SPH4U SOLUTIONS (UNIT 2)

Unit 2 Energy and Momentum ARE YOU READY? (Pages 174–175)

Knowledge and Understanding 1.

Some forms of energy are the chemical potential energy stored in the gasoline, kinetic energy, gravitational potential energy, thermal energy (heat from the motor), and sound. 2. (a) mass (b) energy (c) force (d) power 3. joule – James Joule newton – Isaac Newton watt – James Watt 4. The work you do on the box depends on its mass, m, the gravitational field strength, g  strength,  g , and the vertical height it is raised, ∆ y.  y.

Inquiry and Communication 5. (a)

29.5 − 25.1

×100% = 14.9% 29.5 The percentage of initial kinetic energy lost is 14.9%. (b) The energy did not cease to exist, but was converted into other forms (likely heat and sound). 6. (a) The measure the efficiency of this incline, you would need to know the distance up the ramp, the force applied, the mass of the sled and child and the vertical height. If you knew the angle that the hill was compared to the horizontal, you could calculate the vertical height from the length up the slope using trigonometry. (b) For 100% efficiency, there could be no friction between the sled and the the ground. 7. The conservation of energy refers to the inability to create or destroy energy during an interaction. Conserving energy is the reduction of our use of energy converted from fossil fuels and other sources to allow them to last longer.

Making Connections 8. (a) The large mass of the bus help to reduce injury because because the acceleration will be small. The bumpers and body panels also help to absorb energy to minimize the energy that will be absorbed by passengers. (b) We could increase the safety of buses by adding more padding around passenger passenger compartments, seat belts, and perhaps even air bags. (c) School busses currently have excellent excellent safety ratings, and the added features in part (b) are expensive to install. 9. (a) The ski jumper will acquire acquire gravitational potential energy as he is transported to the top of the hill. This gravitational energy will be converted to kinetic energy as the jumper slides down the hill, and then reconverted back into some gravitational energy as he jumps from the end of the ramp. All of the energy will be converted to kinetic as he reaches the bottom of the hill (if we ignore the loss due to friction). (b) The law of conservation conservation of energy is what what governs the energy transformations. (c) The main safety feature of this sport is the careful construction of the hill so that the ski jumper does not experience a large normal force from landing. Also the protection of the head by a helmet and suits minimize injury in case of a fall.

Math Skills 10. (a) W (b)  EP

∝ ∆d 

∝

(c)  P  ∝

x2

1 t 

(d)  EK  ∝ v 2

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Unit 2 Are You Ready?  Ready?  231

The graphs for (a)–(d) are shown below.

11. (a) x-components: F K , F    A cos α   y-components: F  N , F A sin α , F g (b)  x-components: F K , F    A cos α ,  F g sin β   y-components: F  N, F A sin α , F g cos β  12.

Technical Skills and Safety 13. (a) Place the air pucks on the table while it is running to see if they regularly move in one direction. (b) The determination of speeds depends on the assumption that the speeds are constant. For a table that is not level, a component of the force of gravity will be accelerating the objects and changing their speed and direction. (c) Other sources of error can be vibration from motion, and air currents from a location near a ventilation duct for the school environmental controls. You must also check equipment to make sure there are no rough edges that might catch. 14. Most sensors are electronic and have a shock hazard so the electrical equipment should be kept away from water. This is also wise to protect the equipment. All experiments should be run slowly in a trial form to make sure that any electrical connectors or cables do not interfere with data collection or damage the electronic devices.

CHAPTER 4 WORK AND ENERGY Reflect on Your Learning (Page 176)

1.

A compression spring is designed to be at its maximum length when there is no force, and resist being squeezed together. An extension spring is designed to be at its minimum length when there is no force, and resist being stretched. Some examples of compression springs are the springs in a car and a retractable ballpoint pen. Some examples of extension springs are bungee cords and storm door springs. 2. During the bounce of the ball, most of the kinetic energy is stored as elastic potential energy to be converted back into kinetic energy after the bounce. Some energy is “lost” (converted) to unwanted forms such as heat and sound. 3. A grandfather clock uses the gravitational potential energy to operate a pendulum with a constant period. 4. (a) The work done on each cart is the same. (b) The work done on each box would be the same since it only depends on the size of the component of force in the direction of motion and the displacement. For each box, these values are still the same. (c) If F 2 is not enough to overcome the frictional force, then the box will not move and no work will be done.

Copyright © 2003 Nelson

Unit 2 Are You Ready?  232

5.

6.

The shock absorber would be a spring with a large force constant to absorb lots of energy quickly, combined with a slow air release to prevent too much bouncing. The size would be suited to fit into the length of the forks. The forces involved would mean the choice of a strong material, such as steel, for strength.

Try This Activity: Which Ball Wins? (Page 177)

(a) The ball on track Y will win because it converts its gravitational energy into kinetic energy more quickly, and will speed up more quickly than the ball on track X. (b) The total energy of the balls remains constant down the track. The energy is just transformed from one type to another.

4.1 WORK DONE BY A CONSTANT FORCE PRACTICE (Pages 181–182)

Understanding Concepts 1.  F 1 will do more work than F 2 because the component of F 2 in the direction of motion is smaller than F 1. 2. No. The force of kinetic friction is always acting opposite to the direction of motion. Since negative work is always opposite the direction of motion, the kinetic friction will always do negative work. 3. Yes. The force of gravity can move an object toward itself, and therefore does positive work on that object. 4. m = 2.75 kg (a) ∆d  = 1.37 m W  = ? W = ( F cosθ )∆d  =

( mg cos θ )∆d 

=

(2.75 kg)(9.80 N/kg)(cos 0°)(1.37 m)

W  = 36.9 J The work done to move the plant 1.37 m up is 36.9 J. (b) ∆ y = 1.07 m  µK  = 0.549 W  = ? First we must calculate the normal force acting on the potted plant: Σ F y = ma y = 0  F N

F g

=

0

 F N

=

F g

=

mg 

=

(2.75 kg)(9.80 N/kg)

=

26.95 N

−

 F   N

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Chapter 4 Work and Energy   233

Let F A be the applied force to move the potted plant horizontally: Σ F x = max = 0  FA

− F K  = 0  FA = F K  = µ  F  N = (0.549)(26.95 N )  F A = 14.796 N

To calculate the work done on the potted plant: W = ( FA cos θ )∆ d 

= (14.796 N)(cos0 °)(1.07 m) W  = 15.8 J 5.

The work done to move the plant 1.07 m across the shelf is 15.8 J. m = 24.5 kg  F  = 14.2 N [22.5° below the horizontal] ∆d  = 14.8 m W  = ? 

We only need to consider the component of force in the direction of motion: W = ( F cosθ )∆ d 

= (14.2 N)(cos22.5°)(14.8 m) W  = 194 J The work done by the force is 194 J. 

6.

 F T

= 12.5 N [19.5° above the horizontal]

W  = 225 J ∆d  = ? W

= ( F cosθ )∆d 

∆d  = =

W   F cos θ  225 J (12.5 N)(cos19.5°)

∆d  = 19.1 m The toboggan moves 19.1 m. 7. (a) We will calculate the area using the formula for a rectangle  A = lw

= (4.0 N)(2.0 m)  A = 8.0 J The area represents the work done on the object. (b) First calculate the area of the second portion of the graph:  A = lw

= (−4.0 N)(6.0 m − 2.0 m)  A = −8.0 J The total work is 8.0 + (–8.0) = 0.0 J (c) One situation could be pushing a box across a table and pulling it back.

Applying Inquiry Skills 8.

You could set the pen on the paper and pull the paper across the desk. The force of static friction between the paper and the pen is in the direction of motion, doing positive work on the pen.

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Chapter 4 Work and Energy   234

Understanding Concepts 9.

Four different situations are: • a book sitting on a desk (∆d = 0) • a student carrying a book at a constant speed ( θ  = 90º) • a teacher whirling a putty pat in a circle at the end of a string • a toy car travelling in a circular path 10. (a) A box being pulled by string at an angle involves the forward component doing positive work, and the vertical component doing zero work.

(b) A box being pulled up a ramp involves the parallel component of gravity doing negative work, and the perpendicular component of gravity doing zero work.

Section 4.1 Questions (Page 183)

Understanding Concepts 1.

2. 3. 4.

The everyday use of the word “work” is different from the usage in physics when it is referring to employment, or a duty to perform. “Working” as a teacher involves very little physical work. The physics definition of work means the transfer of energy to an object to move it a certain distance. Many types of employment or daily activities involve physical work. For example, the sentence “Loading the cement bags onto the truck was a lot of work,” uses the word “work” similar to the physics definition of work. The centripetal force is always directed toward the centre of the circle, and is by definition perpendicular to the motion of the object. The 90º angle means that work is not done on the object by the centripetal force. As you push on a wall, you are exerting a force, which involves the use of energy. Even though no physical work is being done, your muscles are still burning your body’s fuel, causing you to become tired. Assuming the classroom to be 4 m tall, and the student to have a mass of 70 kg: W = ( F cosθ )∆d 

= ( mg cosθ ) ∆d  = (70 N)(9.80 N/kg)(cos 0°)(4.0 m) 3 W  = 3.0 × 10 J It would take about 3.0



3

 10  J, or 3.0 kJ of work to climb the ladder.

 F A  = 75 N [22° below the horizontal] 

5.

 F T = 75 N [32° above the horizontal] 

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Chapter 4 Work and Energy   235

(a) The work done by the boy (W B): WB = ( FT cos θ ) ∆d 

= (75 N)(cos 32°)(13 m) W B = 826.8 J The work done by the girl (W G): WG = ( FA cos θ )∆d  W G

= (75 N)(cos 22°)(13 m) = 904.0 J

The total work done: Wtotal

= WB + W G = 828.8 J + 904.0 J 3 W total = 1.73 × 10 J  –3

The total amount of work done on the crate is 1.7  10  J. (b) The crate is moving at a constant speed, so it is not gaining any energy. This means that the crate will have the same amount of energy before and after the move, so it must have work done on it opposite the direction of motion. 3 Therefore, the work done on the crate by the floor is –1.7  10  J. 2 6. W  = 9.65  10  J ∆d  = 45.3 m 

 F   = 24.1 N [parallel to the handle of the sleigh] θ  = ? W = ( F cosθ )∆d  cos θ  =

W   F ∆d 

 W       F ∆d      965 J = cos−1     (24.1 N)(45.3 m)   θ  = 27.9° θ  = cos −1 

The angle between the snowy surface and the handle is 27.9º. 7. (a) ∆d  = 38 m m = 66 kg

 F A = 58 N [18° above the horizontal] 

 F  N = ?  µK  = ?

First we must calculate the normal force: Σ F y = ma y  F N

Copyright © 2003 Nelson

=0

+ FA sin18° − F g = 0  F N = Fg − F A sin18° = mg − F A sin18° = (66 kg)(9.80 N/kg) − (58 N)sin18°  F   N = 628.88 N

Chapter 4 Work and Energy   236

We then calculate the force of gravity: Σ F x = max

=0

 FA cos18° − F K  = 0

= F A cos18° = (58 N) cos18°  F K  = 55.161 N  FK

The coefficient of kinetic friction is:  F  µ K  = K   F   N

=

55.161 N

628.88 N µ K  = 0.088 2

The normal force is 6.3  10  N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088. (b) W  = ? W = ( F cosθ )∆ d 

= (55.161 N)(cos180°)(38 m) 3 W  = −2.1× 10 J 3

The work done by kinetic friction is –2.1  10  J. (c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan. (d) ∆d  = 25 m W  = ? W = ( F cos θ )∆ d 

= (58 N)(cos18°)(25 m) W  = 1.4 × 103 J The work done by the parent is 1.4

3

 10  J.



Applying Inquiry Skills 8.

As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for angles equal to 90º, and negative for angles between 90º and 180º.

Making Connections 9.

Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermal energy.

4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM PRACTICE (Pages 186–187)

Understanding Concepts 1.

The kinetic energy of a moving object is related to both the mass and the velocity. If the mass of the truck is large enough, a slow moving truck can have more kinetic energy than a fast moving car.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   237

2.

The kinetic energy is proportional to the speed, so if the speed increases by (a) 2, the kinetic energy will increase by a factor of 22, or 4 (b) 3, the kinetic energy will increase by a factor of 32, or 9 2 (c) 37%, the kinetic energy will increase by a factor of 1.37 , or 1.9 3. Assume a 75-kg student running at 8.0 m/s: 1  EK  = mv 2 2 1 = (75 kg)(8.0 m/s)2 2  E K  = 2.4 × 103 J 4.

The kinetic energy at maximum speed is 2.4 kJ. m = 45 g = 4.5  10 –2 kg vi = 0 m/s vf  = 43 m/s (a) W  = ? W = ∆E K 

= =

1 2 1

− vi2 )

m(vf2

(4.5 × 10−2 kg)

((43 m/s )

2

2 W  = 41.6025 J, or42 J The work done by the club is 42 J.  –2 (b) ∆d  = 2.0 cm = 2.0  10  m  F  = ? W = F ∆d   F  =

=

− (0 m/s )2 )

W 

∆d  41.6025 J 2.0 × 10 −2 m

 F  = 2.1× 103 N 3 The average force exerted by the club is 2.1 × 10  N.  –1 5. m = 27 g = 2.7  10  kg  F  = 95 N  –1 ∆d  = 31 cm = 3.1  10  m vf  = ? W

= ∆E K 

 F ∆d

=

vf 2

=

vf 

= =

vf  6.

1

mvf 2 (since the initial speed is zero) 2 2 F ∆d  m 2 F ∆d  m 2(95 N)(3.1× 10

−1

m)

−1

2.7 × 10 kg

= 47 m/s

The final speed of the arrow is 47 m/s. 4 m = 4.55  10  kg 4 vi = 1.22  10  m/s  F  = 3.85  105 N 6 ∆d  = 2.45  10  m vf  = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   238

W

= ∆E K 

 F ∆d

=

1

mvf2

=

vf

=

2

= vf  7.

1 2

mvf2

1

−

mvi2

2 1 F ∆d + mvi2 2 2 F ∆d  m

+ vi2

2(3.85 × 105 N)(2.45 × 106 m) 4.55 × 10 kg 4

= 1.38 × 10 4

The final speed of the probe is 1.38 m = 28.0 kg

+ (1.22 × 104 m/s) 2

m/s 4

 10  m/s.



 F A = 95.6 N [35° above the horizontal]  F K  = 75.5 N vi = 0 m/s ∆d  = 0.750 m vf  = ? 

The total work done on the box will become kinetic energy. Since the initial speed is zero: 1 W = mvf 2 2 1  FA cos 35.0°∆d + FK cos180°∆ d = mvf 2 2 2∆d  2 vf = ( FA cos 35.0° + F K  cos180° ) m vf

= =

vf  8.

2∆d  m

( FA cos 35.0° + F K  cos180° )

2(0.750 m) 20.8 kg

((95.6 N)(0.81952) + (75.5 N)(− 1) )

= 0.45 m/s

The final speed of the box is 0.45 m/s. W  = 1.47(cos 38º ) = 1.16 The toboggan would have increased its kinetic energy by 16%.

Applying Inquiry Skills 9.

W

= F ∆d  =  N ⋅ m kg ⋅ m = 2 ⋅m

 EK  =

2

kg ⋅ m 2 s

2

mv 2

m = kg ⋅       s   kg ⋅ m 2  E K  = 2

s

W  =

1

2

s

The base units for both are the same.

Making Connections 3

10. m = 6.85  10  kg 3 vA = 2.81  10  m/s 3 vB = 8.38  10  m/s W  = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   239

(a) The work done is equal to the change in kinetic energy W = ∆E K 

= = =

1 2 1 2 1 2

mvB2

− 12 mvA2

2

m(vB

− vA2 )

(

(6.85 × 10 kg) (8.38 × 10 m/s) 3

3

2

− (2.81× 103 m/s) 2 )

W  = 2.13 × 1011 J (b) The work done by Earth to move the satellite from A to B is 2.13 × 1011 J. W = ∆E K 

= = =

1 2 1 2 1 2

mvA2

−

2

m(vA

1 2

mvB2

− vB2 )

(

(6.85 × 10 kg) (2.81× 10 m/s) 3

3

2

− (8.38× 103 m/s) 2 )

W  = −2.13 × 1011 J 11 The work done by Earth to move the satellite from B to A is –2.13 × 10  J.

Section 4.2 Questions (Page 188)

Understanding Concepts 1.

The first doubling will require much less energy than the second doubling of the speed. This can clearly be shown using: 1 2 2 W = m( vf − vi ) 2 W

∝ vf2 − vi2

W

= vf2 − vi2 = (2v)2 − (v)2 = 4v 2 − v 2 = 3v2

To go from v to 2v:

W To go from 2v to 4v: W

= vf2 − vi2 = (4v )2 − (2v )2 = 16v 2 − 4v 2 = 11v 2

W The first doubling of speed will require work proportional to 3 times the square of the original speed. The second doubling will require work proportional to 11 times the square of the original speed.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   240

2.

3

m = 1.50  10  kg v  = 18.0 m/s [E]  E K  = ? 

 EK  =

=

1 2 1 2

mv 2 (1.5 × 103 kg)(18.0 m/s)2

 E K  = 2.43 ×105 J The kinetic energy of the car is 2.43  105 J. 3. (a) v = 1.150 × 18.0 = 20.7 m/s  E K =   ? 1 2  EK  = mv 2 1 = (1.5 ×103 kg)(20.7 m/s)2 2  E K  = 3.21× 105 J The new kinetic energy of the car is 3.21 (b) The increase in E K is:  



5

 10  J.

3.21× 105 J

= 1.32 2.43 × 105 J We can verify this with (1.15) 2 = 1.32 This represents an increase in the kinetic energy of 32%. (c) W  = ? W = ∆E K  = 3.21 ×105 J − 2.43× 105 J 4 W  = 7.8 × 10 J The work done to speed up the car was 7.8 4. m = 55 kg 3  E K  = 3.3  10  J v = ? 1  EK  = mv 2 2 v=

=



4

 10  J.

2 E K  m 2(3.3 × 10 J) 3

55 kg

v = 11 m/s The speed of the sprinter is 11 m/s. 5. v = 12 m/s  E K  = 43 J m = ? 1 2  EK  = mv 2 2 E  m = 2K  v 2(43 J)

=

(12 m/s)

2

m = 0.60 kg The mass of the basketball is 0.60 kg.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   241

6.

m = 0.353 kg ∆d  = 89.3 cm = 0.893 m (a) W  = ? W

= ( mg cosθ )∆d   = (0.353 kg)(9.80 N/kg)(cos 0°)(0.893 m) W  = 3.0892 J

The work done by gravity is 3.09 J. (b) Using the work energy theorem, W

= ∆E K  .

Since vi = 0: 1

W

=

vf 

=

2W 

=

2(3.0892 J )

vf 

2

2

mvf 

m 0.353 kg

= 4.18 m/s

The speed of the plate just before it hits the floor is 4.18 m/s. 7. m = 61 kg θ  = 23°  F K  = 72 N vi = 3.5 m/s ∆d  = 62 m vf  = ? The component of gravity along the slope is mg  sin 23º. Using the work energy theorem: 1 1 mg sin 23°(cos 0 °) ∆d + FK (cos180°) ∆d = mvf2 − mvi2 2 2 1 1 mvf2 = mg sin 23°(1)∆d + FK (−1)∆ d + mvi2 2 2 2 F ∆d  2 2 + vi vf = 2 g sin 23°∆d − K  m vf

vf 

2 FK ∆d 

=

2 g sin 23°∆d −

=

2(9.8 m/s 2 )(sin 23°)(62 m) −

m

+ vi2 2(72 N)(62 m) 61 kg

+ (3.5 m/s)2

= 18 m/s

The speed of the skier after travelling 62 m downhill is 18 m/s. 8. m = 55.2 kg ∆d  = 4.18 m  µK  = 0.27 vi = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   242

Using the FBD to calculate F  N, Σ F y

= ma y = 0  F N − mg  = 0  F N = mg  = (55.2 kg)(9.80 N/kg)  F   N = 540.96 N

To calculate F K :

= µ K F N = (0.27)(540.96 N )  F K  = 146.06 N  FK

Using the work-energy theorem: W

= ∆E K  1

 FK (cos180°) ∆d

=

 FK (cos180°) ∆d

=−

2

m( vf2 − vi2 )

Since vf  = 0,

vi

= =

vi

1 2

mvi2

−2 FK (cos180°)∆d  m

−2(146.06 N)(−1)(4.18 m) 55.2 kg

= 4.7 m/s

The initial speed of the skater was 4.7 m/s.

Applying Inquiry Skills 9. (a)

Car Speed (m/s)

Car Energy (J)

10.0 20.0 30.0 40.0

6.0 × 10 5 2.4 × 10 5 5.4 × 10 5 9.6 × 10

4

Truck Speed (m/s)

Truck Energy (J)

10.0 20.0 30.0 40.0

2.5 × 10 8 1.0 × 10 8 2.2 × 10 8 4.0 × 10

To calculate the energy of the car and the truck, use the equation  EK  =

7

1 2

mv 2 .

To convert tonnes to kilograms, multiply by 1000: 3 m = 1.2 t = 1.2  10  kg (for the car) 2 5 m = 5.0  10  t = 5.0  10  kg (for the truck) (b)

Copyright © 2003 Nelson

Chapter 4 Work and Energy   243

(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kinetic energy increases proportional to the square of the speed.

Making Connections 10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat. (b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts. The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit, causing death.

4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACE PRACTICE (Page 191)

Understanding Concepts 1.

The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the way up, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you could argue that because the ∆d  = 0 (25 cm – 25 cm), the work done must be equal to zero. 2. m = 62.5 kg ∆ y = 346 m (using the ground as y = 0)  E g = ? ∆ Eg = mg ∆y

∆ E g

= (62.5 kg)(9.80 N/kg)(346 m) = 2.12 ×105 J

Relative to the ground, the gravitational potential energy is 2.12  105 J. 3. m = 58.2 g = 5.82  10 –2 kg ∆ y = 1.55 m (a) ∆ E g = ? At 1.55 m above the court: ∆ Eg = mg ∆y

∆ E g

= (5.82× 10−2 m)(9.80 N/kg)(1.55 m) = 0.884 J

At the court height:

∆ Eg = mg ∆y = (5.82× 10−2 m)(9.80 N/kg)(0.00 m) ∆ E g = 0.00 J The gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is 0.00 J. (b) W  = ? W = ( F cosθ )∆ d 

= (mg cosθ )∆d  = (5.82 × 10−2 kg)(9.80 N/kg)(cos0°)(1.55 m) W  = 0.884 J At the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball. (c) The work done in (b) is equal to the change in kinetic energy of the ball. 4. m = 68.5 kg 3 m = 2.56 km = 2.56  10  m θ  = 13.9°  E g = ?

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Chapter 4 Work and Energy   244

First, calculate the vertical lift:

∆ y 2.56 × 103 m ∆ y = 2.56 ×10 −3 m(sin 13.9°) ∆ y = 614.98 m

sin13.9° =

Then calculate the gravitational potential energy: ∆ Eg = mg ∆y

∆ E g 5.

= (68.5 kg)(9.80 N/kg)(614.98 m) = 4.13 ×105 J

The skier’s gravitational potential energy at the top of the mountain is 4.13 ∆ y = 2.36 m 3 ∆ E g = –1.65  10  J m = ?

5

 10  J.



The ground is 2.36 m down from the pole, therefore: ∆ Eg = mg ∆y m=

=

∆ E g  g ∆y −1.65 × 103 J (9.80 N/kg)(− 2.36 m)

m = 71.3 kg The mass of the jumper is 71.3 kg. 6. (a) The first coin does not need to be lifted, so no work is done on it. Each successive coin will need to be raised one more  y coin thickness, t , than the previous. The thickness of each coin will be t  = . The work done on each coin will be  N  equal to its increase in gravitational potential energy. WT = W1 + W2 +  + WN  

= mg ∆y1 + mg ∆y2 +  + mg∆ y N = mg (∆y1 + ∆y2 +  + ∆ yN ) = mg (0 + t + 2t +  Nt)   = mgt (0 + 1 + 2 +  N )   WT

=

mgy  N 

(1 + 2 +  + N )

The sum of an arithmetic series is: Sn

t + t    = n  1 n    2  

For the series (1 + 2 + ··· + N ):  1 + N    S n = N     2  

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Chapter 4 Work and Energy   245

Substituting in the original equation to find the amount of work that must be done on the last coin: mgy  1 + N     WT = N     N   2   WT

1 + N    = mgy     2  

(b) The energy stored is equal to the work done, therefore: 1 + N    ∆ Eg = mgy     2  

Applying Inquiry Skills 7.

