thus we have I = V/Rw = VA/L. The magnetic field strength at the center of the coil is B = 0NI/2R = 0(L/2pR)(VA/L)/2R = 0VA/4pR2. Thus we see that the magnetic field strength is unchanged. 61. Although the currents are equal, we label them to show the individual fields. The magnitudes of the individual fields are equal: B1 = B2 = B3 = B4 = (0/2p)I/(¬/v2) = 0I/¬pv2. From the symmetry of the diagram we see that the resultant field will be in the – x-direction, with magnitude B = 4B1 cos 45° = 4(0I/¬pv2)(1/v2) = 20I/¬p (left).

y I1 

 I2 B2 B4

C

¬

x

B1 B3

I4

62. Although the current is the same, we label them so we can designate a segment of the wire. We choose the direction into the page as positive. We find the magnetic field from each of the segments of the wire. Because the point P is along the line of the I1 and I5 : B1 = B5 = 0. The magnetic field from the other three segments will be into the page. For segments 2 and 4 we use the result from Problem 37: B2 = B4 = 0Ia/4p(!a)[a2 + (!a)2]1/2 = 0I/2pa(1 + #)1/2 = 0I/pav5. For segment 3 we use the result from Problem 36:

I3

a I3 a

a I1

I4

I2

I5 P

Chapter 28 p.21

B3 = 0Ia/2pa[a2 + (2a)2]1/2 = 0I/2pa(1 + 4)1/2 = 0I/2pav5. Thus the net field is B = B2 + B3 + B4 = (0I/pa)[(2/v5) + (1/2v5)] =

0Iv5/2pa into the page.

63. At any instant the currents will be in opposite directions, with Irms = P/V = (40106 W)/(10103 V) = 4.0103 A. The maximal current will be Imax = Irmsv2. We show the directions of each field on the diagram, with tan  = d/2H, and sin  = d/2r, where r2 = H2 + #d2. Because the magnitudes of the two fields are the same, the net field will be down with a magnitude given by Bmax = 2B1 sin  = 2[(0/4p)2Imax/r](d/2r) = 2(0/4p)Imaxd/r2 = 2(10–7 T · m/A)(4.0103 A)v2(3 m)/[(30 m)2 + #(3 m)2] = 410–6 T. When we compare to the Earth’s field, we get Bmax/BE = (410–6 T)/(510–5 T) ˜ 0.1, so it is about 10% of the Earth’s field.

I2

d

I1



r

H



P B2

B1



Chapter 29 Page 1

CHAPTER 29 – Electromagnetic Induction and Faraday’s Law 1.

The average induced emf is å = – N ?B/?t = – (2)[(+58 Wb) – (– 80 Wb)]/(0.72 s) =

– 3.8102 V.

2.

Because the plane of the coil is perpendicular to the magnetic field, the initial flux through the loop is maximal. The magnitude of the average induced emf is å = – ?B/?t = – A ?B/?t = – p(0.13 m)2(0 – 0.90 T)/(0.15 s) = 0.32 V.

3.

As the coil is pushed into the field, the magnetic flux increases into the page. To oppose this increase, the flux produced by the induced current must be out of the page, so the induced current is counterclockwise.

B

4.

As the magnet is pushed into the coil, the magnetic flux increases to the right. To oppose this increase, the flux produced by the induced current must be to the left, so the induced current in the resistor will be from right to left.

S

N

B

R

5.

6.

The magnitude of the average induced emf is å = – ?B/?t = – A ?B/?t = – p(0.036 m)2(0 – 1.3 T)/(0.20 s) =

0.026 V.

We choose up as the positive direction. The average induced emf is å = – ?B/?t = – A ?B/?t = – p(0.046 m)2[(– 0.25 T) – (+ 0.63 T)]/(0.15 s) = 3.910–2 V =

39 mV.

Chapter 29 Page 2

7.

8.

9.

(a) The increasing current in the wire will cause an increasing field into the page through the loop. To oppose this increase, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise. (b) The decreasing current in the wire will cause a decreasing field into the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise. (c) Because the current is constant, there will be no change in flux, so the induced current will be zero. (d) The increasing current in the wire will cause an increasing field into the page through the loop. To oppose this increase, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise. (a) As the resistance is increased, the current in the outer loop will decrease. Thus the flux through the inner loop, which is out of the page, will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise. (b) If the small loop is placed to the left, the flux through the small loop will be into the page and will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux into the page, so the direction of the induced current will be clockwise. As the solenoid is pulled away from the loop, the magnetic flux to the right through the loop decreases. To oppose this decrease, the flux produced by the induced current must be to the right, so the induced current is counterclockwise as viewed from the solenoid.

(b) I decreasing

(a) B

B I increasing (c)

B

B

(d)

I constant

I increasing

I

B

I

I B

v

10. (a) The average induced emf is å = – ?B/?t = – A ?B/?t = – p(0.10 m)2[(– 0.45 T) – (+ 0.52 T)]/(0.180 s) = 0.17 V. (b) The positive result for the induced emf means the induced field is away from the observer, so the induced current is clockwise.

Chapter 29 Page 3

11. (a) The magnetic flux through the loop is into the paper and decreasing, because the area is decreasing. To oppose this decrease, the induced current in the loop will produce a flux into the paper, so the direction of the induced current will be clockwise. (b) We choose into the paper as the positive direction. The average induced emf is å = – ?B/?t = – B ?A/?t = – pB ?(r2)/?t = – p(0.75 T)[(0.030 m)2 – (0.100 m)2]/(0.50 s) = 4.310–2 V = 43 mV. (c) We find the average induced current from I = å/R = (43 mV)/(2.5 ) = 17 mA. 12. If  is the angle between the magnetic field and the normal to the plane of the loop, the flux through the loop is B = BA cos  = BA cos t, so the induced emf is å = – dB /dt = BA sin t. For a sinusoidal variation, the rms value is årms = BA/v2; 7010–3 V = B(0.15 m)2[2p rad/(4510–3 s)]/v2. which gives B = 3.210–2 T. 13. As the loop is pulled from the field, the flux through the loop decreases. We find the induced emf from å = – ?B/?t = – B ?A/?t = – (0 – 0.40 T)p(0.10 m)2/(10010–3 s) = 0.126 V. The dissipated energy is 1.110–5 J. Energy = P ?t = (å2/R) ?t = [(0.126 V)2/(150 )](10010–3 s) = 14. As the loop is pulled from the field, the flux through the loop will decrease. We find the induced emf from å = – ?B/?t = – B ?A/?t = – B¬ ?x/?t = – B¬ (– v) = B¬v. FB Because the inward flux is decreasing, the induced flux will be into the page, so the induced current is clockwise, given by I = å/R. B Because this current in the left hand side of the loop is in a downward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, which we find from F = I¬B = (å/R)¬B = B2¬2v/R = (0.450 T)2(0.350 m)2(3.40 m/s)/(0.230 ) =

¬

F

x

0.367 N.

15. For the resistance of the loop, we have R = L/A = pD/#pd2 = 4D/d2. The induced emf is å = – ?B/?t = – #pD2 ?B/?t; so the induced current is I = å/R = – (pDd2/16) ?B/?t. In the time ?t the amount of charge that will pass a point is Q = I ?t = – (pDd2/16) ?B = – [p(0.156 m)(2.0510–3 m)2/16(1.6810–8  · m)](0 – 0.550 T) =

4.21 C.

Chapter 29 Page 4

16. The flux through a loop is B = [(8.8 s–1)t – (0.51 s–3)t3]10–2 T · m2. (a) The induced emf is å = – N dB/dt = – (60)[(8.8 s–1) – (1.53 s–3)t2]10–2 T · m2 = [– 528 + (91.8 s–2)t2]10–2 V = – 5.3 V+ (0.92 V · s–2)t2. (b) At t = 1.0 s we have å1 = – 5.28 V + (0.918 V · s–2)(1.0 s)2 = – 4.4 V. At t = 5.0 s we have å5 = – 5.28 V + (0.918 V · s–2)(5.0 s)2 = 18 V. 17. (a) For the resistance of the loop, we have R = L/A = (1.6810–8  · m)(20)p(0.350 m)/p(1.010–3 m)2 = 0.118 . The induced emf is å = – ?B/?t = – NA ?B/?t = – (20)p(0.175 m)2(3.2010–3 T/s) = – 6.1610–3 V. Thus the induced current is I = å/R = (6.1610–3 V)/(0.118 ) = 5.210–2 A. (b) Thermal energy is produced in the wire at the rate of P = I 2R = (5.210–2 A)2(0.118 ) = 3.210–4 W = 0.32 mW. 18. Inside the solenoid, the magnetic field is B = 0nI, so the flux through the loop is B = A1B = A10nI0 cos t. The induced emf is å = – dB/dt = A10nI0 sin t. 19. The flux through the loop is B = BA. The induced emf is å = – dB/dt = – B dA/dt = – (0.48 T)(– 3.5010–2 m2/s) = 1.710–2 V. Because the area changes at a constant rate, this is the induced emf for both times. 20. The flux through the loop when it has area A is B = BA = Bpr2. The induced emf is å = – dB/dt = – B2pr dr/dt. At t = 0 the radius is r0 = (A0/p)1/2 = (0.285 m2/p)1/2 = 0.301 m. At a later time t the radius is r = r0 + (dr/dt)t, so the induced emf is å = – 2pB[r0 + (dr/dt)t] dr/dt. At t = 0 we have å0 = – 2p(0.48 T)[0.301 m + (0.0700 m/s)(0)](0.0700 m/s) = – 0.064 V. At t = 1.00 s we have å1 = – 2p(0.48 T)[0.301 m + (0.0700 m/s)(1.00 s)](0.0700 m/s) = – 0.078 V.

Chapter 29 Page 5

21. At a distance r from the wire, the magnetic field is directed into the paper with magnitude B = 0I/2pr. Because the field is not constant over the square, we find the magnetic flux by integration. We choose a differential element parallel to the wire at position r with area a dr. The magnetic field through this element is constant, so the flux is a+ a 0 I  Ia 0Ia  = B · d A = B d A = a d r = 0 l n 2a = l n 2. a 2r 2 2 a

22. Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from å = B¬v = (0.750 T)(0.190 m)(0.250 m/s) 35.6 mV. = 3.5610–2 V =

23. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from å = B¬v = (0.35 T)(0.240 m)(1.8 m/s) = 0.15 V. (b) We find the induced current from I = å/R = (0.151 V)/(2.2  + 26.0 ) = 5.410–3 A. (c) The induced current in the rod will be down. Because this current is in an outward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, which we find from F = I¬B = (5.410–3 A)(0.240 m)(0.35 T) = 4.510–4 N.

B a a

I a

dr

r

B

v

B

¬

v

¬

I

24. If we assume that the movable rod starts at the bottom of the U, in a time t it will have moved a distance x = vt. For the resistance of the U, we have R = L/A = (2vt + ¬)/A. The induced emf is å = B¬v; so the induced current is I = å/R = B¬v/[(2vt + ¬)/A] = B¬vA/(2vt + ¬).

Chapter 29 Page 6

25. (a) Although there is an induced emf, there is no current because there is no closed circuit. Thus the rod will move at constant speed. (b) When the circuit is completed, there will be a current. The induced emf is B¬v. Because the only resistance is from the rod, the current is I = B¬v/R. The induced current in the rod will be down. Because this current is in an outward magnetic field, there will be a magnetic force to the left, which will produce an acceleration: F = – IB¬ = – B2¬2v/R = ma = m dv/dt, or – (B2¬2/mR) dt = dv/v. We integrate to get the speed: t 0

–

B

v

¬

v B2 2 dv ; dt = mR v0 v 2

2 22 – B t = l n vv , or v = v 0 e – B t/ mR. mR 0 The speed eventually goes to zero. This is an application of Lenz’s law.

26. (a) If the polarity of the source of emf is as indicated in the figure, the current in the rod will be up, which will produce a force in the outward magnetic field F that is to the right. For a constant current, the B – constant force will produce a constant acceleration: å + F = I¬B = ma. I Thus the speed of the rod is v = v0 + at = 0 + (I¬B/m)t = I¬Bt/m. Note that there will be an induced emf in the rod, so the source must compensate to maintain the constant current. (b) With a constant source emf å0 , the current will be determined by the two emfs in the circuit. Because the initial motion is to the right, the induced emf will be down, opposing the source. Thus we have I = (å0 – B¬v)/R. The force produced by this current provides the acceleration: F = I¬B = (å0 – B¬v)B¬/R = m dv/dt, or (B¬/mR) dt = dv/(å0 – B¬v). When we integrate, we get v

t

v

B2 2 B dt = dv , or dt = – 0 mR 0 mR 0 å0 – Bv 0 t

¬

– B dv ; å0 –B v

2 2 2 2 å –B v å – B t = ln 0 , or v = 0 1 – e – B t/ mR . mR å0 B (c) When the current is constant, we see that the speed continues to increase, so no terminal speed is reached. With a constant å0 , as t  8, the exponential term goes to 0, so there is a terminal speed:

Chapter 29 Page 7

vterm = å0/B¬.

27. (a) At a distance r from the long wire, the magnetic field is directed into the paper with magnitude B B = 0I/2pr. Because the field is not constant over the short section, we a b find the induced emf by integration. We choose a differential r dr element dr a distance r from the long wire. The induced emf in this element is då = Bv dr toward the long wire. v We find the total emf by integrating: a +b a +b  Iv  Iv dr = 0 l n a + b t owa rd l ong wi re. Bv d r = 0 å= r 2 b 2 b b (b) If the current is in the opposite direction to I, the only change will be in the direction of the emf:  Iv å = 0 l n a + b a wa yfrom l ong wi re. 2 b 28. The maximum induced emf is å = NBA. If the only change is in the rotation speed, for the two conditions we have å2/å1 = 2/1 ; å2/(12.4 V) = (2500 rpm)/(950 rpm) which gives å2 = 32.6 V. 29. The induced emf is å = NBA sin t. For the rms value of the output, we have Vrms = [(å2)av]1/2 = [(NBA)2(sin2 t)av]1/2 = NBA [(sin2 t)av]1/2. The sin2 t function varies from 0 to 1, with an average value of 1/2, so we get Vrms = NBA(1/v2) = NBA/v2. 30. We find the rotation speed from åpeak = NBA; 120 V = (420 turns)(0.350 T)(0.210 m)2, which gives  = 18.5 rad/s =

2.95 rev/s.

31. We find the peak emf from åpeak = NBA = (350 turns)(0.45 T)p(0.050 m)2(60 rev/s)(2p rad/rev) = 466 V. The rms voltage output is Vrms = åpeak/v2 = (466 V)/v2 = 330 V = 0.33 kV. If only the rotation frequency changes, to double the rms voltage output, we must double the rotation speed, so we have f2 = 2f1 = 2(60 rev/s) = 120 rev/s. 32. We find the counter emf from å – åback = IR; 120 V – åback = (9.20 A)(3.75 ), which gives åback =

86 V.

I

Chapter 29 Page 8

33. Because the counter emf is proportional to the rotation speed, for the two conditions we have åback2/åback1 = 2/1 ; åback2/(72 V) = (2500 rpm)/(1800 rpm), which gives åback2 = 100 V. 34. Because the counter emf is proportional to both the rotation speed and the magnetic field, for the two conditions we have åback2/åback1 = (2/1)(B2/B1); (75 V)/(100 V) = [(2500 rpm)/(1000 rpm)](B2/B1), which gives B2/B1 = 0.30. 35. Because the counter emf is proportional to the rotation speed, we find the new value from åback2/åback1 = 2/1 ; åback2/(108 V) = (1/2), which gives åback2 = 54 V. We find the new current from å – åback = IR; 120 V – 54 V = I(5.0 ), which gives I = 13 A. 36. (a) We find the emf of the generator from the load conditions: V = å – IarmatureRarmature; 200 V = å – (50 A)(0.40 ), which gives å = 220 V. When there is no load on the generator, the current is zero, so the voltage will be the emf: 220 V. Note that no-load means little torque required to turn the generator. (b) Because the generator emf is proportional to the rotation speed, we find the new value from å2/å1 = 2/1 ; å2/(220 V) = (800 rpm/1000 rpm), which gives å2 = 176 V. We find the new load voltage from V2 = å2 – IarmatureRarmature = 176 V – (50 A)(0.40 ) = 156 V. Note that the power output is reduced to 7.8 kW. 37. We find the number of turns in the secondary from VS/VP = NS/NP ; (8500V)/(120 V) = NS/(500 turns), which gives NS =

3.54104 turns.

38. Because NS < NP , this is a step-down transformer. We find the ratio of the voltages from VS/VP = NS/NP = (120 turns)/(720 turns) = 0.167. With 100% efficiency, for the ratio of currents we have IS/IP = NP/NS = (720 turns)/(120 turns) = 6.00. 39. With 100% efficiency, the power on each side of the transformer is the same: IPVP = ISVS , so we have IS/IP = VP/VS = (22 V)/(120) = 0.18. 40. We find the ratio of the number of turns from NS/NP = VS/VP = (12103 V)/(220 V) = 55. If the transformer is connected backward, the role of the turns will be reversed: VS/VP = NS/NP ; VS/(220 V) = 1/55, which gives VS = 4.0 V. 41. (a) We assume 100% efficiency, so we have IPVP = ISVS ;

Chapter 29 Page 9

(0.65 A)(120 V) = (15 A)VS , which gives VS = 5.2 V. (b) Because VS < VP , this is a step-down transformer. 42. (a) We assume 100% efficiency, so we find the input voltage from P= IPVP ; 100 W = (26 A)VP , which gives VP = 3.8 V. Because VS > VP , this is a step-up transformer. (b) For the voltage ratio we have VS/VP = (12 V)/(3.84 V) = 3.1. 43. We find the output voltage of this step-up transformer from VS/VP = NS/NP ; VS/(120 V) = (1510 turns)/(330 turns), which gives VS = 549 V. We find the input current from IS/IP = NP/NS ; (15.0 A)/IP = (330 turns)/(1510 turns), which gives IP = 68.6 A. 44. (a) We can find the current in the transmission lines from the output emf: Pout = IVout ; 30106 W = I(45103 V), which gives I = 667 A. We find the input emf from åin = Vout + IRlines = 45103 V + (667 A)(3.0 ) = 47103 = 47 kV. (b) The power loss in the lines is Ploss = I 2Rlines = (667 A)2(3.0 ) = 1.33106 W = 1.33 MW. The total power is Ptotal = Pout + Ploss = 30 MW + 1.33 MW = 31.3 MW, so the fraction lost is (1.33 MW)/(31.3 MW) = 0.042 (4.2%). 45. Without the transformers, we can find the delivered current, which is the current in the transmission lines, from the delivered power: Pout = IVout ; 65103 W = I(120V), which gives I = 542 A. The power loss in the lines is PL0 = I 2Rlines = (542 A)2(2)(0.100 ) = 5.87104 W = 58.7 kW. With the transformers, to deliver the same power at 120 V, the delivered current from the step-down transformer is still 542 A. If the step-down transformer is 99% efficient, we have (0.99)IP2VP2 = IS2VS2 ; (0.99)IP2(1200 V) = (542 A)(120 V), which gives IP2 = 54.7 A. Because this is the current in the lines, the power loss in the lines is PL2 = IP22Rlines = (54.7 A)2(2)(0.100 ) = 5.99102 W = 0.599 kW. For the 1% losses in the transformers, we approximate the power in each transformer: PLt = (0.01)(65 kW + 65.6 kW) = 1.31 kW. The total power loss using the transformers is PL = 0.599 kW + 1.31 kW = 1.9 kW. The power saved by using the transformers is Psaved = PL0 – PL = 58.7 kW – 1.9 kW = 56.8 kW. 46. We can find the delivered current, which is the current in the transmission lines, from the delivered

Chapter 29 Page 10

power: Pout = IVout ; 300106 W = I(600103 V), which gives I = 500 A. The 2% loss means Pout = 0.98 Pin . Thus the power loss in the lines is PL = (0.02)Pin = (0.02)Pout /(0.98)= I 2Rlines ; (0.02)(300106 W)/(0.98) = (500 A)2Rlines , which gives Rlines = 24.5 . We find the radius of the 2 lines, each 200 km long, from R = L/A ; 24.5  = (2.6510–8  · m)(2)(200103 m)/pr2 , which gives r = 1.17410–2 m. Thus the diameter of the lines should be 2.35 cm. 47. We find the electric field from åinduced = B¬v = ý E · d¬ = E¬, which gives E = Bv = (0.750 T)(0.250 m/s) = 0.188 V/m. 48. The induced emf around a circle is å = – dB /dt = – A dB/dt. Because A and dB/dt are the same for the two circular paths, å is the same. Even though E is greater in the region of the outer circle, the integral ý E · d¬ is the same. Note that for part of the path the component of E is parallel to d¬ and for part it is antiparallel.

B

49. (a) The changing magnetic field through the area defined by the circular vacuum tube produces a circular electric field along the axis of the tube. This electric field along the electron path creates the acceleration to increase the speed of the electron. (b) The magnetic force on the electron provides the centripetal acceleration. To get a force toward the center, the current must be in on the right and out on the left, that is, counterclockwise. Because electrons have a negative charge, they must be moving clockwise. (c) The electric field along the electron path is given by the rate of change of the magnetic flux: ý E · d¬ = – dB/dt = – A dB/dt. If we choose clockwise (the direction of the electron motion) for the positive direction, the electric field must be negative, so dB/dt > 0. B (d) To provide the radial acceleration, the magnetic field must be down (positive). To provide the tangential T acceleration, the magnetic field must be increasing. In a sinusoidal variation, these two conditions exist T/4 only for # of the cycle.

t

Chapter 29 Page 11

50. We find the electric field along the electron’s path at a radius R by applying Faraday’s law: ý E · d¬ = – dB/dt = – A dBav/dt; E2pR = – pR2 dBav/dt, or E = – !R dBav/dt. Thus for the tangential acceleration we have atan = dv/dt = – eE/m = !(eR/m) dBav/dt. The magnetic force at the position of the electron orbit provides the centripetal acceleration: evB0 = mv2/R, which gives v = eB0R/m. If we differentiate this we get dv/dt = (eR/m) dB0/dt. When we compare this to the previous result, we see that dB0/dt = ! dBav/dt, at any moment. This is satisfied at any time if B0 = !Bav . 51. (a) Because the current in the rod is constant, the potential difference across the rod must be constant: V = IR. The net electric field is the potential gradient along the rod: Enet = V/¬ = IR/¬ (constant). (b) With a constant source emf å0 , the potential difference across the rod will be V = IR = å0 – B¬v. If we use the result for the velocity from Problem 26, we have 2 2 2 2 V= å 0 – B v = å 0 – å 0 1 – e – B t/ mR = å 0 e – B t/ mR . The net electric field is the potential gradient along the rod: å 0 – B 2 2t/ mR E n et = V/ = e .

52. (a) The clockwise current in the left-hand loop produces a magnetic field which is into the page within the loop and out of the page outside the loop. Thus the right-hand loop is in a magnetic field out of the page. Before the current in the left-hand loop reaches its steady state, there will be an induced current in the right-hand loop that will produce a magnetic field into the page to oppose the increase of the field from the left-hand loop. Thus the induced current will be clockwise. (b) After a long time, the current in the left-hand loop will be constant, so there will be no induced current. (c) If the second loop is pulled to the right, the magnetic field out of the page from the left-hand loop through the second loop will decrease. During the motion, there will be an induced current in the right-hand loop that will produce a magnetic field out of the page to oppose the decrease of the field from the left-hand loop. Thus the induced current will be counterclockwise. 53. We find the number of turns from åpeak = NBA; 24.0 V = N(0.420 T)(0.070 m)2(60 rev/s)(2p rad/rev), which gives N = 54. The average induced emf is å = – ?B/?t = – A ?B/?t = – (0.240 m)2(0 – 0.755 T)/(0.0400 s) = 1.087 V. The average current is I = å/R = (1.087 V)/(6.50 ) = 0.167 A. The energy dissipated is 7.2710–3 J. Energy = I2R ?t = (0.167 A)2(6.50 )(0.0400 s) =

31 turns.

Chapter 29 Page 12

55. A side view of the rail and bar is shown in the figure. The FN component of the velocity of the bar that is perpendicular to the magnetic field is v cos , so the induced emf is B v å = B¬v cos . FB If we assume the resistance of the copper bar can be neglected, s this produces a current in the bar mg  I = å/R = (B¬v cos )/R into the page. Because the current is perpendicular to the magnetic field, the force on the bar from the magnetic field will be horizontal, as shown, with magnitude FB = I¬B = (B2¬2v cos )/R. For the bar to slide down at a steady speed, the net force must be zero. If we consider the components along the rail, we have FB cos  – mg sin  = 0, or [(B2¬2v cos )/R] cos  = (B2¬2v cos2 )/R = mg sin  (0.55 T)2(0.30 m)2v(cos2 5.0°)/(0.60 ) = (0.040 kg)(9.80 m/s2) sin 5.0°, which gives v = 0.76 m/s.  56. (a) The voltage drop across the lines is ?V = 2IR = 2(700 A)(0.80 ) = 1.12103 V = 1.12 kV. Thus the voltage at the other end is Vout = Vin – ?V = 42 kV – 1.12 kV = 41 kV. (b) The power input is Pin = IVin = (0.700 kA)(42 kV) = 29.4 MW. (c) The power loss in the lines is Ploss = 2I 2R = 2(0.700 kA)2(0.80 ) = 0.78 MW. (d) The power output is Pout = IVout = (0.700 kA)(40.9 kV) = 28.6 MW. Note that this is Pin – Ploss . 57. If we assume perfect transformers, the only losses will be the energy dissipated in the transmission lines. The resistance of the lines is R = (0.10 /km)(100 km) = 10 . If Pin is the power supplied to the town, we have Pin/(Pin + Iline2R) = 0.985; (50106 W)/[50106 W + Iline2(10 )] = 0.985, which gives Iline = 276 A. For the perfect transformer at the town end of the line, we have Pin = IlineV2 ; 50103 kW = (276 A)V2 , which gives V2 = 181 kV. The voltage drop in the transmission line is Vdrop = IlineR = (0.276 kA)(10 ) = 2.76 kV. Thus the voltage at the station must be stepped up to V1 = V2 + Vdrop = 181 kV + 2.76 kV = 184 kV.

58. The magnetic field through the loop is B = 0nI inside the solenoid only, so the flux through the loop is B = NBA = N0nIprsolenoid2. The current induced in the loop is

I I'

Chapter 29 Page 13

I = å/R = – (1/R) dB /dt = – (N0nprsolenoid2/R) dI/dt = – [(150 turns)(4p10–7 T · m/A)(200102 turns/m)p(0.045 m)2/(11.0 )][(2.0 A)/(0.10 s)] = – 4.410–2 A (opposite to the direction of the current in the solenoid). This assumes the windings of the coil and the solenoid are in the same direction. 59. The magnitude of the average induced emf is A å = ?B/?t, so the average current is I = å/R = (1/R) ?B/?t. t The charge moving past a fixed point is q = I ?t = ?B/R = (B/R) ?A. The number of electrons is N = q/e = (B/eR) ?A. The maximum number of electrons occurs for the maximum change in area in a 90° rotation. The area perpendicular to the field varies sinusoidally. From the plot we see that the greatest change occurs when the coil moves from an angle of 45° with the field to an angle of 45° on the other side. Thus we have Nmax = [(0.15 T)p(0.030 m)2/(1.6010–19 C)(0.025 )][cos 45° – (– cos 45°)] = 1.51017. 60. We find the peak emf from åpeak = NBA = (125 turns)(0.200 T)(0.0660 m)2(120 rev/s)(2p rad/rev) = 82.1 V. 61. (a) Because we have direct coupling, the torque provided by the motor balances the torque of the friction force: NIAB = Fr; (300 turns)I(0.10 m)(0.15 m)(0.60 T) = (250 N)(0.25 m), which gives I = 23 A. (b) To maintain the speed, we have a force equal to the friction force, so the power required is Fv = (250 N)(30 km/h)/(3.6 ks/h) = 2.08103 W. This must be provided by the net power from the motor, which is Pnet = IVin – I 2R = I(Vin – IR) = Iåback = Fv; (23 A)åback = 2.08103 W, which gives åback = 90 V. (c) The power dissipation in the coils is Ploss = Pin – Pnet = (23 A)(10)(12 V) – 2.08103 W = 6.9102 W. (d) The useful power percentage is (Pnet/Pin)(100) = (Iåback/IVin)(100) = (90 V/120 V)(100) = 75%. 62. The average induced emf is å = – ?B/?t = – NA ?B/?t = – NA[(– B) – (+ B)]/?t = 2NAB/?t. The average current is I = å/R = 2NAB/R ?t, so the total charge that passes through the galvanometer is Q = I ?t = (2NAB/R ?t) ?t = 2NAB/R, or B = QR/2NA. 63. (a) From the efficiency of the transformer, we have PS = 0.80PP. For the power input to the transformer, we have PP = IPVP ; (75 W)/0.80 = IP(110 V), which gives IP = 0.85 A. (b) We find the secondary voltage from PS = VS2/RS ; 75 W = VS2/(2.4 ), which gives VS = 13.4 V. We find the ratio of the number of turns from NP/NS = VP/VS = (110 V)/(13.4 V) = 8.2.

Chapter 29 Page 14

64. The induced emf is å = – dB/dt = – Npr2 dB/dt = (Npr2B0/)e – t/. Thus the rate at which energy is dissipated is P = I2R = å2/R = [(Npr2B0/)2/R]e – 2t/. Because this rate is a function of time, we find the energy dissipated by integrating:

Energy =

N r 2B0 2R 

2

t

2 e – 2t/  2 d t = N r B0  2R  0

10(0.100 m) 2 (0.50 T) = 2(2.0 )(0.10 s)

2

2

1 – e – 2t/ 

1 – e – 2t/ (0.10 s) =

65. We choose a differential element dr a distance r from the center of rotation. The speed of this element is v = r, so the differential induced emf is då = Bv dr = Br dr. We integrate to find the total emf: å=

L 0

Bv d r =

L 0

Br d r =

1 2

6.2  10

–2

–1 J 1 – e – 20 s t .

dr

B

BL 2 t owa rd t h ecent er.

r v

66. (a) Because VS < VP , this is a step-down transformer. (b) We assume 100% efficiency, so we find the current in the secondary from P= ISVS ; 40 W = IS(12 V), which gives IS = 3.3 A. (c) We find the current in the primary from P= IPVP ; 40 W = IP(120 V), which gives IP = 0.33 A. (d) We find the resistance of the bulb from VS = ISRS ; 12 V = (3.33 A)RS , which gives RS = 3.6 . 67. We can find the current in the transmission lines from the power transmitted to the user: PT = IV, or I = PT/V. The power loss in the lines is PL = I 2RL = (PT/V)2RL = (PT)2RL/V2.



Chapter 29 Page 15

68. (b) At start up there will be no induced emf in the armature. Because the line voltage is across each resistor, we find the currents from Ifield0 = å/Rfield = (115 V)/(36.0 ) = 3.19 A; Iarmature0 = å/Rarmature = (115 V)/(3.00 ) = 38.3 A. We use the junction condition to find the total current: I0 = Ifield0 + Iarmature0 = 3.19 A + 38.3 A = 41.5 A. (c) At full speed, the back emf is maximum. Because the line voltage is across the field resistor, we find the field current from Ifield = å/Rfield = (115 V)/(36.0 ) = 3.19 A. We find the armature current from å – åback = IarmatureRarmature ; 115 V – 105 V = Iarmature(3.00 ), which gives Iarmature = 3.33 A. Thus the total current is I0 = Ifield + Iarmature = 3.19 A + 3.33 A = 6.52 A.

(a)

R field

Ifield0

R armature Iarmature0 –

+ I0

å Starting R field R armature

Ifield

–

+

å back

Iarmature –

+ I

å Full Speed

69. If we apply Faraday’s law to a counterclockwise path around the square, we have ý E · d¬ = – dB/dt = 0, because the electric field is static. We can express the integral as ý E · d¬ = ?bottom E · d¬ + ?sides E · d¬ + ?top E · d¬ = 0. On the bottom segment E and d¬ are parallel, so ?bottom E · d¬ > 0. If there were no fringing, ?sides E · d¬ = 0 because E and d¬ are perpendicular, and

?top E · d¬ = 0 because E = 0. However, we must have ?sides E · d¬ + ?top E · d¬ < 0, so there must be a fringing field outside the edge of the plates. We see that on the side and the top segments there will be a component of E antiparallel to d¬.

E

Chapter 29 Page 16

70. We find the flux through the area swept out by a radial line on the disk: B = B(/2p)pR2 = !BtR2. The magnitude of the induced emf is  å = dB/dt = !BR2. The magnetic flux through the area is increasing out of the page, so the induced flux will be in, which means the induced emf along the radial line is radially out from the axis, and the rim is at the higher potential.

71. We choose a differential element dr a distance r from the center of rotation. The speed of this element is v = r, so the radial differential induced emf is då = Bv dr = Br dr. The electric field is the potential gradient: E = då/dr = Br, radially out from the axis.



B r

 R

B

dr r

 R



Chapter 29 Page 17

72. We approximate the small area as a square with side D. Because the area A = D2 is small, the speed of the area is v = ¬. Thus the induced radial emf is å = BDv = BD¬. This emf produces a radial current in the area. If we neglect the resistance of the outside return path of the current, the approximate resistance is R = D/Dd = /d. The current is approximately I = å/R = BD¬d/. This current in the magnetic field produces a tangential force and thus a torque about the axis:  = F¬ = IBD¬ = B2D2¬2d/ = B2A¬2d/.

R

¬ B D

D



Chapter 30, p. 1

CHAPTER 30 – Inductance; and Electromagnetic Oscillations 1.

The magnetic field of the long solenoid is essentially zero outside the solenoid. Thus there will be the same linkage of flux with the second coil and the mutual inductance will be the same: M = 0N1N2A/¬.

2.

(a) We find the mutual inductance from M = N1N2A/¬ = (2000)(4p10–7 T · m/A)(300 turns)(100 turns)p(0.0200 m)2/(2.44 m) 38.8 mH. = 3.8810–2 H = (b) The induced emf in the second coil is å = – M ?I1/?t = – (3.8810–2 H)(0 – 12.0 A)/(0.0980 s) = 4.75 V.

3.

The magnetic field inside the outer solenoid is B = 0n1I1. The magnetic flux linked with the inner solenoid is 21 = 0n1I1A2. Thus the mutual inductance with the N2 turns of the inner solenoid is M21 = N221/I1 = n2¬0n1A2 , so the mutual inductance per unit length is M/¬ = 0n1n2pr22.

4.

We find the mutual inductance of the system by finding the mutual inductance of the small coil. The magnetic field inside the long solenoid is along the axis with a magnitude B1 = 0n1I1 = 0N1I1/¬. We find the magnetic flux through the coil by using the area perpendicular to the field: 21 = (0N1I1/¬)A2 sin . Thus the mutual inductance with the N2 loops of the coil is M21 = N221/I1 = (0N1N2A2/¬) sin .

5.

We find the mutual inductance of the system by finding the mutual inductance of the loop. The magnetic field of the long wire depends only on the distance from the wire. To find the magnetic flux through the loop, we choose a strip a distance x from the wire with width dx:  2  I 0 w d x =  0Iw ln 2 .  = B · d A = 2x 2 1 1 The mutual inductance is M = B/I = (0w/2p) ln(¬2/¬1).

x

B

I ¬1

6.

We find the induced emf from å = – L ?I/?t = – (180 mH)(38.0 mA – 20.0 mA)/(340 ms) = – 9.53 mV. The emf is opposite to the direction of the current, to oppose the increase in the current.

7.

We estimate the inductance by using the inductance of a solenoid: L = 0N2A/¬ = (4p10–7 T · m/A)(20,000 turns)2p(1.8510–2 m)2/(0.45 m) =

8.

dx

1.2 H.

Because the current in increasing, the emf is negative. We find the self-inductance from å = – L ?I/?t; – 8.50 V = – L[23.0 mA – (– 22.0 mA)]/(21.0 ms), which gives L = 3.97 H.

¬2

w

Chapter 30, p. 2

9.

We use the result from Ex. 30–4: L = (0¬/2p) ln(r2/r1) = (0¬/2p) ln(D2/D1) = (210–7 T · m/A)(22.0 m) ln(3.5 mm/2.0 mm) =

10. We find the current from å = – L ?I/?t; – 35 V = – (150 mH)(I0 – 0)/(3.0 ms), which gives I0 =

2.510–6 H.

0.70 A.

11. We use the result for the inductance from Ex. 30–4: L/¬ = (0/2p) ln(r2/r1); (210–7 T · m/A) ln(3.0 mm/r1) = 4010–9 H/m, which gives

r1 = 2.5 mm.

12. (a) The magnetic flux through coil 1 is due to its own current and the current in the other coil. From the definition of self-inductance, the flux from its own current is 11 = L1I1. From the definition of mutual inductance, the flux from the current in the other coil is 12 = M12I2. Thus the total magnetic flux is 1 = 11 + 12 = L1I1 + MI2. Similarly for the other coil we have 2 = 22 + 21 = L2I2 + MI1. (b) We find the induced emf from the rate of change of the flux: å1 = – d1/dt = – L1 dI1/dt – M dI2/dt. å2 = – d2/dt = – L2 dI2/dt – M dI1/dt. 13. If D represents the diameter of the solenoid, the length of the wire is N(pD). Because this is constant, we have N1pD1 = N2pD2 , or N2/N1 = D1/D2 = @. The solenoid is tightly wound, so the length of the solenoid is ¬ = Nd, where d is the diameter of the wire. Thus we have ¬2/¬1 = N2/N1 = @. We use the inductance of a solenoid: L = 0AN2/¬, and form the ratio of inductances for the two conditions, so we have L2/L1 = (D2/D1)2(N2/N1)2/(¬2/¬1) = (3)2(@)2/(@) =

3.

14. We find the induced emf from åinduced = – L ?I/?t = – (0.418 H)(4.50 A/s) = – 1.88 V. The negative sign indicates a direction opposite to the current. If we start at point b and add the potential changes, we get Vb + IR +  åinduced  = Va , or Vab = (5.00 A)(2.70 ) + 1.88 V = 15.4 V. 

R – Iincreasing b

+

å induced a

15. (a) For two inductors placed in series, the current through each inductor is the same. This current is also the current through the equivalent inductor, so the total emf is å = å1 + å2 – Lseries dI/dt = (– L1 dI/dt) + (– L2 dI/dt) = – (L1 + L2) dI/dt, which gives Lseries = L1 + L2 . (b) For two inductors placed in parallel, the potential difference across each inductor, which is the emf, is the same:    å = å1 = å2 = – L1 dI1/dt = – L2 dI2/dt = – Lparallel dI/dt.

Chapter 30, p. 3

The total current through the equivalent inductor is I = I1 + I2 , so we have dI/dt = dI1/dt + dI2/dt; – å/Lparallel = – å/L1 – å/L2 , which gives 1/Lparallel = (1/L1) + (1/L2), or 16. From the symmetry of the toroid, the magnetic field is circular. We apply Ampere’s law to a circular path to find the magnetic field inside the toroid: ı B · ds = 0Ienclosed; B2pr = 0NI, which gives B = 0NI/2pr. To find the magnetic flux through one turn of the toroid, we integrate over the rectangular cross-section. For a differential element, we choose a vertical strip at a radius r with width dr:  r 2  NI  NIh r 0  = B · dA = h dr = 0 ln r2 . 2r 2 1 r

Lparallel = L1L2/(L1 + L2).

dr r1 r r2

1

The flux through the entire toroidal winding is N times this, so the self-inductance is L = N/I = (0N2h/2p) ln(r2/r1). 17. The magnetic energy in the field is U = uBV = !(B2/0)¬pr2 = p(1.0510–2 m)2 =

15.9 J.

18. For the energy stored in the inductor we have U = !LI 2 = !(0.400 H)(9.0 A)2 = 16 J. 19. (a) For the energy densities we have uE = !Å0E2 = !(8.8510–12 F/m)(1.0104 V/m)2 = uB = !B2/0 = !(2.0 T)2/(4p10–7 T · m/A) =

4.410–4 J/m3;

1.6106 J/m3.

We see that uB » uE. (b) We find the magnitude of the electric field from uE = !Å0E2; 1.6106 J/m3 = !(8.8510–12 F/m)E2, which gives

E = 6.0108 V/m.

20. The magnetic field at the center of the loop is B = 0I/2R. The energy density of the magnetic field is uB = !B2/0 = 0I 2/8R2 = (4p10–7 T · m/A)(30 A)2/8(0.280 m)2 =

1.810–3 J/m3.

21. The magnetic field at the surface of the wire is B = 0I/2pR. The energy density of the magnetic field is uB = !B2/0 = 0I 2/8p2R2 = (4p10–7 T · m/A)(25 A)2/8p2(1.510–3 m)2 = If V is the potential along a length ¬ of the wire, the electric field is E = V/¬ = IR/¬ = I/A = I/pR2. The energy density of the electric field is uE = !Å0E2 = !Å0(I/pR2)2

4.4 J/m3.

h

Chapter 30, p. 4

= !(8.8510–12 F/m)[(25 A)(1.6810–8  · m)/p(1.510–3 m)2]2 =

22. From the symmetry of the toroid, the magnetic field is circular. We apply Ampere’s law to a circular path to find the magnetic field inside the toroid: ı B · ds = 0Ienclosed; B2pr = 0NI, which gives B = 0NI/2pr. The energy density of this field is u = !B2/0 = 0N2I 2/8p2r2. To find the total energy stored in the magnetic field, we integrate over the volume. For a differential element, we choose a ring at a radius r with width dr and height h:

U=

u dV =

r2 r1

0 N 2 I 2 2rh d r = 8 2r 2

1.610–14 J/m3.

dr r1 r r2

 0N 2I 2h l n r2 . r1 4

23. From Example 30–4, the magnetic field is B = 0I/2pr. The energy density of this field is u = !B2/0 = 0I 2/8p2r2. To find the total energy stored in the magnetic field, we integrate over the volume. For a differential element, we choose a ring at a radius r with width dr and length ¬:

U=

u dV =

r2 r1

 0I 2 2r d r = 8 2r 2

 0 I 2 l n r2 . r1 4

This is the same as found in Example 30–5. 24. For an LR circuit, we have I = Imax(1 – e – t/), which we can write as e – t/ = 1 – (I/Imax), or t/ = – ln[1 – (I/Imax)] = – ln[(Imax – I)/Imax]. (a) ta/ = – ln(0.10), which gives ta/ = 2.3. 4.6. (b) tb/ = – ln(0.010), which gives tb/ = 6.9. (c) tc/ = – ln(0.0010), which gives tc/ = 25. The potential difference across the resistor in Fig. 30–6 is VR = IR = ImaxRe – t/. We find the time to drop to 1.0 percent of its maximum value from 0.010 = e – t/, which gives t/ = 4.6. 26. Power is being dissipated as thermal energy in the resistor. At the instant the current is 0.20 A in Example 30–6, the rate at which energy is being dissipated is PR = I2R = (0.20 A)2(30 ) = 1.2 W. This is the difference between the 2.4 W delivered by the battery and the 1.2 W rate at which energy is being stored in the inductor.

h

Chapter 30, p. 5

In general, the current in the circuit is I = Imax(1 – e – t/) = (V0/R)(1 – e – t/). The power dissipated in the resistor is PR = I2R = (V02/R)(1 – e – t/)2 = (V02/R)(1 – 2e – t/ + e – 2t/). The rate at which energy is being stored in the inductor is PL = LI dI/dt = (LV02/R2)(1 – e – t/)(e – t//) = (V02/R)(e – t/ – e – 2t/). The sum is PR + PL = (V02/R)(1 – 2e – t/ + e – 2t/ + e – t/ – e – 2t/) = (V02/R)(1 – e – t/) = IV0 = PB. Thus energy is conserved at any time. 27. At t = 0 there is no voltage drop across the resistor, so we have V0 = L(dI/dt)0 , or (dI/dt)0 = V0/L. The maximum value of the current is reached after a long time, when there is no voltage across the inductor: V0 = ImaxR, or Imax = V0/R. We find the time to reach maximum if the initial rate were maintained from V0/R = (dI/dt)0t = (V0/L)t, which gives t = L/R = . 28. (a) For an LR circuit, we have I = Imax(1 – e – t/);

! = 1 – e – t/, or – t/ = – (2.56 ms)/ = ln !, which gives  =

(b) We find the resistance from  = L/R; 3.6910–3 s = (310 H)/R, which gives R = 8.39104  =

3.69 ms.

83.9 k.

29. (a) The current in the circuit is I = Imax(1 – e – t/) = (V/R)(1 – e – t/). The energy stored in the inductor is UL = !LI2 = !(LV2/R2)(1 – e – t/)2 = (LV2/2R2)(1 – 2e – t/ + e – 2t/). (b) We find the time for the stored energy to reach 99 percent of the maximum from 0.99 = (1 – e – t/)2 =, which gives e – t/ = 0.00501, or t/ = 5.3. 30. (a) Immediately after the switch is closed, the induced emf in L R3 the inductor is maximum while the current in the inductor is zero, so we have I3 I3 = 0. 2 For loop 1, we have R2 I1 = I2 = å/(R1 + R2). a (b) After a long time, the currents will be constant, and there will be I2 no induced emf in the inductor. For the junction at point a, we have 1 I1 = I2 + I3 . For loop 1, we have å – I1R1 – I2R2 = 0. S For loop 2, we have + I2R2 – I3R3 = 0, or I2 = (R3/R2)I3. When we use this in the junction equation, we get I1 = [1 + (R3/R2)]I3 = [(R2 + R3)/R2]I3. From the loop 1 equation we get å = [(R2 + R3)/R2]I3R1 + (R3/R2)I3R2 , which gives I3 = R2å/(R1R2 + R1R3 + R2R3).

b R1 I1

å

Chapter 30, p. 6

Then I1 = (R2 + R3)å/(R1R2 + R1R3 + R2R3), I2 = R3å/(R1R2 + R1R3 + R2R3). (c) Immediately after the switch is opened, the current in R1 is zero, so we have I1 = 0, and the current I3 starts to decay in the LR circuit. Thus we have – I2 = I3 = R2å/(R1R2 + R1R3 + R2R3). (d) After a long time, all currents will be zero: I1 = I2 = I3 = 0.

31. (a) The resonant frequency is given by f02 = (1/2p)2(1/LC). When we form the ratio for the two stations, we get (f02/f01)2 = C1/C2 ; (1600 kHz/550 kHz)2 = (1800 pF)/C2 , which gives C2 = 213 pF. (b) We find the inductance from the first frequency: f01 = (1/2p)(1/LC1)1/2; 550103 Hz = (1/2p)[1/L(180010–12 F)]1/2, which gives L = 4.6510–5 H =

46.5 H.

32. (a) To have maximum current at t = 0, we can write the sinusoidal variation as I = I0 cos t. Because there is a 90° phase difference between the current and the charge, we have Q = Q0 sin t. To determine Q0 we differentiate: I = dQ/dt = Q0 cos t, so Q0 = I0/: Q = (I0/) sin t. (b) We could set up these initial conditions by having an inductor and a switch connected to a battery. When the switch is thrown to disconnect the battery and connect the inductor to a capacitor, the initial charge on the capacitor is zero. 33. We reduce the units to their basic elements. From Q = CV, for the farad we have F = C/V = C2/J = C2/(kg · m2/s2) = C2 · s2/kg · m2. From L = /I = BA/I = FA/QvI, we have H = N · m2/(C · m · C/s2) = (kg · m/s2) · m2/(C2 · m/s2) = kg · m2/C2. Thus LC has the units HF = [kg · m2/C2][C2 · s2/kg · m2] = s2, so 1/(LC)1/2 has the units of s–1. 34. (a) The resonant frequency is f0 = (1/2p)(1/LC)1/2 = (1/2p)[1/(17510–3 H)(76010–12 F)]1/2 = 1.38104 Hz = 13.8 kHz. (b) The maximum charge on the capacitor is Q0 = CV, so the peak value of the current is I0 = Q0 = CV = (76010–12 F)(135 V)2p(1.38104 Hz) = 8.9010–3 A = 8.90 mA. (c) The maximum energy stored in the inductor is ULmax = !LI02 = !(17510–3 H)(8.9010–3 A)2 = 6.9310–6 J. 35. (a) The initial energy is the energy stored in the capacitor at t = 0: U0 = !Q02/C. When the capacitor has half the energy, we have

Chapter 30, p. 7

UC/U0 = Q2/Q02 = !, which gives Q = Q0/v2. (b) The charge on the capacitor is Q = Q0 cos t = Q0/v2. This gives t = p/4; (2p/T)t = p/4, which gives t = T/8. 36. If R « (4L/C)1/2, the motion is underdamped. The charge on the capacitor is Q = Q0e – Rt/2L cos (t + ). The energy stored in the capacitor and inductor can be expressed in terms of the amplitude of the cosine function: U = Q2/2C = Q02e – Rt/L/2C = U0e – Rt/L. When this is half its initial value, we have !U0 = U0e – Rt/L, which gives Rt/L = ln 2, or t = (L/R) ln 2. 37. We assume underdamping, with  ˜ 0 = 1/(LC)1/2, and T = 2p/0 = 2p(LC)1/2. The charge on the capacitor is Q = Q0e – Rt/2L cos (t + ). The energy stored in the capacitor and inductor can be expressed in terms of the amplitude of the cosine function: U = Q2/2C = Q02e – Rt/L/2C = U0e – Rt/L. In one period the energy is reduced by 5.5 percent, so we have 0.945U0 = U0e – RT/L, which gives ln (1/0.945) = RT/L = 2pR(C/L)1/2; ln (1/0.945) = 2pR[(1.0010–6 F)/(6510–3 H)]1/2, which gives R = 2.30 . We can check to see if we have underdamping: R2 = (2.30 )2 = 5.3 2; 4L/C = 4(6510–3 H)/(1.0010–6 F) = 2.6105 2. Thus R2 « 4L/C. 38. (a) The differential equation for the charge on the capacitor is 2 d Q dQ Q L 2 +R + = 0. C dt dt Because I = – dQ/dt, when we differentiate we get 2 2 – L d I2 – R dI – I = 0, or L d I2 + R dI + I = 0. dt C dt C dt dt (b) If Q = Q0 at t = 0, we have Q = Q0e – Rt/2L cos t. Because the differential equation for the current has the same form as the one for the charge, the solution for the current in an underdamped circuit is I = I0e – Rt/2L cos (t + ). At t = 0, I = 0, so we can find the phase constant: 0 = I0 cos , which gives  = ± p/2. Because I = – dQ/dt, we choose  = – p/2, and write the current as I = I0e – Rt/2L sin t. (c) This is the same result as in Example 30–8. (d) If R ˜ (L/C)1/2, the circuit would be critically or over damped. The solution would no longer be sinusoidal, but the charge and current would decay to zero.

Chapter 30, p. 8

39. The charge is given by Q = Q0e – Rt/2L cos t. We find the current by differentiating: I = – dQ/dt = – Q0(– R/2L)e – Rt/2L cos t + Q0 e – Rt/2L sin t = + Q0 e – Rt/2L [ sin t + (R/2L) cos t]. We form the triangle shown so we can express the terms in the bracket as a combination of trig functions:

[ 2 + (R/2 L)2 ]1/2 R/2 L

 

I = Q0 e – Rt/2L [2 + (R/2L)2]1/2({/[2 + (R/2L)2]1/2} sin t + {(R/2L)/[2 + (R/2L)2]1/2} cos t) = Q0 e – Rt/2L [2 + (R/2L)2]1/2(cos  sin t + sin  cos t), where tan  = R/2L. If we use the expression for , we find [2 + (R/2L)2]1/2 = [(1/LC) – (R/2L)2 + (R/2L)2]1/2 = 1/(LC)1/2. Thus we have I = [Q0/(LC)1/2] e – Rt/2L sin (t + ), with  = tan–1 (R/2L).

40. If R « (4L/C)1/2, the motion is underdamped, with  = [(1/LC) – (R/2L)2]1/2 ˜ 1/(LC)1/2 = 0. If I = 0 when t = 0, the charge on the capacitor is Q = Q0e – Rt/2L cos (t + ). We find the current by differentiating: I = – dQ/dt = – Q0(– R/2L)e – Rt/2L cos (t + ) + Q0 e – Rt/2L sin (t + ) = + Q0 e – Rt/2L [ sin (t + ) + (R/2L) cos (t + )]. At t = 0 we have 0 = Q0[ sin  + (R/2L) cos ], or cot  = – (2L/R) = – (4L/R2C)1/2, or  = – cot–1 (4L/R2C)1/2. 41. The frequency of the LC circuit is 0 = 1/(LC)1/2. When the resistance is added the frequency becomes  = [(1/LC) – (R2/4L2)]1/2 = (1/(LC)1/2)[1 – (R2C/4L)]1/2 = 0[1 – (R2C/4L)]1/2. Thus we see that the frequency will decrease. If we write the new frequency as  = (1 – x)0, we have 1 – (R2C/4L) = (1 – x)2 ˜ 1 – 2x, or R2(180010–12 F)/4(30010–3 H) = 2(0.1010–2), which gives R = 1.15103  =

1.15 k.

Chapter 30, p. 9

42. (a) If a battery is added to the LRC circuit, a loop equation gives V0 – (Q/C) – IR – L dI/dt = 0. L We see from the direction of the current in the circuit diagram that I = dQ/dt, so this becomes V0 I (L d2Q/dt2) + (R dQ/dt) + Q/C = V0 . _ + From the theory of differential equations, the general solution is the sum of the solution to the equation with V0 = 0 and a particular solution. By inspection a particular solution is Q = CV0 . To find the solution when V0 = 0, we let Q = a e bt. When we use this in the differential equation, we get Lb2a e bt + Rba e bt + (a/C) e bt = 0, or Lb2 + Rb + (1/C) = 0. This is a quadratic equation for b, with the solutions b = (1/2L){– R ± [R2 – (4L/C)]1/2} = – (R/2L) ± (R/2L)[1 – (4L/R2C)]1/2 = (R/2L){– 1 ± [1 – (4L/R2C)]1/2}. 2 When R » 4L/C, this becomes b ˜ (R/2L){– 1 ± [1 – (2L/R2C)]} = – 1/RC, and – (R/2L)[2 – (2L/R2C)] ˜ – R/L. The general solution for the charge is Q = a1 e – t/RC + a2 e –Rt/L + CV0 . The current solution is I = dQ/dt = – (a1/RC) e – t/RC – (a2R/L) e – Rt/L. At t = 0, Q = 0, and I = 0, which allows us to determine a1 and a2: 0 = a1 + a2 + CV0 ; 0 = – (a1/RC) – (a2R/L), which gives a2 = – (L/R2C)a1. When we use this in the previous equation, we get a1 = – CV0/[1 – (L/R2C)] ˜ – CV0 . Thus a2 = – (L/R2C){– CV0/[1 – (L/R2C)]} = LV0/[R2 – (L/C)] ˜ LV0/R2. Our result for the current is I = (CV0/RC) e – t/RC – [(LV0/R2)R/L] e – Rt/L

R C +

–

Chapter 30, p. 10

= (V0/R)(e – t/RC – e – Rt/L). (b) We scale the current by plotting IR/V0 and scale the time by plotting Rt/L. (c) In an RC circuit the charge is Q = CV0(1 – e – t/RC), so the current is I = dQ/dt = CV0(1/RC)e – t/RC = (V0/R)e – t/RC. In an RC circuit the current starts at maximum and decays. For the LRC circuit there is an additional term of opposite sign. The current starts at zero but the magnitude of the additional term decreases rapidly, so there is initial rapid growth and then decay. Eventually the current coincides with the result for an RC circuit.

IR/ V 0 1.0

RC circuit

0.8 LRC circuit

0.6

R 2 = 25(4L/ C)

0.4 0.2 0

0

5

43. We use the inductance of a solenoid: L = 0AN2/¬ = (4p10–7 T · m/A)p(1.2510–2 m)2(3000 turns)2/(0.282 m) = 1.9710–2 H = If we form the ratio of inductances for the two conditions, we have L2/L1 = (/0)(N2/N1)2; 1 = (1000)[N2/(3000 turns)]2, which gives N2 = 95 turns. 44. The initial energy stored in the inductor is U0 = !LI 02 = !(60.010–3 H)(50.010–3 A)2 = 7.5010–5 J. For the increase in energy, we have U/U0 = (I /I 0)2; 10 = (I /50.0 mA)2, which gives I = 158 mA. We find the time from ?I /?t = 100 mA/s = (158 mA – 50.0 mA)/?t, which gives ?t =

Rt / L

10

20 mH.

1.08 s.

45. (a) The frequency of oscillation is the resonance frequency. We find the inductance from f0 = (1/2p)(1/LC)1/2; 20103 Hz = (1/2p)[1/L(300010–12 F)]1/2, which gives L = 2.1110–2 H = 21 mH. (b) The energy initially stored in the capacitor will oscillate between the capacitor and the inductor. We find the maximum current, when all of the energy is stored in the inductor, by equating the maximum energies: !CV02 = !LImax2; (300010–12 F)(120 V)2 = (2.1110–2 H)Imax2, which gives Imax = 4.5210–2 A = 45 mA. (c) The maximum energy stored in the inductor is ULmax = !LImax2 = !(2.1110–2 H)(4.5210–2 A)2 = 2.210–5 J. 46. We find the mutual inductance by finding the linkage of the magnetic field of loop 1 with loop 2. Because the loops are small, we can assume the field from loop 1 at loop 2 is constant and equal to the magnitude at the center of the loop: B1 = 0I1r12/2(r12 + ¬2)3/2. Because r1 « ¬, this becomes B1 ˜ 0I1r12/2¬3. The flux linking with loop 2 is 21 = B1pr22 cos , so the mutual inductance is

Chapter 30, p. 11

M = 21/I1 =

(0pr12r22/2¬3) cos .

47. We use the inductance of a solenoid: L = 0AN2/¬; 2510–3 H = (4p10–7 T · m/A)p(1.110–2 m)2N2/(0.170 m), which gives N = If we form the ratio of inductances for the two conditions, we have L2/L1 = (/0)(N2/N1)2; 1 = (103)[N2/(3.0103 turns)]2, which gives N2 = 95 turns.

3.0103 turns.

48. (a) From the symmetry of the toroid, the magnetic field is circular. We apply Ampere’s law to a circular path along the axis of the toroid to find the magnetic field inside the toroid: ı B · ds = 0Ienclosed; B2pr0 = 0NI, which gives B = 0NI/2pr0 . If we assume that the field is uniform inside the toroid (because the field varies with the distance from the center of the toroid, this is not actually true, but will be a good approximation if r0 » r), the magnetic flux through one turn of the toroid is  ˜ B#pd2 = 0NId2/8r0 . Thus the self-inductance is L = N/I = 0N2d2/8r0 . The length of the toroid is ¬ = 2pr0 , and the cross-sectional area is #pd2, so we have L = 0AN2/¬, which is consistent with that of a solenoid, as it should be. (b) For the given data we have L = 0N2d2/8r0 = (4p10–7 T · m/A)(550 turns)2(2.010–2 m)2/8(0.25 m) = 7.610–5 H = 76 H. 49. We find the inductance by finding the flux linking the two wires, which we assume can be treated as two long wires. Because the currents are in opposite directions, the magnetic field a distance x from one wire is B = (0/2p)(I/x) + (0/2p)I/(¬ – x) = (0/2p)I¬/x(¬ – x). To find the magnetic flux through the area between a length D of the wires, we integrate over the rectangular cross-section. For a differential element,

I B x ¬ I

Chapter 30, p. 12

we choose a strip at a distance x with width dx parallel to the wires:  –r 0I  I D – r dx D dx = 0  = B · dA = 2 r x( – x) 2x( – x) r

 0I D 1 – x – r = – 0ID ln 2 – ln x 2 r Thus the inductance per unit length is L/D = /ID = (0/p) ln[(¬ – r)/r]. =

r – ln –r

–r r

=

 0ID  ln

–r . r

50. The potential difference across the coil is V = L dI/dt + IR. For the two conditions we have 2.55 V = L(34010–3 A/s) + (36010–3 A)R; 1.82 V = L(– 18010–3 A/s) + (42010–3 A)R. When we solve these two equations for the two unknowns, we get L = 2.00 H, R = 5.19 .

51. (a) For two coils placed in series, the current through each coil is the same and is also the current through the equivalent inductor. The total induced emf is å = å1 + å2 + å21 + å12 – Lseries dI/dt = (– L1 dI/dt) + (– L2 dI/dt) ± [(– M dI/dt) + (– M dI/dt)] = – (L1 + L2 ± 2M) dI/dt, which gives Lseries = L1 + L2 ± 2M. The sign of the mutual inductance term is determined by the orientation of one winding with respect to the other. The upper sign is used when the emf from the mutual inductance is in the same sense as the emf produced by the self-inductance. The lower sign is used when the emf from the mutual inductance is in the opposite sense as the emf produced by the self-inductance. (b) To reduce the value of M the flux linkage must be reduced. If the separation cannot be increased, the coils can be oriented so the magnetic field of one coil is parallel to the other coil. This can be achieved by positioning one coil perpendicular to the other. (c) For two coils placed in parallel, the potential difference across each coil, which is the emf, is the same. With no mutual inductance we have    å = å1 = å2 = – L1 dI1/dt = – L2 dI2/dt = – Lparallel dI/dt. The total current through the equivalent inductor is I = I1 + I2 , so we have dI/dt = dI1/dt + dI2/dt; – å/Lparallel = – å/L1 – å/L2 , which gives 1/Lparallel = (1/L1) + (1/L2), or Lparallel = L1L2/(L1 + L2). If we cannot ignore the mutual inductance, we have å = å1 = – L1 dI1/dt — M dI2/dt = å2 = – L2 dI2/dt — M dI1/dt. When these equations are combined, we get dI1/dt = – (L2 — M)å/(L1L2 – M2), and dI2/dt = – (L1 — M)å/(L1L2 – M2). Because I = I1 + I2 , we have dI/dt = dI1/dt + dI2/dt;

Chapter 30, p. 13

– å/Lparallel = – (L2 — M)å/(L1L2 – M2) – (L1 — M)å/(L1L2 – M2), which gives Lparallel = (L1L2 – M2)/(L1 + L2 — 2M). The sign of the mutual inductance term in the denominator is determined by the orientation of one winding with respect to the other. The upper sign is used when the emf from the mutual inductance is in the same sense as the emf produced by the self-inductance. The lower sign is used when the emf from the mutual inductance is in the opposite sense as the emf produced by the self-inductance. 52. The magnetic energy in the field is U = uBV = !(B2/0)h4pr2 = 4p(6.38106 m)2 =

5.11015 J.

53. For lightly damped motion, R « (4L/C)1/2,  ˜  = 1/(LC)1/2, and T ˜ 2p/ = 2p(LC)1/2. The charge on the capacitor is Q = Q0e – Rt/2L cos (t + ). The energy stored in the capacitor and inductor can be expressed in terms of the amplitude of the cosine function: U = Q2/2C = Q02e – Rt/L/2C = U0e – Rt/L. In one cycle the exponent is RT/L = 2pR/L = 2pR(C/L)1/2, which for lightly damped motion is « 1. If we use the energy lost from t = 0 to t = T, we have ?U/U0 = (U0 – U)/U0 = 1 – e – RT/L ˜ 1 – (1 – RT/L) = RT/L = 2pR/L = 2p/Q, where Q = L/R. 54. (a) We use the inductance of a solenoid: L = 0AN2/¬. Because they are tightly wound, the number of turns is determined by the diameter of the wire: N = ¬/d. If we form the ratio of inductances for the two conditions, we have L1/L2 = (N1/N2)2 = (d2/d1)2 = 22 = 4. (b) The length of wire used for the turns is ¬wire = N(pD), where D is the diameter of the solenoid. Thus for the ratio of resistances, we have R1/R2 = (¬wire1/¬wire2)(d2/d1)2 = (N1/N2)(d2/d1)2 = (d2/d1)3. For the ratio of the time constants, we get 1/2 = (L1/L2)(R2/R1) = (L1/L2)(d1/d2)3 = (4)(!)3 = !. 55. (a) For lightly damped motion, R « (4L/C)1/2,  ˜  = 1/(LC)1/2, T = 2p/ ˜ 2p(LC)1/2, and R/L ˜ R(C/L)1/2 « 1. The charge on the capacitor is Q = Q0e – Rt/2L cos (t + ). We find the current by differentiating: I = – dQ/dt = – Q0(– R/2L)e – Rt/2L cos (t + ) + Q0 e – Rt/2L sin (t + ) = + Q0 e – Rt/2L [ sin (t + ) + (R/2L) cos (t + )].

Because the values for the sine and cosine functions range over ± 1, and  » R/2L, we neglect the second term to get

Chapter 30, p. 14

I ˜ Q0 e – Rt/2L  sin (t + ). The total energy is U = UE + UB = !(Q2/C) + !LI2 = !(Q02/C)e – Rt/L cos2 (t + ) + !LQ022 e – Rt/L sin2 (t + ). When we use 2 ˜ 1/LC, we get U = !(Q02/C)e – Rt/L [cos2 (t + ) + sin2 (t + )] = !(Q02/C)e – Rt/L. (b) The rate at which the stored energy changes is dU/dt = !(Q02/C)(– R/L)e – Rt/L = – !(Q02R/LC)e – Rt/L. The rate at which thermal energy is produced in the resistor is I2R = Q022R e – Rt/L sin2 (t + ) = (Q02R/LC) e – Rt/L sin2 (t + ). For lightly damped motion, the amplitude does not change significantly over a period, while the average value for sin2 (t + ) is !. Thus we get I2R = !(Q02R/LC) e – Rt/L = – dU/dt. Thus the loss of stored energy becomes the thermal energy produced in the resistor. 56. Because we need a final current, we use an inductor in series with the device for an LR circuit. The current in the circuit is I = Imax(1 – e – Rt/L), or e – Rt/L = 1 – (I/Imax), which can be written as – Rt/L = ln [1 – (I/Imax)]. For the limiting current after 100 s, we have – (150 )(10010–6 s)/L = ln [1 – (7.5 mA/50 mA)], which gives L = 9.210–2 H = 92 mH. We note that a larger value of L will cause the current to rise more slowly, so we need an inductor with L = 92 mH.

57. From the circuit we see that Vin = IR + L dI/dt, or RI/L + dI/dt = Vin /L. We multiply both sides by the integrating factor e Rt/L: e Rt/LRI/L + e Rt/L dI/dt = d(I e Rt/L)/dt = e Rt/LVin /L. If we assume I = 0 when t = 0, when we integrate we get t

t

d I e Rt / L = I eRt / L = 1 eRt / LV in dt . L 0 0 If t « L/R, the exponent is small, e Rt/L ˜ 1 and e Rt/L ˜ 1, so we have t  I = 1 V in dt . L 0 Thus the output voltage is t Vout = IR = R V in dt . L 0 This is the area under the curve when Vin is plotted against Rt/L. For the square wave input, we get

L

V in

I

R

V out

Chapter 30, p. 15

V in t

V out t

Chapter 31 page 1

CHAPTER 31 – AC Circuits 1.

(a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(7.210–6 F) = 3.7102 . (b) For the new frequency we have XC2 = 1/2pf2C = 1/2p(1.0106 Hz)(7.210–6 F) = 2.210–2 .

2.

We find the frequency from XL = L = 2pfL; 660  = 2pf(22.010–3 H), which gives f = 4.77103 Hz =

3.

4.

We find the frequency from XC = 1/2pfC; 6.70103  = 1/2pf(2.4010–6 F), which gives f =

4.77 kHz.

9.90 Hz.

The impedance is Z = XC = 1/2pfC. X C (k)

3.0

1.5

0 5.

f (Hz) 0

1000

500

The impedance is Z = XL = L = 2pfL. X L ()

300

150

0 0

5000

f (Hz) 10,000

Chapter 31 page 2

6.

We find the impedance from Z = XL = L = 2pfL = 2p(33.3 kHz)(36.010–3 H) = For the rms current we have Irms = Vrms/XL = (750 V)/(7.53 k) = 99.6 mA.

7.53 k.

7.

If there is no current in the secondary, there will be no induced emf from the mutual inductance. We find the impedance from Z = XL = Vrms/Irms = (110 V)/(2.2 A) = 50 . We find the inductance from XL = 2pfL; 50  = 2p(60 Hz)L, which gives L = 0.13 H.

8.

(a) We find the impedance from Z = XC = 1/2pfC = 1/2p(600 Hz)(0.03610–6 F) = 7.37103  = 7.4 k. (b) We find the peak value of the current from Ipeak = v2 Irms = v2(Vrms /Z) = v2(22 kV)/(7.37 k) = 4.2 A. The frequency of the current will be the frequency of the line: 600 Hz.

9.

(a) If the voltage is V = V0 sin t, the charge on the capacitor is Q = CV = CV0 sin t. Thus the current is I = dQ/dt = CV0 cos t = CV0 sin (t + 90°). (b) If the voltage is V = V0 sin t, for the circuit we have V = V0 sin t = L dI/dt, or dI/dt = (V0 /L) sin t. When we integrate this we get I = – (V0/L) cos t = (V0/L) sin (t – 90°).

10. The average power dissipation is P = Irms2R = !Ipeak2R = !(1.80 A)2(260 ) =

ICmax

90°

421 W.

11. Because the capacitor and resistor are in parallel, their currents are IC = V/XC , and IR = V/R. The total current is I = IC + IR . (a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(0.3510–6 F) = 7.58103  = 7.58 k. For the fraction of current that passes through C, we have fraction1 = IC1/(IC1 + IR) = (1/XC1)/[(1/XC1) + (1/R)] = R/(R + XC1) = (0.400 k)/(0.400 k + 7.58 k) = 0.050 = 5.0%. (b) The reactance of the capacitor is XC2 = 1/2pf2C = 1/2p(60,000 Hz)(0.3510–6 F) = 7.58 . For the fraction of current that passes through C, we have fraction2 = IC2/(IC2 + IR) = (1/XC2)/[(1/XC2) + (1/R)] = R/(R + XC2) = (400 )/(400  + 7.58 ) = 0.98 = 98%. Thus most of the high-frequency current passes through the capacitor.

V0

t 90°

ILmax

Chapter 31 page 3

12. (a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(6.810–6 F) = 390  = 0.39 k. The impedance of the circuit is Z1 = (R2 + XC12)1/2 = [(1.20 k)2 + (0.39 k)2]1/2 = 1.26 k. (b) The reactance of the capacitor is XC2 = 1/2pf2C = 1/2p(60,000 Hz)(6.810–6 F) = 0.39  = 0.00039 k. The impedance of the circuit is Z2 = (R2 + XC22)1/2 = [(1.20 k)2 + (0.00039 k)2]1/2 = 1.20 k. 13. (a) The reactance of the inductor is XL1 = 2pf1L = 2p(50 Hz)(26.010–3 H) = 8.17 . The impedance of the circuit is Z1 = (R2 + XL12)1/2 = [(9.0 k)2 + (0.00817 k)2]1/2 = 9.0 k. (b) The reactance of the inductor is XL2 = 2pf2L = 2p(30 kHz)(26.010–3 H) = 4.9 k. The impedance of the circuit is Z2 = (R2 + XL22)1/2 = [(9.0 k)2 + (4.9 k)2]1/2 = 10.2 k. 14. We find the impedance from Z = Vrms/Irms = (120 V)/(70 mA) =

1.7 k.

15. (a) The reactance of the capacitor is XC = 1/2pfC = 1/2p(60 Hz)(0.8010–6 F) = 3.32103  = 3.32 k. The impedance of the circuit is Z = (R2 + XC2)1/2 = [(6.0 k)2 + (3.32 k)2]1/2 = 6.86 k. The rms current is Irms = Vrms/Z = (120 V)/(6.86 k) = 18 mA. (b) We find the phase angle from cos  = R/Z = (6.0 k)/(6.86 k) = 0.875. – 29°. In an RC circuit, the current leads the voltage, so  = (c) The power dissipated is P = Irms2R = (17.510–3 A)2(6.0103 ) = 1.8 W. (d) The rms readings across the elements are VR = IrmsR = (17.5 mA)(6.0 k) = 105 V; VC = IrmsXC = (17.5 mA)(3.32 k) = 58 V. Note that, because the maximal voltages occur at different times, the two readings do not add to the applied voltage of 120 V.

Chapter 31 page 4

16. (a) The reactance of the inductor is XL = 2pfL = 2p(60 Hz)(0.250 H) = 94.2 . The impedance of the circuit is Z = (R2 + XL2)1/2 = [(765 )2 + (94.2 )2]1/2 = 770.8 . The rms current is Irms = Vrms/Z = (120 V)/(770.8 ) = 0.156 A. (b) We find the phase angle from cos  = R/Z = (765 )/(770.8 ) = 0.993. + 7.02°. In an RL circuit, the current lags the voltage, so  = (c) The power dissipated is P = Irms2R = (0.156 A)2(765 ) = 18.5 W. (d) The rms readings across the elements are VR = IrmsR = (0.156 A)(765 ) = 119 V; VL = IrmsXL = (0.156 A)(94.2 ) = 15 V. Note that, because the maximal voltages occur at different times, the two readings do not add to the applied voltage of 120 V. 17. The reactances and impedance in the circuit are XL = 2pfL = 2p(60 Hz)(3510–3 H) = 13.2 . XC = 1/2pfC = 1/2p(60 Hz)(2010–6 F) = 133 . Z = [R2 + (XL – XC)2]1/2 = [(2.0 )2 + (13.2  – 133 )2]1/2 = 119 . (a) The rms current is Irms = Vrms/Z = (45 V)/(119 ) = 0.38 A. (b) We find the phase angle from cos  = R/Z = (2.0 )/(119 ) = 0.0168. – 89°. Because XC > XL , the current leads the voltage, so  = (c) The power dissipated is P = Irms2R = (0.38 A)2(2.0 ) = 0.29 W. 18. For the current and voltage to be in phase, the net reactance of the capacitor and inductor must be zero. Thus we have XL = XC; 2pfL = 1/2pfC, or f = (1/2p)(1/LC)1/2; 4.9 F. 360 Hz = (1/2p)[1/(4010–3 H)C]1/2, which gives C = 4.910–6 F = 19. We find the resistance from Z = [R2 + XL2]1/2 ; 335  = [R2 + (45.5 )2]1/2, which gives R =

332 .