The units for gravitational potential energy:  m   mg ∆y = ( kg )  2  m  s  

= kg ⋅ m 2 /s2 The units for work:  F ∆d  = ( N )( m ) m   =  kg 2   (m )  s   = kg ⋅ m 2 /s2 The units for kinetic energy: 1 2

2

mv

2

m = ( kg )       s   = kg ⋅ m 2 /s 2

Therefore, all three units are the same.

Making Connections 8.

∆ E g =

9

6.1  10  J (a) ∆ y = ? Assume 920 students in the school. Assume an average mass of 70.0 kg per student. 4 m = 70.0 kg  920 = 6.44  10  kg ∆ Eg = mg ∆y

∆ y = =

∆ E g mg  6.1× 109 J (6.44 ×104 kg)(9.80 N/kg)

∆ y = 9.472 ×10 4

m The energy from one barrel of oil could raise the students 9.472 (b) There are 158.987 L in a barrel of oil 6.1× 109 J



4

 10  m above ground level.

= 3.8 × 107

J/L 158.987 L 7 There are 3.8  10  J stored in each litre of oil.

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Chapter 4 Work and Energy   246

PRACTICE (Page 193)

Understanding Concepts 9. (a) The Sun’s radiant energy is converted to thermal energy in the water, which is then converted into gravitational  potential energy as the water rises. The gravitatio nal energy converts into kinetic energy as the water falls and t urns the turbine. The kinetic energy of the turbine is converted into electrical energy by the generator. (b) A run-of-the-river generating station does not dam the water to create a large vertical drop in a short area, but rather uses the natural drop of the land over a certain distance, and diverts part of the water to flow down this path.

Making Connections (c) The main difference students will find is that most run-of-the-river generating stations in Canada are much smaller. (d) (i) rainfall, rivers, and glacier/mountain snow melting (ii) Bhutan relies heavily on its environment for exports from logging and energy. (e) possible points: cost, limited suitable locations, low population density, long-term effects

Section 4.3 Questions (Page 194)

Understanding Concepts 1. 2.

As the construction worker raises the wood, the wood’s gravitational energy increases. m = 63 kg ∆ y = 3.4 m (a) On Earth, g  = 9.80 N/kg ∆ E g = ? ∆ Eg = mg ∆y

∆ E g

= (63 kg)(9.80 N/kg)(3.4 m) = 2.1× 103 J

The astronaut’s gravitational potential energy is 2.1  103 J. (b) On the Moon, g  = 1.6 N/kg ∆ E g = ? ∆ E g  = mg ∆y

= (63 kg)(1.6 N/kg)(3.4 m) ∆ E  g  = 3.4 × 102 J 2

The astronaut’s gravitational potential energy is 3.4  10  J. 3. m = 125 g = 0.125 kg ∆ y = 3.50 m (a) ∆ E g = ? (of the pear relative to the ground) The pear on the branch: ∆ Eg

∆ E g

= mg ∆y = (0.125 kg)(9.80 N/kg)(3.50 m) = 4.29 J

The pear at ground level: ∆ Eg = mg ∆y

∆ E g

= (0.125 kg)(9.80 N/kg)(0.00 m) = 0.00 J

The gravitational potential energy of the pear on the branch relative to the ground is 4.29 J. The gravitational potential energy of the pear on the ground relative to the ground is 0.00 J.

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Chapter 4 Work and Energy   247

(b) ∆ E g = ? (of the pear relative to the branch) The pear on the branch: ∆ Eg

∆ E g The pear on the ground: ∆ Eg

∆ E g 4.

= mg ∆y = (0.125 kg)(9.80 N/kg)(0.00 m) = 0.00 J = mg ∆y = (0.125 kg)(9.80 N/kg)(− 3.50 m) = −4.29 J

The gravitational potential energy of the pear on the branch relative to the branch is 0.00 J. The gravitational potential energy of the pear on the ground relative to the branch is –4.29 J. m = 0.15 kg ∆ E g = 22 J ∆ y = ? ∆ Eg = mg ∆y

∆ y = =

∆ E g mg  22 J (0.15 kg)(9.80 N/kg)

∆ y = 15 m The ball’s maximum height is 15 m above the point where it was hit. 5. Let the subscript g represent gravity, and W represent the weightlifter. m = 15 kg ∆ y = 66 cm = 0.66 (a) W g = ? Wg = ( F cos θ )∆ d 

W g

= (mg cosθ )∆d  = (15 kg)(−9.80 N/m)(cos180°)(0.66 m) = −97 J

The amount of work done by gravity on the mass is –97 J. (b) W W = ? WW = ( F cos θ ) ∆d 

W W

= (mg cosθ )∆d  = (15 kg)(9.80 N/kg)(cos 0°)(0.66 m) = 97 J

The amount of work done by the weightlifter on the mass is 97 J. (c) ∆ E g = ? ∆ Eg = mg ∆y

∆ E g

= (15 kg)(9.80 N/kg)(0.66 m) = 97 J

The gravitational potential energy of the mass increases by 97 J.

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Chapter 4 Work and Energy   248

Applying Inquiry Skills 6.

Making Connections 7. (a) V  = 32.8 km3 = 3.28 × 10 10 m3 ∆ y = 23.1 m  ρ = 1.00  103 kg/m3 ∆ E g = ? First we must determine the mass of the water: m = ρ V 

= (1.00 × 103 kg/m 3 )(3.28 × 1010 m3 ) m = 3.28 × 1013 kg To calculate the gravitational potential energy: ∆ Eg = mg ∆y

∆ E g

= (3.28 × 1013 kg)(9.80 N/kg)(23.1 m) = 7.43 ×1015 J

The gravitational potential energy of the lake relative to the turbines is 7.43 7.43 × 10 J 15

(b) This value is approximately

1.14 × 1015 J



 1015 J.

= 6.52  times the annual energy output of the Chukha plant in Bhutan.

4.4 THE LAW OF CONSERVATION OF ENERGY PRACTICE (Page 197)

Understanding Concepts 1. 2. 3.

The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease in kinetic energy. This can only be done with negative work. If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Mass doesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity. ∆ y = 59.4 m vi = 0.0 m/s vf  = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   249

Using conservation of energy, we will have no kinetic energy at the top, and no gravitational energy at the bottom.  ET1 = E T2 mgy / 1

=

vf

=

1 2

2

mv / f 

2 gy1

= 2(9.80 m/s2 )(59.4 m) vf  = 34.1 m/s To convert to km/h:

 34.1 m  3600 s   1 km  =  123 km/h       s  h   1000 m     4.

The roller coaster reaches a maximum speed of 123 km/h at the bottom of the hill. vi = v1 = 9.7 m/s ∆ y = 4.2 m vf  = v2 = ? We will use y = 0 at the point of contact on the hill.  ET1 = E T2 1 2

2

mv1

v12

+ mgy1 = 12 mv22 + mgy2 + 2 gy1 = v22 + 2 gy2 2 2 v2 = v1 + 2 gy1 − 2 gy2

Since y2 = 0, 2 gy2 = 0, therefore: v2

=

v12

+ 2 gy1

= (9.7 m/s)2 + 2(9.80 m/s2 )( 4.2 m) v2 = 13 m/s 5.

The skier’s speed upon touching the hillside is 13 m/s. ∆ y = 4.4  102 m v2 = 93 m/s v1 = ?  ET1 = E T2 1 2

mv12 2

v1

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy 2 2 2 v1 = v 2 + 2 gy 2 − 2 gy1 v1

=

v 22

+ 2 g ( y2 − y1 )

= (93 m/s)2 + 2(9.8 m/s2 )(0 − 440 m) v1 = 5.0 m/s 6.

The speed of the water at the top of the waterfall is 5.0 m/s. v1 = 9.7 m/s ∆ y = 4.7 m v2 = ?

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Chapter 4 Work and Energy   250

 ET1 1 2

mv12 2

v1

7.

= E T2

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy2 2 2 v2 = v1 + 2 gy1 − 2 gy2 v2

=

v2

= (9.7 m/s)2 + 2(9.8 m/s2 )(0 − 4.7 m) = 1.4 m/s

v12

+ 2 g ( y1 − y2 )

The cyclist crests the hill at a speed of 1.4 m/s. v2 = ?

First determine how high the pendulum is vertically raised: h12

+ 24.5 cm 2 = 85.5 cm 2 = 85.5 cm 2 − 24.5 cm 2 h1 = 81.915 cm h1

h1 + h2

= 85.5 cm h2 = 85.5 cm − h1 = 85.5 cm − 81.915 cm h2 = 3.585 cm

Using the value h2 = 3.585 cm, or 0.03585 m, calculate the maximum speed:  ET1 = E T2 1 2

mv12

+ mgy1 =

1

2 since v1 = 0

mv22

+ mgy2

v2

= v22 + 2 gy2 = 2 gy1 − 2 gy2 = 2 g ( y1 − y2 )

v2

= 2(9.80 m/s2 )(0.03585 m − 0) = 0.838 m/s

2 gy1 2

v2

The maximum speed of the pendulum bob is 0.838 m/s.

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Chapter 4 Work and Energy   251

Applying Inquiry Skills 8.

∆y  (m)

∆E g (J)

E K (J)

E T (J)

8.00 6.00 4.00 2.00 0.00

392 294 196 98.0 0.00

0.00 98.0 196 294 392

392 392 392 392 392

Making Connections 9.

A wrecking ball works as a pendulum that is slowly pulled back, increasing its gravitational energy. When it is released, the gravitational potential energy is converted into kinetic energy which is used to destroy buildings.

PRACTICE (Page 200)

Understanding Concepts 10. (a) The energy supplied becomes sound and thermal energy through friction. (b) The energy supplied still produces sound and thermal energy, but some is also converted into kinetic energy. 11.  F K  = 67 N ∆d  = 3.5 m (a) W  = ? W = ( F cos θ )∆ d 

= (67 N)(cos180°)(3.5 m) 2 W  = −2.3× 10 J The amount of work done by friction is –2.3 (b)  E th = ?  Eth = FK ∆d   E th



= (67 N)(3.5 m) = 2.3 × 102 J

The amount of thermal energy produced is 2.3 12.  E th = 0.620 J  F K  = 0.83 N ∆d  = ?

Copyright © 2003 Nelson

 102 J.



2

 10  J.

Chapter 4 Work and Energy   252

The 0.620 J of energy comes from the work done by friction, therefore:  Eth = FK ∆d 

∆d  =

 E th  F K 

=

0.620 J

0.83 N ∆d  = 0.75 m The plate slides 0.75 m. 13. m = 22.0 kg  F  = 98 N  F K  = 87 N vi = 0.0 m/s ∆d  = 1.2 m vf  = ? W = Eth ( F cos θ ) ∆d

= FK ∆d +

v22

=

v2

=

1

mv22

2 ( F cos θ ) ∆d − FK ∆d  1 m 2 ( F cos θ ) ∆d

− FK ∆d 

0.5m (98 N)(cos0 °)(1.2 m) − 87 N(1.2 m)

= v2

+ E k 

0.5(22.0 kg)

= 1.1 m/s

The speed of the cabinet after it moves 1.2 m is 1.1 m/s. 14. m = 0.057 kg ∆d  = 25 cm = 0.25 m  –2 vf  = 5.7 cm/s = 5.7  10  m/s  F K  = 0.15 N vi = ?  EK1 = Eth + E K2 1 2

2

mv1

2

v1

= FK ∆d + =

 FK ∆d + 1 2

v1

= =

v1

1 2 1 2

2

mv2

mv22

m

 FK ∆d +

1

2 0.5m

mv22

(0.15 N)(0.25 m) +

1

(0.057 kg)(5.7 × 10−2 m/s)2

2 0.5(0.057 kg)

= 1.1 m/s

The initial speed of the pen is 1.1 m/s.

Applying Inquiry Skills 15. (a) The law of conservation of energy could be verified by observing how close to its original position the pendulum would return after a swing.

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Chapter 4 Work and Energy   253

(b) Some sources of error would be loss of energy due to friction at the point of attachment and air friction of the moving  pendulum. Some error may be observed if the string is somewhat elasti c.

Making Connections 16. The oil circulation system is an attempt to minimize the frictional forces between the moving parts of the engine, and thereby reduce the loss (and damage) due to the thermal energy caused by friction. The water (coolant) circulation is used to absorb thermal energy from the engine and dissipate it rapidly into the air to prevent damage from overheating.

Section 4.4 Questions (Pages 201–202)

Understanding Concepts 1.

Roller coasters are gravity rides that have an initial input of gravitational potential energy that is converted to kinetic (and  back into gravitational) energy throughout the ride. T o give them this initial energy, they must be pulled up the largest hill at the beginning. 2. m = 0.052 kg ∆ y = 11 cm = 0.11 m  y = 0 (a) ∆ E g = ? ∆ Eg = mg ∆y

∆ E g

= (0.052 kg)(9.80 m/s2 )(0.11 m) = 0.056 J

The initial gravitational potential energy of the egg’s contents is 0.056 J. (b) ∆ E g = ? ∆ Eg = mg ∆y

∆ E g

= (0.052 kg)(9.80 m/s2 )(0.0 m) = 0.0 J

The final gravitational potential energy of the egg’s contents is 0.0 J. (c) ∆ E g = 0.0 – 0.056 = –0.056 J The change in gravitational potential energy as the egg’s contents fall is –0.056 J. (d)  E K =   ? v = ? The kinetic energy will be equal to the loss of gravitational potential, so  E K  = 0.056 J. 1  EK  = mv2 2 v=

=

3.

2 E K  m 2(0.056 J) 0.052 kg

v = 1.5 m/s The kinetic energy is 0.056 J. The speed of the egg’s contents just before hitting the pan is 1.5 m/s. ∆ y = 1.2 m vf  = 9.9 m/s vi = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   254

 ET1 1 2

mv12 2

v1

= E T2

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy2 2 2 v1 = v 2 + 2 gy 2 − 2 gy1 v1

=

v 22

+ 2 g ( y2 − y1 )

= (9.9 m/s)2 + 2(9.8 m/s2 )(1.2 m − 0 m) v1 = 11 m/s 4.

The initial speed of the ball was 11 m/s. v1= 3.74 m/s ∆ y = 8.74 m 4 m = 7.12  10  kg (a) ∆ E g = ? ∆ Eg = mg ∆y

∆ E g

= (7.12 × 104 kg)(9.80 m/s2 )(8.74 m) = 6.10 × 106 J

The gravitational potential energy of the mass of water at the top of the waterfall is 6.10 (b) v2 = ?  ET1 = E T2 1 2

mv12 2

v1

+ mgy1 =

1 2

mv22

6

 10  J.



+ mgy2

+ 2 gy1 = v22 + 2 gy2 2 2 v2 = v1 + 2 gy1 − 2 gy2 v2

=

v2

= (3.74 m/s)2 + 2(9.80 m/s2 )(8.74 m − 0 m) = 13.6 m/s

v12

+ 2 g ( y1 − y2 )

The speed of the water at the bottom of the waterfall is 13.6 m/s. 5. (a)

First, determine the vertical height above the bottom of the swing: h1 cos48 ° = 3.7 m h1 = 3.7 m ( cos48° ) h1

Copyright © 2003 Nelson

= 2.476 m

Chapter 4 Work and Energy   255

h1 + h2

= 3.7 m h2 = 3.7 m − h1 = 3.7 m − 2.476 m h2 = 1.224 m

 Now we can calculate the acrobat’s speed:  ET1 = E T2 1 2

mv12

+ mgy1 =

1 2

mv22 + mgy 2

Since v1 = 0:

v2

= v22 + 2 gy2 = 2 gy1 − 2 gy 2 = 2 g ( y1 − y2 )

v2

= 2(9.80 m/s2 )(1.224 m − 0 m) = 4.9 m/s

2 gy1 2

v2

The acrobat’s speed at the bottom of the swing is 4.9 m/s. (b) Due to conservation of energy, the maximum height on the other side is equal to the starting height of the acrobat. 6. m = 55 kg ∆d  = 3.7 m  F K  = 41.5 N v1 = 65.7 cm/s = 0.657 m/s v2 = 7.19 m/s φ  = ? Relating φ  to h: sin φ  =

h

11.7 h = 11.7sin φ 

Using conservation of energy:  ET1  EK1 + ∆Eg 1 2

mv12

= E T2 = EK2 + E th

+ mgh = mgh =

1 2 1 2 1

h= 2 1 11.7sin φ  = 2

mv22

+ FK ∆ d  

mv22

+ FK ∆d −

m( v22

1 2

mv12

− v12 ) + FK ∆ d

 

mg 

(

2

(55.0 kg) ( 7.19 m/s )

− ( 0.657 m/s )2 ) + (41.5 N)(11.7 m) 2

(55.0 kg)(9.80 m/s )

sin φ  = 0.30054

φ  = 17.5° The angle is 17.5 °.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   256

7.

v1 = 0.0 m/s v2 = 6.8 m/s Since the skateboarder starts even with the height of the centre of the circle, the total vertical drop at the bottom will be equal to the radius of the circle. r  = y1 = ?  ET1 1 2

= E T2 1

+ mgy1 =

2

mv1

v12

2

2

mv2

+ mgy2

+ 2 gy1 = v22 + 2 gy2

Since v1 = 0, and y2 = 0: 2 gy1  y1

= v22 = =

 y1

v22 2 g  6.8 m/s

2 2

2(9.8 m/s )

= 2.4 m

The radius of the half-pipe is 2.4 m.  –2 8. m = 55 g = 5.5  10  kg v1 = 1.9 m/s ∆d  = 54 cm = 0.54 m v2 = 0.0 m/s (a) µ K  = ? First, solve for F K:   ET1 = E T 2 1 2

 EK

= E th

mv 2

= FK ∆ d  

 F K  =

=

mv

2

2∆d  (5.5 × 10−2 kg )(1.9 m/ s)2 2(0.54 m)

 F K  = 0.1838 N  Now calculate the normal force: Σ F y = ma y  F N

=0

− mg  = 0  F N = mg  = (5.5 × 10−2 kg)(9.80 m/s2 )  F   N = 0.539 N

Solve for µK :

µ K  =

=

 F K   F   N 0.1838 N

0.539 N µ  K  = 0.34 The coefficient of kinetic friction is 0.34.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   257

(b) µ K  =? First, solve for the acceleration:

= vi2 + 2a∆d 2 2 vf − vi a= 2∆d  2

vf

= a

 

2 2 (1.9 m/s ) − ( 0 m/s )

2(0.54 m)

= 3.3426 m/s2

Using the FBD, calculate the magnitude of kinetic friction: Σ F x = max  FK  = ma x

= (5.5 ×10−2 kg)(3.3426 m/s 2 )  F K  = 0.1838 N Calculate the normal force: Σ F y

= ma y = 0  F N − mg  = 0  F N = mg  = (5.5 × 10−2 kg)(9.80 m/s2 )  F   N = 0.539 N

Calculate the coefficient of kinetic friction:  F  µ K  = K   F   N

=

0.1838 N

0.539 N µ K  = 0.34 The coefficient of kinetic friction is 0.34. (c) The kinetic energy is converted into thermal energy. 9. vi = 85 km/h = 23.61 m/s vf  = 0.0 m/s ∆d  = 47 m 3  F K  = 7.4  10  N (a)  E th = ?  Eth = FK ∆d 

 E th

= (7.4 × 103 N)(47 = 3.5 × 105 J

m) 5

The amount of thermal energy produced is 3.5  10  J. (b) Before the skid, the thermal energy was kinetic energy.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   258

(c) m = ?  ET1  EK 1 2

mv

2

= E T2 = E th = FK ∆ d  

m=

=

2 FK ∆d  v2 2(7.4 × 103 N )(47 m ) 2

( 23.61 m/s )

m = 1.2 ×103 kg 3 The mass of the car is 1.2  10  kg. (d) µ K  = ? First calculate the normal force: Σ F y = ma y  F N

=0

− mg  = 0  F N = mg  = (1.248 × 103 kg)(9.80 m/s2 )  F   N = 12 228

To calculate the coefficient of kinetic friction:  F  µ K  = K   F   N

=

7.4 ×103 J

12228 J µ K  = 0.61 The coefficient of kinetic friction is 0.61. 10. m = 22 kg ∆d  = 2.5 m θ  = 44°  F K  = 79 N (a) W  = ? W = ( FK  cos θ )∆d 

= (79 N)(cos180°)(2.5 m) 2 W  = −2.0 × 10 J The work done by friction is –2.0 (b)  E K =   ?

Copyright © 2003 Nelson

2

 10  J.



Chapter 4 Work and Energy   259

First, determine the vertical drop: h sin44° = 2.5 h = 2.5sin 44° h = 1.737 m To calculate the final kinetic energy:  ET1 = E T2

∆ Eg = EK + E th mg ∆y = EK + FK ∆d    EK = mg ∆y − FK ∆ d  = (9.8 N/kg)(22 kg)(1.737 m) − (79 N)(2.5 m)  E K  = 1.8 ×102 J The box’s final kinetic energy is 1.8 (c)  E th = ?  Eth = FK ∆d   E th

2

 10  J.



= (79 N)(2.5 m) = 2.0 ×102 J

The thermal energy produced is 2.0



2

 10  J.

Applying Inquiry Skills 11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energy  produced. (b) The equation needed would be:  ET1 = E T2

∆ Eg1 = ∆Eg2 + E th mgy1 = mgy2 + Eth   Eth = mgy1 − mgy2  Eth = mg ( y1 − y2 ) (c) You could determine the height at each point by standing a known distance from the base of the ride and measure the angle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energy  produced.

Making Connections 12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potential energy is converted into the kinetic energy of the projectile.

4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTION PRACTICE (Pages 206–207)

Understanding Concepts 1.

The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring A would be more difficult to stretch than spring B. 2. The spring would exert a southward force on you. 3. k  = 25 N/m (a)  x = 16 cm = 0.16 m  F x = kx

= (25 N/m)(0.16 m)  F   x = 4.0 N Copyright © 2003 Nelson

Chapter 4 Work and Energy   260

 x = 32 cm = 0.32 m

= kx = (25 n/m)(0.32 m)  F   x = 8.0 N  F x

The magnitude of force would be 4.0 N for a stretch of 16 cm, and 8.0 N for a stretch of 32 cm. (b)  x = 16 cm = 0.16m  F x = − kx

= −(25 N/m)(0.16 m)  F   x = −4.0 N  x = 32 cm = 0.32 m

= −kx = −(25 N/m)(0.32 m)  F   x = −8.0 N  F x

The magnitudes of the forces are 4.0 N and 8.0 N, respectively. 2 4. k  = 3.2  10  N/m  –2  x = 2.0 cm = 2.0  10  m  F x = − kx

= −(3.2 × 102 N/m)(2.0 × 10−2 m)  F   x = −6.4 N The magnitude of the force applied by the air is 6.4 N. 5. m = 1.37 kg k  = 5.20  102 N/m (a)  x = ? Σ F y = ma = 0  F x

− mg  = 0 kx = mg    x =

=

mg  k  (1.37 kg)(9.80 N/kg)

5.20 × 102 N/m  x = 0.0258 m The spring stretches 0.0258 m. (b)  x = 1.59 cm = 0.0159 m Σ F  y = ?

Σ F y = Fx − mg  = kx − mg   = (5.20 × 102 N/m)(0.0159 m) − (1.37 kg)(9.80 N/kg) Σ F  y = −5.16 N The net force on the fish is 5.16 N [down]. (c)  x = 2.05 cm = 0.0205 m a = ? Σ F y = ma  F x

− mg = ma kx − mg   a= m

=

(5.20 × 102 N/m)(0.0205 m) − (1.37 kg)(9.80 N/kg)

a = −2.02 m/s 2 The acceleration of the fish is 2.02 m/s  [down] Copyright © 2003 Nelson

1.37 kg 2

Chapter 4 Work and Energy   261

Applying Inquiry Skills 6. (a)

(b) The slope of the line is negative.

Making Connections 7. (a)

Mass (kg) 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00

Stretch (m) 0.122 0.245 0.368 0.490 0.612 0.735 0.858 0.980

(b)

(c) The mass value may not be correct if the value of g  is different then where it was calibrated. The weight value would  be accurate anywhere (even on the Moon).

Copyright © 2003 Nelson

Chapter 4 Work and Energy   262

PRACTICE (Page 211)

Understanding Concepts 8. (a) The graph shown is the force applied by the spring. Since the spring is being stretched to the right (positive x), the force exerted by the spring will be to the left (negative). This is what is shown on the graph. (b) The force constant is the slope of the graph. Since the equation F  x = kx relates the force exerted by the spring, we must change the direction (i.e., the sign) of the force, − F  k  =  x  x −(−15 N)

=

0.40 m k  = 38 N/m The force constant of the spring is 38 N/m. (c)  x = 35 cm = 0.35 m The energy stored is the area between the curve and the x-axis. 1  Ee = A = bh 2 1 = (0.35 m)(13 N) 2  E e = 2.3 J 9.

The elastic potential energy is 2.3 J. 3 k  = 9.0  10  N/m (a)  x = 1.0 cm = 0.010 m  E e = ? 1  Ee = kx 2 2 1 = (9.0×103 N/m)(0.010 m)2 2  E e = 0.45 J The elastic potential energy stored by the spring is 0.45 J. (b)  x = –2.0 cm = –0.020 m  E e = ? 1  Ee = kx 2 2 1 = (9.0 ×103 N/m)(−0.020 m)2 2  E e = 1.8 J

The elastic potential energy stored in the spring is 1.8 J. 10. m = 7.8 g = 0.0078 kg 2 k  = 3.5  10  N/m  x = –4.5 cm = –0.045 m (a)  E e = ? 1  Ee = kx 2 2 1 = (3.5 × 102 N/m)(−0.045 kg) 2 2  E e = 0.35 J The elastic potential energy of the spring is 0.35 J.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   263

(b) v = ?  Ee

= E T2 = E K 

kx 2

=

v

=

 ET1 1 2

=

1 2

mv 2 kx

2

m (3.5 × 10 N/m)(− 0.045 m) 2

2

0.0078 kg

v = 9.5 m/s The speed of the dart as it leaves the toy is 9.5 m/s.  –3 11. m = 3.5  10  kg k  = 9.5 N/m (a) ∆ y = 5.7 cm = 0.057 m  x = ?  ET1 = E T2 1 2

 Ee

= ∆E g

kx 2

= mg ∆y

 x =

=

2mg ∆y k  −3

2(3.5 × 10 kg)(9.8 N/kg)(0.057 m) 9.5 N/m

 x = 0.020 m The spring must be compressed by 0.020 m, or 2.0 cm. (b) If friction was not negligible, the amount of compression would need to be increased. The energy supplied by the compressed spring would now become both gravitational potential and thermal energy. In order to have the same final gravitational energy, the spring would need to be compressed more. 12. m = 0.20 kg k  = 55 N/m (a) ∆ y = 1.5 cm = 0.015 m v = ?  ET1 = E T2

∆ Eg = EK + E e mg ∆y mv

2

=

1 2

mv 2

+

1 2

kx 2

= 2mg ∆y − kx 2

v=

=

2 mg ∆y − kx

2

m 2(0.20 kg)(9.8 N/kg)(0.015 m) − (55 N/m)(0.015 m)2 0.20 kg

v = 0.48 m/s The speed of the mass is 0.48 m/s.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   264

(b) For the fall, ∆ y = x. At maximum fall, there will be no kinetic energy.  ET1 = E T2

∆ Eg = E e mg ∆y

=

1 2

kx

2

− 2mg ∆y = 0 (55) x 2 − 2(0.20)(9.8) x = 0 (55) x 2 − (3.92 ) x = 0  x(55 x − 3.92) = 0 55 x − 3.92 = 0 kx

2

 x =

or

x=0

3.92

55  x = 0.071m or x = 0 m The value x = 0 refers to the moment of release, so the maximum stretch will be 0.071 m. 13. k  = 12 N/m ∆ y = 93.0 cm = 0.93 m  –3 m = 8.3  10  kg  x = 4.0 cm = 0.040 m ∆d  = ? Analyze as a projectile motion question. First, determine horizontal speed at launch:  ET1 = E T2  Ee 1 2

kx

2

= E K  =

1 2

mv

2

kx 2

v=

m (12 N/m)(− 0.040 m)

2

=

8.3 × 10 −3kg

v = 1.521 m/s Choosing down as positive, the time for the marble to drop 0.93 m vertically is: 1 ∆ y = vi ∆t + a(∆t ) 2 2 Since vi = 0:

∆ y = ∆t  = =

1 2

a (∆t )2 2∆ y a 2(0.93 m)

9.8 m/s ∆t  = 0.4357 s

2

The horizontal distance travelled during this time is: ∆d = v∆t 

= (1.521 m/s)(0.4357 s) ∆d  = 0.66 m The marble travels 0.66 m horizontally before hitting the floor.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   265

Applying Inquiry Skills 14. (a) The measurement needed would be the mass of the largest friend. (b) Assuming a largest mass of 115 kg, Σ F y = ma = 0  F x

− mg  = 0 kx = mg   k  =

=

mg   x (115 kg)(9.80 N/kg) 0.75 m

k  = 1.5 × 103 N/m The approximate force constant is 1.5  103 N/m.

Making Connections 2  –7 15. m = 2.0  10 µ g = 2.0  10  kg ∆ y = 65 mm = 0.065 m  x = 75 cm = 0.75 m  E e = ?  ET1 = E T2

 Ee

 E e

= x (∆E g ) =  x(mg ∆y ) = 0.75 m(2.0 × 10−7 kg)(9.8 N/kg)(0.065 m) = 9.6 × 10−8 J

The initial quantity of elastic potential energy is 9.6  10 –8 J.

PRACTICE (Pages 214–215)

Understanding Concepts 16. (a) The maximum displacement from the rest position will be at the top and bottom of the bounce. (b) The speed is a maximum at the rest position. (c) The speed will be zero at the top and the bottom of the bounce. (d) The acceleration will be a maximum at the top and the bottom of the bounce. (e) The acceleration will be zero at the rest position. 17. T  = ?  f  = ? (a) number of vibrations = 12 t  = 48 s total time T  = number of complete vibrations

=

48

12 T  = 4.0 s  f  =

=

number of complete vibrations total time 12

48  f  = 0.25 Hz

Copyright © 2003 Nelson

Chapter 4 Work and Energy   266

Alternatively, you could calculate frequency by using the equation: 1  f  = T  1

=

4.0 s  f  = 0.25 Hz The period is 4.0 s, and the frequency is 0.25 Hz. (b) number of vibrations = 210 t  = 1 min = 60 s total time T  = number of complete vibrations

=

60

210 T  = 0.29 s  f  =

number of complete vibrations total time

=

210

=

5.0

60  f  = 3.5 Hz The period is 0.29 s, and the frequency is 3.5 Hz. (c) number of vibrations = 2200 t  = 5.0 s total time T  = number of complete vibrations 2200

T  = 2.3 × 10 −3 s  f  =

=

number of complete vibrations total time 2200 5.0

 f  = 4.4 ×10 2 Hz  –3 The period is 2.3  10  s, and the frequency is 4.4 18. m = 0.25 kg  A = 8.5 cm 2 k  = 1.4  10  N/m (a) d  = ?



2

 10  Hz.

During each cycle, the mass moves 4 amplitudes, or 4 A. In 5 cycles, the mass will move: d = 5× 4A

= 5 × 4(8.5 cm) 2 d  = 1.7 × 10 cm The mass moves 1.7 (b) T  = ?



2

 10  cm in the first five cycles.

T  = 2π 

= 2π 

m k  0.25 kg 1.4 × 10 N/m 2

T  = 0.27 s The period of vibration is 0.27 s. Copyright © 2003 Nelson

Chapter 4 Work and Energy   267

19. m = 0.10 kg  f  = 2.5 Hz k  = ?  f  =

1

k 

2π 

m

k 

2π  f  =

m

= 4π 2 f 2 m = 4π 2 (2.5 Hz)2 (0.10 kg) k  = 25 N/m k

The force constant of the spring is 25 N/m. 2 20. k  = 1.4  10  N/m T  = 0.85 s m = ? m

T  = 2π  T 2π 

m

=

m=

=

k 

k  T 2 k  4π 2 (0.85 s) 2 (1.4 × 102 N/m) 2 4π 

m = 2.6 kg The mass would have to be 2.6 kg.

Applying Inquiry Skills 21. Examining the base SI units for each: m  x a

= =

 x a

m

k 

 m    s2     

=

=

s2

=s

= m k 

kg

 N    m      kg ⋅ m  kg ⋅ m    s 2      s

2

=s

Therefore, they are dimensionally equivalent.

Making Connections 22. number of vibrations = 6.0 t  = 8.0 s (a) k  = ? First we must calculate the frequency: number of complete vibrations  f  = total time 6.0

=

8.0 s  f  = 0.75 Hz

Copyright © 2003 Nelson

Chapter 4 Work and Energy   268

Using a mass of 75 kg:  f  = 2π  f  =

1

k 

2π 

m

k  m

= 4π 2 f 2 m = 4π 2 (0.75 Hz)2 (75 kg) k  = 1.7 × 103 N/m k

The force constant is 1.7  103 N/m. (b) No, you are not undergoing SHM. When you leave the trampoline, there is a period of time when it is not exerting any force on you. SHM requires that the force be proportional to the displacement.

PRACTICE (Pages 217–218)

Understanding Concepts 23. (a) The speed is zero at lengths of 12 cm and 38 cm. (b) The maximum speed will be at the rest position. The rest position will be halfway between the minimum and maximum extensions: 12 + 38 = 25 cm 2 (c) The amplitude is 38 – 25 = 13 cm. 24.  E e = 5.64 J m = 0.128 kg k  = 244 N/m (a)  x = ? The maximum energy is constant, and all elastic potential at either end of the system, at the maximum amplitude. 1 2  Ee = kx 2  x =

=

2 E e k  2(5.64 J)

244 N/m  x = 0.215 m The amplitude of the vibration is 0.215 m. (b) v = ? Approach 1: All of the energy will be kinetic as it passes through the rest position.

 EK  = mv 2 v=

=

2 E K  m 2(5.64 J) 0.128 kg

v = 9.39 m/s

Copyright © 2003 Nelson

Chapter 4 Work and Energy   269

Approach 2: All of the energy stored in the spring at maximum compression will be converted to kinetic energy.  ET1 = E T2

1 2

kx 2

=

1 2

mv 2 kx

v=

2

m (244 N/m)(0.215 m)

=

2

0.128 kg

v = 9.39 m/s The speed is 9.39 m/s regardless of the approach used. (c)  x2 = 15.5 cm = 0.155 m v = ?  ET1 = E T2 1 2

2 kxmax

=

1

v2

=

kxmax

2

mv 2 2

v=

=

+

1 2

kx22

− kx22

m 2 k ( xmax

− x22 )

m

(

2

(244 N/m) ( 0.215 m )

− ( 0.155 m )2 )

0.128 kg

v = 6.51 m/s The speed of the mass is 6.51 m/s. 25.  x = 0.18 m m = 58 g = 0.058 kg k  = 36 N/m (a)  E e = ? v = ? Maximum energy during maximum stretch/compression of the spring: 1  Emax = kx 2 2 1 = (36 N/m)(0.18 m)2 2  E max = 0.58 J This will all be kinetic energy at the rest position. 1 2  EK  = mv 2 v=

=

2 E K  m 2(0.58 J) 0.058 kg

v = 4.5 m/s The maximum energy of the system is 0.58 J. The maximum speed of the mass is 4.5 m/s.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   270

(b)  x = ? Double the energy would be: ′ = 2 × E e  Emax 1

= 2×

kx 2

2

= (36 N/m)(0.18 m)2 ′ = 1.1664 J  E max This is all stored as elastic potential energy at the full amplitude. 1 2  Ee = kx 2 2 E e

 x =

k  2(1.1664 J)

=

36 N/m  x = 0.25 m The amplitude of vibration required would be 0.25 m. (c) v = ? 1  EK  = mv 2 2 2 E K 

v=

m 2(1.1664 J)

=

0.058 kg

v = 6.3 m/s The maximum speed of the mass is 6.3 m/s. 26. For the maximum speed, all of the elastic energy will be converted to kinetic:  ET1 = E T2 1 2

kx 2

=

1 2

mv 2 kx

v=

2

m k 

v=x

m

For a SHM system, x = A, therefore:  f  =

1

k 

2π 

m

k 

2π  f  =

m

Substituting above: v=x

k 

m =  A(2π  f )

v = 2π  fA

Copyright © 2003 Nelson

Chapter 4 Work and Energy   271

Applying Inquiry Skills

27. (a)

 N    m   k  2   (m2 − m2 ) 2 (A − x ) =

m

kg

 kg ⋅ m    2   =  m ⋅ s  (m2 ) kg

=

m2 s2

= m/s The dimensions are m/s. (b) This expression is to calculate the speed of an object at a location during SHM.

Making Connections 28. (a) Tuning fork prongs have slow damping to produce a long tone. (b) The voltmeter needle has fast damping to stabilize the reading quickly. (c) A guitar string has slow damping to play the note as long as possible. (d) Saloon doors have medium damping to keep the doors from swinging too much, but they still swing back and forth. (e) The string of the bow has fast damping to prevent dangerous vibration.

Section 4.5 Questions (Pages 218–219)

Understanding Concepts 1.

When the two students pull on either end of the spring, it will not stretch as much as when it is pulled by both students while attached to the wall. When they both pull on it from the same side, the wall pulls back with equal force. When they  pull from opposite ends, the force will be hal f as much, and the stretch will also be half as much. 2. The elastic potential energy is the same when the spring is stretched or compressed 2.0 cm. The amount of energy stored only depends on the magnitude of the distortion, not the direction. 3. Harmonic means that it is regularly repeated, symmetrical motion. 1 4. (a) Period is inversely proportional to the frequency, T  ∝ .  f  (b) The acceleration is directly proportional to the displacement, a ∝ x . (c) The period is inversely proportional to square root of the force constant, T  ∝

1 k 

.

(d) The maximum speed is directly proportional to the amplitude, v ∝ A . 5. m = 62 kg 3 k  = 2.4  10  N/m  x = ? Σ F y = ma = 0 6 F x

− mg  = 0 6kx = mg    x =

=

mg  6k  (62 kg)(9.8 N/kg) 6(2.4 × 103 N/m)

 x = 0.042 m The compression of each spring is 0.042 m.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   272

6.

k  = 78 N/m  x = 2.3 cm = 0.023 m  F  x = ?

= kx = (78 N/m)(0.023 m)  F   x = 1.8 N  F x

The magnitude of force is 1.8 N. 7.  x1 = 1.85 cm  F  x1 = 85.5 N  x2 = 4.95 cm = 0.0495 m  F  x2 = ? First we must calculate k :  F x1

= kx1

k  =

=

 F   x1  x1 85.5 N

0.0185 m k  = 4621.6 N/m Solve for x2:

= kx2 = (4621.6 N/m)(0.0495 m)  F   x 2 = 229 N  F x 2

8.

The force required is 229 N. m = 97 kg k  = 2.2  103 N/m 2 a = 0.45 m/s  x = ? Ignoring friction to calculate the applied force: Σ F x = ma  FA  F A

= ma = (97 kg)(0.45 m/s2 ) = 43.65 N

This force is the force exerted on the spring:  F x = kx  x =

=

 F   x k  43.65 N

2.2 × 10 N/m  x = 0.020 m The spring stretches 0.020 m, or 2.0  10 –2 m. 9. m = 289 g = 0.289 kg k  = 18.7 N/m (a)  x = 10.0 cm = 0.100 m Σ F  y =  ? 3

a = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   273

To calculate the net force: Σ F y = Fx

Σ F  y

− mg  = kx − mg   = (18.7 N/m)(0.100 m) − (0.289 kg)(9.80 N/kg) = −0.962 N

To calculate acceleration: Σ F y = ma a

= =

Σ F  y m −0.962 N 0.289 kg

a = −3.33 m/s 2 The net force is 0.962 N [down], and the acceleration is 3.33 m/s 2 [down]. (b)  x = ? Σ F y = ma = 0  F x

− mg  = 0 kx = mg    x =

=

mg  k  (0.289 kg)(9.80 N/kg)

18.7 N/m  x = 0.151 m The spring will be stretched by 0.151 m. 10. m = 64.5 kg ∆ y = 48.0 m –12.5 m = 35.5 m k  = 65.5 N/m  x = 35.5 m – 10.1 m = 25.4 m. v = ? ∆ Eg = Ee + E K  mg ∆y mv

2

=

1 2

kx

2

+

1 2

mv

2

= 2mg ∆y − kx 2

v=

=

2 mg ∆y − kx 2 m 2(64.5 kg)(9.80 N/kg)(35.5 m) − (65.5 N/m)(25.4 m)

2

64.5 kg

v = 6.37 m/s The jumper’s speed at a height of 12.5 m above the water is 6.37 m/s. 11.  F  x = 8.6 N  x = 9.4 cm = 0.094 m (a) k  = ?  F x = kx k  =

=

 F   x  x 8.6 N

0.094 m k  = 91 N/m The force constant of the spring is 91 N/m.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   274

(b)  E e = ?  Ee

= =

1 2 1

kx 2 (91.49 N/m)(0.094 m) 2

2  E e = 0.40 J

The maximum energy of the spring is 0.40 J.  –7 12.  x = 1.0  10  m  E e = 1.0  10–13 J k  = ? 1  Ee = kx 2 2 2 E  k  = 2e  x

=

2(1.0 × 10 −13 J) (1.0 × 10−7 m) 2

k  = 2.0 × 101 N/m 1 The force constant is 2.0  10  N/m. 13. m = 22 kg θ  = 29° k  = 8.9  102 N/m  x = 0.30 m d  = ? To calculate the total vertical drop, ∆ y:  ET1 = E T2

∆ Eg = E e mg ∆y

=

∆ y = =

1 2

kx

kx

2

2

2mg  (8.9× 102 N/m)(0.30 m)2 2(22 kg)(9.8 N/kg)

∆ y = 0.1858 m To calculate the distance: sin θ  = d  =

=

∆ y d  ∆ y sin θ  0.1858 m

sin 29° d  = 0.38 m The crate slides 0.38 m along the ramp. 14. m = 0.20 kg k  = 28 N/m ∆ y = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   275

For fall, ∆ y = x. At maximum fall, there will be no kinetic energy.  ET1 = E T2

∆ Eg = E e mg ∆y

=

1 2

kx 2

− 2mg ∆y = 0 2 28 x − 2(0.20)(9.8) x = 0 2 28 x − 3.92 x = 0  x (28 x − 3.92) = 0 28 x − 3.92 = 0 kx

2

 x =

or

x=0

3.92

28  x = 0.14 m or x = 0 Since x = 0 refers to the moment of release, the maximum stretch will be 0.14 m.

Applying Inquiry Skills 15. (a) (i) Line A (total energy is constant) (ii) Line B (no kinetic energy at maximum stretch) (iii) Line C (maximum elastic potential energy a maximum stretch) (b) By observation from the graph, A = 10.0 cm = 0.100 m (c) At the end, the total energy of 5.0 J is all elastic potential energy, 1 2  Ee = kx 2 2 E  k  = 2e  x 2(5.0 J)

=

(0.100 m)

2

k  = 1.0 ×10 N/m The force constant of the spring is 1.0  103 N/m. (d) Maximum kinetic energy is 5.0 J when there is no elastic potential energy. 1  EK  = mv 2 2 3

v=

=

2 E K  m 2(5.0 J)

0.12 m v = 9.1 m/s The maximum speed of the mass is 9.1 m/s. 16. (a) If the spring were cut in two, it would take more force to stretch the spring the same amount because each coil would need to be moved twice as far. This will cause the force constant to double for the remaining two half springs. If a mass hung from two identical springs attached together caused them to stretch 36 cm, than each of them would stretch half of that amount. Since each bears the same weight, the force constant of each would only allow them to stretch 18 cm under the same force, indicating a force constant that is twice as great. (b) Answers will vary.

Making Connections 17. m = 5.5  102 kg number of vibrations = 6.0 t  = 3.5 s k  = ?

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Chapter 4 Work and Energy   276

Calculate the frequency:  f  =

=

number of complete vibrations total time 6.0

3.5 s  f  = 1.7143 Hz To calculate the force constant:  f  = 2π  f  =

1

k 

2π 

m

k  m

= 4π  2 f 2 m = 4π  2 (1.7143 Hz) 2 (5.5 × 102 k  = 6.4 × 104 N/m k

The force constant of each spring is 6.4 18. a = 25 g   f  = 8.9 Hz  A = ? T  = 2π 

 A

1

 A

 f 1

= 2π 

a

 A

 A =

a

=

4

 10  N/m.



a

=

2π  f

kg)

a 4π  2 f 2 2

25(9.8 m/s ) 4π  2 (8.9 Hz) 2

 A = 0.078 m  –2 The minimum amplitude is 0.078 m, or 7.8  10  m.

CHAPTER 4 LAB ACTIVITIES Activity 4.4.1: Applying the Law of Conservation of Energy (Page 220) (a) Energy conversions would be gravitational potential to kinetic to move the hands. (b) Different types include water clocks, hour-glass clocks, and pendulum clocks. (c) – (f) Answers will vary based on the students’ choice of design.

Investigation 4.5.1: Testing Real Springs (Pages 220–221)

Questions (i) By graphing the stretch as a function of the applied force we can learn the relationship is linear. (ii) The force constant of two combined springs is always less than the force constant for either spring individually.

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Chapter 4 Work and Energy   277

Hypothesis (a) The graph of F  x versus x should be a straight line with a positive slope. k k  (b) The combined force constant will be 1 2 . k1 + k 2

Prediction (c)

(d) k total

=

k1k 2 k1 + k 2

Procedure 1 – 3. See sample data table below.

Mass (g)

Force (N)

0 50 100 150 200

0 0.49 0.98 1.47 1.96

Stretch Spring 1 (cm) 0 1.10 2.22 3.29 4.43

Stretch Spring 2 (cm) 0 0.72 1.43 2.14 2.86

Stretch Spring 3 (cm) 0 0.56 1.11 1.66 2.22

Stretch Spring 1 + Spring 2 (cm) 0 1.82 3.63 5.44 7.26

Analysis (e)

(f) Spring 1 rise slope = run 1.47 N − 0.49 N

Spring 2 rise slope = run 1.47 N − 0.49 N

Spring 3 rise slope = run 1.47 N − 0.49 N

0.0329 m − 0.0110 m slope = 44.5 N/m

0.0214 m − 0.0072 m slope = 69.0 N/m

0.0166 m − 0.0056 m slope = 89.1 N/m

=

=

=

These slopes represent the force constant for the spring.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   278

(g) The extrapolation is not feasible because the spring is not capable of supporting that much mass. (h) When two springs are connected linearly, each spring must support the full weight (if we ignore the mass of the lower spring), so the total stretch will be equal to the sum of the individual stretches, (i.e.,  xtotal = x1 + x2).

= kt ot al xt ot al = ktotal ( x1 + x2 )  F F    = k total  +    k1 k 2    1 1    F = Fk total  +    k1 k 2    k + k 1   1 = k total  2    k1k 2    F

k total

=

k1k 2 k1 + k 2

Evaluation (i) Answers will vary based on the hypothesis and predictions made. (j) Sources of error could be improper zero for measuring, parallax, bend in support apparatus, and permanent distortion of the spring. These can be avoided by careful calibration at lab setup, caution to avoid parallax during measurement, and rigid apparatus for the support stand. The value of g  in the area, as well as the accuracy of the stamped value of the masses can contribute to error.

Synthesis (k) To calculate the slope of the line and have it represent the force constant, the force had to be plotted on the vertical axis. (l) A real spring will heat up when stretched and has an elastic limit beyond which it will not act as a Hooke’s law spring any longer. Ideal springs do not have these limitations.

Investigation 4.5.2: Analyzing Forces and Energies in a Mass-Spring System (Pages 222–223)

Questions (i) The action of a real mass–spring system supports the law of conservation of energy. (ii) The damping is slow for a real vertical mass–spring system.

Hypothesis (a) A real vibrating spring will undergo damped harmonic motion. The presence of friction within the spring and within the air will transform some of the kinetic and elastic potential energy into thermal energy. (b) The damping of a real spring will likely be very slow.

Prediction (c)

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Chapter 4 Work and Energy   279

(d)

Procedure 1.

Suspend a mass from the spring and measure the stretch (do several times to get a precise value). mg  k  =  x (0.200 kg)(9.80 N/kg) =

2. 3.

0.0126 m k  = 156 N/m Using a 200.0-g mass, maximum stretch is 2.51 cm. Maximum vertical displacement is 2.51 cm, 2.36 cm, 2.22 cm, 2.08 cm, 1.96 cm, 1.84 cm, 1.73 cm, 1.63 cm, 1.53 cm, 1.44 cm. Total time for 10 cycles is 2.25 s.