20. The power dissipated is the power delivered to the circuit: P = IV = I0V0 sin t sin (t + ). If we expand the second term, we get P = I0V0 sin t (sin t cos  + cos t sin ) = I0V0 (cos  sin2 t + sin  sin t cos t ) = I0V0(cos  sin2 t + sin  ! sin 2t ). When we average over a cycle, the average of a sin2 function = !, and the average of a sine function = 0.

Chapter 31 page 5

Thus we have P = !I0V0 cos .

21. (a) The average value over a cycle is T

V=

0

V 0 si n t dt T 0

2

=

0

V 0 si n t d(t ) 2

dt

0

=

V 0 ( – cos t ) 2

d( t)

2 0

=

0.

(b) We assume the half-cycle is from zero to !T: T/ 2

V 1/ 2 =

0

V 0 si n t d t T/ 2 0

dt



=

0

V 0 si n  t d ( t)  0

=

V 0 ( – cos t)

d ( t)



 0

=

2V .  0

When we compare this to Vrms , we get 2V V 1/ 2 =  0 = 2 2 , so V 1/ 2 = 2 2 V rms.   1 V V rms 0 2 22. We find the rms current from P = Irms2R; 15.5 W = Irms2(150 ), which gives Irms = 0.321 A. We find the impedance from Z = Vrms/Irms = (60.0 V)/(0.321 A) = 187 . From this we can find the reactance of the coil: Z2 = R2 + XL2; (187 )2 = (150 )2 + XL2, which gives XL = 112 . We find the frequency that will produce this reactance from XL = 2pfL ; 112  = 2pf(0.040 H), which gives f = 446 Hz. 23. The reactance of the capacitor is XC = 1/2pfC = 1/2p(10.0103 Hz)(500010–12 F) = 3.18103  = 3.18 k. The reactance of the inductor is XL = 2pfL = 2p(10.0 kHz)(0.0320 H) = 2.01 k. The impedance of the circuit is Z = [R2 + (XL – XC)2]1/2 = [(8.70 k)2 + (2.01 k – 3.18 k)2]1/2 = 8.78 k. We find the phase angle from tan  = (XL – XC)/R = (2.01 k – 3.18 k)/(8.70 k) = – 0.134, so  = – 7.66°. The rms current is Irms = Vrms/Z = (800 V)/(8.78 k) = 91.1 mA. 24. The resonant frequency is f0 = (1/2p)(1/LC)1/2 = (1/2p)[1/(26.010–6 H)(320010–12 F)]1/2 = 25. The resonant frequency is f0 = (1/2p)(1/LC)1/2 = (1/2p)[1/(30.010–3 H)(12.010–6 F)]1/2 =

5.52105 Hz.

265 Hz.

Chapter 31 page 6

At the resonant frequency, Z = R, so the rate at which energy is supplied by the generator is P = Irms2R = Vrms2/R = (90.0 V)2/(25.0 ) = 324 W.

26. (a) We find the capacitance from the resonant frequency: f0 = (1/2p)(1/LC)1/2; 33.0103 Hz = (1/2p)[1/(4.1510–3 H)C]1/2, which gives C = 5.6010–9 F = (b) At resonance the impedance is the resistance, so the current is I0 = V0/R = (136 V)/(220 ) = 0.618A.

5.60 nF.

27. We find the capacitance from the new resonant frequency: f0 = (1/2p)(1/LC)1/2; 2(33.0103 Hz) = (1/2p)[1/(4.1510–3 H)C]1/2, which gives C = 1.4010–9 F. At the applied frequency the reactances are XL = 2pfL = 2p(33.0103 Hz)(4.1510–3 H) = 860  = 0.860 k. XC = 1/2pfC = 1/2p(33.0103 Hz)(1.4010–9 F) = 3.44103  = 3.44 k. The impedance is Z = [R2 + (XL – XC)2]1/2 = [(0.220 k)2 + (0.860 k – 3.44 k)2]1/2 = 2.59 k. The peak current is I0 = V0/Z = (136 V)/(2.59 k) = 52.5 mA. 28. Because the circuit is in resonance, we find the inductance from f0 = (1/2p)(1/LC)1/2; 18.0103 Hz = (1/2p)[1/L(22010–6 F)]1/2, which gives L = 3.5510–7 H. If r is the radius of the coil, the number of turns is N = ¬wire/2pr. If d is the diameter of the wire, for closely-wound turns, the length of the coil is ¬ = Nd. Thus the inductance of the coil is L = 0AN2/¬ = 0pr2(¬wire/2pr)2/Nd = 0¬wire2/4pNd; 3.5510–7 H = (4p10–7 T · m/A)(12.0 m)2/4pN(1.110–3 m), which gives N =

3.68104 turns.

29. (a) We can express the charge amplitude in terms of the impedance: Q0 = CVC0 = CXCI0 = (1/)(V0/Z) = V0/[R2 + (L – 1/C)2]1/2 = V0/[(R)2 + (2L – 1/C)2]1/2. (b) We find the frequency to make Q0 a maximum by setting the first derivative equal to zero: dQ0/d = V0(– !)[2R2 + 2(2L – 1/C)(2L)]/[(R)2 + (2L – 1/C)2]3/2 = 0, or R2 + 2L(2L – 1/C) = 0, which gives R2/2L = (1/C) – 2L, or 2 = [(1/LC) – (R2/2L2)]. (c) For the forced harmonic oscillator, the amplitude is A0 = F0/m[(2 – 02)2 + b22/m2]1/2 = F0/[(b)2 + (2m – k)2]1/2. The frequency that produces the maximum amplitude is 2 = [(k/m) – (b2/2m2)]. This has the same form as the result for an LRC circuit, with k  1/C, m  L, b  R.

Chapter 31 page 7

30. The current in an LRC circuit is I = V0/Z, so the maximum occurs for Z = Zmin = R: I0 = V0/R, with 0 = 1/(LC)1/2. To get I = !I0 , we have Z = 2Zmin , or Z2 = 4R2 = R2 + (L – 1/C)2, which gives 3R2 = (L – 1/C)2, or v3R = ± (L – 1/C), which is a quadratic equation for : ± L2 – v3R — 1/C = 0, or L2 — v3R – 1/C = 0, with solutions ( must be positive)  = ± v3R/2L + [(3R2/4L2) + (1/LC)]1/2. Thus the width of the peak is ? = + – – = v3R/L. 31. (a) The average power dissipated is P = IV = Irms2Z cos  = Irms2R = !I02R. If we use I0 = V0/Z, we get P = V02R/2Z2 = V02R/2[R2 + (L – 1/C)2]. (b) We find the frequency to make P a maximum by setting the first derivative equal to zero: 2 d P dZ = 0, or equivalently, = 0; d d 2 d Z = 2 L – 1 L + 12 = 0. d C  C Because (L + 1/2C) > 0, we have L = 1/C, or 2 = 1/LC. (c) The peak for the average power, P0 = V02R/2Z02, occurs when Z0 = R. To get P = V02R/2Z2 = !P0 = V02R/4Z02, we have Z2 = R2 + (L – 1/C)2 = 2Z02 = 2R2, or R = ± (L – 1/C), which is a quadratic equation for : ± L2 – R — 1/C = 0, or L2 — R – 1/C = 0, with solutions ( must be positive)  = ± R/2L + [(R2/4L2) + (1/LC)]1/2. Thus the width of the peak is ? = + – – = R/L. 32. Because the loudspeaker is connected to the secondary side of the transformer, we have ZP/ZS = (NP/NS)2; (35103 )/(8.0 ) = (NP/NS)2, which gives NP/NS = 66. 33. We find the equivalent resistance of the two speakers in parallel from 1/Req = (1/R1) + (1/R2) = (1/8 ) + (1/8 ), which gives Req = 4 . Thus the speakers should be connected to the 4 terminals.

Chapter 31 page 8

34. Because the capacitor and resistor are in series, the impedance of the circuit is Z = (R2 + XC2)1/2, so the current is I = V/Z, and the voltage across the resistor is VR = IR. (a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(1.210–6 F) = 2.21103  = 2.21 k. The impedance of the circuit is Z1 = (R2 + XC12)1/2 = [(0.800 k)2 + (2.21 k)2]1/2 = 2.35 k. The current is I1 = V/Z1 = (130 mV)/(2.35 k) = 55.3 A. The voltage across the resistor is VR1 = I1R = (55.3 A)(0.800 k) = 44 mV. Thus the signal is reduced but not eliminated. (b) The reactance of the capacitor is XC2 = 1/2pf2C = 1/2p(60,000 Hz)(1.210–6 F) = 2.21 . The impedance of the circuit is Z2 = (R2 + XC22)1/2 = [(800 )2 + (2.21 )2]1/2 = 800 . The current is I2 = V/Z2 = (130 mV)/(800 ) = 0.163 mA. The voltage across the resistor is VR2 = I2R = (0.163 mA)(800 ) = 130 mV. Thus the high-frequency signal passes to the resistor. 35. For the current and voltage to be in phase, the net reactance of the capacitor and inductor must be zero. Thus we have XL = XC; 2pfL = 1/2pfC, or f = (1/2p)(1/LC)1/2; 9.76 nF. 3360 Hz = (1/2p)[1/(0.230 H)C]1/2 , which gives C = 9.7610–9 F = 36. Because the current lags the voltage by less than 90°, resistance must be present so this must be an

Chapter 31 page 9

RL circuit. We find the impedance of the circuit from Z = Vrms/Irms = (120 V)/(5.6 A) = 21.4 . We find the resistance from the phase: cos  = R/Z; 13.8 . cos 50° = R/(21.4 ), which gives R = We find the reactance from the phase: sin  = XL/Z; sin 50° = XL/(21.4 ), which gives XL = 16.4 . We find the inductance from XL = 2pfL; 16.4  = 2p(60 Hz)L, which gives L = 4.3510–2 H = 37. We find the reactance from XL = Vrms/Irms = (240 V)/(22.8 A) = 10.5 . We find the inductance from XL = 2pfL; 10.5  = 2p(60 Hz)L, which gives L = 0.0279 H =

y

V



VL

t

43.5 mH.

27.9 mH.

38. (a) We find the impedance from Z = XC = 1/2pfC = 1/2p(700 Hz)(0.03810–6 F) = 5.98103  = (b) We find the peak value of the current from Ipeak = v2 Irms = v2(Vrms/Z) = v2(4.0 kV)/(6.0 k) = 0.95 A.

6.0 k.

39. At 60 Hz, the reactance of the inductor is XL1 = 2pf1L = 2p(60 Hz)(0.620 H) = 234 . The impedance of the circuit is Z1 = (R2 + XL12)1/2 = [(3.5 k)2 + (0.234 k)2]1/2 = 3.51 k. Thus the impedance at the new frequency is Z2 = 2Z1 = 2(3.51 k) = 7.02 k. We find the new reactance from Z2 = (R2 + XL22)1/2; 7.02 k = [(3.5 k)2 + XL22]1/2, which gives XL2 = 6.09 k. We find the new frequency from XL2 = 2pf2L; 6.09 k = 2pf2(0.620 H), which gives f2 = 1.6 kHz. 40. (a) The reactance of the capacitor is XC = 1/2pfC = 1/2p(60.0 Hz)(1.8010–6 F) = 1.47103  = 1.47 k. The impedance of the circuit is Z = (R2 + XC2)1/2 = [(8.80 k)2 + (1.47 k)2]1/2 = 8.92 k. The rms current is Irms = Vrms/Z = (120 V)/(8.92 k) = 13.5 mA. (b) We find the phase angle from cos  = R/Z = (8.80 k)/(8.92 k) = 0.986. – 9.48°. In an RC circuit, the current leads the voltage, so  = (c) The power dissipated is P = Irms2R = (13.510–3 A)2(8.80103 ) = 1.59 W. (d) The rms readings across the elements are

II VR x

Chapter 31 page 10

VR = IrmsR = (13.5 mA)(8.80 k) = 119 V; VC = IrmsXC = (13.5 mA)(1.47 k) = 19.8 V. Note that, because the maximum voltages occur at different times, the two readings do not add to the applied voltage of 120 V. 41. We find the resistance of the coil from the dc current: R = Vdc/Idc = (36 V)/(2.5 A) = 14 . We find the impedance from the ac current: Z = Vrms/Irms = (120 V)/(3.8 A) = 31.6 . We find the reactance from Z = [R2 + XL2]1/2 ; 31.6  = [(14.4 )2 + XL2]1/2, which gives XL = 28.1 . We find the inductance from XL = 2pfL; 28.1  = 2p(60 Hz)L, which gives L = 0.075 H = 75 mH.

42. (a) At resonance we have 2pf0 = (1/LC)1/2. The Q factor is Q = VC/VR = IrmsXC/IrmsR = 1/2pf0CR = (LC)1/2/CR = (1/R)(L/C)1/2. (b) We find the inductance from the resonant frequency: 2pf0 = (1/LC)1/2; 2p(1.0106 Hz) = [1/L(0.01010–6 F)]1/2, which gives L = 2.510–6 H = 2.5 H. We find the resistance required from Q = (1/R)(L/C)1/2; 1.610–2 . 1000 = (1/R)[(2.510–6 H)/(0.01010–6 F)]1/2, which gives R = (c) For the circuit of Example 31–3, we have Q = (1/R)(L/C)1/2 = (1/25.0 )[(30.010–3 H)/(12.010–6 F)]1/2 = 2.00. 43. We find the impedance from the power factor: cos  = R/Z ; 0.17 = (200 )/Z , which gives Z = 1.18103 . We get an expression for the reactances from Z2 = R2 + (XL – XC)2; (1.18103 )2 = (200 )2 + (XL – XC)2, which gives XL – XC = ± 1.16103 . When we express this in terms of the inductance and capacitance, we get 2pfL – (1/2pfC) = ± 1.16103 ; 2pf(0.020 H) – [1/2pf(5010–9 F)] = ± 1.16103 , which reduces to two quadratic equations: 0.126f 2 ± (1.16103 Hz)f – 3.183106 Hz2 = 0, 2.2103 Hz, 1.1104 Hz. which have positive solutions of f = 44. (a) If we choose V = V0 sin t and assume XL > XC , we have the diagram shown.

Chapter 31 page 11

(b) We take projections on the y-axis. From the diagram we see that I = – I0 sin ( – t) = I0 sin (t – ), where cos  = VR0/V0 = R/Z; Z2 = R2 + (XL – XC)2; and I0 = V0/Z.

45. (a) From the expression for V, we see that V0 = 0.95 V, and 2pf = 754 s–1. For the reactances, we have XC = 1/2pfC = 1/(754 s–1)(0.3010–6 F) = 4.42103  = 4.42 k. XL = 2pfL = (754 s–1)(0.0220 H) = 16.6  = 0.0166 k. The impedance of the circuit is Z = [R2 + (XL – XC)2]1/2 = [(23.2 k)2 + (0.0166 k – 4.42 k)2]1/2 = 23.6 k. We find the phase angle from tan  = (XL – XC)/R = (0.0166 k – 4.42 k)/(23.2 k) = – 0.191, which gives  = – 10.8°. (b) The power dissipated is P = Irms2R = (Vrms/Z)2R = (V0/Zv2)2R = !(V0/Z)2R = !(0.95 V/23.6103 )2(23.2103 ) = 1.8810–5 W. (c) The rms current is Irms = I0/v2 = V0/Zv2 = (0.95 V)/(23.6103 )v2 = 2.84610–5 A = The rms readings across the elements are VR = IrmsR = (2.8510–5 A)(23.6103 ) = 0.66 V; VL = IrmsXL = (2.8510–5 A)(16.6 ) = 4.710–4 V. VC = IrmsXC = (2.8510–5 A)(4.42103 ) = 0.126 V. 46. (a) With XL > XC , and R = 0, the impedance of the circuit to ac current is Z = [(XL – XC)2]1/2 = XL – XC . The ac current amplitude is I0 = V20/Z = V20/(XL – XC) = V20/(L – 1/C). We find the phase angle from tan  = (XL – XC)/R = (XL – XC)/0 = + 8, so  = 90°.

2.810–5 A.

I L V in

C

V out

Chapter 31 page 12

Thus the current is I = I0 sin (t – 90°) = – I0 cos t. There will be no dc current through the capacitor. (b) We can treat the emf as a superposition of ac and dc components. For the dc component V1 , the voltage across the inductor is V1L = L dI/dt = 0, so V1out = V1C = V1 , and Q1 = CV1. At any instant the charge on the capacitor is Q = CVout = CV1 + CV2out , and the current through the capacitor is I = dQ/dt = C dV2out/dt = – I0 cos t. When we integrate this, we get V 2 ou t

t

I0 cos t d t ; C 0 0 I V 20 V 20 V 2out = – 0 si n t = – si n t = si n ( t – 180°). C C( L – 1/  C) (  2LC – 1) (c) If we ignore the sign change, the ratio of output ac voltage to input ac voltage is V2out/V20 = 1/C(L – 1/C) = XC/(XL – XC). d V 2out =

–

This will be least when the denominator is maximal, XC « XL , and the ratio will be

V2out/V20 ˜ XC/XL « 1. (d) Because there is no dc current,

V1out = V1 .

47. The impedance of the circuit to ac current is Z = [R2 + XC2]1/2 = [R2 + (1/C)2]1/2. The ac current amplitude is I0 = V20/Z = V20/[R2 + (1/C)2]1/2. We find the phase angle from tan  = (XL – XC)/R = – XC/R.

I R V in

C

V out

If R » XC , then tan  ˜ 0, and  ˜ 0. Thus the current is I = I0 sin t. There will be no dc current through the capacitor, and thus no appreciable dc voltage drop across the resistor, so V1out = V1C = V1 , and Q1 = CV1. At any instant the charge on the capacitor is Q = CVout = CV1 + CV2out , and the current through the capacitor is I = dQ/dt = C dV2out/dt = I0 sin t. When we integrate this, we get t V2 out I0 dV 2out = sin t  dt ; 0 0 C I X V V 2out = – 0 cos  t = 2 C 20 sin (t – 90°). 2 1/ 2 C R +X C

If R » XC , then Z = [R2 + (1/C)2]1/2 ˜ R, and the current will be I0 = V20/Z = V20/[R2 + (1/C)2]1/2 ˜ V20/R, which will be low.

Chapter 31 page 13

If we form the ratio of output ac voltage to input ac voltage, we get V2out/V20 = XC/[R2 + XC2]1/2 ˜ XC/R « 1.

48. (a) The current in the resistor is in phase with the voltage across the resistor: IR = V/R = (V0/R) sin t, R = 0°. (b) The voltage across the inductor is V = V 0 sin t I VL = V0 sin t = L dIL/dt. R L When we integrate this, we get IL

C

t

V0 si n t d t ; 0 0 L V V I L = – 0 cos t = 0 si n (t – 90°). L XL (c) The charge on the capacitor is QC = CV = CV0 sin t. Thus the current in the capacitor is IC = dQC/dt = CV0 cos t = CV0 sin (t + 90°) = (V0/XC) sin (t + 90°). (d) The total current is I = IR + IL + IC = (V0/R) sin t + (V0/L) sin (t – 90°) + (CV0) sin (t + 90°) = V0[(1/R) sin t + (C – 1/L) cos t]. [(1/ R) 2 + (C – 1/ L) 2 ]1/2 We form the triangle shown with the following values: cos  = (1/R)/[(1/R)2 + (C – 1/L)2]1/2; sin  = (C – 1/L)/[(1/R)2 + (C – 1/L)2]1/2;  tan  = (C – 1/L)/(1/R) = R(C – 1/L). This lets us express the current as a combination of 1/ R trig functions: I = V0[(1/R)2 + (C – 1/L)2]1/2(cos  sin t + sin  cos t) d IL =

C – 1/ L

Chapter 31 page 14

= V0[(1/R)2 + (C – 1/L)2]1/2 sin (t + ), with  = tan–1 [R(C – 1/L)]. (e) The impedance is Z = V0/I0 = [(1/R)2 + (C – 1/L)2]–1/2 = R/[1 + (RC – R/L)2]1/2. (f) Because power is dissipated only by the resistance, the average power is P = (I0R2/2)R = (I0R2/2)Z(power factor), or, in terms of the voltage: [(V0/R)2/2]R = [(V0/Z)2/2]Z(power factor), or V02/2R = (V02/2Z)(power factor). Thus we see that power factor = Z/R = [1 + (RC – R/L)2]–1/2 = cos . 49. We find the equivalent values for each type of element: R = R1 + R2 ; C = C1C2/(C1 + C2), so XC = (C1 + C2)/C1C2 ; L = L1 + L2 , assuming the windings are symmetric, so XL = (L1 + L2). The impedance of the circuit is Z = [R2 + (XL – XC)2]1/2 = {(R1 + R2)2 + [(L1 + L2) – (C1 + C2)/C1C2]2}1/2. 50. If there is no current in the secondary, there will be no induced emf from the mutual inductance. We find the inductance from Z = XL = Vrms/Irms = 2pfL; (220 V)/(5.8 A) = 2p(60 Hz)L, which gives L = 0.10 H.

51. We can find the resistance from the power dissipated: P = Irms2R; 300 W = (4.0 A)2R, which gives R = 19 . The impedance of the electromagnet is Z = Vrms/Irms = (120 V)/(4.0 A) = 30 . We can find the inductance from Z = (R2 + XL2)1/2 = [R2 + (2pfL)2]1/2;

30  = {(19 )2 + [2p(60 Hz)L]2}1/2, which gives L = 6.210–2 H =

62 mH.

52. For the inductor circuit we have V = V0 sin t = L dI/dt + RI. If we let I = I0 sin (t – ), this becomes V0 sin t = LI0 cos (t – ) + I0R sin (t – ) = LI0(cos t cos  + sin t sin ) + I0R(sin t cos  – cos t sin ) = (I0R cos  + I0L sin ) sin t – (I0R sin  – I0L cos ) cos t. Because this must be satisfied for any t, the equation must be satisfied separately by the sin t and cos t terms. Thus we get two equations from which we get I0 and :

Chapter 31 page 15

I0R sin  – I0L cos  = 0, which gives tan  = L/R; V0 = I0R cos  + I0L sin  . We form the triangle shown with the following values: cos  = R/(R2 + 2L2)1/2; sin  = L/(R2 + 2L2)1/2; tan  = L/R. Thus I0 = V0/(R cos  + L sin )

= V0/{R[R/(R2 + 2L2)1/2] + L[L/(R2 + 2L2)1/2]} =

53. For the junction equation, we have I = IC + IL. For the loop equations, we have left loop: V0 sin t – IR – Q/C = 0; outside loop: V0 sin t – IR – L dIL/dt = 0. Using a trial solution, I = I0 sin (t + ), we find the capacitor current from Q = C[V0 sin (t) – I0R sin (t + )]; IC = dQ/dt = + CV0 cos t – CI0R cos (t + ) = + (V0/XC) cos t – (I0R/XC) cos (t + ). We find the inductor current from dIL/dt = (V0/L) sin t – (I0R/L) sin (t + );

[R 2 + (L)2 ]1/2

L  R V0/(R2 + 2L2)1/2.

R I

C V = V 0 sin t

= ? (dIL/dt) dt = – (V0/L) cos t + (I0R/L) cos (t + ) + constant = – (V0/XL) cos t + (I0R/XL) cos (t + ) + constant. By using the appropriate expansion of the trigonometric functions, we have IL

IL

IC

L

Chapter 31 page 16

I = I0[sin t cos  + cos t sin ], IC = + (V0/XC) cos t – (I0R/XC)(cos t cos  – sin t sin ), IL = – (V0/XL) cos t + (I0R/XL)(cos t cos  – sin t sin ) + constant. For the junction equation to be satisfied for arbitrary t, the coefficients of sin t and cos t must satisfy the junction equation, which gives constant = 0, + I0 cos  = I0R[(1/XC) – (1/XL)] sin . (i) + I0 sin  = V0[(1/XC) – (1/XL)] – I0R[(1/XC) – (1/XL)] cos , (ii) From (i) we get tan  = XCXL/(XL – XC)R, which gives Xeq = XCXL/(XL – XC). This is the equivalent reactance of the two parallel elements. Because the capacitor current leads the voltage and the inductor current lags the voltage, there is a p phase difference between the two, which gives the negative sign in the denominator. The impedance of the circuit is Z = (R2 + Xeq2)1/2 = {R2 + [XCXL/(XL – XC)]2}1/2, which can also be obtained from (ii). When we use these results in the current equations, we obtain I0 = V0/Z, so the three currents are I = (V0/Z) sin (t + ), IC = (V0/XC)[– (R/Z) cos (t + ) + cos t], IL = (V0/XL)[(R/Z) cos (t + ) – cos t], where Z = {R2 + [XCXL/(XL – XC)]2}1/2, and tan  = XCXL/(XL – XC)R.

54. We write the applied voltage as V0 sin t. For the capacitor branch we have V = V 0 sin t V0 sin t = Q/C + I1R1, with Z12 = R12 + XC2. Thus I1 = I10 sin (t – 1), with tan 1 = – XC/R1 . For the inductor branch we have V0 sin t = I2R2 + L dI2/dt, with Z22 = R22 + XL2. Thus I2 = I20 sin (t – 2), , with tan 2 = XL/R2 . We can find the potential difference Vab from the two paths: Vab = – I1R1 + I2XL ; Vab = I1XC – I2R2 . When we eliminate I1 , we get [1 + (R1/XC)]Vab = [XL – (R1R2/XC)]I2 . Thus Vab = 0 if R1R2 = XCXL = L/C, independent of the frequency.

C a R1

I1

R2 b L

c

I2

Chapter 32 p. 1

CHAPTER 32 – Maxwell’s Equations and Electromagnetic Waves 1.

The electric field between the plates depends on the voltage: E = V/d, so 9.2  104 V/m · s. dE/dt = (1/d) dV/dt = (1/1.3  10–3 m)(120 V/s) =

2.

The displacement current is ID = Å0A (dE/dt) = (8.85  10–12 C2/N · m2)(0.038 m)2(2.0  106 V/m · s) =

3.

The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from ID = Å0A (dE/dt); 1.8 A = (8.85  10–12 C2/N · m2)(0.0160 m)2 dE/dt, which gives dE/dt = 7.9  1014 V/m · s.

4.

The current in the wires is the rate at which charge is accumulating on the plates and must also be the displacement current in the capacitor. Because the location is outside the capacitor, we can use the expression for the magnetic field of a long wire: B = (0/4p)2ID/R = (10–7 T · m/A)2(35.0  10–3 A)/(0.100 m) = 7.00  10–8 T. After the capacitor is fully charged, all currents will be zero, so the magnetic field will be zero.

5.

The electric field between the plates depends on the voltage: E = V/d, so dE/dt = (1/d) dV/dt. Thus the displacement current is ID = Å0A (dE/dt) = (Å0A/d)(dV/dt) = C dV/dt.

6.

(a) The current in the wires is the rate at which charge is accumulating on the plates and must also be the displacement current in the capacitor: Imax = ID,max = 35 A. (b) At any instant, the charge on the plates is Q = åC, so the current is I = dQ/dt = C då/dt = – å0C sin t, or Imax = å0Å0pR2/d; 35  10–6 A = 2p(96.0 Hz)å0(8.85  10–12 C2/N · m2)p(0.025 m)2/(2.0  10–3 m), which gives å0 = 6.7  103 V. (c) We can find the maximum value of dE/dt from the maximum value of the displacement current: ID,max = Å0 (dE/dt)max; 35  10–6 A = (8.85  10–12 C2/N · m2)(dE/dt)max , which gives (dE/dt)max = 4.0  106 V · m/s.

7.

Gauss’s law for electricity and Ampere’s law will not change. From the analogy to Gauss’s law for electric fields, where Q is the source, Qm would be the source of the magnetic field, so we have ı B · dA = 0Qm. From the analogy to Ampere’s law, we have an additional “current” contribution to Faraday‘s law. ı E · d¬ = 0 dQm/dt – dB/dt. The dQm/dt term corresponds to an electric field created by the “current” of magnetic monopoles.

8.

The electric field is E0 = cB0 = (3.00108 m/s)(17.510–9 T) =

5.25 V/m.

2.6  10–8 A.

Chapter 32 p. 2

9.

We find the magnetic field from E0 = cB0; 0.4310–4 V/m = (3.00108 m/s)B, which gives B =

1.410–13 T.

10. The frequency of the two fields must be the same: 80.0 kHz. The rms strength of the electric field is Erms = cBrms = (3.00108 m/s)(6.7510–9 T) = 2.03 V/m. The electric field is perpendicular to both the direction of travel and the magnetic field, so the electric field oscillates along the horizontal north-south line. 11. (a) If we write the argument of the cosine function as kz + t = k(z + ct), we see that the wave is traveling in the – z-direction. Because E and B are perpendicular to each other and to the direction of propagation, and E is in the x-direction, B can have only a y-component, with magnitude B0 = E0/c. Because a rotation of E into B must give the direction of propagation, – z-direction, B must be in the – y-direction. (b) The wave is traveling in the – z-direction. 12. The frequency of the microwave is f = c/ = (3.00108 m/s)/(1.8010–2 m) =

1.671010 Hz.

13. (a) The wavelength of the radar signal is  = c/f = (3.00108 m/s)/(27.75109 Hz) = 1.0810–2 m = (b) The frequency of the X-ray is f = c/ = (3.00108 m/s)/(0.1010–9 m) = 3.01018 Hz. 14. The frequency of the wave is f = c/ = (3.00108 m/s)/(85010–9 m) = 3.521014 Hz. This frequency is just outside the red end of the visible region, so it is

1.08 cm.

infrared.

15. The wavelength of the wave is  = c/f = (3.00108 m/s)/(9.561014 Hz) = 3.1410–7 m = 314 nm. This wavelength is just outside the violet end of the visible region, so it is ultraviolet. 16. The distance that light travels in one year is d = (3.00108 m/s)(3.156107 s/yr) = 9.471015 m. 17. (a) If we assume the closest approach of Mars to Earth, we have ?t = L/c = [(227.9 – 149.6)109 m]/(3.00108 m/s) = 2.6102 s = (b) If we assume the closest approach of Saturn to Earth, we have ?t = L/c = [(1427 – 149.6)109 m]/(3.00108 m/s) = 4.3103 s =

4.3 min. 71 min.

18. The length of the pulse is ?d = c ?t, so the number of wavelengths in this length is N = (c ?t)/ = (3.00108 m/s)(3010–12 s)/(106210–9 m) = 8.5103 wavelengths. The time for the length of the pulse to be one wavelength is 3.54 fs. ?t = /c = (106210–9 m)/(3.00108 m/s) = 3.5410–15 s = 19. The energy per unit area per unit time is S = !cÅ0E02

Chapter 32 p. 3

= !(3.00108 m/s)(8.8510–12 C2/N · m2)(36.510–3 V/m)2 =

1.7710–6 W/m2.

20. The energy per unit area per unit time is S = cBrms2/0 = (3.00108 m/s)(32.510–9 T)2/(4p10–7 T · m/A) = 0.252 W/m2. We find the time from t = U/AS = (335 J)/(1.0010–4 m2)(0.252 J/s · m2) = 1.32107 s = 0.419 yr =

5.03 months.

21. The energy per unit area per unit time is S = cÅ0Erms2 = (3.00108 m/s)(8.8510–12 C2/N · m2)(28.610–3 V/m)2 = 2.1710–6 W/m2. We find the energy transported from U/t = AS = (1.0010–4 m2)(2.1710–6 W/m2)(3600 s/h) = 7.8210–7 J/h. 22. Because the wave spreads out uniformly over the surface of the sphere, the intensity is I = P/A = (1000 W)/4p(10.0 m)2 = 0.796 W/m2. This intensity is represented by the Poynting vector. We find the rms value of the electric field from S = cÅ0Erms2; 0.796 W/m2 = (3.00108 m/s)(8.8510–12 C2/N · m2)Erms2, which gives Erms = 17.3 V/m. 23. For the energy density, we have u = !Å0E02 = S/c = (1350 W/m2)/(3.00108 m/s) = 4.5010–6 J/m3. The radiant energy is U = uV = (4.5010–6 J/m3)(1.00 m3) = 4.5010–6 J. 24. The energy per unit area per unit time is S = P/A = cÅ0Erms2; (12.810–3 W)/p(1.0010–3 m)2 = (3.00108 m/s)(8.8510–12 C2/N · m2)Erms2, which gives Erms = 1.24103 V/m. The rms value of the magnetic field is Brms = Erms/c = (1.24103 V/m)/(3.00108 m/s) = 4.1310–6 T. 25. The radiation from the Sun has the same intensity in all directions, so the rate at which it reaches the Earth is the rate at which it passes through a sphere centered at the Sun: P = S4prSE2 = (1350 W/m2)4p(1.51011 m)2 = 3.81026 W. 26. (a) The energy emitted in each pulse is U = Pt = (2.51011 W)(1.010–9 s) = 2.5102 J. (b) We find the rms electric field from S = P/A = cÅ0Erms2; (2.51011 W)/p(2.210–3 m)2 = (3.00108 m/s)(8.8510–12 C2/N · m2)Erms2, which gives Erms = 2.5109 V/m.

Chapter 32 p. 4

27. (a) We choose the axis of the circular plates as the z-axis. d We find the electric field from the charge density: E = (/Å0)k = (Q/pR2Å0)k. From the cylindrical symmetry, we know that the magnetic field will be circular, centered on the z-axis. We choose a B circular path with radius r < R to apply Ampere’s law: ı B · ds = 0Ienclosed + 0Å0 (d/dt) ı E · dA; E B2pr = 0 + 0Å0 (d/dt)[(Q/pR20)pr2], which gives B = (0r/2pR2) dQ/dt circular, for r = R. The Poynting vector, S = (1/0)E  B, will be directed toward the axis of the plates: S = – (1/0)(Q/pR2Å0)(0r/2pR2)(dQ/dt )® = – (Qr/2p2R4Å0)(dQ/dt )®. (b) The energy stored in the electric field is U = uEpR2d = !Å0E2pR2d = !Å0(Q/pR2Å0)2pR2d = !(d/pR2Å0)Q2. The rate at which this energy is being stored is dU/dt = (Qd/Å0pR2) dQ/dt. To find the energy flow into the capacitor through the cylinder at r = R, we note that S is perpendicular to the surface and has a constant magnitude. Thus P = ? S · dA = S(surface area) = (Qr/2p2Å0R4)(dQ/dt )(2pRd) = (Qd/Å0pR2) dQ/dt, which is the rate at which energy is being stored in the electric field.

R I

28. We assume the light does not reflect from the surface. We find the radiation pressure from F/A = S/c = P/4pR2c = (100 W)/4p(0.080 m)2(3108 m/s) = 4.110–6 N/m2. If we estimate an area 1 cm on a side for the fingertip, the force will be F = (F/A)A = (4.110–6 N/m2)(0.01 m)2 ˜ 410–10 N. 29. The force exerted by the radiation on a cross-section of a spherical particle is Fradiation = (pressure)A = (S/c)A = (P/4prS2)pr2/c. If we want the force from the radiation to be greater than the gravitational force at the surface of the Sun, we have Pr2/4rS2c > GmMS/rS2 = G)pr3MS/rS2, or P/c > 16GprMS/3; (3.81026 W)/(3108 m/s) > 16(6.6710–11 N · m2/kg2)(2.0103 kg/m3)pr(2.01030 kg)/3, which gives r < 310–7 m. 30. If a particle with momentum p collides completely inelastically with a large object, the final momentum of the particle will be zero, so the momentum change is p. If the collision takes place in a time ?t, the force on the particle is F1 = p/?t. From Newton’s third law, there will be a force of equal magnitude on the object. If a particle with momentum p collides completely elastically with a large object, the final momentum of the particle will be p in the opposite direction, so the momentum change is 2p. If the collision takes place in a time ?t, the force on the particle is F2 = 2p/?t.

Chapter 32 p. 5

From Newton’s third law, there will be a force of equal magnitude on the object. Thus the ratio of the pressures is equal to the ratio of the forces acting on the object, which is F2/F1 = 2. 31. The resonant frequency is given by f02 = (1/2p)2(1/LC). When we form the ratio for the two stations, we get (f02/f01)2 = C1/C2 ; (1550 kHz/550 kHz)2 = (2400 pF)/C2 , which gives C2 =

302 pF.

32. We find the capacitance from the resonant frequency: f0 = (1/2p)(1/LC)1/2; 96.1106 Hz = (1/2p)[1/(1.810–6 H)C]1/2, which gives C = 1.510–12 F =

1.5 pF.

33. We find the inductance for the first frequency: f01 = (1/2p)(1/L1C)1/2; 88106 Hz = (1/2p)[1/L1(84010–12 F)]1/2, which gives L1 = 3.8910–9 H = 3.89 nH. For the second frequency we have f02 = (1/2p)(1/L2C)1/2; 108106 Hz = (1/2p)[1/L2(84010–12 F)]1/2, which gives L2 = 2.5910–9 H = 2.59 nH. Thus the range of inductances is 2.59 nH = L = 3.89 nH. 34. (a) The minimum value of C corresponds to the higher frequency, so we have f01 = (1/2p)(1/LC1)1/2; 15.0106 Hz = (1/2p)[1/L(9210–12 F)]1/2, which gives L = 1.2210–6 H = 1.22 H. (b) The maximum value of C corresponds to the lower frequency, so we have f02 = (1/2p)(1/LC2)1/2; 14.0106 Hz = (1/2p)[1/(1.2210–6 H)C2]1/2, which gives C2 = 1.0610–10 F = 106 pF. 35. (a) The wavelength of the AM station is  = c/f = (3.00108 m/s)/(680103 Hz) = 441 m. (b) The wavelength of the FM station is  = c/f = (3.00108 m/s)/(100.7106 Hz) = 2.979 m. 36. The frequencies are 940 kHz on the AM dial and 94 MHz on the FM dial. From c = f, we see that the lower frequency will have the longer wavelength: the AM station. When we form the ratio of wavelengths, we get 2/1 = f1/f2 = (94106 Hz)/(940103 Hz) = 100. 37. The wavelength of Channel 2 is 2 = c/f2 = (3.00108 m/s)/(54.0106 Hz) = 5.56 m. The wavelength of Channel 69 is 69 = c/f69 = (3.00108 m/s)/(806106 Hz) = 0.372 m. 38. After the change occurred, we would find out when the change in radiation reached the Earth: ?t = L/c = (1.501011 m)/(3.00108 m/s) = 5.00102 s = 8.33 min.

Chapter 32 p. 6

39. (a) The time for a signal to travel to the Moon is 1.28 s. ?t = L/c = (3.84108 m)/(3.00108 m/s) = (b) The time for a signal to travel to Mars at the closest approach is 4.3 min. ?t = L/c = (78109 m)/(3.00108 m/s) = 260 s = 40. The time consists of the time for the radio signal to travel to Earth and the time for the sound to travel from the loudspeaker: t = tradio + tsound = (dradio/c) + (dsound/vsound) = (3.84108 m/3.00108 m/s) + (50 m/343 m/s) = 1.43 s. Note that about 10% of the time is for the sound wave. 41. The light has the same intensity in all directions, so the energy per unit area per unit time over a sphere centered at the source is S = P0/A = P0/4pr2 = !cÅ0E02 = !c(1/c20)E0 2, which gives E0 = (0cP0/2pr2)1/2. 42. The light has the same intensity in all directions, so the energy per unit area per unit time over a sphere centered at the source is S = P0/A = P0/4pr2 = (100 W)/4p(2.00 m)2 = 1.99 W/m2. We find the electric field from S = !cÅ0E02; 1.99 W/m2 = !(3.00108 m/s)(8.8510–12 C2/N · m2)E02, which gives E0 = The magnetic field is B0 = E0/c = (38.7 V/m)/(3.00108 m/s) = 1.2910–7 T.

38.7 V/m.

43. The radiation from the Sun has the same intensity in all directions, so the rate at which it passes through a sphere centered at the Sun is P = S4pR2. The rate must be the same for the two spheres, one containing the Earth and one containing Mars. When we form the ratio, we get PMars/PEarth = (SMars/SEarth)(RMars/REarth)2; 1 = (SMars/1350 W/m2)(1.52)2, which gives SMars = 584.3 W/m2. We find the rms value of the electric field from SMars = cÅ0Erms2; 584.3 W/m2 = (3.00108 m/s)(8.8510–12 C2/N · m2)Erms2, which gives Erms = 469 V/m. 44. If we curl the fingers of our right hand from the direction of the electric field (south) into the direction of the magnetic field (west), our thumb points down, so the direction of the wave is downward. We find the electric field from S = !cÅ0E02; 500 W/m2 = !(3.00108 m/s)(8.8510–12 C2/N · m2)E02, which gives E0 = The magnetic field is B0 = E0/c = (614 V/m)/(3.00108 m/s) = 2.0510–6 T.

614 V/m.

45. If we ignore the time for the sound to travel to the microphone, the time difference is ?t = tradio – tsound = (dradio/c) – (dsound/vsound) = (3000103 m/3.00108 m/s) – (50 m/343 m/s) = – 0.14 s, so the person at the radio hears the voice 0.14 s sooner. 46. To produce the voltage over the length of the antenna, we have

Chapter 32 p. 7

Erms = Vrms/d = (1.010–3 V)/(1.80 m) = 5.610–4 V/m. The rate of energy transport is S = cÅ0Erms2 = (3.00108 m/s)(8.8510–12 C2/N · m2)(5.5610–4 V/m)2 =

8.210–10 W/m2.

47. (a) The radio waves have the same intensity in all directions, so the energy per unit area per unit time over a sphere centered at the source with a radius of 100 m is S = P0/A = P0/4pr2 = (50103 W)/4p(100 m)2 = 0.398 W/m2. Thus the power through the area is P = SA = (0.398 W/m2)(1.0 m2) = 0.40 W. (b) We find the rms value of the electric field from S = cÅ0Erms2; 0.398 W/m2 = (3.00108 m/s)(8.8510–12 C2/N · m2)Erms2, which gives Erms = 12 V/m. (c) If the electric field is parallel to the antenna, the voltage over the length of the antenna is Vrms = Ermsd = (12 V/m)(1.0 m) = 12 V. 48. (a) The radio waves have the same intensity in all directions, so the energy per unit area per unit time over a sphere centered at the source with a radius of 100 km is S = P0/A = P0/4pr2 = (50103 W)/4p(100103 m)2 = 3.9810–7 W/m2. Thus the power through the area is P = SA = (3.9810–7 W/m2)(1.0 m2) = 3.9810–7 W = 0.40 W. (b) We find the rms value of the electric field from S = cÅ0Erms2; 3.9810–7 W/m2 = (3.00108 m/s)(8.8510–12 C2/N · m2)Erms2, which gives Erms = 0.012 V/m. (c) The voltage over the length of the antenna is Vrms = Ermsd = (0.012 V/m)(1.0 m) = 0.012 V. 49. The energy per unit area per unit time is S = !cÅ0E02; = !(3.00108 m/s)(8.8510–12 C2/N · m2)(3106 V/m)2 = 1.201010 W/m2. The power output is P = S4pr2 = (1.201010 W/m2)4p(1.0 m)2 = 1.51011 W. 50. We find the magnetic field from S = !(c/0)B02; 1.010–4 W/m2 = ![(3.00108 m/s)/(4p10–7 T · m/A)]B02, which gives B0 = 9.1510–10 T. Because this field oscillates through the coil at  = 2pf, the maximum emf is å0 = NAB0 = (380 turns)p(0.011 m)2(9.1510–10 T)2p(810103 Hz) = 6.7210–4 V = 0.672 mV. The rms emf is årms = å0/v2 = (0.672 mV)/v2 = 0.48 mV.

Chapter 32 p. 8

51. (a) We see from the diagram that all positive plates are d connected to the positive side of the battery, and that all negative plates are connected to the negative side of the + battery, so the 11 capacitors are connected in parallel. V – (b) For parallel capacitors, the total capacitance is the sum, so we have Cmin = 11(Å0Amin/d) = 11(8.8510–12 C2/N · m2)(1.010–4 m2)/(1.110–3 m) = 8.8510–12 F = 8.9 pF; Cmax = 11(Å0Amax/d) = 11(8.8510–12 C2/N · m2)(9.010–4 m2)/(1.110–3 m) = 79.710–12 F = 80 pF. Thus the range is 8.9 pF = C = 80 pF. (c) The lowest resonant frequency requires the maximum capacitance. We find the inductance for the lowest frequency: f01 = (1/2p)(1/L1Cmax)1/2; 550103 Hz = (1/2p)[1/L1(79.710–12 F)]1/2, which gives L1 = 1.0510–3 H = 1.05 mH. We must check to make sure that the highest frequency can be reached. We find the resonant frequency using this inductance and the minimum capacitance: f0max = (1/2p)(1/L1Cmin)1/2 = (1/2p)[1/(1.0510–3 H)(8.8510–12 F)]1/2 = 1.65106 Hz = 1650 kHz. Because this is greater than the highest frequency desired, the inductor will work. We could also start with the highest frequency. We find the inductance for the highest frequency: f02 = (1/2p)(1/L2Cmin)1/2; 1600103 Hz = (1/2p)[1/L2(8.8510–12 F)]1/2, which gives L2 = 1.1210–3 H = 1.12 mH. We must check to make sure that the lowest frequency can be reached. We find the resonant frequency using this inductance and the maximum capacitance: f0min = (1/2p)(1/L2Cmax)1/2 = (1/2p)[1/(1.1210–3 H)(79.710–12 F)]1/2 = 533105 Hz = 533 kHz. Because this is less than the lowest frequency desired, this inductor will also work. Thus the range of inductances is 1.05 mH = L = 1.12 mH. 52. (a) We choose the direction of current as the z-axis. We find the electric field inside and on the surface from the current density: E = J/ = (I/A)k = (I/pr2)k = (IR/L)k, where R is the resistance of a length L. (b) From the cylindrical symmetry, we know that the magnetic field will be circular, centered on the axis of the wire. We choose a circular path with radius r1 > r to apply Ampere’s law:

ı B · ds = 0Ienclosed;

B2pr1 = 0I, which gives

r

I B

E

Chapter 32 p. 9

B = 0I/2pr1 circular, for r1 > r. (c) The Poynting vector on the surface of the wire at r1 = r, S = (1/0)E  B, will be directed toward the axis of the wire (radially inward): S = – (1/0)(I/pr2)(0I/2pr)® = – (I 2/2p 2 r3)®. (d) To find the energy flow into a length L of the conductor at r1 = r, we note that S is perpendicular to the surface and has a constant magnitude. Thus P = ? S · dA = S(surface area) = (I 2/2p 2 r3)(2prL) = I 2L/pr2 = I 2R, which is the rate at which energy is being dissipated in the conductor. 



Ch. 33

p. 1

CHAPTER 33 – Light: Reflection and Refraction 1.

2.

3.

(a) The speed in ethyl alcohol is v = c/n = (3.00  108 m/s)/(1.36) = (b) The speed in lucite is v = c/n = (3.00  108 m/s)/(1.51) =

2.21  108 m/s. 1.99  108 m/s.

We find the index of refraction from v = c/n; 2.29  108 m/s = (3.00  108 m/s)/n, which gives n =

The time for light to travel from the Sun to the Earth is ?t = L/c = (1.501011 m)/(3.00108 m/s) = 5.00102 s =

4.

We convert the units: d = (4.2 ly)(3.00108 m/s)(3.16107 s/yr) =

5.

The length in space of a burst is ?s = c ?t = (3.00108 m/s)(10–8 s) =

6.

1.31. 8.33 min.

4.01016 m.

3 m.

The speed in water is vwater = c/nwater. We find the index of refraction in the substance from v = c/n; 0.92vwater = 0.92c/nwater = c/n, which gives n = 1.33/0.92 =

1.45.

7.

The eight-sided mirror would have to rotate 1/8 of a revolution for the succeeding mirror to be in position to reflect the light in the proper direction. During this time the light must travel to the opposite mirror and back. Thus the angular speed required is  = ?/?t = (2p rad/8)/(2L/c) = (p rad)c/8L = (p rad)(3.00108 m/s)/8(35103 m) = 3.4103 rad/s (3.2104 rev/min).

8.

For a flat mirror the image is as far behind the mirror as the object is in front, so the distance from object to image is do + di = 1.8 m + 1.8 m = 3.6 m.

9.

We show some of the rays from the tip of the arrow that form the three images. Single reflections form the two side images. Double reflections form the third image. The two reflections have reversed the orientation of the image.

I3

I2

I1

Object

Ch. 33

p. 2

10. The angle of incidence equals the angle of reflection. Thus we have tan  = (H – h)/L = h/x; (1.54 m – 0.40 m)/(2.30 m) = (0.40 m)/x, which gives x = 0.81 m = 81 cm.

Mirror

H

 h

 x L

11. From the triangle formed by the mirrors and the first reflected ray, we have  +  +  = 180°; 5°. 40° + 135° +  = 180°, which gives  =







12. Because the rays entering your eye are diverging from the image position behind the mirror, the diameter of the area on the mirror and the diameter of your pupil must subtend the same angle at the image: Dmirror/di = Dpupil/(do + di); Dmirror/(80 cm) = (5.0 mm)/(80 cm + 80 cm), which gives Dmirror = 2.5 mm. Thus the area on the mirror is Amirror = #pDmirror2 = #p(2.5  10–3 m)2 = 4.9  10–6 m2. 13. For the first reflection at A the angle of incidence 1 is the angle of reflection. For the second reflection at B the angle of incidence 2 is the angle of reflection. We can relate these angles to the angle at which the mirrors meet, , by using the sum of the angles of the triangle ABC:  + (90° – 1) + (90° – 2) = 180°, or  = 1 + 2 . In the same way, for the triangle ABD, we have  + 21 + 22 = 180°, or  = 180° – 2(1 + 2) = 180° – 2. At point D we see that the deflection is  = 180° –  = 180° – (180° – 2) = 2.

B

 C

      

D 

A

14. (a) The velocity of the wave, which specifies the direction of the light wave, is in the direction of the ray. If we consider a single reflection, because the angle of incidence is equal to the angle of reflection, the component of the velocity perpendicular to the mirror is reversed, while the component parallel to the mirror is unchanged. When we have three mirrors at right angles, we can choose the orientation of the mirrors for our axes. In each of the three reflections, the component of velocity perpendicular to the mirror will reverse. Thus after three reflections, all three components of the velocity will have been reversed, so the wave will return along the line of the original direction. (b) Assuming the mirrors are large enough, the incident ray will make two reflections only if its direction is parallel to one of the mirrors. In that case, it has only two components, both of which will be reversed, so the ray will be reversed along the line of the original direction.

Ch. 33

p. 3

15. The rays from the Sun will be parallel, so the image will be at the focal point. The radius is r = 2f = 2(18.2 cm) = 36.4 cm. 16. To produce an image at infinity, the object will be at the focal point: do = f = r/2 = (22.0 cm)/2 = 11.0 cm. 17.

Object

Object F

C Image

C

F

Image

18. The ball is a convex mirror with a focal length f = r/2 = (– 4.5 cm)/2 = – 2.25 cm. We locate the image from (1/do) + (1/di) = 1/f; [1/(25.0 cm)] + (1/di) = 1/(– 2.25 cm), which gives di = – 2.06 cm. The image is 2.06 cm behind the surface, virtual. The magnification is m = – di/do = – (– 2.06 cm)/(25.0 cm) = + 0.082. Because the magnification is positive, the image is upright. 19. We find the image distance from the magnification: m = hi/ho = – di/do ; + 3 = – di/(1.5 m), which gives di = – 4.5 m. We find the focal length from (1/do) + (1/di) = 1/f; [1/(1.5 m)] + [1/(– 4.5 m)] = 1/f, which gives f = 2.25 m. The radius of the concave mirror is r = 2f = 2(2.25 m) = 4.5 m. 20. We find the image distance from the magnification: m = hi/ho = – di/do ; + 5.0 = – di/(2.00 cm), which gives di = – 10.0 cm. We find the focal length from (1/do) + (1/di) = 1/f; [1/(2.00 cm)] + [1/(– 10.0 cm)] = 1/f, which gives f = 2.5 cm. Because the focal length is positive, the mirror is concave with a radius of r = 2f = 2(2.5 cm) = 5.0 cm.

Ch. 33

p. 4

21. To produce an upright image, we have di < 0. A smaller image means di  < do , so f < 0, which means the mirror is We find the image distance from the magnification: m = hi/ho = – di/do ; 0.33 = – di/(20.0 m), which gives di = – 6.6 m. We find the focal length from (1/do) + (1/di) = 1/f; [1/(20.0 m)] + [1/(– 6.6 m)] = 1/f, which gives f = – 9.85 m. The radius of the mirror is r = 2f = 2(– 9.85 m) = – 20 m.

convex.

22. (a) We see from the ray diagram that the image is behind the mirror, so it is virtual. We estimate the image distance as – 7 cm.

O

(b) If we use a focal length of – 10 cm, we locate the image from (1/do) + (1/di) = 1/f; [1/(20 cm)] + (1/di) = 1/(– 10 cm), which gives di = – 6.7 cm. (c) We find the image size from the magnification: m = hi/ho = – di/do ; hi/(3.0 mm) = – (– 6.7 cm)/(20 cm), which gives hi = 1.0 mm. 23. (a) With di = do , we locate the object from (1/do) + (1/di) = 1/f; (1/do) + (1/do) = 1/f, which gives do = 2f = r. The object should be placed at the center of curvature. (b) Because the image is in front of the mirror, di > 0, it is real. (c) The magnification is m = – di/do = – do/do = – 1. Because the magnification is negative, the image is inverted. (d) As found in part (c), m = – 1. 24. We take the object distance to be 8, and find the focal length from (1/do) + (1/di) = 1/f; (1/8) + [1/(– 14.0 cm)] = 1/f, which gives f = – 14.0 cm. Because the focal length is negative, the mirror is convex. The radius is r = 2f = 2(– 14.0 cm) = – 28.0 cm.

I

F

C

Ch. 33

p. 5

25. We find the image distance from (1/do) + (1/di) = 1/f = 2/r, which we can write as di = rdo/(2do – r). The magnification is m = – di/do = – r/(2do – r). If do > r, then (2do – r) > r, so  m  = r/(2do – r) = r/(> r) < 1. If do < r, then (2do – r) < r, so  m  = r/(2do – r) = r/(< r) > 1. 26. From the ray that reflects from the center of the mirror, we have tan  = ho/do = hi/di ;  m  = hi/ho = di/do . Because the magnification is positive and the image distance on the ray diagram is negative, we get m = – di/do .

27. From the ray diagram, we see that tan  = ho/do = hi/di ; tan  = ho/(do + r) = hi/(r – di ). When we divide the two equations. we get (do + r)/do = (r – di )/di ; 1 + (r/do) = (r/di) – 1, or (r/do) – (r /di) = – 2; (1/do) – (1/di) = – 2/r = – 1/f, with f = r/2. From the ray diagram, we see that di < 0. If we consider f to be negative, we have (1/do) + (1/di) = 1/f.

ho

 

do

O

ho O

28. We find the image distance from the magnification: m = hi/ho = – di/do ; + 0.55 = – di/(3.2 m), which gives di = – 1.76 m. We find the focal length from (1/do) + (1/di) = 1/f; [1/(3.2 m)] + [1/(– 1.76 m)] = 1/f, which gives f =

do



di

 

hi I



di

C

F

hi  I

F

– 3.9 m.

29. (a) To produce a smaller image located behind the surface of the mirror requires a (b) We find the image distance from the magnification: m = hi/ho = – di/do ; (3.5 cm)/(4.5 cm) = – di/(28 cm), which gives di = – 21.8 cm. As expected, di < 0. The image is located 22 cm behind the surface.

convex mirror.

C

Ch. 33

p. 6

(c) We find the focal length from (1/do) + (1/di) = 1/f; [1/(28 cm)] + [1/(– 21.8 cm)] = 1/f, which gives f = (d) The radius of curvature is r = 2f = 2(– 98 cm) = – 196 cm.

– 98 cm.

30. (a) To produce a larger image requires a concave mirror. (b) The image will be erect and virtual. (c) We find the image distance from the magnification: m = hi/ho = – di/do ; 1.3 = – di/(20.0 cm), which gives di = – 26.0 cm. We find the focal length from (1/do) + (1/di) = 1/f; [1/(20.0 cm)] + [1/(– 26.0 cm)] = 1/f, which gives f = 86.7 cm. The radius of curvature is r = 2f = 2(86.7 cm) = 173 cm. 31. For a mirror we have (1/do) + (1/di) = 1/f, or di = dof/(do – f); and m = hi/ho = – di/do = – f/(do – f). ¬ If we place the object so one end is at do1 and the other is at do2 = do1 – ¬, the image distances to the ends will be do1 di1 di1 = do1f/(do1 – f) and di2 = do2f/(do2 – f) = di1 – ¬. Thus the length of the image is ¬ = di1 – di2 = [do1f/(do1 – f)] – [(do1 – ¬)f/(do1 – ¬ – f)] = [(do1 – ¬ – f)do1f – (do1 – ¬)(do1 – f)f]/(do1 – f)(do1 – ¬ – f) = – ¬f 2/(do1 – f)(do1 – ¬ – f).

With ¬ « do1 , this becomes ¬ ˜ – ¬f 2/(do1 – f)2. Thus the longitudinal magnification is ¬/¬ = – m2. The negative sign indicates that the image is reversed front to back, as shown in the diagram.

32. We find the angle of refraction in the glass from n1 sin 1 = n2 sin 2 ; (1.00) sin 63° = (1.50) sin 2 , which gives 2 =

36°.

33. We find the angle of refraction in the air from n1 sin 1 = n2 sin 2 ; (1.33) sin 32.5° = (1.00) sin 2 , which gives 2 =

45.6°.

34. We find the incident angle in the water from n1 sin 1 = n2 sin 2 ; (1.33) sin 1 = (1.00) sin 76.0°, which gives 1 =

46.8°.

35. We find the incident angle in the air from n1 sin 1 = n2 sin 2 ;

¬

Ch. 33

p. 7

(1.00) sin 1 = (1.33) sin 43.0°, which gives 1 = 65.1°. Thus the angle above the horizon is 90.0° – 1 = 90.0° – 65.1° = 24.9°.

36. For the refraction at the first surface, we have nair sin 1 = n sin 2 ; (1.00) sin 45.0° = (1.50) sin 2 , which gives 2 = 28.1°. We find the angle of incidence at the second surface from (90° – 2) + (90° – 3) + A = 180°, which gives 3 = A – 2 = 60.0° – 28.1° = 31.9°. For the refraction at the second surface, we have n sin 3 = nair sin 4 ; (1.50) sin 31.9° = (1.00) sin 4 , which gives    4 = 52.4° from the normal. 37. We find the angle of incidence from the distances: tan 1 = L1/h1 = (2.7 m)/(1.3 m) = 2.076, so 1 = 64.3°. For the refraction from air into water, we have nair sin 1 = nwater sin 2 ; (1.00) sin 64.3° = (1.33) sin 2 , which gives 2 = 42.6°. We find the horizontal distance from the edge of the pool from L = L1 + L2 = L1 + h2 tan 2 = 2.7 m + (2.1 m) tan 42.6° = 4.6 m.

A 1 2

4

3

n

n air

1

h1 h2

L1

L2

38. The angle of reflection is equal to the angle of incidence: refl = 1 = 22 . For the refraction we have nair sin 1 = nglass sin 2 ; (1.00) sin 22 = (1.52) sin 2 . We use a trigonometric identity for the left-hand side: sin 22 = 2 sin 2 cos 2 = (1.52) sin 2 , or cos 2 = 0.760, so 2 = 40.5°. Thus the angle of incidence is 1 = 22 = 81.0°. 39. Because all of the surfaces are parallel, the angle of refraction from one surface is the angle of incidence at the next: n1 sin 1 = n2 sin 2 ; n2 sin 2 = n3 sin 3 . Thus we have n1 sin 1 = n3 sin 3 , which is the same as refraction from the first medium directly into the third medium.

n water

2

n1

n2

n3 3

2 1

2

Ch. 33

p. 8

40. (a) We find the angle in the glass from the refraction at the air–glass surface: n2 n3 n1 n1 sin 1 = n2 sin 2 ; (1.00) sin 43.5° = (1.58) sin 2 , which gives 2 = 25.8°. 3 (b) Because the surfaces are parallel, the refraction angle 2 from the first surface is the incident angle at the second 2 surface. We find the angle in the water from the refraction 1 at the glass–water surface: n2 sin 2 = n3 sin 3 ; (1.58) sin 25.8° = (1.33) sin 3 , which gives 3 = 31.2°. (c) If there were no glass, we would have n1 sin 1 = n3 sin 3; (1.00) sin 43.5° = (1.33) sin 3, which gives 3 = 31.2°. Note that, because the sides are parallel, 3 is independent of the presence of the glass. 41. Because the surfaces are parallel, the angle of refraction from the first surface is the angle of incidence at the second, Thus for the refractions, we have n1 sin 1 = n2 sin 2 ; n2 sin 2 = n1 sin 3 . When we add the two equations, we get n1 sin 1 = n1 sin 3 , which gives 3 = 1 . Because the ray emerges in the same index of refraction, it is undeviated.

42. Because the glass surfaces are parallel, the exit beam will be traveling in the same direction as the original beam. We find the angle inside the glass from nair sin  = n sin  . If the angles are small, we use cos  ˜ 1, and sin  ˜ , where  is in radians. (1.00)  = n, or  = /n. We find the distance along the ray in the glass from L = t/cos  ˜ t. We find the perpendicular displacement from the original direction from d = L sin ( – ) ˜ t( – ) = t[ – (/n)] = t(n – 1)/n.

n2

n1

n1 3

2

2

1

d

n –

 





L



n air = 1.00

t

n air = 1.00

Ch. 33

p. 9

43. We find the speed of light from the index of refraction, v = c/n. For the change, we have (vred – vviolet)/vviolet = [(c/nred) – (c/nviolet)]/(c/nviolet) = (nviolet – nred)/nred = (1.662 – 1.613)/(1.613) = 0.030 = 3.0%. 44. We find the speed of light from the index of refraction, v = c/n. For the change, we have (vblue – vred)/vred = [(c/nblue) – (c/nred)]/(c/nred) = (nred – nblue)/nblue = (1.613 – 1.643)/(1.643) = – 0.018 = – 1.8%.

45. We find the angles of refraction in the glass from n1 sin 1 = n2 sin 2 ; (1.00) sin 60.00° = (1.4820) sin 2,450 , which gives 2,450 = 35.76°; (1.00) sin 60.00° = (1.4742) sin 2,700 , which gives 2,700 = 35.98°. Thus the angle between the refracted beams is 2,700 – 2,450 = 35.98° – 35.76° = 0.22°. 46. For the refraction at the first surface, we have nair sin a = n sin b ; A (1.00) sin 45° = (1.643) sin b1 , which gives b1 = 25.49°; (1.00) sin 45° = (1.617) sin b2 , which gives b2 = 25.93°. We find the angle of incidence at the second surface from a (90° – b) + (90° – c) + A = 180°, which gives b c1 = A – b1 = 60.00° – 25.49° = 34.51°; c2 = A – b2 = 60.00° – 25.93° = 34.07°. n For the refraction at the second surface, we have n sin c = nair sin d ; (1.643) sin 34.51° = (1.00) sin 1 , which gives 1 = 68.6° from the normal; (1.617) sin 34.07° = (1.00) sin 2 , which gives 2 = 64.9° from the normal.

c

d

47. When the light in the material with a higher index is incident at the critical angle, the refracted angle is 90°: nlucite sin 1 = nwater sin 2 ; (1.51) sin 1 = (1.33) sin 90°, which gives 1 = 61.7°. Because lucite has the higher index, the light must start in lucite. 48. When the light in the liquid is incident at the critical angle, the refracted angle is 90°: nliquid sin 1 = nair sin 2 ; nliquid sin 51.3° = (1.00) sin 90°, which gives nliquid = 1.28.

Ch. 33

p. 10

49. We find the critical angle for light leaving the water: n sin 1 = sin 2 ; (1.33) sin C = sin 90°, which gives C = 48.75°. If the light is incident at a greater angle than this, it will totally reflect. We see from the diagram that R > H tan C = (82.0 cm) tan 48.75° = 93.5 cm.

air

R

n H

1 1

50. The ray reflects at the same angle, so each segment makes a 15° angle with the side. We find the distance L between reflections from d = L sin ; 5.41  10–4 m. 1.40  10–4 m = L sin 15°, which gives L =

51. We find the angle of incidence from the distances: tan 1 = L/h = (7.0 cm)/(8.0 cm) = 0.875, so 1 = 41.2°. For the maximum incident angle for the refraction from liquid into air, we have nliquid sin 1 = nair sin 2 ; nliquid sin 1max = (1.00) sin 90°, which gives sin 1max = 1/nliquid . Thus we have sin 1 = sin 1max = 1/nliquid ; sin 41.2° = 1/nliquid , or nliquid = 1.5. 52. For the refraction at the first surface, we have nair sin 1 = n sin 2 ; (1.00) sin 1 = n sin 2 , which gives sin 2 = sin 1/n. A We find the angle of incidence at the second surface from (90° – 2) + (90° – 3) + A = 180°, which gives 1 3 = A – 2 = 75° – 2 . For the refraction at the second surface, we have 2 n sin 3 = nair sin 4 = (1.00) sin 4. The maximum value of 4 before internal reflection takes n place at the second surface is 90°. Thus for internal reflection to occur, we have n sin 3 = n sin (A – 2) = 1. When we expand the left-hand side, we get n(sin A cos 2 – cos A sin 2) = 1. If we use the result from the first surface to eliminate n, we get sin 1 (sin A cos 2 – cos A sin 2)/(sin 2) = sin 1[(sin A/tan 2) – cos A] = 1, or 1/tan 2 = [(1/sin 1) + cos A]/sin A = [(1/sin 45°) + cos 75°]/sin 75° = 1.732, which gives tan 2 = 0.577, so 2 = 30°.

3

4

Ch. 33

p. 11

From the result for the first surface, we have nmin = sin 1/sin 2max = sin 45°/sin 30° = 1.414, so

n = 1.414.

53. For the refraction at the side of the rod, we have n2 sin  = n1 sin .  n1 The minimum angle for total reflection min occurs when  = 90°: a n2 sin min = (1.00)(1) = 1, or sin min = 1/n2 . 90° We find the maximum angle of refraction at the end of the rod from  max = 90° – min .  Because the sine function increases with angle, for the  refraction at the end of the rod, we have n2 n1 sin max = n2 sin max ; (1.00) sin max = n2 sin (90° – min) = n2 cos min . If we want total internal reflection to occur for any incident angle at the end of the fiber, the maximum value of  is 90°, so n2 cos min = 1. When we divide this by the result for the refraction at the side, we get tan min = 1, or min = 45°. Thus we have n2 = 1/sin min = 1/sin 45° = 1.414.

54. (a) The ray enters normal to the first surface, so there is no deviation there. The angle of incidence is 45° at the second surface. When there is air outside the surface, we have n1 sin 1 = n2 sin 2 ; n1 sin 45° = (1.00) sin 2 . For total internal reflection to occur, sin 2 = 1, so we have n1 = 1/sin 45° = 1.414. (b) When there is water outside the surface, we have n1 sin 1 = n2 sin 2 ; (1.50) sin 45° = (1.33) sin 2 , which gives sin 2 = 0.80. Because sin 2 < 1, the prism will not be totally reflecting. (c) For total reflection when there is water outside the surface, we have n1 sin 1 = n2 sin 2 ; n1 sin 45° = (1.33) sin 2 . For total internal reflection to occur, sin 2 = 1, so we have n1 = 1.33/sin 45° = 1.88. 55. We find the location of the image of a point on the bottom from the refraction from water to glass (with R = 8): (n1/do1) + (n2/di1) = (n2 – n1)/R = 0; h2 [(1.33)/(12.0 cm)] + [(1.50)/di1] = 0, which gives di1 = – 13.5 cm. Using this as the object for the refraction from glass to air, we find h1 the location of the final image (with R = 8): (n2/do2) + (n3/di2) = (n3 – n2)/R = 0; [(1.50)/(25.5 cm)] + [(1.00)/di2] = 0, which gives di2 = – 17.0 cm. Thus the bottom appears to be 17.0 cm below the surface of the glass.

n2 45°

n1

90°

45°

n3 n2 I2

O I1

n1

Ch. 33

p. 12

56. (a) When the fish is at the center of the bowl, the light has an incident angle of 0°, so there is no deviation. Thus the image of the fish will be at the center. Because the glass is thin, we can treat the refraction as if it was from water to air. We find the location of the image of the fish from (n1/do1) + (n2/di1) = (n2 – n1)/R; [(1.33)/(25.0 cm)] + [(1.00)/di1] = (1.00 – 1.33)/(– 25.0 cm), which gives di1 = – 25.0 cm (at the center of the bowl). (b) Because the glass is thin, we can treat the refraction as if it was from water to air. We find the location of the image of the fish from (n1/do2) + (n2/di2) = (n2 – n1)/R; [(1.33)/(20.0 cm)] + [(1.00)/di2] = (1.00– 1.33)/(– 25.0 cm), which gives di2 = – 18.8 cm (behind the glass).

57. From the diagram for a convex surface with n2 < n1 , we have A 2 =  + , 1 =  + ,    2 1 AB = BC tan  = OB tan  = IB tan , where  BC = R, OB = do , and IB = – di (from the sign convention). When the angles are small, sin  ˜ tan  ˜ ,    so we have B I O AB = R = do = – di . n 1 n 2 < n1 For the refraction at the surface, we have n1 sin 1 = n2 sin 2 . For small angles, this becomes n11 = n22 ; n1( + ) = n2( + ); n1[(AB/R) + (AB/do)] = n2[(AB/R) – (AB/di)], which gives (n1/do) + (n2/di) = (n2 – n1)/R. Figure 33–36 shows the result for a convex surface with n2 > n1 . From the diagram for a concave surface with n2 > n1 , we have

C

Ch. 33

p. 13

2 =  – , 1 =  – ,  2 AB = BC tan  = OB tan = IB tan , where BC = – R, OB = do , and IB = – di  (from the sign convention). A When the angles are small, sin  ˜ tan  ˜ , so we have 1 AB = – R = do = – di . For the refraction at the surface,    we have B I O C n1 sin 1 = n2 sin 2 . n1 n 2 > n1 For small angles, this becomes n11 = n22 ; n1( – ) = n2( – ); n1[(– AB/R) – (AB/do)] = n2[(– AB/R) – (– AB/di)], which gives (n1/do) + (n2/di) = (n2 – n1)/R. From the diagram for a concave surface with n2 < n1 , we have  2 2 =  – , 1 =  – ,  AB = BC tan  A = OB tan = IB tan , where BC = – R, OB = do , and IB = – di 1 (from the sign convention). When the angles are small,    sin  ˜ tan  ˜ , so we have B O I C AB = – R = do = – di . For the refraction at the surface, n1 n 2 < n1 we have n1 sin 1 = n2 sin 2 . For small angles, this becomes n11 = n22 ; n1( – ) = n2( – ); n1[(AB/do) – (– AB/R)] = n2[(– AB/di) – (– AB/R)], which gives (n1/do) + (n2/di) = (n2 – n1)/R. 58. The angle of refraction is 45°, so we find the angle of the ray y in the water from  n air = 1 n sin  = nair sin  = sin  ; d D (1.33) sin  = sin 45° , which gives  = 32.1°.  r This ray passes through the surface of the water at  x = D = H tan  = (1.00 m) tan 32.1° = 0.627 m; H y = H = 1.00 m. n  d To locate the image we must find the apparent source of a cone of rays reaching the viewer. We consider a second ray from x the coin at an incident angle of  + d. We can find the angle between the refracted rays from n sin  = sin  ; n cos  d = cos  d, which gives d = n cos  d/cos . We find the separation of the two rays at the surface from dD = H sec2  d. This distance subtends the angle d at the location of the image, a distance r from the point where the rays

Ch. 33

p. 14

pass through the surface. We use the distance perpendicular to the ray to find r: dD cos  = r d; H sec2  d cos  = rn cos  d/cos ; which gives r = H cos2 /n cos3  = (1.00 m)cos2 45°/(1.33) cos3 (32.1°) = 0.618 m. The location of the image is x = D – r sin  = 0.627 m – (0.618 m) sin 45° = 0.19 m; y = H – r cos  = 1.00 m – (0.618 m) cos 45° = 0.56 m. Thus the image is 0.56 m above and 0.19 m toward the viewer. 59. For a plane mirror each image is as far behind the mirror as the object is in front. Each reflection produces a front-to-back reversal. We show the three images and the two intermediate images that are not seen.

D

I3

I2

I1

d

L

I1 

I2 

(a) The first image is from a single reflection, so it is d1 = 2D = 2(1.5 m) = 3.0 m away. The second image is from two reflections, so it is d2 = L + d + D = 2.0 m + 0.5 m + 1.5 m = 4.0 m away. The third image is from three reflections, so it is d3 = 2L + D + D = 2(2.0 m) + 1.5 m + 1.5 m = 7.0 m away. (b) We see from the diagram that the first image is facing toward you; the second image is facing away from you; the third image is facing toward you.

60. We find the angle of incidence for the refraction from water into air: nwater sin 1 = nair sin 2 ; (1.33) sin 1 = (1.00) sin (90.0° – 14.0°), which gives 1 = 46.8°. We find the depth of the pool from tan 1 = x/h; tan 46.8° = (5.50 m)/h, which gives h = 5.16 m.

2

h

n air n water

1

x

Ch. 33

p. 15

61. For both mirrors the image is virtual (behind the mirror), so the image distances are negative. The image distance for the plane mirror is di1 = – do , and the image is upright and the same size. Because the angle subtended by the image is small, it is 1 = h1/(do – di1) = h/2do . The image distance for the convex mirror is di2 , and the image is upright and smaller. Because the angle subtended by the image is small, it is 2 = h2/(do – di2) = 1/2 = h/4do , or h2/h = (do – di2)/4do . We find the image distance from the magnification: m2 = h2/h = – di2/do = (do – di2)/4do , which gives di2 = – do/3. We find the required focal length of the convex mirror from (1/do) + (1/di2) = 1/f2 ; (1/do) + [1/(– do/3)] = 1/f2 , which gives f2 = – do/2. Thus the radius of curvature is R2 = 2f2 = 2(– do/2) = – do = – 3.80 m.

1

h O

h1 I1

do 2

O

h2 I2

62. At the critical angle, the refracted angle is 90°. For the refraction from plastic to air, we have nplastic sin plastic = nair sin air ; nplastic sin 37.3° = (1.00) sin 90°, which gives nplastic = 1.65. For the refraction from plastic to water, we have nplastic sin plastic = nwater sin water ; (1.65) sin plastic = (1.33) sin 90°, which gives plastic = 53.7°. 63. The two students chose different signs for the magnification, i. e., one upright and one inverted. The focal length of the concave mirror is f = R/2 = (40 cm)/2 = 20 cm. We relate the object and image distances from the magnification: m = – di/do ; ± 3 = – di/do , which gives di = — 3do . When we use this in the mirror equation, we get (1/do) + (1/di) = 1/f; (1/do) + [1/(— 3do)] = 1/f, which gives do = 2f/3, 4f/3 = 13.3 cm, 26.7 cm. The image distances are = – 40 cm (virtual, upright), and + 80 cm (real, inverted).

64.

Ch. 33

p. 16

I3 I4 I2

I1

30° 30°

30°

30° 30° 30°

30°

I5 Object

65. For the refraction at the second surface, we have n sin 3 = nair sin 4; (1.56) sin 3 = (1.00) sin 4. The maximum value of 4 before internal reflection takes place at the second surface is 90°. Thus for internal reflection not to occur, we have (1.56) sin 3 = 1.00; sin 3 = 0.641, so 3 = 39.9°. We find the refraction angle at the second surface from (90° – 2) + (90° – 3) + A = 180°, which gives 2 = A – 3 = 72° – 3 . Thus 2 = 72° – 39.9° = 32.1°. For the refraction at the first surface, we have nair sin 1 = n sin 2 ; (1.00) sin 1 = (1.56) sin 2 ; which gives sin 1 = (1.56) sin 2 . For the limiting condition, we have sin 1 = (1.56) sin 32.1° = 0.830, so 1 = 56.1°.

 1 2

n

3

4

Ch. 33

p. 17

66. For the refraction at the first surface, we have sin i = n sin 2 . We find the angle of incidence at the second surface from 3 =  – 2 . For the refraction at the second surface, we have n sin 3 = sin f . The total deviation angle is the sum of the deviations that take place at each surface: m = (f – 3) + (i – 2) = i + f –  = sin–1(n sin 2) + sin–1[n sin( – 2)] – . This gives the deviation as a function of 2 . To find the angle 2 that minimizes the deviation, we set dm/d2 = 0: d  m n cos  2 n cos( –  2 ) = – + 0 = 0, or 2 2 d 2 1 – (n sin  2) 1 – n sin(  –  2 ) n cos  2 n cos(  –  2 ) = . 2 2 1 – (n sin  2) 1 – n sin(  –  2 ) We can see by inspection that we have 2 =  – 2 , or 2 = !.

 i

 2

 3

f

m

n

Thus 3 =  – 2 = ! and i = f , so the internal ray is perpendicular to the bisector of . The minimum deviation angle is m = 2i – , or i = !(m +  ). From the first refraction equation, we have n = sin !(m +  )/sin !. sin i = n sin 2 , or

Ch. 33

p. 18

67. (a) The ray travels from point A, a distance H1 from the surface, to point B, a distance H2 from the surface, after reflecting from the surface. The component of the distance between A and B parallel to the surface is L. If x is the component of the distance between A and the point P, where the ray meets the surface, parallel to the surface, the time of travel is tAB = tAP + tPB

L x

P i r

H1 A

= + + {[(L – + }. We find the value of x for the minimum time from dtAB/dx = [(1/c)!(2x)/(x2 + H12)1/2] + {(1/c)!(– 2)(L – x)/[(L – x)2 + H22]1/2} = 0, which reduces to x/(x2 + H12)1/2 = (L – x)/[(L – x)2 + H22]1/2. From the diagram, we see that x/(x2 + H12)1/2 = sin i and (L – x)/[(L – x)2 + H22]1/2 = sin r , so we have sin i = sin r , or i = r . (b) The ray travels from point A in the top medium, a distance L A H1 from the surface, to point B in the bottom medium, a distance H2 from the surface. The component of the distance  H1 between A and B parallel to the surface is L. If x is the P component of the distance between A and the point P, where the ray meets the surface, parallel to the surface, the time x of travel is   tAB = tAP + tPB [(x2

H12)1/2/c]

x)2

H2 B

H22]1/2/c

= [(x2 + H12)1/2/(c/n1)] + {[(L – x)2 + H22]1/2/(c/n2)}. We find the value of x for the minimum time from dtAB/dx = [(n1/c)!(2x)/(x2 + H12)1/2] + {(n2/c)!(– 2)(L – x)/[(L – x)2 + H22]1/2}= 0, which reduces to n1x/(x2 + H12)1/2 = n2(L – x)/[(L – x)2 + H22]1/2. From the diagram, we see that x/(x2 + H12)1/2 = sin 1 and (L – x)/[(L – x)2 + H22]1/2 = sin 2 , so we have n1 sin 1 = n2 sin 2 .

n1 n2 H2 B

Ch. 33

p. 19

68. (a) For the refraction at the side of the rod, we have n2 sin 3 = n1 sin 4. n1 The minimum angle for total reflection 3min occurs 90° when 4 = 90°: 3 (1.54) sin 3min = (1.00)(1), which gives 3min = 40.5°. 2 We find the maximum angle of refraction at the end of the rod from 1 n2 2max = 90° – 3min = 90° – 40.5° = 49.5°. Because the sine function increases with angle, for the refraction at the end of the rod, we have n1 sin 1max = n2 sin 2max; (1.00) sin 1max = (1.54) sin 49.5°, which gives sin 1max = 1.17. The maximum value of the sine function is 1, so 1max = 90°, and total internal reflection will occur for all angles of incidence. (b) If the rod were in water, we would have n2 sin 3 = n1 sin 4 ; (1.54) sin 3min = (1.33)(1), which gives 3min = 59.7°. We find the maximum angle of refraction at the end of the rod from 2max = 90° – 3min = 90° – 59.7° = 30.3°. For the refraction at the end of the rod, we have n1 sin 1max = n2 sin 2max; (1.33) sin 1max = (1.54) sin 30.3°, which gives 1max = 35.7°. All incident angles between 0° and 35.7° will internally reflect inside the rod. 69. For the refraction at the convex surface, for which R > 0, we have (n1/do) + (n2/di) = (n2 – n1)/R; [(1.33)/(20 cm)] + [(1.50)/di] = (1.50 – 1.33)/(2.0 cm), which gives 70. For the refraction from the fiber to air we have n sin 1 = nair sin 2 . To have total internal reflection at the surface, n sin 1 > nair sin 90° = 1, or sin 1 > 1/n. The smallest angle of incidence, and thus the smallest value for sin 1 , occurs for rays parallel to the axis that are next to the inner side of the bend. From the diagram we see that sin 1 = r/ (r + d), so r/ (r + d) > 1/n, or r > d/(n – 1).

di = 81 cm (inside the glass).

2

n

n air 1

d r

r d

4

Ch. 34 p. 1

CHAPTER 34 – Lenses and Optical Instruments 1.

(a) From the ray diagram, the object distance is about 3& focal lengths, or

250 mm.

F I O

F

(b) We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/88.0 mm) = 1/65.0 mm, which gives do = 249 mm =

24.9 cm.

2.

To form a real image from a real object requires a converging lens. We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/285 cm) + (1/48.3 cm) = 1/f, which gives f = + 41.3 cm. Because di > 0, the image is real.

3.

(a) The power of the lens is P = 1/f = 1/0.275 m = 3.64 D, converging. (b) We find the focal length of the lens from P = 1/f; – 6.25 D = 1/f, which gives f = – 0.160 m = – 16.0 cm, diverging.

4.

(a) We locate the image from (1/do) + (1/di) = 1/f; (1/18 cm) + (1/di) = 1/24 cm, which gives di = – 72 cm. The negative sign means the image is 72 cm behind the lens (virtual). (b) We find the magnification from m = – di/do = – (– 72 cm)/(18 cm) = + 4.0.

5.

(a) Because the Sun is very far away, the image will be at the focal point. We find the size of the image from m = hi/ho = – di/do ; hi/(1.4  106 km) = – (28 mm)/(1.5  108 km), which gives hi = – 0.26 mm. (b) For a 50 mm lens, we have hi/(1.4  106 km) = – (50 mm)/(1.5  108 km), which gives hi = – 0.47 mm. (c) For a 200 mm lens, we have hi/(1.4  106 km) = – (200 mm)/(1.5  108 km), which gives hi = – 1.9 mm.

6.

We find the object distance from the required magnification (which is negative for a real object and a real image): m = hi/ho = – di/do ; – (2.70  103 mm)/(36 mm) = – (8.00 m)/do , which gives do = 0.107 m. We find the focal length of the lens from (1/do) + (1/di) = 1/f;

Ch. 34 p. 2

(1/0.107 m) + (1/8.00 m) = 1/f, which gives f = 0.105 m = 7.

+ 10.5 cm.

(a) We find the image distance from (1/do) + (1/di) = 1/f; (1/10.0  103 mm) + (1/di) = 1/80 mm, which gives di = 81 mm. (b) For an object distance of 3.0 m, we have 82 mm. (1/3.0  103 mm) + (1/di) = 1/80 mm, which gives di = (c) For an object distance of 1.0 m, we have 87 mm. (1/1.0  103 mm) + (1/di) = 1/80 mm, which gives di = (d) We find the smallest object distance from (1/domin) + (1/120 mm) = 1/80 mm, which gives domin = 240 mm =

24 cm.

8.

We find the image distance from (1/do) + (1/di) = 1/f; (1/0.125 m) + (1/di) = – 6.0 D, which gives di = – 0.071 m = – 7.1 cm (virtual image behind the lens). We find the height of the image from m = hi/ho = – di/do ; hi/(1.0 mm) = – (– 7.1 cm)/(12.5 cm), which gives hi = 0.57 mm (upright).

9.

(a) We see that the image is behind the lens, so it is virtual. (b) From the ray diagram we see that we need a converging lens. (c) We find the image distance from the magnification: m = – di/do ; + 2.5 = – di/(8.0 cm), which gives di = – 20 cm. We find power of the lens from (1/do) + (1/di) = 1/f = P, when f is in meters; (1/0.080 m) + [1/(– 0.20 m)] = P = 7.5 D.

I

F

O

F

10. We can relate the image and object distance from the magnification: m = – di/do , or do = – di/m. We use this in the lens equation: (1/do) + (1/di) = 1/f; – (m/di) + (1/di) = 1/f, which gives di = (1 – m)f. (a) If the image is real, di > 0. With f > 0, we see that m < 1; thus m = – 2.00. The image distance is di = [1 – (– 2.00)](50.0 mm) = 150 mm. The object distance is do = – di/m = – (150 mm)/(– 2.00) = 75.0 mm. (b) If the image is virtual, di < 0. With f > 0, we see that m > 1; thus m = + 2.00. The image distance is di = [1 – (+ 2.00)](50.0 mm) = – 50 mm. The object distance is do = – di/m = – (– 50 mm)/(+ 2.00) = 25.0 mm.

Ch. 34 p. 3

11. (a) We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/37.5 cm) + [1/(– 8.20 cm)] = 1/f, which gives f = The image is in front of the lens, so it is virtual. (b) We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/37.5 cm) + [1/(– 46.0 cm)] = 1/f, which gives f = The image is in front of the lens, so it is virtual.

– 10.5 cm (diverging).

+ 203 cm (converging).

12. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/1.10  103 mm) + (1/di) = 1/135 mm, which gives di = 154 mm (real, behind the lens). We find the height of the image from m = hi/ho = – di/do ; hi/(2.20 cm) = – (154 mm)/(1.10  103 mm), which gives hi = – 0.308 cm (inverted). (b) We find the image distance from (1/do) + (1/di) = 1/f; (1/1.10  103 mm) + (1/di) = 1/(– 135 mm), which gives di = – 120 mm (virtual, in front of the lens). We find the height of the image from m = hi/ho = – di/do ; hi/(2.20 cm) = – (– 120 mm)/(1.10  103 mm), which gives hi = + 0.240 cm (upright). 13. The sum of the object and image distances must be the Screen distance between object and screen: do + di = L = 76.0 cm. For the lens we have F I (1/do) + (1/di) = 1/f; (1/do) + [1/(76.0 cm – do)] = 1/(16.0 cm), O F which gives a quadratic equation: do2 – (76.0 cm)do + 1216 cm2 = 0, with the solutions do = 22.9 cm, 53.1 cm. Note that the sum of the distances is 76.0 cm, the distance between the object and the screen. This corresponds to interchanging the object and image. In general the screen must be at least 4f from the object for an image to be formed on the screen. 14. For a real object and image, both do and di must be positive, so the magnification will be negative: m = – di/do ; – 2.75 = – di/do , or di = 2.75do . We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/2.75do) = 1/(+ 75 mm), which gives do = 102 cm. The image distance is di = 2.75do = 2.75(102 cm) = 281 cm. The distance between object and image is L = do + di = 102 cm + 281 cm = 383 cm.

Ch. 34 p. 4

15. We get an expression for the image distance from the lens equation: (1/do) + (1/di) = 1/f; 1/di = (1/f) – (1/do), or di = fdo/(do – f). The magnification is m = – di/do = – f/(do – f). If the lens is converging, f > 0. For a real object, do > 0. When do > f, we have (do – f) > 0, so all factors in the expressions for di and m are positive; thus di > 0 (real), and m < 0 (inverted). When do < f, we have (do – f) < 0, so the denominator in the expressions for di and m are negative; thus di < 0 (virtual), and m > 0 (upright). For an object beyond the lens, do < 0. When – do > f, we have (do – f) < 0, so both numerator and denominator in the expression for di are negative; thus di > 0, so the image is real. The numerator in the expression for m is negative; thus m > 0, so the image is upright. When 0 < – do < f, we have (do – f) < 0, so we get the same result: real and upright. 16. (a) For the thin lens we have O (1/do) + (1/di) = 1/f; [1/( f + x)] + [1/(f + x)] = 1/f, which can be written as 2f + x + x = ( f + x)( f + x)/f x = f + (x + x) + (xx/f), or xx = f 2. (b) For the standard form we have (1/do) + (1/di) = 1/f; (1/45.0 cm) + (1/di) = 1/32.0 cm, which gives di = + 110.8 cm. (c) For the Newtonian form we have xx = f 2; (45.0 cm – 32.0 cm)x = (32.0 cm)2, which gives x = 78.8 cm. Thus the distance from the lens is di = x + f = 78.8 cm + 32.0 cm = 110.8 cm.

x

F F

I

17. We find the image formed by the refraction of the first lens: (1/do1) + (1/di1) = 1/f1 ; (1/35.0 cm) + (1/di1) = 1/27.0 cm, which gives di1 = + 118.1 cm. This image is the object for the second lens. Because it is beyond the second lens, it has a negative object distance: do2 = 16.5 cm – 118.1 cm = – 101.6 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; [1/(– 101.6 cm)] + (1/di2) = 1/27.0 cm, which gives di2 = + 21.3 cm. Thus the final image is real, 21.3 cm beyond second lens. The total magnification is the product of the magnifications for the two lenses: m = m1m2 = (– di1/do1)(– di2/do2) = di1di2/do1do2 = (+ 118.1 cm)(+ 21.3 cm)/(+ 35.0 cm)(– 101.6 cm) = – 0.708 (inverted).

Ch. 34 p. 5

18. The image of an infinite object formed by the refraction of the first lens will be at the focal point: di1 = f1 = + 20.0 cm. This image is the object for the second lens. Because it is beyond the second lens, it has a negative object distance: do2 = 14.0 cm – 20.0 cm = – 6.0 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; [1/(– 6.0 cm)] + (1/di2) = 1/(– 31.5 cm), which gives di2 = + 7.4 cm. Thus the final image is real, 7.4 cm beyond second lens. 19. (a) The image from the first lens (c) is the same: di1 = + 30.0 cm. This image is the object for the second lens. Because it is beyond the second lens, it has F1 F2 a negativeobject distance: O do2 = 20.0 cm – 30.0 cm = – 10.0 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; [1/(– 10.0 cm)] + (1/di2) = 1/25.0 cm, which gives di2 = + 7.14 cm. (b) The total magnification is the product of the magnifications for the two lenses: m = m1m2 = (– di1/do1)(– di2/do2) = di1di2/do1do2 = (+ 30.0 cm)(+ 7.14 cm)/(+ 60.0 cm)(– 10.0 cm) = – 0.357 (inverted). 20. We see from the ray diagram that the image from the first lens will be a virtual image at its focal point. This is a real object for the second lens, and must be at the focal point of the second lens. If L is the separation of the lenses, the focal length of the first lens is f1 = L – f2 = 24.0 cm – 31.0 cm = – 7.0 cm.

f1 < 0

I2 I1 F1

f2 > 0

f1 f2

21. We find the focal length by finding the image distance for an object very far away. For the first converging lens, we have (1/do1) + (1/di1) = 1/fC ;

(1/8) + (1/di1) = 1/fC , or, as expected, di1 = fC . The first image is the object for the second lens. If the first image is real, the second object distance is negative: do2 = – di1 = – fC . For the second diverging lens, we have (1/do2) + (1/di2) = 1/fD ; [1/(– fC)] + (1/di2) = 1/fD . Because the second image must be at the focal point of the combination, we have (– 1/fC) + (1/fT ) = 1/fD , which gives 1/fD = (1/fT) – (1/fC).

F2

Ch. 34 p. 6

22. We find the focal length of the lens from 1/f = (n – 1)[(1/R1) + (1/R2)]

= (1.58 – 1){[1/(– 33.4 cm)] + [1/(– 23.8 cm)]}, which gives

f = – 24.0 cm.

23. We find the index from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)]; 1/28.9 cm = (n – 1)[(1/31.0 cm) + (1/31.0 cm)], which gives n =

1.54.

24. When the surfaces are reversed, we get 1/f = (n – 1)[(1/R1) + (1/R2)]

= (1.51 – 1){[1/(– 18.4 cm)] + (1/8)}, which gives f = – 36.0 cm, which is the result from Example 34–7.

25. We find the radius from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)];

1/17.5 cm = (1.46 – 1)[(1/R1) + (1/8)], which gives R1 =

8.1 cm.

26. We find the radius from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)];

1/(– 21.5 cm) = (1.50 – 1)[(1/R1) + (1/8)], which gives R1 = The negative sign indicates concave.

27. We find the radius from the lensmaker’s equation: 1/f = (n – 1)[(1/R1) + (1/R2)]; + 2.50 D = (1.56 – 1)[(1/0.200 m) + (1/R2)], which gives R2 =

– 10.8 cm.

– 1.87 m (concave).

28. We find the focal length of the lens from 1/f = (n – 1)[(1/R1) + (1/R2)]

= (1.56 – 1){[1/(– 22.0 cm)] + [1/(+18.5 cm)]}, which gives f = 207.7 cm. We find the image distance from (1/do) + (1/di) = 1/f; (1/90.0 cm) + (1/di) = 1/207.7 cm, which gives di = – 159 cm (in front of the lens). The magnification is m = – di/do = – (– 159 cm)/(90.0 cm) = + 1.76 (upright).

29. The refraction equations are all based on n1 sin 1 = n2 sin 2 . The lensmaker’s equation is derived assuming air (n = 1.00) on the left-hand side. If we have some other material with n different from 1.00, we can make the equation equivalent to this by using an effective index: sin 1 = (n2/n1) sin 2 = neff sin 2 , where neff = n2/n1 . Thus we have Pair = 1/fair = (nglass – 1)[(1/R1) + (1/R2)]; Pwater = 1/fwater = [(nglass/nwater) – 1][(1/R1) + (1/R2)]. If we divide the two equations, we get Pwater/Pair = [(nglass/nwater) – 1]/(nglass – 1); Pwater/(+ 4.5 D) = [(1.50/1.33) – 1]/(1.50 – 1), which gives Pwater = + 1.15 D.

Ch. 34 p. 7

30. From the definition of the f-stop, we have f-stop = f/D; 1.4 = (45 mm)/D1.4 , which gives D1.4 = 32 mm; 22 = (45 mm)/D22 , which gives D22 = 2.0 mm. Thus the range of diameters is 2.0 mm = D = 32 mm. 31. We find the f-number from f-stop = f/D = (14 cm)/(6.0 cm) =

f/2.3.

32. The exposure is proportional to the area and the time: Exposure  At  D2t  t/(f-stop)2. Because we want the exposure to be the same, we have t1/(f-stop1)2 = t2/(f-stop2 )2; [(1/60) s]/(16)2 = [(1/1000) s]/(f-stop2)2, which gives f-stop2 = 33. From the similar triangles on the ray diagram, we find the separation of the circles: H/L1 = h/L2 ; (2.0 cm)/(100 cm) = h/(7.0 cm), which gives h = 0.14 cm = 1.4 mm. If we use rays from one object passing through the pinhole, we find the diameter of the circle from D/d = (L1 + L2)/L1 ; D/(1.0 mm) = (100 cm + 7.0 cm)/(100 cm), which gives D = 1.07 mm. Because this is less than h, the two circles do not overlap.

f/4.

H

h L1

L2

34. We find the effective f-number for the pinhole: f-stop2 = f/D = (70 mm)/(1.0 mm) = f/70. The exposure is proportional to the area and the time: Exposure  At  D2t  t/(f-stop)2. Because we want the exposure to be the same, we have t1/(f-stop1)2 = t2/(f-stop2 )2; [(1/250) s]/(11)2 = t2/(70)2, which gives t2 = (1/6) s. 35. The converging camera lens will form a real, inverted image. For the magnification, we have m = hi/ho = – di/do ; – (24  10–3 m)/(32 m) = – di/(55 m), or di = 4.13  10–2 m. We find the focal length of the lens from (1/do) + (1/di) = 1/f; (1/55 m) + (1/4.13  10–2 m) = 1/f, which gives f = 4.1  10–2 m = 41 mm. 36. The length of the eyeball is the image distance for a far object, i. e., the focal length of the lens. We find the f-number from f-stop = f/D = (20 mm)/(5.0 mm) = f/4.0. 37. The actual near point of the person is 50 cm. With the lens, an object placed at the normal near point, 25 cm, or 23 cm from the lens, is to produce a virtual image 50 cm from the eye, or 48 cm from the lens. We find the power of the lens from

Ch. 34 p. 8

(1/do) + (1/di) = 1/f = P, when distances are in m; (1/0.23 m) + (1/– 0.48 m) = P = + 2.3 D. 38. With the lens, the screen placed 50 cm from the eye, or 48.2 cm from the lens, is to produce a virtual image 120 cm from the eye, or 118.2 cm from the lens. We find the power of the lens from (1/do) + (1/di) = 1/f = P, when distances are in m; (1/0.482 m) + (1/– 1.182 m) = P = + 1.2 D. 39. With the contact lens, an object at infinity would have a virtual image at the far point of the eye. We find the power of the lens from (1/do) + (1/di) = 1/f = P, when distances are in m; (1/8) + (1/– 0.17 m) = P = – 5.9 D. To find the new near point, we have (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.12 m) = – 5.9 D, which gives do = 0.41 m. Glasses would be better, because they give a near point of 32 cm from the eye.

40. (a) The lens is diverging, so it produces images closer than the object, thus the person is (b) We find the far point by finding the image distance for an object at infinity: (1/do) + (1/di) = 1/f = P;

nearsighted.

(1/8) + (1/di) = – 4.0 D, which gives di = – 0.25 m = – 25 cm. Because this is the distance from the lens, the far point without glasses is 25 cm + 2.0 cm =

27 cm.

41. (a) We find the power of the lens for an object at infinity: (1/do) + (1/di) = 1/f = P;

– 1.33 D. (1/8) + (1/– 0.75 m) = P = (b) To find the near point with the lens, we have (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.25 m) = – 1.33 D, which gives do = 0.38 m =

38 cm.

42. The 2.0 cm of a normal eye is the image distance for an object at infinity, thus it is the focal length of the lens of the eye. To find the length of the nearsighted eye, we find the image distance (distance from lens to retina) for an object at the far point of the eye: (1/do) + (1/di) = 1/f; (1/17 cm) + (1/di) = 1/2.0 cm, which gives di = 2.27 cm. Thus the difference is 2.27 cm – 2.0 cm = 0.3 cm longer. 43. We find the far point of the eye by finding the image distance with the lens for an object at infinity: (1/do1) + (1/di1) = 1/f1 ;

(1/8) + (1/di1) = 1/– 25.0 cm, which gives di1 = – 25.0 cm from the lens, or 26.8 cm from the eye. We find the focal length of the contact lens from (1/do2) + (1/di2) = 1/f2 ; (1/8) + (1/– 26.8 cm) = 1/f2 , which gives f2 =

– 26.8 cm.

44. (a) We find the focal length of the lens for an object at infinity and the image on the retina: (1/do1) + (1/di1) = 1/f1 ; (1/8) + (1/2.0 cm) = 1/f1 , which gives f1 =

2.0 cm.

Ch. 34 p. 9

(b) With the object 30 cm from the eye, we have (1/do2) + (1/di2) = 1/f2 ; (1/30 cm) + (1/2.0 cm) = 1/f2 , which gives f2 =

1.9 cm.

45. We find the object distance for an image at her eye’s near point: (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.100 m) = – 4.0 D, which gives do = 1.7  10–1 m = 17 cm. We find the object distance for an image at her eye’s far point: (1/do) + (1/di) = 1/f = P; (1/do) + (1/– 0.200 m) = – 4.0 D, which gives do = 1.0 m = 100 cm. 46. The magnification with the image at infinity is M = N/f = (25 cm)/(11 cm) = 2.3. 47. We find the focal length from M = N/f; 3.0 = (25 cm)/f, which gives f =

8.3 cm.

48. (a) We find the focal length with the image at the near point from M = 1 + N/f1 ; 2.5 = 1 + (25 cm)/f1 , which gives f1 = 17 cm. (b) If the eye is relaxed, the image is at infinity, so we have M = N/f2 ; 2.5 = (25 cm)/f2 , which gives f2 = 10 cm. 49. Maximum magnification is obtained with the image at the near point. We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/– 25.0 cm) = 1/8.50 cm, which gives do = 6.3 cm from the lens. The magnification is M = 1 + N/f = 1 + (25.0 cm)/(8.50 cm) = 3.9. 50. (a) The angular magnification with the image at the near point is M = 1 + N/f = 1 + (25.0 cm)/(9.00 cm) = 3.78. (b) Because the object without the lens and the image with the lens are at the near point, the angular magnification is also the ratio of widths: M = hi/ho; 3.78 = hi/(3.70 mm), which gives hi = 14.0 mm. (c) We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/– 25.0 cm) = 1/9.00 cm, which gives do = 6.62 cm from the lens.

hi



ho

I

O N

ho

 O

Ch. 34 p. 10

51. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/5.35 cm) + (1/di) = 1/6.00 cm, which gives di = – 49.4 cm. (b) From the diagram we see that the angular magnification is M =  / = (ho/do)/(ho/N) = N/do = (25 cm)/(5.35 cm) = 4.7.

hi



ho

I

O N ho

 O

52. (a) We find the image distance from (1/do) + (1/di) = 1/f; (1/8.5 cm) + (1/di) = 1/9.5 cm, which gives di = – 81 cm. (b) The linear magnification is m = – di/do = – (– 81 cm)/(8.5 cm) = 9.5. (c) The angular magnification is M =  / = (ho/do)/(ho/N) = N/do = (25 cm)/(8.5 cm) = 2.9. Note that this is less than the linear magnification because the image is farther away. 53. We find the focal length of the lens from M = N/f; 3.0 = (25 cm)/f, which gives f = 8.3 cm. (a) The magnification with the image at infinity is M1 = N1/f = (60 cm)/(8.3 cm) = 7.2. (b) The magnification with the image at infinity is M2 = N2/f = (18 cm)/(8.3 cm) = 2.2. Without the lens, the closest an object can be placed is the near point. A farther near point means a smaller angle subtended by the object without the lens, and thus greater magnification when the lens is used. 54. The magnification of the telescope is given by M = – fo/fe = – (72 cm)/(2.8 cm) = – 26. For both object and image far away, the separation of the lenses is L = fo + fe = 72 cm + 2.8 cm = 75 cm. 55. We find the focal length of the eyepiece from the magnification: M = – fo/fe ; – 25 = – (80 cm)/fe , which gives fe = 3.2 cm. For both object and image far away, the separation of the lenses is L = fo + fe = 80 cm + 3.2 cm = 83 cm.

Ch. 34 p. 11

56. We find the focal length of the objective from the magnification: M = fo/fe ; 8.0 = fo/(3.0 cm) , which gives fo = 24 cm. 57. We find the focal length of the eyepiece from the power: fe = 1/P = 1/35 D = 0.0286 m = 2.86 cm. The magnification of the telescope is given by M = – fo/fe = – (95 cm)/(2.86 cm) = – 33. 58. For both object and image far away, we find the focal length of the eyepiece from the separation of the lenses: L = fo + fe ; 76.0 cm = 74.5 cm + fe , which gives fe = 1.5 cm. The magnification of the telescope is given by M = – fo/fe = – (74.5 cm)/(1.5 cm) = – 50.

59. For both object and image far away, we find the (negative) focal length of the eyepiece from the separation of the lenses: L = fo + fe ; 33.0 cm = 36.0 cm + fe , which gives fe = – 3.0 cm. The magnification of the telescope is given by M = – fo/fe = – (36.0 cm)/(– 3.0 cm) = 12. 60. The reflecting mirror acts as the objective, with a focal length fo = r/2 = (6.2 m)/2 = 3.1 m. The magnification of the telescope is given by M = – fo/fe = – (310 cm)/(2.8 cm) = – 110. 61. We find the focal length of the mirror from M = – fo/fe ; – 120 = – fo/(3.3 cm), which gives fo = 4.0  102 cm = The radius is r = 2fo = 2(4.0 m) = 8.0 m. 62. For the magnification we have M = – fo/fe = – 130, or fo = 130fe . For both object and image far away, we have L = fo + fe ; 1.25 m = 130fe + fe , which gives fe = 9.54  10–3 m = The focal length of the objective is fo = 130fe = 130(9.54  10–3 m) = 1.24 m.

4.0 m.

9.54 mm.

Ch. 34 p. 12

63. We assume a prism binocular so the magnification is positive, doi fe but simplify the diagram by ignoring the prisms. We find  the focal length of the eyepiece from the design magnification:  h  M = fo/fe ; 7.0 = (26 cm)/fe , which gives fe = 3.71 cm. I We find the intermediate image formed by the objective: (1/doo) + (1/doi) = 1/fo ; (1/400 cm) + (1/doi) = 1/26 cm, which gives doi = 27.8 cm. With the final image at infinity (relaxed eye), the intermediate image will be at the focal point of the eyepiece lens. From the diagram the angle subtended by the object is  = h/doi , while the angle subtended by the image is   = h/fe . Thus the angular magnification is M =  / = (h/fe)/(h/doi) = doi/fe = (27.8 cm)/(3.71 cm) = 7.5. 64. The magnification of the microscope is M = N¬/fo fe = (25 cm)(17.5 cm)/(0.65 cm)(1.6 cm) =

420.

65. We find the focal length of the eyepiece from the magnification of the microscope: M = N¬/fo fe ; 650 = (25 cm)(17.5 cm)/(0.40 cm) fe , which gives fe = 1.7 cm.

66. (a) The total magnification is M = MoMe = (56.0)(12.0) = 672. (b) With the final image at infinity, we find the focal length of the eyepiece from Me = N/fe ; 12.0 = (25.0 cm)/fe , which gives fe = 2.08 cm. Because the image from the objective is at the focal point of the eyepiece, the image distance for the objective is di = ¬ – fe = 20.0 cm – 2.08 cm = 17.9 cm. We find the object distance from the magnification of the objective: Mo = di/do ; 56.0 = (17.9 cm)/do , which gives do = 0.320 cm. We find the focal length of the objective from the lens equation: (1/do) + (1/di) = 1/fo ; (1/0.320 cm) + (1/17.9 cm) = 1/fo , which gives fo = 0.314 cm. (c) We found the object distance: do = 0.320 cm. 67. (a) Because the image from the objective is at the focal point of the eyepiece, the image distance for the objective is dio = ¬ – fe = 16.0 cm – 1.8 cm = 14.2 cm. We find the object distance from the lens equation for the objective: (1/doo) + (1/dio) = 1/fo ; (1/doo) + (1/14.2 cm) = 1/0.80 cm, which gives doo = 0.85 cm. (b) With the final image at infinity, the magnification of the eyepiece is Me = N/f = (25.0 cm)/(1.8 cm) = 13.9. The magnification of the objective is

Ch. 34 p. 13

Mo = dio/doo = (14.2 cm)/(0.85 cm) = 16.7. The total magnification is M = MoMe = (16.7)(13.9) = 230. 68. (a) We find the object distance from the lens equation for the eyepiece: (1/doe) + (1/die) = 1/fe ; (1/doe) + (1/– 25 cm) = 1/1.8 cm, which gives doe = 1.7 cm. The image distance for the objective is dio = ¬ – doe = 16.0 cm – 1.7 cm = 14.3 cm. We find the object distance from the lens equation for the objective: (1/doo) + (1/dio) = 1/fo ; (1/doo) + (1/14.3 cm) = 1/0.80 cm, which gives doo = 0.85 cm. (b) With the final image at the near point, the magnification of the eyepiece is Me = 1 + N/f = 1 + (25.0 cm)/(1.8 cm) = 14.9. The magnification of the objective is Mo = dio/doo = (14.3 cm)/(0.85 cm) = 16.9. The total magnification is M = MoMe = (16.9)(14.9) = 250.

69. (a) We find the image distance from the lens equation for the objective: (1/doo) + (1/dio) = 1/fo ; (1/0.790 cm) + (1/dio) = 1/0.740 cm, which gives dio = 11.7 cm. For the relaxed eye, the image from the objective is at the focal point of the eyepiece: doe = 2.70 cm. The distance between lenses is ¬ = dio + doe = 11.7 cm + 2.70 cm = 14.4 cm. (b) With the final image at infinity, the magnification of the eyepiece is Me = N/f = (25.0 cm)/(2.70 cm) = 9.26. The magnification of the objective is Mo = dio/doo = (11.7 cm)/(0.790 cm) = 14.8. The total magnification is M = MoMe = (14.8)(9.26) = 137. 70. (a) When lenses are in contact, the powers add: P = P1 + P2 = (1/– 0.28 m) + (1/0.23 m) = + 0.776 D. It is a positive lens, and thus converging. (b) The focal length is f = 1/P = 1/0.776 D = 1.3 m.

Ch. 34 p. 14

71. (a) We find the incident angle from sin 1 = h1/R = (1.0 cm)/(12.0 cm) = 0.0833, so 1 = 4.78°. For the refraction at the curved surface, we have sin 1 = n sin 2 ; sin 4.78° = (1.50) sin 2 , which gives sin 2 = 0.0556, so 2 = 3.18°. We see from the diagram that 3 = 1 – 2 = 4.78° – 3.18° = 1.60°. For the refraction at the flat face, we have n sin 3 = sin 4 ; (1.50) sin 1.60° = sin 4 , which gives sin 4 = 0.0419, so 4 = 2.40°.

1

Ch. 34 p. 15

h

 1 –  2 = 3

A 2

B

3 h 1 If we use the law of sines for the triangle ABC, we have AB/sin (90° – 1) = h1/sin 2 ; C n AB/sin 85.22° = h1/sin 3.18°, which gives AB = 17.96h1. R We see from the diagram that h1 = h1 – AB sin 3 ; h1 = 1.0 cm – 17.96h1 sin 1.60°, which gives h1 = 0.666 cm. Thus the distance from the flat face to the point where the ray crosses the axis is d1 = h1/tan 4 = (0.666 cm)/tan 2.40° = 15.9 cm. (b) We find the incident angle from sin 1 = h2/R = (6.0 cm)/(12.0 cm) = 0.500, so 1 = 30.00°. For the refraction at the curved surface, we have sin 1 = n sin 2 ; sin 30.00° = (1.50) sin 2 , which gives sin 2 = 0.333, so 2 = 19.47°. We see from the diagram that 3 = 1 – 2 = 30.00° – 19.47° = 10.53°. For the refraction at the flat face, we have n sin 3 = sin 4 ; (1.50) sin 10.53° = sin 4 , which gives sin 4 = 0.274, so 4 = 15.91°. If we use the law of sines for the triangle ABC, we have AB/sin (90° – 1) = h2/sin 2 ; AB/sin 60.00° = h2/sin 19.47°, which gives AB = 2.598h2. We see from the diagram that h2 = h2 – AB sin 3 ; h2 = 6.0 cm – 2.598h2 sin 10.53°, which gives h2 = 4.07 cm. Thus the distance from the flat face to the point where the ray crosses the axis is d2 = h2/tan 4 = (4.07 cm)/tan 15.91° = 14.3 cm. (c) The separation of the “focal points” is ?d = d1 – d2 = 15.9 cm – 14.3 cm = 1.6 cm. (d) When h2 = 6.0 cm, the rays focus closer to the lens, so they will form a circle at the “focal point” for h1 = 1.0 cm. We find the radius of this circle from similar triangles: h2/d2 = r/(d1 – d2); (4.07 cm)/(14.3 cm) = r/(1.6 cm), which gives r = 0.46 cm.

72. For an object at infinity, the image will be in the focal plane, so we have d1 = f = 135 mm. When the object is at 1.40 m, we locate the image from (1/do) + (1/di) = 1/f; (1/1400 mm) + (1/di) = 1/135 cm, which gives di = 149 mm. Thus the distance from the lens to the film must change by di – f = 149 mm – 135 mm = 14 mm. 73. We find the object distances from (1/do) + (1/di) = 1/f;

(1/do1) + (1/200.0 mm) = 1/200 mm, which gives do1 = 8; (1/do2) + (1/206.0 mm) = 1/200 mm, which gives do2 = 6.87  103 mm = 6.87 m.

Thus the range of object distances is

6.87 m = do = 8.

4

4

d

Ch. 34 p. 16

74. When an object is very far away, the image will be at the focal point di = f. Thus the magnification is m = – di/do = – f/do , that is, proportional to f. 75. If we recognize that the image is inverted, for the magnification we have m = hi/ho = – di/do = – 1, so di = do . We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/do) = 1/50 mm, which gives do = 100 mm. The distance between the object and the film is d = do + di = 100 mm + 100 mm = 200 mm. 76. The magnification for a relaxed eye is M = N/f = NP = (0.25 m)(+ 8.0 D) =

2.0.

77. We find the focal lengths of the lens for the two colors: 1/fred = (nred – 1)(1/R1 + 1/R2)

= (1.5106 – 1)[(1/18.4 cm) + (1/8)] which gives fred = 36.04 cm. 1/fyellow = (nyellow – 1)(1/R1 + 1/R2)

= (1.5226 – 1)[(1/18.4 cm) + (1/8)] which gives fyellow = 35.21 cm. We find the image distances from (1/do) + (1/dired) = 1/fred ; (1/66.0 cm) + (1/dired) = 1/36.04 cm, which gives dired = 79.4 cm. (1/do) + (1/diyellow) = 1/fyellow ; (1/66.0 cm) + (1/diyellow) = 1/35.21 cm, which gives diyellow = 75.5 cm. The images are 3.9 cm apart, an example of chromatic aberration. 78. We can relate the object and image distances from the magnification of the inverted image: m = hi/ho = – di/do ; (– 8.2510–3 m)/(1.75 m) = – di/do , or di = (4.7110–3)do . We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + [1/(4.7110–3)do] = 1/20010–3 m, which gives do = 42.7 m. Thus the reporter was standing 42.7 m from his subject. 79. We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/7.50  103 mm) = 1/100 mm, which gives do = 101 mm = We find the size of the image from m = hi/ho = – di/do ; hi/(0.036 m) = – (7.50 m)/(0.101 m), which gives hi = – 2.7 m.

0.101 m.

80. We find the focal length by finding the image distance for an object very far away. For the first lens, we have (1/do1) + (1/di1) = 1/f1 ;

(1/8) + (1/di1) = 1/(12.0 cm), or, as expected, di1 = 12.0 cm. The first image is the object for the second lens. The first image is real, so the second object is

Ch. 34 p. 17

virtual: do2 = – di1 = – 12.0 cm. For the second lens, we have (1/do2) + (1/di2) = 1/f2 ; [1/(– 12.0 cm)] + (1/di2) = 1/(– 20.0 cm), which gives di2 = + 30.0 cm. Because the second image must be at the focal point of the combination, we have f = + 30.0 cm (converging). 81. We get an expression for the image distance from the lens equation: (1/do) + (1/di) = 1/f; 1/di = (1/f) – (1/do), or di = fdo/(do – f). If the lens is diverging, f < 0. If we write f = –  f , we get di = –  f do/(do +  f ). For a real object, do > 0. All factors in the expression for di are positive, thus di < 0, so the image is always virtual. We can have a real image, di > 0, if do < 0 (the object is formed by a preceding device and is behind the lens), and  do  <  f , so the denominator is still positive. Thus to have a real image from a diverging lens, the condition is 0 < – do < – f. 82. (a)

f1

f2 F1

O

F1

I1

F2 F2

I2

We see that the image is real and upright, and estimate that it is 20 cm beyond the second lens. (b) We find the image formed by the refraction of the first lens: (1/do1) + (1/di1) = 1/f1 ; (1/30 cm) + (1/di1) = 1/15 cm, which gives di1 = + 30 cm. This image is the object for the second lens. Because it is in front of the second lens, it is a real object, with an object distance of do2 = 50 cm – 30 cm = 20 cm. We find the image formed by the refraction of the second lens: (1/do2) + (1/di2) = 1/f2 ; (1/20 cm) + (1/di2) = 1/10 cm, which gives di2 = + 20 cm. Thus the final image is real, 20 cm beyond second lens. The total magnification is the product of the magnifications for the two lenses: m = m1m2 = (– di1/do1)(– di2/do2) = di1di2/do1do2 = (+ 30 cm)(+ 20 cm)/(+ 30 cm)(+ 20 cm) = + 1.0. Thus the final image is upright, same size as object.

83. (a) We use the lens equation with do + di = dT : (1/do) + (1/di) = 1/f; (1/do) + [1/(dT – do) = 1/f.

Ch. 34 p. 18

When we rearrange this, we get a quadratic equation for do : do2 – dTdo + dT f = 0, which has the solution do = ![dT ± dT2 – 4dT f]. If dT > 4 f, we see that the term inside the square root is positive: dT2 – 4dT f > 0, and dT2 – 4dT f < dT , so we get two real, positive solutions for do . (b) If dT < 4 f, we see that the term inside the square root is negative: dT2 – 4dT f < 0, so there are no real solutions for do . (c) When there are two solutions, the distance between them is dT2 – 4dT f. ?d = do1 – do2 = ![dT + dT2 – 4dT f] – ![dT – dT2 – 4dT f] = The image positions are given by di = dT – do = ![dT — dT2 – 4dT f]. The ratio of image sizes is the ratio of magnifications: m = m2/m1 = (di2/do2)/(di1/do1) = (di2/do2)(do1/di1) =

1 2 1 2

dT +

d T2 – 4dT f

dT –

dT – 4dT f

2

2

=

2

dT +

d T – 4d T f

dT –

dT – 4dT f

2

2

.

84. (a) We find the focal length by finding the image distance for an object very far away. For the first lens, we have (1/do1) + (1/di1) = 1/f1 ;

(1/8) + (1/di1) = 1/f1 , or, as expected, di1 = f1 . The first image is the object for the second lens. If the first image is real, the second object is virtual: do2 = – di1 = – f1 . For the second lens, we have (1/do2) + (1/di2) = 1/f2 ; [1/(– f1)] + (1/di2) = 1/f2 . Because the second image must be at the focal point of the combination, we have (– 1/f1) + (1/fT ) = 1/f2 , which gives 1/fT = (1/f1) + (1/f2). When we solve for fT , we get fT = f1 f2/(f1 + f2). (b) If we use the intermediate result 1/fT = (1/f1) + (1/f2), we see that P = P1 + P2 .

85. The refraction equations for the two surfaces are all based on n sin 1 = n sin 2 , and n sin 3 = n sin 4 .

Ch. 34 p. 19

Eq. 34–4 was derived based on having air (n = 1) outside the lens. We can make these equivalent to having air outside the lens, if we write them as sin 1 = (n/n) sin 2 , and (n/n) sin 3 = sin 4 . Thus we see that we have an effective index, neff = n/n, to use in Eq. 34–4: 1 = (n/ n) – 1 1 + 1 . f R1 R2 (n/ n) – 1 The focal length in air is 1 = n – 1 1 + 1 ; so we have 1 = . f R1 R2 f (fn – 1) Eq. 34–2 is derived using the focal points. Thus, if we use the above focal length, the derivation is the same, so we have the same equation: 1 + 1 = 1 , wh ere 1 = (n / n) – 1 . do d i f f f(n – 1) Eq. 34–3 is derived by comparing heights and distances, so it is unchanged: m = hi/ho = – di/do . 86. (a) With a positive (converging) lens, Sam is farsighted. (b) The focal length of the lens is f = 1/P = 1/2.5 D = 0.40 m = 40 cm. (c) The lens produces a virtual image at his near point: (1/do1) + (1/di1) = 1/f = P; (1/0.23 m) + (1/di1) = + 2.5 D, which gives di1 = – 0.54 m, so his near point is (d) For Pam, we find the object distance that will have an image at her near point: (1/do2) + (1/di2) = 1/f = P; (1/do2) + (1/– 0.23 m) = + 2.5 D, which gives do2 = 0.15 m, which is 17 cm

56 cm.

from her eyes.

87. To find the new near point, we have (1/do1) + (1/di1) = 1/f1 = P1 , when distances are in m; (1/0.35 m) + (1/di1) = + 2.5 D, which gives di1 = – 2.8 m. To give him a normal near point, we have (1/do2) + (1/di2) = 1/f2 = P2 ; (1/0.25 m) + (1/– 2.8 m) = P2 = + 3.6 D. 88. The actual far point of the person is 150 cm. With the lens, an object far away is to produce a virtual image 150 cm from the eye, or 148 cm from the lens. We find the power of the lens from (1/do1) + (1/di1) = 1/f1 = P1 , when distances are in m;

– 0.68 D (upper part). (1/8) + (1/– 1.48 m) = P1 = The actual near point of the person is 40 cm. With the lens, an object placed at the normal near point, 25 cm, or 23 cm from the lens, is to produce a virtual image 40 cm from the eye, or 38 cm from the lens. We find the power of the lens from (1/do2) + (1/di2) = 1/f2 = P2 ; (1/0.23 m) + (1/– 0.38 m) = P2 = + 1.7 D (lower part).

89. The maximum magnification is achieved with the image at the near point: M1 = 1 + N1/f = 1 + (15.0 cm)/(8.0 cm) = 2.9. For the normal eye of an adult we have M2 = 1 + N2/f = 1 + (25.0 cm)/(8.0 cm) = 4.1.

90. The exposure is proportional to the intensity, the area and the time:

Ch. 34 p. 20

Exposure  IAt  ID2t  It/(f-stop)2. With the same shutter speed, the time is constant. Because we want the exposure to be the same, we have I1/(f-stop1)2 = I2/(f-stop2 )2; I1/(5.6)2 = I2/(22)2, which gives I2 = 16I1. Note that we have followed convention to use multiples of 2. 91. (a) The magnification of the telescope is given by M = – fo/fe = – Pe/Po = – (5.0 D)/(2.0 D) = – 2.5. (b) To get a magnification greater than 1, for the eyepiece we use the lens with the smaller focal length, or greater power: 5.0 D. 92. (a) When an object is very far away, the image will be at the focal point di1 = f. From the magnification of the inverted image, we have m = hi/ho = – di/do , so we see that hi is proportional to di . When the object and image are the same size, we get hi/ho = – 1 = – di2/do2 , so do2 = di2 . From the lens equation, we get (1/do2) + (1/di2) = 1/f; (1/di2) + (1/di2) = 1/f, which gives di2 = 2f. The required exposure time is proportional to the area of the image on the film: t  A  (hi)2  (di)2. When we form the ratio, w get t2/t1 = (di2/di1)2 = (2f/f)2 = 4. (b) For the increased object distances, we have (1/do3) + (1/di3) = 1/f; (1/4f) + (1/di3) = 1/f, which gives di3 = 4f/3; and t3/t1 = (4/3)2 = 1.78. (1/do4) + (1/di4) = 1/f; (1/5f) + (1/di4) = 1/f, which gives di4 = 5f/4; and t4/t1 = (5/4)2 = 1.56. These increased exposures are less than the minimal adjustment on a typical camera, so they are negligible. 93. The focal length of the eyepiece is fe = 1/Pe = 1/25 D = 4.010–2 m = 4.0 cm. For both object and image far away, we find the focal length of the objective from the separation of the lenses: L = fo + fe ; 85 cm = fo + 4.0 cm, which gives fo = 81 cm. The magnification of the telescope is given by M = – fo/fe = – (81 cm)/(4.0 cm) = – 20.

Ch. 35 p. 1

CHAPTER 35 – The Wave Nature of Light; Interference 1.

We draw the wavelets and see that the incident wave fronts are parallel, with the angle of incidence 1 being the angle between the wave fronts and the surface. The reflecting wave fronts are parallel, with the angle of reflection 2 being the angle between the wave fronts and the surface. Both sets of wave fronts are in the same medium, so they travel at the same speed. The perpendicular distance between wave fronts is BC = AD = c ?t. From the triangles, we see that AB = BC/sin 1 = AD/sin 2 . Thus we have sin 1 = sin 2 , or 1 = 2 .

c²t

c²t D

C

2 2

1 1

2

A

1 B

2.

For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … . For the fifth order, we have 540 nm. (1.610–5 m) sin 9.8° = (5), which gives  = 5.410–7 m =

3.

For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … . For the third order, we have d sin 18° = (3)(61010–9 m), which gives d = 5.910–6 m = 5.9 m.

4.

For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … . We find the location on the screen from y = L tan . For small angles, we have sin  ˜ tan , which gives y = L(m/d) = mL/d. For adjacent fringes, ?m = 1, so we have ?y = L?m/d; 0.62 m. 0.065 m = (5.00 m)(1)/(0.04810–3 m), which gives  = 6.2410–7 m = The frequency is f = c/ = (3.00108 m/s)/(6.2410–7 m) = 4.81014 Hz.

5.

For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … . We find the location on the screen from y = L tan . For small angles, we have sin  ˜ tan , which gives y = L(m/d) = mL/d. For adjacent fringes, ?m = 1, so we have ?y = L?m/d;

Ch. 35 p. 2

= (3.6 m)(65610–9 m)(1)/(0.06010–3 m) = 3.910–2 m =

3.9 cm.

6.

For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … . We find the location on the screen from y = L tan . For small angles, we have sin  ˜ tan , which gives y = L(m/d) = mL/d. For the fourth order we have 3810–3 m = (2.0 m)(68010–9 m)(4)/d, which gives d = 1.410–4 m = 0.14 mm.

7.

For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … . We find the location on the screen from y = L tan . For small angles, we have sin  ˜ tan , which gives y = L(m/d) = mL/d. For the second order of the two wavelengths, we have ?y = mL ?/d = 2(1.0 m)[(720 – 660)10–9 m]/(0.5810–3 m) = 2.0710–4 m =

0.21 mm.

8.

The 180° phase shift produced by the glass is equivalent to a path length of !. For constructive interference on the screen, the total path difference is a multiple of the wavelength: !d sin  = m, m = 0, ± 1, ± 2, ± 3, … ; or d sin  = (m – !), m = 0, ± 1, ± 2, ± 3, … . For destructive interference on the screen, the total path difference is !d sin  = (m + !), m = 0, ± 1, ± 2, ± 3, … ; or d sin  = m, m = 0, ± 1, ± 2, ± 3, … . Thus the pattern is just the reverse of the usual double-slit pattern.

9.

For constructive interference of the second order for the blue light, we have d sin  = mb = (2)(460 nm) = 920 nm. For destructive interference of the other light, we have d sin  = (m + !), m = 0, 1, 2, 3, … . When the two angles are equal, we get 920 nm = (m + !), m = 0, 1, 2, 3, … . For the first three values of m, we get 920 nm = (0 + !), which gives  = 1.84103 nm; 920 nm = (1 + !), which gives  = 613 nm; 920 nm = (2 + !), which gives  = 368 nm. The only one of these that is visible light is 613 nm.

10. The presence of the water changes the wavelength: water = /nwater = 480 nm/1.33 = 360 nm. For constructive interference, the path difference is a multiple of the wavelength in the water: d sin  = mwater , m = 0, 1, 2, 3, … . We find the location on the screen from y = L tan . For small angles, we have sin  ˜ tan , which gives y = L(mwater/d) = mLwater/d. For adjacent fringes, ?m = 1, so we have ?y = Lwater?m/d; = (0.400 m)(36010–9 m)(1)/(6.0010–5 m) = 2.4110–3 m = 2.41 mm.

Ch. 35 p. 3

11. To change the center point from constructive interference to destructive interference, the phase shift produced by the introduction of the plastic must be an odd multiple of half a wavelength, corresponding to the change in the number of wavelengths in the distance equal to the thickness of the plastic. The minimum thickness will be for a shift of a half wavelength: N = (t/plastic) – (t/) = (tnplastic/) – (t/) = (t/)(nplastic – 1) = !; [t/(640 nm)](1.60 – 1) = !, which gives t =

533 nm.

12. If E0 is the amplitude of the electric field at the center of the screen from the light of one slit, for the constructive interference when both slits are open the amplitude is 2E0. Thus the ratio of intensities is I01/I02 = (E0/2E0)2 =

#.

13. The intensity of the pattern is I = I0 cos2 [(pd sin )/]. At the central maximum, we have  = 0 and I = I0. We find the angle where the intensity is half its maximum value from I = !I0 = I0 cos2 [(pd sin 1/2)/], or cos [(pd sin 1/2)/] = 1/v2; (pd sin 1/2)/ = p/4, which gives 1/2 = sin–1 (/4d). If  « d, the angle will be small, so 1/2 ˜ sin 1/2. The angular width is twice this angle: 1/2 ˜ 2/4d = /2d.

14. A doubling of the intensity means the electric field amplitude increases by a factor of v2. If E0 is the amplitude of the electric field of one slit, for the other slit it will be E0v2. At an angle where the phase difference is  = 2pd (sin )/, the resultant electric field is E = E0 sin(t) + E0v2 sin(t + ). If we expand the trig function, we get E = E0 sin(t) + E0v2 [sin t cos  + cos t sin ] = E0 (1 + v2 cos ) sin t + E0v2 sin  cos t. The square of this is E2 = E02 (1 + v2 cos )2 sin2 t + 2E02 sin2  cos2 t + 2v2E02 (1 + 2v2 cos ) sin  sin t cos t. The intensity is the time average, so the contribution of the last term is zero and the average for sin2 t and cos2 t is !. Thus we have E2 = !E02 [(1 + v2 cos )2 + 2 sin2 ] = !E02 (3 + 2v2 cos ). Thus we have I/I0 = !E02 (3 + 2v2 cos )/!E02 (3 + 2v2) = (3 + 2v2 cos )/(3 + 2v2), with  = (2pd/) sin .

Ch. 35 p. 4

15. (a) If the sources have equal intensities, their electric fields E30 will have the same magnitude: E10 = E20 = E30 = E0.   From the symmetry of the phasor diagram we see that  = , E 0 where  = 2pd sin / is the phase difference between E20 adjacent slits. Thus the amplitude of the resultant field is E0 = E10 cos  + E20 + E30 cos  = E0(1 + 2 cos ).  At the center where  = 0,  = 0, we have  E00 = 3E0. t E10 The ratio of intensities is I/I0 = E02(1 + 2 cos )2/(3E0)2 = (1 + 4 cos  + 4 cos2 )/9. (b) The intensity will be maximal when 1 + 2 cos  is maximal, which will be when cos  = 1. This occurs when the three phasors are parallel. Thus we have  = …, – 2p, 0, 2p, 4p, … = 2mp, m = 0, ±1, ±2, … . The angles for the maxima are given by E20 sin max = 2mp/2pd = m/d, m = 0, ±1, ±2, … . The intensity will be minimal when 1 + 2 cos  is   minimal (zero), which will be when cos  = – !. E30 E10 This occurs when the three phasors form an E10 equilateral triangle, as shown. Thus we have E30  = …, – 2p/3, 2p/3, 4p/3, 8p/3, 10p/3, … ; or  = (m + @k)2p; k = 1, 2; m = 0, ±1, ±2, … . E20 The angles for the minima are given by sin min = (m + @k)/d; k = 1, 2; m = 0, ±1, ±2, … . 16. If the sources have equal intensities, their electric fields will have the same magnitude: E10 = E20 = E30 = E40 = E0. From the symmetry of the phasor diagram we see that  = *, where  = 2pd sin / is the phase difference between adjacent slits. Thus the amplitude of the resultant field is E0 = E10 cos * + E20 cos ! + E30 cos ! + E40 cos * = 2E0(cos ! + cos *) = 4E0 cos  cos !. At the center where  = 0,  = 0, we have E00 = 4E0. The ratio of intensities is I/I0 = (4E0)2(cos  cos !)2/(4E0)2 = cos2  cos2 !.

E40 E 0



/2



E30

 /2 

 t E10

The intensity will be maximal when cos  = ±1 and cos ! = ±1. This occurs when the four phasors are parallel. Thus we have ! = …, – 2p, – p, 0, p, 2p, 3p, … = mp, m = 0, ±1, ±2, … ; or  = 2mp, m = 0, ±1, ±2, … . The angles for the maxima are given by

E20

Ch. 35 p. 5

sin max = 2mp/2pd = m/d, m = 0, ±1, ±2, … . The intensity will be minimal when cos  = 0 or cos ! = 0. This occurs when the four phasors form a square or adjacent phasors are antiparallel. Thus we have ! = …, – *p, – !p, !p, *p, (p, … or  = …, – *p, – !p, !p, *p, (p, … , which can be combined:  = …, – *p, – p, – !p, !p, p, *p, (p, … = (m + #k)2p; k = 1, 2, 3; m = 0, ±1, ±2, … . The angles for the minima are given by sin min = (m + #k)/d; k = 1, 2, 3; m = 0, ±1, ±2, … .

E20

E30

E10

E40

E40 E20

E30 E10

17. We equate a path difference of one wavelength with a phase 1 = š difference of 2p. With respect to the incident wave, the wave 2 = (2t / film )2š + 0 that reflects at the top surface from the higher index of the soap bubble has a phase change of 1 = p. t n With respect to the incident wave, the wave that reflects from the air at the bottom surface of the bubble has a phase change due to the additional path-length but no phase change on reflection: 2 = (2t/film)2p + 0. For constructive interference, the net phase change is  = (2t/film)2p – p = m2p, m = 0, 1, 2, …; or t = !film(m + !), m = 0, 1, 2, … . The wavelengths in air that produce strong reflection are given by  = nfilm = 2nt/(m + !) = 4(1.34)(120 nm)/(2m + 1) = (643 nm)/(2m + 1). Thus we see that, for the light to be in the visible spectrum, the only value of m is 0:  = (643 nm)/(0 + 1) = 643 nm, which is an orange-red. 18. Between the 25 dark lines there are 24 intervals. When we add the half-interval at the wire end, we have 24.5 intervals, so the separation is 26.5 cm/24.5 intervals = 1.08 cm. 19. We equate a path difference of one wavelength with a phase 1 = š difference of 2p. With respect to the incident wave, the wave 2 = (2t / film )2š + 0 that reflects at the top surface from the higher index of the soap bubble has a phase change of 1 = p. t n With respect to the incident wave, the wave that reflects from the air at the bottom surface of the bubble has a phase change due to the additional path-length but no phase change on reflection: 2 = (2t/film)2p + 0. For destructive interference, the net phase change is  = (2t/film)2p – p = (m – !)2p, m = 0, 1, 2, …; or t = !filmm = !(/n)m, m = 0, 1, 2, … . The minimum non-zero thickness is tmin =  = 179 nm. 20. With respect to the incident wave, the wave that reflects from the top surface of the coating has a phase change of 1 = p. With respect to the incident wave, the wave that reflects from the glass (n ˜ 1.5) at the bottom surface of the coating has a phase change due to the additional path-length and a phase change of p on reflection:

1 = š 2 = (2t / film )2š + š t

Ch. 35 p. 6



2 = (2t/film)2p + p. For constructive interference, the net phase change is  = (2t/film)2p + p – p = m2p, m = 1, 2, 3, …; or t = !filmm = !(/nfilm)m, m = 1, 2, 3, … . The minimum non-zero thickness occurs for m = 1: tmin = /2nfilm = (570 nm)/2(1.28) = 223 nm. 570 nm is in the middle of the visible spectrum. The transmitted light will be stronger in the wavelengths at the ends of the spectrum, so the lens would emphasize the red and violet wavelengths.

21. The phase difference for the reflected waves from the 1 = 0 path-length difference and the reflection at the 2 = (2t / film )2š + š bottom surface is  = (2t/)2p + p. t For the dark rings, this phase difference must be an odd multiple of p, so we have    = (2t/)2p + p = (2m + 1)p, m = 0, 1, 2, …; or t = !m, m = 0, 1, 2, … . Because m = 0 corresponds to the dark center, m represents the number of the ring. Thus the thickness of the lens is the thickness of the air at the edge of the lens: t = !(28)(650 nm) = 9.1103 nm = 9.1 m. 22. There is a phase difference for the reflected waves from the path-length difference, (2t/)2p, and the reflection at the bottom surface, p. For destructive interference, this phase difference must be an odd multiple of p, so we have  = (2t/)2p + p = (2m + 1)p, m = 0, 1, 2, …; or t = !m, m = 0, 1, 2, … . Because m = 0 corresponds to the edge where the glasses touch, m + 1 represents the number of the fringe. Thus the thickness of the foil is d = !(24)(670 nm) = 8.04103 nm = 8.04 m. 23. With respect to the incident wave, the wave that reflects from the air at the top surface of the air layer has a phase change of 1 = 0. With respect to the incident wave, the wave that reflects from t the glass at the bottom surface of the air layer has a phase change due to the additional path-length and a change on reflection: 2 = (2t/)2p + p.  For constructive interference, the net phase change is  = (2t/)2p + p – 0 = m2p, m = 1, 2, 3, …; or t = !(m – !), m = 1, 2, 3, … .  The minimum thickness is tmin = !(480 nm)(1 – !) = 120 nm. For destructive interference, the net phase change is

t d L

1 = 0 2 = (2t / )2š + š

Ch. 35 p. 7



 = (2t/)2p + p – 0 = (2m + 1)p, m = 0, 1, 2, …; or t = !m, m = 0, 1, 2, … . The minimum non-zero thickness is tmin = !(480 nm)(1) = 240 nm.

24. With respect to the incident wave, the wave that reflects 1 = š from the top surface of the alcohol has a phase change of 2 = (2t / film )2š + š 1 = p. With respect to the incident wave, the wave that reflects n1 t from the glass at the bottom surface of the alcohol has a phase change due to the additional path-length and n2 a phase change of p on reflection: 2 = (2t/film)2p + p. For constructive interference, the net phase change is  = (2t/1film)2p + p – p = m12p, m1 = 1, 2, 3, …; or t = !1film(m1) = !(1/nfilm)(m1), m1 = 1, 2, 3, … . For destructive interference, the net phase change is  = (2t/2film)2p + p – p = (2m2 + 1)p, m2 = 0, 1, 2, …; or t = #(2/nfilm)(2m2 + 1), m2 = 0, 1, 2, … . When we combine the two equations, we get !(1/nfilm)(m1) = #(2/nfilm)(2m2 + 1), or (2m2 + 1)/2m1 = 1/2 = (640 nm)/(512 nm) = 1.25 = 5/4. If we choose the smallest integers, we see that m1 = m2 = 2, and the thickness of the film is t = !(1/nfilm)(m1) =  =

471 nm.

25. At a distance r from the center of the lens, the thickness of the air space is y, and the phase difference for the reflected waves from the path-length difference and the reflection at the bottom surface is  = (2y/)2p + p. R For the dark rings, we have    = (2y/)2p + p = (2m + 1)p, m = 0, 1, 2, …; or y = !m, m = 0, 1, 2, … . Because m = 0 corresponds to the dark center, m represents the number of the ring. From the triangle in the diagram, we have y r2 + (R – y)2 = R2, or r2 = 2yR – y2 ˜ 2yR, when y « R, which becomes r r2 = 2(!m)R = mR, m = 0, 1, 2, … . When the apparatus is immersed in the liquid, the same analysis holds, if we use the wavelength in the liquid. If we form the ratio for the two conditions, we get (r1/r2)2 = 1/2 = n, so

Ch. 35 p. 8

n = (2.92 cm/2.60 cm)2 =

1.26.

26. At a distance r from the center of the lens, the thickness of the air space is y, and the phase difference for the reflected waves from the path-length difference and the reflection at the bottom surface is  = (2y/)2p + p. For the bright rings, we have    = (2y/)2p + p = m2p, m = 1, 2, 3, …; or y = !(m – !), m = 1, 2, 3, … , where m represents the number of the ring. From the triangle in the diagram, we have r2 + (R – y)2 = R2, or r2 = 2yR – y2 ˜ 2yR, when y « R, which becomes r2 = 2[!(m – !)]R = (m – !)R. For the 48th ring, we have R = 10.5 m. (1.710–2 m)2 = (58010–9 m)(48 – !)R, which gives We find the focal length of the lens from 1/f = (n – 1)[(1/R1) + (1/R2)] = (1.51 – 1)[(1/10.5 m) + (1/8)], which gives 27. At a distance r from the center of the lens, the thickness of the air space is y, and the phase difference for the reflected waves from the path-length difference and the reflection at the bottom surface is  = (2y/)2p + p. For the dark rings, we have    = (2y/)2p + p = (2m + 1)p, m = 0, 1, 2, …; or y = !m, m = 0, 1, 2, … . Because m = 0 corresponds to the dark center, m represents the number of the ring. From the triangle in the diagram, we have r2 + (R – y)2 = R2, or r2 = 2yR – y2 ˜ 2yR, when y « R, which becomes r2 = 2(!m)R = mR, m = 0, 1, 2, … ; or r = (mR)1/2. 28. From Problem 27 the radius of a dark ring is

R

y r

f = 20.6 m.

R

y r

Ch. 35 p. 9

r = (mR)1/2. Thus the separation of adjacent rings is ?r = [(m + 1)R]1/2 – (mR)1/2 = [(m + 1)1/2 – m1/2](R)1/2 = {[1 + (1/m)]1/2 – 1}(mR)1/2. If m » 1, [1 + (1/m)]1/2 ˜ 1 + (1/2m), so we have ?r ˜ [ 1 + (1/2m) – 1](mR)1/2 = (R/4m)1/2.

29. We assume n1 < n2 < n3 and most of the light is transmitted. Thus the electric field amplitude of a reflected wave is rE0 ,

1 = š n1

2 = (2t / film )2š + š where r « 1. With respect to the incident wave, the wave that reflects from the top surface of the coating has a phase n2 t change of 1 = p. n3 With respect to the incident wave, the wave that reflects from the glass at the bottom surface of the coating has a phase change due to the additional path-length and a phase change of p on reflection: 2 = (2t/coating)2p + p. For destructive interference, the net phase change is  = (2t/coating)2p + p – p = (2m + 1)p, m = 0, 1, 2, …; or t = #(/n2)(2m + 1), m2 = 0, 1, 2, … . If we assume the thinnest coating for the 550 nm light, we have t = 0/4n2 . The phase difference for a different wavelength will be  = 2(0/4n2)[1/(/n2)]2p = (0/)p. The resultant electric field of the reflected light is E = rE0 sin(t) + rE0 sin(t + ). This is similar to the analysis of the double-slit interference, so the amplitude of the reflected light as a result of the interference is Er = 2rE0 cos !. With no coating, the amplitude is rE0 , so the ratio of intensities is Ir/Ir0 = (2rE0)2(cos2 !)/(rE0)2 = 4 cos2 ! = 4 cos2 [!(0/)p]. For the two wavelengths we have Ir1/Ir0 = 4 cos2 [!(550 nm/450 nm)p] = 0.47; Ir2/Ir0 = 4 cos2 [!(550 nm/650 nm)p] = 0.23. Note that if both wavelengths are present, the reflected light will have a purplish hue.

Ch. 35 p. 10

30. One fringe shift corresponds to a change in path length of . The number of fringe shifts produced by a mirror movement of ?L is N = 2 L/; 727 nm. 344 = 2(0.12510–3 m)/, which gives  = 7.2710–7 m = 31. One fringe shift corresponds to a change in path length of . The number of fringe shifts produced by a mirror movement of ?L is N = 2 L/; 0.221 mm. 750 = 2 L/(58910–9 m), which gives L = 2.2110–4 m = 32. One fringe shift corresponds to an effective change in path length of . The actual distance has not changed, but the number of wavelengths in the depth of the cavity has. If the cavity has a depth d, the number of wavelengths in vacuum is d/, and the number with the gas present is d/gas = ngasd/. Because the light passes through the cavity twice, the number of fringe shifts is N = 2[(ngasd/) – (d/)] = 2(d/)(ngas – 1); 186 = 2[(1.3010–2 m)/(61010–9 m)](ngas – 1), which gives ngas = 1.00436.

33. The two fringe patterns overlap but do not interfere with each other. When the bright fringe of one occurs where there is a dark fringe of the other, there will be a region without fringes. When the next region occurs, the mirror movement must produce an integer number of fringe shifts for each wavelength: N1 = 2 L/1 ; N2 = 2 L/2 ; and the difference in the number of fringe shifts must be 1. Thus we have N12 = N21 ; N1(589.6 nm) = (N1 + 1)(589.0 nm), which gives N1 = 982. We find the mirror movement from N1 = 2 L/1 ; 982 = 2 L/(589.0 nm), which gives L = 2.89105 nm = 0.289 mm. 34. We assume the luminous intensity of the Sun is the same in all directions. The luminous flux is F = EA = (105 lm/m2)4p(1.51011 m)2 = 31028 lm. The luminous intensity is I = F/(4p sr) = (31028 lm)/(4p sr) = 21027 lm/sr = 21027 cd. 35. (a) The luminous efficiency is luminous efficiency = F/P = (1700 lm)/(100 W) = 17 lm/W. (b) If half the luminous flux provides the illuminance of the floor, we have E = !NP(luminous efficiency)/A; 156. 250 lm/m2 = !N(40 W)(60 lm/W)/(25 m)(30 m), which gives N =

Ch. 35 p. 11

36. There will be a phase difference between the waves at the two slits because the wave at the upper slit will have traveled farther. The path difference at the two slits for the incident wave is d sin i . The path difference between the two slits for the diffracted wave is d sin . When the net path difference is a multiple of a wavelength, there will be maxima given by (d sin i) – (d sin ) = m, m = 0, ± 1, ± 2, … ; or sin  = sin i ± (m/d), where m = 0, 1, 2, … .  37. The wavelength of the signal is  = v/f = (3.00108 m/s)/(75106 Hz) = 4.00 m. (a) There is a phase difference between the direct and reflected signals from the path difference, (h/)2p, and the reflection, p. The total phase difference is  = (h/)2p + p = [(118 m)/(4.00 m)]2p + p = 30(2p). Thus the interference is constructive. (b) When the plane is 22 m closer to the receiver, the phase difference is  = [(h – y)/]2p + p = [(118 m – 22 m)/(4.00 m)]2p + p = 24(2p) + p. Thus the interference is destructive.

 i

i 

Signal

y h Signal Receiver

38. The wavelength of the signal is  = v/f = (3.00108 m/s)/(102.1106 Hz) = 2.94 m. Because measurements are made far from the antennae, we can use the analysis for the double slit. For constructive interference, the path difference is a multiple of the wavelength: d sin  = m, m = 0, 1, 2, 3, … ; (7.0 m) sin 1max = (1)(2.94 m), which gives 1max = 25°; (7.0 m) sin 2max = (2)(2.94 m), which gives 2max = 57°; (7.0 m) sin 3max = (3)(2.94 m), which gives sin 3max > 1, so there is no third maximum. Because the interference pattern will be symmetrical above and below the midline and on either side of the antennae, the angles for maxima are 25°, 57°, 123°, 155° above and below the midline. For destructive interference, the path difference is an odd multiple of half a wavelength: d sin  = (m – !), m = 1, 2, 3, … ; or (7.0 m) sin 1min = (1 – !)(2.94 m), which gives 1min = 12°; (7.0 m) sin 2min = (2 – !)(2.94 m), which gives 2min = 39°; (7.0 m) sin 3min = (3 – !)(2.94 m), which gives sin 3min > 1, so there is no third minimum. Because the interference pattern will be symmetrical above and below the midline and on either side of the antennae, the angles for minima are 12°, 39°, 141°, 168° above and below the midline. 39. For constructive interference, the path difference is a multiple of the wavelength:

Ch. 35 p. 12

d sin  = m, m = 0, 1, 2, 3, … . We find the location on the screen from y = L tan . For small angles, we have sin  ˜ tan , which gives y = L(m/d) = mL/d. For the second-order fringes we have y1 = 2L1/d; y2 = 2L2/d. When we subtract the two equations, we get ?y = y1 – y2 = (2L/d)(1 – 2); (1.13 mm)(110–3 m/mm) = [2(1.50 m)/(0.6010–3 m)](690 nm – 2), which gives 2 =

464 nm.