Analysis (e)

 x  (cm) ∆y  (cm) ∆E g (J)

E e (J) E K (J) E T (J)

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top 0

middle 1.26

bottom 2.51

2.51

1.26

0

0.0492

0.0246

0

0

0.0124

0.0491

0

0.0122

0

0.0492

0.0492

0.0491

Chapter 4 Work and Energy   280

(f)  EK  = v

=

=

v

=

1 2

mv 2

2 E K  m 2(0.0122 J ) 0.200 kg 0.349 m/s

(g)

(h) The energy is lost to thermal energy through friction. (i) (i) The action of a real mass system supports the law of conservation of energy for short time intervals (one bounce), but seems to refute it for longer time intervals (several bounces). (ii) A real spring system slowly damps the motion of a vertically vibrating system.

Evaluation (j) Answers will vary depending on the hypotheses and predictions made. (k) The largest source of error is difficulty reading the location of a moving object with accuracy. A slow motion film, or a spring with a lower force constant could reduce the speed.

Synthesis (l) A stiff spring will undergo faster damping because the object will move faster. The higher speed will increase the amount of air friction in each cycle, damping the motion more quickly.

Activity 4.5.1: Achieving a Smooth and Safe Ride (Page 223)

(a) Energy will be converted into elastic potential energy and then into thermal energy, and the spring undergoes damped harmonic motion. (b) - capable of absorbing large quantities of energy - small amounts of energy are absorbed quickly to provide a smooth ride - often the spring has a changing spring constant to accomplish the above ideas - for safety, shock absorber must be capable of damping violent motion quickly - shock absorber must be able to damp both small and large stored energies (c) The system uses a spring to absorb the energy, and then the shock absorber damps the motion quickly to prevent danger  bounce in the vehicle. (d) Lower friction designs last longer because they do not produce as much heat. The heat can break down the seals and increase the chances of warping. (e) Answers will vary based on the resources students use. (f) Answers will vary based on the initial hypothesis made.

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Chapter 4 Work and Energy   281

CHAPTER 4 SUMMARY Make a Summary (Page 224)

1.

Determine the spring constant by suspending a known mass (and, therefore, weight) from it and measuring the stretch. mg  k  =  x (0.100 kg)(9.80 N/kg)

=

0.0356 m k  = 27.5 N/m Use projectile analysis with the known launch angle and range to determine the launch speed. For this problem, use θ  = 34º and ∆d  x = 2.6 m, and ∆d  y = 0.0 m.

In horizontal direction:

∆d x = vx ∆t  2.6 = (v cos 34°) ∆t  v∆t  =

2.6

(Equation 1)

cos34°

In vertical direction:

∆d y = v yi ∆t +

1 2

a(∆t )   2

0 = (v sin 34° )∆t +

1

(− 9.8)(∆ t ) 2 0 = ∆t (v sin 34° − 4.9∆ t )

2

Ignore the case where ∆t  = 0 since it refers to the launch: 0 = v sin 34° − 4.9∆t 

∆t  =

v sin34° 4.9

(Equation 2)

Substitute Equation 2 into Equation 1:  v sin 34°  = 2.6 v    4.9   cos 34°

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2.6 ( 4.9 )

v

=

v

= 5.2 m/s

(sin34°)(cos34°)

Chapter 4 Work and Energy   282

Use conservation of energy to determine the amount of stretch required in the spring. Assume the spring mass is 18 g.  ET1 = E T2 1 2

 Ee

= ∆E g

kx 2

=

1 2

mv

 x =

2

k  (0.018 kg)(5.2 m/s)2

= 2.

mv 2

27.5 N/m

 x = 0.13 m The height could be determined by analyzing the vertical component of velocity.

= vi2 + 2a∆d v 2 − v2 ∆d  = f i 2

vf

 

2a

=

(0 m/s)2

− (5.2 m/s)2

2(−9.8 m/s2 )

∆d  = 1.4 m 3. 4.

5. 6.

The three forms of friction would use up some of the elastic potential energy intended to launch the spring. To compensate, the spring would need to be stretched an extra amount. The energy to stretch the string would come from food you ate. The energy stored in the spring would be converted to kinetic energy as it left. The kinetic energy would be partially converted into gravitational potential as it rises, and than reconverted back to kinetic as it lands. As it lands on the floor and slides to a stop, the kinetic energy is converted into sound and thermal energy. The spring would cause a mass to move up and down with decreasing amplitude until it came to rest. The concepts and equations can be used to design and manufacture vehicles, beds, and a wide variety of other products.

CHAPTER 4 SELF QUIZ (Page 225)

True/False 1. 2. 3. 4. 5. 6. 7. 8. 9.

T F The work done by gravity is zero. F The work you do on the backpack is negative. F The gravitational potential energy decreases in proportion to the distance fallen. F This does not refute the law of conservation of energy because some energy is converted into other forms, such as heat (thermal energy) in the ball and the floor. T T F Maximum speed occurs at the equilibrium position, but elastic potential energy is at a minimum at the point of minimum extension of the spring. F A long damping time would not be appropriate for a bathroom scale. It would be appropriate for a “jolly-jumper” toy.

Multiple Choice 10. 11. 12. 13. 14. 15. 16.

(c) (c) (e) (d) (e) (a) (d) W W

= ( F cosθ )∆d  = (mg sin θ )(cos180°) L = − mgL sin θ 

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Chapter 4 Work and Energy   283

CHAPTER 4 REVIEW (Pages 226–229)

Understanding Concepts 1. (a) No work is done because the force is perpendicular to the displacement. (b) The work is negative because the student is exerting a force opposite the direction of motion. (c) The work is negative because gravity is exerting a force opposite the direction of motion. (d) Assuming a level roadway, the work done is zero because the force is perpendicular to the displacement. (e) No work is done because the electrical force is perpendicular to the displacement. (f) No work is done because the tension is perpendicular to the displacement. 2. The force must be applied perpendicular to the object for it to do no work on the object. 3. The normal force can do work on an object. For example, when you jump, you push down on the ground and the normal force pushes up on you and accelerates you up, giving you kinetic energy. 4. (a) No work is being done on the swimmer because the balancing forces forward and backward produce no motion. (b) Work is done on the student to speed him up, but after that, there is no work being done on the student. Once the student reaches the speed of the current, the only force is the upward buoyant force, which is perpendicular to the waters surface. Technically, a small amount of work is being done by gravity as they whole river/student system is  pulled closer to the earth. 5. (a) The velocity will be changing because Newton’s second law states that if a net force acts on an object, it will accelerate (i.e., change its velocity). (b) It is possible the speed is constant if the particle is travelling in a circle. (c) The kinetic energy is proportional to the square of the speed and the mass. Since both can remain constant under these conditions, the kinetic energy may be constant. 6. Agree. In the absence of friction (including air resistance), all of the gravitational potential energy will be converted into kinetic energy. The mass will cancel in each term. 7. (a) Damped vibrations are useful in the suspension of a vehicle. (b) Damped vibrations are not useful in a pendulum clock. 8. It is not possible to have a motion that is not damped. Such a device would be a perpetual motion machine that cannot exist. The force of friction within the system cannot be avoided. 9. m = 0.425 kg ∆ y = 11.8 m (a) W  = ? W = ( F cosθ )∆ d 

= (mg cosθ )∆y = (0.425 kg)(9.80 m/s2 )(cos180°)(11.8 m) W  = −49.1 J The work gravity does on the ball on the way up is –49.1 J. (b) W  = ? W = ( F cosθ )∆ d 

= (mg cosθ )∆y = (0.425 kg)(9.80 m/s2 )(cos 0°)(11.8 m) W  = 49.1 J The work gravity does on the ball on the way down is 49.1 J. 10.  F  = 9.3 N W  = 87 J ∆d  = 11 m θ  = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   284

W

= ( F cosθ )∆d 

cos θ  =

W   F ∆d 

   F ∆d         87 J = cos −1     (9.3 N)(11 m)   θ  = 32° θ  = cos

−1  W 

The angle between the applied force and the horizontal is 32 °. 11. Let the subscript C represent the child, and TO represent the toboggan. mC = 25.6 kg mTO = 4.81 kg ∆ y = 27.3 m (a) W C = ?

First, find the actual distance: 27.3 sin φ  = ∆d  27.3 ∆d  = sin φ  The applied force, F A, will be equal to the component of gravity down the hill, mg sin φ  WC = ( F cos θ ) ∆d 

= (mg sin φ )(cosθ )∆d   27.3   = (mg sin φ )(cosθ )     sin φ    = (mg cosθ )(27.3) = (4.81 kg)(9.80 N/kg)(cos 0°)(27.3) 3 W C = 1.29 × 10 J 3

The total work done by the child is 1.29  10  J. (b) From part (a), the angle doesn’t matter, therefore W  = 1.29 × 103 J. (c) The total work on the child and toboggan during the slide will be equal to the work done to take them up the hill. Using the equation derived in part (a), WT = ( mg cos θ )(27.3) W T

= (25.6 kg + 4.81 kg)(9.80 N/kg)(cos0°)(27.3) = 8.14 × 103 J

The total work on the child and the toboggan is 8.14 12. m = 73 kg θ  = 9.3° vi = 4.2 m/s ∆d  = ?

Copyright © 2003 Nelson

 103 J.



Chapter 4 Work and Energy   285

The relation between the distance along the slope and the vertical height is: ∆ y sin9.3° = ∆d  ∆ y = ∆d sin9.3° To calculate the change in distance:  ET1 = E T2 1 2

mv 2 v

2

= mg ∆y = 2 g ( ∆d sin 9.3°)

∆d  = =

v

2

2 g (sin 9.3°) (4.2 m/s)2 2(9.8 m/s)(sin9.3°)

∆d  = 5.6 m The skier would travel 5.6 m along the hill before stopping. 13. An increase of 50% is equivalent of multiplying by 1.5: 1 mv22  E K2 2 =  E K1 1 2 mv1 2 1.50 EK1  E K1

=

1.5 =

v22 2

v1

(v1 + 2.00) 2 v12

= v12 + 4v1 + 4 2 0.5v1 − 4v1 − 4 = 0 1.5v12

v1

= =

−(−4) ± (− 4)2 − 4(0.5)(− 4) 2(0.5) 4 ± 4.899

1 v1 = 8.90 m/s, or

− 0.899 m/s

Since the negative value is not admissible, the original speed of the object was 8.90 m/s. 14. m = 7.0  109 kg ∆ y = 36 m (a) ∆ E g = ? ∆ Eg = mg ∆y

∆ E g

= (7.0× 109 kg)(9.8 N/kg)(36 m) = 2.5 × 1012 J 12

The gravitational potential energy is 2.5  10  J. (b) The work done by on the pyramid by one person in 20 years is:

 5 ×106 J   40 d   (0.20 )  ( 20 a ) = 8 ×108 J/person      d   a   To calculate the total number of people: 2.5 × 10

12

= 3 ×103

8 ×10 3 There were 3  10  workers involved. 8

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people

Chapter 4 Work and Energy   286

15. m = 45 kg ∆d  = 66 cm = 0.66 m (a) W  = ?

= ( F cosθ )∆d  = (mg cosθ )∆d  = (45 kg)(9.80 N/kg)(cos180°)(0.66 m) 2 W  = −2.9 × 10 J W

The work done by gravity on the mass is –2.9 (b) W  = ? W = ( F cos θ )∆ d 

2

 10  J.



= (mg cosθ )∆d  = (45 kg)(9.80 N/kg)(cos0°)(0.66 m) 2 W  = 2.9 × 10 J The work done by the weightlifter on the mass is 2.9 (c) ∆ E g = ? ∆ Eg = mg ∆y

= (45 kg)(9.80 N/kg)(0.66 m) = 2.9 × 102 J

∆ E g

The change in gravitational potential energy is 2.9 16. m = 47 g = 0.047 kg (a) ∆ y = 4.3 m v2 = ?  ET1 = E T2 1 2

2

 10  J.



mv12

+ mgy1 =

1 2



2

 10  J.

mv22 + mgy 2

Since v1 = 0:

= v22 + 2 gy2 v22 = 2 gy1 − 2 gy 2 v2 = 2 g ( y1 − y2 )

2 gy1

= 2(9.8 m/s2 )(4.3 m) v2 = 9.2 m/s The speed of the stick just before it hits the ground is 9.2 m/s. (b) If air resistance was included, the answer in part (a) would be slightly smaller. The energy from gravity would be shared between the thermal energy and the kinetic energy. 17.  y1 – 27 m v1 = 18 m/s θ  = 37° (a) v2 = ?  ET1 = E T2 1 2

mv12 v12

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy2 2 2 v2 = v1 + 2 gy1 − 2 gy2 v2

=

v2

= (18 m/s) 2 + 2(9.8 m/s2 )(27 m − 0 m) = 29 m/s

v12

+ 2 g ( y1 − y2 )

The speed of the stick just before it hits the ground is 29 m/s. (b) The angle the stick is thrown at does not appear in the equation, so the answer will still be 29 m/s.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   287

18.  F  = 1.5  10  N [22° below the horizontal] m = 18 kg ∆d  = 1.6 m  µk  = 0.55 2

(a)  F  N = ?  F K =   ? To calculate the normal force:

Σ F  y = 0  F N − FA sin 22° − mg  = 0  F N = FA sin 22° + mg  = (1.5 × 102 N)(si n 22°) + (18 kg )(9.8 N/ kg ) = 232.6 N 2  F   N = 2.3 × 10 N To calculate the force of friction:  FK = µ K F N

= (0.55)(232.6 N ) = 127.9  F K  = 1.3 × 102 N The normal force on the box is 2.3 (b) v = ?

2

 10  N. The force of friction on the box is 1.3



2

 10  N.



If the box is starting from rest, there is no initial kinetic energy. W = Eth + E K  ( F cos θ ) ∆d mv

= FK ∆d +

1 2

mv2

= 2( F cos θ )∆d − 2 FK ∆d   2( F cos θ )∆d − 2 FK ∆d  v= 2

m

=

2(1.5×10 N)(cos22°)(cos0°)(1.6 m) − 2(127.9 N)(1.6 m) 2

18 kg

v = 1.4 m/s The final speed of the box is 1.4 m/s. (c)  E th = ?  Eth = FK ∆d   E th

= (127.9 N)(1.6 m) = 2.0 × 102 J

The amount of thermal energy produced is 2.0

Copyright © 2003 Nelson



2

 10  J.

Chapter 4 Work and Energy   288

3

19. m = 1.2  10  kg v1 = 9.5  103 m/s  F  = 9.2  104 N 3 ∆d  = 86 km = 86  10  m v2 = ? W

= ∆E K 

 F ∆d

=

1

mvf2

=

vf

=

2

1 2

−

1

mvi2

2 1 F ∆d + mvi2 2

= vf 

mvf2

2 F ∆d  m

+ vi2

2(9.2 ×10 4 N)(86 × 103 m) 1.2 × 10 kg 3

+ (9.5 × 103 m/s)2

= 1.0 ×104 m/s

The final speed of the probe is 1.0  104 m/s. 20.  y1 = 1.15 m  y2 = 4.75 m v1 = ?  ET1 = E T2  EK1 + Eg1 1 2

2

mv1

= E g2

+ mgy1 = mgy2 = 2( gy2 − gy1 ) v1 = 2 g ( y2 − y1 )

v12

= 2(9.80 m/s2 )(4.75 m − 1.15 m) v1 = 8.40 m/s The speed with which the gymnast leaves the trampoline is 8.40 m/s. 21. The work in the area under the graph. Consider the area in three parts: W = Atriangle 1 + Arectangle + Atriangle 2

= =

1 2 1

b1h1 + lw +

1 2

b2 h2

(1.0 m )(12.0 N) + ( 2.0 m − 1.0 m )(12.0 N )+

2 W  = 42 J The person does 42 J of work. 22.  x = 0.418 m  F  = 1.00  102 N (a) k  = ?  F x = kx k  =

=

1 2

(6.0 m − 2.0 m)(12.0 N)

 F   x  x 1.00 × 102 N

0.418 m k  = 239 N/m The force constant of the spring is 239 N/m. (b)  x = 0.150 m  F  x = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   289

= kx = (239 N/m)(0.150 m) = 35.9 N

 F x  F   x

The force required to stretch the spring is 35.9 N. (c) To stretch it 0.150 m: 1  Ee = kx 2 2 1 = (239 N/m)(0.150 m)2 2  E e = 2.69 J To compress it 0.300 m:  Ee

1

=

2 1

=

kx

2

(239 N/m)(−0.300 m)2

2  E e = 10.8 J

The work required is 2.69 J to stretch it, and 10.8 J to compress it. 23. k  = 22 N/m  –3 m = 7.5  10  kg  –2  F K  = 4.2  10  N  x = 3.5 cm = 0.035 m ∆d  = ?  ET1 = E T2 1 2

 Ee

= E th

kx 2

= FK ∆ d  

∆d  = =

kx 2 2 F K  (22 N/m)(0.035 m)2 −

2(4.2 × 10 2 N)

∆d  = 0.32 m The eraser will slide 0.32 m along the desk. 24. k  = 75 N/m  A = 0.15 m v = 1.7 m/s  x = 0.12 m m = ? For SHM, all of the energy is E e when at the maximum amplitude, A:  ET1 = E T2  Ee max 1 2

kA

2

= Ee + E K =

1 2

kx

2

+

 

1 2

mv

2

= kA − kx 2 k ( A2 − x 2 ) m= 2

mv

2

2

v

=

(

2

75 N/m ( 0.15 m )

− ( 0.12 m )2 )

(1.7 m/s)

2

m = 0.21 kg The mass of the block is 0.21 kg. Copyright © 2003 Nelson

Chapter 4 Work and Energy   290

25. m = 0.42 kg k  = 38 N/m  A = 5.3 cm = 0.053 m (a)  E e = ? Maximum energy occurs at full amplitude. 1 2  Ee = kx 2 1 = (38 N/m)(0.053 m) 2 2  E e = 0.053 J The maximum energy of the mass-spring system is 0.053 J. (b) v = ? Maximum speed occurs when there is no elastic potential energy:  ET1 = E T2  Ee max 1 2

kA

2

mv

2

= E K =

1 2

 

mv

2

= kA2 2

kA

v=

m 38 N/m(0.053 m) 2

=

0.42 kg

v = 0.50 m/s The maximum speed of the mass is 0.50 m/s. (c)  x = 4.0 cm = 0.040 m v = ?  ET1 = E T2  Ee max 1 2

kA

2

mv

2

= Ee + E K =

1 2

kx

2

+

 

1 2

mv

2

= kA − kx 2 2

v=

=

k ( A2

− x2 )

m

(

2

38 N/m ( 0.053 m )

− ( 0.040 m )2 )

0.42 kg

v = 0.33 m/s The speed of the mass is 0.33 m/s. (d)  x = 4.0 cm = 0.040 m  E T = ?  ET = Ee + E K 

= =

1 2 1

kx 2

+

1 2

mv 2

(38 N/m)(0.040 m)2

2  E T = 0.053 J

+

1 2

(0.42 kg)(0.33 m/s)2

The total energy is 0.053 J. The results are the same.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   291

Applying Inquiry Skills 26. (a) One text weighs 2.0 kg and has a thickness of 3.6 cm. The first one doesn’t need to be raised at all. W = Eg1 + E g2 + Eg3 + Eg4 + Eg5  

= mgy1 + mgy 2 + mgy3 + mgy 4 + mgy5 = mg ( y1 + y2 + y3 + y4 + y5 ) = ( 2.0 kg )(9.8 N/ kg )(0 m + 0.036 m + 0.072 m + 0.108 m + 0.144 m) W  = 7.1 J You would have to do 7.1 J of work. (b) Some errors would be determining the mass of the text. Also, the thickness may compress the books on the bottom.  Not all texts may have the same mass. 27. The calculator would need to know the distance the box was pushed. 28.

29. The tractor seat would need to have a strong spring to absorb large bumps, and a shock absorber to prevent “launching” the driver. There would have to be smaller springs on top of that to absorb small vibrations. Damping would be important to prevent resonance.

Making Connections 30. Roller coasters are often shut down in high winds because of the loss of energy that may occur due to increased air resistance. Cold can cause parts to shrink and increase frictional forces beyond safe limits. 31. (a) The ball on track B will arrive first. Shortly after the start, the vertical drop of the ball on track B causes an increase in the speed, which it will enjoy for the majority of the race. At the end, it will slow down to the same speed that the ball on track A has just accelerated to. (b) Racing cyclists use this in a variety of ways. The sprinters stay high on the track until ready to make a break for it, converting all of their stored gravitational energy into kinetic. Team racers use the change to minimize the work of the rider. When the leader is ready to give up his spot, he rides up the hill a bit, converting some of his kinetic energy into gravitational potential energy, reducing his speed. Once the last team mate has passed, he can drop down again, gaining the gravitational potential energy back as kinetic energy without needing to supply it from his own body power. 32. (a) v1 = 0 m/s v2 = ?  ET1 = E T2 1 2

mv12 v12

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy 2 2 v2 = 2 gy1 − 2 gy 2 v2 = 2 g ( y1 − y2 ) = 2(9.80 m/s2 )(37.8 m − 17.8 m) v2 = 19.8 m/s

The speed of the coaster at position C is 19.8 m/s.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   292

(b) v1 = 5.00 m/s v2 = ?  ET1 1 2

mv12 2

v1

= E T2

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy2 2 2 v2 = v1 + 2 gy1 − 2 gy2 v2

=

v2

= (5.00 m/s)2 + 2(9.80 m/ s2 )(37.8 m − 17.8 m) = 20.4 m/s

v12

+ 2 g ( y1 − y2 )

The speed of the coaster at position C is 20.4 m/s. 33. (a) The mass is not necessary because it cancels out of the mathematical equations. (b) You might expect the speed would be 5.0 m/s more at the second point, but the reality is that the small kinetic energy supplied by having a speed going over the first hill is negligible compared with the large amount of gravitational  potential energy. 34. a = 12 g  [upward] (a)  F A = ? Σ F y = ma  FA

− mg = ma  FA = ma + mg  = m( a + g ) = m(0.12 g + g )  FA = 1.12 mg 

The force required is 1.12mg . (b) W  = ? W = ( F cos θ )∆ d  W

= (1.12mg )(cos 0°)(∆y ) = 1.12 mg ∆y

The work done is 1.12mg ∆ y. 35. m = 1.5 kg 3 k  = 2.1  10  N/m ∆ y = 0.37 m  x = ?

Copyright © 2003 Nelson

Chapter 4 Work and Energy   293

 ET1 mg ∆y

= E T 2 =

(1.5)(9.8)(0.37 + x ) =

1 2 1 2

kx

2

(2.1× 103 ) x 2

5.439 + 14.7 x = 1050 x 2 1050 x 2 − 14.7 x − 5.439 = 0 Using the quadratic equation:

−(−14.7) ± (−14.7)2 − 4(1050)(− 5.439)

 x =

2(1050) 14.7 ± 151.9

=

2100 = 0.079 m or  x = −0.065 m (negative value inadmissable)  x = 0.079 m The maximum distance the spring is compressed is 0.079 m.  –4 36. m = 0.55 g = 5.5  10  kg ∆d  = 95 cm = 0.95 m ∆d   x = 3.7 m  E e = ? First we must calculate the time required for the vertical drop (vi = 0): 1 ∆d = vi ∆t + a(∆t ) 2  2 1 ∆d = a (∆t )2 2 2∆d 

∆t  =

a 2(−0.95 m)

=

−9.8 m/s 2

∆t  = 0.44 s To calculate the horizontal speed: ∆d x = v x ∆t  v x

= =

∆d  x ∆t  3.7 m

0.44 s v x = 8.4 m/s We know that initial kinetic energy came from elastic potential energy, therefore:  ET1 = E T 2  Ee

= E K  = =

1 2 1

mv

2

−4

(5.5 × 10 kg )(8.4 m/s) 2  E e = 0.019 J

2

The elastic potential energy stored was 0.019 J.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   294

37.  y1 = 16 m  y2 = 9.0 m  H  = ? First we must solve for the launch speed if v1 = 0:  ET1 = E T2 1 2

mv12

+ mgy1 =

1 2

mv22

+ mgy2

v2

= v22 + 2 gy2 = 2 gy1 − 2 gy2 = 2 g ( y1 − y2 )

v2

= 2(9.8 m/s2 )(16 m − 9.0 m) = 11.71 m/s

2 gy1 2

v2

Resolve vector components:

As just clearing the wall, the horizontal component will be the same, and the vertical component must produce a 30º angle to the vertical.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   295

To calculate the vertical speed: tan30° = v y

= =

v2 cos45 ° v y v2 cos45 ° tan30° (11.71 m/s) cos 45°

tan30° v y = 14.35 m/s [down] Determine how far the skier has dropped from the launch point:

= vi2 + 2a∆d v2 − v2 ∆d  = f i vf2

 

2a

=

( −14.35 m/s) 2

− ((11.71 m/s )( sin 45° )2 )

2( −9.8 m/s 2 )

∆d  = −7.0 m Therefore, the wall must be 9.0 – 7.0 = 2.0 m tall. 38.  y1 = 2.5 m v1 = 9.0 m/s  y2 = 3.0 m v2 = ?  ET1 = E T2 1 2

mv12 2

v1

+ mgy1 =

1 2

mv22

+ mgy2

+ 2 gy1 = v22 + 2 gy 2 2 2 v2 = v1 + 2 gy1 − 2 gy 2 v2

=

v12

+ 2 g ( y1 − y2 )

= (9.0 m/s)2 + 2(9.8 m/s2 )(2.5 m − 3.0 m) v2 = 8.4 m/s The speed of the ball when it “swishes” through the hoop is 8.4 m/s. 39.  g  = 2.0 v1 = 6.0  y2 = 5.0 S  = ? Solve for the launch speed at the top of the ramp where y1 = 0:  ET1 = E T2 1 2

mv12

+ mgy1 = 2

v1

1 2

mv22

+ mgy2

+ 0 = v22 + 2 gy2 v22 = v12 − 2 gy2 v2

=

v12

− 2 gy2

= (6.0)2 − 2( 2.0)(5.0) v2 = 4.0 units

Copyright © 2003 Nelson

Chapter 4 Work and Energy   296

Analyze projectile motion to find the change in time:

∆d y = v yi ∆t +

1 2

a( ∆ t )   2

−5.0 = (4.0sin 30°)∆t + −5 = 2∆t − (∆t ) (∆t ) 2 − 2∆t − 5 = 0

1 2

(− 2.0)(∆ t )2

2

Using the quadratic equation:

∆t  = =

−(−2) ± (−2)2 − 4(1)(− 5) 2(1) 2 ± 4.9

2 = 3.45 units or

− 1.45 units (dismiss negative answer)

∆t  = 3.45 units To calculate horizontal range (S ): S = v x ∆t 

= (4.0cos30°)(3.45) S  = 12 units The shuttle lands a distance of 12 units from the ramp.