40. We equate a path difference of one wavelength with a phase 1 = š difference of 2p. With respect to the incident wave, the 2 = (2t / film )2š + (š or 0) wave that reflects at the top surface of the film has a phase change of t 1 = p. If we assume that the film has an index less than glass, the wave that reflects from the glass has a phase change due to the additional path-length and a phase change on reflection: 2 = (2t/film)2p + p. For destructive interference, the net phase change is  = (2t/film)2p + p – p = (m – !)2p, m = 1, 2, …; or t = !film(m – !) = !(/n)(m – !), m = 1, 2, … . For the minimum thickness, m = 1, we have no film with n < nglass is possible. 150 nm = , which gives n = 1, so If we assume that the film has an index greater than glass, the wave that reflects from the glass has a phase change due to the additional path-length and no phase change on reflection: 2 = (2t/film)2p + 0. For destructive interference, the net phase change is  = (2t/film)2p – p = (m – !)2p, m = 1, 2, …; or t = !filmm = !(m/n), m = 1, 2, … . For the minimum thickness, m = 1, we have 2.00. 150 nm = !(1)(600 nm)/n, which gives n = 41. With respect to the incident wave, the wave that reflects at 1 = š the top surface of the film has a phase change of 2 = (2t / film )2š 1 = p. The wave that reflects from the bottom surface has a phase t change due to the additional path-length and no phase change on reflection: 2 = (2t/film)2p + 0. For destructive interference, the net phase change is  = (2t/film)2p – p = (m – !)2p, m = 1, 2, …; or t = !filmm = !(m/n), m = 1, 2, … . For the two wavelengths we have t = !(m11/n) = !(m22/n), or m1/m2 = 2/1 = 680 nm/510 nm = 1.333 = 4/3. Thus m1 = 4, and m2 = 3. For the thickness we have t = !(m11/n) = !(4)(510 nm)/1.58 =

646 nm.

42. For destructive interference, the path difference is d sin  = (m – !), m = 1, 2, 3, … ; or

Ch. 35 p. 13

sin  = (m – !)(3.5 cm)/(6.0 cm) = (m – !)(0.583), m = 1, 2, 3, … . The angles for the first three regions of complete destructive interference are sin 1 = (m – !)/d = (1 – !)(0.583) = 0.29, 1 = 17°; sin 2 = (m – !)/d = (2 – !)(0.583) = 0.87, 2 = 61°; sin 3 = (m – !)/d = (3 – !)(0.583) = 1.46, therefore, no third region. We find the locations at the end of the tank from y = L tan ; y1 = (2.0 m) tan 17° = 0.61 m; y2 = (2.0 m) tan 61° = 3.6 m. Thus you could stand 0.61 m, or 3.6m away from the line perpendicular to the board midway between the openings. 43. With respect to the incident wave, the wave that reflects 1 = š from the top surface of the coating has a phase change of 2 = (2t / film )2š + š 1 = p. With respect to the incident wave, the wave that reflects t from the glass (n ˜ 1.5) at the bottom surface of the coating has a phase change due to the additional path-length and a phase change of p on reflection: 2 = (2t/film)2p + p. For destructive interference, this phase difference must be an odd multiple of p, so we have  = (2t/film)2p + p – p = (2m + 1)p, m = 0, 1, 2, …; or t = #(2m + 1)film , m = 0, 1, 2, … . Thus the minimum thickness is tmin = #/n. (a) For the blue light we get tmin = #(450 nm)/(1.38) = 81.5 nm. (b) For the red light we get tmin = #(700 nm)/(1.38) = 127 nm.

44. If we consider the two rays shown in the diagram, we see that the second ray has reflected twice. If nfilm < nglass , the first reflection from the glass produces a shift equivalent n film t to !film , while the second reflection from the air produces no shift. When we compare the two rays at the film-glass n glass surface, we see that the second ray has a total shift of d1 = 0 d2 = 2t + (0 or  film /2) d2 – d1 = 2t + !film . For maxima, we have t = !(m – !)/nfilm , m = 1, 2, 3, … (maxima, nfilm < nglass). 2t + !film = mfilm , m = 1, 2, 3, … ; or For minima, we have t = !m/nfilm , m = 0, 1, 2, 3, … (minima, nfilm < nglass). 2t + !film = (m + !)film , m = 0, 1, 2, 3, … ; or

We see that for a film of “zero” thickness, that is, t « film , there will be a minimum. If nfilm > nglass , the first reflection from the glass produces no shift, while the second reflection from the air

Ch. 35 p. 14

also produces no shift. When we compare the two rays at the film-glass surface, we see that the second ray has a total shift of d2 – d1 = 2t. For maxima, we have t = !m/nfilm , m = 0, 1, 2, 3, … (maxima, nfilm > nglass). 2t = mfilm , m = 0, 1, 2, 3, … ; or For minima, we have t = !(m – !)/nfilm , m = 1, 2, 3, … (minima, nfilm > nglass). 2t = (m – !)film , m = 1, 2, 3, … ; or We see that for a film of “zero” thickness, that is, t « film , there will be a maximum. Because there will be some decrease in the electric field amplitude at each reflection, the transmitted minima will not be zero and the transmitted maxima will be less than the incident light.

45. The phase difference caused by the path back and forth through the coating must correspond to half a wavelength to produce destructive interference: 2t = /2, so t = /4 = (2 cm)/4 = 0.5 cm. 46. To maximize reflection, we want the three rays shown on the diagram to be in phase. We first compare rays 2 and 3. We want them to be in phase when leaving the boundary between n1 and n2 . 1 Ray 2 reflects from n2 > n1 , so there will be a phase shift of p. 2 Ray 3 will have a phase change due to the additional path-length 3 n1 n2 n1 and no phase change on reflection from the next n1 layer: 3 = (2d2/2)2p + 0. d1 d2 For constructive interference, the net phase change is  = (2d2/2)2p – p = m2p, m = 0, 1, 2, … ; or d2 = !2(m + !) = !(/n2)(m + !), m = 0, 1, 2, … . Thus for the minimum thickness (m = 0), we get d2 = #(/n2). We want rays 1 and 2 to be in phase when leaving the first surface. Ray 1 reflects from n1 > 1 , so there will be a phase shift of p. Ray 2 will have a phase change due to the additional path-length and a phase change on reflection from the n2 layer: 2 = (2d1/1)2p + p. For constructive interference, the net phase change is  = (2d1/1)2p + p – p = m2p, m = 1, 2, 3, … ; or d1 = !1m, m = 1, 2, 3, … . Thus for the minimum thickness (m = 1), we get d1 = !(/n1).

47. From the diagram we see that sin  = v/vp = c/nvp = (3.00108 m/s)/(1.52)(2.21108 m/s) = 0.893, so  = 63.3°.

v +

wavefront

 vp

Ch. 35 p. 15

48. With respect to the incident wave, the wave that reflects from the air at the top surface of the air layer has a phase change of 1 = 0. With respect to the incident wave, the wave that reflects from t the glass at the bottom surface of the air layer has a phase change due to the additional path-length and a change on reflection: 2 = (2t/)2p + p. For destructive interference, the net phase change is  = (2t/)2p + p – 0 = (2m + 1)p, m = 0, 1, 2, …; or t = !m, m = 0, 1, 2, … .  The minimum non-zero thickness is tmin = !(640 nm)(1) = 320 nm.  For constructive interference, the net phase change is  = (2t/)2p + p – 0 = m2p, m = 1, 2, 3, …; or t = !(m – !), m = 1, 2, 3, … .  The minimum thickness is tmin = !(640 nm)(1 – !) = 160 nm.

1 = 0 2 = (2t / )2š + š

49. The reflected wave appears to be coming from the virtual image, so this corresponds to a double slit, with the separation being d = 2S. The reflection from the mirror produces a p phase shift, however, so the maxima and minima are interchanged: sin max = (m + !)/2S, m = 0, 1, 2, …; sin min = m/2S, m = 0, 1, 2, … . 50. The phase difference from the path-length difference is  = (L/)2p = {[(/2) sin ]/}2p = p sin . When the signals are 180° out of phase, there will be an additional phase shift of p:  = p(1 + sin ). The radiated intensity is I = I0 cos2 (/2) = I0 cos2 [p(1 + sin )/2]. The intensity will be maximum when sin  = ± 1,  = 90° and 270°. The intensity will be minimum when sin  = 0,  = 0° and 180°. When the signals are in phase, the phase difference is from the path-length difference:  = p sin . The radiated intensity is I = I0 cos2 (/2) = I0 cos2 [(p sin )/2]. The intensity will be maximum when sin  = 0,  = 0° and 180°. The intensity will be minimum when sin  = ± 1,  = 90° and 270°. The maxima and minima are interchanged by the 180° phase difference.

y

²L





/2

51. When the mirror is moved a distance x, the path length changes by 2x. Thus the additional phase shift is  = (2x/)2p. The ratio of the new intensity to the bright maximum is I/I0 = cos2 (/2) = cos2 (2px/).

x

Ch. 36 p. 1

CHAPTER 36 – Diffraction and Polarization 1.

We find the angle to the first minimum from sin 1min = m/a = (1)(68010–9 m)/(0.034510–3 m) = 0.0197, so 1min = 1.13°. Thus the angular width of the central diffraction peak is ?1 = 21min = 2(1.13) = 2.26°.

2.

The angle from the central maximum to the first minimum is 18.5°. We find the wavelength from a sin 1min = m; (3.0010–6 m) sin (18.5°) = (1), which gives  = 9.5210–7 m =

952 nm.

3.

For constructive interference from the single slit, the path difference is a sin  = (m + !), m = 1, 2, 3, … . For the first fringe away from the central maximum, we have (3.5010–6 m) sin 1 = (*)(55010–9 m), which gives 1 = 13.7°. We find the distance on the screen from y1 = L tan 1 = (10.0 m) tan 13.7° = 2.4 m.

4.

The angle from the central maximum to the first bright fringe is 19°. For constructive interference from the single slit, the path difference is a sin  = (m + !), m = 1, 2, 3, … . For the first fringe away from the central maximum, we have a sin (19°) = (*)(68910–9 m), which gives a = 3.1710–6 m = 3.2 m.

5.

Because the angles are small, we have tan 1min = !(?y1)/L = sin 1min . The condition for the first minimum is a sin 1min = !a ?y1/L = . If we form the ratio of the expressions for the two wavelengths, we get ?y1b/?y1a = b/a ; ?y1b/(8.0 cm) = (400 nm)/(550 nm), which gives ?y1b = 5.8 cm.

6.

(a) There will be no diffraction minima if the angle for the first minimum is greater than 90°. Thus the limiting condition is a sin 1min = m; amax sin 90° = (1), or amax = . (b) Visible light has wavelengths from 400 nm to 750 nm, so the maximum slit width for no diffraction minimum for all of these wavelengths is the one for the smallest wavelength: 400 nm.

7.

We find the angle to the first minimum from sin 1min = m/a = (1)(40010–9 m)/(0.065510–3 m) = 6.1110–3, so 1min = 0.350°. We find the distance on the screen from y1 = L tan 1 = (3.50 m) tan 0.350° = 2.1410–2 m = 2.14 cm. Thus the width of the peak is ?y1 = 2y1 = 2(2.14 cm) = 4.28 cm.

Ch. 36 p. 2

8.

9.

The path-length difference between the top and bottom of the slit for the incident wave is a sin i. The path-length difference between the top and bottom of the slit for the diffracted wave is a sin . When the net path-length difference is a multiple of a wavelength, there will be an even number of segments of the wave which will have a path-length difference of /2; there will be minima given by (a sin i) – (a sin ) = m, m = ± 1, ± 2, … , or sin  = sin 30° – (m/a), where m = ± 1, ± 2, … .  When  = 30°, the net path-length difference is zero, and there will be constructive interference. There is a “central maximum” at 30° to the normal.

Slit, width a  i

i

(a) If we consider the slit made up of N wavelets of amplitude ?E0 , the total amplitude at the central maximum, where they are all in phase, is N ?E0. Doubling the size of the slit doubles the number of wavelets and thus the total amplitude. Because I ~ E02, the intensity at the central maximum is increased by a factor of 4. (b) The first minimum occurs at sin  = /a, or  = /a for small angles. Doubling a reduces the size of the central maximum by !. Thus the average intensity over the central maximum is 4(!) = 2, in agreement with the doubling of the incident intensity. This will be true for all fringes, so energy is conserved.

10. (a) The intensity pattern for a single slit is I = I0[(sin !)/(!)]2 = 4I0 (sin2 !)/2. The maxima for this function do not occur for the maxima of sin !, which are ! = (m + !)p. (b) We find the location of the maxima from dI/d = 0: dI/d = 4I0{[(sin ! cos !)/2] – [(2 sin2 !)/3]} = 0, which gives

!max2 = 7.725 rad,

max2 = 15.450 rad = 885.22°. The minima of the pattern are given by sin  = m/a, m = ± 1, ± 2, … ; We find the values of  corresponding to the midpoint between minima from  = (m + !)2p, m = ± 1, ± 2, … . The first two values are 1 = 3p = 9.425 rad = 540.00°; 2 = 5p = 15.707 rad = 900.00°.

1st solution

2nd solution

20

/2 (rad), tan (  /2)

cos ! = 2 sin !/, or tan ! = !. Note that the minima (0) are given by sin ! = 0. (c) A numerical solution of tan ! = ! (remember that  is in rad) gives !max1 = 4.493 rad, max1 = 8.986 rad = 514.86°;

16 12 8 4 0

90

180 270 /2  degrees

360

450

Ch. 36 p. 3

Thus the percent differences are 4.88%. % diff1 = (1 – max1)/max1 = (9.425 rad – 8.986 rad)/8.986 rad = % diff2 = (2 – max2)/max2 = (15.707 rad – 15.450 rad)/15.450 rad = 1.66%. 11. To find the angular width at half-maximum, we find the phase at half-maximum: I = I0[(sin !h)/!h]2 = !I0 , or h2 = 8 sin2 !h. This equation can be solved graphically or numerically to get h = 2.783 rad. We find the corresponding angle from h = (2pa/) sin h . Thus the angular width at half-maximum is ?h = 2h = 2 sin–1 ( h/2pa) = 2 sin–1 (0.443/a). The angle will be small, so we have ?h ˜ 2(0.443/a) = 0.886/a. For the given data we have 10.7°. ?h = 0.886(55010–9 m)/(2.6010–6 m) = 0.187 rad =

h 1

12. If the central diffraction peak contains 15 fringes, there will be 7 on each side of the center. Thus the eighth maximum of the double slit must coincide with the first minimum of the diffraction pattern. The maxima of the double slit are given by sin  = m/d, m = 0, ± 1, ± 2, … . The minima of the single-slit pattern are given by sin  = ms/a, ms = ± 1, ± 2, … . Thus we have d = 8a. 8/d = /a, or 13. If the central diffraction peak contains 7 fringes, there will be 3 on each side of the center. Thus the fourth maximum of the double slit must coincide with the first minimum of the diffraction pattern. The maxima of the double slit are given by sin  = m/d, m = 0, ± 1, ± 2, … . The minima of the single-slit pattern are given by sin  = ms/a, ms = ± 1, ± 2, … . Thus we have d = 4a. 4/d = /a, or 14. The phase for the double-slit pattern is  = (2pd sin )/, and the phase for the single-slit pattern is  = (2pa sin )/. Thus if a = d,  = . The intensity of the pattern is I = I0[(sin !)/(!)]2 cos2 ! = I0 [(sin ! cos !)/!]2 = I0 [(sin )/]2. This is the intensity for a single slit with a phase of   = 2 = 2(2pa sin )/ = [2p(2a) sin )]/, which corresponds to a single slit with width 2a. 

Ch. 36 p. 4

15. (a) The maxima of the double slit are given by sin  = m/d, m = 0, ± 1, ± 2, … . The distance of a fringe on the screen from the center of the pattern is y = L tan . If the angles are small, we have y ˜ L sin  = Lm/d. The separation of adjacent interference fringes will be 1.8 cm. ?y = (L ?m)/d = (1.0 m)(55010–9 m)(1)/(0.03010–3 m) = 0.018 m = (b) The minima of the single-slit pattern are given by sin  = ms/a, ms = ± 1, ± 2, … . The distance of the first minimum on the screen from the center of the pattern is y = L tan . If the angle is small, we have y ˜ L sin  = L/a = (1.0 m)(55010–9 m)/0.01010–3 m) = 0.055 m = 5.5 cm. Thus the distance between the first minima on either side of the center is 2y = 11.0 cm. 16. The maxima of the double slit are given by sin  = m/d, m = 0, ± 1, ± 2, … . The minima of the single-slit pattern are given by sin  = ms/a, ms = ± 1, ± 2, … . Thus at the angle for the first minimum of the single slit we have m/d = /a, or m = d/a. (a) For d = 2.00a, we have m = d/a = 2.00. Thus the m = 2 interference fringe will be missing, so there will be 3 fringes, including the central fringe. (b) For d = 12.0a, we have m = d/a = 12.0. Thus the m = 12 interference fringe will be missing, so there will be 23 fringes, including the central fringe. (c) For d = 4.50a, we have m = d/a = 4.50. Thus the single-slit minimum coincides with the minimum of the interference beyond the m = 4 fringe, so there will be 9 fringes, including the central fringe. (d) For d = 7.20a, we have m = d/a = 7.20. Thus the m = 7 interference fringe will be incomplete, so there will be almost 15 fringes, including the central fringe.

Ch. 36 p. 5

17. The phase for the double-slit pattern is  = (2pd sin )/, and the phase for the single-slit pattern is  = (2pa sin )/. The phasor for each slit will be an arc with the curvature determined by . The relation between the phasors from the first slit and the second slit is determined by . At the center we have At the first interference minimum we have  =  = 0:  = p,  = pa/d:

E0 slit 2 slit 1

slit 2

slit 1

E0 = 0

At the next interference maximum we have When sin  = /a (the diffraction minimum), we have  = 2p,  = 2pa/d:  = 2p,  = 2pd/a:

slit 1 E0 slit 2 E0 = 0 slit 2 slit 1

18. (a) If a ˜ , the central maximum of the diffraction pattern will be very wide. Thus we need consider only the interference between slits. We construct a phasor diagram for the interference, E30 with  = 2pd sin / the phase difference between adjacent slits. The electric fields of the slits will   have the same magnitude: E 0 E10 = E20 = E30 = E0. E20 From the symmetry of the phasor diagram we see that  = ,  where  = 2pd sin / is the phase difference between adjacent slits. Thus the amplitude of the resultant field is  E 0 = E10 cos  + E20 + E30 cos  = E0(1 + 2 cos ). t E10 At the center where  = 0,  = 0, we have E00 = 3E0. The ratio of intensities is I/I0 = E02(1 + 2 cos )2/(3E0)2 = (1 + 4 cos  + 4 cos2 )/9, or 

Ch. 36 p. 6

I = I0(1 + 4 cos  + 4 cos2 )/9. (b) We find the locations of the maxima and minima by setting the first derivative equal to zero: dI/d = I0(– 4 sin  – 8 sin  cos ) = – 4I0(1 + 2 cos ) sin  = 0. For sin  = 0 we get  = 0, p, 2p; so cos  = 1, – 1, 1; and I = I0 , I0/9, I0 ; which are maxima. For cos  = – 1/2 we get  = 2p/3, 4p/3; and I = 0, which are minima. Thus  = 0, 2p correspond to principal maxima;  = p corresponds to a secondary maximum. Thus we see that there is only one secondary maximum between principal peaks. 19. The minimum angular resolution is  = 1.22/D = (1.22)(50010–9 m)/(100 in)(0.0254 m/in) =

2.410–7 rad = (1.410–5)° = 0.050.

20. The resolution of the telescope is  = 1.22/D = (1.22)(55010–9 m)/(0.90 m) = 7.4610–7 rad. The separation of the stars is d = L = (10 ly)(9.461015 m/ly)(7.4610–7 rad) = 7.11010 m. 21. The minimum angular resolution is  = 1.22/D. The distance between lines is the resolving power: RP = f = 1.22f/D= 1.22(f-stop). For f/2 we have RP1 = (1.22)(50010–9 m)(2) = 1.2210–6 m = 1.2210–3 mm, so the resolution is 1/RP1 = 1/(1.2210–3 mm) = 820 lines/mm. For f/16 we have RP2 = (1.22)(50010–9 m)(16) = 9.7610–6 m = 9.7610–3 mm, so the resolution is 1/RP2 = 1/(9.7610–3 mm) = 102 lines/mm. 22. The angular resolution of the eye, which is the required resolution using the telescope, is eye = deye/Leye = (0.1010–3 m)/(2510–2 m) = 4.010–4 rad. The resolution without the telescope is  = d/L = (7.0 km)/(3.84105 km) = 1.8210–5 rad. If we ignore the inversion of the image, the magnification is M = eye/ = fo/fe ; (4.010–4 rad)/(1.8210–5 rad) = (2.0 m)/fe , which gives fe = 0.091 m = 9.1 cm. The resolution limit is  = 1.22/D = (1.22)(50010–9 m)/(0.110 m) = 5.5510–6 rad. This is a distance of 2.1 km on the surface of the Moon. 23. We find the angle for the third order from d sin  = m; (1.3510–5 m) sin  = (3)(44010–9 m), which gives sin  = 9.7810–2, so  = 24. We find the slit separation from d sin  = m; d sin 13.0° = (3)(65010–9 m), which gives d = 8.6710–6 m = 8.6710–4 cm. The number of lines/cm is 1/d = 1/(8.6710–4 cm) = 1.15103 lines/cm.

5.61°.

Ch. 36 p. 7

25. Because the angle increases with wavelength, to have a complete order we use the largest visible wavelength. The maximum angle is 90°, so we have d sin  = m; [1/(6600 lines/cm)](10–2 m/cm) sin 90° = m(75010–9 m), which gives m = 2.02. Thus only two full orders can be seen on each side of the central white line. 26. We find the wavelength from d sin  = m; [1/(3500 lines/cm)](10–2 m/cm) sin 22.0° = 3, which gives  = 3.5710–7 m =

357 nm.

27. We find the wavelengths from d sin  = m; [1/(10,000 lines/cm)](10–2 m/cm) sin 29.8° = (1) , which gives  = 4.9710–7 m = [1/(10,000 lines/cm)](10–2 m/cm) sin 37.7° = (1) , which gives  = 6.1210–7 m = [1/(10,000 lines/cm)](10–2 m/cm) sin 39.6° = (1) , which gives  = 6.3710–7 m = [1/(10,000 lines/cm)](10–2 m/cm) sin 48.9° = (1) , which gives  = 7.5410–7 m =    

497 nm; 612 nm; 637 nm; 754 nm.

28. We find the angles for the first order from d sin  = m = ; [1/(7800 lines/cm)](10–2 m/cm) sin 400 = (40010–9 m), which gives sin 400 = 0.312, so 400 = 18.2°; [1/(7800 lines/cm)](10–2 m/cm) sin 750 = (75010–9 m), which gives sin 750 = 0.585, so 750 = 35.8°. The distances from the central white line on the screen are y400 = L tan 400 = (2.80 m) tan 18.2° = 0.92 m; y750 = L tan 750 = (2.80 m) tan 35.8° = 2.02 m. Thus the width of the spectrum is y750 – y400 = 2.02 m – 0.92 m = 1.10 m. 29. Because the angle increases with wavelength, we compare the maximum angle for the second order with the minimum angle for the third order: d sin  = m, or sin  = m/d; sin 2max = (2)(750 nm)/d; sin 3min = (3)(400 nm)/d. When we divide the two equations, we get sin 3min/sin 2max = (1200 nm)/(1500 nm) = 0.80. Because the value of the sine increases with angle, this means 3min < 2max , so the orders overlap. To determine the overlap, we find the second-order wavelength that coincides with 3min: (2)2 = (3)(400 nm), which gives 2 = 600 nm. We find the third-order wavelength that coincides with 2max from (2)(750 nm) = (3)3 , which gives 3 = 500 nm. Thus 600 nm to 750 nm of the second order overlaps with 400 nm to 500 nm of the third order. 30. Because the angles on each side of the central line are not the same, the incident light is not normal to the grating. We use the average angles: 1 = (26°38' + 26°18')/2 = 26°28' = 26.47°; 2 = (41°02' + 40°27')/2 = 40°45.5' = 40.76°. We find the wavelengths from d sin  = m;

Ch. 36 p. 8

[1/(9550 lines/cm)](10–2 m/cm) sin 26.47° = (1) , which gives  = 4.6710–7 m = [1/(9550 lines/cm)](10–2 m/cm) sin 40.76° = (1) , which gives  = 6.8410–7 m =

467 nm; 684 nm.

 

31. We have the same average angles, but the path differences causing the interference must be measured in terms of the wavelengths in water. Thus the wavelengths calculated in Problem 30 are those in water. The wavelengths in air are air = nwater = (467 nm)(1.33) = 621 nm; air = nwater = (684 nm)(1.33) = 909 nm. Note that the second wavelength is not visible.

32. We find the slit separation from d sin  = m; d sin 15.5° = (1)(58910–9 m), which gives d = 2.2010–6 m = 2.20 m. We find the angle for the third order from d sin  = m; 3 = 53.4°. (2.2010–6 m) sin 3 = (3)(58910–9 m), which gives sin 3 = 0.803, 33. The phase difference for waves through adjacent slits is due to the path-length difference. For a maximum, the path-length difference is a multiple of the wavelength. From the diagram, the additional distance to the grating is d sin  for the incident wave and d sin  for the diffracted wave. For the maxima, we have d sin  – d sin  = m, m = 0, ± 1, ± 2, … . For diffracted rays at an angle  on the other side of the normal to the grating, we have d sin  + d sin  = m, m = 0, ± 1, ± 2, … . Thus we have d(sin  ± sin ) = m, m = 0, ± 1, ± 2, … .



d



34. (a) The interference maxima of the grating are given by sin  = m1/d, m1 = 0, ± 1, ± 2, … . The minima of the single-slit pattern are given by sin  = m2/a, m2 = ± 1, ± 2, … . If d = 2a, the angles for the maxima are given by sin  = m1/2a, = (m1/2)/a, m1 = 0, ± 1, ± 2, … . Thus whenever m1/2 = an integer, this will be at a minimum of the single-slit pattern and the fringe will be missing: m1/2 = 1, 2, 3, … ; or m1 = 2, 4, 6, … . (b) For a maximum of the interference pattern to coincide with a minimum of the diffraction pattern, we must have sin  = m1/d = m2/a, or d/a = m1/m2. (c) If d = a, the large number of lines of the grating means that there will be an effective opening much larger than the wavelength. Thus  ˜ 0; there will be negligible diffraction. There will just be an image of the slit.

Ch. 36 p. 9

35. (a) The maximum angle is 90°, so we have d sin  = m; (1200 nm) sin 90° = m(580 nm), which gives m = 2.07. Thus there are two orders on each side of the central maximum. (b) The half width of an order is ?m = /Nd cos m. Because Nd = L, the width of the grating, the full width is 2 ?m = 2/L cos m. For the principal maxima, we have m = 0: 0 = 0; 2 ?0 = 2(58010–9 m)/(1.8010–2 m) cos 0° = 6.4410–5 rad = 13.3. m = 1: sin 1 = (1)(580 nm)/1200 nm) = 0.483, 1 = 28.9°; 2 ?1 = 2(58010–9 m)/(1.8010–2 m) cos 28.9° = 7.3610–5 rad = 15.2. m = 2: sin 2 = (2)(580 nm)/1200 nm) = 0.967, 2 = 75.2°; 2 ?2 = 2(58010–9 m)/(1.8010–2 m) cos 75.2° = 2.5210–4 rad = 52.0. 36. We find the number of orders from d sin  = m; [1/(6500 lines/cm)](10–2 m/cm) sin 90° = m(62410–9 m), which gives m = 2.5. Thus there are two orders. We find the resolving power from R = mN; R1 = 1(6500 lines/cm)(3.61 cm) = 2.35104; R2 = 2(6500 lines/cm)(3.61 cm) = 4.69104. For the minimum-wavelength separation, we have  = /R; 1 = (624 nm)/(2.35104) = 0.0266 nm; 2 = (624 nm)/(4.69104) = 0.0133 nm. Thus the limiting wavelength difference that can be resolved in any order is 1 = 0.0266 nm. The higher order m=2 gives the better resolution.

Ch. 36 p. 10

37. (a) We find the resolving power from R = mN; 1.60104; R1 = 1(16,000 lines) = R2 = 2(16,000 lines) = 3.20104. (b) For the minimum-wavelength resolution, we have  = /R; 1 = (410 nm)/(1.60104) = 0.026 nm; 4 2 = (410 nm)/(3.2010 ) = 0.013 nm. 38. The interference maxima of the grating are given by sin  = m/d, m = 0, ± 1, ± 2, … ; or m = (d sin )/. The resolving power is R = mN = (Nd sin )/. Thus if  and  are fixed, R  Nd.

39. The frequency is f = c/. If we approximate the changes as differentials, we have ?f = – (c/2) ?. Disregarding the negative sign, we get ?f = (c/)(?/) = fR = f/mN. 40. For the diffraction from the crystal, we have m = 2d sin ; m = 1, 2, 3, … . For the first maximum, we get (1)(0.138 nm) = 2(0.265 nm) sin , which gives  =

15.1°.

41. (a) For the diffraction from the crystal, we have m = 2d sin ; m = 1, 2, 3, … . When we form the ratio for the two orders, we get m2/m1 = (sin 2)/(sin 1); 2/1 = (sin 2)/(sin 26.2°), which gives 2 = 62.0°. (b) We find the wavelength from m1 = 2d sin 1 ; (1) = 2(0.24 nm) sin 26.2°, which gives  = 0.21 nm. 42. For the diffraction from the crystal, we have m = 2d sin ; m = 1, 2, 3, … ; (1) = 2d sin 1 ; (2) = 2d sin 2 ; (3) = 2d sin 3 . We see that each equation contains the ratio /d, so the wavelength and lattice spacing cannot be separately determined. 43. If the initial intensity is I0 , through the two sheets we have I 1 = !I 0 , I2 = I1 cos2  = !I0 cos2 , which gives I2/I0 = ! cos2  = ! cos2 75° =

0.033.

Ch. 36 p. 11

44. If I0 is the intensity passed by the first Polaroid, the intensity passed by the second will be I0 when the two axes are parallel. To reduce the intensity by half, we have I = I0 cos2  = !I0 , which gives  = 45°. 45. Because the light is coming from air to glass, we find the angle from the vertical from tan p = nglass = 1.56, which gives p = 57.3°. 46. Because the light is coming from water to diamond, we find the angle from the vertical from tan p = ndiamond/nwater = 2.42/1.33 = 1.82, which gives p = 61.2°. 47. If the original intensity is I0 , the first Polaroid sheet will reduce the intensity of the original beam to I 1 = !I 0 . If the axis of the second Polaroid sheet is oriented at an angle , the intensity is I2 = I1 cos2  = !I0 cos2 . (a) I2 = !I0 cos2  = @I0 , which gives  = (b) I2 = !I0

cos2

 = 0.10I0 , which gives  =

35°. 63°.

48. For the refraction at the critical angle, with n2 the higher index, we have n1 sin 1 = n2 sin 2 ; n1 sin 90° = n2 sin 52°, which gives n2/n1 = 1.27. If the light is coming from lower index to higher index, we find the angle from tan p = n2/n1 = 1.27, which gives p = 52°. If the light is coming from higher index to lower index, we find the angle from tan p = n1/n2 = 1/1.27, which gives p = 38°. 49. If the light is coming from water to air, we find Brewster’s angle from tan p = nair/nwater = 1.00/1.33 = 0.752, which gives p = 36.9°. For the refraction at the critical angle from water to air, we have nair sin 1 = nwater sin 2 ; (1.00) sin 90° = (1.33) sin c , which gives c = 48.8°. If the light is coming from air to water, we find Brewster’s angle from tan p = nwater/nair = 1.33/1.00 = 1.33, which gives p = 53.1°. Thus p + p = 90.0°. 50. Through the successive sheets we have I1 = I0 cos2 1 , I2 = I1 cos2 2 , which gives I2 = I0 cos2 1 cos2 2 = I0 (cos2 17.0°)(cos2 34.0°) = 0.629I0 . Thus the reduction is 37.1%. 51. Through the successive polarizers we have I 1 = !I 0 ; I2 = I1 cos2 2 = !I0 cos2 2 ; I3 = I2 cos2 3 = !I0 cos2 2 cos2 3 ; I4 = I3 cos2 4 = !I0 cos2 2 cos2 3 cos2 4 ; I5 = I4 cos2 5 = !I0 cos2 2 cos2 3 cos2 4 cos2 5 = !I0 (cos2 45)4 =

I0 /32.

Ch. 36 p. 12

52. If we have N polarizers, we set the angle between adjacent polarizers as , so that N = 90°. Through the successive polarizers, we have I1 = I0 cos2 , I2 = I1 cos2  = I0 cos2  cos2  = I0 cos4 ; I3 = I2 cos2  = I0 cos4  cos2  = I0 cos6 ; … . Thus for N polarizers, we have IN = I0 cos2N  = I0 cos2N (90°/N) = 0.90I0 . By using a numerical method, such as a spreadsheet, or trial and error, we find N = 24,  = 3.75°. Thus we use 24 polarizers, with each at an angle of 3.75° with the previous one.

53. From the definition of the percent polarization we have Emax p = P/100 = (Imax – Imin)/(Imax + Imin), polarization which can be written  direction Imin = (1 – p)Imax/(1 + p). The components of the electric field amplitudes along the Emin polarization direction are Emax cos , and Emin sin . Because the intensity is proportional to E2 and the intensities add, the intensity transmitted by the polarizer is I = Emax2 cos2  + Emin2 sin2  = Imax cos2  + Imin sin2  = Imax{cos2  + [(1 – p)/(1 + p)] sin2 } = [(1 + p) cos2  + (1 – p) sin2 ]Imax/(1 + p) = [ cos2  + sin2  + p(cos2  – sin2 )]Imax/(1 + p) = [(1 + p cos 2)/(1 + p)]Imax. 54. We find the angle to the first minimum from the distances: tan 1min = !(9.20 cm)/(255 cm) = 0.0180 = sin 1min , because the angle is small. We find the slit width from a sin 1min = m; a (0.0180) = (1)(41510–9 m), which gives a = 2.3010–5 m = 0.0230 mm. 55. The wavelength of the sound is  = v/f = (343 m/s)/(750 Hz) = 0.457 m. We find the angles of the minima from a sin  = m, m = 1, 2, 3, … ; (0.88 m) sin 1 = (1)(0.457 m), which gives sin 1 = 0.520, so 1 = 31°; (0.88 m) sin 2 = (2)(0.457 m), which gives sin 2 = 1.04, so there is no 2 . Thus the whistle would not be heard clearly at angles of 31° on either side of the normal.

Ch. 36 p. 13

56. The lines act like a grating. Assuming the first order, we find the separation of the lines from d sin  = m; d sin 50° = (1)(46010–9 m), which gives d = 6.010–7 m = 600 nm. 57. Because the angle increases with wavelength, to miss a complete order we use the smallest visible wavelength. The maximum angle is 90°. We find the slit separation from d sin  = m; d sin 90° = (2)(40010–9 m), which gives d = 8.0010–7 m = 8.0010–5 cm. The number of lines/cm is 1/d = 1/(8.0010–5 cm) = 12,500 lines/cm. 58. We find the angles for the first order from d sin  = m = ; [1/(7500 lines/cm)](10–2 m/cm) sin 1 = 4.410–7 m, which gives sin 1 = 0.330, so 1 = 19.3°; [1/(7500 lines/cm)](10–2 m/cm) sin 2 = 6.310–7 m, which gives sin 2 = 0.473, so 2 = 28.2°. The distances from the central white line on the screen are y1 = L tan 1 = (2.5 m) tan 19.3° = 0.87 m; y2 = L tan 2 = (2.5 m) tan 28.2° = 1.34 m. Thus the separation of the lines is y2 – y1 = 1.34 m – 0.87 m = 0.47 m.

59. The path difference between the top and bottom of the slit for the incident wave is a sin i . The path difference between the top and bottom of the slit for the diffracted wave is a sin . When  = i , the net path difference is zero, and there will be constructive interference. There is a central maximum at  = 20°. When the net path difference is a multiple of a wavelength, there will be minima given by (a sin i) – (a sin ) = m, m = ± 1, ± 2, … , or sin  = sin 20°– (m/a), where m = ± 1, ± 2, … . 

Slit, width a  i

60. We find the angles for the first order from the distances: tan 1 = y1/L = (3.32 cm)/(60.0 cm) = 0.0553, so 1 = 3.17°; tan 2 = y2/L = (3.71 cm)/(60.0 cm) = 0.0618, so 2 = 3.54°. We find the separation of lines from d sin 1 = m1 ; d sin 3.17° = (1)(58910–9 m), which gives d = 1.06610–5 m = 1.06610–3 cm. For the second wavelength we have d sin 2 = m2 ; (1.06610–5 m) sin 3.54° = (1)2 , which gives 2 = 6.5810–7 m = 658 nm. The number of lines/cm is 1/d = 1/(1.06610–3 cm) = 938 lines/cm.

i

Ch. 36 p. 14

61. The maximum angle is 90°, so we have d sin  = m; [1/(6000 lines/cm)](10–2 m/cm) sin 90° = m(63310–9 m), which gives m = 2.63. Thus two orders can be seen on each side of the central white line. 62. Because the angle increases with wavelength, to have a full order we use the highest visible wavelength. The maximum angle is 90°, so we find the minimum separation from d sin  = m; dmin sin 90° = (2)(75010–9 m), which gives dmin = 1.5010–6 m = 1.5010–4 cm. The maximum number of lines/cm is 1/dmin = 1/(1.5010–4 cm) = 6.67103 lines/cm. 63. We find the angles for the first order from d sin  = m = ; [1/(7600 lines/cm)](10–2 m/cm) sin  = 65610–9 m, which gives sin  = 0.499, so  = 29.9°; [1/(7600 lines/cm)](10–2 m/cm) sin  = 41010–9 m, which gives sin  = 0.312, so  = 18.2°. Thus the angular separation is  –  = 29.9° – 18.2° = 11.7°. 64. We find the wavelengths from d sin  = m; [1/(9850 lines/cm)](10–2 m/cm) sin 31.2° = (1) , which gives  = 5.2610–7 m = [1/(9850 lines/cm)](10–2 m/cm) sin 36.4° = (1) , which gives  = 6.0210–7 m = [1/(9850 lines/cm)](10–2 m/cm) sin 47.5° = (1) , which gives  = 7.4910–7 m =   

526 nm; 602 nm; 749 nm.

65. (a) The resolution of the eye is  = 1.22/D = (1.22)(50010–9 m)/(5.010–3 m) = 1.2210–4 rad. We find the maximum distance from d = L ; 16 km. 2.0 m = L(1.2210–4 rad), which gives L = 1.6104 m = (b) The angular separation is the resolution:  = 1.2210–4 rad = (6.9910–3)° = 0.42'. Our answer is less than the real resolution, which includes the effects of aberrations. 66. Because the light is coming from air to water, we find the angle from the vertical from tan p = nwater = 1.33, which gives p = 53.1°. Thus the angle above the horizon is 90.0° – 53.1° = 36.9°. 67. (a) If the initial intensity is I0 , through the two sheets we have I 1 = !I 0 ; I2 = I1 cos2  = !I0 cos2 90° = 0. (b) With the third polarizer inserted, we have I 1 = !I 0 ; I2 = I1 cos2 1 = !I0 cos2 60°; I3 = I2 cos2 2 = !I0 cos2 60° cos2 30° = 0.094I0. (c) If the third polarizer is placed in front of the other two, we have the same situation as in (a), with I0 being less. Thus no light gets transmitted.

Ch. 36 p. 15

68. (a) Through the successive polarizers we have I 1 = !I 0 ; I2 = I1 cos2 2 = !I0 cos2 2 ; I3 = I2 cos2 3 = !I0 cos2 2 cos2 3 ; I4 = I3 cos2 4 = !I0 cos2 2 cos2 3 cos2 4 = !I0 cos2 30° cos2 30° cos2 30° = (b) If we remove the second polarizer, we get I 1 = !I 0 ;

0.21I0 .

I3 = I1 cos2 3 = !I0 cos2 3; I4 = I3 cos2 4 = !I0 cos2 3 cos2 4 = !I0 cos2 60° cos2 30° = 0.094I0 . Thus we can decrease the intensity by removing either the second or third polarizer. (c) If we remove the second and third polarizers, we will have two polarizers with their axes perpendicular, so no light will be transmitted. 69. If the original intensity is I0 , the first Polaroid sheet will reduce the intensity of the original beam to I 1 = !I 0 . If the axis of the second Polaroid sheet is oriented at an angle , the intensity is I2 = I1 cos2 . (a) I2 = I1 cos2  = 0.75I1 , which gives  = 30°. (b) I2 = I1 cos2  = 0.90I1 , which gives  = 18°. 2 (c) I2 = I1 cos  = 0.99I1 , which gives  = 5.7°. 70. If the initial intensity is I0 , through the two sheets we have I1 = I0 cos2 1 ; I2 = I1 cos2 2 = I0 cos2 1 cos2 2 ; 0.15I0 = I0 cos2 1 cos2 40°, which gives 1 = 60°. 71. We can write the electric field amplitude as E = E1i + E2j, so E2 = E12 + E22. Because the intensity is proportional to E2, we have I = I1 + I2, so the intensities add, with no interference. 72. For the minimum aperture the angle subtended at the lens by the smallest feature is the angular resolution:  = d/L = 1.22/D ; 34 cm. (510–2 m)/(25103 m) = (1.22)(55010–9 m)/D, which gives D = 0.34 m = 73. We find the spacing from m = 2d sin ; m = 1, 2, 3, … . (2)(0.0973 nm) = 2d sin 23.4°, which gives d =

0.245 nm.

Ch. 36 p. 16

74. We find the angles for the Bragg scattering from 2d sin  = m; 2(0.25 nm) sin 1 = (1)(0.10 nm), which gives 1 = 11.5 °;     2(0.25 nm) sin 2 = (2)(0.10 nm), which gives 2 = 23.6 °. The radii of the diffraction rings are R1 = L tan 21 = (10 cm) tan 2(11.5 °) = 4.2 cm; R2 = L tan 22 = (10 cm) tan 2(23.6 °) = 11 cm.

X-ray

Crystal 2

Screen

Chapter 37

p.1

CHAPTER 37 – Special Theory of Relativity [1 – (v/c)2]1/2 = {1 – [(20,000 m/s)/(3.00108 m/s)]2}1/2 = [1 – (v/c)2]1/2 = [1 – (0.0100)2]1/2 = 0.99995. [1 – (v/c)2]1/2 = [1 – (0.100)2]1/2 = 0.995. [1 – (v/c)2]1/2 = [1 – (0.900)2]1/2 = 0.436. [1 – (v/c)2]1/2 = [1 – (0.990)2]1/2 = 0.141. [1 – (v/c)2]1/2 = [1 – (0.999)2]1/2 = 0.0447.

1.

(a) (b) (c) (d) (e) (f)

1.00.

2.

You measure the contracted length. We find the rest length from L = L0[1 – (v/c)2]1/2; 28.2 m = L0[1 – (0.750)2]1/2, which gives L0 = 42.6 m.

3.

We find the lifetime at rest from ?t = ?t0/[1 – (v2/c2)]1/2;

4.7610–6 s = ?t0/{1 – [(2.70108 m/s)/(3.00108 m/s)]2}1/2, which gives ?t0 =

4.

You measure the contracted length: L = L0[1 – (v/c)2]1/2

= (100 ly){1 – [(2.50108 m/s)/(3.00108 m/s)]2}1/2 =

5.

2.0710–6 s.

We determine the speed from the time dilation: ?t = ?t0/[1 – (v2/c2)]1/2; 4.1010–8 s = (2.6010–8 s)/[1 – (v/c)2]1/2, which gives v =

55.3 ly.

0.773c.

6.

We determine the speed from the length contraction: L = L0[1 – (v/c)2]1/2; 25 ly = (75 ly)[1 – (v/c)2]1/2, which gives v = 0.94c.

7.

For a 1.00 per cent change, the factor in the expressions for time dilation and length contraction must equal 1 – 0.0100 = 0.9900: [1 – (v/c)2]1/2 = 0.9900, which gives v = 0.141c.

8.

In the Earth frame, the clock on the Enterprise will run slower. (a) We find the elapsed time on the ship from ?t = ?t0/[1 – (v2/c2)]1/2; 5.0 yr = ?t0/[1 – (0.84)2]1/2, which gives ?t0 = 2.7 yr. (b) We find the elapsed time on the Earth from ?t = ?t0/[1 – (v2/c2)]1/2 = (5.0 yr)/[1 – (0.84)2]1/2 = 9.2 yr.

9.

(a) To an observer on Earth, 95.0 ly is the rest length, so the time will be tEarth = L0/v = (95.0 ly)/0.960c = 99.0 yr. (b) We find the dilated time on the spacecraft from ?t = ?t0/[1 – (v2/c2)]1/2; 99.0 yr = ?t0/[1 – (0.960)2]1/2, which gives ?t0 = 27.7 yr. (c) To the spacecraft observer, the distance to the star is contracted: L = L0[1 – (v/c)2]1/2 = (95.0 ly)[1 – (0.960)2]1/2 = 26.6 ly. (d) To the spacecraft observer, the speed of the spacecraft is v = L/?t = (26.6 ly)/27.7 yr = 0.960c, as expected.

Chapter 37

p.2

10. (a) You measure the contracted length. We find the rest length from L = L0[1 – (v/c)2]1/2; 4.80 m = L0[1 – (0.660)2]1/2, which gives L0 = 6.39 m. Distances perpendicular to the motion do not change, so the rest height is (b) We find the dilated time in the sports vehicle from ?t = ?t0/[1 – (v2/c2)]1/2; 20.0 s = ?t0/[1 – (0.660)2]1/2, which gives ?t0 = 15.0 s. (c) To your friend, you moved at the same relative speed: 0.660c. (d) She would measure the same time dilation: 15.0 s.

1.25 m.

11. In the Earth frame, the average lifetime of the pion will be dilated: ?t = ?t0/[1 – (v2/c2)]1/2. The speed as a fraction of the speed of light is v/c = d/c ?t = d[1 – (v2/c2)]1/2/c ?t0 ; v/c = (15 m)[1 – (v2/c2)]1/2/(3.00108 m/s)(2.610–8 s), 0.89c = 2.7108 m/s. which gives v = 12. With the standard orientation of the reference frames, the Galilean transformation is x = x + vt, y = y, z = z. (a) For the given data we have x = 25 m + (30 m/s)(2.5 s) = 100 m; y = 20 m; z = 0. Thus the coordinates of the person are (100 m, 20 m, 0). (b) For the given data we have x = 25 m + (30 m/s)(10.0 s) = 325 m; y = 20 m; z = 0. Thus the coordinates of the person are (325 m, 20 m, 0). 13. With the standard orientation of the reference frames, the Lorentz transformation is x = (x – vt), y = y, z = z; with  = 1/[1 – (v/c)2]1/2 = 1/{1 – [(1.80108 m/s)/(3.00108 m/s)]2}1/2 = 1.25. (a) For the given data we have 25 m = (1.25)[x – (1.80108 m/s)(2.510–6 s)], which gives x = 470 m; y = 20 m; z = 0. Thus the coordinates of the person are (470 m, 20 m, 0). (b) For the given data we have 25 m = (1.25)[x – (1.80108 m/s)(10.010–6 s)]. which gives x = 1820 m; y = 20 m; z = 0. Thus the coordinates of the person are (1820 m, 20 m, 0). 14. We choose the Earth for the S frame and the rocket for the S frame. The speed of the meteor in the S frame is 0.50c. We find the speed of the meteor in the Earth frame from the velocity transformation: u = (u + v)/[1 + (vu/c2)] 0.80c. = (0.50c + 0.50c )/[1 + (0.50c )(0.50c)/c2] = 15. (a) We choose the Earth for the S frame and spaceship 2 for the S frame, so v = – 0.50c. The speed of spaceship 1 in the S frame is 0.50c. We find the speed of spaceship 1 in S from the velocity transformation: u = (u – v)/[1 – (vu/c2)] 0.80c. = [0.50c – (– 0.50c )]/[1 – (– 0.50c )(0.50c)/c2] = (b) We could redefine our reference frames, but we know that the velocity of spaceship 2 relative to spaceship 1 must be – 0.80c.

Chapter 37

p.3

16. We take the positive direction in the direction of the first spaceship. (a) In the reference frame of the Earth, the first spaceship is moving at + 0.71c, and the second spaceship is moving at + 0.87c relative to the first. Thus the speed of the second spaceship relative to the Earth is u = (u + v)/(1 + vu/c2) = (+ 0.87c + 0.71c)/[1 + (0.71)(0.87)] = 0.98c. (b) In the reference frame of the Earth, the first spaceship is moving at + 0.71c, and the second spaceship is moving at – 0.87c relative to the first. Thus the speed of the second spaceship relative to the Earth is u = (u + v)/(1 + vu/c2) = (– 0.87c + 0.71c)/[1 + (0.71)(– 0.87)] = – 0.42c. 17. With the standard orientation of the reference frames, the Galilean velocity transformation is ux = ux + v, uy = uy, uz = uz. For the given data we have ux = 25.0 m/s + 30 m/s = 55 m/s; uy = 25.0 m/s; uz = 0. The magnitude of the velocity is u = (ux2 + uy2)1/2 = [(55 m/s)2 + ( 25.0 m/s)2]1/2 = 60 m/s. We find the angle the velocity makes with the x-axis from 24°. tan  = uy/ux = (25 m/s)/(55 m/s) = 0.455, so  = 18. With the standard orientation of the reference frames, the Lorentz velocity transformation is ux = (ux + v)/[1 + (uxv/c2)], uy = uy[1 – (v/c)2]1/2/[1 + (uxv/c2)], uz = uz[1 – (v/c)2]1/2/[1 + (uxv/c2)]. For the given data we have ux = (2.0108 m/s + 1.80108 m/s)/[1 + (2.0108 m/s)(1.80108 m/s)/(3.00108 m/s)2] = 2.71108 m/s; uy = (2.0108 m/s){1 – [(1.80108 m/s)/(3.00108 m/s)]2}1/2/ [1 + (2.0108 m/s)(1.80108 m/s)/(3.00108 m/s)2] = 1.14108 m/s; uz = 0. The magnitude of the velocity is u = (ux2 + uy2)1/2 = [(2.71108 m/s)2 + (1.14108 m/s)2]1/2 = 2.9108 m/s. We find the angle the velocity makes with the x-axis from 23°. tan  = uy/ux = (1.14108 m/s)/(2.71108 m/s) = 0.421, so  = 19. We choose the Earth for the S frame and the spaceship for the S frame. The velocity components of the module in the S frame are ux = 0, uy = 0.82c. We find the velocity components of the module in the Earth frame from the velocity transformation: ux = (ux + v)/[1 + (uxv/c2)] = (0 + 0.66c)/(1 + 0) = 0.66c; uy = uy[1 – (v/c)2]1/2/[1 + (uxv/c2)] = (0.82c)[1 – (0.66)2]1/2/(1 + 0) = 0.616c. The magnitude of the velocity is u = (ux2 + uy2)1/2 = [(0.66c)2 + ( 0.616c)2]1/2 = 0.90c = 2.7108 m/s. We find the angle the velocity makes with the x-axis from 43°. tan  = uy/ux = (0.616c)/(0.66c) = 0.933, so  = 20. The velocity components of the particle in the S frame are ux = u cos , uy = u sin . We find the velocity components of the particle in the S frame from the velocity transformation: ux = (ux – v)/[1 – (uxv/c2)]; uy = uy[1 – (v/c)2]1/2/[1 – (uxv/c2)].

Chapter 37

p.4

We find the angle the velocity makes with the x-axis from tan  = uy/ux = uy[1 – (v/c)2]1/2/(ux – v) = (u sin )[1 – (v/c)2]1/2/(u cos  – v) = (sin )[1 – (v/c)2]1/2/(cos  – v/u). 21. (a) In the frame S, the x-component of the stick will be contracted: Lx = (L0 cos )[1 – (v/c)2]1/2. The y-component is unchanged: Ly = L0 sin . The magnitude of the length in S is L = (Lx2 + Ly2)1/2 = {(L0 cos )2[1 – (v/c)2] + (L0 sin )2}1/2

= L0{(cos2 )[1 – (v/c)2] + sin2 }1/2 = L0[1 – (v/c)2 cos2 ]1/2. (b) We find the angle that the stick makes with the x-axis from tan  = Ly/Lx = (L0 sin )/(L0 cos )[1 – (v/c)2]1/2 = tan /[1 – (v/c)2]1/2. Thus we have  = tan–1{tan /[1 – (v/c)2]1/2}.

22. (a) We choose the train as frame S and the Earth as frame S. Thus for the two firings we have ?x = xB – xA = 50 m, and ?t = tB – tA = 0. We find the separation of the two firings in the S frame from ?x = (?x + v ?t) = (?x + 0), or ?x = ?x/, which is the contraction of the length. We find the time interval between firings in the S frame from ?t = (?t + v ?x/c2) = (0 + v ?x/c2) = v ?x/c2 = v ?x/c2. Because ?x > 0, we see that ?t > 0, so B fires after A in the S frame; A fires first. (b) The time difference between firings in the Earth frame is ?t = v ?x/c2 = (50 m/s)(50 m)/(3.00108 m/s)2 = 2.810–14 s. (c) If u is the speed of a bullet in the Earth frame and T represents the time at which a gunfighter is struck, we have TA = (?x/u) + tB , and TB = (?x/u) + tA , or TB – TA = tA – tB = – ?t. Thus we see that ?T < 0, or B is struck first in the Earth frame, as expected, because A fired first. In frame S we have ?T = (?T – v ?x/c2) = (– ?t – v ?x/c2). Because ?t > 0, and ?x > 0, we see that ?T < 0, so B is hit first in the S frame also. 23. To an observer in the barn reference frame, if the boy runs fast enough, the measured contracted length of the pole will be less than 13.0 m, so the observer can say that the two ends of the pole were inside the barn simultaneously. We find the necessary speed for the contracted pole to fit inside the barn from Lpole = L0pole[1 – (v/c)2]1/2; 10.0 m = (13.0 m)[1 – (v/c)2]1/2, which gives v = 0.64c. To the boy, the barn is moving and thus the length of the barn, as he would measure it, is less than the length of the pole: Lbarn = L0barn[1 – (v/c)2]1/2 = (10.0 m)[1 – (0.64)2]1/2 = 7.7 m. However, simultaneity is relative. Thus when the two ends are simultaneously inside the barn to the barn observer, those two events are not simultaneous to the boy. Thus he would claim that the observer in the barn determined that the ends of the pole were inside the barn at different times, which is also what the boy would say. It is not possible in the boy’s frame to have both ends of the pole inside the barn simultaneously.

Chapter 37

p.5

24. The momentum of the proton is p = mv/[1 – (v2/c2)]1/2 = (1.6710–27 kg)(0.85)(3.00108 m/s)/[1 – (0.85)2]1/2 = 25. We find the speed from mrel = m/[1 – (v2/c2)]1/2; 2m = m/[1 – (v2/c2)]1/2, which gives v =

8.110–19 kg · m/s.

0.866c.

26. For the momentum to be doubled we have p2 = 2p1 ; mv2/[1 – (v22/c2)]1/2 = 2mv1/[1 – (v12/c2)]1/2; or (v2/c)2/[1 – (v2/c)2] = 4(v1/c)2/[1 – (v1/c)2] = 4(0.20)2/[1 – (0.20)2], which gives

v2 = 0.38c.

27. The two expressions for the momentum are prel = mv/[1 – (v2/c2)]1/2, and pc = mv. Thus the error is

(prel – pc)/prel = ({mv/[1 – (v2/c2)]1/2} – mv)/{mv/[1 – (v2/c2)]1/2} = 1 – [1 – (v2/c2)]1/2. (a) For the given speed we have 0.5%. (prel – pc)/prel = 1 – [1 – (v2/c2)]1/2 = 1 – [1 – (0.10)2]1/2 = 0.005 = (b) For the given speed we have 13%. (prel – pc)/prel = 1 – [1 – (v2/c2)]1/2 = 1 – [1 – (0.50)2]1/2 = 0.13 =

28. The fractional change in momentum is

(p2 – p1)/p1 = ({mv2/[1 – (v22/c2)]1/2} – {mv1/[1 – (v12/c2)]1/2})/{mv1/[1 – (v12/c2)]1/2}

= {v2[1 – (v12/c2)]1/2/v1[1 – (v22/c2)]1/2} – 1. (a) For the given speeds we have 310%. (p2 – p1)/p1 = {0.90c[1 – (0.45)2]1/2/0.45c[1 – (0.90)2]1/2} – 1 = 3.1 = Note that this is about 3 what use of the classical expression would give. (b) For the given speeds we have 140%. (p2 – p1)/p1 = {0.98c[1 – (0.90)2]1/2/0.90c[1 – (0.98)2]1/2} – 1 = 1.4 = Note that this is about 15 what use of the classical expression would give.

29. We find the increase in mass from ?m = ?E/c2 = (4.82104 J)/(3.00108 m/s)2 = 5.3610–13 kg. Note that this is so small, most chemical reactions are considered to have mass conserved. 30. We find the loss in mass from ?m = ?E/c2 = (200 MeV)(1.6010–13 J/MeV)/(3.00108 m/s)2 =

3.5610–28 kg.

31. The rest energy of the electron is E = mc2 = (9.10910–31 kg)(2.998108 m/s)2 = 8.1910–14 J –14 –13 0.511 MeV. = (8.18710 J)/(1.60210 J/MeV) = 32. The mass of the proton is m = E/c2 = mc2/c2 = (1.6710–27 kg)(3.00108 m/s)2/(1.6010–13 J/MeV)c2 = 33. We find the necessary mass conversion from ?m = ?E/c2 = (81019 J)/(3.00108 m/s)2 = 34. We find the energy equivalent of the mass from

9102 kg.

939 MeV/c2.

Chapter 37

p.6

E = mc2 = (1.010–3 kg)(3.00108 m/s)2 = 9.01013 J. If this energy increases the gravitational energy, we have E = Mgh; 9.01013 J = M(9.80 m/s2)(100 m), which gives M = 9.21010 kg. 35. If the kinetic energy is equal to the rest energy, we have K = {mc2/[1 – (v2/c2)]1/2} – mc2 = mc2, or 1/[1 – (v2/c2)]1/2 = 2, which gives v = 0.866c. 36. If the kinetic energy is 25% of the rest energy, we have K = {mc2/[1 – (v2/c2)]1/2} – mc2 = 0.25mc2, or 1/[1 – (v2/c2)]1/2 = 1.25, which gives v = 0.60c. 37. (a) We find the work required from

W = K = mc2({1/[1 – (v/c)2]1/2} – 1)

= (939 MeV)({1/[1 – (0.997)2]1/2} – 1) = 11.2103 MeV = 11.2 GeV (1.7910–9 J). (b) The momentum of the proton is p = mv/[1 – (v/c)2]1/2 = (1.6710–27 kg)(0.997)(3.00108 m/s)/[1 – (0.997)2]1/2 = 6.4510–18 kg · m/s. 38. The speed of the proton is v = (2.60108 m/s)/(3.00108 m/s) = 0.867c. The kinetic energy is K = mc2({1/[1 – (v/c)2]1/2} – 1)

= (939 MeV)({1/[1 – (0.867)2]1/2} – 1) = 943 MeV (1.5110–10 J). The momentum of the proton is p = mv{1/[1 – (v/c)2]1/2} = (1.6710–27 kg)(2.60108 m/s){1/[1 – (0.867)2]1/2} = 8.7010–19 kg · m/s. 39. The total energy of the proton is E = K + mc2 = 750 MeV + 939 MeV = 1689 MeV. The relation between the momentum and energy is (pc)2 = E2 – (mc2)2; p2(3.00108 m/s)2 = [(1689 MeV)2 – (939 MeV)2](1.6010–13 J/MeV)2, 7.4910–19 kg · m/s. which gives p = 40. The kinetic energy acquired by the proton is K = qV = (1 e)(95 MV) = 95 MeV. We find the speed from K = mc2({1/[1 – (v/c)2]1/2} – 1);

95 MeV = (939 MeV)({1/[1 – (v/c)2]1/2} – 1), which gives v =

0.42c.

41. We find the speed from

K = mc2({1/[1 – (v/c)2]1/2} – 1);

1.00 MeV = (0.511 MeV)({1/[1 – (v2/c2)]1/2} – 1), which gives v =

42. The kinetic energy acquired by the electron is K = qV = (1 e)(0.025 MV) = 0.025 MeV.

0.941c.

Chapter 37

p.7

We find the speed from

K = mc2({1/[1 – (v/c)2]1/2} – 1);

0.025 MeV = (0.511 MeV)({1/[1 – (v2/c2)]1/2} – 1), which gives v =

0.302c.

43. If M is the mass of the new particle, for conservation of energy we have 2(K + mc2) = Mc2; 2m/[1 – (v2/c2)]1/2. 2mc2/[1 – (v2/c2)]1/2 = Mc2, which gives M = Because energy is conserved, there was no loss. The final particle is at rest, so the kinetic energy loss is the initial kinetic energy of the two colliding particles: Kloss = 2K = (M – 2m)c2 =

2mc2({1/[1 – (v2/c2)]1/2} – 1).

44. The total energy of the proton is E = K + mc2; 2K = K + mc2, which gives K = mc2. We find the speed from K = {mc2/[1 – (v2/c2)]1/2} – mc2 = mc2, or 1/[1 – (v2/c2)]1/2 = 2, which gives v = 0.866c. 45. We find the speed from K = {mc2/[1 – (v2/c2)]1/2} – mc2 = mc2, or 1/[1 – (v2/c2)]1/2 = 2, which gives v = 0.866c. The momentum of the electron is p = mv/[1 – (v /c)2]1/2 = (9.1110–31 kg)(0.866)(3.00108 m/s)(2) = 4.7310–22 kg · m/s. 46. (a) The kinetic energy is

K = mc2({1/[1 – (v/c)2]1/2} – 1)

= (27,000 kg)(3.00108 m/s)2({1/[1 – (0.21)2]1/2} – 1) = 5.541019 J = (b) When we use the classical expression, we get Kc = !mv2 = !(27,000 kg)[(0.21)(3.00108 m/s)]2 = 5.351019 J. The error is (5.35 – 5.54)/(5.54) = – 0.03 = – 3%.

5.51019 J.

47. The speed of the proton is v = (8.4107 m/s)/(3.00108 m/s) = 0.280c. The kinetic energy is K = mc2({1/[1 – (v/c)2]1/2} – 1)

= (939 MeV)({1/[1 – (0.280)2]1/2} – 1) = 39 MeV (6.310–12 J). The momentum of the proton is p = mv/[1 – (v/c)2]1/2 = (1.6710–27 kg)(8.4107 m/s){1/[1 – (0.280)2]1/2} = 1.4610–19 kg · m/s = From the classical expressions, we get

1.510–19 kg · m/s.

Chapter 37

p.8

Kc = !mv2 = !(1.6710–27 kg)(8.4107 m/s)2 = 5.910–12 J, with an error of (5.9 – 6.3)/(6.3) = – 0.06 = – 6%. p = mv = (1.6710–27 kg)(8.4107 m/s) = 1.4010–19 kg · m/s, with an error of (1.40 – 1.46)/(1.46) = – 0.04 = – 4%. 48. If we ignore the recoil of the neptunium nucleus, the increase in kinetic energy is the kinetic energy of the alpha particle, which equals the loss in mass: K = [mAm – (mNp + m)]c2; 5.5 MeV = [241.05682 u – (mNp + 4.00260 u)]c2(931.5 MeV/uc2), which gives mNp = 237.04832 u.

49. The increase in kinetic energy comes from the decrease in potential energy: K = – ?U = mc2({1/[1 – (v/c)2]1/2} – 1);

– (– 5.6010–14 J) = (9.1110–31 kg)(3.00108 m/s)2({1/[1 – (v/c)2]1/2} – 1), which gives v = 0.804c. K = [p2c2 + (mc2)2]1/2 – mc2

50. (a)

(b)

K

K = pc.

K K = pc – mc2 K = pc K = p2 /2 m Nonzero mass

Zero mass p

p

51. The total energy of the proton is E = mrelc2 = K + mc2 = 900 GeV + 0.938 GeV = 901 GeV, so the relativistic mass is 901 GeV/c2. We find the speed from mrel = m/[1 – (v2/c2)]1/2; 901 GeV/c2 = (0.938 GeV/c2)/[1 – (v2/c2)]1/2, which gives [1 – (v2/c2)]1/2 = 1.0410–3, so v ˜ 1.00c. The speed is constant so the relativistic mass is constant. The magnetic force provides the radial acceleration: qvB = mrelv2/r, or B = mrelv/qr = mv/qr[1 – (v2/c2)]1/2 = (1.6710–27 kg)(3.00108 m/s)/(1.610–19 C)(1.0103 m)(1.0410–3) = 3.0 T. 52. Because the total energy of the muons becomes electromagnetic energy, we have E = K1 + m1c2 + K2 + m2c2 = mc2/[1 – (v12/c2)]1/2 + mc2/[1 – (v22/c2)]1/2 = (105.7 MeV)/[1 – (0.33)2]1/2 + (105.7 MeV)/[1 – (0.50)2]1/2 = 234 MeV. 53. The magnetic force provides the radial acceleration: qvB = mrelv2/r, or mrel = qBr/v = E/c2. With v ˜ c, and q = 1 e, we get E (eV) = (1)Brc2/c = Brc. Note that the relativistic mass is constant during the revolution.

Chapter 37

p.9

54. The total energy is E = K + mc2 and is related to the momentum by E2 = p2c2 + m2c4 = (K + mc2)2 = K2 + 2Kmc2 + m2c4, which gives p2c2 = K2 + 2Kmc2, or p = (K2 + 2Kmc2)1/2/c.

55. (a) We let m be the mass of the particle. Its velocity has an x-component only, ux , so its momentum components in frame S are px = mux/[1 – (ux2/c2)]1/2, py = 0, pz = 0. The energy of the particle in frame S is E = K + mc2 = mc2/[1 – (ux2/c2)]1/2. The velocity in frame S is ux = (ux – v)/[1 – (uxv/c2)], uy = 0, uz = 0; so the particle’s momentum is px = mux/[1 – (ux2/c2)]1/2, py = 0 = py , pz = 0 = pz . If we consider the denominator, we have 2 2 2  u 2 ux – v 1 2 2 – u x – 2u x v + v 1 – x2 = 1– = 1 – u v/ c x 2 1 – u xv/ c 2 c c2 1 – (u xv / c2 c 2 1 = 1 – 2 u xv/ c 2 + u xv/ c 2 2– u x 2/ c 2 + 2u xv/ c 2 – v 2 / c 2 1 – u x v/ c2 1 = 1 – u x 2/ c 2 1 – v 2/ c 2 . 1 – u x v/ c2 When we use this and the velocity transformation in the expression for the momentum, we have mu x m u x – v 1 – u xv/ c 2 mu x – mv  1 p x = = = 2 2 2 2 2 2 2 2 2 1 – u x v/ c 1 – ux / c 1 – v / c 1 – u x 2 / c2 1– v / c 1 – u x / c

px – vE/ c 2 . 1 – v 2/ c 2 For the transformation of the energy we have mc 2 1 – u xv/ c 2 mc 2 – mu x v  1 mc 2 E = = = 2 2 2 2 2 2 2 2 1– v / c 1 – ux / c 1 – v / c 1 – u x 2/ c 2 1 – u x / c E – pxv = . 1 – v 2 / c2 (b) To simplify the expressions, we use  = 1/[1 – (v2/c2)]1/2. The transformations are x =  [x – (v/c)ct], y = y, z = z, ct =  [ct – (vx/c)]; px =  [px – (v/c)E/c], py = py , pz = pz , E/c =  [(E/c) – (vpx/c)]. Thus we see that px , py , pz , E/c transform in the same way as x, y, z, ct. =

56. For a source moving away from us the Doppler shift is f = f0[(c – v)/(c + v)]1/2. Thus we have (f0 – f)/f0 = 1 – [(c – v)/(c + v)]1/2;

0.797 = 1 – {[1 – (v/c)]/[1 + (v/c)]}1/2, which gives v =

0.921c.

Chapter 37

p.10

57. From c = f, we see that f = f0/3.0, so the quasar is moving away from us. (a) For the Doppler shift we have f = f0[(c – v)/(c + v)]1/2; f0/3.0 = f0{[1 – (v/c)]/[1 + (v/c)]}1/2, which gives v = 0.80c. (b) For the “classical” Doppler shift, the wavelength from the quasar is  = (v + c)/f0 , and the received frequency is f = c/ = f0c/(v + c); f0/3.0 = f0/[1 + (v/c)], which gives v = 2.0c.

58. For a source moving toward the Earth the Doppler shift is f = f0[(c + v)/(c – v)]1/2 = f0{[1 + (v/c)]/[1 – (v/c)]}1/2 = (95.0 MHz)[(1 + 0.80)/(1 – 0.80]1/2 = 285 MHz. 59. For a source moving away from us the Doppler shift is  = 0{[1 + (v/c)]/[1 – (v/c)]}1/2. If v « c, we can use the approximation 1/(1 – x) ˜ 1 + x:  ˜ 0{[1 + (v/c)]2}1/2 = 0[1 + (v/c)]. Thus the fractional change is ?/0 = ( – 0)/0 = 1 + (v/c) – 1 = v/c.

60. The electrostatic force provides the radial acceleration: ke2/r2 = mv2/r. Thus we find the speed from v2 = (9.0109 N · m2/C2)(1.610–19 C)2/(9.1110–31 kg)(0.510–10 m), which gives v = 2106 m/s. Because this is less than 0.1c, the electron is not relativistic. 61. Because the North Pole has no tangential velocity, the clock there will measure a year (3.16107 s). The clock at the equator has the tangential velocity of the equator: v = rE = (6.38106 m)(2p rad)/(24 h)(3600 s/h) = 464 m/s. The clock at the equator will run slow: tequator = tNorth[1 – (v2/c2)]1/2 ˜ tNorth[1 – !(v/c)2]. Thus the difference in times is tNorth – tequator = tNorth!(v/c)2 = (3.16107 s)![(464 m/s)/(3.00108 m/s)]2 = 3.810–5 s. 62. (a) To travelers on the spacecraft, the distance to the star is contracted: L = L0[1 – (v/c)2]1/2 = (4.3 ly)[1 – (v/c)2]1/2. Because the star is moving toward the spacecraft, to cover this distance in 4.0 yr, the speed of the star must be v = L/t = (4.3 ly/4.0 yr)[1 – (v/c)2]1/2 = (1.075c)[1 – (v/c)2]1/2, which gives v = 0.73c. Thus relative to the Earth-star system, the speed of the spacecraft is 0.73c. (b) According to observers on Earth, clocks on the spacecraft run slow: tEarth = t/[1 – (v2/c2)]1/2 = (4.0 yr)/[1 – (0.73)2]1/2 = 5.9 yr. Note that this agrees with the time found from distance and speed: tEarth = L0/v = (4.3 ly)/(0.73c) = 5.9 yr. 63. The dependence of the relativistic mass on the speed is

Chapter 37

p.11

mrel = m/[1 – (v2/c2)]1/2. If we consider a box with sides x0 , y0 , and z0 , dimensions perpendicular to the motion, which we take to be the x-axis, do not change, but the length in the direction of motion will contract: x = x0[1 – (v/c)2]1/2. Thus the density is  = mrel/xyz = m/[1 – (v2/c2)]1/2x0[1 – (v2/c2)]1/2y0z0 = 0/[1 – (v2/c2)]. 64. We convert the speed: (1500 km/h)/(3.6 ks/h) = 417 m/s. The flight time as observed on Earth is tEarth = 2prE/v = 2p(6.38106 m)/(417 m/s) = 9.62104 s. The clock on the plane will run slow: tplane = tEarth[1 – (v2/c2)]1/2 ˜ tEarth[1 – !(v/c)2]. Thus the difference in times is tEarth – tplane = tEarth!(v/c)2 = (9.62104 s)![(417 m/s)/(3.00108 m/s)]2 =

9.310–8 s.

65. (a) We find the speed from

K = mc2({1/[1 – (v/c)2]1/2} – 1);

10,000mc2 = mc2({1/[1 – (v2/c2)]1/2} – 1), which gives [1 – (v2/c2)]1/2 = 1.0010–4, or (v/c)2 = 1 – 1.0010–8. When we take the square root, we get v/c = (1 – 1.0010–8)1/2 ˜ 1 – !(1.0010–8) = 1 – 0.5010–8. Thus the speed is 1.5 m/s less than c. (b) The contracted length of the tube is L = L0[1 – (v/c)2]1/2 = (3.0 km)(1.0010–4) = 3.010–4 km = 66. We find the mass change from the required energy: E = Pt = mc2; (100 W)(3.16107 s) = m(3.00108 m/s)2, which gives m =

30 cm.

3.510–8 kg.

67. The minimum energy is required to produce the pair at rest: Emin = 2mc2 = 2(0.511 MeV) = 1.02 MeV (1.6410–13 J). 68. (a) Because the spring is at rest on the spaceship, its period is T = 2p(m/k)1/2 = 2p[(1.68 kg)/(48.7 N/m)]1/2 = 1.17 s. (b) The oscillating mass is a clock. According to observers on Earth, clocks on the spacecraft run slow: TEarth = T/[1 – (v2/c2)]1/2 = (1.17 s)/[1 – (0.900)2]1/2 = 2.68 s. 69. The speed is constant so the relativistic mass is constant. The magnetic force provides the radial acceleration: qvB = mrelv2/r, or r = mrelv/qB = mv/qB[1 – (v2/c2)]1/2 = (9.1110–31 kg)(0.92)(3.00108 m/s)/(1.610–19 C)(1.8 T)[1 – (0.92)2]1/2 = 2.210–3 m = 2.2 mm. 70. We take the positive direction in the direction of the Enterprise. In the reference frame of the alien vessel, the Earth is moving at – 0.60c, and the Enterprise is moving at + 0.90c relative to the Earth. Thus the speed of the Enterprise relative to the alien vessel is u = (v + u)/(1 + vu/c2) = (– 0.60c + 0.90c)/[1 + (– 0.60)(+ 0.90)] = 0.65c. Note that the relative speed of the two vessels as seen on Earth is 0.90c – 0.60c = 0.30c.

Chapter 37

p.12

71. The kinetic energy comes from the decrease in mass: K = [mn– (mp + me + m)]c2 = [1.008665 u – (1.00728 u + 0.000549 u + 0)]c2(931.5 MeV/uc2) = 72. (a) We find the rate of mass loss from m/t = (E/t)/c2 = (41026 W)/(3108 m/s)2 = 4109 kg/s. (b) We find the time from ?t = ?m/rate = (5.981024 kg)/(4.4109 kg/s)(3.16107 s/yr) = (c) We find the time for the Sun to lose all of its mass at this rate from ?t = ?m/rate = (2.01030 kg)/(4.4109 kg/s)(3.16107 s/yr) =

0.78 MeV.

4107 yr. 11013 yr.

73. The speed of the particle is v = (2.24108 m/s)/(3.00108 m/s) = 0.747c. We use the momentum to find the rest mass: p = mv/[1 – (v /c)2]1/2; 3.0710–22 kg · m/s = m(0.747)(3.00108 m/s)/[1 – (0.747)2]1/2, which gives m = 9.1110–31 kg. Because the particle has a negative charge, it is an electron. 74. The binding energy is the energy required to provide the increase in mass: E = [(2mp + 2mn) – mHe]c2 = [2(1.00783 u) + 2(1.00867 u) – 4.00260 u ]c2(931.5 MeV/uc2) = 28.3 MeV. Note that the two electron masses included in the hydrogen atoms and the helium atom cancel. 75. We convert the speed: (110 km/h)/(3.6 ks/h) = 30.6 m/s. Because this is much smaller than c, the relativistic mass of the car is mrel = m/[1 – (v2/c2)]1/2 ˜ m[1 + !(v/c)2]. The fractional change in mass is (mrel – m)/m = [1 + !(v/c)2] – 1 = !(v/c)2 = ![(30.6 m/s)/(3.00108 m/s)]2 = 5.1910–15 =

5.1910–13 %.

76. (a) The magnitudes of the momenta are equal: p = mv/[1 – (v/c)2]1/2 = (1.6710–27 kg)(0.935)(3.00108 m/s)/[1 – (0.935)2]1/2 = 1.3210–18 kg · m/s. (b) Because the protons are moving in opposite directions, the sum of the momenta is 0. (c) In the reference frame of one proton, the laboratory is moving at 0.935c. The other proton is moving at + 0.935c relative to the laboratory. Thus the speed of the other proton relative to the first is u = (v + u)/(1 + vu/c2) = [+ 0.935c + (+ 0.935c)]/[1 + (+ 0.935)(+ 0.935)] = 0.998c. The magnitude of the momentum of the other proton is p = mu/[1 – (u/c)2]1/2 = (1.6710–27 kg)(0.998)(3.00108 m/s)/[1 – (0.998)2]1/2 = 7.4510–18 kg · m/s. 77. The relation between energy and momentum is

Chapter 37

p.13

E = (m2c4 + p2c2)1/2 = c(m2c2 + p2)1/2. For the momentum, we have p = mv/[1 – (v/c)2]1/2 = Ev/c2, or v = pc2/E = pc/(m2c2 + p2)1/2.

78. The speed of light in the medium at rest is c/n. We find the speed in the medium according to the observer from the addition of velocities: v = [(c/n) + v)/[1 + (c/n)v/c2] = [(c/n) + v]/[1 + (v/cn)] = (c/n)[1 + (vn/c)]/[1 + (v/cn)]. The thickness of the glass to the observer is contracted: d  = d[1 – (v/c)2]1/2. Thus the time for light to go from A to B is t  = [(L – d)/c] + (d/v) = (L/c) – d[(1/c) – (1/v)] = (L/c) – d[1 – (v/c)2]1/2[(1/c) – (n/c)(1 + v/cn)/(1 + vn/c)] = (L/c) + [1 – (v/c)2]1/2(d/c)(n – 1)(1 – v/c)/(1 + vn/c). If v = 0, we have t  = (L/c) + (d/c)(n – 1) = [(L – d)/c] + [d/(c/n)], which is the time for light to travel through the air plus the time to travel through the glass. If n = 1, we have t  = L/c, which is the same as the time in the case where the glass is not present. If v = c, we have t  = L/c, which agrees with the fact that the velocity c transforms into c. 79. From the Lorentz transformation we have x =  [x – (v/c)ct], t  =  [t – (vx/c2)], or ?x =  [?x – (v/c)c ?t], c?t  =  [c ?t – (v ?x/c)]. Thus we have (c ?t )2 – (?x)2 =  2[(c ?t)2 – 2v ?x ?t + v2 (?x)2/c2] –  2[(?x)2 – 2v ?x ?t + v2 (c ?t)2/c2] =  2[(c ?t)2(1 – v2/c2) – (?x)2(1 – v2/c2)] =  2(1 – v2/c2)[(c ?t)2 – (?x)2] = (c ?t)2 – (?x)2. 80. Because the speed achieved is 0.1c, we can assume the mass converted is a small fraction of the mass of the Enterprise. We find the mass converted from K = mc2({1/[1 – (v/c)2]1/2} – 1) = ?m c2;

m({1/[1 – (0.1)2]1/2} – 1) = ?m, which gives

Chapter 37

p.14

?m = 0.005m = (0.005)(5109 kg) =

3107 kg.

81. (a) The velocity components of the light in the S frame are ux = 0, uy = c. We find the velocity components in the Earth frame from the velocity transformation: ux = (ux + v)/[1 + (uxv/c2)] = (0 + v)/(1 + 0) = v; uy = uy(1 – v2/c2)1/2/[1 + (uxv/c2)] = c[1 – v2/c2]1/2/(1 + 0) = c(1 – v2/c2)1/2. We find the angle the velocity makes with the x-axis from  = tan–1 [(c2/v2) – 1]1/2. tan  = uy/ux = c(1 – v2/c2)1/2/v = [(c2/v2) – 1]1/2, so (b) The magnitude of the velocity is u = (ux2 + uy2)1/2 = [v2 + c2(1 – v2/c2)]1/2 = c. (c) Classically we have ux = ux + v = 0 + v = v; uy = uy = c. The angle would be  = tan–1 (c/v). The magnitude would be u = (ux2 + uy2)1/2 = [v2 + c2]1/2.

Chapter 38 p. 1

CHAPTER 38 – Early Quantum Theory and Models of the Atom Note:

1.

At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.6310–34 J · s)(3.00108 m/s)(10–9 nm/m)/(1.6010–19 J/eV); E = (1.24103 eV · nm)/.

We find the temperature for a peak wavelength of 440 nm: T = (2.9010–3 m · K)/P = (2.9010–3 m · K)/(44010–9 m) =

6.59103 K.

2.

(a) The temperature for a peak wavelength of 25.0 nm is T = (2.9010–3 m · K)/P = (2.9010–3 m · K)/(25.010–9 m) = 1.16105 K. (b) We find the peak wavelength from P = (2.9010–3 m · K)/T = (2.9010–3 m · K)/(2800 K) = 1.0410–6 m = 1.04 m. Note that this is not in the visible range.

3.

Because the energy is quantized, E = nhf, the difference in energy between adjacent levels is 5.410–20 J = 0.34 eV. ?E = hf = (6.6310–34 J · s)(8.11013 Hz) =

4.

(a) We find the peak wavelength from P = (2.9010–3 m · K)/T = (2.9010–3 m · K)/(273 K) = 1.0610–5 m = 10.6 m. This wavelength is in the infrared. (b) We find the peak wavelength from P = (2.9010–3 m · K)/T = (2.9010–3 m · K)/(3300 K) = 8.7910–7 m = 879 nm. This wavelength is in the near infrared. (c) We find the peak wavelength from P = (2.9010–3 m · K)/T = (2.9010–3 m · K)/(4 K) = 7.2510–4 m = 0.73 mm. This wavelength is in the far infrared.

5.

(a) The potential energy on the first step is U1 = mgh = (58.0 kg)(9.80 m/s2)(0.200 m) = (b) The potential energy on the second step is U2 = mg2h = 2U1 = 2(114 J) = 228 J. (c) The potential energy on the third step is U3 = mg3h = 3U1 = 3(114 J) = 342 J. (d) The potential energy on the nth step is Un = mgnh = nU1 = n(114 J) = 114n J. (e) The change in energy is ?E = U2 – U6 = (2 – 6)(114 J) = – 456 J.

6.

114 J.

We use a body temperature of 98°F = 37°C = 310 K. We find the peak wavelength from P = (2.9010–3 m · K)/T = (2.9010–3 m · K)/(310 K) = 9.410–6 m = 9.4 m.

Chapter 38 p. 2

7.

(a) To find the wavelength when I(, T) is maximal at constant temperature, we set dI/d = 0: 2hc 2  –5 d I = d d  d  e hc/ kT – 1 –5 – hc/  2 kT e hc/ kT  –6 –  = 2 hc 2 hc/– 5kT 2 e –1 e hc/ kT – 1 2 5 – (hc/  kt) e hc/ kT – 5 = – 2hc = 0 , which gives 6 2  e hc/ kT – 1 5 – (hc/  P kT) ehc/ P kT – 5 = 0, or 5 – (hc/  PkT) = 5e –hc/  PkT . This equation will have a solution PT = constant, which is the Wien displacement law. (b) To find the value of the constant, we let x = hc/PkT, 5 so the transcendental equation is 5–x 4 5 – x = 5e–x. 3 One way to solve this equation is to plot each side against x. We see from the plot that the solution is very close to x = 5. 2 If we let ? = 5 – x, we get 5e–x 1 ? = 5e(? – 5) ˜ 5e – 5 = 0.034, so x = 4.966. 0 Thus we have 0 2 4 PT = hc/xk ; –3 8 –23 2.9010 m · K = h(3.0010 m/s)/(4.966)(1.3810 J/K), h = 6.6310–34 J · s. which gives (c) For the rate at which energy is radiated per unit surface area for all wavelengths we have   –5 2hc 2  d . I( , T) d  = hc/ kT 0 –1 0 e

x 6

If we change variable to x = hc/kT, then d = – (hc/kTx2) dx, so we have  0

I( , T) d  = 2 kT 3 h c2

4

 0

4

x 3 d x = 2k 3 –1 h c2

ex

 0

x 3 d x T 4 = (constant )T 4 . –1

ex

8.

The energy of the photons with wavelengths at the ends of the visible spectrum are E1 = hf1 = hc/1 = (1.24103 eV · nm)/(400 nm) = 3.10 eV; E2 = hf2 = hc/2 = (1.24103 eV · nm)/(750 nm) = 1.65 eV. Thus the range of energies is 1.65 eV < E < 3.10 eV.

9.

The energy of the photon is E = hf = (6.6310–34 J · s)(88.5106 Hz) = 5.8710–26 J =

3.6710–7 eV.

10. We find the wavelength from  = c/f = hc/E = (1.24103 eV · nm)/(300103 eV) = 4.110–3 nm. Significant diffraction occurs when the opening is on the order of the wavelength. Thus there would be insignificant diffraction through the doorway. 11. The photon energy must be at least 0.1 eV. We find the minimum frequency from Emin = hfmin ; (0.1 eV)(1.6010–19 J/eV) = (6.6310–34 J · s)fmin , which gives fmin = 2.41013 Hz. The maximum wavelength is max = c/fmin = (3.00108 m/s)/(2.41013 Hz) = 1.210–5 m. 12. At the threshold frequency, the kinetic energy of the photoelectrons is zero, so we have Kmax = hf – W0 = 0;

Chapter 38 p. 3

hfmin = W0 ; (6.6310–34 J · s)fmin = 4.310–19 J, which gives fmin =

6.51014 Hz.

13. At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have Kmax = hf – W0 = 0; hc/max = W0 , or max = hc/W0 = (1.24103 eV · nm)/(3.10 eV) = 400 nm. 14. The photon of visible light with the maximum energy has the least wavelength: hfmax = (1.24103 eV · nm)/min = (1.24103 eV · nm)/(400 nm) = 3.10 eV. Electrons will not be emitted if this energy is less than the work function. The metals with work functions greater than 3.10 eV are copper and iron. 15. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have Kmax = hf – W0 = 0; W0 = hc/max = (1.24103 eV · nm)/(570 nm) = 2.18 eV. (b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy: Kmax = eV0 = hf – W0 ; (1 e)V0 = [(1.24103 eV · nm)/(400 nm)] – 2.18 eV = 3.10 eV – 2.18 eV = 0.92 eV, so the stopping voltage is 0.92 V. 16. The photon of visible light with the maximum energy has the minimum wavelength: hfmax = (1.24103 eV · nm)/min = (1.24103 eV · nm)/(400 nm) = 3.10 eV. The maximum kinetic energy of the photoelectrons is Kmax = hf – W0 = 3.10 eV – 2.48 eV = 0.62 eV. 17. The energy of the photon is hf = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(255 nm) = 4.86 eV. We find the work function from Kmax = hf – W0 ; 1.40 eV = 4.86 eV – W0 , which gives W0 = 3.46 eV. 18. The threshold wavelength determines the work function: W0 = hc/max = (1.24103 eV · nm)/(350 nm) = 3.54 eV. (a) The energy of the photon is hf = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(280 nm) = 4.43 eV. The maximum kinetic energy of the photoelectrons is Kmax = hf – W0 = 4.43 eV – 3.54 eV = 0.89 eV. (b) Because the wavelength is greater than the threshold wavelength, the photon energy is less than the work function, so there will be no ejected electrons. 19. The energy required for the chemical reaction is provided by the photon: E = hf = hc/ = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(660 nm) = 1.88 eV. Each reaction takes place in a molecule, so we have E = (1.88 eV/molecule)(6.021023 molecules/mol)(1.6010–19 J/eV)/(4186 J/kcal) = 43.3 kcal/mol. 20. The reverse voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy:

Chapter 38 p. 4

Kmax = eV0 = hf – W0 ; (1 e)(1.64 V) = [(1.24103 eV · nm)/(230 nm)] – W0 , which gives W0 = 21. (a) h/mec = (6.6310–34 J · s)/(9.1110–31 kg)(3.00108 m/s) = (b) h/mpc = (6.6310–34 J · s)/(1.6710–27 kg)(3.00108 m/s) = (c) For the energy of the photon we have E = hf = hc/ = hc/(h/mc) = mc2.

3.75 eV.

2.4310–12 m. 1.3210–15 m.

22. We find the Compton wavelength shift for a photon scattered from an electron:  –  = (h/mec)(1 – cos ). From the previous problem, h/mec = 2.4310–12 m for an electron. (a) a –  = (2.4310–12 m)(1 – cos 45°) = 7.1210–13 m. 2.4310–12 m. (b) b –  = (2.4310–12 m)(1 – cos 90°) = –12 4.8610–12 m. (c) c –  = (2.4310 m)(1 – cos 180°) = 23. (a) The energy of a photon is E = hf = hc/. For the fractional loss, we have (E – E)/E = [(1/) – (1/)]/(1/) = ( – )/. For 45° we get 5.9010–3. (E – Ea)/E = (a – )/a = (7.1210–13 m)/(0.12010–9 m + 7.1210–13 m) = For 90° we get 1.9810–2. (E – Eb)/E = (b – )/b = (2.4310–12 m)/(0.12010–9 m + 2.4310–12 m) = For 180° we get 3.8910–2. (E – Ec)/E = (c – )/c = (4.8610–12 m)/(0.12010–9 m + 4.8610–12 m) = (b) The energy of the incident photon is E = hf = hc/ = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(0.120 nm) = 10.3103 eV. From conservation of energy, the energy given to the scattered electron is the energy lost by the photon: K = (E – E) = [(E – E)/E]E. For 45° we get Ka = [(E – Ea)/E]E = (5.9010–3)(10.3103 eV) = 60.8 eV. For 90° we get Kb = [(E – Eb)/E]E = (1.9810–2)(10.3103 eV) = 204 eV. For 180° we get Kc = [(E – Ec)/E]E = (3.8910–2)(10.3103 eV) = 401 eV. 24. (a) The wavelength shift for Compton scattering is  –  = (h/mc)(1 – cos ). The fractional energy loss of the scattered electron is – ?E/E = [(hc/) – (hc/)]/(hc/) = ( – )/ = ?/. If  ˜ , this is – ?E/E = ?/. (b) For the approximation to be accurate to 0.1%, we have [(?/) – (?/)]/(?/) = ( – )/ = 0.001. This can be written as (h/mc)(1 – cos ) = 0.001. Thus for a given incident wavelength, this puts a restriction on the scattering angle: cos  = 1 – 0.001mc/h.

Chapter 38 p. 5

25. For the conservation of momentum, we have Before After x: h/ + 0 = (h/) cos  + p cos , or y p cos  = (h/) – (h/) cos ; pe y: 0 + 0 = (h/) sin  – p sin , or p sin  = (h/) sin .  If we square and add, to eliminate , we get – – x p2 = (h/)2 + (h/)2 – (2h2/) cos .  h/  For energy conservation, we have h/   (hc/) + mc2 = (hc/) + [p2c2 + (mc2)2]1/2, or (hc/) + mc2 – (hc/) = [p2c2 + (mc2)2]1/2. When we substitute the result from momentum conservation and square, we get (hc/)2 + (mc2)2 + (hc/)2 + 2(h2c2/) – 2(mc2hc/) – 2(mc2hc/) = (hc/)2 + (hc/)2 – (2h2c2/) cos  + (mc2)2, which reduces to 2mc2hc[(1/) – (1/)] = (2h2c2/)(1 – cos ), or  –  = (h/mc)(1 – cos ).

After Before 26. (a) For the conservation of momentum for the hc/   K + mc 2 hc/  mc 2 energy head-on collision, we have h/ + 0 = – (h/) + p, or h/ = p – (h/). p h/  momentum – h/  0 For energy conservation, we have – – = (hc/) + K + mc2. (hc/) + mc2 When we use the result for momentum conservation, we get K = 2(hc/) – pc. The momentum of the electron is related to its kinetic energy: pc = (K2 + 2mc2K)1/2. Thus we have 2(hc/) – K = pc = (K2 + 2mc2K)1/2. After squaring and reducing, we get [mc2 + 2(hc/)]K = 2(hc/)2; [(0.511106 eV) + 2(1.24103 eV · nm)/(0.100 nm)]K = 2[(1.24103 eV · nm)/(0.100 nm)]2, which gives K = 574 eV. (b) We find the wavelength of the recoiling photon from hc/ = (hc/) – K; (1.24103 eV · nm)/ = (1.24103 eV · nm)/(0.100 nm) – 584 eV, 0.105 nm. which gives  = 27. The kinetic energy of the pair is K = hf – 2mc2 = 2.84 MeV – 2(0.511 MeV) =

1.82 MeV.

28. The photon with the longest wavelength has the minimum energy to create the masses: hfmin = hc/max = 2mc2; (6.6310–34 J · s)/max = 2(1.6710–27 kg)(3.00108 m/s)2, which gives max = 6.6210–16 m. 29. The photon with minimum energy to create the masses is hfmin = 2mc2 = 2(207)(0.511 MeV) = 212 MeV.

Chapter 38 p. 6

The wavelength is  = (1.24103 eV · nm)/(212106 eV) = 5.8510–6 nm =

5.8510–15 m.

30. The energy of the photon is hf = 2(K + mc2) = 2(0.345 MeV + 0.511 MeV) = 1.71 MeV. The wavelength is  = (1.24103 eV · nm)/(1.71106 eV) = 7.2410–4 nm = 7.2410–13 m. 31. We find the wavelength from  = h/p = h/mv = (6.6310–34 J · s)/(0.21 kg)(0.10 m/s) =

3.210–32 m.

32. We find the wavelength from  = h/p = h/mv = (6.6310–34 J · s)/(1.6710–27 kg)(5.5104 m/s) = Note that v « c.

7.210–12 m.

33. We find the speed from  = h/p = h/mv; 0.2810–9 m = (6.6310–34 J · s)/(9.1110–31 kg)v, which gives v = 2.60106 m/s. Because this is much less than c, we can use the classical expression for the kinetic energy. The kinetic energy is equal to the potential energy change: eV = K = !mv2 = !(9.1110–31 kg)(2.60106 m/s)2 = 3.0810–18 J = 19.2 eV. Thus the required potential difference is 19 V. 34. The kinetic energy is equal to the potential energy change: K = eV = (1 e)(30.0103 V) = 30.0103 eV = 0.0300 MeV. Because this is 6% of mc2, the electron is relativistic. We find the momentum from E2 = [K + mc2]2 = p2c2 + m2c4, or p2c2 = K2 + 2Kmc2; p2c2 = (0.0300 MeV)2 + 2(0.0300 MeV)(0.511 MeV), which gives pc = 0.178 MeV, or p = (0.178 MeV)(1.6010–13 J/MeV)/(3.00108 m/s) = 9.4710–23 kg · m/s. The wavelength is  = h/p = (6.6310–34 J · s)/(9.4710–23 kg · m/s) = 7.010–12 m. diffraction effects are negligible. Because  « 5 cm, 35. Because all the energies are much less than mc2, we can use K = p2/2m, so  = h/p = h/(2mK)1/2 = hc/(2mc2K)1/2. 0.39 nm. (a)  = hc/(2mc2K)1/2 = (1.24103 eV · nm)/[2(0.511106 eV)(10 eV)]1/2 = 2 1/2 3 6 1/2 0.12 nm. (b)  = hc/(2mc K) = (1.2410 eV · nm)/[2(0.51110 eV)(100 eV)] = 0.039 nm. (c)  = hc/(2mc2K)1/2 = (1.24103 eV · nm)/[2(0.511106 eV)(1.0103 eV)]1/2 = 36. With K = p2/2m , we have  = h/p = h/(2mK)1/2. If we form the ratio for the two particles with equal kinetic energies, we get p/e =(me/mp)1/2. Because mp > me , p < e . 37. With K = p2/2m, we have  = h/p = h/(2mK)1/2.

Chapter 38 p. 7

If we form the ratio for the two particles with equal wavelengths, we get 1 = (meKe/mpKp)1/2, or Ke/Kp = mp/me = (1.6710–27 kg)/(9.1110–31 kg) =

1.84103.

38. We find the speed from !mv2 = *kT; !(32 u)(1.6610–27 kg/u)v2 = *(1.3810–23 J/K)(300 K), which gives v = 484 m/s. The wavelength is  = h/p = h/mv = (6.6310–34 J · s)/(32 u)(1.6610–27 kg/u)(484 m/s) = 2.610–11 m.

39. For diffraction, the wavelength must be of the order of the opening. We find the speed from  = h/p = h/mv; 3.310–38 m/s. 10 m = (6.6310–34 J · s)/(2000 kg)v, which gives v = Not a good speed if you want to get somewhere. no diffraction. At a speed of 30 m/s,  « 10 m, so there will be 40. The kinetic energy of the electron is K = p2/2m = h2/2m2 = (6.6310–34 J · s)2/2(9.1110–31 kg)(0.1010–9 m)2 = 2.4110–17 J = 150 eV. Because this must equal the potential energy change, the required voltage is 150 V. 41. The wavelength of the electron is  = h/p = h/(2mK)1/2 = hc/(2mc2K)1/2 = (1.24103 eV · nm)/[2(0.511106 eV)(2250 eV)]1/2 = 2.5910–2 nm. If we neglect aberrations, the maximum possible resolution is on the order of the wavelength: 0.026 nm. 42. The energy of a level is En = – (13.6 eV)/n2. absorption, because the final state, (a) The transition from n = 1 to n = 3 is an higher energy. The energy of the photon is hf = En – En = – (13.6 eV)[(1/32) – (1/12)] = 12.1 eV. emission, because the initial state, (b) The transition from n = 6 to n = 2 is an higher energy. The energy of the photon is hf = – (En – En) = (13.6 eV)[(1/22) – (1/62)] = 3.0 eV. absorption, because the final state, (c) The transition from n = 4 to n = 5 is an higher energy. The energy of the photon is hf = En – En = – (13.6 eV)[(1/52) – (1/42)] = 0.31 eV. n = 1 to n = 3 has the largest energy. The photon for the transition from

n = 3, has a

n = 6, has a

n = 5, has a

43. To ionize the atom means removing the electron, or raising it to zero energy: Eion = 0 – En = (13.6 eV)/n2 = (13.6 eV)/22 = 3.4 eV. 44. From ?E = hc/, we see that the third longest wavelength comes from the transition with the third n = 6 to n = 3. smallest energy: 45. Doubly ionized lithium is like hydrogen, except that there are three positive charges (Z = 3) in the nucleus. The square of the product of the positive and negative charges appears in the energy term for

Chapter 38 p. 8

the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2: En = – Z2(13.6 eV)/n2 = – 32(13.6 eV)/n2 = – (122 eV)/n2. We find the energy needed to remove the remaining electron from E = 0 – E1 = 0 – [– (122 eV)/(1)2] = 122 eV.

46. (a) For the jump from n = 4 to n = 2, we have  = (1.24103 eV · nm)/(E4 – E2) = (1.24103 eV · nm)/[– 0.85 eV – (– 3.4 eV)] = (b) For the jump from n = 3 to n = 1, we have  = (1.24103 eV · nm)/(E3 – E1) = (1.24103 eV · nm)/[– 1.5 eV – (– 13.6 eV)] = (c) The energy of the n = 5 level is E5 = – (13.6 eV)/52 = – 0.54 eV. For the jump from n = 5 to n = 2, we have  = (1.24103 eV · nm)/(E5 – E2) = (1.24103 eV · nm)/[– 0.54 eV – (– 3.4 eV)] =

486 nm. 102 nm.

434 nm.

47. For the Rydberg constant we have R = e4m/8Å02h3c = (1.60217710–19 C)4(9.10939010–31 kg)/ 8(8.85418710–12 C2/N · m2)2(6.62607610–34 J · s)3(2.997925108 m/s) = 1.0974107 m–1. 48. The longest wavelength corresponds to the minimum energy, which is the ionization energy:  = (1.24103 eV · nm)/Eion = (1.24103 eV · nm)/(13.6 eV) = 91.2 nm. Note that shorter wavelengths would give the ejected electron some kinetic energy. 49. The energy of the photon is hf = Eion + K = 13.6 eV + 10.0 eV = 23.6 eV. We find the wavelength from  = (1.24103 eV · nm)/hf = (1.24103 eV · nm)/(23.6 eV) =

52.5 nm.

50. Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. The square of the product of the positive and negative charges appears in the energy term for the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2: En = – Z2(13.6 eV)/n2 = – 22(13.6 eV)/n2 = – (54.4 eV)/n2. We find the energy of the photon from the n = 6 to n = 2 transition: E = E6 – E2 = – (54.4 eV)[(1/62) – (1/22)] = 12.1 eV. Because this is the energy difference for the n = 1 to n = 3 transition in hydrogen, the photon can be absorbed by a hydrogen atom which will jump from n = 1 to n = 3.

Chapter 38 p. 9

0 – 3.4 – 6.0

Continuum

– 13.6

n= n=5 n=4 n=3 n=2

Energy (eV)

51. Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. The square of the product of the positive and negative charges appears in the energy term for the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2: En = – Z2(13.6 eV)/n2 = – 22(13.6 eV)/n2 = – (54.4 eV)/n2.

n=1

– 54.4

0 – 7.65 – 13.6

Continuum

– 30.6

n= n=5 n=4 n=3 n=2

Energy (eV)

52. Doubly ionized lithium is like hydrogen, except that there are three positive charges (Z = 3) in the nucleus. The square of the product of the positive and negative charges appears in the energy term for the energy levels. We can use the results for hydrogen, if we replace e2 by Ze2: En = – Z2(13.6 eV)/n2 = – 32(13.6 eV)/n2 = – (122.4 eV)/n2.

– 122.4

53. The potential energy for the ground state is U = – e2/4pÅ0r1 = – (9.00109 N · m2/C2)(1.6010–19 C)2/(0.52910–10 m) = – 4.3610–18 J = The kinetic energy is K = E1 – U = – 13.6 eV – (– 27.2 eV) = +13.6 eV.

n=1

– 27.2 eV.

Chapter 38 p. 10

54. We find the value of n from rn = n2r1 ; 1.0010–3 m = n2(0.52910–10 m), which gives n = 4.35103. The energy of this orbit is E = – (13.6 eV)/n2 = – (13.6 eV)/(4.35103)2 = – 7.210–7 eV. Note that energy differences will be very small. 55. We find the velocity from the quantum condition: mvr1 = nh/2p; (9.1110–31 kg)v(0.52910–10 m) = (1)(6.6310–34 J · s)/2p, which gives v = 2.18106 m/s = 7.310–3c. The relativistic factor is [1 – (v/c)2]1/2 ˜ 1 – !(v/c)2 = 1 – 2.710–5. Because this is essentially 1, the use of nonrelativistic formulas is

justified.

56. If we compare the two forces: Fe = e2/4pÅ0r2, and Fg = Gmemp/r2, we see that we can use the hydrogen expressions if we replace e2/4pÅ0 with Gmemp . For the radius we get r1 = h2/4p2Gme2mp = (6.6310–34 J · s)2/4p2(6.6710–11 N · m2/kg2)(9.1110–31 kg)2(1.6710–27 kg) = 1.201029 m. Note that this is many times intergalactic distances. The ground state energy is E1 = – 2p2G2me3mp2/h2 = – 2p2(6.6710–11 N · m2/kg2)2(9.1110–31 kg)3(1.6710–27 kg)2/(6.6310–34 J · s)2 = – 4.2110–97 J. 57. The potential energy for the nth state is U = – e2/4pÅ0rn . The Coulomb force provides the radial acceleration, so we have e2/4pÅ0rn2 = mvn2/rn , or mvn2 = e2/4pÅ0rn . Thus the kinetic energy is K = !mvn2 = !e2/4pÅ0rn = ! U . 58. For the difference in radius for adjacent orbits, we have ?r = rn – rn – 1 = [n2 – (n – 1)2]r1 = (2n – 1)r1 .

When n » 1, we get ?r ˜ 2nr1 = 2rn/n. In the classical limit, the separation of radii (and energies) should be very small. We see that letting n  8 does this. If we substitute the expression for rn , we see that ?r  h2, so letting h  0 also makes the separation of radii small.

Chapter 38 p. 11

59. (a) We find the frequency of the radiation for a jump from level n to level n – 1 from hf = En – En – 1; f

= [(Z2e4m/8Å02h2)/h]{[1/(n – 1)2] – (1/n2)}

= (Z2e4m/8Å02h3){[n2 – (n – 1)2]/n2(n – 1)2} ˜ (Z2e4m/8Å02h3)(2n/n4) = Z2e4m/4Å02h3n3. From the quantum condition we have mvrn = nh/2p, or v = nh/2pmrn . Thus we get v/2prn = nh/4p2mrn2 = nh/4p2m(n2h2Å0/pmZe2)2 = Z2e4m/4Å02h3n3, which is the same as the above frequency. (b) From the classical theory for an electron revolving in a circular orbit, the time for one revolution is T = 2prn/v, so the frequency is f = 1/T = v/2prn . (c) Classically an accelerated charge radiates. For circular motion, the frequency of the radiation is the orbital frequency. This agrees with the Bohr prediction for large values of n, consistent with the correspondence principle. 60. We find the peak wavelength from P = (2.9010–3 m · K)/T = (2.9010–3 m · K)/(2.7 K) = 1.110–3 m =

1.1 mm.

61. To produce a photoelectron, the hydrogen atom must be ionized, so the minimum energy of the photon is 13.6 eV. We find the minimum frequency of the photon from Emin = hfmin ; (13.6 eV)(1.6010–19 J/eV) = (6.6310–34 J · s)fmin , which gives fmin = 3.281015 Hz. 62. Because the energy is much less than mc2, we can use K = p2/2m, so the wavelength of the electron is  = h/p = h/(2mK)1/2 = hc/(2mc2K)1/2 = (1.24103 eV · nm)/[2(0.511106 eV)(85 eV)]1/2 = 0.133 nm. We find the spacing of the planes from 2d sin  = n; 2d sin 38° = (1)(0.133 nm), which gives d = 0.108 nm. 63. For the energy of the photon, we have E = hf = hc/ = (6.6310–34 J · s)(3.00108 m/s)/(1.6010–19 J/eV) = (1.2410–6 eV · m)/. 64. The energy of the photon is E = hf = hc/ = (6.6310–34 J · s)(3.00108 m/s)/(12.210–2 m) = 1.6310–24 J. Thus the rate at which photons are produced in the oven is N = P/E = (760 W)/(1.6310–24 J) = 4.661026 photons/s. 65. The energy of the photon is hf = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(550 nm) = 2.25 eV. We find the intensity of photons from Iphotons = I/hf = (1000 W/m2)/(2.25 eV)(1.6010–19 J/eV) = 2.781021 photons/s · m2.

Chapter 38 p. 12

66. The impulse on the wall is due to the change in momentum of the photons: F ?t = ?p = np = nh/, or n/?t = F/h = (5.510–9 N)(63310–9 m)/(6.6310–34 J · s) = 5.31018 photons/s. 67. The energy of the photon is hf = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(550 nm) = 2.25 eV. Because the light radiates uniformly, the intensity at a distance L is I = P/4pL2, so the rate at which energy enters the pupil is E/t = Ipr2 = Pr2/4L2. Thus the rate at which photons enter the pupil is n/t = (E/t)/hf = Pr2/4L2hf = (0.030)(100 W)(2.010–3 m)2/4(1.0103 m)2(2.25 eV)(1.6010–19 J/eV) = 8.3106 photons/s. 68. Because the energies of the photons are equal, their momenta have equal magnitudes. Thus the total momentum of the system before and after the collision is zero. This means the momentum of the electron has the same magnitude as the momentum of the positron. Because they have the same mass, their energies must be equal. From conservation of energy we have 2hf = 2(K + mc2), or hf = K + mc2; 0.90 MeV = K + 0.511 MeV, which gives K = 0.39 MeV. 69. We find the scattering angle from  –  = (h/mc)(1 – cos ); (7.3510–11 m) – (7.1110–11 m) = [(6.6310–34 J · s)/(9.1110–31 kg)(3.00108 m/s)](1 – cos ), which gives  = 89.4°. cos  = 0.0107, 70. The required momentum is p = h/, or pc = hc/ = (1.24103 eV · nm)/(5.010–3 nm) = 2.48105 eV = 0.248 MeV. (a) For the proton, pc « mc2, so we can find the required kinetic energy from K = p2/2m = (pc)2/2mc2 = (0.248 MeV)2/2(938 MeV) = 3.310–5 MeV = 33 eV. The potential difference to produce this kinetic energy is V = K/e = (33 eV)/(1 e) = 33 V. (b) For the electron, pc is of the order of mc2, so we can find the required kinetic energy from K = [(pc)2 + (mc2)2]1/2 – mc2 = [(0.248 MeV)2 + (0.511 MeV)2]1/2 – 0.511 MeV = 0.057 MeV = 57 keV. The potential difference to produce this kinetic energy is V = K/e = (57 keV)/(1 e) = 57 kV. 71. If we ignore the recoil motion of the gold mucleus, at the closest approach the kinetic energy of both particles is zero. The potential energy of the two charges must equal the initial kinetic energy of the  particle: K = ZZAue2/4pÅ0rmin ; (4.8 MeV)(1.6010–13 J/MeV) = (2)(79)(1.6010–19 C)2/4p(8.8510–12 C2/N · m2)rmin , which gives rmin = 4.710–14 m. 72. The decrease in mass occurs because of the loss in energy when a photon is emitted: ?m/m = (hf/c2)/m = ?E/mc2 = (– 13.6 eV)[(1/12) – (1/32)]/(939106 eV) = – 1.2910–8. 73. The collision must be elastic as long as the electron does not have enough energy to raise the hydrogen

Chapter 38 p. 13

atom to the n = 2 level, so no energy is transferred to the atom. Thus we have K < E2 – E1 = (– 13.6 eV)[(1/22) – (1/12)] = 10.2 eV.

74. The Coulomb force provides the radial acceleration, so we have e2/4pÅ0rn2 = mvn2/rn , or vn = (e2/4pÅ0mrn)1/2. For the angular velocity, we get n = vn/rn = (e2/4pÅ0mrn3)1/2 = (p2m2e8/4Å04n6h6)1/2 = pme4/2Å02n3h3. where we have used the expression rn = n2h2Å0/pme2 for the radius of the orbit. For the frequency, we get fn = n/2p = me4/4Å02n3h3. (a) For the ground state, we get 1 = p(9.1110–31 kg)(1.6010–19 C)4/2(8.8510–12 C2/N · m2)2(1)3(6.6310–34 J · s)3 = 4.111016 rad/s. The frequency is f1 = 1/2p = (4.111016 rad/s)/2p = 6.541015 Hz. (b) For the first excited state, we get 2 = p(9.1110–31 kg)(1.6010–19 C)4/2(8.8510–12 C2/N · m2)2(2)3(6.6310–34 J · s)3 = 5.141015 rad/s. The frequency is f2 = 2/2p = (5.141015 rad/s)/2p = 8.171014 Hz. 75. The ratio of the forces is Fg/Fe = (Gmemp/r2)/(e2/4pÅ0r2) = 4pÅ0Gmemp/e2 = 4p(8.8510–12 C2/N · m2)(6.6710–11 N · m2/kg2)(9.1110–31 kg)(1.6710–27 kg)/ 4.410–40. (1.6010–19 C)2 = Yes, the gravitational force may be safely ignored. 76. (a) Because the energy is quantized, E = nhf, the difference in energy between adjacent levels is 5.010–34 J. ?E = hf = (6.6310–34 J · s)(0.75 Hz) = (b) The total energy will be the maximum potential energy, so we have E = mgh = nhf; 1.81035. (20 kg)(9.80 m/s2)(0.45 m) = n(5.010–34 J), which gives n = (c) The fractional change in energy is 5.610–36, not measurable. ?E/E = hf/nhf = 1/n = 1/(1.81035) = 77. The potential difference produces a kinetic energy of 12.3 eV, so it is possible to provide this much energy to the hydrogen atom through collisions. From the ground state, the maximum energy of the atom is – 13.6 eV – (– 12.3 eV)= – 1.3 eV. From the energy level diagram, we see that this means the atom could be excited to the n = 3 state, so the possible transitions when the atom returns to the ground state are n = 3 to n = 2, n = 3 to n = 1, and n = 2 to n = 1. For the wavelengths we have 3  2 = (1.24103 eV · nm)/(E3 – E2) = (1.24103 eV · nm)/[– 1.5 eV – (– 3.4 eV)] = 653 nm; 3 3 3  1 = (1.2410 eV · nm)/(E3 – E1) = (1.2410 eV · nm)/[– 1.5 eV – (– 13.6 eV)] = 102 nm; 2  1 = (1.24103 eV · nm)/(E2 – E1) = (1.24103 eV · nm)/[– 3.4 eV – (– 13.6 eV)] = 122 nm. 78. The energy levels are En = – Z2e4m/8Å02n2h2 = – (82)2e4(207)me/8Å02n2h2 = (82)2(207)(– 13.6 eV)/n2 = – (1.89107 eV)/n2 = – (18.9 MeV)/n2. For the n = 2 to n = 1 transition, the photon energy is

Chapter 38 p. 14

hf = E2 – E1 = (– 18.9 MeV)[(1/22) – (1/12)] = Note that this would be a gamma-ray photon.

14.2 MeV.

79. The energy of the ultraviolet photon is hf1 = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(290 nm) = 4.28 eV. The stopping potential is the potential difference that gives a potential energy change equal to the maximum kinetic energy: Kmax1 = eV01 = hf1 – W0 ; (1 e)(2.10 V) = 4.28 eV– W0 , which gives W0 = 2.18 eV. The energy of the blue photon is hf1 = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(440 nm) = 2.82 eV. We find the new stopping potential from Kmax2 = eV02 = hf2 – W0 ; (1 e)V02 =2.82 eV– 2.18 eV, which gives V02 = 0.64 V. 80. (a) The electron has a charge e, so the potential difference produces a kinetic energy of eV. The shortest wavelength photon is produced when all the kinetic energy is lost and a photon emitted: hfmax = hc/0 = eV, which gives 0 = hc/eV. (b) 0 = hc/eV = (1.24103 eV · nm)/(30103 eV) = 0.041 nm. 81. We find the momentum from E2 = (K + mc2)2 = p2c2 + m2c4, or p2c2 = K2 + 2mc2K. The wavelength is  = h/p = hc/pc = hc/(K2 + 2mc2K)1/2. 82. The kinetic energy of a thermal neutron is K = *kT = *(1.3810–23 J/K)(300 K) = 6.2110–21 J = 0.039 eV. We find the speed from K = !mv2; 6.2110–21 J = !(1.6710–27 kg)v2, which gives v = 2.73103 m/s. The wavelength is  = h/p = h/mv = (6.6310–34 J · s)/(1.6710–27 kg)(2.73103 m/s) = 1.510–10 m =

0.15 nm.

83. We find the momentum from E2 = (K + mc2)2 = p2c2 + m2c4, or p2c2 = K2 + 2mc2K. The wavelength is  = h/p = hc/pc = hc/(K2 + 2mc2K)1/2 = (1.24103 eV · nm)/[(60103 eV)2 + 2(0.511106 eV)(60103 eV)]1/2 = 4.910–3 nm. The theoretical resolution limit is of the order of the wavelength, or 510–12 m. 84. The energy of a photon in terms of the momentum is E = hf = hc/ = pc. If the sail is perpendicular to the sunlight, the rate at which photons are striking the sail is N/?t = IA/E = IA/pc. Because the photons reflect from the sail, the change in momentum of a photon is

Chapter 38 p. 15

?p = 2p. The impulse on the sail is due to the change in momentum of the photons: F ?t = N ?p , or F = (N/?t) ?p = (IA/pc)(2p) = 2IA/c = 2(1000 W/m2)(1103 m)2/(3.00108 m/s) = 85. We find the number of photons from N = E/hf = E/(hc/) = (10–18 J)/[(1.24103 eV · nm)/(550 nm)](1.6010–19 J/eV) ˜

7 N. 3.

86. The energy of the photon is hf = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(300 nm) = 4.13 eV. The maximum kinetic energy of the photoelectrons is Kmax = hf – W0 = 4.13 eV – 2.2 eV = 1.93 eV. Because this is much less than the rest energy of the electron, Kmax = pmax2/2m. Thus the shortest wavelength of the electron is min = h/pmax = h/(2mKmax)1/2 = (6.6310–34 J · s)/[2(9.1110–31 kg)(1.93 eV)(1.6010–19 J/eV)]1/2 = 8.810–10 m = 0.88 nm.

Chapter 39 p. 1

CHAPTER 39 – Quantum Mechanics Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.6310–34 J · s)(3108 m/s)(109 nm/m)/(1.6010–19 J/eV); E = (1.24103 eV · nm)/. 1.

We find the wavelength of the neutron from  = h/p = h/(2mK)1/2 = (6.6310–34 J · s)/[2(1.6710–27 kg)(0.025 eV)(1.610–19 J/eV)]1/2 = 1.8110–10 m. The peaks of the interference pattern are given by d sin  = m, m = 1, 2, … . and the positions on the screen are y = L tan . For small angles, sin ˜ tan , so we have y = mL/d. Thus the separation is 1.810–7 m. ?y = L/d = (1.0 m)(1.8110–10 m)/(1.010–3 m) =

2.

We find the wavelength of the bullet from  = h/p = h/mv = (6.6310–34 J · s)/(2.010–3 kg)(120 m/s) = 2.810–33 m. The half-angle for the central circle of the diffraction pattern is given by sin  = 1.22/D. For small angles, sin ˜ tan , so we have r = L tan  ˜ L sin  = 1.22L/D; 4.51027 m. 0.5010–2 m = 1.22L(2.810–33 m)/(3.010–3 m), which gives L = Diffraction effects are negligible for macroscopic objects.

3.

We find the uncertainty in the momentum: ?p = m ?v = (1.6710–27 kg)(0.024105 m/s) = 4.0010–24 kg · m/s. We find the uncertainty in the proton’s position from ?x = ˙/?p = (1.05510–34 J · s)/(4.0010–24 kg · m/s) = 2.610–11 m. Thus the accuracy of the position is ± 1.310–11 m.

4.

We find the minimum uncertainty in the energy of the state from ?E = ˙/?t = (1.05510–34 J · s)/(10–8 s) = 1.110–26 J = 710–8 eV.

5.

We find the uncertainty in the momentum: ?p = ˙/?x = (1.05510–34 J · s)/(1.610–8 m) = 6.5910–27 kg · m/s. We find the uncertainty in the velocity from ?p = m ?v; 7.2103 m/s. 6.5910–27 kg · m/s = (9.1110–31 kg) ?v, which gives ?v =

6.

(a) We find the wavelength of the bullet from  = h/p = h/mv 3.710–34 m. = (6.6310–34 J · s)/(1210–3 kg)(150 m/s) = (b) We find the uncertainty in the momentum component perpendicular to the motion: 1.910–32 kg · m/s. ?py = ˙/?y = (1.05510–34 J · s)/(0.5510–2 m) = (c) We find the possible uncertainty in the y-position at the target from y/L = ?vy/vx = ?py/px ; y/(300 m) = (1.910–32 kg · m/s)/(1210–3 kg)(150 m/s), which gives y = 3.210–30 m.

Chapter 39 p. 2

7.

The uncertainty in the velocity is ?v = (0.065/100)(75 m/s) = 0.0488 m/s. For the electron, we have ?x = ˙/m ?v = (1.05510–34 J · s)/(9.1110–31 kg)(0.0488 m/s) = 2.410–3 m. For the baseball, we have ?x = ˙/m ?v = (1.05510–34 J · s)/(0.150 kg)(0.0488 m/s) = 1.410–32 m. The uncertainty for the electron is greater by a factor of 1.71029.

8.

We use the radius as the uncertainty in position for the electron. We find the uncertainty in the momentum from ?p = ˙/?x = (1.05510–34 J · s)/(10–15 m) = 1.05510–19 kg · m/s. If we assume that the lowest value for the momentum is the least uncertainty, we estimate the lowest possible energy as E = K + mc2 = (p2c2 + m2c4)1/2 = [(?p)2c2 + m2c4]1/2 = [(1.05510–19 kg · m/s)2(3.00108 m/s)2 + (9.1110–31 kg)2(3.00108 m/s)4]1/2 = 3.1710–11 J ˜ 200 MeV.

9.

(a) We find the minimum uncertainty in the energy of the state from 6.610–8 eV. ?E = ˙/?t = (1.05510–34 J · s)/(10–8 s) = 1.110–26 J = Note that, because the ground state is stable, we associate the uncertainty with the excited state. (b) The transition energy is E = – (13.6 eV)[(1/22) – (1/12)] = 10.2 eV, so we have ?E/E = (6.610–8 eV)/(10.2 eV) = 6.510–9. (c) The wavelength of the line is  = (1.24103 eV · nm)/E = (1.24103 eV · nm)/(10.2 eV) = 122 nm. If we treat the width of the line as a differential, we have ? = – (1.24103 eV · nm) ?E/E 2. If we ignore the sign, we get 7.910–7 nm. ? =  (?E/E) = (122 nm)(6.510–9) =

10. The momentum of the electron is p = (2mK)1/2 = [2(9.1110–31 kg)(2.50 keV)(1.6010–16 J/keV)]1/2 = 2.7010–23 kg · m/s. Because the changes are small, we can treat them as differentials. Thus the change in momentum is ?p = !(2m/K)1/2 ?K, or ?p/p = ! ?K/K. We find the uncertainty in the electron’s position from ?x = ˙/?p = 2˙/p(?K/K) = 2(1.05510–34 J · s)/(2.7010–23 kg · m/s)(0.0100) = 7.8210–10 m. 11. The electron has an initial momentum px and a wavelength  = h/px . For the maxima of the double-slit interference we have d sin  = m, m = 0, 1, 2, … . If the angles are small, the separation of maxima is ?max = /d, so the angle between a maximum and a minimum is ? = /2d. The separation on the screen will be H = L ? = L/2d. The uncertainty in the y-position at the slits of d/2 produces an uncertainty in the y-momentum of ?py = h/2p ?y = h/pd. This produces an uncertainty of the y-position at the screen of ?H = (?py/px)L = (h/pd)L/(h/) = L/pd. This is on the order of the separation of maxima and minima, so the pattern is destroyed.

Chapter 39 p. 3

12. The Schrödinger equation for a free particle is – (˙2/2m) ?2/?x2 + U0 = i˙ ?/?t. (a) For the proposed solution  = A e i(kx – t) we have ?/?t = – iA e i(kx – t) = – i; ?2/?x2 = (ik)2A e i(kx – t) = – k2. If we substitute these in the equation, we get – (˙2/2m)(– k2) + U0 = – i2˙ = ˙. Because both sides have the same functional dependence, the solution is valid. For the proposed solution  = A cos (kx – t) we have ?/?t = + A sin (kx – t); ?2/?x2 = – Ak2 cos (kx – t) . If we substitute these in the equation, we get – (˙2/2m)[– Ak2 cos (kx – t)] + U0A cos (kx – t) = i˙A sin (kx – t); [(˙2k2/2m) + U0]A cos (kx – t) = i˙A sin (kx – t). Because cosine and sine functions are different functions of x and t, the solution is not valid. For the proposed solution  = A sin (kx – t) we have ?/?t = – A cos (kx – t); ?2/?x2 = – Ak2 sin (kx – t) . If we substitute these in the equation, we get – (˙2/2m)[– Ak2 sin (kx – t)] + U0A sin (kx – t) = – i˙A cos (kx – t); [(˙2k2/2m) + U0]A sin (kx – t) = – i˙A cos (kx – t). Because cosine and sine functions are different functions of x and t, the solution is not valid. (b) For the valid solution we have – (˙2/2m)(– k2) + U0 = – i2˙ = ˙, or ˙ = (˙2k2/2m) + U0 . 13. The wave function for a free particle is  = A sin kx + B cos kx. (a) For the free electron we have k = p/˙ = mv/˙ = (9.1110–31 kg)(4.0105 m/s)/(1.05510–34 J · s) = 3.5109 m–1. Thus we have  = A sin (3.5109 m–1)x + B cos (3.5109 m–1)x . (b) For the free proton we have k = p/˙ = mv/˙ = (1.6710–27 kg)(4.0105 m/s)/(1.05510–34 J · s) = 6.31012 m–1. Thus we have  = A sin (6.31012 m–1)x + B cos (6.31012 m–1)x . 14. The wave function is  = A sin (1.01010 m–1)x = A sin kx. (a) We find the wavelength from  = 2p/k = 2p/(1.01010 m–1) = 6.310–10 m = 0.63 nm. (b) The momentum is p = ˙k = (1.05510–34 J · s)(1.01010 m–1) = 1.110–24 kg · m/s. (c) The speed is v = p/m = (1.110–24 kg · m/s)/(9.1110–31 kg) = 1.2106 m/s. (d) The kinetic energy is K = p2/2m = (1.110–24 kg · m/s)2/2(9.1110–31 kg) = 6.110–19 J =

3.8 eV.

Chapter 39 p. 4

15. We form the wave packet from  = 1 + 2 = A sin k1x + A sin k2x, where k1 = 2p/1 , and k2 = 2p/2 . If we use a trigonometric identity we get  = A(sin k1x + sin k2x) = 2A sin [!(k1 + k2)x ] cos [!(k1 – k2)x ]. If the wavelengths are almost equal, we have ?k = k1 – k2 , and k ˜ !(k1 + k2); thus

 = 2A cos [!(?k)x ] sin kx . The width of the packet corresponds to the distance from one zero to the next zero of the cosine function: !(?k) ?x = p, which gives ?x = 2p/?k. The momentum is p = ˙k, so ?k = ?p/˙. When we use this in the expression for the width, we have ?x = 2p/(?p/˙), or ?x ?p = h. 16. The energy levels for the particle in a rigid box are En = n2h2/8mL2 = pn2/2m. Thus the momentum is pn = nh/2L. The wavefunctions for the rigid box are n = A sin knx, with kn = np/L. The wavelength is n = 2p/kn = 2L/n = h/pn , which is the deBroglie wavelength. 17. The minimum speed corresponds to the minimum momentum or the maximum wavelength. This is the lowest energy state, so max = 2L. Thus we have pmin = h/max; mvmin = h/2L; (9.1110–31 kg)vmin = (6.6310–34 J · s)/2(0.1010–9 m), which gives vmin = 3.6106 m/s. 18. The energy levels for the electron in a rigid box are En = n2h2/8mL2. For a transition from n3 = 3 to n1 = 1, we have hf = hc/ = ?E = (n32 – n12)h2/8mL2; [(1.24103 eV · nm)/(240 nm)](1.6010–19 J/eV) = (32 – 12)(6.6310–34 J · s)2/8(9.1110–31 kg)L2, 0.76 nm. which gives L = 7.610–10 m = 19. (a) The energy levels for the electron in an infinite square well are En = n2h2/8mL2 = n2E1. The longest wavelength photon has the least energy, so the transition must be from n2 = 2 to n1 = 1. Thus we have hf = hc/ = ?E = (n22 – n12)E1; (1.24103 eV · nm)/ = (22 – 12)(8.0 eV), which gives  = 52 nm. (b) We find the width of the well from the ground state energy: E1 = h2/8mL2; (8.0 eV)(1.6010–19 J/eV) = (6.6310–34 J · s)2/8(9.1110–31 kg)L2, which gives L = 2.210–10 m = 0.22 nm.

Chapter 39 p. 5

20. The longest wavelength photon has the least energy, so the transition must be from n2 = 2 to n1 = 1. Thus we have hf = hc/ = ?E = (n22 – n12)h2/8mL2; [(1.24103 eV · nm)/(690 nm)](1.6010–19 J/eV) = (22 – 12)(6.6310–34 J · s)2/8(9.1110–31 kg)L2, 0.79 nm. which gives L = 7.910–10 m = 21. The energy level for the ground state of a particle in a rigid box is E1 = h2/8mL2 = p12/2m. Thus the momentum is p1 = h/2L. Because the direction is not known, we have ?p ˜ 2p1 = h/L. Because the particle can be anywhere in the box, we have ?x ˜ L. Thus we get ?p ?x ˜ (h/L)L = h, which is consistent with the uncertainty principle. 22. The wavefunctions for the rigid box are n = A sin knx, with kn = np/L. Thus we have n 2 = A2 sin2 (npx/L). (a) The maximal value of n 2 is A2, which occurs n times. We find the positions of the maxima from sin (npxmax/L) = ± 1; npxmax/L = !p, *p, … ; or xmax = (L/n)(m – !), m = 1, 2, … , n. (b) The minimal value of n 2 is 0, which occurs n + 1 times. We find the positions of the minima from sin (npxmin/L) = 0; npxmin/L = 0, p, 2p, … ; or xmin = (L/n)m, m = 0, 1, 2, … , n. 23. The energy levels for the electron in an infinite potential well are En = n2h2/8mL2 = n2(6.6310–34 J · s)2/8(9.1110–31 kg)(2.010–9 m)2(1.6010–19 J/eV) = (9.4210–2 eV)n2. The wave functions are n = (2/L)1/2 sin (npx/L) = (2/2.0 nm)1/2 sin (npx/2.0 nm) = (1.00 nm–1/2) sin (1.57 nm–1 nx). Thus we have E1 = (9.4210–2 eV)(1)2 = 0.094 eV; 1 = (1.00 nm–1/2) sin (1.57 nm–1 x); –2 2 E2 = (9.4210 eV)(2) = 0.38 eV; 2 = (1.00 nm–1/2) sin (3.14 nm–1 x); E3 = (9.4210–2 eV)(3)2 = 0.85 eV; 3 = (1.00 nm–1/2) sin (4.71 nm–1 x); –2 2 E4 = (9.4210 eV)(4) = 1.51 eV; 4 = (1.00 nm–1/2) sin (6.28 nm–1 x).

Chapter 39 p. 6

24. (a) The energy levels for the electron in a rigid box are En = n2h2/8mL2 = n2(6.6310–34 J · s)2/8(9.1110–31 kg)(0.5010–9 m)2(1.6010–19 J/eV) = (1.51 eV)n2. The wave functions are n = (2/L)1/2 sin (npx/L) = (2/0.50 nm)1/2 sin (npx/0.50 nm) = (2.0 nm–1/2) sin (2.0 nm–1 npx). Thus we have E1 = (1.51 eV)(1)2 = 1.51 eV; 1 = (2.0 nm–1/2) sin (2.0 nm–1 px); 2 E2 = (1.51 eV)(2) = 6.03 eV; 2 = (2.0 nm–1/2) sin (4.0 nm–1 px); E3 = (1.51 eV)(3)2 = 13.6 eV; 3 = (2.0 nm–1/2) sin (6.0 nm–1 px); 2 E4 = (1.51 eV)(4) = 24.1 eV; 4 = (2.0 nm–1/2) sin (8.0 nm–1 px). (b) The wavelength of a photon is determined by the energy change:  = (1.24103 eV · nm)/?E. For all possible transitions we have 21 = (1.24103 eV · nm)/(6.03 eV – 1.51 eV) = 274 nm; 31 = (1.24103 eV · nm)/(13.6 eV – 1.51 eV) = 103 nm; 32 = (1.24103 eV · nm)/(13.6 eV – 6.03 eV) = 164 nm; 41 = (1.24103 eV · nm)/(24.1 eV – 1.51 eV) = 55 nm; 42 = (1.24103 eV · nm)/(24.1 eV – 6.03 eV) = 69 nm; 43 = (1.24103 eV · nm)/(24.1 eV – 13.6 eV) = 118 nm. 25. If we consider a one-dimensional system, the energy level of the ground state is E = h2/8mL2. (a) For an electron we have E1e = h2/8mL2 = (6.6310–34 J · s)2/8(9.1110–31 kg)(10–14 m)2(1.6010–19 J/eV) = 3.8109 eV = 4 GeV. (b) For a proton we have E1p = h2/8mL2 = (6.6310–34 J · s)2/8(1.6710–27 kg)(10–14 m)2(1.6010–19 J/eV) = 2.1106 eV = 2 MeV. (c) For a neutron we have E1n = h2/8mL2 = (6.6310–34 J · s)2/8(1.6710–27 kg)(10–14 m)2(1.6010–19 J/eV) = 2.1106 eV = 2 MeV.

Chapter 39 p. 7

26. Because the origin of the x-axis is at the center of the well, the wavefunctions will alternate between sines and cosines: 1 1 2 1 = (2/L)1/2 cos (px/L);

2 = (2/L)1/2 sin (2px/L); 3 = (2/L)1/2 cos (3px/L); 4 = (2/L)1/2 sin (4px/L).

2

2 2

3

3 2

4

4 2

27. (a) Because the wavefunction is normalized, the probability is L L/ 4 L  L/ 4 2 2 P=  1  d x +  1 d x = 2 sin 2 x d x + 2 sin 2 x d x . L L L L 0 3L/ 4 0 3L/ 4 If we change variable to  = px/L, so d = (p/L) dx, we have / 4   2 2 P=2 L sin  d  + 2 L sin  d  L  0 L  3/ 4 / 4

= 2 12  – 14 sin 2  0



+2 1  – 14 sin 2 = 2  –1  2  4 2 3/ 4

= 0.18. (b) For the n = 4 state the probability is L L/ 4 L  L/ 4 2 2 P=  4  d x +  4 d x = 2 sin 2 4x d x + 2 sin 2 4x d x . L 0 L 3L/ 4 L L 0 3L/ 4 If we change variable to  = 4px/L, so d = (4p/L) dx, we have  4  2 2 P=2 L sin  d  + 2 L sin  d  L 4 0 L 4 3 

4

= 2 1  – 1 sin 2 + 2 1  – 1 sin 2 = 1 + 4 2 4 4 2 4 2 2 2 0 3 = 0.50. (c) Classically the electron has equal probability of being anywhere in the well. Thus the classical prediction is 0.50. [(#L – 0) + (L – &L)]/L = We see that the probability approaches the classical value for large n. 28. Because the wavefunction is normalized, the probability is x2  x2 2 P=  n  dx = 2 sin 2 nx dx . L L x1 x1 If we change variable to  = npx/L, so d = (np/L) dx, we have 0.60n  0.60n  P= 2 L sin 2  d  = 2 1  – 1 sin 2 . n 2 L n 0.40n 4 0.40n

Chapter 39 p. 8

(a) For the n = 1 state we have P = (2/p)[!(0.60p – 0.40p) – #(sin 1.20p – sin 0.80p)] = 0.39. (b) For the n = 5 state we have P = (2/5p){![(0.60)(5)p – (0.40)(5)p] – #[sin (1.20)(5)p – sin (0.80)(5)p]} = 0.20. (c) For the n = 20 state we have P = (2/20p){![(0.60)(20)p – (0.40)(20)p] – #[sin (1.20)(20)p – sin (0.80)(20)p]} = 0.20. (d) Classically the electron has equal probability of being anywhere in the well, so the probability is P = ?x/L = (0.20 nm)/(1.00 nm) = 0.20. We see that the probabilities approach the classical value for large n.

29.

4

4 2

5

5 2

30. We choose the zero of potential energy at the bottom of the well. In free space outside the well the potential energy is U0 = 50 eV. We find the total energy from K = E – U0; 100 eV = E – 50 eV, which gives E = 150 eV. (a) In free space we have Ka = pa2/2m = h2/2ma2; (100 eV)(1.6010–19 J/eV) = (6.6310–34 J · s)2/ 2(9.1110–31 kg)a2, 0.12 nm. which gives a = 1.210–10 m = (b) Over the well we have Kb = pb2/2m = h2/2mb2; (150 eV)(1.6010–19 J/eV) = (6.6310–34 J · s)2/ 2(9.1110–31 kg)b2, 0.10 nm. which gives b = 1.010–10 m =

(c)



31. The wavefunction outside the well (negative x) is I = Ce Gx, where G2 = 2m(U0 – E)/˙2. We approximate the energy as that of the ground state of an infinite well:

U E

U0 x 0

L

Chapter 39 p. 9

E ˜ h2/8mL2 = (6.6310–34 J · s)2/8(9.1110–31 kg)(0.1010–9 m)2(1.6010–19 J/eV) = 38 eV. Because this is much less than U0 , this should be a good approximation. We find G from G2 = 2(9.1110–31 kg)(1.00103 eV – 38 eV)(1.6010–19 J/eV)/(1.05510–34 J · s)2, which gives G = 1.591011 m–1. Because of the continuity of the wavefunction, the value at the wall (x = 0) is wall = C. Thus we have I/wall = e Gx; 0.01 = e Gx, or ln (0.01) = (1.591011 m–1)x, which gives x = – 2.910–11 m, so  x  = 0.03 nm.

32. (a) We assume the lowest 3 states are bound in the well.

U

3 U0 E3

2

E2 E1 0 I

L II

x

1

III

(b) In region I, x < 0, I = 0; In the well, region II, 0 = x = L, II = A sin kx, where k2 = 2mE/˙2; In region III, x > L, III = B e –Gx, where G2 = 2m(U0 – E)/˙2. 33. From the transmission we can find the value of G: T = e – 2GL, or ln T = – 2GL; ln (0.010) = – 2G(0.5010–9 m), which gives G = 4.6109 m–1. We find the energy from G2 = 2m(U0 – E)/˙2; (4.6109 m–1)2 = 2(9.1110–31 kg)(10.0 eV – E)(1.6010–19 J/eV)/(1.05510–34 J · s)2, which gives E = 9.2 eV. 34. We find the value of G from G2 = 2m(U0 – E)/˙2 = 2(1.6710–27 kg)(2.0 MeV – 1.0 MeV)(1.6010–13 J/MeV)/(1.05510–34 J · s)2, which gives G = 2.21014 m–1.

Chapter 39 p. 10

For the transmission coefficient we have T = e – 2GL, or ln T = – 2GL; ln T = – 2(2.21014 m–1)(2.010–13 m), which gives T = 8.610–39. Thus the transmitted current is I = I0T = (1.010–3 A)(8.610–39) = 8.610–42 A. 35. (a) We find the small change in G from a small change in the barrier height U0 : G2 = 2m(U0 – E)/˙2; 2G dG = (2m/˙2) dU0 , or dG = (mU0/G˙2) dU0/U0 . This produces a small change in the transmission coefficient: T = e – 2GL; dT = – 2L e – 2GL dG, or dT/T = – 2L dG = – 2(L2mU0/GL˙2) dU0/U0 ; dT/T = – 2[(0.1010–9 m)2(9.1110–31 kg)(70 eV)(1.6010–19 J/eV)/ (2.3)(1.05510–34 J · s)2](0.01) = – 0.08. Thus T decreases by 8%. (b) We find the small change in T from a small change in the barrier thickness L: T = e – 2GL; dT = – 2G e – 2GL dL, or dT/T = – 2G dL = – 2GL (dL /L) = – 2(2.3)(0.01) = – 0.05. Thus T decreases by 5%. 36. The wavefunction inside the barrier is  = A e – Gx + B e + Gx. If the barrier is high or thick, (x = L) ˜ 0, so B ˜ 0. Thus we have  = (0)e – Gx, so (x = L)2/(0)2 = e – 2Gx. 37. (a) We assume the potential energy of the alpha particle (charge = 2e) is produced by the electric field of the remaining nuclear charge of 90e. Thus the potential energy at a radius of 8 fm is UC = qQ/4pÅ0r0 = (2)(90)(1.6010–19 C)2/4p(8.8510–12 C2/N · m2)(810–15 m)(1.6010–19 J/eV) 32 MeV. = 3.2107 eV = (b) The kinetic energy of the free alpha particle is also its total energy. At the exit from the barrier, the kinetic energy is zero, so we find the radius where the alpha particle leaves from E = Ur = qQ/4pÅ0r = (qQ/4pÅ0r0)(r0/r); 4 MeV = (32 MeV)(8 fm/r), which gives r = 64 fm. Thus the width of the barrier at an energy of 4 MeV is 64 fm – 8 fm = 56 fm. (c) If the potential energy is zero inside the barrier, we find the speed of the alpha particle from E = K = !mv2; 1.4107 m/s. (4 MeV)(1.6010–13 J/MeV) = !(4)(1.6710–27 kg)v2, which gives v = Because there is a barrier on both sides of the nucleus, the frequency with which the alpha particle hits a barrier is f = v/2r0 = (1.4107 m/s)/2(810–15 m) = 8.81020 s–1. To find the transmission coefficient, we assume a square barrier. We find the value of G from G2 = 2m(U0 – E)/˙2 = 2(4)(1.6710–27 kg)(32 MeV – 4 MeV)(1.6010–13 J/MeV)/(1.05510–34 J · s)2, which gives G = 2.41015 m–1 = 2.4 fm–1. If we take the height as the value of the Coulomb barrier, the thickness will be less than that found in part (b). The Coulomb potential varies as 1/r, so the assumed thickness of a square barrier will be less than half that found in part (b). If we use @ of the width, L = 18.7 fm, we have

Chapter 39 p. 11

–1 T = e – 2GL = e – 2(2.4 fm )(18.7 fm ) = 1 .0  10 – 39. Each time the alpha particle hits the barrier, T represents the probability it gets through. If the lifetime is , the number of hits is f, which has a probability of one: 1 = f T, or  = 1/fT;  = 1/(8.81020 s–1)(1.010–39) = 1.11018 s ˜ 1010 yr. Note that the result is very sensitive to the value of L. 

38. We find the uncertainty in the momentum: ?p = ˙/?x = (1.05510–34 J · s)/(2.010–8 m) = 5.2810–27 kg · m/s. We find the uncertainty in the velocity from ?p = m ?v; 5.8103 m/s. 5.2810–27 kg · m/s = (9.1110–31 kg) ?v, which gives ?v = 39. We use the radius as the uncertainty in position for the neutron. We find the uncertainty in the momentum from ?p = ˙/?x = (1.05510–34 J · s)/(1.010–15 m) = 1.05510–19 kg · m/s. If we assume that the lowest value for the momentum is the least uncertainty, we estimate the lowest possible kinetic energy as E = (?p)2/2m = (1.05510–19 kg · m/s)2/2(1.6710–27 kg) = 3.3310–12 J = 21 MeV. 40. We find the average lifetime of the particle from ?t = ˙/?E = (1.05510–34 J · s)/(2.5 GeV)(1.6010–10 J/GeV) =

2.610–25 s.

41. We find the uncertainty in the rest energy of the muon from ?E = ˙/?t = (1.05510–34 J · s)/(2.2010–6 s) = 4.810–29 J = 3.0010–10 eV. Thus the uncertainty in the mass is ?m = ?E/c2 = 3.0010–10 eV/c2. 42. We find the uncertainty in the rest energy of the free neutron from ?E = ˙/?t = (1.05510–34 J · s)/(900 s) = 1.1710–37 J. Thus the uncertainty in the mass is ?m = ?E/c2 = (1.1710–37 J)/(3.00108 m/s)2 = 1.3010–54 kg. 43. We can relate the momentum to the radius of the orbit from the quantum condition: L = mvr = pr = n˙, so p = n˙/r = ˙/r1 for the ground state. If we assume that this is the uncertainty of the momentum, the uncertainty of the position is ?x = ˙/?p = ˙/(˙/r1) = r1 , which is the Bohr radius. 44. (a) The energy levels for the neutron in an infinite well are En = n2h2/8mL2 = n2(6.6310–34 J · s)2/8(1.6710–27 kg)(2.010–15 m)2(1.6010–13 J/MeV) = (51.4 MeV)n2. Thus we have E1 = (51.4 MeV)(1)2 = 51 MeV; E2 = (51.4 MeV)(2)2 = 210 MeV; 2 E3 = (51.4 MeV)(3) = 460 MeV; E4 = (51.4 MeV)(4)2 = 820 MeV. (b) The wave functions are n = (2/L)1/2 sin (npx/L) = (2/2.0 fm)1/2 sin (npx/2.0 fm) = (1.0 fm–1/2) sin (0.50 fm–1 npx). Thus we have 1 = (1.0 fm–1/2) sin (0.50 fm–1 px); 2 = (1.0 fm–1/2) sin (1.00 fm–1 px);

Chapter 39 p. 12

3 = (1.0 fm–1/2) sin (1.50 fm–1 px); 4 = (1.0 fm–1/2) sin (2.00 fm–1 px). (c) The energy of the photon is the energy change: hf = ?E = E2 – E1 = 205 MeV – 51 MeV = 150 MeV. The wavelength of the photon is 21 = (1.24103 eV · nm)(10–6 MeV/eV)(106 fm/nm)/?E = (1.24103 MeV · fm)/?E . = (1.24103 MeV · fm)/(150 MeV) = 8.3 fm, (gamma ray). 45. If we assume the ground state, the kinetic energy is K = E1 = n2h2/8mL2 = (1)2(6.6310–34 J · s)2/8(4)(1.6710–27 kg)(10–14 m)2(1.6010–19 J/eV) = 5.1105 eV = 0.5 MeV. We find the speed from K = p2/2m = mv2/2; 5106 m/s. (5.1105 eV)(1.6010–19 J/eV)= (4)(1.6710–27 kg)v2/2, which gives v =

46.

(a) (b) The Schrödinger equation is – (˙2/2m) d2/dx2 + !Cx2 = E. 2 For the proposed solution  =  A e – Bx , we have d  2 = – 2ABx e – Bx ; dx 2 d  2 2 2 = – 2AB e – Bx + 4AB x 2 e – Bx . dx 2 If we substitute these in the equation, we get 2 2 2 2 2 – h – 2AB e – Bx + 4A B2 x 2 e – Bx + 1 Cx 2A e – Bx = EA e – Bx . 2m 2 When we cancel common factors, we get 2 2 2 Bh 4B h – + 1 C x 2 + m – E = 0. 2m 2 Thus we have a solution if B2 = mC/4˙2, and E = B˙2/m = !˙, if  = 2˙B/m.