Copyright © 2003 Nelson

Chapter 4 Work and Energy   297

CHAPTER 5 MOMENTUM AND COLLISIONS Reflect on Your Learning (Page 230)

1. (a) The expression refers to once you start playing well, it is easier to keep playing well (or scoring, or winning). (b) The physics meaning refers to a specific quantity. The everyday use of the word momentum can mean continuing to do well and inertia. 2. (a) The momentum of car A is less than the momentum of car B. (b) The momentum of bicycle and rider A is less than the momentum of bicycle and rider B. (c) The momentum of the large truck A is greater than the momentum of car B. 3. (a) The phrase “follow-through” refers to continue to swing the racket/club even after contact with the ball is made. This allows the racket/club to be in contact with the ball for a longer period of time. (b) “Follow-through” affects the momentum (or change in it). 4. (a) One technique that could be used is to break the total area up into eight different rectangles. The height of each would  be an estimate. Each rectangle’s area could be determined and the sum would give the total area. kg ⋅ m kg ⋅ m (b)  F ⋅ t  =  N ⋅ s = ×s = 2 s s 5. The car should be designed with crunch zones to absorb the energy. If elastic bumpers were used, the car would rebound and be able to impact other objects.

Try This Activity: Predicting the Bounces (Page 231)

(a) Prediction: The balls will bounce to the same height. (b) Ball A and ball B both seem to be made from the same material. They possess approximately equal density and are approximately the same size. Since the balls appear to be identical, they will possess the same amount of elasticity and will bounce to the same height. (c) The balls did not bounce to the same height. Our prediction and hypothesis were not supported by the evidence. (d) Much of the kinetic energy of one ball is conserved in the bounce. The kinetic energy of the other ball is not conserved. It is transformed into other forms. This ball loses kinetic energy in the bounce and does not bounce to the same height. Bales of hay placed on the sides of ski-racing runs transform the kinetic energy of the skier into other forms, slowing the skier down and reducing the chance for bodily harm.

5.1 MOMENTUM AND IMPULSE PRACTICE (Page 233)

Understanding Concepts 1. (a) m = 7.0 × 103 kg v  = 7.9 m/s  p  = ? 





 p



=

mv

=

(7.0 ×103 kg)(7.9 m/s)

=

5.5 × 10 kg ⋅ m/s [E]



 p

4

The momentum of the African elephant is 5.5 × 104 kg ⋅ m/s [E] .

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  299

(b) m = 19 kg v  = 726 m/s  p = ? 



 p = mv 



= (19 kg)(26 m/s)  p = 4.9 × 102 kg ⋅ m/s [S] The momentum of the mute swan is 4.9 ×102 kg ⋅ m/s [S] .  –31 m = 9.1 × 10  kg v  = 1.0 × 107 m/s  p = ?  p = mv = (9.1× 10−31 kg)(1.0 × 107 m/s)  p = 9.1× 10 −24 kg ⋅ m/s [forward] The momentum of the electron is 9.1× 10−24 kg ⋅ m/s [forward] . 

(c)











2.

m = 405 kg

 p = 5.02 × 103 kg ⋅ m/s 

v  = ? 

 p = mv 



 p 

v



= =

v



m 5.02 × 103 kg ⋅ m/s 405 kg

= 12.4 m/s [W]

The velocity of the craft is 12.4 m/s [W] . 3.

 p = 4.5 kg ⋅ m/s 

v  = 9.0 × 102 m/s m = ? 

 p = mv 



 p 

m=

=

v 4.5 kg ⋅ m/s 

9.0 × 10 2 m/s

m = 5.0 × 10−3 kg The mass of the bullet is 5.0 × 10−3 kg . 4. (a) Assuming a mass of 58 kg and a top speed of 8.0 m/s.  p = mv

= (58 kg)(8.0 m/s)  p = 4.6 × 10 2 kg ⋅ m/s The magnitude of the momentum of an average student would be 4.6 ×102 kg ⋅ m/s . (b) Assume a typical compact car to have a mass of 1200 kg.  p = mv v=

=

 p m 4.6 ×10 kg ⋅ m/s 2

1200 kg

v = 0.39 m/s A typical compact car would have to travel at a velocity of 0.39 m/s to achieve the same momentum.

300 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

PRACTICE (Page 237)

Understanding Concepts 5.

impulse  F ∆t  =  N ⋅ s

=  F ∆t  =

kg ⋅ m 2

s kg ⋅ m

change in momentum ∆ p = m(∆v)

×s

m   = kg     s   kg ⋅ m ∆ p =

s s Therefore, the units of impulse and change in momentum are equivalent. 6. (a) m = 0.065 kg vi y = –3.8 m/s  pi y = ?  pi y  pi y

= mviy = (0.065 kg)(−3.8 m/s) = −0.25 kg ⋅ m/s

The momentum of the snowball before hitting the ground is –0.25 kg ⋅m/s. (b) vf  y = 0.0 m/s  pf  y = ?  pf y  pf  y

= mvf y = (0.065 kg)(0.0 m/s) = 0.0 kg ⋅ m/s

The momentum of the snowball after hitting the ground is 0.0 kg ⋅m/s. (c) ∆ p y = pfy − piy

∆ p y

= 0.0 kg ⋅ m/s − (−0.25 kg ⋅ m/s) = 0.25 kg ⋅ m/s

The change in momentum is 0.25 kg ⋅m/s. 7. (a)

(b)

Σ F  x = 4.8 ×103 N ∆t  = 3.5 s impulse = Σ F x ∆t  = ? Σ F x ∆t  = (4.8 ×103 N)(3.5 s) Σ F x ∆t  = 1.7 ×104 N ⋅ s [W] The impulse on the truck over this time interval is 1.7 × 104  pi x = 5.8 × 104 kg ⋅ m/s

N ⋅ s [W] .

 pf  x  = ?

Σ F x ∆t = ∆px Σ F x ∆t = pfx − pix  pf x = Σ Fx ∆t + pix = 1.7 × 104 N ⋅ s + 5.8 × 104  pf  x = 7.5 × 104 kg ⋅ m/s [W] The final momentum of the truck is 7.5 × 104 kg ⋅ m/s [W] .

Copyright © 2003 Nelson

kg ⋅ m/s

Chapter 5 Momentum and Collisions  301

8.

m = 0.27 kg vi x = 2.7 m/s vf  x = 0 m/s Σ F  x  = 33 N [W]

∆t = ? Choosing the initial direction of the ball (east) as positive: Σ F x ∆t = ∆px

Σ F x ∆t = m( vfx − vix ) m( v − v ) ∆t  = f x ix Σ F  x =

0.27 kg(0 m/s − 2.7 m/s)

−33 N

∆t  = 0.022 s, or 22 ms 9.

The ball is in contact with the net for 22 ms. v = 75.5 m/s [11.1° below the horizontal] m = 1.24 × 105 kg  p x = ?  p y = ?

=p = mv = (1.24× 105 kg)(75.5 m/s)  p = 9.362 × 106 kg ⋅ m/s

 p 

cos11.1 ° =  p x

 p x

 p x  p

= p cos11.1° = (9.362 × 106 kg ⋅ m/s)(cos11.1°) = 9.19 × 10 6 kg ⋅ m/s

sin11.1° =

 p y  p

 p y

=

p sin11.1°

 p y

= (9.362 × 106 kg ⋅ m/s)(sin11.1°) = 1.80 ×106 kg ⋅ m/s

The horizontal component of the plane’s momentum is 9.19 × 106 kg ⋅ m/s  and the vertical component is

1.80 ×106 kg ⋅ m/s .

Applying Inquiry Skills 10. (a) Using the area of a triangle: impulse =  A =

=

1 2

bh

1

(0.5 s)(4.0 N) 2 impulse = 1.0 N ⋅ s [E] The impulse imparted is 1.0 N ⋅s [E].

302 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

(b) Estimate the shape from 0 s – 0.2 s to be a triangle. Estimate the shape from 0.2 s – 0.4 s to be a trapezoid. Estimate the shape from 0.4 s – 1.0 s to be a trapezoid. The sum of all three is the total impulse: 1  h + h2  b +  h1′ + h2′  b′   impulse =  A = bh +  1      2  2   2   

=

1 2

 30 N + 40 N  (0.4 s − 0.2 s) +  40 N + 60 N  (1.0 s −  0.4 s)      2 2      

(0.2 s)(30 N) + 

impulse = 4.0 × 10 N ⋅ s [S] 1

The impulse imparted is 4.0 × 10  N⋅s [S]. 1

Making Connections 11. (a) Padded gloves increase the duration of time that the moving hand (punch) comes to rest. This causes a decrease in the size of the force applied to the head (and the hand), reducing fractures. (b) By “rolling with a punch,” a boxer increases the contact time, and reduces the applied force as the punch is applied.

Section 5.1 Questions (Page 238)

Understanding Concepts 1.

The net force on an object is the rate of change in momentum. ∆ p Σ F  = ∆t  Impulse is most useful when there is only a single net force acting on an object. 



2.

Σ F  = 24 N [E] ∆t  = 3.2 s Σ F ∆t  = ? 

3. (a)



Σ F ∆t  = (24 N)(3.2 s) Σ F ∆t  = 77 N ⋅ s [E] The impulse exerted is 77 N ⋅ s [E] . Σ F  = 1.2 × 102 N [forward] ∆t  = 9.1 ms Σ F ∆t  = ? Σ F ∆t  = (1.2 ×102 N)(9.1× 10−3 Σ F ∆t  = 1.1 N ⋅ s [forward] The impulse exerted is 1.1 N ⋅ s [forward] . 





(b)





s)



(c) m = 12 kg  g  = 9.8 m/s2 [down] 

∆t  = 3.0 s Σ F ∆t  = ? 

Σ F ∆t = mg ∆t  = (12 kg)(9.8 m/s2 )(3.0 s) Σ F ∆t  = 3.5 × 102 N ⋅ s [down] The impulse exerted is 3.5 × 102 N ⋅ s [down] . 



Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  303

(d) Solve for the area under the graph. Estimate the geometric shapes shown.

Σ F ∆t = 

=

A1 + A2 1 2

+ A3  1.0 N + 3.7 N  (0.020 s − 0.010 s) +  3.9 N + 5.0 N  (0.040 s  − 0.020 s)      2 2      

(0.010 s)(0.81 N) + 

Σ F ∆t = 0.12 N ⋅ s [S] 

The impulse exerted is 0.12 N ⋅ s [S] . 4.

msystem

= 41 kg + 21 kg = 62 kg

∆t  = 2.0 s Σ F  x = 75 N [W] vi x = 0 m/s vf  x = ? Σ F x ∆t = ∆px Σ F x ∆t = mvfx − mvix mvf x = ΣFx ∆t + mvix Σ F ∆t + mvix vf  x =  x = = vf  x

m Σ F x ∆t 

+ vi x m (75 N)(2.0 s)

( 41 kg + 21 kg )

+ 0 m/s

= 2.4 m/s [W]

The final velocity of the cart and the child will be 2.4 m/s [W]. 5.

m = 1.1× 103 kg

= 22 m/s vf  x = 0 m/s ∆t  = 1.5 s Σ F  x = ? vi x

304 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Σ F x ∆t = ∆px Σ F x ∆t = m(vfx − vix ) m( v − v ) Σ F  x = f x ix ∆t  (1.1×103 kg)(0 m/s − 22 m/s) = 1.5 s

Σ F  x = −1.6 × 10

4

N [E], or 1.6 × 104 N [W]

The average force required to stop the car is 1.6 × 104 N [W] . 6. (a) m = 0.17 kg

= 2.1 m/s [right] vf  x = 1.8 m/s [left] ∆ p x = ? vi x

Choose right as the positive direction. ∆ p x = m(vfx − vix )

(b)

= (0.17 kg)(−1.8 m/s − 2.1 m/s) = −0.66 kg ⋅ m/s [right] ∆ p x = 0.66 kg ⋅ m/s [left] The change in momentum of the ball is 0.66 kg ⋅ m/s [left] . Σ F x ∆t  = ? The impulse is equal to the change in momentum, but the units are written in a different form. Σ F x ∆t = m(vfx − vix )

7. (a)

(b)

= (0.17 kg)(−1.8 m/s − 2.1 m/s) = −0.66 N ⋅ s [right] Σ F x ∆t  = 0.66 N ⋅ s [left] The impulse given to the ball by the cushion is 0.66 N ⋅ s [left] . m = 0.16 kg vf  x = 11 m/s vi x = 18 m/s ∆ p x = ? ∆ p x = m(vfx − vix ) = (0.16 kg)(11 m/s − 18 m/s) = −1.1 kg ⋅ m/s [forward] ∆ p x = 1.1 kg ⋅ m/s [backward] The change in momentum of the puck is 1.1 kg ⋅ m/s [backward] . Σ F x ∆t  = ? Σ F x ∆t = m(vfx − vix ) = (0.16 kg)(11 m/s − 18 m/s) = −1.1 N ⋅ s [forward] Σ F x ∆t  = 1.1 N ⋅ s [backward] The impulse exerted by the snow on the puck is 1.1 N ⋅ s [backward] .

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  305

(c)

∆t  = 2.5 s  F K  = ? Σ F x ∆t = m( vfx − vix ) m( vf x − vix )  F K  = ∆t  =

(0.16 kg)(11 m/s − 18 m/s)

2.5 s = −0.45 N [forward]  F K  = 0.45 N [backward] 8.

The average frictional force exerted by the snow on the puck is 0.45 N [backward]. m = 0.50 kg

∆t  = 1.5 s Σ F  x = 1.4 N [W] vi x = 2.4 m/s vf  x = ? Choose east as the positive direction. Σ F x ∆t = ∆px

Σ F x ∆t = mvfx − mvix mvf x = ΣFx ∆t + mvix Σ F ∆t + mvix vf  x =  x = = vf  x 9.

m Σ F x ∆t 

+ vi x m ( −1.4 N)(1.5 s) (0.50 kg)

+ 2.4 m/s

= −1.8 m/s [E] = 1.8 m/s [W]

The final velocity of the disk is 1.8 m/s [W]. m = 2.0 kg

∆t  = 0.50 s Σ F  x = 6.0 N [N] vf  x = 4.5 m/s vi x = ? Choose north as the positive direction. Σ F x ∆t = ∆px

Σ F x ∆t = mvfx − mvix mvi x = mvf x − Σ Fx ∆t   mvf  x − ΣFx ∆t   vi x = = vf  x −

m Σ F x ∆t  m

= 4.5 m/s − vi x

(6.0 N)(0.50 s) (2.0 kg)

= 3.0 m/s [N]

The initial velocity of the skateboard is 3.0 m/s [N].

306 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

10. (a) m = 0.61 kg

= 9.6 m/s [down] vf  x = 8.5 m/s [up] ∆ p x = ? vi x

Choose up as positive. ∆ p x

(b)

= m(vfx − vix ) = (0.61 kg)(8.5 m/s − (−9.6 m/s)) ∆ p x = 11 kg ⋅ m/s [up] The change in momentum of the basketball is 11 kg ⋅ m/s [up] . ∆t  = 6.5 ms = 6.5 × 10−3 s  F floor  = ? Choose up as positive. Σ F x ∆t

= m(vfx − vix ) m(vf x − vix )  F floor  = ∆t  =

(0.61 kg)(8.5 m/s − (− 9.6 m/s)) 6.5 × 10−3 s

 F floor  = 1.7 × 103 N [up] The average force exerted on the basketball by the floor is 1.7 × 103 N [up] . 11. The force of gravity can be ignored in question 10 because it is so small compared to the force exerted by the floor (approximately 6.0 N compared with 1700 N).

Applying Inquiry Skills 12. By using a high-speed digital camera, take rapid photos during the swing. For each athlete, compare the distance that the racket is in contact with the ball. The longer the distance, the greater the follow-through.

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  307

Making Connections 13. The change in momentum when you land is the same for either type of landing. When you bend your knees, the time that the force is applied is longer, so the required force is less. This smaller force is less likely to cause pain. 14. (a) The analyst could determine the initial speed of the bumper component from its skid marks. To estimate the initial speeds of each vehicle, you would need to know the mass of each vehicle and how far each one skidded after the collision. Using this information, you can also determine approximate coefficients of friction and solve for speeds immediately after the collision. This would provide enough information to calculate the change in momenta of each vehicle. (b) Answers will vary.

5.2 CONSERVATION OF MOMENTUM IN ONE DIMENSION PRACTICE (Pages 243–244)

Understanding Concepts 1. 2.

The net force on the system must be zero for momentum to be conserved. For the centre of mass of the system, the statement is equivalent to Newton’s first law. With no net force acting on the system, the centre of mass of the system will not experience any change in velocity. However, the individual parts of the system will undergo various changes in speed and direction. 3. (a) Earth will exert a downward force on the hairbrush. (b) The hairbrush will exert an upward force on Earth. (c) The forces in (a) and (b) are the same in magnitude. (d) The net force of the system containing Earth and the hairbrush is zero. (e) Momentum of this system will be conserved. (f) Earth will move up as the hairbrush falls down. (g) Choose up as positive. m1 = 0.0598 kg

= 5.98 × 10 24 kg v1 y = 10 m/s [down] v2 y = ? m2

= p′y 0 = m1v1 y + m2 v2 y

 p y

v2 y

=− =−

m1v1 y m2 (0.0598 kg)(−10 m/s) 5.98 × 1024 kg

= 1× 10−25 m/s [up] Earth’s speed at this time is 1× 10−25 m/s [up] . m1 = 45 kg m2 = 33 kg v1 x = 1.9 m/s [E] v2 x = ? v2 y

4.

308 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Choose east as positive.  p x = p ′x

0 = m1v1 x + m2v2 x v2 x = −

=−

m1v1 x m2

(45 kg)(1.9 m/s) 33 kg

= −2.6 m/s [E] v2 x =

5.

2.6 m/s [W]

The velocity of the raft relative to the water is 2.6 m/s [W]. m1 = 56.9 kg v1 x =

3.28 m/s

v2 x = −3.69 m2 =

m/s

?  p x = p ′x

0 = m1v1 x + m2 v2 x m2 = −

=− m2 =

6.

m1v1 x v2 x

(56.9 kg)(3.28 m/s) −3.69 m/s

50.6 kg

The mass of the other skater is 50.6 kg. m1 = 11 kg m2 =

24 kg

v1 x =

95 m/s

v2 x =

?  p x = p x′

0 = m1v1 x + m2 v2 x v2 x = −

=−

m1v1 x m2

(11 kg)(95 m/s) 24 kg

v2 x = −44

m/s

The speed of the 24-kg piece is 44 m/s in the opposite direction. 7.

4

kg

4

kg

m1 = 1.37 × 10

m2 = 1.12 × 10 v1 =

20.0 km/h [N]

v1′ = v2′ = v ′ = 18.3 v2 =

m/s

(when coupled)

?

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  309

Since v1′ = v2′ = v ′, m1v1 + m2 v2 = m1v1′ + m2 v2′ m1v1 + m2 v2 = (m1 + m2 )v ′ m2 v2 = (m1 + m2 )v ′ − m1v1 v2 =

(m1 + m2 ) v′ − m1v1 m2 4

=

4

4

(1.37 × 10 kg + 1.12 ×10 kg )(18.3 km/h ) − (1.37 × 10 kg)( 20.0 km/h ) 4

1.12 ×10 kg v2 = 16.2

8.

km/h [N]

The initial velocity of the second car is 16.2 km/h [N]. m1 = 0.15 kg m2 =

0.045 kg

v1 =

56 m/s

v2 =

0

v2′ =

67 m/s

v1′  =

? m1v1 + m2 v2 = m1v1′ + m2 v2′ m1v1 + 0 = m1v1′ + m2 v2′ m1v1′ = m1v1 − m2v 2′ m1v1 − m2 v2′

v1′  =

m1

= v1 −

m2 v2′ m1

= 56 m/s − v1′  =

(0.045 kg)(67 m/s) 0.15 kg

36 m/s

The head of the driver has a speed of 36 m/s immediately after the collision.

Applying Inquiry Skills 9. (a) Cart A moves toward the stationary cart B. This causes the two carts to collide, imparting speed to cart B. There is no data to tell what happens to cart A during the last part of the collision and after. (b) Cart B has no speed, so the system momentum will be equal to the momentum of cart A.  psystem = pA

= mA vA = (0.40 kg)(0.60 m/s)  psystem =

0.24 kg ⋅ m/s [E]

The momentum of the system of carts before the collision is 0.24 kg ⋅ m/s [E] . (c)

mA =

0.40 kg

mB =

0.80 kg

vA =

0.60 m/s [E]

vB =

0

′ = vB

0.35 m/s [E]

′ = vA

?

310 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Choosing east as positive.  p = p ′

′ + mBvB ′ mA vA + mBvB = mA vA ′ + mB vB ′ mA vA + 0 = mA vA ′ = mA vA − mB vB ′ mA vA ′ = vA

′ mA vA − mBvB

= vA −

mA

′ mB vB mA

= 0.60 m/s − ′ = −0.10 vA

(0.80 kg)(0.35 m/s) 0.40 kg

m/s [E], or 0.10 m/s [W]

The velocity of cart A after the collision is 0.10 m/s [W]. (d)

Making Connections 10. Assuming the astronaut has something to throw, he could throw it away from the spaceship. This would cause him to move toward the ship.

Section 5.2 Questions (Pages 244–245)

Understanding Concepts 1. (a) F The forces are the same (Newton’s third law). (b) F The magnitude of the changes in momenta will be equal. (c) T (d) F The magnitude of the changes in momenta will be equal. 2. Yes. An example would be two dynamics carts on a track that have had a spring bumper explosion. 3. (a) Momentum is conserved. There is no net external force of the system of objects. (b) Momentum is conserved. There is no net external force of the system of objects. (c) Momentum is not conserved. There is an external force applied by the stove on the pan. 4. It is not possible to react against the vacuum of space. A rocket ship works by pushing exhaust gasses out the back, and the exhaust gases push back on the rocket ship. There is not involvement of the air (or not).

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  311

5.

m1 =

57 kg

m2 =

27 kg

v1 = v2 = v =

3.2 m/s [forward] (when together)

v1′  =

3.8 m/s [forward]

v2′ =

?

Choose forward as positive. m1v1 + m2 v2 = m1v1′ + m2v 2′

(m1 + m2 ) v = m1v1′ + m2v2′ m2 v2′ = ( m1 + m2 )v − m1v1′ v2′ =

=

(m1 + m2 )v − m1v1′ m2

(57 kg + 27 kg)(3.2 m/s) − (57 kg)(3.8 m/s) 27 kg

v2′ = 1.9

6.

m/s [forward]

The final velocity of the cart is 1.9 m/s [forward]. m1 = 65 kg + 35 kg = 100 kg (treating the hiker and raft as a single object) m2 = 19 v1 =

0

v2 =

0

v1′  = 1.1 v2′ =

kg

m/s [S]

?