U U = ! Cx 2 E 0

x

47. (a) When the pencil, assumed to be a uniform rod, makes an angle  with the vertical, the torque from its weight about the bottom point (assumed fixed) creates the angular acceleration:  = I; Mg!¬ sin  = @M¬2 d2/dt2. Classically, if  = 0, there is no torque and thus no rotation. From the uncertainty principle we have (?L)(?) = ˙, or ? = ˙/I ?. Thus  will not always be zero. The resulting torque will cause the pencil to fall. (b) We need the equation of motion to determine the time of fall. Most of the time will be taken while the angle is small. Thus while  « 1, we have d2/dt2 = (3g/2¬), for which the solution is  = Ae + kt + Be – kt.

Chapter 39 p. 13

Because  increases with time, we use  = Ae kt, where k = (3g/2¬)1/2 = [3(9.80 m/s2)/2(0.20 m)]1/2 = 8.6 s–1. If we assume  has the minimum value of ? at t = 0, then ? = A. From the uncertainty principle, the initial angular velocity will be 0 = kA = ? = ˙/I ? = ˙/(@M¬2)A, which gives A2 = 3˙/M¬2k = 3(1.05510–34 J · s)/(1010–3 kg)(0.20 m)2(8.6 s–1), which gives A = 3.010–16 rad. We find the time to fall through 0.1 rad from  = Ae kt, or ln (/A) = kt; 4 s. ln [(0.1 rad)/(3.010–16 rad)] = (8.6 s–1)t, which gives t = Because most of the time will be taken while the angle is small, this should be within a factor of 2. 48. For the average value of the position of a particle in a rigid box, we have L  L x= x  n  2 d x = 2 x sin 2 nx d x . L L 0 0 If we change variable to  = npx/L, so d = (np/L) dx, we have n

n

 2 –  si n 2 – cos 2  si n 2  d  = 2L 2 2 4 8 0 n 4 0 2 2 1 1 n  – 0– + = 2L = L. 2 8 8 n 2 2 4 This is expected from the symmetry of the probability distributions. x= 2 L L n

2

49. The potential difference between the surface and the tip creates vacant energy levels on one side of the barrier so tunneling can occur. The change does not appreciably affect the shape of the barrier, so we assume a square barrier of height U0 . If the work function is the energy required to raise the electron to the top of the barrier, then W0 = U0 – E. We find the value of G from G2 = 2m(U0 – E)/˙2; = 2(9.1110–31 kg)(2.28 eV)(1.6010–19 J/eV)/(1.05510–34 J · s)2, which gives G = 7.7109 m–1. We use the distance from the sodium surface to the tip as the width of the barrier, so the transmission coefficient is T = e – 2GL. If we find the ratio of the coefficient after the rise of the tip to that before the rise, we have +9 –9 T2  e – 2GL 2 –1 = = e – 2G L 2 – L1 = e – 2(7.7  10 m )(0 .010  10 m) = 0.86. T1 e – 2GL 1 The fractional change is (T2 – T1)/T1 = 0.86 – 1 = – 0.14. Thus the small change in separation (about a tenth of atomic size) produces a 14% decrease in the tunneling current. 50. From energy conservation, the speed after falling a height H or to rise to a height H is v = (2gH)1/2. If H0 is the initial height and the height after the first bounce is H1 = bH0 , the height after n bounces is Hn = bnH0. Thus the speed after the nth bounce is vn = (2gbnH0)1/2. If we take the height after the nth bounce as the uncertainty in the position of the ball, we find the uncertainty in the speed from ?p ?x = ˙; m ?v Hn = ˙, or ?v = ˙/mHn .

Chapter 39 p. 14

The uncertainty principle will play a role when the uncertainty in the speed is on the order of the speed: ?v ˜ vn = (2gbnH0)1/2 = ˙/mbnH0 , or b3n = ˙2/2gH03m2; (0.80)3n = (1.05510–34 J · s)2/2(9.80 m/s2)(2.0 m)3(1.010–6 kg)2, which gives n = 200.

Chapter 40 p. 1

CHAPTER 40 – Quantum Mechanics of Atoms Note:

At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.6310–34 J · s)(3108 m/s)(109 nm/m)/(1.6010–19 J/eV); E = (1.24103 eV · nm)/.

1.

The value of ¬ can range from 0 to n – 1. Thus for n = 6, we have ¬ = 0, 1, 2, 3, 4, 5.

2.

The value of m¬ can range from – ¬ to + ¬. Thus for ¬ = 3, we have m¬ = – 3, – 2, – 1, 0, 1, 2, 3. – !, + !. The possible values of ms are

3.

The value of ¬ can range from 0 to n – 1. Thus for n = 4, we have ¬ = 0, 1, 2, 3. For each ¬ the value of m¬ can range from – ¬ to + ¬, or 2¬ + 1 values. For each of these there are two values of ms . Thus the total number for each ¬ is 2(2¬ + 1). The number of states is N = 2(0 + 1) + 2(2 + 1) + 2(4 + 1) + 2(6 + 1) = 32 states. We start with ¬ = 0, and list the quantum numbers in the order (n, ¬, m¬ , ms); (4, 0, 0, – !), (4, 0, 0, + !), (4, 1, –1, – !), (4, 1, –1, + !), (4, 1, 0, – !), (4, 1, 0, + !), (4, 1, 1, – !), (4, 1, 1, + !), (4, 2, – 2, – !), (4, 2, – 2, + !), (4, 2, –1, – !), (4, 2, –1, + !), (4, 2, 0, – !), (4, 2, 0, + !), (4, 2, 1, – !), (4, 2, 1, + !), (4, 2, 2, – !), (4, 2, 2, + !), (4, 3, – 3, – !), (4, 3, – 3, + !), (4, 3, – 2, – !), (4, 3, – 2, + !), (4, 3, – 1, – !), (4, 3, – 1, + !), (4, 3, 0, – !), (4, 3, 0, + !), (4, 3, 1, – !), (4, 3, 1, + !), (4, 3, 2, – !), (4, 3, 2, + !), (4, 3, 3, – !), (4, 3, 3, + !).

4.

The value of m¬ can range from – ¬ to + ¬, so we have ¬ = 3. The value of ¬ can range from 0 to n – 1. Thus we have n = ¬ + 1 (minimum 4). m s = – !, + !. There are two values of ms :

5.

n = 5. The value of ¬ can range from 0 to n – 1. Thus for ¬ = 4, we have For each ¬ the value of m¬ can range from – ¬ to + ¬: m¬ = – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4. m s = – !, + !. There are two values of ms :

6.

The magnitude of the angular momentum depends on ¬ only: L = ˙[¬(¬ + 1)]1/2 = (1.05510–34 J · s)[(2)(2 + 1)]1/2 = v6˙ = 2.5810–34 kg · m2/s.

7.

(a) The principal quantum number is n = 6. (b) The energy of the state is E6 = – (13.6 eV)/n2 = – (13.6 eV)/62 = – 0.378 eV. (c) From spdfgh, we see that the “g” subshell has ¬ = 4. The magnitude of the angular momentum depends on ¬ only: L = ˙[¬(¬ + 1)]1/2 = (1.05510–34 J · s)[(4)(4 + 1)]1/2 = v20˙ = 4.7210–34 kg · m2/s. (d) For each ¬ the value of m¬ can range from – ¬ to + ¬: m¬ = – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4.

Chapter 40 p. 2

8.

(a) For each ¬ the value of m¬ can range from – ¬ to + ¬, or 2¬ + 1 values. For each of these there are two values of ms . Thus the total number of electrons allowed in a subshell is N = 2(2¬ + 1). (b) For the values of ¬ we have ¬ = 0, N = 2[2(0) + 1] = 2; ¬ = 1, N = 2[2(1) + 1] = 6; ¬ = 2, N = 2[2(2) + 1] = 10; ¬ = 3, N = 2[2(3) + 1] = 14; ¬ = 4, N = 2[2(4) + 1] = 18; ¬ = 5, N = 2[2(5) + 1] = 22; ¬ = 6, N = 2[2(6) + 1] = 26.

9.

From Problem 8 the total number of electrons allowed in a subshell with quantum number ¬ is 2(2¬ + 1). For a state with quantum number n the value of ¬ can range from 0 to n – 1. Thus we have n–1 n–1 (n – 1)n  n–1 N = 2  (2 + 1) = 4  + 2  1 = 4 + 2n = 2n 2. 2 =0 =0 =0

10. (a) The 6d subshell has ¬ = 2. Allowed transitions will be to any lower value of n with ?¬ = ± 1. Thus the final value of ¬ can be 1 (p) or 3 (f ), so we have 5f, 5p, 4f, 4p, 3p, 2p. (b) If we ignore fine structure, the energy change depends only on the two values of n. The allowed transitions from n = 6 are to n = 5, 4, 3, 2, so there are 4 wavelengths. 11. To see if the ground-state wave function is normalized, we integrate the radial probability density over all radii:   2 4r 2 100 d r = 4r 2 13 e – 2r/ r0 d r. 0 r0 0 We change variable to x = 2r/r0 and use the result for the integration in Ex. 40–4:  1  2 0



x 2e – x d x = – 12 x 2 – x – 1 e – x = 0 – ( – 1 e 0) = 1. 0

12. The ground-state wave function is 100 = (a) 100(r = r0) = [1/(pr03)1/2] e – 1. (b) 100(r = r0)2 = (1/pr03) e – 2. (c) The radial probability density is Pr = 4pr2100(r)2. At r = r0 we have Pr(r = r0) = (4/r0) e – 2.

1 e – r/ r 0 . . r03

1 2 – rr e – r/ 2r 0. 0 32r30 (a) 200(r = 5r0) = [1/(32pr03)1/2](2 – 5) e – 5/2 = – [3/(32pr03)1/2] e – 5/2. (b) 200(r = 5r0)2 = (9/32pr03) e – 5. (c) The radial probability density is Pr = 4pr2200(r)2. At r = 5r0 we have Pr(r = 5r0) = 4p(5r0)2(9/32pr03) e – 5 = (225/8r0) e – 5.

13. The n = 2, ¬ =0 wave function is 200 =

Chapter 40 p. 3

14. To see if the wave function is normalized, we integrate the radial probability density over all radii:   2 2 4r 2 200 d r = 4r 2 1 3 2 – rr e – r/ r0 d r. 0 32r 0 0

0

We change variable to x = r/r0 and expand:  1  2 0

x 2e – x d x – 12





x 3 e – x d x + 18 x 4e – x d x = 12 (2!) – 12 (3!) + 18 (4!) = 1 – 3 + 3 = 1. 0 0

15. We form the ratio of the radial probability densities: 4r02 –2r / r e 0 0 Pr (r0 ) r 30 e– 2 = = = 1.85. Pr (2r0 ) 4  2r 2 4e – 4 0 –4r / r e 0 0 r 30 16. Because the change in r is small, we can use the radial probability density at r = r0 : 4r 20 –2r0 / r 0 P = Pr (r 0) r = e (0.02r0 )= 0.08 e–2 = 0.011 = 1.1%. r03 17. The n = 2, ¬ =0 wave function is 200 =

1 2 – rr e – r/ 2r 0, 0 32r30

The radial probability density is  r2 2 2 Pr (r) = 4r 2 200 = 3 2 – rr e – r/ r 0. 0 8r0 To find the probability we integrate the radial probability density:

P=

5 .00r 0

r2 r 2 – r/ r 0 d r. 3 2 – r0 e 8r 4.00r0 0

We change variable to x = r/r0 and expand: 5.00

P=

1 8

4.00

5 .00

4x 2 – 4x 3 + x 4 e– x dx = – 12 x 2 – x – 1 – 18 x 4 e – x = 0.173 = 4 .00

17.3%.

18. (a) To find the probability we integrate the radial probability density:  r0 4r 2 e –2r/ r0 d r. P= 3 0 r0 We change variable to x = 2r/r0 and use the result for the integration in Ex. 40–4: 2 2 P = 12 x 2 e– x d x = – 12 x 2 – x – 1 e – x = ( – 2 – 2 – 1) e –2 – ( – 1) = 0.32 = 32%. 0 0

(b) To find the probability we integrate the radial probability density:  2 r0 4r 2 e –2r/ r0 dr. P= r03 r 0

We change variable to x = 2r/r0 and use the result for the integration in Ex. 40–4: 4 4 P = 12 x 2 e– x d x = – 12 x 2 – x – 1 e – x = ( – 8 – 4 – 1) e –4 – ( – 2 – 2 – 1) e– 2 = 0.44 = 2 2

44%.

Chapter 40 p. 4

19. To find the probability for the electron to be within a sphere of radius r, we integrate the radial probability density:  r 4r 2 –2r/ r 0 d r. P= 3 e 0 r0 We change variable to x = 2r/r0 and use the result for the integration in Ex. 40–4: x x P = 12 x 2 e – x  d x = – 12 x 2 – x – 1 e – x  = 1 – 12 x 2 + x + 1 e– x. 0 0

(a) For the probability to be 50% we have (!x2 + x + 1)e –x = 0.50. A numerical calculation gives x = 2.68, so r = (b) For the probability to be 90% we have (!x2 + x + 1)e –x = 0.10. A numerical calculation gives x = 5.32, so r = (c) For the probability to be 99% we have (!x2 + x + 1)e –x = 0.01. A numerical calculation gives x = 8.40, so r =

1.34r0 .

2.7r0 .

4.2r0 .

20. (a) Because r « r0 , we assume the probability density is constant and equal to the value at r = 0: P = (1/pr03) e – 0()pr3) = )(r/r0)3 = )[(1.110–15 m)/(0.52910–10 m)]3 = 1.210–14. (b) From the Bohr theory, we have r0 ~ 1/m. Thus Pr ~ 1/r03 ~ m3. If we form the ratio we have Pmuon/Pelectron = (mmuon/melectron)3; Pmuon/(1.210–14) = (207)3, which gives Pmuon = 1.110–7. 21. The three wave functions are x + iy – r/ 2r0 x – iy – r/ 2r0 z e– r/ 2r0, 211 = e , 21–1 = e . 210 = 32r05 64r05 64r05 The radial probability density is r2 2 1 2 2 2 Pr = 4r 2 13 210 + 13 211 + 13 21– 1 = z + 2 (x + i y)(x – i y) + 12 (x – i y)(x + i y) e– r/ r0 24r50 r2 2 r 4 – r/ r0 = z + x 2 + y 2 e– r/ r0 = e . 5 24r0 24r05 22. From the result for Problem 21 the radial probability density is r 4 – r/ r 0. Pr = e 24r 50 We find the most probable distance by setting the first derivative equal to zero: 4 4 d Pr 1 = 4r3 – rr e – r/ r 0 = 0; 4r 3 = rr , which gives r = 4r 0 . 5 dr 0 0 24r0 23. We find the mean value of r from

  2 r= r 100 4r2 d r = 0



0

4r 3 1 3 e – 2r/ r 0 d r. r 0

We change variable to x = 2r/r0 :

Chapter 40 p. 5



r=

r0  3 – x r0 x e dx = 3! = 32 r0. 4 0 4

1 2 – rr e – r/ 2r 0. 0 32r30

24. (a) The n = 2, ¬ =0 wave function is 200 =

The radial probability density is  r2 2 2 Pr (r) = 4r 2 200 = 3 2 – rr e – r/ r 0. 0 8r0 To find the probability for 0 < r < r0 , we integrate the radial probability density:

P=

r0

r2 r 2 – r/ r 0 dr. 3 2 – r0 e 8r 0 0

We change variable to x = r/r0 and expand:

P=

11 0 8

1

1

4x 2 – 4x 3 + x 4 e– x d x = – 2 x 2 – x – 1 – 8x 4 e – x

1 0

1

(b) From the result for Problem 21 the radial probability density is r 4 – r/ r 0. Pr = e 24r 50 To find the probability for 0 < r < r0 , we integrate the radial probability density:

P=

r0

r 4 – r/ r 0 dr. 5e 0 24r0

We change variable to x = r/r0: 1 1 4 –x 1 x e dx = – x 4 – 4x 3 – 12x 2 – 24x 24 0 24 1 =1+ – 1 – 4 – 12 – 24 – 24 e– 1 = 0.0037 = 24

P=

– 24 e– x 0.37%.

1 0

1

= 1 + – 2 – 1 – 1 – 8 e – 1 = 0.034 = 3.4%.

Chapter 40 p. 6

25. The Schrödinger equation is 2  2 2  2  2  – h + + – e  = E, 4r 0r z2 2m x 2  y 2 where r2 = x2 + y2 + z2. For later use we find a partial derivative: 2r ?r/?x = 2x, or ?r/?x = x/r. We find the partial derivatives of the ground-state wave function 100 =

 d  r 1 = =– e – r/ r0 xr ; x d r x r0 r03 2 1 1 =+ e – r/ r0 r xr – e– r/ r0 1 – x2 r 2 3 3 r r x x 2 x r0 r0 r0 r0 2 2 x x 1 1 =+ e– r/ r0 – 1+ 2 = – e – r/ r0 1 – x 2 1 + rr0 rr0 r rr0 r03 rr0 r03 By exchanging x with y and z, we have 2  2  – r/ r 0 1 – y 2 1 + 1 ;   = – 1 1 = – e e– r / r 0 1 rr0 r 2 y 2 z2 rr r 3 rr r 3

1 e – r/ r 0 : r03

1 . r2

– z 2 1 + 12 . rr0 r 0 0 0 0 2 2 2 2 When we put these in the Schrödinger equation and use r = x + y + z , we get 2  h2 1 e – r/ r0 = E 1 e – r / r0 . e – r/ r0 3 – r 2 1 + 12 – e 3 rr r 4Å0 r r03 2mrr0 r0 0 r03 If we cancel common factors, we have 2 2 2 – h 2 + h – e = E. 2mr0 mrr0 4Å0 r If we look at the second and third terms and use r0 = h2Å0/pme2, we get 

h 2

2

me 2

–

e 2 = 0. 4Å0 r

4 2mr h Å0/ Thus we have 2 4  h E= – = – me 2 2. 2 8h Å0 2m h 2Å0 / me 2 26. From Fig. 40–8 we see that the first peak is from r = 0 to r = 2r0 . The n = 2, ¬ =0 wave function is 1 2 – rr e – r/ 2r 0. 200 = 0 32r30 The radial probability density is  r2 2 2 Pr (r) = 4r 2 200 = 3 2 – rr e – r/ r 0. 0 8r0 To find the probability for 0 < r < 2r0 , we integrate the radial probability density:

P=

2 r0 0

r2 2 2 – rr e – r/ r 0 d r. 0 8r03

We change variable to x = r/r0 and expand:

Chapter 40 p. 7

2 0

1 8

2 1 1 4x 2 – 4x 3 + x 4 e – x d x = – 2 x 2 – x – 1 – 8 x 4 e – x = 1– 0.947 = 0.053 = 0

5.3%.

27. (a) The n = 3, ¬ =0 wave function is (b) r0 Pr 2  1 2r 2r – r/ 3r 0. 300 = 1 – + e 3r 0 27r20 27r30 0.10 The radial probability distribution is 2 2 4r 2 2 Pr (r) = 4r 2 300 = 1 – 2r + 2r 2 e – 2r/ 3r0 . 3 3r0 27r 0 27r0 0.05 (c) To find the most probable distance from the nucleus, we find the maxima by setting the first derivative equal to zero. If we change variable to x = r/r0 , we have 0 2 2 4 0 10 20 2x 2 – 2 x/ 3 2x Pr (x) = 1– + x e . 3 27 27r0 If we suppress the constant in front, the derivative is 2 2 2 2 d Pr = 1 – 2x + 2x 2x – 2x e – 2x/ 3 + 2x 2e – 2x/ 3 1 – 2x + 2x – 2 + 4 x 3 27 dx 3 27 3 3 27 2 2 2x 2x 2x 2x 2 4x x – 2x/ 3 = 2xe 1– + 1– 1– + +x – + 3 27 3 27 3 3 27 2 2 3 2x 2x 5x 12x 2x – 2x/ 3 = 2xe 1– + 1– + – = 0. 3 27 3 27 81 Thus we have x = 0, which is the minimum at r = 0; 2 1 – 2x + 2x = 0. 3 27 Because this is a factor in , its two solutions are the other minima at x = 1.90, 7.10. 2 3 1 – 5x + 12x – 2x = 0, which can be solved numerically for the three solutions, 3 27 81 which are x = 0.74, 4.19, 13.07. These correspond to the three peaks in the distribution. The highest peak is the most probable distance: r = 13r0 .

30

r/ r0

28. We start with the n = 1 shell, and list the quantum numbers in the order (n, ¬, m¬ , ms): (1, 0, 0, – !), (1, 0, 0, + !), (2, 0, 0, – !), (2, 0, 0, + !), (2, 1, – 1, – !), (2, 1, – 1, + !), (2, 1, 0, – !). Note that, without additional information, there are other possibilities for the last three electrons. 29. (a) We start with the n = 1 shell, and list the quantum numbers in the order (n, ¬, m¬ , ms): (1, 0, 0, – !), (1, 0, 0, + !), (2, 0, 0, – !), (2, 0, 0, + !), (2, 1, – 1, – !), (2, 1, – 1, + !). Note that, without additional information, there are other possibilities for the last two electrons. (b) We start with the n = 1 shell, and list the quantum numbers in the order (n, ¬, m¬ , ms): (1, 0, 0, – !), (1, 0, 0, + !), (2, 0, 0, – !), (2, 0, 0, + !), (2, 1, – 1, – !), (2, 1, – 1, + !), (2, 1, 0, – !), (2, 1, 0, + !), (2, 1, 1, – !), (2, 1, 1, + !), (3, 0, 0, – !), (3, 0, 0, + !). 30. The number of electrons in the subshell is determined by the value of ¬. For each ¬ the value of m¬ can range from – ¬ to + ¬, or 2¬ + 1 values. For each of these there are two values of ms . Thus the total number for ¬ = 3 is N = 2(2¬ + 1) = 2[2(3) + 1] = 14 electrons.

Chapter 40 p. 8

31. (a) Selenium has Z = 34: 1s22s22p63s23p63d104s24p4. (b) Gold has Z = 79: 1s22s22p63s23p63d104s24p64d104f145s25p65d106s1. (c) Uranium has Z = 92: 1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p65f36d17s2.

32. (a) (b) (c) (d) (e)

The 4p  3p transition is The 2p  1s transition is The 3d  2d transition is The 4d  3s transition is The 4s  2p transition is

forbidden, because ?¬ ? ± 1. allowed, ?¬ = – 1. forbidden, because ?¬ ? ± 1. forbidden, because ?¬ ? ± 1. allowed, ?¬ = + 1.

33. From the Bohr formula for the radius, we see that r  1/Z, so rU = r1/Z = (0.52910–10 m)/(92) = 5.810–13 m. The innermost electron would “see” a nucleus with charge Ze. Thus we use the energy of the hydrogen atom: E1 = – (13.6 eV)Z2/n2 = – (13.6 eV)(92)2/12 = – 1.15105 eV, so the binding energy is 0.115 MeV. 34. The third electron in lithium is in the 2s subshell, which is outside the more tightly bound filled 1s shell. This makes it appear as if there is a “nucleus” with a net charge of + 1e. Thus we use the energy of the hydrogen atom: E2 = – (13.6 eV)/n2 = – (13.6 eV)/22 = – 3.4 eV, so the binding energy is 3.4 eV. Our assumption of complete shielding of the nucleus by the 1s electrons is probably not correct. 35. In a filled subshell, we have 2(2¬ + 1) electrons. All of the m¬ values – ¬, – ¬ + 1, … , 0, … , ¬ – 1, ¬ are filled, so their sum is zero. For each m¬ value, both values of ms are filled, so their sum is also zero. Thus the total angular momentum is zero. 36. The energy levels of the infinite square well of width L are En = h2n2/8mL2. An electron in the well has two quantum numbers: n and ms = ± !. To be consistent with the Pauli exclusion principle, a maximum of two electrons can be in each level. The lowest energy state will have two electrons in the n = 1 state, two electrons in the n = 2 state, and one electron in the n = 3 state. The total energy is E = 2E1 + 2E2 + E3 = 2(h212/8mL2) + 2(h222/8mL2) + (h232/8mL2) = 19h2/8mL2. 37. The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the electron in the tube:  = (1.24103 eV · nm)/E = (1.24103 eV · nm)/(30103 eV) = 0.041 nm. The longest wavelength of the continuous spectrum would be at the limit of the X-ray region of the electromagnetic spectrum, generally on the order of 1 nm. 38. The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the electron

Chapter 40 p. 9

in the tube: E = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(0.029 nm) = 4.3104 eV = 43 keV. Thus the operating voltage of the tube is 43 kV. 39. The energy of the photon with the shortest wavelength must equal the maximum kinetic energy of an electron: hf0 = hc/0 = eV, or 0 = hc/eV = (6.6310–34 J · s)(3.00108 m/s)(109 nm/m)/(1.6010–19 J/eV)(1 e)V = (1.24103 V · nm)/V.

40. With the shielding provided by the remaining n = 1 electron, we use the energies of the hydrogen atom with Z replaced by Z – 1. The energy of the photon is hf = ?E = – (13.6 eV)(27 – 1)2[(1/22) – (1/12)] = 6.90103 eV. The wavelength of the photon is  = (1.24103 eV · nm)/?E = (1.24103 eV · nm)/(6.90103 eV) = 0.18 nm. 41. With the shielding provided by the remaining n = 1 electron, we use the energies of the hydrogen atom with Z replaced by Z – 1. The energy of the photon is hf = ?E = – (13.6 eV)(26 – 1)2[(1/22) – (1/12)] = 6.38103 eV. The wavelength of the photon is  = (1.24103 eV · nm)/?E = (1.24103 eV · nm)/(6.38103 eV) = 0.19 nm. 42. If we assume that the shielding is provided by the remaining n = 1 electron, we use the energies of the hydrogen atom with Z replaced by Z – 1. The energy of the photon is hf = ?E = – (13.6 eV)(42 – 1)2[(1/32) – (1/12)] = 2.03104 eV. The wavelength of the photon is  = (1.24103 eV · nm)/?E = (1.24103 eV · nm)/(2.03104 eV) = 0.061 nm. We do not expect perfect agreement because there is some partial shielding provided by the n = 2 shell, which was ignored when we replaced Z by Z – 1. 43. The K line is from the n = 2 to n = 1 transition. We use the energies of the hydrogen atom with Z replaced by Z – 1. Thus we have hf = ?E  (Z – 1)2, so   1/(Z – 1)2. When we form the ratio for the two materials, we get X/iron = (Ziron – 1)2/(ZX – 1)2; (229 pm)/(194 pm) = (26 – 1)2/(ZX – 1)2, which gives ZX = 24, so the material is chromium. 44. For the emission of a photon, energy and momentum are conserved in any reference frame. If we consider the frame in which the electron is initially at rest, its momentum is 0 and its energy is mc2. The emitted photon will have momentum h/ = hf/c and energy hf. The final momentum of the electron will be opposite to that of the photon with magnitude p. Thus we have momentum conservation: 0 = p – hf/c; mc2 = (p2c2 + m2c4)1/2 + hf. energy conservation: When we combine the two equations to eliminate p, we get (mc2)2 – 2hfmc2 + (hf)2 = (hf)2 + m2c4, which gives hf = 0. Thus no photon is emitted. If a third body is present, some momentum can be transferred to it; a photon can be emitted.

Chapter 40 p. 10

45. For the Bohr magneton we have B = e˙/2m = (1.6010–19 C)(1.05510–34 J · s)/2(9.1110–31 kg) = 9.2710–24 J/T.

46. (a) The additional energy term in a magnetic field is +2  B U = – zB = – Bm¬B. +  B Thus the separation of energy levels is n = 3, ¬ = 2 0 ?U = ?m¬ BB = (1)(9.2710–24 J/T)(2.0 T) –  B = 1.910–23 J = 1.210–4 eV. (b) The 3d level, with ¬ = 2, will split into 5 levels. –2 B The 2p level, with ¬ = 1, will split into 3 levels. With the restriction that ?m¬ = 0, ± 1, there will +  B be only 3 wavelengths from the 9 transitions, n = 2, ¬ = 1 as shown on the diagram. 0 (c) If we assume the hydrogen energy levels, we find the –  B wavelength for the n = 3 to n = 2 transition, which is ² m = –1 0 +1 also the wavelength for ?m¬ = 0, from 1/0 = R[(1/n2) – (1/n2)] = (1.0974107 m–1)[(1/22) – (1/32)], which gives 0 = 656.1010–9 m = 656.10 nm. Because the splitting of the levels is much smaller than their separation, we can find the wavelength change from E = hc/; ?E = – (hc/2) ? = – (hc/)(?/); ± 1.210–4 eV = – [(1.24103 eV · nm)/(656.10 nm)] ?/(656.10 nm), which gives ? = ± 0.04 nm. Thus the wavelengths are 656.06 nm, 656.10 nm, 656.14 nm. 47. (a) We take the original direction of the atomic beam as the x-axis and the direction of the force from the magnetic field gradient as the z-axis. The force is Fz = z dB/dz = – gBms dB/dz. This constant force will produce a constant acceleration. The time to traverse the field is t = x/v0 = (4.010–2 m)/(700 m/s) = 5.7110–5 s. so the deflection of one of the beams is z = !at2 = !(Fz/m)t2 = ![(gBms dB/dz)/m]t2 = 2 = 1.2610–4 m. Thus the separation is 2z = 2.5210–4 m = 0.25 mm. (b) Because the separation is proportional to g, we have 2z = (1/2.0023)(2.5210–4 m) = 1.2610–4 m = 0.13 mm. 48. (a) Aluminum has Z = 13. We start with the n = 1 shell, and list the quantum numbers in the order

Chapter 40 p. 11

(n, ¬, m¬ , ms):

(b) (c) (d) (e)

(1, 0, 0, – !), (1, 0, 0, + !), (2, 0, 0, – !), (2, 0, 0, + !), (2, 1, – 1, – !), (2, 1, – 1, + !), (2, 1, 0, – !), (2, 1, 0, + !), (2, 1, 1, – !), (2, 1, 1, + !), (3, 0, 0, – !), (3, 0, 0, + !), (3, 1, 0, – !). Note that, without additional information, there are other possibilities for the last electron. The filled subshells are n = 1, ¬ = 0; n = 2, ¬ = 0; n = 2, ¬ = 1; n = 3, ¬ = 0. The possible values of the total angular momentum of the last electron are j=¬+s=1±!= ! , *. The 3p electron is the only electron not in a filled subshell. The angular momentum of a filled subshell is zero, so the total angular momentum of the atom is the angular momentum of the 3p electron. When the beam passes through the magnetic field gradient, the deflecting force will be proportional to mj . If j = !, the values of mj are ± !, and there will be two lines. If j = *, the values of mj are ± ! , ± *, and there will be four lines. Thus the

number of lines shows the value of mj .

49. (a) For the 3p state, ¬ = 1 and s = !. Thus the values of j are j=¬±s=1±!= ! , *. The values of J are J = [j(j + 1)]1/2˙ = [(1/2)(3/2)]1/2˙ = !v3 ˙; 1/2 1/2 J = [j(j + 1)] ˙ = [(3/2)(5/2)] ˙ = !v15 ˙. (b) For the 4f state, ¬ = 3 and s = !. Thus the values of j are j=¬±s=3±!= 5/2, 7/2. The values of J are J = [j(j + 1)]1/2˙ = [(5/2)(7/2)]1/2˙ = !v35 ˙; 1/2 1/2 J = [j(j + 1)] ˙ = [(7/2)(9/2)] ˙ = !v63 ˙. (c) For the 4d state, ¬ = 2 and s = !. Thus the values of j are j=¬±s=2±!= 3/2, 5/2. The values of J are J = [j(j + 1)]1/2˙ = [(3/2)(5/2)]1/2˙ = !v15 ˙; 1/2 1/2 J = [j(j + 1)] ˙ = [(5/2)(7/2)] ˙ = !v35 ˙. 50. For the 5g state, ¬ = 4 and s = !. Thus the values of j are j=¬±s=4±!= 7/2, 9/2. For j = 7/2, we have mj = – 7/2, – 5/2, – 3/2, – 1/2, 1/2, 3/2, 5/2, 7/2; J = [j(j + 1)]1/2˙ = [(7/2)(9/2)]1/2˙ = *v7 ˙; Jz = mj˙ = – 7˙/2, – 5˙/2, – 3˙/2, – ˙/2, ˙/2, 3˙/2, 5˙/2, 7˙/2. For j = 9/2, we have mj = – 9/2, – 7/2, – 5/2, – 3/2, – 1/2, 1/2, 3/2, 5/2, 7/2, 9/2; J = [j(j + 1)]1/2˙ = [(9/2)(11/2)]1/2˙ = *v11 ˙; Jz = mj˙ = – 9˙/2, – 7˙/2, – 5˙/2, – 3˙/2, – ˙/2, ˙/2, 3˙/2, 5˙/2, 7˙/2, 9˙/2. 51. (a) The additional energy term for the spin-orbit interaction is U = – sBn = – gBmsBn . Thus the separation of energy levels is ?U = ?ms gBBn ; 0.4 T. (510–5 eV)(1.6010–19 J/eV) = [! – (– !)](2.0023)(9.2710–24 J/T)Bn , which gives Bn = (b) If we consider the nucleus to be a charge e revolving in a circle of radius r, the effective current is

Chapter 40 p. 12

I = e/(2pr/v) = ev/2pr = meevr/2pmer2 = eL/2pmer2, where L is the orbital angular momentum of the electron, because the nucleus has the same v and r. The electron is at the center of this circular current, so the magnetic field is B = 0I/2r = 0eL/4pmer3. If we use the Bohr quantization, L = n˙, and r = n2r0 , we have B = 0en˙/4pmen6r03 = (0/4p)e˙/men5r03 = (10–7 T · m/A)(1.6010–19 C)(1.05510–34 J · s)/(9.1110–31 kg)(2)5(0.52910–10 m)3 = 0.4 T. This is consistent with the result from part (a). 52. The energy of a pulse is E = P ?t = (0.65 W)(2510–3 s) = 0.016 J. The energy of a photon is hf = hc/ = (1.24103 eV · nm)/ = (1.24103 eV · nm)/(640 nm) = 1.94 eV. Thus the number of photons in a pulse is N = E/hf = (0.016 J)/(1.94 eV)(1.610–19 J/eV) = 5.21016 photons.

53. We find the angular half width of the beam from ? = 1.22/d = 1.22(69410–9 m)/(3.010–3 m) = 2.810–4 rad, 5.610–4 rad. so the angular width is  = The diameter of the beam when it reaches the satellite is D = r = (300103 m)(5.610–4 rad) = 1.7102 m. 54. In thermal equilibrium we find the fraction of atoms relative to those in E0 from N2 – 19 – 23 – 37 = e – E/ kT = e – (2 .2 eV)(1.60  10 J/ eV)/ (1.38  10 J/ K)(300 K) = 1.2  10 ; N0 N1 – 19 – 23 – 31 = e – E/ kT = e – (1 .8 eV)(1.60  10 J/ eV)/ (1.38  10 J/ K)(300 K) = 6.1  10 . N0 55. We find the temperature from N2 = e – E/ kT ; N0 1 = e – (2.2 eV)(1.60  10– 19 J/ eV)/ (1.38  10– 23 J/ K)T , wh i chgi ves T= 2

4

3.7  10 K.

56. If we assume the population follows the Boltzmann distribution, we have N 2 = e – E/ kT ; or – E = ln N 2/ N 1 . N1 kT If N2/N1 > 1, then ln (N2/N1) > 0. Thus T < 0. There is no contradiction because the system is not in thermal equilibrium. 57. (a) Boron has Z = 4, so the outermost electron has n = 2. We use the Bohr result with an effective Z: E2 = – (13.6 eV)(Zeff)2/n2; – 8.26 eV = – (13.6 eV)(Zeff)2/22, which gives Zeff = 1.56. Note that this indicates some shielding by the second electron in the n = 2 shell. (b) We find the average radius from r = n2r1/Zeff = 22(0.52910–10 m)/(1.56) = 1.410–10 m. 58. From spdfg, we see that the “g” subshell has ¬ = 4, so the number of electrons is N = 2(2¬ + 1) = 2[2(4) + 1] = 18 electrons.

Chapter 40 p. 13

59. (a) For Z = 27 we start with hydrogen and fill the levels as indicated in the periodic table: 1s22s22p63s23p63d74s2. Note that the 4s2 level is filled before the 3d level is completed. (b) For Z = 36 we have 1s22s22p63s23p63d104s24p6. (c) For Z = 38 we have 1s22s22p63s23p63d104s24p65s2. Note that the 5s2 level is filled before the 4d level is started. 60. The value of ¬ can range from 0 to n – 1. Thus for n = 5, we have ¬ = 4. The largest magnitude of L is L = ˙[¬(¬ + 1)]1/2 = (1.05510–34 J · s)[(4)(4 + 1)]1/2 = 4.7210–34 kg · m2/s. The smallest value of L is 0.

61. (a) We find the quantum number for the orbital angular momentum from L = MearthvR = Mearth2pR2/T = ˙[¬(¬ + 1)]1/2; (5.981024 kg)2p(1.501011 m)2/(3.16107 s) = (1.05510–34 J · s)[¬(¬ + 1)]1/2, which gives ¬ = 2.51074. (b) The value of m¬ can range from – ¬ to + ¬, or 2¬ + 1 values, so the number of orientations is N = 2¬ + 1 = 2(2.51074) + 1 = 5.01074. 62. The wavelength of the photon emitted for the transition from n to n is 1/ = (Z2e4m/8Å02h3c)[(1/n2) – (1/ n2)] = RZ2[(1/n2) – (1/ n2)]. The K line is from the n = 2 to n = 1 transition, and the other n = 1 electron shields the nucleus, so the effective Z is Z – 1: 1/ = RZeff2[(1/n2) – (1/ n2)] = R(Z – 1)2[(1/12) – (1/22)] = &R(Z – 1)2. Thus we have 1/v = (&R)1/2(Z – 1), which is the equation of a straight line, as in the Moseley plot, with b = 1. The value of a is a = (&R)1/2 = [&(1.0974107 m–1)]1/2 = 2869 m–1/2. 63. The n = 2, ¬ = 0 wave function is 1 2 – rr e – r/ 2r 0. 200 = 0 32r30 The radial probability density is  r2 2 2 Pr (r) = 4r 2 200 = 3 2 – rr e – r/ r 0. 0 8r0 To find the most probable distance from the nucleus, we find the maxima by setting the first derivative equal to zero. If we change variable to x = r/r0 , we have 1 2 2 Pr (x) = x 2 – x e – x. 8r 0 If we suppress the constant in front, the derivative is d Pr 2 2 = 2x 2 – x e – x – 2x 2 2 – x e – x – x 2 2 – x e– x dx = x 2 – x x 2 – 6x + 4 e– x = 0.

Chapter 40 p. 14

Thus we have x = 0, which is the minimum at r = 0; 2 – x = 0. Because this is a factor in , this is the other minimum at x = 2. x2 – 6x + 4 = 0, with solutions x = 0.76, 5.24. These correspond to the two peaks in the distribution. The highest peak is the most probable distance: r = 5.24r0 .

64. The magnitude of the angular momentum is given by L = [¬(¬ + 1)]1/2˙, and the z-component is given by Lz = m¬˙. Thus the angle between L and the z-axis is given by cos  = Lz/L = m¬˙/[¬(¬ + 1)]1/2˙ = m¬/[¬(¬ + 1)]1/2. For each ¬ the value of m¬ can range from – ¬ to + ¬, so there are 2¬ + 1 values for . (a) For ¬ = 1 the magnitude of L is L = [(1)(1 + 1)]1/2˙ = (2)1/2˙. The angles for the 3 values of m¬ are cos 1,3 = (± 1)/(2)1/2 = ± 1/(2)1/2, or 1 = 45°, 3 = 135°. cos 2 = (0)/(2)1/2 = 0, or 2 = 90°. Thus the possible values for ¬ = 1 are 45°, 90°, 135°. (b) For ¬ = 2 the magnitude of L is x L = [(2)(2 + 1)]1/2˙ = (6)1/2˙. The angles for the 5 values of m¬ are cos 1,5 = (± 2)/(6)1/2 = ± (2/3)1/2, or 1 = 35.3°, 5 = 144.7°; cos 2,4 = (± 1)/(6)1/2 = ± (1/6)1/2, or 2 = 65.9°, 4 = 114.1°; cos 3 = (0)/(6)1/2 = 0, or 3 = 90°. Thus the possible values for ¬ = 2 are 35.3°, 65.9°, 90°, 114.1°, 144.7°. (c) For ¬ = 3 the magnitude of L is L = [(3)(3 + 1)]1/2˙ = (12)1/2˙. The angles for the 7 values of m¬ are cos 1,7 = (± 3)/(12)1/2 = ± (3/4)1/2, or 1 = 30.0°, 7 = 150.0°; cos 2,6 = (± 2)/(12)1/2 = ± (1/3)1/2, or 2 = 54.7°, 6 = 125.3°; cos 3,5 = (± 1)/(12)1/2 = ± (1/12)1/2, or 3 = 73.2°, 5 = 106.8°; cos 4 = (0)/(12)1/2 = 0, or 4 = 90°.

z

Lz

L



y

Chapter 40 p. 15

Thus the possible values for ¬ = 3 are 30.0°, 54.7°, 73.2°, 90°, 106.8°, 125.3°, 150.0°. (d) The minimum value of  has the largest value of m¬ . For ¬ = 100, we have cos min = (100)/[(100)(100 + 1)]1/2 = 0.995, or min = 5.71°. 6 For ¬ = 10 , we have 0.0573°. cos min = (106)/[(106)(106 + 1)]1/2 = 1/(1 + 10–6)1/2 ˜ 1 – !(10–6) = 0.9999995, or min = This is consistent with the correspondence principle; L could align with the z-axis when n  8 65. (a) From ?Lz ? = ˙, we see that if ?Lz = 0, ?  8. Thus the angle  is unknown. (b) If  is unknown, then the direction of the component of L perpendicular to the z-axis is unknown, so Lx and Ly are unknown. (c) From L2 = Lx2 + Ly2 + Lz2, we have ¬(¬ + 1)˙2 = Lx2 + Ly2 + m¬2˙2, or (Lx2 + Ly2)1/2 = [¬(¬ + 1) – m¬2]1/2˙. 66. We find the uncertainty in the momentum component perpendicular to the motion, when the width of the beam is constrained to a dimension a: ?py = ˙/?y = ˙/a. The half angle of the beam is given by the direction of the velocity (or momentum): sin  = py/p. We assume that the angle is small: sin  ˜ , and we take the minimum uncertainty to be the perpendicular momentum, so we have  ˜ py/p = (˙/a)/(h/) = /2pa. The angular spread is 2 ˜ /pa ˜ /a. 67. (a) The additional energy term in a magnetic field is U = – sB = – gBmsB. Thus the separation of energy levels is ?U = ?ms gBB = [! – (– !)](2.0023)(9.2710–24 J/T)(1.0 T) = 1.910–23 J = 1.210–4 eV. (b) We find the wavelength from  = (1.24103 eV · nm)/?U = (1.24103 eV · nm)/(1.210–4 eV) = 1.07107 nm = 1.1 cm. (c) From the periodic table we see that the participating electron is in the 5s1 state. Thus the splitting for both atoms is of a single s-state electron, so there will be no difference. 68. We find the angular half width of the beam from ? = 1.22/d = 1.22(69410–9 m)/(4.010–3 m) = 2.110–4 rad, 4.210–4 rad. so the angular spread is  = (a) The diameter of the beam when it reaches the satellite is D = r = (1000103 m)(4.210–4 rad) = 4.2102 m. (b) The diameter of the beam when it reaches the Moon is D = r = (3.84108 m)(4.210–4 rad) = 1.6105 m = 160 km. 69. (a) Because the Boltzmann factor compares two single energy levels, to find the fraction in the 8 n = 2 levels relative to the 2 n = 1 levels, we multiply be 8/2: N 2 8 – E/ kT – 19 – 23 = e = 4e – [(– 3.4 eV) – (– 13.6 eV)](1.60  10 J/ eV)/ (1.38  10 J/ K)(300 K) = 3  10 – 171. N1 2 The n = 3 level has ¬ = 0, 1, 2, so the number of states is 2 + 6 + 10 = 18. Thus we have N 3 18 – E/ kT – 19 – 23 = e = 9e – [(– 1.5 eV) – (– 13.6 eV)](1.60  10 J/ eV)/ (1.38  10 J/ K)(300 K) = 7  10 – 203. N1 2 (b) At the new temperature we have

Chapter 40 p. 16

N 2 8 – E/ kT – 19 – 23 = e = 4e – [(– 3.4 eV) – (– 13.6 eV)](1.60  10 J/ eV)/ (1.38  10 J/ K)(6000 K) = N1 2 N 3 18 – E/ kT – 19 – 23 = e = 9e – [(– 1.5 eV) – (– 13.6 eV)](1.60  10 J/ eV)/ (1.38  10 J/ K)(6000 K) = 2 N1 (c) We find the number of hydrogen atoms in 1.0 g from N = (1.0 g)(6.021023 atoms/mol)/(1.0 g/mol) = 6.021023 atoms. Because the fraction in the excited states is very small, we use this for N1 : N2 = (1.110–8)(6.021023) = 6.61015; –10 23 N3 = (6.310 )(6.0210 ) = 3.81014. (d) If the lifetime of the excited state is 10–8 s, the rate at which photons are emitted is n2 = N2/ = (6.61015)/(10–8 s) ˜ 71023 photons/s; n3 = N3/ = (3.81014)/(10–8 s) ˜ 41022 photons/s.

1.1  10

–8

6.3  10

;

– 10

.

70. (a) The energy levels for a rigid box are given by En = n2h2/8mL2. There are two electrons in each level, corresponding to the two different spin states. With N electrons, the quantum number of the highest filled state is nhighest = N/2. At T = 0 K, the highest energy of an electron in the ground state is the energy of the highest filled state: EF = N2h2/32mL2. (b) The smallest energy absorption will raise an electron to the next unfilled state: ?E = Enhighest + 1 – Enhighest = {[(N/2) + 1]2 – (N/2)2}h2/8mL2 = (N + 1)h2/8mL2 ˜ Nh2/8mL2 =

4EF/N.

71. Each shell with quantum number n can contain 2n2 electrons. Thus the maximum number of electrons in the shells from n = 1 to n = 6 is

N =

6

 2n 2 = 2 16 n n + 1 n= 1

2n + 1 = 1 6 6 + 1 12 + 1 = 182. 3 Because each electron corresponds to one proton in an atom, there would be a maximum of elements. 72. The binding energy of the state is – En = (13.6 eV)/n2 = (13.6 eV)/(50)2 = 5.410–3 eV. The radius of the orbit is r = n2r0 = (50)2(0.52910–10 m) = 1.3210–7 m. The effective cross-section is  = pr2 = p(1.3210–7 m)2 = 5.510–14 m2. 73. The additional energy term in a magnetic field is U = – sB = – gBmsB. For resonance the separation of energy levels is the energy in the photon: hf = hc/ = ?ms gBB; [(1.24103 eV · nm)/(2.0 cm)(107 nm/cm)](1.6010–19 J/eV) = [! – (– !)]g(9.2710–24 J/T)(0.476 T), which gives g =

2.25.

182

Chapter 41

p. 1

CHAPTER 41 – Molecules and Solids Note:

At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.6310–34 J · s)(3.00108 m/s)(109 nm/m)/(1.6010–19 J/eV); E = (1.24103 eV · nm)/. A factor that appears in the analysis of electron energies is e2/4pÅ0 = (1.6010–19 C)2/4p(8.8510–12 C2/N · m2) = 2.3010–28 J · m.

1.

With the reference level at infinity, the binding energy of the two ions is Binding energy = – U = e2/4pÅ0r = (2.3010–28 J · m)/(0.2810–9 m) = 8.2110–19 J =

5.1 eV.

2.

We convert the units: 1 kcal/mol = (1 kcal/mol)(4186 J/kcal)/(6.021023 molecules/mol)(1.6010–19 J/eV) = 0.0435 eV/molecule. For KCl we have (4.43 eV/molecule)[(1 kcal/mol)/(0.0435 eV/molecule)] = 102 kcal/mol.

3.

With the repulsion of the electron clouds, the binding energy is Binding energy = – U – Uclouds ; 4.43 eV = 5.1 eV – Uclouds , which gives Uclouds = 0.7 eV.

4.

When the electrons are midway between the protons, each electron will have a potential energy due to the two protons: Uep = – (2)(0.33)e2/4pÅ0(r/2) = – (4)(0.33)(2.3010–28 J · m)/(0.07410–9 m)(1.6010–19 J/eV) = – 25.6 eV. The protons have a potential energy: Upp = + e2/4pÅ0r = + (2.3010–28 J · m)/(0.07410–9 m)(1.6010–19 J/eV) = + 19.4 eV. When the bond breaks, each hydrogen atom will be in the ground state with an energy E1 = – 13.6 eV. Thus the binding energy is Binding energy = 2E1 – (2Uep + Upp) = 2(–13.6 eV) – [2(– 25.6 eV) + 19.4 eV] = 4.6 eV.

5.

The neutral He atom has two electrons in the ground state, n = 1, ¬ = 0, m¬ = 0. Thus the two electrons have opposite spins, ms = ± !. If we try to form a covalent bond for the two atoms, we see that an electron from one of the atoms will have the same quantum numbers as one of the electrons on the other atom. From the exclusion principle, this is not allowed, so the electrons cannot be shared. We consider the He2+ molecular ion to be formed from a neutral He atom and an He+ ion. If the electron on the ion has a certain spin value, it is possible for one of the electrons on the atom to have the opposite spin. Thus the electron can be in the same spatial region as the electron on the ion, so a bond can be formed.

6.

The units of ˙2/I are (J · s)2/(kg · m2) = J2/(kg · m/s2)m = J2/(N · m) = J2/J = J.

Chapter 41

p. 2

7.

The reduced mass of the molecule is  = m1m2/(m1 + m2). (a) Using data from the periodic table, for NaCl we have  = mNamCl/(mNa + mCl) = (22.9898 u)(35.4527 u)/(22.9898 u + 35.4527 u) = 13.941 u. (b) For N2 we have  = mNmN/(mN + mN) = (14.0067 u)(14.0067 u)/(14.0067 u + 14.0067 u) = 7.0034 u. (c) For HCl we have  = mHmCl/(mH + mCl) = (1.00794 u)(35.4527 u)/(1.00794 u + 35.4527 u) = 0.9801 u.

8.

The reduced mass of the N2 molecule is N = mNmN/(mN + mN) = (14.0067 u)(14.0067 u)/(14.0067 u + 14.0067 u) = 7.0034 u. We find the bond length from ˙2/2I = ˙2/2Nr2 (2.4810–4 eV)(1.6010–19 J/eV) = (1.05510–34 J · s)2/2(7.0034 u)(1.6610–27 kg/u)r2, 1.1010–10 m. which gives r =

9.

(a) The reduced mass of the CO molecule is  = mCmO/(mC + mO) = (12.0 u)(16.0 u)/(12.0 u + 16.0 u) = 6.86 u. (b) We find the effective spring constant from f = (k/)1/2/2p; 6.421013 Hz = [k/(6.86 u)(1.6610–27 kg/u)]1/2/2p, which gives k = This is 3.4 the constant for H2.

1.85103 N/m.

10. The total energy change for the absorption or emission is ?E = ?Evib + ?Erot = hf + ?Erot . If ?E = hf, we have ?Erot = 0. When L changes by ?L, we get ?Erot = (˙2/2I)[(L + ?L)(L + ?L + 1) – L(L + 1)] = (˙2/2I )(2L + 1 + ?L) ?L. Thus ?Erot = 0 requires ?L = 0, which is a violation of the selection rule ?L = ± 1.

mO mO 11. (a) The moment of inertia of O2 about its CM is CM 2 2 I = 2mO(r/2) = mOr /2.  We find the characteristic rotational energy from r ˙2/2I = ˙2/mOr2 = (1.05510–34 J · s)2/(16.0 u)(1.6610–27 kg/u)(0.12110–9 m)2 = 2.8610–23 J = 1.7910–4 eV. (b) The rotational energy is Erot = L(L + 1)(˙2/2I). Thus the energy of the emitted photon from the L = 2 to L = 1 transition is hf = ?Erot = [(2)(2 + 1) – (1)(1 + 1)](˙2/2I) = 4(˙2/2I) = 4(1.7910–4 eV) = 7.1610–4 eV. The wavelength is  = c/f = hc/hf = (1.24103 eV · nm)/(7.1610–4 eV) = 1.73106 nm = 1.73 mm.

Chapter 41

p. 3

12. The moment of inertia of H2 about its CM is mH I = 2mH(r/2)2 = mHr2/2. We find the characteristic rotational energy from ˙2/2I = ˙2/mHr2 = (1.05510–34 J · s)2/(1.6710–27 kg)(0.07410–9 m)2 = 1.2210–21 J = 7.6110–3 eV. The rotational energy is Erot = L(L + 1)(˙2/2I). Thus the energy of the emitted photon from the L to L – 1 transition is hf = ?Erot = [(L)(L + 1) – (L – 1)(L)](˙2/2I) = 2L(˙2/2I). (a) For the L = 1 to L = 0 transition, we get hf = 2L(˙2/2I) = 2(1)(7.6110–3 eV) = 1.5210–2 eV. The wavelength is  = c/f = hc/hf = (1.24103 eV · nm)/(1.5210–2 eV) = 8.16104 nm = (b) For the L = 2 to L = 1 transition, we get hf = 2L(˙2/2I) = 2(2)(7.6110–3 eV) = 3.0410–2 eV. The wavelength is  = c/f = hc/hf = (1.24103 eV · nm)/(3.0410–2 eV) = 4.08104 nm = (c) For the L = 3 to L = 2 transition, we get hf = 2L(˙2/2I) = 2(3)(7.6110–3 eV) = 4.5610–2 eV. The wavelength is  = c/f = hc/hf = (1.24103 eV · nm)/(4.5610–2 eV) = 2.72104 nm =

CM



r

0.082 mm.

0.041 mm.

0.027 mm.

13. We find the energies for the transitions from ?E = hf = hc/ = (1.24103 eV · nm)/: ?E1 = (1.24103 eV · nm)/(23.1106 nm) = 5.3710–5 eV; ?E2 = (1.24103 eV · nm)/(11.6106 nm) = 10.710–5 eV; ?E3 = (1.24103 eV · nm)/(7.71106 nm) = 16.110–5 eV. The rotational energy is Erot = L(L + 1)(˙2/2I). Thus the energy of the emitted photon from the L to L – 1 transition is hf = ?Erot = [(L)(L + 1) – (L – 1)(L)](˙2/2I) = 2L(˙2/2I). Because ?E3 = 3 ?E1 , and ?E2 = 2 ?E1 , the three transitions must be from the L = 1, 2, and 3 states. We find the moment of inertia about the center of mass from ?E3 = (16.110–5 eV)(1.6010–19 J/eV) = 2(3)(1.05510–34 J · s)2/2I, which gives I = 1.2910–45 kg · m2. The reduced mass of the NaCl molecule is  = mNamCl/(mNa + mCl) = (22.9898 u)(35.4527 u)/(22.9898 u + 35.4527 u) = 13.946 u. We find the bond length from I = r2; 2.3610–10 m. 1.2910–45 kg · m2 = (13.946 u)(1.6610–27 kg/u)r2, which gives r =

mH

Chapter 41

p. 4

14. The vibrational energy levels are Evib = ( + !)hf. The rotational energy levels are Erot = L(L + 1)(˙2/2I). For an absorption from level , L to  + 1, L + 1, the absorbed energy is ?E = ?Evib + ?Erot = [( + 1 + !) – ( + !)]hf + [(L + 1)(L + 2) – (L)(L + 1)](˙2/2I) = hf + 2(L + 1)(˙2/2I) = hf + (L + 1)(˙2/I), L = 0, 1, 2, … . For an absorption from level , L to  + 1, L – 1, the absorbed energy is ?E = ?Evib + ?Erot = [( + 1 + !) – ( + !)]hf + [(L – 1)(L) – (L)(L + 1)](˙2/2I) = hf – 2L(˙2/2I) = hf – L(˙2/I), L = 1, 2, 3, … . 15. The ionic cohesive energy is U0 = – (e2/4pÅ0r)[1 – (1/m)] = – [(1.75)(2.3010–28 J · m)/(0.2810–9 m)(1.6010–19 J/eV)][1 – (1/8)] = 16. From the figure we see that the distance between nearest neighbor Na ions is the diagonal of the cube: D = dv2 = (0.24 nm)v2 = 0.34 nm.

Na

+

– 7.9 eV.

Cl

– d

Cl

–

+

Na

17. Because each ion occupies a cell of side s, a molecule occupies two cells. Thus the density is  = mNaCl/2s3; 2.165103 kg/m3 = (58.44 u)(1.6610–27 kg/u)/2s3, which gives s = 2.8310–10 m = 0.283 nm. 18. Because each ion occupies a cell of side s, a molecule occupies two cells. Thus the density is  = mKCl/2s3; 1.99103 kg/m3 = (39.1 u + 35.5 u)(1.6610–27 kg/u)/2s3, which gives s = 3.1510–10 m =

0.315 nm.

Chapter 41

p. 5

19. (a) The potential energy is 2 U = –  e + B . 4Å0r rm The separation at equilibrium occurs at the minimum in the energy. We can relate this distance r0 to the value of B by setting dU/dr = 0: 2 e 2 = mB , which gives B = e 2r0m – 1 . dU =  e – mB = 0, or +1 dr 4Å0 r2 r m + 1 4Å0 r02 r m 4Å0m 0 The ionic cohesive energy is the value of U at the equilibrium distance: 2 2 e 2r0m – 1  U0 = –  e + = – e 1 – 1 . m m 4Å0r 0 4Å0 mr 0 4Å0 r0 (b) For NaI we have – 28 (1.75)(2.30  10 J  m) U0 = – 1– 1 = – 6.9 eV. – 19 –9 10 (1.60  10 J/ eV)(0.33  10 m ) (c) If we assume the same value for the Madelung constant, for MgO we have – 28 (1.75)(2.30  10 J  m) U0 = – 1– 1 = – 10.8 eV. – 19 –9 10 (1.60  10 J/ eV)(0.21  10 m ) (d) If we use a new value for m, the fractional change is 1– 1 – 1 – 1 U 0 8 10 = – 0.028 = = – 3%. U0 1– 1 10 20. If we select a charge in the middle + – + – + of the chain, there will be two r charges of opposite sign a distance r away, two charges of the same sign a distance 2r away, etc. The potential energy of the charge is 2 2 2 2 2 U = – 2e + 2e – 2e + 2e – … = – 2e 1 – 1 + 1 – 1 + … . 2 3 4 4Å0r 4Å0(2r) 4Å0 (3r) 4Å0(4r) 4Å0 r If we consider the expansion of ln (1 + x), which is 2 3 4 ln 1 + x = x – x + x – x + … , 2 3 4 we see that ln 1 + 1 = 1 – 1 + 1 – 1 + … . 2 3 4 Thus we have 2 U = – 2e ln 1 + 1 , so  = 2 ln 2. 4Å0r

–

21. The density of energy states in a small energy interval is dn = g(E) dE. Because the energy range is small we assume a constant density at the middle of the range:

+

–

Chapter 41

p. 6

V8 2m 3/ 2 1/ 2 E E 3 h –6 – 31 1.0  10 m 3 8 2 9.11  10 kg

N = V g(E) E = = =

3/ 2

6.95 eV

6.63  10

– 34

J s

1/ 2

0.10 eV 1.60  10

– 19

J/ eV

3/ 2

3

1.8  10 21.

22. We find the density of molecules in an ideal gas from PV = nRT = NkT, or N/V = P/kT = (1.013105 Pa)/(1.3810–23 J/K)(300 K) = 2.41025 m–3. If we assume each copper atom contributes one free electron, the density of free electrons is n = NA/M = (8.89103 kg/m3)(6.021023/mol)/(63.510–3 kg/mol) = 8.41028 m–3. Thus the ratio is (N/V)/n = (2.41025 m–3)/(8.41028 m–3) = 310–4. 23. (a) The value of kT is kT = (1.3810–23 J/K)(300 K)/(1.6010–19 J/eV) = 0.0259 eV. We find the energy from f= 1/ e (E – EF)/ kT + 1 ;

0.90 = 1/ e (E – 7.0 eV)/ (0.0259 eV) + 1 , wh i chgi ves E = 6.9 eV. (b) The value of kT is kT = (1.3810–23 J/K)(1200 K)/(1.6010–19 J/eV) = 0.1035 eV. We find the energy from f= 1/ e (E – EF)/ kT + 1 ; 0.90 = 1/ e (E – 7.0 eV)/ (0.1035 eV) + 1 , wh i chgi ves E =

6.8 eV.

24. (a) The value of kT is kT = (1.3810–23 J/K)(300 K)/(1.6010–19 J/eV) = 0.0259 eV. We find the energy from f= 1/ e (E – EF)/ kT + 1 ;

0.10 = 1/ e (E – 7.0 eV)/ (0.0259 eV) + 1 , wh i chgi ves E = 7.1 eV. (b) The value of kT is kT = (1.3810–23 J/K)(1200 K)/(1.6010–19 J/eV) = 0.1035 eV. We find the energy from f= 1/ e (E – EF)/ kT + 1 ; 0.10 = 1/ e (E – 7.0 eV)/ (0.1035 eV) + 1 , wh i chgi ves E =

7.2 eV.

25. The value of kT is kT = (1.3810–23 J/K)(300 K)/(1.6010–19 J/eV) = 0.0259 eV. We find the occupancy probability from f= 1/ e (E – EF)/ kT + 1 ;

= 1/ e (1.010 – 1.000)(7.0 eV)/ (0.0259 eV) + 1 = 0.063 = 26. The density of energy states in a small energy interval is

6.3%.

Chapter 41

p. 7

dn = g(E) dE. Because the energy range is small we assume a constant density at the middle of the range: V8 2m 3/ 2 1/ 2 N = V g(E) E = E E h3 –6

=

1.0  10

– 31

m 3 8 2 9.11  10

kg

3/ 2

(0.995)(5.48 eV)

1/ 2

6.63  10 – 34 J  s

3

0.01)(5.48 eV 1.60  10

– 19

J/ eV

20

8.7  10 .

=

27. Because each sodium atom contributes one conduction electron, the density of conduction electrons is n = NA/M = (0.97103 kg/m3)(6.021023/mol)/(23.010–3 kg/mol) = 2.5391028 m–3. We find the Fermi energy from

h 2 3n 2/ 3 8m  2 2/ 3 6.63  10 – 34 J  s 3 2.539  10 28 m – 3 = = 5.05  10 – 19 J = 3.2 eV.  8 9.11  10 – 31 kg We find the Fermi speed as the speed which gives a kinetic energy equal to the Fermi energy: EF = !mvF2; EF =

5.0510–18 J = !(9.1110–31 kg)vF2, which gives vF =

1.05106 m/s.

28. (a) Because each zinc atom contributes two free electrons, the density of free electrons is twice the density of atoms: n = 2NA/M = 2(7.1103 kg/m3)(6.021023/mol)/(65.3810–3 kg/mol) = 1.311029 m–3. (b) We find the Fermi energy from 2 h 3n 2/ 3 EF = 8m  2

2/ 3

6.63  10– 34 J  s 3 1.31  10 29 m – 3 – 18 = 1.50  10 J= 9.4 eV.  8 9.11  10 – 31 kg (c) We find the Fermi speed as the speed which gives a kinetic energy equal to the Fermi energy: EF = !mvF2; =

1.5010–18 J = !(9.1110–31 kg)vF2, which gives vF =

1.81106 m/s.

29. (a) We find the density of free electrons from 2 h 3n 2/ 3 EF = ; 8m  2 6.63  10 – 34 J  s 3n 2/ 3 – 19 29 11.63 eV 1.60  10 J/ eV = , wh i chgi ves n = 1.79  10 m 3. – 31 8 9.11  10 kg  (b) If we let v be the valence, we can relate the density of free electrons to the atomic density: n = vNA/M; 1.791029 m–3 = v(2.70103 kg/m3)(6.021023/mol)/(27.010–3 kg/mol), which gives v = 3. 30. We find the density of neutrons from n = MS/mnV = 2(2.01030 kg)/(1.6710–27 kg))p(10103 m)3 = 5.721044 m–3. We find the Fermi energy from

3/ 2

Chapter 41

p. 8

2

h 3n 2/ 3 8m  2 2/ 3 6.63  10– 34 J  s 3 5.72  10 44 m – 3 – 11 = = 2.18  10 J=  8 1.67  10 – 27 kg

EF =

2

1.4  10 MeV.

31. The maximum energy of an electron at T = 0 K is EF. All states above EF are unoccupied, while all states with E = EF are occupied. We find the average energy of an electron from



E tot E= = N

EF

0

En o (E) d E

EF

0

n o(E) d E

V 8 2 m 3/ 2/ h

3

0

= V 8 2 m 3/ 2/ h

EF

3

EF

0

E 3/ 2 d E

5/ 2

= E

1/ 2

dE

(2/ 5)E F

3/ 2

(2/ 3)E F

= 3 E F. 5

32. The Fermi–Dirac distribution gives the probability that a state is occupied. If E = EF , we have f= (E – E 1)/ kT = 01 = 1. 2 F e +1 e +1 33. We find the occupation probability from f= 1/ e (E –EF)/ kT + 1 . The value of kT is kT = (1.3810–23 J/K)(300 K)/(1.610–19 J/eV) = 0.0259 eV. (a) For E – EF = 0.10 eV, we have f = 1/[e (0.10 eV)/(0.0259 eV) + 1] = 1/(e 3.86 + 1) = 0.021. This is reasonable. For a good conductor, higher states should have low occupancy. (b) For E – EF = – 0.10 eV, we have f = 1/[e (– 0.10 eV)/(0.0259 eV) + 1] = 1/(e – 3.86 + 1) = 0.979. (c) Because a state must be either occupied or unoccupied, we have Punoccupied = 1 – f = 1 – 0.979 = 0.021. Note that this is the same as the occupation probability for a state 0.10 eV above the Fermi energy. 34. The energy levels for a one-dimensional potential well are E = (h2/8mL2)n2, n = 1, 2, 3, … . The quantum number n also gives the number of levels with energies between 0 and E. n = (8mL2/h2)1/2E1/2. The number of levels with energies between E and dE is dn = !(8mL2/h2)1/2E–1/2 dE. Because there can be two electrons with opposite spins in each state, the number of electron states per unit energy interval is gL(E) = 2 dn/dE = (8mL2/h2E)1/2. 35. If we consider the cube to be a three-dimensional infinite well, we can apply the boundary conditions separately to each dimension. Each dimension gives a quantum number which we label n1 , n2 , n3 . Thus there is a contribution to the energy from each dimension equal to the energy from the onedimensional well: E1 = (h2/8mL2)n12; E2 = (h2/8mL2)n22; E3 = (h2/8mL2)n32, n1 , n2 , n3 = 1, 2, 3, … .

Chapter 41

p. 9

The energy of a state specified by the three quantum numbers is n3 E = (h2/8mL2)(n12 + n22 + n32). If we create a three-dimensional space with the axes labeled by n1 , n2 , n3 , each state corresponds to a point in the lattice. When we construct a sphere of radius R, where R2 = n12 + n22 + n32, each point within the octant corresponding to positive values for n1 , n2 , n3 represents a state with energy between 0 and E = h2R2/8mL2. Because the density of points is one and there can be two electrons in each state, the number of electrons with energy between 0 and E is n1 N = 2(1/8)()pR3) = pR3/3 = (p/3)(8mL2E/h2)3/2. We find the density of states with energy E from g(E) = dn/dE = (1/V) dN/dE = *(1/L3)(p/3)(8mL2/h2)3/2E1/2 = (8pv2 m3/2/h3)E1/2.

R n2

36. The photon with the minimum frequency for conduction must have an energy equal to the energy gap: Eg = hf = hc/ = (1.24103 eV · nm)/(640 nm) = 1.94 eV.

37. The partially filled shell in Na is the 3s shell, which has 1 electron in it. The partially filled shell in Cl is the 2p shell which has 5 electrons in it. In NaCl the electron from the 3s shell in Na is transferred to the 2p shell in Cl, which results in filled shells for both ions. Thus when many ions are considered, the resulting bands are either completely filled (the valence band) or completely empty (the conduction band). Thus a large energy is required to create a conduction electron by raising an electron from the valence band to the conduction band. 38. (a) In the 2s shell of an atom, ¬ = 0, so there are two states: ms = ± !. When N atoms form bands, each atom provides 2 states, so the total number of states in the band is 2N. (b) In the 2p shell of an atom, ¬ = 1, so there are three states from the m¬ values: m¬ = 0, ± 1; each of which has two states from the ms values: ms = ± !, for a total of 6 states. When N atoms form bands, each atom provides 6 states, so the total number of states in the band is 6N. (c) In the 3p shell of an atom, ¬ = 1, so there are three states from the m¬ values: m¬ = 0, ± 1; each of which has two states from the ms values: ms = ± !, for a total of 6 states. When N atoms form bands, each atom provides 6 states, so the total number of states in the band is 6N. (d) In general, for a value of ¬, there are 2¬ + 1 states from the m¬ values: m¬ = 0, ± 1, … , ± ¬. For each of these there are two states from the ms values: ms = ± !, for a total of 2(2¬ + 1) states. When N atoms form bands, each atom provides 2(2¬ + 1) states, so the total number of states in the band is 2N(2¬ + 1). 39. The photon with the longest wavelength or minimum frequency for conduction must have an energy equal to the energy gap:  = c/f = hc/hf = hc/Eg = (1.24103 eV · nm)/(1.1 eV) = 1.1103 nm = 1.1 m. 40. The minimum energy provided to an electron must be equal to the energy gap: Eg = 0.72 eV. Thus the maximum number of electrons is N = hf/Eg = (710103 eV)/(0.72 eV) = 9.9105. 41. If we consider a mole of pure silicon (28 g or 6.021023 atoms), the number of conduction electrons is

Chapter 41

p. 10

NSi = [(2810–3 kg)/(2330 kg/m3)](1016 electrons/m3) = 1.201011 conduction electrons. The additional conduction electrons provided by the doping is Ndoping = (6.021023 atoms)/106 = 6.021017 added conduction electrons. Thus the density of conduction electrons has increased by Ndoping/NSi = (6.021017)/(1.201011) = 5106. 42. The photon will have an energy equal to the energy gap:  = c/f = hc/hf = hc/Eg = (1.24103 eV · nm)/(1.4 eV) = 8.9102 nm =

0.89 m.

43. The photon will have an energy equal to the energy gap: Eg = hf = hc/ = (1.24103 eV · nm)/(650 nm) = 1.91 eV. 44. From the current-voltage characteristic, we see that a current of 12 mA means a voltage of 0.7 V across the diode. Thus the battery voltage is Vbattery = Vdiode + VR = 0.7 V + (1210–3 A)(760 ) = 9.8 V.

45. The battery voltage is Vbattery = Vdiode + VR ; 2.0 V = V(I) + I(0.100 k), or V(I) = 2.0 V – I(0.100 k). This is a straight line which passes through the points (20 mA, 0 V) and (12 mA, 0.8 V), as drawn in the figure. Because V(I) is represented by both curves, the intersection will give the current, which we see is 13 mA.

I (mA) 30 20 10

0

46.

R (k) 5 4 3 2 1 0

10

20 I (mA)

30

IR

Idiode

0.6 0.2 0.4 V (volts)

0.8

Chapter 41

p. 11

47. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. Thus the average current is Iav = !Vrms/R = !(120 V)/(28 k) = 2.1 mA. (b) For a full-wave rectifier without a capacitor, the current is positive all the time. Thus the average current is Iav = Vrms/R = (120 V)/(28 k) = 4.3 mA. 48. There will be a current in the resistor while the ac voltage varies from 0.6 V to 9.0 V rms. Because the 0.6 V is small, the voltage across the resistor will be almost sinusoidal, so the rms voltage across the resistor will be close to 9.0 V – 0.6 V = 8.4 V. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. If we ignore the short time it takes to reach 0.6 V, this will also be true for the resistor. Thus the average current is Iav = !Vrms/R = !(8.4 V)/(0.150 k) = 28 mA. (b) For a full-wave rectifier without a capacitor, the current is positive all the time. If we ignore the short time it takes to reach 0.6 V, this will also be true for the resistor. Thus the average current is Iav = Vrms/R = (8.4 V)/(0.150 k) = 56 mA.

49. (a) The time constant for the circuit is 1 = RC1 = (18103 )(2510–6 F) = 0.45 s. Because there are two peaks per cycle, the period of the rectified voltage is T = 1/2f = 1/2(60 Hz) = 0.0083 s. Because 1 » T, the voltage across the capacitor will be essentially constant during a cycle, so the voltage will be the peak voltage. Thus the average current is Iav = V0/R = v2(120 V)/(18 k) = 9.4 mA (smooth). (b) The time constant for the circuit is 2 = RC2 = (18103 )(0.1010–6 F) = 0.0018 s. Because 2 < T, the voltage across the capacitor will be rippled, so the average voltage will be close to the rms voltage. Thus the average current is Iav = Vrms/R = (120 V)/(18 k) = 6.7 mA (rippled).