Choose south as positive.  p = p ′ m1v1 + m2 v2 = m1v1′ + m2 v2′

0 = m1v1′ + m2 v2′ v2′ = −

=−

m1v1′ m2

(100 kg)(1.1 m/s) 19 kg

v2′ = −5.8

m/s [S], or 5.8 m/s [N]

The hiker threw the backpack with a velocity of 5.8 m/s [N] relative to the water. 7.

3

kg

3

kg

m1 = 1.13 ×10

m2 = 1.25 × 10 v1 =

25.7 m/s [E]

v2 = 13.8

m/s [W]

v1′ = v2′ = v ′ =

?

(when coupled)

Choose east as positive. m1v1 + m2 v2 = m1v1′ + m2v 2′ m1v1 + m2 v2 = (m1 + m2 )v ′ v ′ =

=

m1v1 + m2 v2 m1 + m2

(1.13 × 103 kg)(25.7 m/s) + (1.25 × 103 kg )(−13.8 m/s) 1.13 ×103 kg + 1.25 × 10 3 kg

v ′  =

4.95 m/s [E] The velocity of the system after the collision is 4.95 m/s [E]. 312 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

8. (a) For the first vehicle: 3

m1 = 1.13 × 10 v1 =

kg

25.7 m/s [E]

v1′ = v2′ = v ′ =

4.95 m/s [E] (when coupled)

∆ p = ? 

∆ p = p2 − p1 

= m1v ′ − m1v1 = m1 (v ′ − v1 ) 3

= (1.13 × 10 kg)(4.95 m/s − 25.7 m/s) 4

∆ p = −2.34 × 10 kg ⋅ m/s [E] 

For the second vehicle: m2 = 1.25 × 10 v2 = 13.8

3

kg

m/s [W]

v1′ = v2′ = v ′ =

4.95 m/s [E] (when coupled)

∆ p = ? 

∆ p = p2 − p1 

= m2 v ′ − m2 v2 = m2 (v ′ − v2 ) 3

= (1.25 × 10 kg )( 4.95 m/s − (−13.8 m/s)) 4

∆ p = 2.34 ×10 kg ⋅ m/s [E] 

(b) These two quantities are equal in magnitude and opposite in direction. (c) The total change in momentum of the two-automobile system is zero. 9. mL = 89 kg vQ =

0

vL =

5.2 m/s

′ = vL′ = v ′ = vQ mQ =

2.7 m/s

? ′ + mL vL′ mQ vQ + mL vL = mQ vQ

0 + mL vL = mQv ′ + mLv ′ mQ v ′ = mL vL − mL v ′ mQ =

=

mL (vL − v ′) v′

(89 kg)(5.2 m/s − 2.7 m/s) 2.7 m/s

mQ =

82 kg

The mass of the quarterback is 82 kg. 10. Choose original direction of 0.25-kg ball as positive. m1 = 0.18 kg m2 =

0.25 kg

v1 = −2.5 v2 = 1.7

m/s

m/s

v2′ = −0.10 v1′ =

m/s

?

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  313

m1v1 + m2 v2 = m1v1′ + m2 v2′ m1v1′ = m1v1 + m2 v 2 − m2v 2′ m1v1 + m2 (v2 − v2′ )

v1′  =

m1

(0.18 kg)(−2.5 m/s) + (0.25 kg)(1.7 m/s − (− 0.10 m/s))

=

0.18 kg v1′  =

0 m/s

The velocity of the 0.18-kg ball after the collision is 0 m/s.

Applying Inquiry Skills 11. Set up the carts as shown in the text on a track. Attach a spark timing recorder to each cart to record location for each time interval. Turn on the timers, and release the carts. Analyze each tape close to the collision to determine the speed of each cart. Mass each cart and calculate the momentum of each cart after the collision. Check to see if each one adds up to zero (as it should). 12.(a) Since the car and SUV came to an immediate halt at the location of the crash, the t otal momentum of the system was zero, before and after the collision. That would mean that the total momentum before and after would have the same magnitude. (but mS = 2mC ) mC vC = mSvS mC vC =

(2 mC )vS

mC vC =

2mC vS

vC =

2vS

This data indicates the car was travelling at twice the speed of the SUV. (b) If both drivers had the same speed, there would have been momentum after the collision in the direction of the original motion of the SUV.

5.3 ELASTIC AND INELASTIC COLLISIONS Try This Activity: Newton’s Cradle (Page 248)

(a) Each collision is successive. Assuming an elastic collision, (but v 2 = 0 and all masses are equal) m1v1 + m2v2 = m1v1′ + m2 v 2′ mv1 + 0 = mv1′ + mv2′ v1 = v1′ + v2′ v2′ = v1 − v1′

(Equation 1)

and 2 1 mv 2 1 1

+

2 1 m2 v2 2

=

2 1 m v′ 2 1 1

2

2

+ 12 m2v2′

2

(but v2 = 0 and all masses are equal)

2

mv1 + 0 = mv1′ + mv2′ 2

2

2

v1 = v1′ + v2′ 2

2

v2′ = v1 − v1′

2

(Equation 2)

Substitute Equation 1 into Equation 2: (v1 − v1′ ) 2 = v12 − v1′2 2

v1 − 2v1v1′ + v1′

2

2 2 = v1 − v1′

2/ v1′2 − 2/ v1v1′ = 0 v1′ ( v1′ − v1 ) =

0

v1′ =

0

or

v1′ − v1 =

0

If v1′ − v1 = 0, then v1′ = v1 (i.e., the speed of the ball after the collision is unchanged). This is not possible, so v1′ = 0. The final speed of the first ball is zero after the collision. 314 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Substituting back into Equation 1 gives: v2′ = v1 − v1′

= v1 − 0 v2′ = v1 Each sphere collision will leave the previous one stationary as the next one moves on. Only one sphere can move out from the end. (b) Similarly, only one sphere can end up with all of the kinetic energy. (c) According to the above calculations, each sphere will “pass” its momentum onto the next. When the last sphere receives the momentum, and therefore, speed, it will rise to the same vertical height as the original sphere was dropped. (d) If two spheres are used, then two spheres will move out from the other side. If three spheres are used, then three spheres will move out from the other side.

PRACTICE (Page 248)

Understanding Concepts 1. 2.

All of the original kinetic energy is transformed to other forms if both objects come to rest after the collision. Yes, we can conclude the collision is completely inelastic. After a “hit-and-stick” collision, no energy is stored as elastic  potential to be returned to either object at the end of the collision. 3. A head on collision is very dangerous because of the high relative velocity between the vehicles and the large (and rapid) change in speed for each one. These large accelerations produce large forces that are capable of inflicting serious damage on a human body.

Applying Inquiry Skills 4.

If the ball tends to return to shape quickly when squeezed, it will have a more elastic collision than one that returns to its original shape more slowly. 5. (a)

(b)

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  315

Making Connections 6. (a) The impact force is reduced by the soft interior because it takes the same impulse, and causes the interaction to last longer. The larger time interval of collision means that a smaller force is applied. A hard interior would have a short duration of collision, and a higher force. (b) If a helmet does not fit properly, the force applied to the head is not evenly distributed to the whole head. The smaller distribution area means the force would be applied to a smaller area, increasing the pressure to that portion of the head. (c) Once a helmet is involved in a collision, it should be replaced. One way a helmet is designed to reduce injury is to absorb some of the impact by breaking. This is only able to help the wearer one time. In future collisions, the safety of the wearer could be compromised. 7. Some possible answers: Flexible • absorb energy • reduce deceleration of train (and passengers) Rigid •  prevent car from collapsing and injuring passengers located at either end

• less likely to have debris flying about the interior (due to crumpling car) PRACTICE (Pages 251–252)

Understanding Concepts 8.

The larger truck would have the larger momentum.  E K  =

= =  E K  =

mv 2 2 2 2

m v

2m (mv )2 2m  p 2 2m

 p =  psmall

Since  EK small

2mE K 

=

2msmall EK small

=

2msmall EK

and

plarge

=

2 mlarge E K large

= E K large ,  psmall

and

plarge

=

2 mlarge E K

 

The mass of the larger vehicle is larger, and the momentum will be too. 9. (a) All objects have mass. If it also has a velocity, then it will have both momentum and kinetic energy. (b) For an isolated system of two objects, it is possible to have a momentum of zero. One example would be two carts released from an internal spring “explosion”. The total momentum is zero, but both of the individual parts have kinetic energy. If the system of objects has momentum, then at least one of them is moving, and there will be kinetic energy. It is not possible to have momentum and zero kinetic energy at the same time. 10. m1 = 0.15 kg

= 0.15 kg v1 = 22 m/s [N] v2 = 22 m/s [S] v1′ = v2′ = v′ = ? m2

(when coupled)

316 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Choose north as positive. m1v1 + m2 v2

= m1v1′ + m2 v2′ m1v1 + m2 v2 = ( m1 + m2 ) v′ m v + m2 v2 v ′ = 1 1 m1 + m2 =

(0.15 kg)(22 m/s) + (0.15 kg)(− 22 m/s) 0.15 kg + 0.15 m/s

v ′ = 0 m/s The velocity of the two-ball system after the collision is 0 m/s. 11. Conservation of Momentum m1v1 + m2 v2 = m1v1′ + m2 v2′ (but v2 = 0 and the masses are equal) mv1 + 0 = mv1′ + mv2′

= v1′ + v2′ v2′ = v1 − v1′ v1

(Equation 1)

Conservation of Energy 1 2

2

m1v1

+ 12 m2 v22 = 12 m1v1′2 + 12 m2 v2′2 (but v2 = 0 and the masses are equal) mv12 + 0 = mv1′2 + mv2′2 v12 = v1′2 + v2′2 2 2 2 v2′ = v1 − v1′ (Equation 2)

Substitute Equation 1 into Equation 2: (v1 − v1′ ) 2

= v12 − v1′2 v12 − 2v1v1′ + v1′2 = v12 − v1′2 2/ v1′2 − 2/ v1v1′ = 0 v1′ (v1′ − v1 ) = 0 v1′ = 0 or v1′ − v1 = 0 If v1′ − v1

= 0, then v1′ = v1

(i.e., the speed of the proton after the collision is unchanged). This is not possible, so v1′ = 0.

The final speed of the first proton is zero after the collision. Substituting back into Equation 1 gives: v2′ = v1 − v1′

= v1 − 0 = v1 v2′ = 815 m/s The final velocity of the second proton after the collision is 815 m/s in the direction of the initial velocity. 12. For an inelastic collision, the vehicles stick together.

= 1.3 × 104 kg m2 = 1.1× 103 kg v1 = 90 km/h [N] v2 = 30 km/h [N] v1′ = v2′ = v′ = ? (when coupled) m1

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  317

Choose north as positive.

= m1v1′ + m2 v2′ when coupled, v1′ = v2′ = v′ m1v1 + m2 v2 = ( m1 + m2 ) v′ m v + m2 v2 v′ = 1 1 m1 + m2 4 3 (1.3 × 10 kg )(90 km/h ) + (1.1× 10 kg )(30 km/h ) = 4 3 1.3 × 10 kg + 1.1× 10 kg v′ = 85 km/h [N] m1v1 + m2 v2

The velocity of the vehicles is 85 km/h [N] just after the collision.

= 1.3 ×104 kg mc = 1.1× 103 kg vt = 90 km/h [N] = 25 m/s [N] vc = 30 km/h [N] = 8.333 m/s [N]  ET = EK truck + E K car

13. mt

= =  E T

1 2 1 2

2

mt vt

1

+

2

  2

mc vc

(1.3 × 104 kg)(25 m/s)2 +

= 4.1× 10

6

1 2

(1.1× 103 kg)(8.333 m/s)2

J

The final velocity of the vehicles after the collision is 85.319 km/h [N] = 23.7 m/s [N].  ET′ = EK′ truck + E K′ car  

= =  E T′

1 2 1 2

mt vt′

2

+

1 2

mc vc′

2

(1.3 × 104 kg)(23.7 m/s) 2

= 4.0 × 10

6

+

1 2

(1.1× 103 kg)(23.7 m/s)2

J

The decrease in kinetic energy is 4.1 × 10  – 4.1 × 10  = 1 × 10  J. 14. Choose the original direction of motion as the positive direction. 6

6

5

= 5.31× 10−26 kg m N = 4.65 × 10 −26 kg vO = 0 vO′ = 4.81×10 2 m/s ′ = −34.1 m/s v N v N = ? mO

+ mO vO = mN vN′ + mO vO′ m N vN + 0 = mN vN′ + mO vO′ m v′ + mO vO′ v N =  N N

m N vN

m N

= v N′ +

mO vO′ m N

= −34.1 m/s + v N

= 5.15 × 102

(5.31× 10

−26

kg)(4.81× 10 m/s)

4.65 × 10

2

−26

kg

m/s

The initial speed of the nitrogen molecule was 5.15 × 102 m/s .

318 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Applying Inquiry Skills 15. (a) Line A represents the total momentum because the total momentum in the system is constant. (b) The collision is an inelastic one. If the collision were elastic, the total kinetic energy before and after would have been the same.

Making Connections 16. (a) The rubber bullet would have the elastic collision and the lead bullet would have the inelastic collision. ∆ pT = −∆pR  (b) 



mT vT2 − mT vT1 = −( mR vR2 − mR vR1 ) 







vT1 

= 0, and vR2 = −vR1 



mT vT2 − 0 = mR vR1 − mR ( −v R1 ) 



= mR vR2 + mR vR1 vT2 = mR vR 2 + mR vR 1  pT′ = pR′ + pR 

mT vT2 

mT















=

vR2 























∆ pT = −∆pL mT vT2 − mT vT1 = − mL vL2 + mL v L1 mT vT2 − 0 = mL vL2 − mL vL2 mT vT2 = mL vL1 − mL vL1  pT′ = pL − pL′ 

vR1











vT1 = 0





The rubber bullet transfers more momentum to the target. (c) Rubbers bullets are preferred in crowd control because they are less likely to kill or permanently injure any of the crowd and they impart a larger backward impulse on the crowd.

Section 5.3 Questions (Page 253)

Understanding Concepts 1. (a) It is not possible for both objects to be at rest. If they were both at rest, the initial momentum of the first object would have violated the law of conservation of t momentum. (b) It is possible for the first object to be at rest after the collision. One example is a curling stone that strikes another and then stops moving. 2. The does not violate the law of conservation of momentum for the system which contains the earth and the snowball. The earth exerts a net external force on the tree/snowball system. This external force negates the conservation of momentum for that system. 3. There momentums will only be the same if they have the same mass. The relationship between momentum and kinetic energy is  E K  =

= =  E K  =  p =

mv 2 2 m2 v 2 2m (mv )2 2m  p 2 2m 2mE K 

When two objects have the same kinetic energy, the object with the larger mass will always have a larger momentum.

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  319

4.

Conservation of Momentum m1v1 + m2 v2

= m1v1′ + m2 v2′ (but v2 = 0) m1v1 + 0 = m1v1′ + m2 v2′ m1v1 − m1v1′ = m2 v2′ m1 ( v1 − v1′ ) = m2 v2′ (Equation 1)

Conservation of Energy 1 2

+ 12 m2 v22 = 12 m1v1′2 + 12 m2 v2′2 (but v2 = 0) m1v12 + 0 = m1v1′2 + m2 v2′2 m1v12 − m1v1′2 = m2 v2′2 m1 (v12 − v1′2 ) = m2 v2′2 (Equation 2) 2

m1v1

Divide Equation 2 by Equation 1:

− v1′2 ) m2 v2′2 = m1 ( v1 − v1′ ) m2 v2′ (v1 + v1′ )(v1 − v1′ ) = v2′ (v1 − v1′ ) v2′ = v1 + v1′ m1 ( v12

(Equation 3)

or

v1′

= v2′ − v1

(Equation 4)

Substitute Equations 3 and 4 back into the original conservation of momentum equation: m1v1 + 0 = m1v1′ + m2 v2′

= m1v1′ + m2 ( v1 + v1′ ) m1v1 = m1v1′ + m2 v1 + m2 v1′ m1v1′ + m2 v1′ = m1v1 − m2 v1 v1′ ( m1 + m2 ) = ( m1 − m2 ) v1  m − m2   v1′ =  1  v1  m1 + m2    0.022 kg − 0.027 kg   =   (3.5 m/s)  0.022 kg + 0.027 kg   v1′ = − 0.36 m/s [forward], or  0.36 m/s [backward] m1v1

The velocity of the 22-g superball after the collision is 0.36 m/s [backward] . m1v1 + 0 = m1v1′ + m2 v2′

= m1 (v2′ − v1 ) + m2 v2′ m1v1 = m1v2′ − m1v1 + m2 v2′ m1v2′ + m2 v2′ = 2 m1v1 v2′ (m1 + m2 ) = 2m1v1  2m1   v2′ =   v1  m1 + m2      2(0.022 kg) =  (3.5 m/s)    0.022 kg + 0.027 kg   v2′ = 3.1 m/s [forward] m1v1

The velocity of the 27-g superball after the collision is 3.1 m/s [forward] .

320 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

5.

Using the equations derived in question 4 ,

 m1 − m2    v1  m1 + m2    m − m2   1 v1 =   v1 3  m + m2   m + m2 = 3(m − m2 ) m + m2 = 3m − 3 m2 4m2 = 2m v1′ = 

m2 6.

=

1 2

m

m = 66 kg  g  = 9.8 m/s 2

∆ y = 25 m v1 = ? Use conservation of energy to solve for the speed of the moving skier at the bottom of the hill.  ET = E T′ mg ∆y

=

v1

=

1 2

2

mv1

2 g ∆y

= 2(9.8 m/s2 )(25 m) v1 = 22.14 m/s = 72 kg v1 = 22.14 m/s v2 = 0 v1′ = v2′ = v′ = ? m2

(when coupled)

= m1v1′ + m2 v2′ v2 = 0, and when coupled, v1′ = v2′ = v′ m1v1 + 0 = ( m1 + m2 ) v′

m1v1 + m2 v2

v ′ =

=

m1v1 m1 + m2 (66 kg)(22.14 m/s) 66 kg + 72 kg

v ′ = 11 m/s The speed of the two-skier system immediately after the collision is 11 m/s.

Applying Inquiry Skills 7. (a) The collision takes place very quickly. The duration of time that the bullet and block are interacting, the strings are vertical, and only balance the gravitational forces. During the collision (not during the swing), momentum is conserved. (b) Let V  be the speed of the bullet and block combination after the collision. m1v1 + m2 v2 = m1v1′ + m2 v2′   (but v2 = 0 and when coupled, v1′ = v2′ = V )   mv + 0 = ( m + M )V   V  =

mv

m + M  (c) The law of conservation of energy can be used to relate the maximum vertical height to the speed just after collision.

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  321

(d)

 ET

= E T′

mg ∆y

=

 gh = h=

1 2 1 2

mv 2 V 2

V 2 2 g  2

 mv    m + M      = 2 g 

h= (e)

h= m2v2

2 2

m v

2 g ( m + M ) 2 m2v2 2 g ( m + M ) 2

= 2 gh(m + M )  2

v=

2 gh( m + M ) m

2

2 2

 m + M       m    m + M    v = 2 gh     m   m = 87 g = 0.0087 kg  M  = 5.212 kg 2  g  = 9.8 m/s h = 6.2 cm = 0.062 m v=? =

(f)

2 gh 

v=

=

 m + M       m  

2 gh 

 0.0087 kg + 5.212 kg     0.0087 kg   

2

2(9.8 m/s )(0.062 m) 

v = 6.6 × 10 m/s 2

The initial speed of the bullet was 6.6 × 10 m/s . (g) Some of the sources of error would be friction of the moving parts of the gun and loss of energy to thermal energy in the spring. The frictional draw of the catch mechanism would use some of the energy that should be converted into gravitational potential. 8. There are a number of advantages of a crumple zone. One is that it converts a significant part of the kinetic energy of the vehicle into thermal energy as the steel become permanently deformed. The crumple zone also causes the vehicle to slow down over a greater distance. This greater distance increases the length of time the vehicle is slowing down, and the average force to stop the vehicle (and its passengers) decreases. 9. Most meteorites burn up in the atmosphere. Larges known one is about 60 metric tons, and the next known one is about 30 metric tons. Large collisions have rarely (ever 300 million years or so) and can cause devastating climactic change. 2

322 Unit 2 Energy and Momentum

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5.4 CONSERVATION OF MOMENTUM IN TWO DIMENSIONS PRACTICE (Pages 257–258)

Understanding Concepts 1.

= 52 kg mc = 26 kg vs = 1.2 m/s [W] vc = 1.2 m/s [S] vs′  = ? vc′  = ?

2. (a) ms 







Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  323

Calculate the momenta:  ps = ms vs 



 ps 

 ps





 pc

= (52 kg)(1.2 m/s) = 62.4 kg ⋅ m/s [S]



 pc

= mc vc = (26 kg)(1.2 m/s) = 31.2 kg ⋅ m/s [S] 

+ pc = 93.6 kg ⋅ m/s [S] 

 ps′

= ms vs′ = (52 kg)(1.0 m/s)  ps′  = 52 kg ⋅ m/s [W] 





Measure length and angle to get:  pc′ 

= 1.1× 102

 pc′

kg ⋅ m/s [61° S of E]

= mc vc′  p ′ vc′  = c









mc

=

1.1 × 10 2 kg ⋅ m/s 26 kg

vc′  = 4.1 m/s [61° S of E] 

The approximate final velocity of the cart is 4.1 m/s [61° S of E]. (b) Using the diagram from part (a), the total momentum after is  pc′ 

=

ps′ 

2

+



ps

+ pc 2 

= (52 kg ⋅ m/s)2 + (93.6 kg ⋅ m/s)2 2  pc′ = 1.1× 10 kg ⋅ m/s 

+ pc tan θ  =  ps′ −1   ps + pc θ  = tan    ps′  93.6   = tan −1     52   θ  = 61° kg ⋅ m/s [61° S of E]. 



 ps









So,  pc′ 

3.

= 1.1×102

       

Diagram is not to scale. Dotted line represents the direction “after.” A completely inelastic collision means they will stick together.

= 1.4 × 103 kg 3 m2 = 1.3 × 10 kg v1 = 45 km/h [S] = 12.5 m/s [S] v2 = 39 km/h [E] = 10.83 m/s [E] ′ =? v12 m1 



324 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

= m1 v1 = (1.4 ×103 kg)(12.5 m/s) = 1.75 × 10 4 kg ⋅ m/s





 p1



 p1

 p1

= m2 v2 = (1.3 ×103 kg)(10.83 m/s) = 1.408 ×10 4 kg ⋅ m/s

′  p12

=





 p2







p1

2

+



p2

2

= (1.75 ×104 kg ⋅ m/s)2 + (1.408 × 104 ′ = 2.246 ×104 kg ⋅ m/s  p12

kg ⋅ m/s) 2



′  p12

= m12 v12′  p ′ ′ = 12 v12





m12

= ′ v12

2.246 ×104 kg ⋅ m/s 1.3 × 10 kg + 1.4 × 10 kg 3

3

= 8.3 m/s 

tan θ  =

 p1 

 p2 −1   p1

  = tan        p2    1.75 ×104 kg ⋅ m/s   = tan −1    4  1.408 ×10 kg ⋅ m/s   θ  = 51° 

θ 



The final speed of the cars is 8.3 m/s [51º S of E], or 3.0 × 10 1 km/h [51º S of E].

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  325

4. (a) Diagram not to scale.

(b)

0 = mvB y 0 = vB y

− mvAy

− vAy

= vAy vB sin 48.9° = vA sin 31.1° sin31.1° vB = vA sin 48.9° vB y

= mvA x + mvBx vA1 = vA x + vBx 2.25 = vA cos 31.1° + vB cos 48.9°  sin31.1° v  cos 48.9° 2.25 = vA cos 31.1 ° +  A    sin 48.9°   2.25 = 1.307vA vA = 1.72 m/s

mvA1

vB

= =

sin31.1° sin 48.9° sin31.1°

sin 48.9° vB = 1.18 m/s

vA

(1.72 m/s)

The velocities of the balls are 1.18 m/s at 48.9 ° and 1.72 m/s at 31.1 °.

326 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

(c)

′  EKT

(d)  EKT

 EKT

2 = 12 mvA1 = 12 m(2.25) 2 = 12 m(5.06)

= 12 mvA2 + 12 mvB2 = 12 m(vA2 + vB2 ) = 12 m(1.722 + 1.182 ) ′ = 12 m(4.35)  EKT

The total kinetic energy after is less than the total kinetic energy before, so the collision is not elastic. 5. (a) The residual nucleus will move in the opposite direction of the combined momentums of the two particles so the total momentum is still zero. (Note: Diagram is not to scale.)