50. The output voltage is the voltage across the resistor: V = iCRC = iBRC = (80)(2.010–6 A)(3.3103 ) =

0.53 V.

51. The output voltage is the voltage across the resistor: V = iCRC = iBRC ; 0.40 V = (100)(1.010–6 A)RC , which gives RC = 4.0103  = 52. (a) The output voltage is the voltage across RC : VC = iCRC , while the input voltage is the voltage across RB : VB = iBRB . Thus the voltage gain is VC/VB = iCRC/iBRB = RC/RB = (70)(6.8 k)/(3.2 k) = (b) The power amplification is

4.0 k.

1.49102.

Chapter 41

p. 12

Poutput/Pinput = iCVC/iBVB = VC/VB = (70)(1.49102) = 53. The output current is iC = VC/RC = (80)VB/RC = (80)(0.080 V)/(15 k) =

1.04104.

0.43 mA.

54. For an electron confined within ?x, we find the uncertainty in the momentum from ?p = ˙/?x, which we take to be the momentum of the particle. The kinetic energy of the electron is K = p2/2m = ˙2/2m(?x)2. When the two electrons are in separated atoms, we get K1 = 2˙2/2m(?x1)2 = 2(1.05510–34 J · s)2/2(9.1110–31 kg)(0.05310–9 m)2 = 4.3510–18 J = 27.2 eV. When the electrons are in the molecule, we get K2 = 2˙2/2m(?x2)2 = 2(1.05510–34 J · s)2/2(9.1110–31 kg)(0.07410–9 m)2 = 2.2310–18 J = 14.0 eV. Thus the binding energy is K1 – K2 = 27.2 eV – 14.0 eV = 13 eV. 55. (a) We find the temperature from K = *kT; (4.5 eV)(1.6010–19 J/eV) = *(1.3810–23 J/K)T, which gives T = (b) We find the temperature from K = *kT; (0.15 eV)(1.6010–19 J/eV) = *(1.3810–23 J/K)T, which gives T =

3.5104 K.

1.2103 K.

56. (a) The potential energy for the point charges is U = – ke2/r = – (2.3010–28 J · m)/(0.2710–9 m) = – 8.5210–19 J = – 5.33 eV = – 5.3 eV. (b) Because the potential energy of the ions is negative, 5.3 eV is released when the ions are brought together. A release of energy means that energy must be provided to return the ions to the state of free atoms. Thus the total binding energy of the KF ions is Binding energy = 5.33 eV + 4.07 eV – 4.34 eV = 5.1 eV. 57. (a) The reduced mass of the HCl molecule is  = mHmCl/(mH + mCl) = (1.00794 u)(35.4527 u)/(1.00794 u + 35.4527 u) = 0.9801 u. (b) We find the effective spring constant from f = (k/)1/2/2p; 482 N/m. 8.661013 Hz = [k/(0.9801 u)(1.6610–27 kg/u)]1/2/2p, which gives k = This is 88% of the constant for H2. 58. The reduced mass of the molecule is x1 x2  = m1m2/(m1 + m2). CM The center of mass does not move. If the equilibrium  separation of the atoms is L and we choose the directions m1 m2 indicated on the diagram, the center of mass relative to L the equilibrium position of m1 is xCM = m2L/(m1 + m2) = [– m1x1 + m2(L + x2)]/(m1 + m2), which gives m1x1 = m2x2 . The stretch of the spring is x = x1 + x2 . Thus for the two atoms we have m1 d2x1/dt2 = – kx, m2 d2x2/dt2 = – kx. This is consistent with the result from the center of mass calculation. If we rearrange and combine the equations, we get

Chapter 41

p. 13

d2x1/dt2 + d2x2/dt2 = – kx(1/m1 + 1/m2) = – kx[(m1 + m2)/m1m2];

d2x/dt2 = – (k/)x. This is the equation for SHM with 2 = k/. Thus the frequency of vibration is f = /2p = (k/)1/2/2p. 59. From the Boltzmann factor the population of a state with energy E is proportional to e –E/kT. The rotational energy of a state is E = (˙2/2I )L(L + 1). The selection rule requires ?L = ± 1. States with higher values of L are less likely to be occupied and thus less likely to absorb a photon. For example, there is a greater probability for absorption from L = 1 to L = 2 than from L = 2 to L = 3. The molecule is not rigid and thus I will depend on L, which will affect the spacing. 60. The moment of inertia of the baton of length d about its center of mass is I = (Md2/12) + [2m(d/2)2] = [(0.200 kg)(0.30 m)2/12] + [2(0.300 kg)(0.15)2] = 0.015 kg · m2. The rotational energy of the baton is E = !I2 = !(0.015 kg · m2)[(1.6 rev/s)(2p rad/rev)]2 = 0.76 J. We find the rotational quantum number from L2 = (I)2 = ¬(¬ + 1)˙2; [(0.015 kg · m2)(1.6 rev/s)(2p rad/rev)]2 = ¬(¬ + 1)(1.05510–34 J · s)2, which gives ¬ = 1.41033. Thus the difference in rotational energy levels is ?Erot = ¬˙2/I = (1.41033)(1.05510–34 J · s)2/(0.015 kg · m2) = 1.110–33 J. No,

because ?Erot « E, we do not consider quantum effects for the baton.

61. From the diagram of the cubic lattice, we see that an atom inside the cube is bonded to the six nearest neighbors. Because each bond is shared by two atoms, the number of bonds per atom is 3. We find the heat of fusion for argon from the energy required to break the bonds: Lfusion = (number of bonds/atom)(number of atoms/kg)Ebond = (3/atom)[(6.021023 atoms/mol)/(39.9510–3 kg/mol)]  (3.910–3 eV)(1.6010–19 J/eV) = 2.8104 J/kg.

62. The Hall voltage is produced by the drifting electrons: åH = vdBL; 1810–3 V = vd(1.6 T)(1.510–2 m), which gives vd = 0.75 m/s. We find the density of drifting electrons from the current: I = neAvd ; 0.2010–3 A = n(1.6010–19 C)(1.510–2 m)(1.010–3 m)(0.75 m/s), which gives n = 1.111020 electrons/m3. The density of silicon atoms is N = [(2.33106 g/m3)/(28 g/mol)](6.021023 atoms/mol) = 5.01028 atoms/m3.

Chapter 41

p. 14

Thus the ratio of electrons to atoms is n/N = (1.111020 electrons/m3)/(5.01028 atoms/m3) =

2.210–9.

63. The photon with the longest wavelength has the minimum energy, so the energy gap must be Eg = hc/ = (1.24103 eV · nm)/(1000 nm) = 1.24 eV. 64. The energy of the photon must be greater than or equal to the energy gap. Thus the longest wavelength that will excite an electron is  = c/f = hc/hf = hc/Eg = (1.24103 eV · nm)/(0.72 eV) = 1.7103 nm = 1.7 m. Thus the wavelength range is  = 1.7 m. 65. To use silicon to filter the wavelengths, we want wavelengths below the IR to be able to cause the electron to be raised to the conduction band, so the photon is absorbed in the silicon. We find the shortest wavelength from  = c/f = hc/hf > hc/Eg = (1.24103 eV · nm)/(1.14 eV) = 1.09103 nm = 1.09 m. Because this is in the IR, the shorter wavelengths of visible light will excite the electron, so silicon could be used as a window. 66. (a) From the definition of the Fermi temperature, we have TF = EF/k =(7.0 eV)(1.610–19 J/eV)/(1.3810–23 J/K) = 8.1104 K. (b) We can write the Fermi distribution as f= 1/ e (E –EF)/ kT + 1 = 1/ e E/ kTe –T F/ T + 1 . If T » TF , the second exponential factor is 1. If E/kT » 1, we can also ignore the 1 to get f ˜ 1/e E/kT = e – E/kT. This is not useful for conductors because TF is much higher than the melting point, so T « TF . 67. In a dielectric, Coulomb’s law becomes F = e2/4pKÅ0r2. Thus where Å0 appears in an equation, we insert K. If the “extra” electron is outside the arsenic ion, the effective Z will be 1, and we can use the hydrogen results. (a) The energy of the electron is E = – Z2e4m/8K2Å02h2n2 = – (13.6 eV)Z2/K2n2 = – (13.6 eV)(1)2/(12)2(1)2 = – 0.094 eV. Thus the binding energy is 0.094 eV. (b) The radius of the electron orbit is r = KÅ0h2n2/pZe2m = Kn2r0/Z = Kn2(0.0529 nm)/Z = (12)(1)2(0.0529 nm)/(1)2 = 0.63 nm. Note that this result justifies the assumption that the electron is outside the arsenic ion. 68. With a full-wave rectifier, there are two peaks for each cycle of the input voltage, so the time between peaks is T = 1/2f = 1/2(60 Hz) = 0.00833 s. The time constant of the rectifier is  = RC = (8.8103 )(3010–6 F) = 0.264 s. Because T « , we assume that the exponential discharge of the capacitor voltage is linear: VC = V0(1 – t/). We approximate the lowest voltage of the capacitor by finding the value reached in the time from one peak to the next: V = V0[1 – (0.00833 s)/(0.264 s)] = 0.968 V0 . Thus the ripple about the mean value is ± 1.6%. ± !(V0 – V)/V0 = ± !(1 – 0.968) = ± 0.016 =

Chapter 41

p. 15

R 69. (a) The current through the load resistor is Iload = Voutput/Rload = (130 V)/(16.0 k) = 8.125 mA. At the minimum supply voltage the current through + the diode will be zero, so the current through R V supply V output R is 8.125 mA, and the voltage across R is load – VR,min = IR,minR = (8.125 mA)(1.80 k) = 14.6 V. The minimum supply voltage is Vmin = VR,min + Voutput = 14.6 V + 130 V = 145 V. At the maximum supply voltage the current through the diode will be 110 mA, so the current through R is 110 mA + 8.125 mA = 118.1 mA, and the voltage across R is VR,max = IR,maxR = (118.1 mA)(1.80 k) = 213 V. The maximum supply voltage is Vmax = VR,max + Voutput = 213 V + 130 V = 343 V. Thus the range of supply voltages is 145 V = V = 343 V. (b) At a constant supply voltage the voltage across R is 200 V – 130 V = 70 V, so the current in R is IR = (70 V)/(1.80 k) = 38.9 mA. If there is no current through the diode, this current must be in the load resistor, so we have Rload = (130 V)/(38.9 mA) = 3.34 k. If Rload is less than this, there will be a greater current through R, and thus the voltage across the load will drop and regulation will be lost. If Rload is greater than 3.34 k, the current through Rload will decrease and there will be current through the diode. The current through the diode is 38.9 mA when Rload is infinite, which is less than the maximum of 110 mA. Thus the range for load resistance is

3.34 k = Rload < 8.

Ch. 42 Page 1

CHAPTER 42 – Nuclear Physics and Radioactivity Note:

A factor that appears in the analysis of energies is e2/4pÅ0 = (1.6010–19 C)2/4p(8.8510–12 C2/N · m2) = 2.3010–28 J · m = 1.44 MeV · fm.

1.

To find the rest mass of an  particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m = MHe – 2me = (4.002603 u)(931.5 MeV/uc2) – 2(0.511 MeV/c2) = 3727 MeV/c2.

2.

We convert the units: m = (139 MeV/c2)/(931.5 MeV/uc2) =

3.

0.149 u.

The  particle is a helium nucleus: r = (1.210–15 m)A1/3 = (1.210–15 m)(4)1/3 =

1.910–15 m

= 1.9 fm.

4.

The radius of a nucleus is r = (1.210–15 m)A1/3. If we form the ratio for the two isotopes, we get r14/r12 = (14/12)1/3 = 1.053. Thus the radius of 14C is 5.3% greater than that for 12C.

5.

(a) The mass of a nucleus with mass number A is A u and its radius is r = (1.210–15 m)A1/3. Thus the density is  = m/V = A(1.6610–27 kg/u)/)pr3 = A(1.6610–27 kg/u)/)p(1.210–15 m)3A = 2.31017 kg/m3, independent of A. (b) We find the radius from M = V; 184 m. 5.981024 kg = (2.31017 kg/m3))pR3, which gives R = (c) For equal densities, we have  = MEarth/)pREarth3 = mU/)prU3; (5.981024 kg)/(6.38106 m)3 = (238 u)(1.6610–27 kg/u)/rU3, which gives rU =

6.

7.

(a) The radius of 64Cu is r = (1.210–15 m)A1/3 = (1.210–15 m)(64)1/3 = 4.810–15 m (b) We find the value of A from r = (1.210–15 m)A1/3; 34. 3.910–15 m = (1.210–15 m)A1/3, which gives A =

2.610–10 m.

= 4.8 fm.

We find the radii of the two nuclei from r = r0A1/3; r = (1.2 fm)(4)1/3 = 1.9fm; rU = (1.2fm)(238)1/3 = 7.4 fm. If the two nuclei are just touching, the Coulomb potential energy must be the initial kinetic energy of the  particle: K = U = ZZUe2/4pÅ0(r + rU) = (2)(92)(1.44 MeV · fm)/(1.9 fm + 7.4 fm) = 28 MeV.

Ch. 42 Page 2

8.

We find the radii of the two nuclei from r = r0A1/3; r = (1.2 fm)(4)1/3 = 1.9fm; rAm = (1.2fm)(243)1/3 = 7.5 fm. We assume that the nucleus is so much heavier than the  particle that we can ignore the recoil of the nucleus. We find the kinetic energy of the  particle from the conservation of energy: Ki + Ui = Kf + Uf ; 0 + ZZAme2/4pÅ0(r + rAm) = Kf + 0; (2)(95)(1.44 MeV · fm)/(1.9 fm + 7.5 fm) = Kf , which gives Kf = 29 MeV.

9.

The radius of a nucleus is r = (1.210–15 m)A1/3. If we form the ratio for the two nuclei, we get rX/rU = (AX/AU)1/3;

! = (AX/238)1/3, which gives AX = 30. From the Appendix, we see that the stable nucleus could be

31 15

P.

10. From Figure 42–1, we see that the average binding energy per nucleon at A = 40 is 8.6 MeV. Thus the total binding energy for 40Ca is (40)(8.6 MeV) = 340 MeV. 11. (a) From Figure 42–1, we see that the average binding energy per nucleon at A = 238 is 7.5 MeV. Thus the total binding energy for 238U is (238)(7.5 MeV) = 1.8103 MeV. (b) From Figure 42–1, we see that the average binding energy per nucleon at A = 84 is 8.7 MeV. Thus the total binding energy for 84Kr is (84)(8.7 MeV) = 7.3102 MeV. 12. Deuterium consists of one proton and one neutron. We find the binding energy from the masses: Binding energy = [M(1H) + m(1n) – M(2H)]c2 = [(1.007825 u) + (1.008665 u) – (2.014102 u)]c2(931.5 MeV/uc2) = 2.22 MeV. 13.

14N

consists of seven protons and seven neutrons. We find the binding energy from the masses: Binding energy = [7M(1H) + 7m(1n) – M(14N)]c2 = [7(1.007825 u) + 7(1.008665 u) – (14.003074 u)]c2(931.5 MeV/uc2) = 104.7 MeV. Thus the binding energy per nucleon is (104.7 MeV)/14 = 7.48 MeV.

14. We find the binding energy of the last neutron from the masses: Binding energy = [M(39K) + m(1n) – M(40K)]c2 = [(38.963707 u) + (1.008665 u) – (39.963999 u)]c2(931.5 MeV/uc2) =

7.80 MeV.

Ch. 42 Page 3

15. (a)

6Li

consists of three protons and three neutrons. We find the binding energy from the masses: Binding energy = [3M(1H) + 3m(1n) – M(6Li)]c2 = [3(1.007825 u) + 3(1.008665 u) – (6.015122 u)]c2(931.5 MeV/uc2) = 32.0 MeV. Thus the binding energy per nucleon is (32.0 MeV)/6 = 5.33 MeV. 208 (b) Pb consists of 82 protons and 126 neutrons. We find the binding energy from the masses: Binding energy = [82M(1H) + 126m(1n) – M(208Pb)]c2 = [82(1.007825 u) + 126(1.008665 u) – (207.976635 u)]c2(931.5 MeV/uc2) = 1636 MeV. Thus the binding energy per nucleon is (1636 MeV)/208 = 7.87 MeV.

less a proton becomes 11B. We find the binding energy of the last proton from the masses: Binding energy = [M(11B) + M(1H) – M(12C)]c2 = [(11.009305 u) + (1.007825 u) – (12.000000 u)]c2(931.5 MeV/uc2) = 16.0 MeV. (b) 12C less a neutron becomes 11C. We find the binding energy of the last neutron from the masses: Binding energy = [M(11C) + m(1n) – M(12C)]c2 = [(11.011434 u) + (1.008665 u) – (12.000000 u)]c2(931.5 MeV/uc2) = 18.7 MeV.

16. (a)

12C

17. We find the binding energy of the last neutron from the masses: Binding energy(23Na) = [M(22Na) + m(1n) – M(23Na)]c2 = [(21.994437 u) + (1.008665 u) – (22.989770 u)]c2(931.5 MeV/uc2) = 12.4 MeV. Binding energy(24Na) = [M(23Na) + m(1n) – M(24Na)]c2 = [(22.989770 u) + (1.008665 u) – (23.990963 u)]c2(931.5 MeV/uc2) = 7.0 MeV. Thus the neutron is more closely bound in 23Na. 18. We find the required energy for separation from the masses. (a) Removal of a proton creates an isotope of nitrogen: Energy(p) = [M(15N) + M(1H) – M(16O)]c2 = [(15.000108 u) + (1.007825 u) – (15.994915 u)]c2(931.5 MeV/uc2) = 12.1 MeV. (b) Removal of a neutron creates another isotope of oxygen: Energy(n) = [M(15O) + m(1n) – M(16O)]c2 = [(15.003065 u) + (1.008665 u) – (15.994915 u)]c2(931.5 MeV/uc2) = 15.7 MeV. The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the proton because there is also the repulsive electric force from the other protons. 19. (a) We find the binding energy from the masses: Binding energy = [2M(4He) – M(8Be)]c2 = [2(4.002603 u) – (8.005305 u)]c2(931.5 MeV/uc2) = – 0.092 MeV. Because the binding energy is negative, the nucleus is unstable. (b) We find the binding energy from the masses: Binding energy = [3M(4He) – M(12C)]c2 = [3(4.002603 u) – (12.000000 u)]c2(931.5 MeV/uc2) = + 7.3 MeV. Because the binding energy is positive, the nucleus is stable.

Ch. 42 Page 4

20. The decay is 31 H  32He + – 01e +  . When we add an electron to both sides to use atomic masses, we see that the mass of the emitted  particle is included in the atomic mass of 3He. Thus the energy released is Q = [M(3H) – M(3He)]c2 = [(3.016049 u) – (3.016029 u)]c2(931.5 MeV/uc2) = 0.0186 MeV = 18.6 keV. 21. The decay is 10 n  11p + –10e +  . We take the electron mass to use the atomic mass of 1H. The kinetic energy of the electron will be maximum if no neutrino is emitted. If we ignore the recoil of the proton, the maximum kinetic energy is K = [m(1n) – M(1H)]c2 = [(1.008665 u) – (1.007825 u)]c2(931.5 MeV/uc2) = 0.782 MeV. 11

10

22. For the decay 6C  5B + 11p , we find the difference of the initial and the final masses: ?m = M(11C) – M(10B) – M(1H) = (11.011434 u) – (10.012937 u) – (1.007825 u) = – 0.009328 u. Thus some additional energy would have to be added. – 23. If 22 11 Na were a  emitter, the resulting nucleus would be 12Mg , which has too few neutrons relative to the  + emitter. number of protons to be stable. Thus we have a +   22 For the reaction 22 11 Na 10 Ne +  +  , if we add 11 electrons to both sides in order to use atomic masses, we see that we have two extra electron masses on the right. The kinetic energy of the  + will be maximum if no neutrino is emitted. If we ignore the recoil of the neon, the maximum kinetic energy is K = [M(22Na) – M(22Ne) – 2m(e)]c2 = [(21.994437 u) – (21.991386 u) – 2(0.00054858 u)]c2(931.5 MeV/uc2) = 1.82 MeV. 22

24. For each decay, we find the difference of the initial and the final masses: (a) ?m = M(236U) – M(235U) – m(1n) = (236.045561 u) – (235.043923 u) – (1.008665 u) = – 0.00703 u. Because an increase in mass is required, the decay is not possible. (b) ?m = M(16O) – M(15O) – m(1n) = (15.994915 u) – (15.003065 u) – (1.008665 u) = – 0.0168 u. Because an increase in mass is required, the decay is not possible. (c) ?m = M(23Na) – M(22Na) – m(1n) = (22.989770 u) – (21.994437 u) – (1.008665 u) = – 0.0133 u. Because an increase in mass is required, the decay is not possible. 25. We find the daughter nucleus by balancing the mass and charge numbers: Z(X) = Z(U) – Z(He) = 92 – 2 = 90; 228 A(X) = A(U) – A(He) = 232 – 4 = 228, so the daughter nucleus is 90Th . If we ignore the recoil of the thorium, the kinetic energy of the  particle is K = [M(232U) – M(228Th) – M(4He)]c2; 5.32 MeV = [(232.037146 u) – M(228Th) – (4.002603 u)]c2(931.5 MeV/uc2), 228.02883 u. which gives M(228Th) = 26. The kinetic energy of the electron will be maximum if no neutrino is emitted. If we ignore the recoil of the sodium, the maximum kinetic energy of the electron is K = [M(23Ne) – M(23Na)]c2 = [(22.9945 u) – (22.9898 u)]c2(931.5 MeV/uc2) = 4.4 MeV. When the neutrino has all of the kinetic energy, the minimum kinetic energy of the electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be from the mass

Ch. 42 Page 5

difference, so the energy range of the neutrino will be

0 = E = 4.4 MeV.

27. (a) We find the final nucleus by balancing the mass and charge numbers: Z(X) = Z(P) – Z(e) = 15 – (– 1) = 16; 32 A(X) = A(P) – A(e) = 32 – 0 = 32, so the final nucleus is 1 6 S. (b) If we ignore the recoil of the sulfur, the maximum kinetic energy of the electron is K = [M(32P) – M(32S)]c2; 1.71 MeV = [(31.973907 u) – M(32S)]c2(931.5 MeV/uc2), 31.97207 u. which gives M(32S) = 218

214

4

28. For alpha decay we have 84Po   82Pb + 2He . The Q value is Q = [M(218Po) – M(214Pb) – M(4He)]c2 = [(218.008965 u) – (213.999798 u) – (4.002603 u)]c2(931.5 MeV/uc2) = 218 218 For beta decay we have 84Po   85At + –10e . The Q value is Q = [M(218Po) – M(218At)]c2 = [(218.008965 u) – (218.00868 u)]c2(931.5 MeV/uc2) = 0.26 MeV.

6.11 MeV.

29. For the electron capture 4Be + –10e  3Li +  , we see that if we add three electron masses to both sides to use the atomic mass for Li, we use the captured electron for the atomic mass of Be. We find the Q value from Q = [M(7Be) – M(7Li)]c2 = [(7.016929 u) – (7.016004 u)]c2(931.5 MeV/uc2) = 0.862 MeV. 7

7

30. The mass number changes only with an  decay for which the change is – 4. If the mass number is 4n, then the new number is 4n – 4 = 4(n – 1) = 4n. Thus for each family, we have 4n  4n – 4  4n; 4n + 1  4n – 4 + 1  4n + 1; 4n + 2  4n – 4 + 2  4n + 2; 4n + 3  4n – 4 + 3  4n + 3. Thus the daughter nuclides are always in the same family. 31. The total kinetic energy of the daughter and the  particle is K + KPb = Q = [M(210Po) – M(206Pb) – M(4He)]c2 = [(209.982857 u) – (205.974449 u) – (4.002603 u)]c2(931.5 MeV/uc2) = 5.41 MeV. If the polonium nucleus is at rest when it decays, for momentum conservation we have p = pPb . The kinetic energy of the lead nucleus is KPb = pPb2/2mPb = p2/2mPb = (m/mPb)K . Thus we have K + (m/mPb)K = [1 + (4 u)/(206 u)]K = 5.41 MeV, which gives K = 5.31 MeV. 

238

234

4

32. The decay is 92U   90Th + 2He . If the uranium nucleus is at rest when it decays, for momentum conservation we have p = pTh . Thus the kinetic energy of the thorium nucleus is KTh = pTh2/2mTh = p2/2mTh = (m/mTh)K = (4 u/234 u)(4.20 MeV) = 0.0718 MeV. The Q value is the total kinetic energy produced: Q = K + KTh = 4.20 MeV + 0.0718 = 4.27 MeV.

Ch. 42 Page 6

33. (a) For the positron-emission process A A + Z + 1X  Z X + e +  , we need to add Z + 1 electrons to the nuclear mass of X to be able to use the atomic mass. On the right-hand side we use Z electrons to be able to use the atomic mass of X. Thus we have 1 electron mass and the -particle mass, which means that we must include 2 electron masses on the righthand side. The Q value will be Q = [MP – (MD + 2me)]c2 = (MP – MD – 2me)c2. (b) The kinetic energy of the  + particle will be maximum if no neutrino is emitted. If we ignore the recoil of the boron, the maximum kinetic energy is K = Q = [M(11C) – M(11B) – 2me]c2 = [(11.011434 u) – (11.009305 u) – 2(0.00054858)]c2(931.5 MeV/uc2) = 0.961 MeV. + The sum of the kinetic energy of the  particle and the energy of the neutrino must be from the mass difference, so the kinetic energy of the neutrino will range from 0.961 MeV to 0. 236 4   232 34. (a) The decay is 92U 90 Th + 2He . The Q value is 236 232 Q = [M( U) – M( Th) – M(4He)]c2 = [(236.045561 u) – (232.038050 u) – (4.002603 u)]c2(931.5 MeV/uc2) = 4.572 MeV. If the uranium nucleus is at rest when it decays, for momentum conservation we have p = pTh . Thus the kinetic energy of the thorium nucleus is KTh = pTh2/2mTh = p2/2mTh = (m/mTh)K . Thus we have Q = K + (m/mTh)K = [1 + (4 u)/(232 u)]K = 4.572 MeV, which gives K = 4.49 MeV. (b) We find the radii of the two nuclei from R = r0A1/3; R = (1.2 fm)(4)1/3 = 1.9fm; 1/3 RTh = (1.2fm)(232) = 7.4 fm. (c) We assume the potential energy of the alpha particle (charge = 2e) is produced by the electric field of the remaining nuclear charge of 90e. Thus the potential energy when the two nuclei are just touching is UC = qQ/4pÅ0(R + RTh) = (2)(90)(1.44 MeV · fm)/(1.9fm + 7.4 fm) = 28 MeV. (d) The radius when the alpha particle leaves the Coulomb barrier is where the Coulomb energy is equal to the final kinetic energy: UB = qQ/4pÅ0RB = K ; (2)(90)(1.44 MeV · fm)/RB = 4.49 MeV, which gives RB = 57.7 fm. Thus the width of the barrier is RB – RTh = 57.7 fm – 7.4 fm ˜ 50 fm.

Ch. 42 Page 7

35. We find the decay constant from N = N0 e – t; 320 decays/min = (1280 decays/min) e – (6 h), which gives  = 0.231/h. Thus the half-life is T1/2 = 0.693/ = 0.693/(0.231/h) = 3.0 h. Note that in 6.0 h the decay rate was reduced to # the original rate. This means the elapsed time was 2 half-lives. 36. (a) We find the decay constant from  = 0.693/T1/2 = 0.693/(4.5109 yr)(3.16107 s/yr) = (b) We find the half-life from  = 0.693/T1/2 ; 8.210–5 s–1 = 0.693/T1/2 , which gives T1/2 = 8.45103 s =

4.910–18 s–1.

2.3 h.

37. The activity of the sample is ?N/?t = N = (0.693/T1/2)N = [0.693/(5730 yr)(3.16107 s/yr)](3.11020) =

1.2109 decays/s.

38. The number of half-lives that elapses is n = (3.0 yr)(12 mo/yr)/(9 mo) = 4. We find the fraction remaining from N/N0 = (!)n = (!)4 = 0.0625. 39. We find the number of nuclei from the activity of the sample: ?N/?t = N; 875 decays/s = [(0.693)/(4.468109 yr)(3.16107 s/yr)]N, which gives N =

1.781020 nuclei.

40. (a) The fraction left is N/N0 = (!)n = (!)4 = 0.0625. (b) The fraction left is N/N0 = (!)n = (!)4.5 = 0.0442. 41. Because only  particle decay changes the mass number (by 4), we have N = (235 – 207)/4 = 7  particles. An  particle decreases the atomic number by 2, while a – particle increases the atomic number by 1, so we have N = [92 – 82 – 7(2)]/(– 1) = 4 – particles. 42. The decay constant is  = 0.693/T1/2 = 0.693/(8.0207 days)(24 h/day)(3600 s/h) = 1.00010–6 s–1. The initial number of nuclei is N0 = [(63210–6 g)/(131 g/mol)](6.021023 atoms/mol) = 2.9041018 nuclei. (a) When t = 0, we get N = N0 e – t = (1.00010–6 s–1)(2.9041018) e 0 = 2.901012 decays/s.

Ch. 42 Page 8

(b) When t = 1.0 h, the exponent is t = (1.00010–6 s–1)(1.0 h)(3600 s/h) = 3.60010–3, so we get N = N0 e – t = (2.901012 decays/s) e – 0.003600 = 2.891012 decays/s. (c) When t = 6 months, the exponent is t = (1.00010–6 s–1)(6 mo)(30 days/mo)(24 h/day)(3600 s/h) = 15.55, so we get N = N0 e – t = (2.901012 decays/s) e – 15.55 = 5.11105 decays/s. 43. The decay constant is  = 0.693/T1/2 = 0.693/(30.8 s) = 0.0225 s–1. (a) The initial number of nuclei is N0 = [(9.810–6 g)/(124 g/mol)](6.021023 atoms/mol) = 4.81016 nuclei. (b) When t = 2.0 min, the exponent is t = (0.0225 s–1)(2.0 min)(60 s/min) = 2.7, so we get N = N0 e – t = (4.81016) e – 2.7 = 3.21015 nuclei. (c) The activity is N = (0.0225 s–1)(3.21015) = 7.21013 decays/s. (d) We find the time from N = N0 e – t; 1 decay/s = (0.0225 s–1)(4.81016) e – (0.0225 /s)t, which gives t = 1.54103 s = 26 min. 44. The number of nuclei is N = [(7.710–6 g)/(32 g/mol)](6.021023 atoms/mol) = 1.451017 nuclei. The activity is N = [(0.693)/(1.23106 s)](1.451017) = 8.21010 decays/s. 45. We find the number of nuclei from Activity = N; 2.65105 decays/s = [(0.693)/(7.56106 s)]N, which gives N = 2.891012 nuclei. The mass is m = [(2.891012 nuclei)/(6.021023 atoms/mol)](35 g/mol) = 1.6810–10 g. 46. (a) The decay constant is  = 0.693/T1/2 = 0.693/(1.59105 yr)(3.16107 s/yr) = 1.3810–13 s–1. (b) The activity is N = (1.3810–13 s–1)(7.501019) = 1.035107 decays/s = 6.21108 decays/min. 47. We find the number of half-lives from (?N/?t)/(?N/?t)0 = (!)n; 1/10 = (!)n, or n log 2 = log 10, which gives n = 3.32. Thus the half-life is T1/2 = t/n = (8.6 min)/3.32 = 2.6 min. 48. Because the fraction of atoms that are 14C is so small, we use the atomic weight of 12C to find the number of carbon atoms in 185 g: N = [(185 g)/(12 g/mol)](6.021023 atoms/mol) = 9.281024 atoms. The number of 14C nuclei is N14 = (1.3/1012)(9.281024) = 1.2071013 nuclei.

Ch. 42 Page 9

The activity is N = [0.693/(5730 yr)(3.16107 s/yr)](1.2071013) =

46 decays/s.

49. We find the number of nuclei from Activity = N; 8.70102 decays/s = [(0.693)/(1.28109 yr)(3.16107 s/yr)]N, which gives N = 5.081019 nuclei. The mass is m = [(5.081019 nuclei)/(6.021023 atoms/mol)](40 g/mol) = 3.410–3 g = 3.4 mg. 50. We assume that the elapsed time is much smaller than the half-life, so we can use a constant decay rate. Because 87Sr is stable, and there was none present when the rocks were formed, every atom of 87Rb that decayed is now an atom of 87Sr. Thus we have NSr = – ?NRb = NRb ?t, or NSr/NRb = (0.693/T1/2) ?t; 0.0160 = [0.693/(4.751010 yr)]?t, which gives ?t = 1.1109 yr. This is ˜ 2% of the half-life, so our original assumption is valid. 51. The decay rate is ?N/?t = N. If we assume equal numbers of nuclei decaying by  emission, we have = T1/2,214/T1/2,218 (?N/?t)218/(?N/?t)214 = 218/214 = (1.610–4 s)/(3.1 min)(60 s/min) =

8.610–7.

52. The decay constant is  = 0.693/T1/2 = 0.693/(53 days) = 0.0131 /day = 1.5210–7 s–1. We find the number of half-lives from (?N/?t)/(?N/?t)0 = (!)n; (10 decays/s)/(250 decays/s) = (!)n, or n log 2 = log 25, which gives n = 4.64. Thus the elapsed time is ?t = nT1/2 = (4.64)(53 days) = 2.5102 days ˜ 8 months. We find the number of nuclei from Activity = N; 250 decays/s = (1.5210–7 s–1)N, which gives N = 1.64109 nuclei. The mass is m = [(1.64109 nuclei)/(6.021023 atoms/mol)](7 g/mol) = 1.910–14 g = 1.910–17 kg. 53. We find the number of half-lives from (?N/?t)/(?N/?t)0 = (!)n; 1.05010–2 = (!)n, or n log 2 = log (1/1.05010–2), which gives n = 6.57. Thus the half-life is T1/2 = t/n = (4.00 h)/6.57 = 0.609 h = 36.5 min. From the Appendix we see that the isotope is

211 82 Pb

.

54. The decay constant is  = 0.693/T1/2 = 0.693/(5730 yr) = 1.20910–4 /yr. Because the fraction of atoms that are 14C is so small, we use the atomic weight of 12C to find the number of carbon atoms in 190 g: N = [(190 g)/(12 g/mol)](6.021023 atoms/mol) = 9.531024 atoms, so the number of 14C nuclei in a sample from a living tree is

Ch. 42 Page 10

N14 = (1.310–12)(9.531024) = 1.241013 nuclei. Because the carbon is being replenished in living trees, we assume that this number produced the activity when the club was made. We determine its age from N = N14 e – t; –4

 10  5.0 decays/s = [(1.20910–4 /yr)/(3.16107 s/yr)](1.241013 nuclei) e– (1.209 4 1.910 yr. which gives t =

/ yr )t ,

55. The number of radioactive nuclei decreases exponentially: N = N0 e – t. Every radioactive nucleus that decays becomes a stable daughter nucleus, so we have ND = N0 – N = N0(1 – e – t). 56. (a) We find the number of parent nuclei from the rate at which parent nuclei are decaying: dNP/dt = – PNP, which gives N P = N P0 e – P t . Each decay of a parent nucleus produces a daughter nucleus. Because the daughter nuclei are also decaying, the rate at which daughter nuclei are increasing is d N D dN D =  P N P –  DN D =  PN P 0 e –  Pt –  DN D , or +  D N D =  P N P0 e – P t . dt dt If we multiply both sides by the integrating factor e Dt , we have  dN D e  Dt +  DN D = d N D e  Dt =  PN P 0 e –  Pt e  Dt , or d N D e  Dt =  PN P0 e – ( P –  D)t d t . dt dt When we integrate this from t = 0 (when ND = 0) to t = t, we get t –  PN P 0 – ( P – D )t t  N D e D t = e ; 0 P –  D 0  N  N N D e D t = P P 0 1 – e – ( P – D )t , or N D = P P0 e –  Dt – e –  Pt . P – D  P – D In terms of the half-lives this is T N N D = D P0 e – 0.693t/ TD – e – 0.693t/ TP . TD – TP (b) We write our result as ND e – 0.693t/ TD – e – 0.693t/ TP = T D. N P0 TD – T P When TP = TD , both numerator and denominator are zero, so we must use l'Hôpital’s rule: d e – 0.693t / TD – e – 0.693t/ TP d TP e – 0.693t / TD – 0.693t 2 T P = TD TD = = 0.693t e – 0.693t / TD . 2 –1 TD d TD – TP d TP TP = T D

Ch. 42 Page 11

Thus we have ND = 0.693t e – 0.693t/ T D. TD N P0 When TP = 3TD , we have ND = 1 e – 0.693t/ 3T D – e – 0.693t / TD . 2 N P0 When TP = @TD , we have ND = 3 e – 0.693t/ TD – e – 3(0.693)t/ 2 N P0

TD

N D/ N P0 0.6

T P = T D/3

0.5 0.4 0.3

TP = TD

0.2 0.1 0

T P = 3T D 0

1

2

3

4

5

t/ T D

57. (a) The fraction of mass is mp/(mp + me) = (1.672610–27 kg)/(1.672610–27 kg + 9.1110–31 kg) = 0.99946. (b) The fraction of volume is 1.210–14. (rnucleus/ratom)3 = [(1.210–15 m)/(0.5310–10 m)]3 = (c) If we take the density of the hydrogen nucleus as the density of nuclear matter, we get  = mp/)pr3 = (1.6710–27 kg)/)p(1.210–15 m)3 = 2.31017 kg/m3. The density of water is 1000 kg/m3, so nuclear matter is 1014 greater. 58. The radius of the iron nucleus is r = (1.210–15 m)A1/3 = (1.210–15 m)(56)1/3 = 4.610–15 m. We use the radius as the uncertainty in position for the nucleon. We find the uncertainty in the momentum from ?p = ˙/?x = (1.05510–34 J · s)/(4.610–15 m) = 2.2910–20 kg · m/s. If we assume that the lowest value for the momentum is the least uncertainty, we estimate the minimum possible kinetic energy as K = (?p)2/2m = (2.2910–20 kg · m/s)2/2(1.6610–27 kg) = 1.5810–13 J ˜ 1 MeV. 59. If the 40K nucleus in the excited state is at rest, the gamma ray and the nucleus must have equal and opposite momenta: pK = p = E/c , or pKc = E = 1.46 MeV. The kinetic energy of the nucleus is K = pK2/2m = (pKc)2/2mc2 = (1.46 MeV)2/2(40 u)(931.5 MeV/uc2)c2 = 2.8610–5 MeV = 28.6 eV. 60. Because the carbon is being replenished in living trees, we assume that the amount of 14C is constant until the wood is cut, and then it decays. We find the number of half-lives from N/N0 = (!)n; 0.090 = (!)n, or n log 2 = log (1/0.090), which gives n = 3.47. Thus the time is t = nT1/2 = (3.47)(5730 yr) = 2.0104 yr. 61. (a) We find the mass number from the radius: r = (1.210–15 m)A1/3; 5.0103 m = (1.210–15 m)A1/3, which gives A = 7.21055. (b) The mass of the neutron star is m = A(1.6610–27 kg/u) = (7.21055 u)(1.6610–27 kg/u) = 1.21029 kg.

Ch. 42 Page 12

Note that this is about 6% of the mass of the Sun. (c) The acceleration of gravity on the surface of the neutron star is g = Gm/r2 = (6.6710–11 N · m2/kg2)(1.21029 kg)/(5.0103 m)2 =

3.21011 m/s2.

62. Because the tritium in water is being replenished, we assume that the amount is constant until the wine is made, and then it decays. We find the number of half-lives from N/N0 = (!)n; 0.10 = (!)n, or n log 2 = log (10), which gives n = 3.32. Thus the time is t = nT1/2 = (3.32)(12.33 yr) = 41 yr. 63. If we assume a body has 70 kg of water, the number of water molecules is Nwater = [(70103 g)/(18 g/mol)](6.021023 atoms/mol) = 2.341027 molecules. The number of protons in a water molecule (H2O) is 2 + 8 = 10, so the number of protons is N0 = 2.341028 protons. If we assume that the time is much less than the half-life, the rate of decay is constant, so we have ?N/?t = N = (0.693/T1/2)N; (1 proton)/?t = [(0.693)/(1032 yr)](2.341028 protons), which gives ?t = 6103 yr. 64. The capture is 11 H + 10 n  21 H + . Because the kinetic energies of the particles are small, the gamma energy is the energy released: Q = [M(1H) + m(1n) – M(2H)]c2 = [(1.007825 u) + (1.008665 u) – (2.014102 u)]c2(931.5 MeV/uc2) = 2.22 MeV. 65. We find the number of half-lives from (?N/?t)/(?N/?t)0 = (!)n; 1.0010–2 = (!)n, or n log 2 = log (100), which gives n = 6.64, so the time is

6.64 T1/2 .

66. We find the number of 40K nuclei from Activity = N40 ; 60 decays/s = [(0.693)/(1.28109 yr)(3.16107 s/yr)]N40 , which gives N40 = 3.51018 nuclei. The mass of 40K is m40 = (3.51018)(40 u)(1.6610–27 kg/u) = 2.3210–7 kg = 0.23 mg. From the Appendix we have N40 = (0.0117%)N, and N39 = (93.2581%)N. Thus the number of 39K nuclei is N39 = [(93.2581%)/(0.0117%)](3.51018 nuclei) = 2.81022 nuclei. The mass of 39K is m39 = (2.81022)(39 u)(1.6610–27 kg/u) = 1.8110–3 kg = 1.8 g. 67. If we use 146 C  147 N + e – +  as the decay, we see that 14C and 14N each have an even number of nucleons with spin !, so their total spin must be an integer. Because e– has spin !,  must have spin ! to conserve angular momentum. 68. If the initial nucleus is at rest when it decays, for momentum conservation we have p = pD . Thus the kinetic energy of the daughter is KD = pD2/2mD = p2/2mD = (m/mD)K = (A/AD)K = (4/AD)K .

Ch. 42 Page 13

Thus the fraction carried away by the daughter is KD/(K + KD) = (4/AD)K/[K + (4/AD)K] = 1/[1 + (AD/4)]. For the decay of 226Ra, the daughter has AD = 222, so we get fractionD = 1/[1 + (222/4)] = 0.018. Thus the  particle carries away 1 – 0.018 = 0.982 = 98.2%. 69. We see from the periodic chart that Sr is in the same column as calcium. If strontium is ingested, the body will treat it chemically as if it were calcium, which means it will be stored by the body in bones, where its radioactivity could affect the bone marrow. We find the number of half-lives to reach a 1% level from N/N0 = (!)n; 0.01 = (!)n, or n log 2 = log (100), which gives n = 6.64. Thus the time is t = nT1/2 = (6.64)(29 yr) = 193 yr. The decay reactions are 90 90 90 0 38Sr  39Y + – 1 e +  , 39 Y is radioactive; 90 90 90 0 39Y  40Zr + – 1 e +  , 40 Zr is stable. 70. (a) We find the daughter nucleus by balancing the mass and charge numbers: Z(X) = Z(Os) – Z(e–) = 76 – (– 1) = 77; A(X) = A(Os) – A(e–) = 191 – 0 = 191, 191 so the daughter nucleus is 77Ir . (b) Because there is only one  energy, the  decay must be to the higher excited state.

191 76 Os

 – (0.14 MeV)  (0.042 MeV)  (0.129 MeV)

191 77 Ir* 191 77 Ir* 191 77 Ir

71. We use an average nuclear radius of 510–15 m. We use the radius as the uncertainty in position for the electron to find the uncertainty in the momentum from ?p = ˙/?x = (1.05510–34 J · s)/2(510–15 m) = 2.110–20 kg · m/s. If we assume that the lowest value for the momentum is the least uncertainty, we estimate the minimum possible energy as E = K + mc2 = (p2c2 + m2c4)1/2 = [(?p)2c2 + m2c4]1/2 = [(2.110–20 kg · m/s)2(3.00108 m/s)2 + (9.1110–31 kg)2(3.00108 m/s)4]1/2 = 6.310–12 J ˜ 40 MeV. Because this is » mc2, the electron is unlikely to be found in the nucleus. 72. From Figure 42–1, we see that the average binding energy per nucleon at A = 29 is 8.6 MeV. If we use the average atomic weight as the average number of nucleons for the two stable isotopes of copper, the total binding energy is (63.5)(8.6 MeV) = 550 MeV. The number of atoms in a penny is N = [(3.0 g)/(63.5 g/mol)](6.021023 atoms/mol) = 2.841022 atoms. Thus the total energy needed is (2.841022)(550 MeV) = 1.571025 MeV = 2.51012 J. 73. (a) ?(4He) = M(4He) – A(4He) = 4.002603 u – 4 u = 0.002603 u 2.425 MeV/c2. = (0.002603 u)(931.5 MeV/uc2) = (b) ?(12C) = M(12C) – A(12C) = 12.000000 u – 12 u = 0. (c) ?(107Ag) = M(107Ag) – A(107Ag) = 106.905093 u – 107 u = – 0.094907 u

Ch. 42 Page 14

= (– 0.094907 u)(931.5 MeV/uc2) = – 88.41 MeV/c2. 235 235 = M( U) – A( U) = 235.043923 u – 235 u = 0.043923 u (d) 40.92 MeV/c2. = (0.043923 u)(931.5 MeV/uc2) = (e) From the Appendix we see that ? = 0 for 1 = Z = 8 and Z = 85; ? < 0 for 9 = Z = 84. ?(235U)

74. (a) The usual fraction of 14C is 1.310–12. Because the fraction of atoms that are 14C is so small, we use the atomic weight of 12C to find the number of carbon atoms in 100 g: N = [(100 g)/(12 g/mol)](6.021023 atoms/mol) = 5.021024 atoms. The number of 14C nuclei in the sample is N14 = (1.310–12)(5.021024) = 6.531012 nuclei. We find the number of half-lives from N/N0 = (!)n; 1/6.531012 = (!)n, or n log 2 = log (6.531012), which gives n = 42.6. Thus the time is t = nT1/2 = (42.6)(5730 yr) = 2.4105 yr. (b) It would take one half-life for the activity of 200 g to decay to the activity of 100 g. This is so much smaller than the time in (a) that there is no change. Thus carbon dating cannot be used for times much greater than the half-life. 75. Because there are so many low-energy electrons available, the reaction e – + p  n +  would turn most of the protons into neutrons, which would eliminate chemistry, and thus life. The Q-value of the reaction is Q = [M(1H) – m(1n)]c2 = [(1.007825 u) – (1.008665 u)]c2(931.5 MeV/uc2) = – 0.782 MeV. The percentage increase in the proton’s mass to make the Q-value = 0 is 0.083%. (?m/m)(100) = [(0.782 MeV/c2)/(938.3 MeV/c2)](100) = 76. (a) From the definition of the mean life we have 



=

0



N (t ) t d t =



N (t) d t

0

0

N 0 e – t t d t



0

N 0 e – t d t



=1 

0

e – u u du

 0

, e – u du

where we have changed variable to u = t. We integrate the numerator by parts: 

0

e – u u du =



0

– u d e– u = – u e–u

 0

+



0

e – u du = 0 +



0

e – u du.

Thus we get

 = 1 



0



0

e – u du e – u du

= 1. 

(b) We find the fraction remaining after one mean life from N/N0 = e –t = e –1 = 0.368. 77. For the , , , ,  sequence we have 232 4 228 228 228 228 228 228 4 224 224 4 220 0 0 90Th  2He + 88Ra , 88Ra  –1 e + 89Ac , 89Ac  –1e + 90Th , 90Th  2He + 88Ra , 88Ra  2 He + 86Rn . 228 228 228 224 220 Thus the daughter nuclei are 88 Ra , 89 Ac , 90 Th , 88 Ra , 86 Rn . For the , , , ,  sequence we have 235 4 231 231 231 231 4 227 227 227 227 4 223 0 0 92U  2He + 90Th , 90Th  –1 e + 91Pa , 91Pa  2He + 89Ac , 89Ac  –1e + 90Th , 90Th  2He + 88Ra . 231 Th , 231 Pa , 227 Ac , 227 Th , 223 Ra . Thus the daughter nuclei are 90 91 89 90 88

Ch. 42 Page 15

78. For the circular motion in the magnetic field, the magnetic force provides the radial acceleration: qvB = mv2/R. The kinetic energy is K = !mv2 = q2B2R2/2m. When we form the ratio for the two particles with the same radii, we get K/K = (q/q)2(m/m) = [(2 e)/(1 e)]2[(9.1110–31 kg)/4(1.6710–27 kg)] = 5.510–4. Note that the particles will have opposite curvatures.

79. (a) If the speed of the  particle within the nucleus is vin , the time to traverse the nucleus is 2R0/vin . The frequency of a collision with the Coulomb barrier is vin/2R0 . From Ch. 39 the probability of passing through the barrier each time is T = e –2GL, where L is the thickness of a square barrier, and G = [2m(U0 – E)/˙2]1/2. The decay constant is the probability of the  particle escaping from the nucleus:  = (vin/2R0)T = (vin/2R0)e –2GL. The Coulomb barrier decreases as 1/r. For a square barrier that approximates this we will use the maximum height found in Problem 34 for the height of the barrier. We find the value of G from G2 = 2m(U0 – E)/˙2 = 2(4)(1.6710–27 kg)(28 MeV – 4.5 MeV)(1.6010–13 J/MeV)/(1.05510–34 J · s)2, which gives G = 2.11015 m–1 = 2.1 fm–1. If we choose the potential inside the well to be zero, the kinetic energy of the  particle inside is the same as the kinetic energy of the  particle after the decay. We find the speed inside the well from K = !mv2; (4.49 MeV)(1.6010–13 J/MeV) = !(4)(1.6710–27 kg)vin2, which gives vin = 1.47107 m/s. The frequency of striking the barrier is vin/2R0 = (1.47107 m/s)/2(7.410–15 m) = 9.91020 s–1. To account for the decrease with r of the Coulomb barrier, we use a square barrier with a thickness less than the width found in Problem 34. (b) If we use a thickness of ! the width of 50 fm, we have

Ch. 42 Page 16 –1 vin e – 2 GL = (9.9  10 20 s – 1 )e – 2(2.1 fm )(25 fm) = 2.5  10– 25 s – 1. 2R 0 Thus the half-life is T1/2 = (ln 2)/ = 0.693/(2.510–25 s–1) = 2.81024 s ˜ 1017 yr.

 =

(c) If we use a thickness of @ the width of 50 fm, we have –1 v  =  in e – 2 GL = (9.9  10 20 s – 1 )e – 2(2.1 fm )(16.7 fm) = 3.4  10– 10 s – 1. 2R 0 Thus the half-life is T1/2 = (ln 2)/ = 0.693/(3.410–10 s–1) = 2.0109 s ˜ 60 yr. Note that the result is very sensitive to the value used for the barrier thickness. (d) To find the width that gives us the known half-life, we have  (ln 2) e 2GL T1/ 2 = (ln 2)/  = , or e 2 GL = vin / 2R0 /( ln 2) T1/ 2 ; v in / 2R0 e 2GL = [(9.91020 s–1)/(ln 2)](2107 yr)(3.16107 s/yr), or 2GL = 83; L = (83)/2(2.1 fm–1) = 20 fm, which is 0.4 of the width found in Problem 34.

Ch. 43 Page 1

CHAPTER 43 – Nuclear Energy; Effects and Uses of Radiation Note:

1.

A factor that appears in the analysis of energies is e2/4pÅ0 = (1.6010–19 C)2 /4p(8.8510–12 C2/N · m2)= 2.3010–28 J · m = 1.44 MeV · fm.

We find the product nucleus by balancing the mass and charge numbers: Z(X) = Z(27Al) + Z(n) = 13 + 0 = 13; 28 A(X) = A(27Al) + A(n) = 27 + 1 = 28, so the product nucleus is 13 Al . If 13Al were a + emitter, the resulting nucleus would be 12Mg , which has too many neutrons relative to – emitter. the number of protons to be stable. Thus we have a 28 28 28 0   14Si + – 1e +  , so the product is The decay is 13Al 14 Si . 28

28

2

3

2.

For the reaction 1H (d , n ) 2He , we find the difference of the initial and the final masses: ?M = M(2H) + M(2H) – m(n) – M(3He) = 2(2.014102 u) – (1.008665 u) – (3.016029 u) = + 0.003510 u. Thus no threshold energy is required.

3.

For the reaction 92U  (n ,  ) 92U with slow neutrons, whose kinetic energy is negligible, we find the difference of the initial and the final masses: ?M = M(238U) + m(n) – M(239U) = (238.050782 u) + (1.008665 u) – (239.054287 u) = + 0.005160 u. Thus no threshold energy is required, so the reaction is possible.

4.

For the reaction 3Li(p , ) 2He , we determine the Q-value: Q = [M(7Li) + M(1H) – M(4He) – M(4He)]c2 = [(7.016004 u) + (1.007825 u) – 2(4.002603 u)]c2(931.5 MeV/uc2) = + 17.35 MeV. Thus 17.35 MeV is released.

5.

For the reaction 4Be ( , n ) 6C , we determine the Q-value: Q = [M(9Be) + M(4He) – m(n) – M(12C)]c2 = [(9.012182 u) + (4.002603 u) – (1.008665 u) – (12.000000 u)]c2(931.5 MeV/uc2) = + 5.701 MeV. Thus 5.701 MeV is released.

6.

23 (a) For the reaction 24 12 Mg (n, d ) 11 Na , we determine the Q-value: Q = [M(24Mg) + m(n) – M(2H) – M(23Na)]c2 = [(23.985042 u) + (1.008665 u) – (2.014102 u) – (22.989770 u)]c2(931.5 MeV/uc2) = – 9.469 MeV. Because (K + Q) > 0, the reaction can occur. (b) The energy released is K + Q = 10.00 MeV – 9.469 MeV = 0.53 MeV.

7.

(a) For the reaction 3Li(p , ) 2He , we determine the Q-value: Q = [M(7Li) + M(1H) – M(4He) – M(4He)]c2 = [(7.016004 u) + (1.007825 u) – 2(4.002603 u)]c2(931.5 MeV/uc2) = + 17.35 MeV. Because Q > 0, the reaction can occur. (b) The kinetic energy of the products is K = Ki + Q = 2.500 MeV + 17.35 MeV = 19.85 MeV.

238

239

7

4

9

12

7

4

Ch. 43 Page 2

8.

(a) For the reaction 7N ( , p ) 8O , we determine the Q-value: Q = [M(14N) + M(4He) – M(1H) – M(17O)]c2 = [(14.003074 u) + (4.002603 u) – (1.007825 u) – (16.999131 u)]c2(931.5 MeV/uc2) = – 1.191 MeV. Because (K + Q) > 0, the reaction can occur. (b) The kinetic energy of the products is K = Ki + Q = 7.68 MeV – 1.191 MeV = 6.49 MeV.

9.

For the reaction 8O (,  )10Ne , we determine the Q-value: Q = [M(16O) + M(4He) – M(20Ne)]c2 = [(15.994915 u) + (4.002603 u) – (19.992440 u)]c2(931.5 MeV/uc2) =

14

16

13

17

20

+ 4.730 MeV.

14

10. For the reaction 6C (d , n ) 7N , we determine the Q-value: Q = [M(13C) + M(2H) – m(n) – M(14N)]c2 = [(13.003355 u) + (2.014102 u) – (1.008665 u) – (14.003074 u)]c2(931.5 MeV/uc2) = + 5.326 MeV. The kinetic energy of the products is K = Ki + Q = 36.3 MeV + 5.326 MeV = 41.6 MeV. 11. (a) We find the product nucleus by balancing the mass and charge numbers: Z(X) = Z(6Li) + Z(2H) – Z(1H) = 3 + 1 – 1 = 3; 7 A(X) = A(6Li) + A(2H) – A(1H) = 6 + 2 – 1 = 7, so the product nucleus is 3 Li . neutron is stripped from the deuteron. (b) It is a “stripping” reaction because a 6 7 (c) For the reaction 3Li(d , p ) 3Li , we determine the Q-value: Q = [M(6Li) + M(2H) – M(1H) – M(7Li)]c2 = [(6.015122 u) + (2.014102 u) – (1.007825 u) – (7.016004 u)]c2(931.5 MeV/uc2) = MeV. Because Q > 0, the reaction is exothermic.

+ 5.025

3He picks up a neutron. 12. (a) It is a “pickup” reaction because the (b) We find the product nucleus by balancing the mass and charge numbers: Z(X) = Z(12C) + Z(3He) – Z(4He) = 6 + 2 – 2 = 6; 11 A(X) = A(12C) + A(3He) – A(4He) = 12 + 3 – 4 = 11, so the product nucleus is 6C . 12 3 11 (c) For the reaction 6C (2He , ) 6C , we determine the Q-value: Q = [M(12C) + M(3He) – M(4He) – M(11C)]c2 = [(12.000000 u) + (3.016029 u) – (4.002603 u) – (11.011434 u)]c2(931.5 MeV/uc2) = + 1.856 MeV. Because Q > 0, the reaction is exothermic.

13. (a) We find the initial nucleus by balancing the mass and charge numbers: Z(X) = Z(32S) – Z(1H) = 16 – 1 = 15; 31 A(X) = A(32S) – A(1H) = 32 – 1 = 31, so the initial nucleus is 15P . 31 32 The reaction is 15 P (p ,  ) 16S . (b) For the reaction, we determine the Q-value: Q = [M(31P) + M(1H) – M(32S)]c2 = [(30.973762 u) + (1.007825 u) – (31.972071 u)]c2(931.5 MeV/uc2) =

+ 8.864 MeV.

Ch. 43 Page 3

13

13

14. For the reaction 6C (p , n ) 7N , we determine the Q-value: Q = [M(13C) + M(1H) – m(n) – M(13N)]c2 = [(13.003355 u) + (1.007825 u) – (1.008665 u) – (13.005739 u)]c2(931.5 MeV/uc2) = – 3.003 MeV. The kinetic energy of the products is Kn + KN = Kp + Q. Because the kinetic energies « mc2, we can use a non relativistic treatment: K = mv2/2 = p2/2m. The least kinetic energy is required when the product particles move together with the same speed. With the target at rest, for momentum conservation we have pp = pn + pN = (mn + mN)v, or Kp = pp2/2mp = [(mn + mN)2/2mp]v2 = [(mn + mN)/mp](Kn + KN), or Kn + KN = [mp/(mn + mN)]Kp . When we use this in the kinetic energy equation, we get [mp/(mn + mN)]Kp = Kp + Q;

{[(1 u)/(1 u + 13 u)] – 1}Kp = – 3.003 MeV, which gives Kp = 3.23 MeV.

15. From the figure we see that a collision will occur if d = (R1 + R2). Thus the area of the effective circle presented by R2 to the center of R1 is  = p(R1 + R2)2.

v R1 d

R2

16. We assume the density of the gas atoms is low enough that we can use the result for a thin target. Thus the fraction of neutrons that interact is R/R0 = n t = (1.71021 m–3)(4010–28 m2)(0.120 m) = 8.210–7. 

17. The initial rate of incident particles is R0 , and the reduced rate a distance x inside the target is Rx . The rate at which particles collide in the next small distance dx is dR = Rxn dx. Because each collision removes a particle from the beam, the change in the incident rate of particles in the beam a distance x inside the target is – dR, or dRx = – Rxn dx. When we integrate this from the initial incidence, we get

R0

Rx x

dx

R x

x dRx Rx = – n  d x ; ln = – n x, or R0 0 R0 R x

Rx = R0 e –n x. If x = t, we get Rt = R0 e –n t, which represents 



the rate at which incident particles pass through the target without scattering.

Ch. 43 Page 4

18. We use the result from Problem 17 in the form ln (Rt /R0) = – n t. We can find the value of n from ln (R1/R0) = ln (0.30) = – n t1 = – n (0.010 m), which gives n = 1.2102 m–1. We find the thickness to reduce the rate by 1/106 from 12 cm. ln (R2/R0) = ln (10–6) = – n t2 = – (1.2102 m–1)t2 , which gives t2 = 0.12 m = 19. We assume a 1% reaction rate allows us to treat the target as thin. The density of Cd atoms is n = [(8650 kg/m3)(103 g/kg)/(113.9 g/mol)](6.021023 atoms/mol) = 4.571028 m–3. (a) From Fig. 43–3 we find that the cross section for 0.1-eV neutrons is 3000 bn. Thus we have R/R0 = n x; 0.01 = (4.571028 m–3)(300010–28 m2)x, which gives x = 710–7 m = 0.7 m. (b) From Fig. 43–3 we find that the cross section for 10-eV neutrons is 2 bn. Thus we have R/R0 = n t; 0.01 = (4.571028 m–3)(210–28 m2)x, which gives x = 110–3 m = 1 mm. 



20. If we neglect the initial kinetic energy of the neutron, the released energy is the Q-value: Q = [M(235U) + m(n) – 12m(n) – M(88Sr) – M(136Xe)]c2 = [(235.043923 u) – 11(1.008665 u) – (87.905614 u) – (135.907220 u)]c2(931.5 MeV/uc2) = MeV. 235

141

92

126.5

21. For the reaction n + 92U  56Ba + 36Kr + 3n , we determine the Q-value: Q = [M(235U) + m(n) – 3m(n) – M(141Ba) – M(92Kr)]c2 = [(235.043923 u) – 2(1.008665 u) – (140.91440 u) – (91.92630 u)]c2(931.5 MeV/uc2) = 173.2 MeV. If the kinetic energy of the incident neutron is very small, the released energy is 173.2 MeV. 22. If we assume 100% efficiency, we have P = E/t; 200 MW = (200 MeV)(1.6010–19 J/eV)(n/t), which gives n/t =

6.31018 fissions/s.

23. With an efficiency of 100% we find the number of fissions from P = E/t; 300 W = (200 MeV)(1.6010–13 J/MeV)n/(3.16107 s), which gives n = 2.961020 fissions. Each fission uses one uranium atom, so the required mass is m = [(2.961020 atoms)/(6.021023 atoms/mol)](235 g/mol) = 0.116 g. 24. We find the number of fissions from P = E/t; (500 MW)/(0.40) = (200 MeV)(1.6010–19 J/eV)n/(3.16107 s), which gives n = 1.231027 fissions. Each fission uses one uranium atom, so the required mass is m = [(1.231027 atoms)/(6.021023 atoms/mol)](235 g/mol) = 4.82105 g = 25. We find the number of collisions from En = E0(!)n; 0.040 eV = (1.0106 eV)(!)n, which gives n =

25.

4.8102 kg.

Ch. 43 Page 5

26. The number of fissions in one second is n = t/?t = (1.0 s)/(1.010–3 s) = 1.0103. For each fission the number of neutrons is 1.0004 times the number from the previous fission. Thus the reaction rate will increase by (1.0004)n = (1.0004)1000 = 1.49. 27. When 236U decays by  emission, the resulting nucleus is 232Th. Thus the radii of the two particles are r = (1.2 fm)(4)1/3 = 1.90 fm; rTh = (1.2 fm)(232)1/3 = 7.37 fm. At the instant of separation, the two particles are in contact, so the Coulomb energy is U = ZZThe2/4pÅ0(r + rTh). If we assume fission into equal parts, the resulting nucleus has A = 118. Thus each radius is rf = (1.2 fm)(118)1/3 = 5.89 fm. At the instant of separation, the two particles are in contact, so the Coulomb energy is Uf = Zf2e2/4pÅ0(2rf). The ratio is U/Uf = ZZTh(2rf)/Zf2(r + rTh) = (2)(90)(2)(5.89 fm)/(46)2(1.90 fm + 7.37 fm) = 0.11. 28. If we assume fission into equal parts, the resulting nucleus has A = 118. Thus each radius is rf = (1.2 fm)(118)1/3 = 5.89 fm. At the instant of separation, the two particles are in contact, so the Coulomb energy is Uf = Zf2e2/4pÅ0(2rf) = (46)2(1.44 MeV · fm)/(2)(5.89 fm) = 259 MeV. This is greater than the fission energy release of 200 MeV, which is probably an indication of the complexity of the fission process. 29. We find the average kinetic energy from K = *kT = *(1.3810–23 J/K)(107 K)/(1.6010–19 J/eV) = 1.29103 eV = 2

3

4

2

2

3

1.3 keV.

30. For the reaction 1H + 1H  2He + n , we determine the Q-value: Q = [M(2H) + M(3H) – m(n) – M(4He)]c2 = [(2.014102 u) + (3.016049 u) – (1.008665 u) – (4.002603 u)]c2(931.5 MeV/uc2) = + 17.59 MeV. Thus 17.59 MeV is released. 31. For the reaction 1H + 1H  2He + n , we determine the Q-value: Q = [M(2H) + M(2H) – m(n) – M(3He)]c2 = [2(2.014102 u) – (1.008665 u) – (3.016029 u)]c2(931.5 MeV/uc2) = + 3.27 MeV. Thus 3.27 MeV is released. 32. For the reaction 1H + 1H  1H + +10 e +  , we must add two electron masses to the left hand side to use atomic masses. We need one electron mass on the right hand side to use the atomic mass for hydrogen. Thus we have two extra electron masses on the right hand side. We determine the Q-value: Q = [M(1H) + M(1H) – M(2H) – 2m(e)]c2 = [2(1.007825 u) – (2.014102 u) – 2(0.0005486 u)]c2(931.5 MeV/uc2) = 0.42 MeV. 1 2 3 For the reaction 1H + 1H  2He +  , we determine the Q-value: Q = [M(1H) + M(2H) – M(3He)]c2 = [(1.007825 u) + (2.014102 u) – (3.016029 u)]c2(931.5 MeV/uc2) = 5.49 MeV. 3 3 4 1 1 For the reaction 2He + 2He  2He + 1H + 1H , we determine the Q-value: Q = [M(3He) + M(3He) – M(4He) – M(1H) – M(1H)]c2 = [2(3.016029 u) – (4.002603 u) – 2(1.007825 u)]c2(931.5 MeV/uc2) = 12.86 MeV. 1

1

2

Ch. 43 Page 6

2

2

3

1

33. For the reaction 1H + 1H  1H + 1H , two atoms of 2H, or 4 u, are used as fuel. The energy release is 6.11023 MeV/g. (4.03 MeV/reaction)/(4 u/reaction)(1.6610–27 kg/u) = 6.11026 MeV/kg = 2 2 3 2 For the reaction 1H + 1H  2He + n , two atoms of H, or 4 u, are used as fuel. The energy release is 4.91023 MeV/g. (3.27 MeV/reaction)/(4 u/reaction)(1.6610–27 kg/u) = 4.91026 MeV/kg = 2 3 4 2 3 For the reaction 1H + 1H  2He + n , one atom of H and one atom of H, or 5 u, are used as fuel. The energy release is 2.11024 MeV/g. (17.59 MeV/reaction)/(5 u/reaction)(1.6610–27 kg/u) = 2.11027 MeV/kg = In the fission reaction, one atom of 235U, or 235 u, is used as fuel. The energy release is 5.11023 MeV/g. (200 MeV/reaction)/(235 u/reaction)(1.6610–27 kg/u) = 5.11026 MeV/kg = Thus most fusion reactions yield more energy per unit mass. Note that 235U is only 0.72% of natural uranium. Thus approximately every gram of 235U would come from 1/(0.0072) = 1.4102 g 238U. On this basis the fission yield is (5.11023 MeV/g 235U)/(1.4102 g 238U/g 235U) = 3.61021 MeV/g 238U. 34. We find the minimum number of fusions by assuming 100% efficiency: P = E/t; 300 W = (3.27 MeV)(1.6010–13 J/MeV)n/(3.16107 s/yr), which gives n = 1.811022 fusions/yr. Each fusion uses two deuterium atoms, so the required mass is m = [(1.811022 atoms/yr)/(6.021023 atoms/mol)](2)(2 g/mol) =

0.120 g/yr.

35. Because the kinetic energies « mc2, we can use a non relativistic treatment: K = mv2/2 = p2/2m. We assume the kinetic energy of the deuterium and the tritium can be neglected, so for momentum conservation we have pHe = pn. The kinetic energy of the particles is Q = KHe + Kn = (pHe2/2mHe) + (pn2/2mn) = pHe2(mn + mHe)/2mHemn ; 17.59 MeV = pHe2(1 u + 4 u)/2(4 u)(1 u), which gives pHe2 = 28.1 MeV · u. The kinetic energy of 4He is KHe = pHe2/2mHe = (28.1 MeV · u)/2(4 u) = 3.5 MeV. The kinetic energy of n is Kn = pn2/2mn = (28.1 MeV · u)/2(1 u) = 14 MeV. This result is not independent of the plasma temperature, which is a measure of the initial kinetic energies. 36. The rate at which input energy is needed by the reactor is P = [(1000106 W)/(0.30)](3600 s/h)/(1.6010–13 J/MeV) = 7.501025 MeV/h. If we assume equal contributions from the two equations, four deuterium nuclei release 4.03 MeV + 3.27 MeV = 7.30 MeV. Each water molecule has two H atoms. If 0.015% of them are deuterium nuclei, the number of water molecules needed for fuel is

Ch. 43 Page 7

N = (7.501025 MeV/h)/[(7.30 MeV)/(4 nuclei)](2 nuclei/molecule)(0.01510–2) = 1.371029 molecules/h. The required mass of water is m = [(1.371029 molecules/h)/(6.021023 molecules/mol)](18 g/mol) = 4.1106 g/h = 4.1103 kg/h.

37. The number of deuterium nuclei in 1.00 kg of water is N = [(1.00103 g)/(18 g/mol)](6.021023 molecules/mol)(2 nuclei/molecule)(0.01510–2) = 1.001022 nuclei. Two deuterium nuclei release 4.03 MeV, so the total energy is E = (1.001022 nuclei)(4.03 MeV)(1.6010–13 J/MeV)/(2 nuclei) = 3.23109 J. the energy from burning 1.0 kg of gasoline. This is 65 38. (a) If we look at the reactions for the carbon cycle, we see that one carbon atom is used in the first reaction and one carbon atom is produced in the last reaction. If we add all of the reactions, we get 1 4 4 1H   2 He + 2 e + + 2 . (b) If we ignore the gamma rays and the pair annihilation, the Q-value is Q = [4M(1H) – M(4He) – 4m(e)]c2 = [4(1.007825 u) – (4.002603 u) – 4(0.0005486 u)]c2(931.5 MeV/uc2) = + 24.69 MeV. When we include the energy from the electron-positron annihilation, the total energy release is 24.69 MeV + 2(1.022 MeV) = 26.73 MeV. 12 1 13 (c) For the reaction 6C + 1H  7N , we determine the Q-value: Q = [M(12C) + M(1H) – M(13N)]c2 = [(12.000000 u) + (1.007825 u) – (13.005739 u)]c2(931.5 MeV/uc2) = + 1.94 MeV. 13 13 For the decay 7N  6 C + e + , we determine the Q-value: Q = [M(13N) – M(13C) – 2m(e)]c2 = [(13.005739 u) – (13.003355 u) – 2(0.0005486 u)]c2(931.5 MeV/uc2) = + 1.20 MeV. 13 1 14 For the reaction 6C + 1H  7N , we determine the Q-value: Q = [M(13C) + M(1H) – M(14N)]c2 = [(13.003355 u) + (1.007825 u) – (14.003074 u)]c2(931.5 MeV/uc2) = + 7.55 MeV. 14 1 15 For the reaction 7N + 1H  8O , we determine the Q-value: Q = [M(14N) + M(1H) – M(15O)]c2 = [(14.003074 u) + (1.007825 u) – (15.003065 u)]c2(931.5 MeV/uc2) = + 7.30 MeV. 15 15 For the decay 8O  7N + e + , we determine the Q-value: Q = [M(15O) – M(15N) – 2m(e)]c2 = [(15.003065 u) – (15.000108 u) – 2(0.0005486 u)]c2(931.5 MeV/uc2) = + 1.73 MeV. 15 1 12 4 For the reaction 7N + 1H  6C + 2 He , we determine the Q-value: Q = [M(15N) + M(1H) – M(12C) – M(4He)]c2 = [(15.000108 u) + (1.007825 u) – (12.000000 u) – (4.002603 u)]c2(931.5 MeV/uc2) = + 4.96 MeV. (d) For the nuclei with higher atomic number, there is a greater repulsion between the positive charges, so higher kinetic energies are required. Thus higher temperatures are required.