= 4.8 × 10−21 kg ⋅ m/s [S] −21 kg ⋅ m/s [E]  pe = 9.0 × 10 θ  = ? 

 pn 

tan θ  =

 pn  pe

  p   = tan −1  n     pe    4.8 × 10−21 kg ⋅ m/s   = tan −1    −21  9.0 × 10 kg ⋅ m/s   θ  = 28° θ 

The nucleus will move in the direction 28º N of W. (b)  pnucleus = ?  pnucleus

=

2

pn

+ pe2

= (4.8 × 10−21 kg ⋅ m/s) 2 + (9.0 × 10−21 kg ⋅ m/s) 2 −20  pnucleus = 1.0 × 10 kg ⋅ m/s The magnitude of the nucleus’s momentum is 1.0 ×10−20 kg ⋅ m/s .

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  327

= 3.6 ×10−25 vnucleus = ?

(c) mnucleus

kg



 pnucleus

= mnucleusvnucleus

vnucleus

= =

 pnucleus mnucleus

1.0 ×10 −20 kg ⋅ m/s 3.6 ×10 −25 kg

= 2.8 × 104 4 vnucleus = 2.8 × 10 vnucleus 

m/s m/s [28° N of W]

The recoil velocity of the nucleus is 2.8 ×104 m/s [28° N of W] .

Applying Inquiry Skills 6. (a) The black car was travelling faster at the moment of impact. There was a larger component of momentum in the original direction of the motion of the black car because there is less deviation in its path than the path of the white car. (b) The investigator could determine more precise information by measuring the angle of deviation for the cars and the total distance that the combined cars slid.

Making Connections 7.

Answers will vary depending on the type of equipment chosen. Most likely there will be some padding included in it that will reduce the force required by increasing the time of collision.

Section 5.4 Questions (Pages 258–239)

Understanding Concepts 1.

328 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

2.

= 1.2 ×10−26 kg ′ = 0.40 km/s [54° to original direction of motion of neutron] vLi vn′ = 2.5 km/s ′ =?  pLi  pn′ = ? θ  = ? mLi

Diagram is not to scale.

′  pLi

= mLi vLi′ = (1.2 × 10−26 kg)(0.40 km/s) ′ = 4.8 × 10−27 kg ⋅ km/s  pLi  pn′

= mn vn′ = (1.7 × 10−27 kg)( 2.5 km/s) −27  pn′ = 4.25 × 10 kg ⋅ km/s In the  y-direction:

= p ′y ′  y − pn′ y 0 =  pLi ′ sin 54° − pn′ sin θ  0 =  pLi  p ′ sin54° sin θ  = Li  pn′ ′ sin54°   −1   pLi θ  = sin      pn′    (4.8 ×10−27 kg ⋅ km/s)(sin 54°)   = sin −1    4.25 × 10 −27 kg ⋅ km/s    θ  = 66° [from original direction of motion of the neutron] The neutron is now travelling at 66° from its original direction of motion.  p y

3.

Diagram is not to scale.

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  329

= 71 kg v1 = 2.3 m/s [12° N of E] v2 = 1.9 kg ⋅ m/s [52° S of W] m2 = ?  p1 = m1v1 = (71 kg)(2.3 m/s)  p1 = 163.3 kg ⋅ m/s m1 

= m2 v2 = m2 (1.9 kg ⋅ m/s)  p2 = 1.9 m2 kg ⋅ m/s  p2

Using the sine law: sin101° 163.3

=

sin 39° 1.9m2

 sin 39°   163.3 kg ⋅ m/s   =       1.9 kg ⋅ m/s   sin101°   m2 = 55 kg m2

4.

The mass of the second skater is 55 kg. Diagram not to scale.

 p1′

= m1v1′ = (0.50 kg)(1.5 m/s)  p1′ = 0.75 kg ⋅ m/s

= m1v1x = (0.50 kg)(2.0 m/s)  p1 x = 1.0 kg ⋅ m/s  p1 x

= p1′x + p2′ x  p1 x = p1′ cos 30° + p2′ cos θ  1.0 = (0.75) cos 30° + (0.30v2′ ) cos θ  (Equation 1) v2′ cos θ  = 1.168  p1 x

 p2′

= m2 v2′ = (0.30 kg)v2′  p2′ = 0.30v2′ kg ⋅ m/s 0 =  p1′ y

− p2′ y 0 =  p1′ sin 30° − p2′ sin θ  0 = (0.75) sin 30° − (0.30v2′ ) sin θ  v2′ sin θ  = 1.25 (Equation 2)

Divide Equation 2 by Equation 1: v2′ sin θ  1.25 v2′ cos θ 

=

1.168

tan θ  = 1.070

= tan −1 (1.070) θ  = 47° θ 

330 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Substitute back into Equation 2: v2′ sin θ  = 1.25 v2′

= =

1.25 sin θ  1.25

sin 47° v2′ = 1.7 m/s The velocity of the second ball after collision is 1.7 m/s [47º S of E].

Applying Inquiry Skills 5. (a) Referring to the diagram: Ignore points A3 and B3 for calculations because they are during the collision. Measure between points A1 and A2, A4 and A5, and divide by 0.50 s to calculate velocities. Repeat for Puck B (using two values for the after calculation because you have them). Measure angles before and after for velocities.



vA



vB

=

2.00

=

1.9

0.50

0.50

= 4.0 cm/s [45° N of E]

vA′

=

= 3.8 cm/s [38° W of N]

vB′

=

Copyright © 2003 Nelson



1.80 0.50

= 3.6 cm/s [N]

1.3 + 1.4 0.50 + 0.50

= 2.7 cm/s [43° E of N]

Chapter 5 Momentum and Collisions  331

Let the  x direction be the east-west direction.  p x = p ′x

− mBvBx = mA vA′ x + mB vB′ x vA′  x = 0 mA vA x − mB vBx = 0 + mB vB′ x mB vB′ x + mB vBx = mA vAx mB (vB′ x + vBx ) = mA vAx mA vA x

mB

= =

mA vA x vB′ x

+ vBx

(0.32 kg)(4.0cos 45°)

2.7 sin 43° + 3.8 sin 38° = 0.2165 kg mB

= 0.22 kg

= 0.32 kg mB = 0.2165 kg vA = 0.040 m/s vB = 0.038 m/s  E K lost = ?

(b) mA

 ET

= =

1 2 1 2

mA vA2

+

1 2

(0.32 kg)(0.040 m/s) 2

 E T

= 4.123 ×10 −4

 ET′

= =

 E T′

1 2 1 2

mB vB2

′2 mA vA

+

1 2

= 2.863 ×10

1

+

1

2

(0.2165 kg)(0.038 m/s)2

J mB vB′2

(0.32 kg)(0.036 m/s) 2 −4

+

2

(0.2165 kg)(0.027 m/s)2

J

= 4.123 ×10−4 J − 2.863 ×10−4 −4  E K lost = 1.3 × 10 J The total kinetic energy lost is 1.3 × 10 −4 J .  E K lost

J

(c) This is an inelastic collision. (d) The most likely source of error is measuring the angles from the diagram.

Making Connections 6. (a) Some possible points: • The perception of a safe vehicle is a common advertising feature of most vehicles. • The idea of protecting loved ones and yourself in a collision is socially desirable. • Insurance companies offer better rates for vehicles with high safety performance in collisions. (b) Some points are: • seat belts • roll bars (roll safety) •  bumpers • crumple zones • air bag • anti-lock braking systems • traction control (4WD, AWD, tires) 332 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

(c) Any of the ones listed in (b). (d) Could compare safety ratings and insurance prices, cost of vehicle (often safer vehicles or options are more expensive), and fuel economy.

CHAPTER 5 LAB ACTIVITIES Investigation 5.2.1: Analyzing One-Dimensional Collisions (Pages 260–262)

Question Observing two objects before, during, and after a collision allows the verification of conservation of momentum and kinetic energy theory.

Prediction (a) For all three categories, the total momentum of the system will always be the same. (b) Category I: The total kinetic energy before and after will be the same. During the collision there will be some loss of kinetic energy as it is stored as some other form in the repulsive device. Category II: The total kinetic energy before the collision will be zero. It will be a maximum at the end of the explosion, increasing as the explosion takes place. Category III: The total kinetic energy will be a maximum before, decreasing throughout the collision and a minimum at the end.

Hypothesis (c) The velocity of both carts before and after the collision (an adhesive collision would be the easiest) can be determined using the ticker-tape timer. Using the known mass of the carts, the conservation of momentum can be used to calculate the unknown mass.

Before the Collision Collision I (a) I (b) I (c) II (a) II (b) II (c) III (a) III (b) III (c) IV

After the Collision

Total  p (kg⋅m/s)

Total E K (J)

m1 (kg)

v 1 (m/s)

m2 (kg)

v 2 (m/s)

m1 (kg)

v 1’ (m/s)

m2 (kg)

v 2’ (m/s)

before

after

before

after

Loss in E K (%)

0.50 1.0 0.50

0.19 0.18 0.15

0.50 0.50 0.50

0.0 0.0 –0.13

0.50 1.0 0.50

0.0 0.060 –0.12

0.50 0.50 0.50

0.17 0.22 0.14

0.095 0.18 0.010

0.085 0.17 0.010

0.0090 0.016 0.0098

0.0072 0.014 0.0085

20 12 13

0.50 1.0 0.50

0.0 0.0 0.0

0.50 0.50 0.50

0.0 0.0 0.0

0.50 1.0 0.50

–0.22 –0.091 –0.10

0.50 0.50 0.50

0.21 0.27 0.12

0.0 0.0 0.0

–0.0050 0.044 0.010

0.0 0.0 0.0

0.023 0.022 0.0061

— — —

0.50 1.0 0.50

0.24 0.20 0.18

0.50 0.50 0.50

0.0 0.0 –0.17

0.50 1.0 0.50

0.11 0.12 0.0

0.50 0.50 0.50

0.11 0.12 0.0

0.12 0.20 0.0050

0.11 0.18 0.0

0.014 0.020 0.015

0.0060 0.011 0.0

57 45 100

0.50

0.22

m2

0.0

0.50

0.088

m2

0.088

—

—

—

—

—

Analysis (d) For each collision, the total momentum before and after was essentially the same. (e) For Category I collisions, there was very little loss of kinetic energy during the collisions. Category II collisions did not start with any, but received the kinetic energy from the spring bumper. Category III collisions lost a significant amount of kinetic energy during the collisions. (f) Category I collisions were all elastic (within experimental error). Category III collisions were all completely inelastic. (g) During Category I collisions, there was a temporary loss of kinetic energy that reappeared after the collision.

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  333

(h)

m1v1 + 0 = m1v1′ + m2v 2′ m2 =

= m2 =

m1v1 − m1v1′ v2′

(0.50 kg)(0.22 m/s) − (0.50 kg)(0.088 m/s) 0.088 m/s 0.75 kg

But m2 = mcart + munknown mass , therefore munknown mass = m2 − mcart

= 0.75 munknown mass =

kg − 0.50 kg

0.25 kg

Evaluation (i) The predictions and hypothesis were correct to within experimental errors. (j) Sources of error during the experiment include: • There was friction between the cart and the track. • There was friction between the tape and the timing recorder. • It is difficult to measure the exact distance between the ticker tape timer dots because the distance is so small.

Synthesis (k) It is important to distinguish between scalar and vector quantities because a system momentum of zero may still involve dangerous speeds or motions. (l) Using electronic values, the refuse bag could be tossed to a stationary astronaut. After catching the bag, the velocities and known mass of the astronaut could be used to calculate the mass of the refuse bag. (m) The friction pads would provide a source of external force t o the system. This would increase the loss of kinetic energy  before and after the collision and change the total momentum before and after as well.

Investigation 5.3.1: Analyzing Two-Dimensional Collisions (Pages 262–265)

Question The laws of conservation of momentum and kinetic energy can be verified through two-dimensional collisions.

Prediction (a) The kinetic energy and the total momentum will be the same before and after the collisions. (b) The change in momentum for each puck will be the same as the change in momentum of the other puck.

Hypothesis (c) I expect that the two pucks will stick together and move along a straight path that will bisect the initial angle of collision. (d) The change in momentum of puck A will be equal to the change in momentum of puck B. The velocities can me measured, to the mass can be calculated. (e) Category I and Category II ′ + pB ′  pA + p B = pA 







′ + mvB ′ mvA + mvB = mvA 







′ + vB ′ vA + vB = vA 







334 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Category III ′ + pB ′  pA + p B = pA 







′ + mvB ′ mvA + mv B = mvA 







′ = vB ′ vA 

(call this v ′)





vA + vB = v ′ + v ′ 





vA + vB = 





2v′ 

Category IV ′ + pB ′  pA + p B = pA 







′ + mBvB ′ mA vA + mBvB = mA vA 







′ = mA vA ′ − mA vA mB vB − mB vB 







′ ) = mA (vA ′ − vA mB (vB − vB 







mB =

)

′ − vA ) mA (vA 



(vB − vB′ ) 



(f) Mark off the four markers to get a known distance over a known time (the distance between each pair of dots should be approximately the same). Measure the angles. The sum of the two velocity vectors before should be equal to the sum of the velocity vectors after for Category I. (g) For category II, the change in each vector should be the same magnitude and opposite direction. (h) Category III should have the two velocity vectors before be equal to the twice the combined after velocity. (i) For Category IV, subtract the two velocity vectors for each puck and substitute the values into the above equation. (j) Measure each length of velocity vectors before and after. Substitute with the mass into the kinetic energy equation and compare the total values before and after. Any lost energy went into thermal energy.

Evaluation (l) Some possible sources of error: • The table surface may not be flat. • There may be some frictional or air current forces acting on the pucks. • The air supply line to the pucks may exert some lateral forces on the pucks. • Masses of the pucks may not be identical. • Frequency of the spark timer may not be perfectly uniform.

Synthesis (m) This would apply to all of the categories because all of them have the total momentum conserved. (n) The momentum would only be conserved if you considered both pucks. (o) It is wish to avoid dots during the impulse and collision because there is a changing speed during these interactions. The analysis of this experiment is not designed to include a puck that is accelerating. (p) It is better to have steel barriers because they will absorb some energy and prevent the cars from bouncing off as much as a rubber barrier would.

CHAPTER 5 SELF QUIZ (Pages 267–268)

True/False 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

F The impulse is the same in magnitude and is in the same direction. T F You have increased the amount of time the same force is applied. F The total momentum before and after is the same. T T F The kinetic energies will be different. F The momentum is conserved in the snowball-earth system, even though the collision is completely inelastic. T F The final kinetic energy can be greater (e.g., an exploding cart system).

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  335

Multiple Choice 11. (e) 2 12. (d) Momentum will increase by 2 × 2, and kinetic enegy by 2 × 2 . 13. (d) 14. (d) 15. (e) 16. (d) Some energy is lost to thermal energy of the collision. 17. (b) The momentum is  p =

2mEK

so p ∝

E K  .

18. (c) M only stops, while T must continue to act on R to give it momentum in the opposite direction. 19. (a) They have both stopped, so the total momentum is zero, but the initial kinetic energy is stored as electric potential. 20. (d) 1   21. (a) mv + (2m)  (−v)  = (m + 2m)v′ 2   v′ =

mv − mv 3m

v′ = 0 22.(d) 0 = mv − (M − m )V   V  =

mv  M

−m

CHAPTER 5 REVIEW (Pages 269–271)

Understanding Concepts 1. 2. 3.

4.

5.

Impulse is the product of the force and the time. A smaller force exerted over a long period of time can impart a larger impulse than a large force for a short time. As the meteor comes to a stop, the kinetic energy is converted into thermal energy which melts the material at the impact site. This observation does not contradict the law of conservation of momentum. The momentum of the earth increases toward the falling object. The amount of change in the earth’s velocity is so small it is imperceptible, so it appears the law of conservation is being contradicted, but it is not. The change in momentum of the ball that bounces is greater than the putty that sticks to the floor. Assuming both have the same mass and are dropped from the same height: For the putty, ∆ p = m(v2 − v1 )

For the ball ∆ p = m(v2 − v1 )

= m(0 − v1 ) ∆ p = − mv1

= m((−v1 ) − v1 ) ∆ p = −2mv1

 E K  =

=  E K  =

6.

mv

2

2 2 2

m v

2m  p 2

2m The momentum of the two players is the same, but the person with the larger mass will have a smaller kinetic energy. It would be better to avoid the faster moving lighter player,  p2 = m2v2. A car crashing into a tree is an example. Momentum is not conserved in the tree-car system because the tree is attached to the earth. The earth supplies a net external force to the system of objects so that momentum is not conserved.

336 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

7.

Diagram is not to scale.

m = 1.3 × 10 2 kg

= 8.7 m/s [44° E of N]  p N = ?  pE = ? v



 p = mv 



= (1.3 × 102 kg)(8.7 m/s)  p = 1131 kg ⋅ m/s [44° E of N] 

 p N  p N  pE  pE

= p cos44° = (1131 kg ⋅ m/s)(cos44°) = 8.1× 102 kg ⋅ m/s = p sin44° = (1131 kg ⋅ m/s)(sin 44°) = 7.9 × 102 kg ⋅ m/s

The northward component of the boat’s momentum is 8.1× 102 kg ⋅ m/s . The eastward component is 7.9 × 102 kg ⋅ m/s . 8.

Diagram is not to scale.

= 2.6 × 104 kg ⋅ m/s 3 m = 1.1× 10 kg θ  = 22° v =?  pE

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  337

 pE

= mv cos22°

v=

=

9.

 pE m cos22 ° 2.6 × 10 kg ⋅ m/s 4

(1.1× 103 kg)(cos 22°)

v = 25 m/s The car is travelling at a speed of 25 m/s. Choose west as positive. m = 1.2 × 10 kg 3

= 53 km/h [W] = 14.72 m/s [W] v2 = 0 ∆t  = 55 ms = 5.5 × 10−2 s Σ F  x = ? Σ F x ∆t = ∆p v1

Σ F  x = =

m(v2

− v1 )

∆t  (1.2 × 103

kg )(0 m/s − 14.72 m/s) 5.5 × 10

−2

s

= −3.2 ×10 N [W] Σ F  x = 3.2 × 105 N [E] 5

The average force exerted on the car by the pole is 3.2 × 105 N [E] . 10. (a) Σ F   x

(b)

= 324 N ∆t  = 5.1 ms = 5.1× 10−3 s Σ F x ∆t = ? Σ F x ∆t  = (324 N)(5.1× 10−3 Σ F x ∆t  = 1.7 N ⋅ s [fwd] The impulse on the ball is 1.7 N ⋅s [fwd]. m = 0.059 kg v1 = 0 v2 = ? Σ F x ∆t = ∆p Σ F x ∆t = m( v2 − v1 ) Σ F x ∆t = mv2 Σ F ∆t  v2 =  x

s)

m

= v2

(324 N)(5.1× 10−3 s) 0.059 kg

= 28 m/s [fwd]

The velocity of the ball just as it leaves the racket is 28 m/s [fwd]. 11. m1 = 0.112 kg

= 0.154 kg v1′ = 1.38 m/s v2′ = ? m2

338 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

 p = p′ 0 = m1v1′ + m2 v2′ v2′

=− =−

v2′

m1v1′ m2 (0.112 kg)(1.38 m/s) 0.154 kg

= −1.00 m/s

The speed of the second car is 1.00 m/s.

= 1.67 ×10−27 kg m = 6.64 × 10 −27 kg vP = 1.57 km/s v =0 vP′ = −0.893 km/s v′ = ?

12. mP α

α

α

= mP vP′ + m v′ mP vP + 0 = mP vP′ + m v′ m v − mP vP′ v′ = P P

mP vP

+m

α

v

α

α

α

α

α

α

m

α

=

− vP′ )

mP ( vP m

α

= v′

α

(1.67 ×10 −27 kg)(1.57 km/s − (−0.893 km/s)) 6.64 × 10−27 kg

= 0.620 km/s

The speed of the alpha particle is 0.620 km/s. 13. m1 = 2.67 kg

= 5.83 kg 2 v1 = 1.70 × 10 m/s [toward Jupiter] v1′ = 185 m/s [toward Jupiter] v2′ = 183 m/s [toward Jupiter] v2 = ? m2 







Choose the direction toward Jupiter as positive. m1v1 + m2 v2 = m1v1′ + m2 v2′

= m1v1′ − m1v1 + m2 v2′ m (v′ − v ) + m2 v2′ v2 = 1 1 1

m2 v2

m2

= =

m1 (v1′ − v1 ) m2

+ v2′

(2.67 kg)(185 m/s − 1.70 × 10 m/s) 2

5.83 kg

+ 183 m/s

= 1.90 × 10 2 m/s 2 v2 = 1.90 × 10 m/s [toward Jupiter] The velocity of the more massive rock is 1.90 × 102 m/s [toward Jupiter] . v2 

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  339

= 6.0 m/s v2i = 0 v1f = v2f =   2.0 m/s

14. (a) v1i

The two objects have the same final velocity. Having the same final velocity is an indication of a ‘hit-and-stick’ collision. All collisions that have the objects with the same speed (in the same direction) after are completely inelastic collisions. (b) v1i = 24 m/s

=0 v1f  = −4.0 m/s v2f  = 14 m/s v2i

 EKi

= =

1 2 1

m1v1i2

+

1 2

2 m2 v2i

 EKf

( 2.0 kg )( 24 m/s )2

2  E Ki = 576 J

+

1 2

( 4.0 kg )( 0 m/s)2

= =

1 2 1

m1v1f2

+

1 2

2 m2 v2f

 

(2.0 kg)(− 4.0 m/s)2 +

2  E Kf  = 408 J

1 2

(4.0 kg)(14 m/s)2

There is some loss of kinetic energy. This is an inelastic collision. (c) v1i = 12 m/s

=0 v1f  = −4.0 m/s v2f  = 8.0 m/s v2i

 EKi

= =

1 2 1

m1v1i2

+

1 2

2 m2 v2i

 EKf

( 2.0 kg )(12 m/ s)2

2  E Ki = 144 J

+

1 2

( 4.0 kg )( 0 m/s)2

= =

1 2 1

m1v1f2

+

1 2

2 m2 v2f

 

(2.0 kg)(−4.0 m/s)2 +

2  E Kf  = 144 J

1 2

(4.0 kg)(8.0 m/s)2

There is no loss of kinetic energy, so this collision is elastic. 15. Choose north as positive. Conservation of Energy

Conservation of Momentum

′ = m1v1′ + m1v12 = m1v1′2 + m2 v2′2 0.253(1.80)2 = 0.253v1′2 + 0.232v2′2 3.24 = v1′2 + 0.917v2′2 (Equation 1) 1 2

2 m1v1

1 2

2

1 2

2 m2 v2

= m1v1′ + m2 v2′ (0.253)(1.80) = 0.253v1′ + 0.232v2′ 1.80 = v1′ + 0.917v2′ (Equation 2) v1′ = 1.80 − 0.917v2′ m1v1

Substitute Equation 2 into Equation 1: 3.24 = (1.80 − 0.917v2′ ) 2 + 0.917v2′2 3.24 = (3.24 − 3.30v2′ + 0.841v2′2 ) + 0.917v2′2 0 = −3.30v2′ + 1.758v2′2 0 = v2′ (−3.30 + 1.758v2′ ) v2′

=0

or

− 3. 30 + 1. 758v2′ = 0

Since v2′  = 0 is not valid (no change in speed), then 1.758v2′  = 3.30 v2′  = 1.88 m/s [N]

340 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Substitute back into Equation 2: v1′ = 1.80 − 0.917(1.88) v1′ = 0.08 m/s [N] The velocity of the first cart is 0.08 m/s [N]. The velocity of the second cart is 1.88 m/s [N]. 16. (a) Diagram not to scale.

(b)

0 = mvB y 0 = vB y

− mvAy

− vAy

= vAy vB sin 46° = vA sin 33° sin33° vB = vA sin 46° vB y

= mvA x + mvBx vA1 = vA x + vBx 5.4 = vA cos 33° + vB cos 46°  sin33° v  cos 46° 5.4 = vA cos 33° +  A    sin46°   5.4 = 1.365vA vA = 3.957 m/s, or 4.0 m/s

mvA1

Substitute back into equation for vB : vB

= =

sin33° sin 46° sin33°

sin 46° vB = 3.0 m/s

vA (3.957 m/s)

The speed of the first puck is 4.0 m/s after the collision. The speed of the second puck is 3.0 m/s after the collision.

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  341

17.

= 67.8° θ B = 30.0° θ A

vB vA mB mA

= 3.30 =?