Ch. 43 Page 8

39. (a) To initiate the reaction, the kinetic energy of the initial nuclei must be sufficient to overcome the Coulomb barrier. The radii of the two initial nuclei for the first reaction of the carbon cycle are rC = (1.2 fm)(12)1/3 = 2.75 fm; rp = (1.2 fm)(1)1/3 = 1.2 fm. We assume the total kinetic energy equals the Coulomb energy when the nuclei are in contact: KC-p = UC-p = ZCZpe2/4pÅ0(rC + rp). The radii of the two initial nuclei for the deuteron-tritium reaction are rd = (1.2 fm)(2)1/3 = 1.51 fm; rt = (1.2 fm)(3)1/3 = 1.73 fm. We assume the total kinetic energy equals the Coulomb energy when the nuclei are in contact: Kd-t = Ud-t = ZdZte2/4pÅ0(rd + rt). Thus the ratio is KC-p/Kd-t = ZCZp(rd + rt)/ZdZt(rC + rp) = (6)(1)(1.51 fm + 1.73 fm)/(1)(1)(2.75 fm + 1.2 fm) = 4.9. (b) Because the kinetic energy is proportional to the temperature, we have KC-p/Kd-t = TC-p/Td-t ; 4.9 = TC-p/(3108 K), which gives TC-p = 1.5109 K. 40. (a) If n is the density of particles, for equal numbers of 2H and 3H atoms we have  = 200103 kg/m3 = !n(2 u)(1.6610–27 kg/u) + !n(3 u)(1.6610–27 kg/u), 4.81031 particles/m3. which gives n = (b) To meet the Lawson criterion, we have n = 31020 s/m3;  = 610–12 s. (4.81031 particles/m3) = 31020 s/m3, which gives 41. Because the quality factor for gamma rays is ˜ 1, we have effective dose (Gy) = effective dose (Sv)/QF = (4.0 Sv)/1 =

4.0 Gy.

42. Because the quality factor for -particle radiation is ˜ 20, we have effective dose (rem) = effective dose (rad)QF; rad20 = radX1; radX = (50 rad)(20) = 1000 rad. 43. Because the quality factor for slow neutrons is ˜ 3 and for fast neutrons is ˜ 10, we have effective dose (rem) = effective dose (rad)QF; radslow3 = radfast10; radslow = (50 rad)(10)/(3) = 167 rad. 44. The energy deposited is E = (50 rad)(1.0010–2 J/kg · rad)(70 kg) =

35 J.

Ch. 43 Page 9

45. If the counter counts 90% of the intercepted  particles, we have n = (0.90)(0.20)(0.02510–6 Ci)(3.71010 decays/s · Ci) = 1.7102 counts/s. 46. If the decay rate were constant, the time required would be t1 = (36 Gy)/(1.010–2 Gy/min)(60 min/h)(24 h/day) = 2.50 days. This time is (2.50 days)/(14.3 days) = 0.175 half-lives. The activity of the source at this time would be (1.010–2 Gy/min)(!)0.175 = 0.88610–2 Gy/min. If we approximate the exponential decay as linear, and use the average activity, we get t2 = (36 Gy)/!(1.010–2 Gy/min + 0.88610–2 Gy/min)(1440 min/day) = 2.65 days

˜ 2.7 days.

47. If we start with the current definition of the roentgen, we get 1 R = (0.87810–2 J/kg)/(1.6010–19 J/eV)(1000 g/kg)(35 eV/pair) = 1.571012 pairs/g ˜ 1.61012 pairs/g. 48. Because each decay gives one gamma ray, the rate at which energy is emitted is P = (1.8510–6 Ci)(3.71010 decays/s · Ci)(122103 eV/decay)(1.6010–19 J/eV) = 1.3410–9 J/s. If 50% of the energy is absorbed by the body, the dose rate is dose rate = (0.50)(1.3410–9 J/s)(86,400 s/day)/(1.00 J/kg · Gy)(70 kg) = 8.210–7 Gy/day. 49. The decay constant is  = 0.693/T1/2 = 0.693/(5730 yr) = 1.20910–4 /yr = 3.8310–12 s–1. We find the number of 14C atoms from Activity = N; (1.0010–6 Ci)(3.71010 decays/s · Ci) = (3.8310–12 s–1)N, which gives N = 9.671015 atoms. The mass is m = [(9.671015 atoms)/(6.021023 atoms/mol)](14 g/mol) = 2.2510–7 g = 0.225 g. 131 0 50. (a) 131 53I  54Xe + – 1 e +  +  . (b) We find the number of half-lives from N/N0 = (!)n; (0.10) = (!)n, or n log 2 = log 10, which gives n = 3.32. Thus the elapsed time is ?t = nT1/2 = (3.32)(8.0 days) = 27 days. (c) We find the number of atoms from Activity = N; (1.0010–3 Ci)(3.71010 decays/s · Ci) = [0.693/(8.0 days)(86,400 s/day)]N, which gives N = 3.691013 atoms. The mass is m = [(3.691013 atoms)/(6.021023 atoms/mol)](131 g/mol) = 8.010–9 g.

51. Because each decay gives one gamma ray, the rate at which energy is emitted is P = (200010–12 Ci/L)(3.71010 decays/s · Ci)(1.5 MeV/decay)(1.6010–13 J/MeV) = 1.7810–11 J/s · L. If 10% of the energy is absorbed by the body for half a year (12 h/day), the total absorbed energy rate is rate = (0.10)(1.7810–11 J/s · L)(0.5 L)!(3.16107 s/yr) = 1.410–5 J/yr. For beta particles and gamma rays, QF = 1. (a) For an adult, the total dose in one year is dose = (1.410–5 J/yr)(1)/(1.00 J/kg · Gy)(50 kg) 0.03 mrem ˜ 0.006% of allowed dose. = 2.810–7 Sv = 2.810–5 rem =

Ch. 43 Page 10

(b) For a baby, the total dose in one year is dose = (1.410–5 J/yr)(1)/(1.00 J/kg · Gy)(5 kg) 0.3 mrem ˜ 0.06% of allowed dose. = 2.810–6 Sv = 2.810–4 rem =

52. (a) We find the daughter nucleus by balancing the mass and charge numbers: Z(X) = Z(222Rn) – Z(4He) = 86 – 2 = 84; 218 A(X) = A(222Rn) – A(4He) = 222 – 4 = 218, so the product nucleus is 84Po . 218 (b) From Figure 42–11, we see that 84Po is radioactive, and decays by both  and  – emission: 218 Po   82Pb + 2He ; 218  o  85At + – 01e +  . 84P The half life for both decays is 3.1 min. (c) Both daughter nuclei are chemically reactive. (d) The decay constant is  = 0.693/T1/2 = 0.693/(3.8 days)(86,400 s/day) = 2.1110–6 s–1. The number of radon atoms in 1.0 ng is N0 = [(1.010–9 g)/(222 g/mol)](6.021023 atoms/mol) = 2.711012 atoms. The initial activity is N0 = (2.1110–6 s–1)(2.711012)/(3.71010 decays/s · Ci) = 1.510–4 Ci = 150 Ci. After 1 month, the number of half-lives is n = (30 days)/(3.8 days) = 7.89. The activity is N = N0 (!)n = (150 Ci)(!)7.89 = 0.63 Ci.

218 84

214

4

53. (a) For parallel rays the object and image will be the same size, so the magnification will be 1. (b) When the film is pressed against the back, the image of the back on the film will be the same size as the back: mback = 1. From the diagram we see that the rays which form the image of the front of the chest will form a larger image. For the magnification of the front we have mfront = h2/h1 = (d1 + d2)/d1 = (15 cm + 25 cm)/(15 cm) = 2.7. Thus the range of magnifications is 1 = m = 2.7, depending on which part of the body is being examined.

d1

d2

h1

h2

54. From the text we know that in a magnetic field of 1.000 T, resonance occurs when the photon has a frequency of 42.58 MHz. The wavelength is  = c/f = (3.00108 m/s)/(42.58106 Hz) = 7.05 m, FM radio wave. 55. (a) We find the other nucleus by balancing the mass and charge numbers:

Ch. 43 Page 11

Z(X) = Z(9Be) + Z(4He) – Z(n) = 4 + 2 – 0 = 6; 12 A(X) = A(9Be) + A(4He) – A(n) = 9 + 4 – 1 = 12, so the other nucleus is 6C . 9 4 12 1 (b) For the reaction 4 Be  + 2 He  6C + 0 n , we determine the Q-value: Q = [M(9Be) + M(4He) – M(12C) – m(n)]c2 = [(9.012182 u) + (4.002603 u) – (12.000000 u) – (1.008665 u)]c2(931.5 MeV/uc2) = 56. We find the conversion from K = kT by converting the units for k: 1.3810–23 J/K = (1.3810–23 J/K)/(1.6010–16 J/keV) =

+ 5.70 MeV.

8.6310–8 keV/K.

57. The average kinetic energy is a function of the temperature: K = !mv2 = *kT. When we form the ratio for the molecules of the two isotopes at the same temperature, we get v235/v238 = (m238/m235)1/2 = {[238 u + 6(19 u)]/[235 u + 6(19 u)]}1/2 = 1.004. 58. (a) We find the number of fissions from n = (20 kt)(51012 J/kt)/(200 MeV/fission)(1.6010–13 J/MeV) = 3.131024 fissions. Each fission uses one uranium atom, so the required mass is M = [(3.131024 atoms)/(6.021023 atoms/mol)](235 g/mol) = 1.2103 g = 1.2 kg. (b) We find the mass transformation from m = E/c2 = (20 kt)(51012 J/kt)/(3.00108 m/s)2 = 1.110–3 kg = 1.1 g. 59. The effective dose in rem = effective dose (rad)QF. For the two radiations, we get 51 mrem/yr. dose (rem) = ? dose (rad)QF = (21 mrad/yr)(1) + (3.0 mrad/yr)(10) = 60. If we assume the oceans cover 70% of the Earth’s surface and the average depth is 3 km, the mass of the ocean water is M = V = (1000 kg/m3)(0.70)4p(6.38106 m)2(3103 m) = 11021 kg. Each water molecule has two H atoms, so the number of deuterium atoms is N = [(11021 kg)(103 g/kg)/(18 g/mol)](6.021023 atoms/mol)(2)(0.01510–2) = 11043 atoms. The mass of deuterium is m = (11043 atoms)(2 g/mol)/(6.021023 atoms/mol) = 31019 g = 31016 kg. If we assume equal contributions from the reactions in Eq. 43–8a and Eq. 43–8b, the average energy released by a deuterium nucleus is (4.03 MeV + 3.27 MeV)/(2 + 2) = 1.8 MeV/nucleus. Thus the total energy released is E = (1.8 MeV/nucleus)(11043 nuclei) = 1.81043 MeV = 31030 J. 61. Because the QF for gamma rays is 1, the dose in rem is the dose in rad. The allowed dose rate is (5.0 rem/yr)/(50 wk/yr)(40 h/wk) = 2.510–3 rem/h = 2.510–3 rad/h. If the dose rate falls off as the square of the distance, we form the ratio: (dose rate)2/(dose rate)1 = (r1/r2)2; (2.510–3 rad/h)/(0.052 rad/h) = [(1.0 m)/r2]2, which gives r2 = 4.6 m. 226

222

4

62. (a) 88Ra  86Rn + 2He . (b) If we ignore the K of the daughter, the K of the  particle is the Q-value: K = Q = [M(226Ra) – M(222Rn) – M(4He)]c2 = [(226.025402 u) – (222.017570 u) – (4.002603 u)]c2(931.5 MeV/uc2) =

4.871 MeV.

Ch. 43 Page 12

(c) From momentum conservation, the momenta of the  particle and the daughter will have equal magnitudes: pRn = p = (2mK)1/2 = [2(4 u)(1.6610–27 kg/u)(4.87 MeV)(1.610–13 J/MeV)]1/2 = 1.0210–19 kg · m/s. (d) The kinetic energy of the daughter is KRn = pRn2/2mRn = (1.0210–19 kg · m/s)2/2(222 u)(1.6610–27 kg/u) = 1.410–14 J = 0.088 MeV. Because this is less than 2% of the Q-value, our approximation is valid. 18

18

63. Because the reaction 8O (p , n ) 9F requires an input of energy, the Q-value is negative: Q = [M(18O) + M(1H) – m(n) – M(18F)]c2; – 2.453 MeV = [(17.999160 u) + (1.007825 u) – (1.008665 u) – M(18F)]c2(931.5 MeV/uc2), 18.000953 u. which gives M(18F) = 64. (a) We find the number of fissions from P = E/t; 3400 MW = (200 MeV)(1.6010–19 J/eV)n/(3.16107 s), which gives n = 3.361027 fissions. Each fission uses one uranium atom, so the required mass is m = [(3.361027 atoms)/(6.021023 atoms/mol)](235 g/mol) = 1.31106 g = 1.31103 kg. (b) We find the activity from the number of Sr atoms produced: Activity = NSr = [(0.693)/(29 yr)(3.16107 s/yr)](0.06)(3.361027)/(3.71010 decays/s · Ci) = 4.1106 Ci. 65. In the net reaction four protons produce 26.2 MeV. Thus the heat of combustion is [(26.2 MeV)/(4 u)](1.610–13 J/MeV)/(1.6610–27 kg/u) = 6.311014 J/kg. 7 This is ˜ 10  the heat of combustion of coal. 66. (a) The energy is radiated uniformly over a sphere, so the total energy rate is P = (1400 W/m2)4p(1.501011 m)2 = 4.01026 J/s = 2.51039 MeV/s. (b) In the net reaction four protons produce 26.2 MeV. (The neutrinos escape with 0.5 MeV.) Thus the consumption of protons is n = (2.51039 MeV/s)/[(26.2 MeV)/4 protons] = 3.81038 protons/s. (c) We find the time from t = N/n = [(2.01030 kg)/(1.6610–27 kg/proton)]/(3.81038 protons/s) = 3.21018 s ˜ 1011 yr. 67. In the net proton-proton cycle, four protons produce two neutrinos. If we use the result from Problem 66 for the rate at which protons are consumed, for the rate at which neutrinos are produced we have n = (3.81038 protons/s)(2 neutrinos/4 protons)(3.16107 s/yr) = 6.01045 neutrinos/yr. The neutrinos are spread uniformly over a sphere centered at the Sun, so the number that would pass through an area of 100 m2 at the Earth that is always perpendicular to the neutrino flux is N0 = [(6.01045 neutrinos/yr)/4p(1.501011 m)2](100 m2) = 2.11024 neutrinos/yr. At a latitude of 40° the ceiling is not always perpendicular to the neutrino flux, so the number is reduced by a factor of cos 40°, which we assume as an average value for the annual variation in the elevation of the Sun. During the daily rotation of the Earth, the neutrino flux will vary sinusoidally from zero to the maximum. (We take both directions through the ceiling as positive.) To average this variation, we use a factor of cos 45°, so we have N = N0 cos 40° cos 45° = (2.11024 neutrinos/yr) cos 40° cos 45° ˜ 11024 neutrinos/yr. 68. The kinetic energy of the products is

Ch. 43 Page 13

Kpr = Kb + Q.

Because the kinetic energies « mc2, we can use a non relativistic treatment: K = mv2/2 = p2/2m. The least kinetic energy is required when the product particles move together with the same speed. With the target at rest, for momentum conservation we have pb = ppr = mprv, or Kb = pb2/2mb = (mpr2/2mb)v2 = (mpr/mb)Kpr , or Kpr = (mb/mpr)Kb . When we use this in the kinetic energy equation, we get (mb/mpr)Kb = Kb + Q; [(mb/mpr) – 1]Kb = Q, which gives Kb = – Qmpr/(mpr – mb).

69. (a) If we assume a thin target, we find the cross section for backward scattering from R/R0 = nx; 1.610–5 = (5.91028 m–3)(4.010–7 m), which gives  = 6.810–28 m2 = 6.8 bn. (b) If we assume the cross section is the area presented by the nucleus, we have  = 6.810–28 m2 = pr2, which gives r = 1.510–14 m, so the diameter is 310–14 m. 12

12

24

70. (a) For the reaction 6C + 6C  12Mg , we determine the Q-value: Q = [M(12C) + M(12C) – M(24Mg)]c2 = [2(12.000000 u) – (23.985042 u)]c2(931.5 MeV/uc2) = + 13.93 MeV. Thus 13.93 MeV is released. (b) If the two nuclei are just touching, the Coulomb potential energy must be the initial kinetic energies of the two nuclei: 2K = U = ZCZCe2/4pÅ0R = (6)(6)(1.44 MeV · fm)/(6.0 fm) = 8.6 MeV. Thus each carbon nucleus has a kinetic energy of 4.3 MeV. (c) We find the temperature from K = *kT; 31010 K. (4.3 MeV)(1.6010–13 J/MeV) = *(1.3810–23 J/K)T, which gives T = 71. (a) The rate of decay is 3.7103 decays/s. Activity = (0.1010–6 Ci)(3.71010 decays/s · Ci) = (b) Because the QF for beta particles is 1, the dose in Sv is the dose in Gy: dose rate = (3.7103 decays/s)(1.4 MeV/decay)(3.16107 s/yr)(1.610–13 J/MeV)/ (50 kg)(1.00 J/kg · Gy) = 5.210–4 Sv/yr ˜ 0.15 background. 72. The activity released into the atmosphere was activity = (2.0107 Ci)(3.701010 Bq/Ci) = 7.41017 Bq. If we assume uniform spreading over the surface of the Earth, we have activity/m2 = (7.41017 Bq)/4p(6.38106 m)2 = 1.45103 Bq/m2.

Ch. 44 Page 1

CHAPTER 44 – Elementary Particles Note:

1.

2.

3.

A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.6310–34 J · s)(3108 m/s)(109 nm/m)/(1.6010–19 J/eV); E = (1.24103 eV · nm)/ = (1.2410–12 MeV · m)/.

The total energy of the proton is E = K + mpc2 = 6.35 GeV + 0.938 GeV =

7.29 GeV.

The total energy of the electron is E = K + mec2 = 35 GeV + 0.511 MeV = 35 GeV. Because the mass is negligible, the momentum is p = E/c. We find the wavelength from  = h/p = hc/E = (1.2410–12 MeV · m)/(35 GeV)(103 MeV/GeV) =

3.510–17 m.

We find the magnetic field from the cyclotron frequency: f = qB/2pm; 2.8107 Hz = (1.6010–19 C)B/2p(1.6710–27 kg), which gives B =

1.8 T.

4.

Very-high-energy protons will have a speed v ˜ c. Thus the time for one revolution is t = 2pr/v = 2p(1.0103 m)/(3.00108 m/s) = 2.110–5 s = 21 s.

5.

The cyclotron frequency is f = qB/2pm; If we form the ratio for the two particles, we get f2/f1 = (q2/q1)(m1/m2); f2/(26 MHz) = (2)(1/4), which gives f2 = 13 MHz.

6.

(a) The maximum kinetic energy is K = !mv2 = q2B2R2/2m. If we form the ratio for the two particles, we get K/Kp = (q/qp)2(mp/m); K/(8.7 MeV) = (2)2(1/4) = 1, which gives K = 8.7 MeV. We find the speed from K = !mv2; (8.7 MeV)(1.6010–13 J/MeV) = !(4)(1.6710–27 kg)v2, which gives v = (b) For deuterons we get Kd/Kp = (qd/qp)2(mp/md); Kd/(8.7 MeV) = (1)2(1/2) = 1/2, which gives Kd = 4.3 MeV. We find the speed from Kd = !mvd2; (4.3 MeV)(1.6010–13 J/MeV) = !(2)(1.6710–27 kg)vd2, which gives vd = Note that the  particle and the deuteron have the same q/m. (c) The cyclotron frequency is f = qB/2pm; If we form the ratio for the two particles, we get fd/f = (qd/q)(m/md) = (1/2)(4/2) = 1, so they require the same frequency. We find the frequency from fd/fp = (qd/qp)(mp/md);

2.0107 m/s.

2.0107 m/s.

Ch. 44 Page 2

fd/(26 MHz) = (1)(1/2), which gives fd = f =

13 MHz.

7.

The size of a nucleon is d ˜ 2(1.210–15 m) = 2.410–15 m. Because 30 MeV « mc2, we find the momentum from K = p2/2m, so the wavelength is  = h/p = h/(2mK)1/2. For the  particle we have  = (6.6310–34 J · s)/[2(4)(1.6710–27 kg)(30 MeV)(1.6010–13 J/MeV)]1/2 = 2.610–15 m ˜ size of nucleon. For the proton we have p = (6.6310–34 J · s)/[2(1.6710–27 kg)(30 MeV)(1.6010–13 J/MeV)]1/2 = 5.210–15 m ˜ 2(size of nucleon).  particle is better. Thus the

8.

The alternating frequency means the proton is accelerated by the voltage twice in each revolution. The number of revolutions is n = E/2eV = (25 MeV)(103 keV/MeV)/2(1 e)(55 kV) = 2.3102 rev.

9.

The total energy of the proton is E = K + mc2 = 900 GeV + 0.938 GeV = 900 GeV. Thus v ˜ c and the momentum is p = E/c. We find the wavelength from  = h/p = hc/E = (1.2410–12 MeV · m)/(900 GeV)(103 MeV/GeV) = 1.410–18 m. This is the minimum resolving distance, or the maximum resolving power.

10. (a) We find the magnetic field from the maximum kinetic energy: K = !mv2 = q2B2R2/2m; (10 MeV)(1.6010–13 J/MeV) = (1.6010–19 C)2B2(1.0 m)2/2(2)(1.6710–27 kg), which gives B = 0.65 T. (b) The cyclotron frequency is f = qB/2pm = (1.6010–19 C)(0.65 T)/2p(2)(1.6710–27 kg) = 4.9106 Hz = 4.9 MHz. (c) The deuteron is accelerated by the voltage twice in each revolution. The number of revolutions is n = K/2eV = (10 MeV)(103 keV/MeV)/2(1 e)(22 kV) = 227 rev = 2.3102 rev. (d) Because the time for each revolution is the same, we have t = nT = n/f = (227)/(4.9106 Hz) = 4.610–5 s = 46 s. (e) If the initial speed is zero, after N revolutions the speed is given by !mvN2 = 2NeV, with !mvmax2 = 2neV = K, or vN = (4NeV/m)1/2 = vmax(N/n)1/2. Because each revolution takes the same time, the average speed is n

vN d N

vmax n 1/ 2 N d N = 23 vmax . n n 3/ 2 0 Thus for the total distance we have d = vavt = %vmaxt = %(2K/m)1/2t vav =

0

=

= %[2(10 MeV)(1.6010–13 J/MeV)/(2)(1.6710–27 kg)]1/2(4.610–5 s) = 9.5102 m = 0.95 km.

Ch. 44 Page 3

11. The number of revolutions is n = ?E/2eV = (900 GeV – 8.0 GeV)(103 MeV/GeV)/(1 e)(2.5 MV) = 3.57105 rev. The total distance traveled is d = n2pR = (3.57105)2p(1.0 km) = 2.2106 km. Very-high-energy protons will have a speed v ˜ c. Thus the time is t = d/v = (2.2109 m)/(3.00108 m/s) = 7.5 s. 12. Very-high-energy protons will have a speed v ˜ c. Thus the time for one revolution is T = 2pR/v = 2p(1.0103 m)/(3.00108 m/s) = 2.0910–5 s. The number of turns is n = t/T = (20 s)/(2.0910–5 s) = 9.57105 turns. We find the energy provided on each turn from ?E/n = (900 GeV – 150 GeV)/(9.57105 turns) = 7.810–4 GeV/turn = 0.78 MeV/turn. 13. In the relativistic limit v ˜ c, and E = pc. For Newton’s law we have F = dp/dt = d[m/(1 – v2/c2)1/2v]/dt = [m/(1 – v2/c2)1/2] dv/dt, because v is constant. The magnetic field provides the radial acceleration: qvB = [m/(1 – v2/c2)1/2]v2/r, or B = [m/(1 – v2/c2)1/2]v/qr = p/qr = E/qrc. Thus the energy is E = qBrc. If the energy is in eV, the charge is 1 e, so we have E(eV) = Brc. 14. Because the kinetic energy is much greater than the rest energy, E = pc and v ˜ c. We use the result from Problem 13: E(eV) = Brc; 3.0 T. 900 GeV = 900109 eV = B(1.0103 m)(3.00108 m/s), which gives B = 15. For the reaction p+  + +  , we determine the Q-value: Q = [m(p+) – m(+)]c2 = [(139.6 MeV/c2) – (105.7 MeV/c2)]c2 = 33.9 MeV. Thus 33.9 MeV is released. 

16. For the reaction 0  n + p0, we determine the Q-value: Q = [m(0) – m(n) – m(p0)]c2 = [(1115.7 MeV/c2) – (939.6 MeV/c2) – (135.0 MeV/c2)]c2 = 41.1 MeV. Thus 41.1 MeV is released. 17. The minimum energy must provide the rest energy of the pair: E = 2mnc2 = 2(939.6 MeV) = 1879 MeV = 1.879 GeV. 18. We estimate the range from mc2 = hc/2pd; (497.7 MeV) = (1.2410–12 MeV · m)/2pd, 0.40 fm. which gives d = 4.010–16 m = 19. Because two protons are present before and after the process, and the total momentum is zero, the minimum kinetic energy will produce all three particles at rest. Thus the total initial kinetic energy must

Ch. 44 Page 4

provide the rest energy of the p0 meson: 2K = mp0c2 = 135 MeV, which gives K =

67.5 MeV.

20. (a) For the decay 0  + + p–, the conservation laws are Charge: 0 = + 1 – 1; Energy (mass): 1314.9 MeV < 1189.4 MeV + 139.6 MeV; Thus this decay is forbidden, energy is not conserved. (b) For the decay –  0 + p– + , the conservation laws are Charge: – 1 = 0 – 1; Energy (mass): 1672.5 MeV > 1192.6 MeV + 139.6 MeV; Baryon number:+ 1 = + 1 + 0 + 0; Lepton number:0 ? 0 + 0 + 1. Thus this decay is forbidden, lepton number is not conserved. 0 0 (c) For the decay    + + , the conservation laws are Charge: 0 = 0 + 0 + 0; Energy (mass): 1192.6 MeV > 1115.7 MeV + 0 + 0; Baryon number:+ 1 = + 1 + 0 + 0; Lepton number:– 1 = – 1 + 0 + 0. Thus this decay is possible. 21. If we use the average mass, we estimate the range from mc2 = hc/2pd; (85 GeV)(103 MeV/GeV) = (1.2410–12 MeV · m)/2pd, 2.310–18 m. which gives d = 22. The energy of the two photons must be the rest energy of the proton and antiproton: 2mpc2 = 2hf = 2hc/; 2(938.3 MeV/c2)c2 = 2(1.2410–12 MeV · m)/, which gives  = 1.3210–15 m. 23. (a) For the reaction e–  e– + , the isolated electron is at rest. For the photon, we have E = pc. For energy conservation we have mec2 = [(pec)2 + (mec2)2]1/2 + E. For momentum conservation we have 0 = pe – p. When we eliminate pe in the energy equation and rearrange, we have mec2 – E = [E2 + (mec2)2]1/2. When we square both sides, we have (mec2)2 + E2 – 2E mec2 = E2 + (mec2)2, which gives E= 0. Thus no photon is emitted. (b) For the photon exchange in Fig. 44–10, the photon exists for such a short time that the uncertainty principle allows energy to not be conserved during the exchange. 24. We assume the momentum of the electron-positron pair is zero. The two photons must have opposite and equal momenta, therefore, equal energies. The energy of the two photons must be the total energy of the pair: 2(K + mec2) = 2hf = 2hc/; 2[(0.420 MeV) + (0.511 MeV/c2)c2] = 2(1.2410–12 MeV · m)/, 1.3310–3 nm. which gives  = 1.3310–12 m =

Ch. 44 Page 5

25. The total kinetic energy after the decay of the stationary p+ is the Q-value: Q = [m(p+) – m(e+) – m()]c2 = [(139.6 MeV/c2) – (0.511 MeV/c2) – 0]c2 = 139.1 MeV. For momentum conservation we have 0 = pe – p, or (p c)2 = (pec)2 = Ee2 – (mec2)2 = Ke2 + 2Kemec2. For energy conservation we have Q = Ke + p c = Ke + pec, or (pec)2 = (Q – Ke)2 = Q2 – 2KeQ + Ke2. When we combine this with the result from momentum conservation, we get Ke = Q2/2(Q + mec2) = (139.1 MeV)2/2(139.1 MeV + 0.511 MeV) = 69.3 MeV. 



26. The minimum initial kinetic energy of the neutron and proton must provide the rest energy of the K+K– pair: Kp + Kn = 2mKc2 = 2(493.7 MeV) = 987.4 MeV. Because the neutron and proton have the same speed but different masses, they have slightly different kinetic energies: K = ({1/[1 – (v/c)2]1/2} – 1)mc2  m. Thus we have Kn/(Kp + Kn) = mn/(mp + mn); Kn/(987.4 MeV) = (939.6 MeV)/(938.3 MeV + 939.6 MeV), which gives Kn = Kp/(Kp + Kn) = mp/(mp + mn); Kp/(987.4 MeV) = (938.3 MeV)/(938.3 MeV + 939.6 MeV), which gives Kp = Note that we have ignored the small total momentum of the system.

494.0 MeV; 493.4 MeV.

27. The total kinetic energy after the decay of the stationary – is the Q-value: K + Kp = Q = [m(–) – m(0) – m(–)]c2 = [(1321.3 MeV/c2) – (1115.7 MeV/c2) – (139.6 MeV/c2)]c2 = 66.0 MeV. For energy conservation we have mc2 = E + Ep , or Ep = mc2 – E . For momentum conservation we have 0 = pp – p, or (pc)2 = (ppc)2 = E2 – (mc2)2 = Ep2 – (mpc2)2 . When we combine this with the result from energy conservation, we get E = [(mc2)2 + (mc2)2 – (mpc2)2]/2mc2 = [(1321.3 MeV)2 + (1115.7 MeV)2 – (139.6 MeV)2]/2(1321.3 MeV) = 1124.3 MeV. For the kinetic energies we have K = E – mc2 = 1124.3 MeV – 1115.7 MeV = 8.6 MeV; Kp = Q – K = 66.0 MeV – 8.6 MeV = 57.4 MeV. 

28.

No, because the kinetic energy of the incoming proton is less than the rest energy of the p+ meson, which is 139.6 MeV. The reaction is p + p  p + n + p+. The minimum initial kinetic energy produces the three particles moving together with the same speed, which we can consider to be a single particle with mass M = mp + mn + mp . For energy conservation we have E p + m pc 2 = E M .

Ch. 44 Page 6

For momentum conservation we have pp + 0 = pM , or (ppc)2 = (pMc)2 = Ep2 – (mpc2)2 = EM2 – (Mc2)2 . When we combine this with the result from energy conservation, we get Ep = [(Mc2)2 – 2(mpc2)2]/2mpc2 = {[(mp + mn + mp)c2]2 – 2(mpc2)2}/2mpc2 = [(938.3 MeV + 939.6 MeV + 139.6 MeV)2 – 2(938.3 MeV)2]/2(938.3 MeV) = 1230.7 MeV. For the kinetic energy we have Kp = Ep – mpc2 = 1230.7 MeV – 938.3 MeV = 292.4 MeV. 29. For the decay  –   e – +  e +   , the electron will have maximum kinetic energy when the two neutrinos have the same momentum and move opposite to the direction of the electron. For a neutrino E = pc. For energy conservation we have mc2 = Ee + 2E , or 2E = Ee – mc2. For momentum conservation we have 0 = pe – 2p, or (2p c)2 = 4E2 = (pec)2 = Ee2 – (mec2)2. When we combine this with the result from energy conservation, we get Ee = [(mc2)2 + (mec2)2]/2mc2 = [(105.7 MeV)2 + (0.511 MeV)2]/2(105.7 MeV) = 52.85 MeV. For the maximum kinetic energy we have Ke = Ee – mec2 = 52.85 MeV – 0.511 MeV = 52.3 MeV. 

30. From Fig. 44–14 we estimate the energy width as 140 MeV. We estimate the lifetime from ?t = h/2p ?E = hc/2pc ?E = (1.2410–12 MeV · m)/2p(3.00108 m/s)(140 MeV) =

4.710–24 s.

31. We estimate the lifetime from ?t = h/2p ?E = hc/2pc ?E = (1.2410–12 MeV · m)/2p(3.00108 m/s)(0.088 MeV) =

7.510–21 s.

32. We estimate the lifetime from ?t = h/2p ?E = hc/2pc ?E = (1.2410–12 MeV · m)/2p(3.00108 m/s)(0.277 MeV) =

2.3710–21 s.

33. We find the energy width from the lifetime given in Table 44–2. (a) For 0 we have ?E = h/2p ?t = hc/2pc ?t = (1.2410–12 MeV · m)/2p(3.00108 m/s)(510–19 s) = 1.310–3 MeV = 1.3 keV. 0 (b) For  we have ?E = h/2p ?t = hc/2pc ?t = (1.2410–12 MeV · m)/2p(3.00108 m/s)(7.410–20 s) = 8.910–3 MeV = 8.9 keV. 34. (a) For B – = b u we have Charge: – 1 = – @ – %; Spin: 0 = + ! – !; Baryon number: 0 = + @ – @; Strangeness: 0 = 0 + 0; Charm: 0 = 0 + 0; Bottomness: – 1 = – 1 + 0; Topness: 0 = 0 + 0. B+ = b u . (b) Because B+ is the antiparticle of B–, we have 0 For B , to make the charge zero, we need the antiquark with the same properties as u , but with a charge of + @. Thus we have

B0 = b d .

Because B 0 is the antiparticle of B0, we have

B0 = b d .

Ch. 44 Page 7

35. (a) For the neutron we must have charge, strangeness, charm, bottomness, and topness = 0. For the baryon number to be 1, we need three quarks: n = ddu. n = d d u . (b) For the antineutron we have (c) For the 0 we must have charge, charm, bottomness, and topness = 0. To get strangeness = – 1, we need an s quark. To get a baryon number of 1, we need three quarks. To get a charge = 0, we have 0 = uds. (d) For the 0 we must have charge, charm, bottomness, and topness = 0. To get strangeness = + 1, we need an s quark. To get a baryon number of – 1, we need three antiquarks. To get a charge = 0, we have

0 = u d s .

36. (a) For uud we have charge +1, baryon number = + 1, while strangeness, charm, bottomness, and topness = 0. Thus we have p. (b) For u u s we have charge = – 1, baryon number = – 1, strangeness = + 1, while charm,  +. bottomness, and topness = 0. Thus we have (c) For u s we have charge = – 1, baryon number = 0, strangeness = – 1, while charm, bottomness, and topness = 0. Thus we have K–. (d) For d u we have charge = – 1, baryon number = 0, strangeness = 0, while charm, bottomness, and topness = 0. Thus we have p–. (e) For cs we have charge = – 1, baryon number = 0, strangeness = – 1, charm = – 1, while bottomness, and topness = 0. Thus we have D –S .

37. For the D0 we have charge, baryon number, strangeness, bottomness, and topness = 0. To get charm = + 1 we need the c quark. To get baryon number = 0, we need an antiquark. Thus we have D0 = c u . 38. For the D S+ we have baryon number, bottomness, and topness = 0. To get charm = + 1 we need the c quark. To get baryon number = 0, we need an antiquark. To get strangeness = + 1 we need the s quark. D S+ = c s . Thus we have 39. (a)

(b) From Table 44–3 we see that the p0 could be

Ch. 44 Page 8

š0 _ u u

d

_ u d š–

either u u or d d . Actually it is a combination of the two, but we will simplify the diagram by assuming that one p0 is u u and the other is d d . š0 _ š0 _ u d u d

n u d

u u d p

We see that the exchanged particle is p+.

_ u _ _ _u d p

u

u d p

We see that the exchanged particle is p. Note that high-energy p p collisions usually produce many particles. 40.

K– _ s u

u u

s _ u K–

u u d p

p d

We see that the exchanged particle is p0. 41. The total energy of the proton is E = K + mc2 = 25 GeV + 0.938 GeV = 26 GeV. We find the momentum from (pc)2 = E2 – (mc2)2 = (26 GeV)2 – (0.938 GeV)2, which gives pc = 26 GeV. We find the wavelength from  = h/p = hc/pc = (1.2410–12 MeV · m)/(26 GeV)(103 MeV/GeV) = 4.810–17 m. 42. (a) At the high energy of 900 GeV the speed of the protons ˜ c. Thus the time for one proton to make a revolution is T = 2pr/c = 2p(1.0103 m)/(3.00108 m/s) = 2.0910–5 s. The current carried by the beam of N protons is I = Ne/T = (51013)(1.6010–19 C)/(2.0910–5 s) = 0.38 A. (b) We find the speed of the car from K = !mv2; 98 m/s (350 km/h). (51013)(900 GeV)(1.6010–10 J/GeV) = !(1500 kg)v2, which gives v =

Ch. 44 Page 9

43. Because 7.0 TeV » mpc2, we have K = E = pc. We use the result from Problem 13: E(eV) = Brc; (7.0 TeV)(1012 eV/TeV) = B(4.25103 m)(3.00108 m/s), which gives B =

5.5 T.

44. (a) We assume the momentum of the electron-positron pair is zero. The released energy is E = 2mec2 = 2(0.511 MeV) = 1.022 MeV. (b) We assume the momentum of the proton-antiproton pair is zero. The released energy is E = 2mpc2 = 2(938.3 MeV) = 1876.6 MeV.

45. (a) For the reaction p– + p  K+ + –, the conservation laws are Charge: – 1 + 1 = + 1 – 1; Spin: Baryon number: 0 + 1 = 0 + 1; Lepton number: Strangeness: 0 + 0 = + 1 – 1. Thus the reaction is possible, through the strong interaction. (b) For the reaction p+ + p  K+ + +, the conservation laws are Charge: + 1 + 1 = + 1 + 1; Spin: Baryon number: 0 + 1 = 0 + 1; Lepton number: Strangeness: 0 + 0 = + 1 – 1. Thus the reaction is possible, through the strong interaction. – (c) For the reaction p + p  0 + K0 + p0, the conservation laws are Charge: – 1 + 1 = 0 + 0 + 0; Spin: Baryon number: 0 + 1 = + 1 + 0 + 0; Lepton number: Strangeness: 0 + 0 = – 1 + 1 + 0. Thus the reaction is possible, through the strong interaction. (d) For the reaction p+ + p  0 + p0, the conservation laws are Charge: + 1 + 1 ? 0 + 0; Thus the reaction is forbidden, charge is not conserved. (e) For the reaction p– + p  p + e– +  e , the conservation laws are Charge: – 1 + 1 = + 1 – 1 + 0; Spin: Baryon number: 0 + 1 = + 1 + 0 + 0; Lepton number: Strangeness: 0 + 0 = 0 + 0 + 0. Thus the reaction is possible, through the weak interaction. – (f) For the reaction p + p  K0 + p + p0, the conservation laws are Charge: – 1 + 1 ? 0 + 1 + 0; Thus the reaction is forbidden, charge is not conserved. (g) For the reaction K– + p  0 + p0, the conservation laws are Charge: – 1 + 1 = 0 + 0; Spin: Baryon number: 0 + 1 = + 1 + 0; Lepton number: Strangeness: – 1 + 0 = – 1 + 0.

0 + ! = 0 + !; 0 + 0 = 0 + 0;

0 + ! = 0 + !; 0 + 0 = 0 + 0;

0 + ! = ! + 0 + 0; 0 + 0 = 0 + 0 + 0;

0 + ! = ! + ! – !; 0 + 0 = 0 + 1 – 1;

0 + ! = ! + 0; 0 + 0 = 0 + 0;

Ch. 44 Page 10

Thus the reaction is possible, through the strong interaction. (h) For the reaction K+ + n  + + p0 + , the conservation laws are Charge: + 1 + 0 = + 1 + 0 + 0; Spin: Baryon number: 0 + 1 = + 1 + 0 + 0; Lepton number: Strangeness: + 1 + 0 ? – 1 + 0 + 0. Thus the reaction is forbidden, strangeness is not conserved. (i) For the reaction K+  p0 + p0 + p+, the conservation laws are Charge: + 1 = 0 + 0 + 1; Spin: Baryon number: 0 = 0 + 0 + 0; Lepton number: Strangeness: + 1 = 0 + 0 + 0. Thus the reaction is possible, through the weak interaction. (j) For the reaction p+  e+ + e , the conservation laws are Charge: Baryon number: Strangeness: Thus the reaction is

+ 1 = + 1 + 0; Spin: 0 = 0 + 0; Lepton number: 0 = 0 + 0. possible, through the weak interaction.

0 + ! = – ! + 0 + 1; 0 + 0 = 0 + 0 + 0.

0 = 0 + 0 + 0; 0 = 0 + 0 + 0;

0 = ! – !; 0 = – 1 + 1;

46. (a) For the decay 0  p + p–, the Q-value is Q = [m(0) – m(p) – m(p–)]c2 = [(1115.7 MeV/c2) – (938.3 MeV/c2) – (139.6 MeV/c2)]c2 = 37.8 MeV. (b) The total kinetic energy after the decay of the stationary 0 is the Q-value: Kp + Kp = Q = 37.8 MeV. For energy conservation we have mc2 = Ep + Ep , or Ep = mc2 – Ep . For momentum conservation we have 0 = pp – pp, or (ppc)2 = (ppc)2 = Ep2 – (mpc2)2 = Ep2 – (mpc2)2. When we combine this with the result from energy conservation, we get Ep = [(mc2)2 + (mpc2)2 – (mpc2)2]/2mc2 = [(1115.7 MeV)2 + (938.3 MeV)2 – (139.6 MeV)2]/2(1115.7 MeV) = 943.7 MeV. For the kinetic energies we have Kp = Ep – mpc2 = 943.7 MeV – 938.3 MeV = 5.4 MeV; Kp = Q – Kp = 37.8 MeV – 5.4 MeV = 32.4 MeV. 

47. We estimate the energy from ?x ˜ h/2p ?p = hc/2p ?pc = hc/2p ?E; 10–18 m = (1.2410–12 MeV · m)/2p ?E , which gives ?E = 2105 MeV = 2102 GeV. This is on the order of the 80 GeV rest energy of the W particles. 48. For energy conservation we have mp–c2 + mpc2 = Ep0 + En= Ep0 + Kn + mnc2; 139.6 MeV + 938.3 MeV = Ep0 + 0.60 MeV + 939.6 MeV, which gives Ep0 = 137.7 MeV. For momentum conservation we have 0 = pp0 – pn, or (pnc)2 = (pp0c)2 = En2 – (mnc2)2 = Ep02 – (mp0c2)2; (0.60 MeV + 939.6 MeV)2 – (939.6 MeV)2 = (137.7 MeV)2 – (mp0c2)2,

Ch. 44 Page 11

which gives mp0 =

133.5 MeV/c2.

49. For the reaction p + p  p + p + p0, the Q-value is Q = [2mp – 2mp – mp0]c2 = – mp0c2 = – 135.0 MeV. For the reaction p + p  p + n + p+, the Q-value is Q = [2mp – mp – mn – mp+]c2 = 938.3 MeV – 939.6 MeV – 139.6 MeV =

– 140.9 MeV.

50. For the reaction p– + p  0 + K0, the Q-value is Q = [m(p–) + m(p) – m(0) – m(K0)]c2 = [(139.6 MeV/c2) + (938.3 MeV/c2) – (1115.7 MeV/c2) – (497.7 MeV/c2)]c2 = – 535.5 MeV. The minimum initial kinetic energy produces the two particles moving together with the same speed, which we can consider to be a single particle with mass M = m + mK . For energy conservation we have E p + m pc 2 = E M . For momentum conservation we have pp + 0 = pM , or (ppc)2 = (pMc)2 = Ep2 – (mpc2)2 = EM2 – (Mc2)2 . When we combine this with the result from energy conservation, we get Ep = [(Mc2)2 – (mpc2)2 – (mpc2)2]/2mpc2 = [(mc2 + mKc2)2 – (mpc2)2 – (mpc2)2]/2mpc2 = [(1115.7 MeV + 497.7 MeV)2 – (139.6 MeV)2 – (938.3 MeV)2]/2(938.3 MeV) = 907.6 MeV. For the threshold kinetic energy we have Kp = Ep – mpc2 = 907.6 MeV – 139.6 MeV = 768.0 MeV.

51. We consider the quarks and leptons as the fundamental fermions. A water molecule consists of two hydrogen atoms and one oxygen atom, or 2 + 8 = 10 protons, 2 + 8 = 10 electrons, and 8 neutrons. Each of the protons and neutrons is made up of three quarks, so the number of fundamental fermions is N = (10 + 8)(3) + 10 = 64. 52. (a) For the decay K0  p– + e+ + e , the positron will have maximum kinetic energy when the pion has no kinetic energy. Note that the neutrino, with essentially zero mass, cannot have zero kinetic energy. The total kinetic energy after the decay of the stationary K0 is the Q-value: Ke + K = Q = [m(K0) – m(p–) – m(e+)]c2 = [(497.7 MeV/c2) – (139.6 MeV/c2) – (0.511 MeV/c2)]c2 = 357.6 MeV. For energy conservation we have mKc2 = mpc2 + Ee + E , or E = (mKc2 – mpc2) – Ee . The positron and neutrino must move in opposite directions. For momentum conservation we have 0 = pe – p, or (pec)2 = Ee2 – (mec2)2 = (pc)2 = E2. When we combine this with the result from energy conservation, we get Ee = [(mKc2 – mpc2)2 + (mec2)2]/2(mKc2 – mpc2) = [(497.7 MeV – 139.6 MeV)2 + (0.511 MeV)2]/2(497.7 MeV – 139.6 MeV) = 179.1 MeV. For the kinetic energies we have K = E = (497.7 MeV – 139.6 MeV) – 179.1 MeV = 179.0 MeV; Ke = Q – K = 357.6 MeV – 179.0 MeV = 178.6 MeV. Alternatively, Ke = Ee – mec2 = 179.1 MeV – 0.511 MeV = 178.6 MeV. (b) For the decay K0  p– + e+ + e , the pion will have maximum kinetic energy when the positron has no kinetic energy. Note that the neutrino, with essentially zero mass, cannot have zero kinetic energy. The total kinetic energy after the decay of the stationary K0 is the Q-value: 



Ch. 44 Page 12

Kp + K = Q = [m(K0) – m(p–) – m(e+)]c2 = [(497.7 MeV/c2) – (139.6 MeV/c2) – (0.511 MeV/c2)]c2 = 357.6 MeV. For energy conservation we have mKc2 = mec2 + Ep + E , or E = (mKc2 – mec2) – Ep . The pion and neutrino must move in opposite directions. For momentum conservation we have 0 = pp – p, or (ppc)2 = Ep2 – (mpc2)2 = (pc)2 = E2. When we combine this with the result from energy conservation, we get Ep = [(mKc2 – mec2)2 + (mpc2)2]/2(mKc2 – mec2) = [(497.7 MeV – 0.511 MeV)2 + (139.6 MeV)2]/2(497.7 MeV – 0.511 MeV) = 268.2 MeV. For the kinetic energies we have K = E = (497.7 MeV – 0.511 MeV) – 268.2 MeV = 229.0 MeV; Kp = Q – K = 357.6 MeV – 229.0 MeV = 128.6 MeV. Alternatively, Kp = Ep – mpc2 = 268.2 MeV – 139.6 MeV = 128.6 MeV. 



53. (a) If the momentum is taken to be the minimum allowed by the uncertainty principle, we have E = pc = c ?p. Thus we estimate the energy from ?x ˜ h/2p ?p = hc/2pc ?p = hc/2pE; 10–30 m = (1.2410–12 MeV · m)/2pE , which gives E = 21017 MeV = 21014 GeV. (b) We find the temperature from E = kT; (21014 GeV)(1.6010–10 J/GeV) = (1.3810–23 J/K)T. which gives T ˜ 1027 K.

54. For the reaction p + p  3p + p the Q-value is Q = [2mp – 4mp]c2 = – 2mpc2. The minimum initial kinetic energy produces the four particles moving together with the same speed, which we can consider to be a single particle with mass M = 4mp . For energy conservation we have E p + m pc 2 = E M . For momentum conservation we have pp + 0 = pM , or (ppc)2 = (pMc)2 = Ep2 – (mpc2)2 = EM2 – (Mc2)2. When we combine this with the result from energy conservation, we get Ep = [(Mc2)2 – 2(mpc2)2]/2mpc2 = [(4mpc2)2 – 2(mpc2)2]/2mpc2 = 7mpc2. For the threshold kinetic energy we have Kp = Ep – mpc2 = 7mpc2 – mpc2 = 6mpc2. This is three times the magnitude of the Q-value. 55. We find the speed of the tau-lepton from its kinetic energy: K = {[1/(1 – v2/c2)1/2] – 1}mc2; 450 MeV = {[1/(1 – v2/c2)1/2] – 1}(1777 MeV), which gives 1/(1 – v2/c2)1/2 = 1.25, and v = 0.603c. In the lab the lifetime of the tau-lepton will be dilated: ?t = ?t0/(1 – v2/c2)1/2. Thus the length of the track will be ?x = v ?t = (0.603)(3.00108 m/s)(2.9110–13 s)(1.25) = 6.5810–5 m.

Ch. 44 Page 13

56. For energy conservation of the particle’s decay, we have mPc2 = ED1 + ED2 = KD1 + mD1c2 + KD2 + mD2c2. The two decay fragments must move in opposite directions. For momentum conservation we have 0 = pD1 – pD2, or (pD1c)2 = (pD2c)2 = ED12 – (mD1c2)2 = ED22 – (mD2c2)2. When we use the result from energy conservation, we get ED12 – (mD1c2)2 = (mPc2 – ED1)2 – (mD2c2)2, which gives ED1 = [(mPc2)2 + (mD1c2)2 – (mD2c2)2]/2mPc2. Thus the kinetic energy is KD1 = ED1 – mD1c2 = {[(mPc2)2 + (mD1c2)2 – (mD2c2)2]/2mPc2} – (2mPc2mD1c2/2mPc2) = [(mPc2 – mD1c2)2 – (mD2c2)2]/2mPc2. 57. From Table 44–2 we see that the p0 has charge = 0, B = 0, L = 0, and S = 0. From Table 44–3 we see that the p0 could be either u u or d d . Actually it is a combination of the two, but we will simplify the analysis by assuming the p0 is d d . Thus we can write the reaction as p + n  p– + p0; u ud + u d d  u d + dd . Thus we see the annihilation of a u u pair. 58. We find the speed of the proton from its kinetic energy: K = {[1/(1 – v2/c2)1/2] – 1}mc2; 7.0106 MeV = {[1/(1 – v2/c2)1/2] – 1}(938 MeV), which gives  = 1/(1 – v2/c2)1/2 = 7.46103. This means the speed is very close to the speed of light. If we rearrange and use the approximation (1 – x)1/2 ˜ 1 – x/2, we get v/c = [1 – (1/ 2)]1/2 ˜ 1 – (1/2 2) = 1 – [1/2(7.46103)2], which gives v/c = 1 – 910–9. 



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CHAPTER 45 – Astrophysics and Cosmology Note: A factor that appears in the analysis of energies is e2/4pÅ0 = (1.6010–19 C)2/4p(8.8510–12 C2/N · m2) = 2.3010–28 J · m = 1.44 MeV · fm. 1.

Using the definition of the parsec, we find the equivalent distance in m: D = d/; 1 pc = (1.501011 m)/[(1.000)/(3600/°)(180°/p rad)] = 3.0941016 m. From the definition of the light-year, we have 1 pc = (3.0941016 m)/(3.00108 m/s)(3.16107 s/yr) = 3.26 ly.

2.

We find the distance from D = 1/ = [1/(0.38)](3.26 ly/pc) =

3.

8.6 ly.

We find the distance from D = 1/ = [1/(0.00019°)(3600/°)](3.26 ly/pc) =

4.

(a) We find the parallax angle from D = 1/; 0.028. 36 pc = 1/, which gives  = (b) We convert this to degrees:  = (0.028)/(3600/°) = (7.710–6)°.

5.

We find the parallax angle from D = 1/; (55 ly)/(3.26 ly/pc) = 1/, which gives  = The distance in parsecs is D = (55 ly)/(3.26 ly/pc) = 17 pc.

4.8 ly.

0.059.

6.

We find the distance in light-years: D = (35 pc)(3.26 ly/pc) = 114 ly. Thus the light takes 114 yr to reach us.

7.

The star farther away will subtend a smaller angle, so the parallax angle will be From D = 1/, we see that D1/D2 = 2 = 2/1 , or 1/2 = !.

8.

(a) The apparent brightness of the Sun is ¬ = 1.3103 W/m2. (b) The absolute luminosity of the Sun is L = ¬A = (1.3103 W/m2)4p(1.51011 m)2 = 3.71026 W.

9.

The apparent brightness is determined by the absolute luminosity and the distance: L = ¬A = ¬4pD2. If we form the ratio for the apparent brightness at the Earth and at Jupiter, we get ¬E/¬J = (DJ/DE)2; (1.3103 W/m2)/¬J = (5.2)2, which gives ¬J = 48 W/m2.

less.

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10. The diameter of our Galaxy is 100,000 ly so the angle subtended by our Galaxy is Galaxy = d1/D1 = (100,000 ly)/(2106 ly) = 0.05 rad = 2.9°. The angle subtended by the Moon at the Earth is Moon = d2/D2 = (3.48106 m)/(3.84108 m) = 9.0610–3 rad = 0.52°. Thus we have Galaxy ˜ 6Moon . 11. If we assume negligible mass change, as a red giant, the density of the Sun will be  = M/V = M/)pr3; = (21030 kg)/)p(1.51011 m)3 = 1.410–4 kg/m3. 12. The angle subtended at the Earth by the Sun as a white dwarf will be  = d/D = (3.48106 m)/(1.51011 m) = 2.310–5 rad = 4.8. 13. The density of the white dwarf is dwarf = M/V = M/)pr3; = (2.01030 kg)/)p(6.38106 m)3 = 1.8109 kg/m3. The ratio of densities will be the ratio of masses: dwarf/Earth = Mdwarf/MEarth = (2.01030 kg)/(5.981024 kg) =

3.3105.

14. The density of the neutron star is ns = Mns/Vns = Mns/)prns3; = (1.5)(21030 kg)/)p(11103 m)3 = 5.41017 kg/m3. From the result for Problem 13 we see that ns = (3108)dwarf. For the density of nuclear matter we calculate the density of a proton: p = mp/)prp3 = (1.6710–27 kg)/)p (1.210–15 m)3 = 2.31017 kg/m3. Thus we see that ns ˜ nuclear matter. 15. For the reaction 2He + 2He  4Be +  , we determine the Q-value: Q = [M(4He) + M(4He) – M(8Be)]c2 = [2(4.002603 u) – (8.005305 u)]c2(931.5 MeV/uc2) = – 0.0922 MeV = – 92.2 keV. 4 8 12 For the reaction 2He + 4Be  6C +  , we determine the Q-value: 4 8 Q = [M( He) + M( Be) – M(12C)]c2 = [(4.002603 u) + (8.005305 u) – (12.000000 u)]c2(931.5 MeV/uc2) = 7.366 MeV. Note that the total Q-value is 7.27 MeV. 4

4

8

16. (a) For the reaction 6C + 6C  12Mg +  , we determine the Q-value: Q = [M(12C) + M(12C) – M(24Mg)]c2 = [2(12.000000 u) – (23.985042 u)]c2(931.5 MeV/uc2) = 13.93 MeV. Thus 13.93 MeV is released. (b) We find the radius of the carbon nucleus from r = (1.2 fm)A1/3 = (1.2 fm)(12)1/3 = 2.75 fm. If the two nuclei are just touching, the Coulomb potential energy must be the initial kinetic energies of the two nuclei: 2K = U = ZCZCe2/4pÅ02r = (6)(6)(1.44 MeV · fm)/2(2.75 fm) = 9.4 MeV. Thus each carbon nucleus has a kinetic energy of 4.7 MeV. 12

12

24

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(c) We find the temperature from K = *kT; (4.7 MeV)(1.6010–13 J/MeV) = *(1.3810–23 J/K)T, which gives T = 16

16

28

41010 K.

4

17. (a) For the reaction 8O + 8O  14Si + 2He , we determine the Q-value: Q = [M(16O) + M(16O) – M(28Si) – M(4He)]c2 = [2(15.994915 u) – (27.976927 u) – (4.002603 u)]c2(931.5 MeV/uc2) = 9.594 MeV. Thus 9.594 MeV is released. (b) We find the radius of the oxygen nucleus from r = (1.2 fm)A1/3 = (1.2 fm)(16)1/3 = 3.02 fm. If the two nuclei are just touching, the Coulomb potential energy must be the initial kinetic energies of the two nuclei: 2K = U = ZOZOe2/4pÅ02r = (8)(8)(1.44 MeV · fm)/2(3.02 fm) = 15.25 MeV. Thus each oxygen nucleus has a kinetic energy of 7.6 MeV. (c) We find the temperature from K = *kT; 61010 K. (7.6 MeV)(1.6010–13 J/MeV) = *(1.3810–23 J/K)T, which gives T = 18. From Wien’s displacement law we can compare the temperatures of the two stars: 1T1 = 2T2 , or T2/T1 = 1/2 = (800 nm)/(400 nm) = 2. If r is the radius of a star, the radiating area is A = 4pr2. From the Stefan-Boltzmann law we have L = ¬4pD2  AT4 = 4pr2T4. If we form the ratio for the two stars, we get (D2/D1)2 = (T2/T1)4 = (2)4, which gives D2 = 4D1 . 19. From Wien’s displacement law we can compare the temperatures of the two stars: 1T1 = 2T2 , or T2/T1 = 1/2 = (500 nm)/(700 nm) = 5/7. If r is the radius of a star, the radiating area is A = 4pr2. From the Stefan-Boltzmann law we have L = ¬4pD2  AT4 = 4pr2T4. If we form the ratio for the two stars, we get ¬2/¬1 = (r2/r1)2(T2/T1)4; 1/0.091 = (r2/r1)2(5/7)4, which gives r2/r1 = 6.5, so d2/d1 = 6.5 . 20. (a) For the vertices of the triangle we choose the North pole and two points on a latitude line on opposite sides of the Earth, as shown on the diagram.

(b) We can get 180° only approximately. For the vertices of the triangle we choose the North pole and two points on a latitude line very close together, so the angle at the North pole is negligible, as shown on the diagram.

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- 0°

180° 90°

90°

90°

90°

21. (a) The Schwarzschild radius for a star with mass equal to that of our Sun is R = 2GMS/c2 = 2(6.6710–11 N · m2/kg2)(1.991030 kg)/(3.00108 m/s)2 = 2.95103 m = 2.95 km. (b) The Schwarzschild radius for a star with mass equal to that of the Earth is R = 2GME/c2 = 2(6.6710–11 N · m2/kg2)(5.971024 kg)/(3.00108 m/s)2 = 8.910–3 m = 8.9 mm. 22. If we use the data for our galaxy, we have R = 2GM/c2 = 2(6.6710–11 N · m2/kg2)(31041 kg)/(3.00108 m/s)2 = 23. If we consider a triangle with its three vertices on a great circle, such as one through the North and South poles as shown in the diagram, we see that the sum of the angles is 540°.

41014 m.

180°

180°

180°

24. To escape from an object with mass M and radius r, the kinetic energy of the body must be greater than the gravitational potential energy at the surface: !mv2 = GMm/r, or vesc2 = 2GM/r. At the Schwarzschild radius we have vesc2 = 2GM/(2GM/c2) = c2, or vesc = c. 25. We find the distance from Hubble’s law: v = Hd; (0.010)(3.00108 m/s) = [(70103 m/s/Mpc)/(106 pc/Mpc)(3.26 ly/pc)]d,

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which gives d =

1.4108 ly.

26. We find the distance from Hubble’s law: v = Hd; 3.5106 m/s = [(70103 m/s/Mpc)/(106 pc/Mpc)(3.26 ly/pc)]d, 1.6108 ly. which gives d = 27. We estimate the speed of the galaxy from Hubble’s law: v = Hd = [(70103 m/s/Mpc)/(3.26 ly/pc)](12103 Mly) = 2.6108 m/s =

0.86c.

28. (a) We find the receding speed of the galaxy from v = Hd = [(70103 m/s/Mpc)/(3.26 ly/pc)](1.0 Mly) = 2.15104 m/s = (7.210–5)c. For the Doppler shift of the wavelength we have /0 = {[1 + (v/c)]/[1 – (v/c)]}1/2; /(656 nm) = [(1 + 7.210–5)/(1 – 7.210–5)]1/2, which gives = 656 nm. (b) We find the receding speed of the galaxy from v = Hd = [(70103 m/s/Mpc)/(3.26 ly/pc)](1.0102 Mly) = 2.15106 m/s = (7.210–3)c. For the Doppler shift of the wavelength we have /0 = {[1 + (v/c)]/[1 – (v/c)]}1/2; /(656 nm) = [(1 + 7.210–3)/(1 – 7.210–3)]1/2, which gives = 661 nm. (c) We find the receding speed of the galaxy from v = Hd = [(70103 m/s/Mpc)/(3.26 ly/pc)](1.0104 Mly) = 2.15108 m/s = 0.72c. For the Doppler shift of the wavelength we have /0 = {[1 + (v/c)]/[1 – (v/c)]}1/2; /(656 nm) = [(1 + 0.72)/(1 – 0.72)]1/2, which gives = 1.63103 nm = 1.63 m. 29. We find the speed from the Doppler shift: /0 = {[1 + (v/c)]/[1 – (v/c)]}1/2;

(610 nm)/(434 nm) = {[1 + (v/c)]/[1 – (v/c)]}1/2, which gives v= 0.328c. We find the distance from Hubble’s law: v = Hd; (0.328)(3.00108 m/s) = [(70103 m/s/Mpc)/(106 pc/Mpc)(3.26 ly/pc)]d, 4.6109 ly. which gives d =

30. The Doppler shift is /0 = {[1 + (v/c)]/[1 – (v/c)]}1/2.

From the binomial expansion for small x, we have (1 – x)–1 ˜ 1 + x. Thus when v/c « 1, we get /0 = {[1 + (v/c)]/[1 – (v/c)]}1/2 ˜ {[1 + (v/c)]2}1/2 = 1 + (v/c). or  ˜ 0 + (v/c)0 . Thus the fractional wavelength change is ( – 0)/0 = ?/0 ˜ v/c.

31. We find the peak wavelength from T = 2.9010–3 m · K;

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(2.7 K) = 2.9010–3 m · K, which gives  = 1.110–3 m =

1.1 mm.

32. We express the critical density in terms of the nucleon density from c = M/V = nmn/V; 10–26 kg/m3 = n(1.6710–27 kg)/V, which gives n/V = 6 nucleons/m3. 33. If d  1/T, when we form the ratio for two different times, we get d/d0 = T0/T, where T0 = 2.7 K is the temperature today. (a) From Fig. 45–24 we estimate the temperature to be 3000 K, so we have d/d0 = T0/T = (2.7 K)/(3000 K) ˜ 10–3. (b) From Fig. 45–24 we estimate the temperature to be 1010 K, so we have d/d0 = T0/T = (2.7 K)/(1010 K) ˜ 10–10. (c) From Fig. 45–24 we estimate the temperature to be 1013 K, so we have d/d0 = T0/T = (2.7 K)/(1013 K) ˜ 10–13. (d) From Fig. 45–24 we estimate the temperature to be 1027 K, so we have d/d0 = T0/T = (2.7 K)/(1027 K) ˜ 10–27. 34. We find the equivalent temperature from E = Mc2 = *kT, or T = %Mc2/k. (a) For the kaon threshold we have T = %Mc2/k = %(500 MeV)(1.6010–13 J/MeV)/(1.3810–23 J/K) = 41012 K. 10–5 s. We estimate the time from Fig. 45–24: t ˜ (b) For the Y threshold we have T = %Mc2/k = %(9500 MeV)(1.6010–13 J/MeV)/(1.3810–23 J/K) = 71013 K. 10–7 s. We estimate the time from Fig. 45–24: t ˜ (c) For the muon threshold we have T = %Mc2/k = %(100 MeV)(1.6010–13 J/MeV)/(1.3810–23 J/K) = 81011 K. 10–4 s. We estimate the time from Fig. 45–24: t ˜ 35. The absolute luminosity depends on the radius of the star and the temperature: L  AT4 = 4pr2T4. (a) For case A we have temperature increases, luminosity is constant, and size decreases. (b) For case B we have temperature is constant, luminosity decreases, and size decreases. (c) For case C we have temperature decreases, luminosity increases, and size increases. 36. For the luminosity we have L = ¬4pd2. If we form the ratio with constant luminosity, we get (dstar/dSun)2 = ¬Sun/¬star;

{dstar/[(1.51011 m)/(9.51015 m/ly)]}2 = 1011, which gives dstar =

5 ly.

37. For the conservation of angular momentum, with constant mass, we have I = I0 0 ;

^Mr2 = ^Mr020 ;

(5 km)2 = (7105 km)2(1 rev/mo), which gives  = 21010 rev/mo = (21010 rev/mo)/(30 day/mo)(24 h/day)(3600 s/h) =

8103 rev/s.

38. We simplify the ratio of rotational kinetic energies by using the conservation of angular momentum:

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K/K0 = !I2/!I0 0 2 = /0 = (21010 rev/mo)/(1 rev/mo) =

21010.

39. The kinetic energy of the neutron star is K = !I2 = !(^Mr2)2 = !(^)(1.5)(2.01030 kg)(5.0103 m)2[(1.0 rev/s)(2p rad/rev)]2 = 5.921038 J. The power output is P = E/t = (10–9 day–1)(5.921038 J)/(86,400 s/day) = 71024 W.

40. If R is the radius of the universe, the volume is V = )pR3. We find the increase in radius from Hubble’s law: ?R = v ?t = HR ?t. Thus the rate at which the universe in expanding is ?V/?t = 4pR2 ?R/?t = 4pR3H = 3VH. The rate at which hydrogen atoms have to be created is given by ?m/?t = mH ?N/?t =  ?V/?t =  3VH. If we find the rate per unit volume, we have (?N/?t)/V = 3 H/mH = 3(10–27 kg/m3) )(103 m/km)3 (70 km/s/Mpc)(3.16107 s/yr/ (3.26106 ly/Mpc)(9.461012 km/ly)(1.6710–27 kg/atom) = 0.13 atoms/km3/yr. Thus hydrogen atoms would need to be created at the rate 1 H atom/km3 every 8 years. 41. We use M for the mass of the universe and n for the number of nucleons. (a) If nucleons provide 2% of the mass, we have 0.02M = nmn , and 0.98M = (109 n)m. When we combine these, we get n(939106 eV)/0.02 = (109 n)m/0.98, which gives m = 46 eV. (b) If nucleons provide 5% of the mass, we have 0.05M = nmn , and 0.95M = (109 n)m. When we combine these, we get n(939106 eV)/0.05 = (109 n)m/0.95, which gives m = 18 eV. 42. We find the temperature of the stars from T = 2.9010–3 m · K; (60010–9 m)T1 = 2.9010–3 m · K, which gives T1 = 4800 K; (40010–9 m)T2 = 2.9010–3 m · K, which gives T2 = 7300 K. From the H–R diagram, we estimate the luminosities: L1 = 1026 W, and L2 = 21027 W.

Ch. 45

Page 8

From the Stefan-Boltzmann law, the absolute luminosity depends on the radius of the star and the temperature: L  AT4 = 4pr2T4. When we form the ratio, we get L2/L1 = (r2/r1)2(T2/T1)4 = (r2/r1)2(1/2)4; (21027 W)/(1026 W) = (r2/r1)2(600 nm/400 nm)4, which gives r2/r1 = 2. 43. We find half the subtended angle from  = 1/30 pc = (1/30)/(3600/°) ˜ (110–5)°. Thus the minimum subtended angle is ˜ (210–5)°. 44. The wavelengths for the same transition in hydrogen-like atoms are given by 1/ = (constant)Z2. Thus we have / = H/He = (ZHe/ZH)2 = (2/1)2 = {[1 + (v/c)]/[1 – (v/c)]}1/2, which gives v = 45. We find the temperature from E = *kT, or T = %Mc2/k = %(1.8 TeV)(1.6010–7 J/TeV)/(1.3810–23 J/K) = From Fig. 45–24 we see that this is the hadron era.

0.88c.

1.41016 K.

46. We find the speed of the gas clouds from the radial acceleration provided by the gravitational attraction: GMm/R2 = mv2/R, or v2 = GM/R, where M = (2109)(2.01030 kg) = 4.01039 kg, and R = (60 ly)(9.461015 m/ly) = 5.71017 m. To account for the gases moving in opposite directions on the two sides of the cloud, we have v/c = ± (1/c)(GM/R)1/2 = ± (1/3.00108 m/s)[(6.6710–11 N · m2/kg2)(4.01039 kg)/(5.71017 m)]1/2 = ± 2.310–3. Because v « c, we use the result from Problem 30: ?/0 ˜ v/c = ± 2.310–3. This will be in addition to the shift from any overall receding of the galaxy. 47. Because Venus has a more negative value, Venus is brighter. We write the logarithmic scale as m = k log ¬. We find the value of k from m2 – m1 = k(log ¬2 – log ¬1) = k log(¬2/¬1); + 5 = k log(1/100), which gives k = – 2.5. For Venus and Sirius, we have mV – mS = k log(¬V/¬S); – 4.4 – (– 1.4) = – 2.5 log(¬V/¬S), which gives ¬V/¬S = 16. 48. For the critical density we have c = M/V = M/)pr3; 10–26 kg/m3 = (2.01030 kg)/)pr3, which gives r = 3.61018 m = When we compare to the Earth-Sun distance we get r/rES = (3.61018 m)/(1.51011 m) = 2.4 107. When we compare to our Galaxy we get r/rGalaxy = (4102 ly)(50,000 ly) = 810–3.

4102 ly.

Ch. 45

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49. We find the equivalent mass of the photon from m = E/c2. To escape from a mass M and radius R, the energy of the photon must be greater than the gravitational potential energy at the surface: mc2 = GMm/R, so the Schwarzschild radius is R = GM/c2.

50. If there are N nucleons, approximately half of them will be protons. Thus there will be !N electrons. From Eqs. 41–12 and 41–13, the average energy of an electron for a system where the energy levels are filled to the Fermi energy is 3EF/5. Thus the total energy of the electrons is 2  3N 2/ 3. E e = 3 EF 1 N = 3 1 N h 5 5 2 8m e 2V 2 For a spherical volume, we get 2 2 9N 2/ 3 = 3N h 9N 2/ 3. Ee = 3 1 N h 2 2 5 2 8m e 8 2R 3 320me R

The Fermi energy for the nucleons will have a similar expression; however, because mn » me , their Fermi energy is negligible. For electric charges, Gauss’s law shows that the electric field at the surface of a spherical charge distribution is the same as that of a point charge with the total charge of the sphere. Thus the potential at the surface is V = q/4pÅ0r = ()pr3)/4pÅ0r = )pr2/4pÅ0. If we compare the Coulomb force qQ/4pÅ0r2 with the gravitational force GmM/r2, we see that the gravitational potential of a spherical mass will be – )pr2G, where we use the negative sign because like masses attract, while like charges repel. When we bring in the next dm as a spherical shell, dm = 4pr2 dr, the increase in potential energy is dEg = – )pr2G dm = – G2[(4p)2/3]r4 dr. To find the total gravitational energy we integrate from r = 0 to r = R: R 2  1 G 2 4 2 r 4 d r = – 1 G 2 4 2 R 5 = – 1 G 3 M 2 4 2 R 5 = – 3 GM . Eg = – 3 15 15 4R 5 R 0 3 Thus the total energy is 2 2 2/ 3 E = Ee + Eg = 3Nh 2 9N2 – 3GM . 5R 320m eR  We find the equilibrium radius from dE/dR = 0: 2 2 2 Nh d E = – 3N h 9N 2/ 3 + 3G M = 0, which gives R = 9N 2/ 3 . 3 2 2 2 2 dR 160m eR  5R 32m eGM  Because the mass of the electrons is negligible, N = M/mn. Thus we have

Ch. 45

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2

9 2/ 3. 2 For a mass equal to that of the Sun, we get R=

R=

h

32m e m n5/ 3GM 1 / 3

6.63  10– 34 J · s – 31

32 9.11  10 6 = 7.2  10 m =

– 27

kg 1.67  10 3 7.2  10 km.

kg

5/ 3

(6.67  10

2 – 11

2

N · m 2/ kg ) 2.0  10

30

kg

1/ 3

9 2/ 3 2

51. If we have only N = M/mn neutrons, the Fermi energy is 2

2

2/ 3 2/ 3 En = 3 EF N = 3 N h 3N = 3N h 2 18N . 2 5 5 8mn V  160m n R The gravitational energy will be the same, so the total energy is 2 2 2/ 3 E = En + Eg = 3N h 2 18N – 3GM . 2 5R 160m n R  We find the equilibrium radius from dE/dR = 0: 2 2 2 Nh d E = – 3N h 18N 2/ 3 + 3GM = 0, which gives R = 9N 3 2 2 2 2 dR 80m n R 5R 16m n GM  In terms of the mass, this is 2 h 18 2/ 3. R= 8/ 3 1/ 3  2 16m n G M For a mass of 1.5 solar masses, we get

6.63  10 – 34 J · s



R=

– 27

8/ 3

16 1.67  10 kg (6.67  10 4 = 1.1  10 m = 11 km .

– 11

N·

m 2/

52. (a) A photon of energy E will have a mass m = E/c2, and a momentum p = E/c = mc. We assume the deflection is small. Using the coordinate system shown in the figure, this means y ˜ R and r2 ˜ x2 + R2. We want the deflection in the y-direction. When the photon is at the position x, the vertical component of the gravitational force will produce an impulse Fy dt in the time dt = dx/c. Thus we have dpy = – (GMm/r2) cos  dt

2/ 3

.

2

2

kg ) (1.5)(2.0  10

30

kg )

1/ 3

18 2/ 3 2

y p

dx = c d t ² r

R



F = GMm / r2

x

Ch. 45

Page 11

= – (GMm/cr2)(R/r) dx = – (GMmR/c)[dx/(x2 + R2)3/2]. To find the total deflection we integrate over the path from x = – 8 to x = + 8: + +  2GMp GMmR dx GMmR x p = – = – = – 2GMm = – 2 . 2 3/ 2 2 2 2 1/ 2 2 cR cR c c x +R R x +R – – For the magnitude of the deflection angle we have ? = – ?py/p = 2GM/c2R. (b) When we use the data for the Sun, we get ? = [2(6.6710–11 N · m2/kg2)(2.01030 kg)/(3.00108 m/s)2(7.0108 m)](180°/p)(3600/°) = 0.87.