0 =  pA′ y

− pB′ y

0 = mA vA sin 67.8° − mB vB sin 30.0° mB vB sin 30.0° = mA vA sin 67.8° mB mA

= =

vA sin 67.8° vB sin30.0° sin 67.8° vB vA

= mB mA

(sin30.0°) sin 67.8°

(3.30)(sin30.0°)

= 0.561

The ratio of the masses of the particles is 0.561. 18. Diagrams not to scale.

342 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Conserve momentum in the y direction.  p1 y + p2 y = − p1′ y + p2′ y 0 + mv2 y v2 y

v2 y

= − mv1′y + mv2′ y = −v1′y + v2′ y = −v1′ sin 69.2° + v2′ sin 62.8° = −(2.37 m/s)(sin 69.2°) + (2.49 m/s)(sin 62.8° ) = 0.00 m/s

Conserve momentum in the x direction.  p1 x + p 2 x = p1′x

− p2′ x mv1 x + mv2 x = mv1′x − mv2′ x v1 x + v 2 x = v1′x − v2′ x v2 x = v1′x − v′2 x − v1x = v1′ cos 69.2° − v2′ cos 62.8° − v1 = (2.37 m/s)(cos69.2°) − (2.49 m/s)(cos62.8°) − 2.70 m/s = −3.00 m/s [E] v2 x = 3.00 m/s [W]

There is no y component for velocity, so the unknown initial velocity is 3.00 m/s [W] 19. (a)  g  = 9.8 m/s 2

∆ y = 0.26 m v=? Use conservation of mechanical energy for the swing. 1 2 mg ∆y = mv 2 v=

2 g ∆y

= 2(9.8 m/s2 )(0.26 m) = 2.2574 m/s v = 2.3 m/s The speed of the ball just before the collision is 2.3 m/s. (b) Conservation of Energy Conservation of Momentum 1 1 1 2 2 2 m1v1 = m1v1′ + m2 v2′ m1v1 = m1v1′ + m2 v2′ 2 2 2 (0.25)(2.2574) = 0.25v1′ + 0.21v2′ m1v12 = m1v1′2 + m2 v2′2 2.2574 = v1′ + 0.84v2′ 2 2 2 0.25(2.2574) = 0.25v1′ + 0.21v2′ v1′ = 2.2574 − 0.84v2′ 5.096 = v1′2 + 0.84v2′2 (Equation 1)

(Equation 2)

Substitute Equation 2 into Equation 1: 5.096 = (2.2574 − 0.84v2′ ) 2

+ 0.84v2′2 5.096 = (5.096 − 3.792v2′ + 0.7056v2′2 ) + 0.84v2′2 0 = −3.792v2′ + 1.5456v2′2 0 = v2′ (−3.792 + 1.5456v2′ ) v2′ = 0 or − 3.792 + 1.5456v2′ = 0

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  343

Since v2′

= 0 is not valid (no change in speed), then: 1.5456v2′  = 3.792 v2′  = 2.5 m/s [fwd]

The speed of the 0.21-kg ball just after the collision is 2.5 m/s [fwd]. 20. (a) m1 = 0.45 kg

= 0.79 kg v1 = 2.2 m/s ′ =? v12 m2

m1v1

′ v12

= (m1 + m2 )v12′ = =

′ v12

m1v1 m1 + m2 (0.45 kg)(2.2 m/s) 0.45 kg + 0.79 kg

= 0.80 m/s

The speed of the ball and the box just after the collision is 0.80 m/s. (b) ∆d  = 0.051 m vi = 0.80 m/s

=0 Σ F  = ? vf 

v +v ∆d =  i f     ∆t   2    2   ∆t =   ∆d   vi + vf    2  (0.051 m) =     0.80 m/s + 0 m/s   ∆t  = 0.1275 s

Σ F ∆t = ∆p Σ F  = =

m(v2

− v1 )

∆t  (0.79 kg + 0.45 kg)(0 m/s − 0.80 m/s) 0.1275 s

Σ F  = −7.8 N The magnitude of the friction force acting on the ball and the box is 7.8 N.

= 1.9 ×104 kg 4 m2 = 1.7 × 10 kg 3 v1′ = 3.5 × 10 km/h 3 v2′ = 3.4 × 10 km/h

21. m1

θ  =

180.0º – 5.1º – 5.9º = 169.0º v = ?

344 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Using the sine law: sin 5.1° m2 v2′ v

= = =

v

sin θ  ( m1 + m2 ) v m2 v2′ sin θ  ( m1 + m2 ) sin 5.1° (1.7 × 104 kg )(3.4 × 103 km/ h) s in 169.0° (1.9 × 10 kg + 1.7 × 10 kg) sin 5.1° 4

= 3.4 × 103

4

km/h

The original speed of the spacecrafts was 3.4 × 103 km/h . 22. The total momentum before is zero, so the total momentum after must also be zero. Diagram not to scale.

= 2.0 kg m2 = 3.0 kg m3 = 4.0 kg v1 = 1.5 m/s [N] v2 = 2.5 m/s [E] v3 = ? m1

= p12 + p22 (m3v3 ) 2 = ( m1v1 ) 2 + (m2 v2 )2 m32 v32 = ( m1v1) 2 + (m2 v2 )2  p32

v3

= =

v3 tan θ  =

( m1v1 ) 2

+ (m2 v2 )2 2

m3

2

(2.0 kg(1.5 m/s))

+ (3.0 kg(2.5 m/s))2

(4.0 kg) 2

= 2.0 m/s m1v1 m2 v2

 (2.0 kg)(1.5 m/s)   = tan −1     (3.0 kg)(2.5 m/s)   θ  = 22° θ 

The final velocity of the third piece is 2.0 m/s [22º S of W].

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  345

Applying Inquiry Skills 23.

24.

25. (a) The coefficient of restitution is a measure of how quickly an object returns to its shape after it has been compressed. The apparatus shown could help determine the coefficient of restitution because the speed the sphere returns to its original shape corresponds to the height the ball bounces to. 26. (a) The bed sheet spreads the force applied over a greater distance (and therefore time) than if the egg was to strike the ground directly. The same impulse is imparted to the egg (i.e. it is brought to rest), so the force required is much smaller. (b) This can be used to cushion the landing of a person who was falling from a high position by reducing the average force needed to stop them. You could test several heights with a sack of potatoes and plot the results on a graph. For a human, you could extrapolate from the graph.

Making Connections 27. It is better for safety to have telephone poles that collapse upon impact. In this way, they absorb some energy and increase the time of collision. Both of these reduce the force imparted to the vehicle and occupants in a collision. 28. High speed photography and spark timers both show the location of an object at fixed time intervals. High speed  photography gives more thorough information because you can see the state of the object s during each stage of the collision. 29. Many arrester cables are connected to a water squeezer that dampens the motion and converts the kinetic energy of the aircraft to kinetic energy of ejected water. 30. The main purpose of the ablation shield is to protect the shuttle during re-entry into the earth’s atmosphere.

346 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Extension 31. (a) Conservation of Momentum

= m1v1′ + m2 v2′ (but v2 = 0) m1v1 + 0 = m1v1′ + m2 v2′   m1v1 − m1v1′ = m2 v2′ m1 ( v1 − v1′ ) = m2 v2′ (Equation 1)

Conservation of Energy 1 1 1 m1v12 + m2 v22 = m1v1′2 2 2 2

m1v1 + m2 v2

and

+

1 2

m2 v2′2

(but v2

= 0)

+ 0 = m1v1′2 + m2 v2′2 2 2 2 m1v1 − m1v1′ = m2 v2′ 2 2 2 m1 (v1 − v1′ ) = m2 v2′ (Equation 2) 2

m1v1

Divide Equation 2 by Equation 1:

− v1′2 ) m2 v2′2 = m1 ( v1 − v1′ ) m2 v2′ (v1 + v1′ )(v1 − v1′ ) = v2′ (v1 − v1′ ) v2′ = v1 + v1′ m1 (v12

(Equation 3) or v1′ = v2′ − v1

(Equation 4)

Substitute Equations 3 and Equation 4 back into the original conservation of momentum equation: m1v1 + 0 = m1v1′ + m2 v2′ m1v1 + 0 = m1v1′ + m2 v2′

= m1v1′ + m2 ( v1 + v1′ ) m1v1 = m1v1′ + m2 v1 + m2 v1′ m1v1′ + m2 v1′ = m1v1 − m2 v1 v1′ ( m1 + m2 ) = (m1 − m2 )v1  m − m2   v1′ =  1  v1  m1 + m2   m1v1

 m1 − m2    v1  m1 + m2   m − m   =   v1  m + m   v1′ = 0  m − m2   v1′ =  1  v1  m1 + m2    m − 0   =  1  v1  m1 + 0   v1′ = v1  m − m2   v1′ =  1  v1  m1 + m2  

(b) v1′ = 

(c)

(d)

 0 − m2   =   v1 + 0 m  2   v1′ = − v1

= m1 (v2′ − v1 ) + m2 v2′ m1v1 = m1v2′ − m1v1 + m2 v2′ m1v2′ + m2 v2′ = 2 m1v1 v2′ (m1 + m2 ) = 2m1v1  2m1   v2′ =   v1  m1 + m2   m1v1

 2m1   =   v1  m1 + m2   2m   =   v1  m + m   v2′ = v1  2m1   v2′ =   v1  m1 + m2    2m   =  1  v1  m1 + 0   v2′ = 2v1  2m1   v2′ =   v1  m1 + m2    2m   =  1  v1  0 + m2   v2′

v2′

=

2m1 m2

v1

= 1.0 ×103 kg mB = 2.0 × 103 kg vP = 50 m/s vB = 0 ∆d  = ?

32. mP

Copyright © 2003 Nelson

Chapter 5 Momentum and Collisions  347

For the collision: mP vP

+ mB vB = (mP + mB )vPB m v + mB vB vPB = P P mP + mB (1.0 ×103 kg)(50 m/s) + (2.0 × 103 = 1.0 × 103 kg + 2.0 × 103 kg vPB = 16.67 m/s

kg)(0)

Calculate the frictional force on the plane and on the barge: For the plane 1  FK  = mg  4 1 = (1.0 × 103 kg )(9.8 m/s2 ) 4  F K  = 2450 N [backward] Acceleration of the plane Σ F = ma a

=

For the barge

By Newton’s third law, F K  = 2450 N [forward]

Acceleration of the barge Σ F = ma

Σ F 

a

m

−2450 N 1.0 × 103 kg 2 a = −2.45 m/s =

= a

Distance plane will travel during collision vf2

=

∆d  = =

+ 2 a ∆d 2 vf − vi2

vi2

=

Σ F  m 2450 N 2.0 × 103 kg

= 1.225 m/s2

Distance barge will travel during collision

= vi2 + 2a∆d vf2 − vi2 ∆d  = vf2

 

2a

 

2a

2

(16.67 m/s)

− (50 m/s)

2

=

2( −2.45 m/s 2 )

∆d  = 453.5 m

(16.67 m/s) 2

− ( 0 m/s)2

2(1.225 m/s 2 )

∆d  = 113.4 m 2

The required length of the barge is 453.5 – 113.4 = 3.4 × 10  m.

348 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

CHAPTER 6 GRAVITATION AND CELESTIAL MECHANICS Reflect on Your Learning (Page 272)

1. (a) The force of gravity exerted by Saturn keeps the rings in their orbit around Saturn. (b) The force of gravity exerted by Earth keeps the Hubble Space Telescope in a stable orbit around Earth. The force exerted by the rockets of the telescope counteracts the small amount of friction of the upper atmosphere. 2. (a) The two probes would have the same minimum speed (escape speed) even though the probes have different masses. Students will discover that the escape speed depends on the mass of Earth, not on the mass of the probe, according to the equation:

v

=

2GM E r 

.

(b) The minimum kinetic energy of the probe of mass 2m would be twice that of the probe of mass m. The minimum kinetic energy of the probe is equal to the gravitational potential energy, which is proportional to the mass of the probe.  EK

= − E 

g

GMm

∴ E 

K  =

3.

r 

Since the gravitational potential energy E g is inversely proportional to the distance from the main body,  E g will approach zero as the distance increases, according to the equation presented later in the chapter:  E g

GMm = −

r 

4.

Students should be able to answer this question based on their study of the law of universal gravitation in Section 3.3.

5.

As students have seen on page 147 of the text, a black hole is an extremely dense celestial body. The gravitational field of a black hole is so strong that nothing, including electromagnetic radiation, can escape from its vicinity. Since light can enter, but not escape, a black hole appears totally black. The surface of the body is called its event horizon because no event can be observed from outside the surface. At the core of a black hole is a dense centre called a singularity. The distance from the centre of the singularity to the event horizon is the Schwartzschild radius. Black holes are thought to form during the course of stellar evolution. When the nuclear fuels are exhausted in the core of the star, the star collapses. If the mass of the core is greater than a critical value that is almost twice as great as the mass of the Sun, the core may collapse into a black hole.

Try This Activity: Drawing and Comparing Ellipses (Page 273)

(a) The eccentricity (e1) of the first ellipse is less than the eccentricity ( e2) of the second ellipse. c1 5.0 cm = = 0.33 e1 = a1 15 cm e2

=

c2 a2

=

7.5 cm 1.5 cm

=

0.60

(b) The Sun is at one focus of the elliptical orbit of each planet. Earth is at one focus of the elliptical orbit of the Moon.

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  349

6.1 GRAVITATIONAL FIELDS PRACTICE (Pages 276–277)

Understanding Concepts 1.

As they are moving at a high speed in their orbits, space vehicles are attracted to Earth by the force of gravity, which keeps them in their orbits. Also, the vehicles have small booster rockets to control the location of the orbit and to counteract low amounts of friction in the upper atmosphere.

2. (a) G = 6.67 × 10−11 N ⋅ m 2 /kg 2

= 5.98 × 1024 kg 22 mMoon = 7.35 × 10 kg 8 r Moon = 3.84 × 10 m

 M E

 F G

= =

 F G

GM E mMoon r Moon 2 (6.67 × 10

−11

N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)(7.35 × 1022 kg) (3.84 × 108 m) 2

= 1.99 × 1020

N [toward Earth's centre]

The gravitational force exerted on the Moon by Earth is 1.99 × 1020 N [toward Earth’s centre]. GM E mMoon (b)  F G = r Moon 2

=  F G

3.

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)(7.35 × 10 22 kg) (3.84 ×108 m) 2

= 1.99 × 1020

N [toward Moon'scentre]

The gravitational force exerted on Earth by the Moon is 1.99 × 1020 N [toward Moon’s centre]. GM E GM E Let g r represent the gravitational field strength at the desired radius r . Thus, g r  = and  g  = .   2 ( r + r E ) r E 2 (a) r

= r E  g r  =

=  g r  =

GM E (rE

+ r E ) 2

GM E 4r E 2  g  4

The magnitude of the gravitational field strength (in terms of g ) at 1.0 Earth radius is (b) r

 g  4

.

= 3r E  g r  =

=  g r  =

GM E (3rE

+ r E )2

GM E 16r E 2  g  16

The magnitude of the gravitational field strength (in terms of g ) at 3.0 Earth radii is

350 Unit 2 Energy and Momentum

 g  16

.

Copyright © 2003 Nelson

(c) r

= 4.2r E GM E

 g r  =

(4.2rE

+ r E )2

GM E

=

27 r E 2  g 

 g r  =

27

The magnitude of the gravitational field strength (in terms of g ) at 4.2 Earth radii is 4.

 g  27

Let g P represent the magnitude of the gravitational field strength of the planet. Thus,  g P

=

. GM E (0.50r E )

2

and  g  =

GM E r E

2

.

4.0GM E

 g P

=

 g P

= 4.0 g 

r E

2

The planet has a surface gravitational field strength of magnitude 4.0 g . 5. (a)  g  = 1.6 N/kg r  = 1.74 ×106 m GM Moon

 g  =  M Moon

=

2

r   gr 2

=

G (1.6 N/kg)(1.74 × 10 m) 6

6.67 × 10

−11

N ⋅ m /kg 2

2

2

= 7.3 × 1022 kg 22 The mass of the Moon is 7.3 × 10  kg.  M Moon = 7.35 × 1022 kg mstudent = 55 kg (example mass of student)  M Moon

(b)

 F G

= =

 F G

GM Moon mstu dent 2

r 

(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(7.35 × 10 22 kg)(55 kg) (1.74 × 106 m) 2

= 89 N

A 55-kg student would have a weight of 89 N on the Moon. (Students can also determine the weight by applying the fact that the ratio of the Moon’s gravitational field strength to Earth’s gravitational field strength is 1.6:9.8.) 6. (a)  g  = 5.42 × 10−9 N/kg m = 1.00kg  FG

  F G

= mg  = (1.00 k g)(5.42 × 10−9 N/kg) = 5.42 × 10−9 N −9

The magnitude of the gravitational force is 5.42 × 10  N. (b) m = 8.91× 105 kg

= mg  = (8.91× 105 kg)(5.42 × 10−9 N/kg) −  F G = 4.83 × 10 3 N − The magnitude of the gravitational force is 4.83 × 10 3 N.  FG

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  351

Applying Inquiry Skills 7.

It is an inverse square relationship. The graphs for (a) and (b) are the same except that the vertical axes are labelled differently.

Making Connections 8. (a) If Earth’s density were greater, then its mass would also be greater. Therefore, Earth’s surface gravitational field strength would also be much greater since it is proportional to Earth’s mass. (b) Since the surface gravitational field strength of Earth would be much greater if its density were greater, a human’s weight would also be greater. To support the increased weight, bones would need to be very sturdy (and heavier). Shorter, thicker bones would be more efficient at supporting the larger weight. (c) Answers will vary. Some effects in nature could be shorter maximum heights of plants and more powerful winds. Examples of effects on human activities include different sizes of equipment used in sports (balls, bats, rackets, etc.) and small sizes of some transportation vehicles, especially aircrafts.

Section 6.1 Questions (Page 277)

Understanding Concepts 1.

The weight of a space probe decreases as the probe travels away from Earth because the gravitational field strength of Earth, g E, decreases. However, as it approaches the Moon, the space probe experiences the gravitational field strength of the Moon, g M. There is a location where g E and g M are equal, but in opposite directions. At this location, the weight of the  probe is zero, although the mass of the space probe does not change. Since Earth is almost 100 ti mes more massive than the Moon, this location is much closer to the Moon. 2. (a) Let the subscript s represent the satellite. m = 225kg

= 5.98 × 1024 kg r s = 8.62 × 106 m r E = 6.38 × 106 m r = rs + r E

 M E

 F G

= =

 F G

GM E m r 2 (6.67 × 10

−11

N ⋅ m /kg )(5.98 × 10 kg)(225 kg) 2

2

24

(8.62 × 106 m + 6.38 × 106 m) 2

= 3.99 × 102

N

The magnitude and direction of the gravitational force on the satellite is 3.99 × 10  N [toward Earth’s centre]. 2

∑ F = ma 

(b)





a



= =

a



 F G m 3.99 × 10 N [toward Earth's centre] 2

225kg

= 1.77 m/s

2

[toward Earth's centre] 2

The resulting acceleration of the satellite is 1.77 m/s  [toward Earth’s centre].

352 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

3.

r  = 7.4 × 107 m  M E

= 5.98 ×1024 kg GM E

 g  =

r 2 (6.67 × 10

=

−11

N ⋅ m /kg )(5.98 × 10 kg) 2

2

24

(7.4 × 107 m) 2

 g  = 7.3 × 10−2 m The magnitude and direction of the gravitational field strength is 7.3 × 10  N/kg [toward Earth’s centre]. 4. (a) Let the subscript s represent the satellite.  g s = 4.5 N/kg  –2

= 6.38 × 106 m  g  = 9.8 N/kg r E

Since  gs

=

GM E r  g

 gs  g  g s

and

2

g  =

GM E   : 2 r E

 GM    r 2   =  2 E        r E   GM E   =

2

r 

r E 2

r = r E

 g   g s

= 6.38 × 106 m

9.8N/kg 4.5 N/kg

r  = 9.42 × 106 m The distance the satellite is above Earth = r  – r E 6 6 = 9.42 × 10  m – 6.38 × 10  m 6 = 3.0 × 10  m Thus, the satellite is 3.0 × 106 m above Earth’s surface. (b) m = 6.2 × 102 kg  FG

 F G

= mg  = (6.2 × 102 kg )( 4.5 N /kg) = 2.8 × 103 N

The gravitational force on the satellite is 2.8 × 10  N. 3

5.

= 2.48 ×107 m  M  N =1.03 × 1026 kg r   N

 g  =

=

GM  N r 2 (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.03 × 10 26 kg) (2.48 ×107 m) 2

 g  = 11.2 N/kg The magnitude of Neptune’s surface gravitational field strength is 11.2 N/kg. The value given in Table 1 is 1.14 × 9.8 N/kg = 11.2 N/kg. Therefore, the values are the same. 6. (a) r  = 2.5 × 10 7 m m = 456kg v = 3.9 km/s = 3.9 × 103m/s

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  353

=

ac

=

v2 r  (3.9 × 10 m/s) 3

2

2.5 × 107 m

= 0.61m/s2

ac

2

The acceleration of the satellite is 0.61 m/s  [toward Earth’s centre]. GM E m (b)  F G = r 2

=  F G

(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(456 kg)(5.98 × 1024 kg) (2.5 × 10 m) 7

= 2.9 × 102

2

N

The gravitational force on the satellite is 2.9 × 102 N [toward Earth’s centre]. 7. (a)  g  = 1.3 N/kg

 M  = 1.3 × 1023 kg  g  =

GM  2

r 

GM 

r  =

 g  (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.3 × 1023 kg)

=

(b)

8.

1.3N/kg

= 2.6 × 106 m r  = 2.6 × 103 km 3 Titan’s radius is 2.6 × 10  km. m = 0.181kg  FG = mg  = (0.181kg)(1.3N/kg)  F G = 0.24 N

The force of gravity on a 0.181-kg rock on Titan is 0.24 N. Let r 2 be the distance from Earth’s centre at which the gravitational field strength has a magnitude,  g 2, of 3.20 N/kg.  g E = 9.80 N/kg at Earth's surface  g 2

= 3.20 N/kg at distance r 2

Using ratio and proportion:  g 2 GmE GmE  g E

= =

r 2

2

r 2

= = =

r2

r2

÷

2

r E

2

r E 2 r 2

2

 g E r E

2

g2  g E r E

2

g2 (9.80 N/kg)(r E 2 ) (3.20 N/m)

= 1.75r E

354 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

= 1.75rE − 1.00r E d = 0.75r E d

The gravitational field strength is 3.20 N/kg at a distance of 0.75 r E above Earth’s surface.

Applying Inquiry Skills 9.

The diagram below shows examples of the required FBDs.

Making Connections 10. Based on the data in Table 1, some astronomers might argue that Pluto should not be considered to be a planet because the gravitational field strength of Pluto is so much smaller than the other planets. On average, Pluto’s gravitational field strength is less than one-tenth the gravitational field strength of the other planets.

6.2 ORBITS AND KEPLER’S LAWS PRACTICE (Page 279)

Understanding Concepts 1.

The Moon does not fall into Earth because it travels at a specific speed around Earth. This keeps the Moon at an approximately constant distance from Earth’s centre called the orbital radius. 2. Since the space probe is in a circular orbit, the direction of the gravitational force is perpendicular to the direction of the instantaneous velocity. Thus, the force of gravity does not do any work on the probe, and there is no change in the kinetic energy (or speed) of the probe. 3. Let the subscript s represent the satellite.  M E = 5.98 × 1024 kg 6 r E = 6.38 × 10  m 5 r s = 525 km = 5.25 × 10  m r  = r E + r s (a) vs

= =

vs

GM E r  (6.67 × 10

−11

N ⋅ m / kg )(5.98 × 10 2

2

24

kg)

(6.38 ×10 m) + (5.25 × 10 m)

= 7.60 × 103

6

5

m/s

The speed of the satellite is 7.60 × 10  m/s. 3

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  355

(b)

v

=

2π  r 

=

2π  r 

T 

= T 

T 

v

(

2π  6.38 ×10 6 m + 5.25 × 105 m

)

7.60 × 10 m/s 3

= 5.71× 10

3

s

 1 h   = 1.59 h.    3600 s  

The period of revolution of the satellite is 5.71 × 10  s or (5.71× 10  s)  3

4.

3

Let the subscript s represent the satellite. 22  M Moon = 7.35 × 10  kg 6 r Moon = 1.74 × 10  m vs

= =

vs

GM Moon r 

(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(7.35 × 10 22 kg) 1.74 × 10 m 6

= 1.68 × 103

m/s

The speed of the satellite is 1.68 × 10  m/s. 3

Applying Inquiry Skills 5. (a) The speed of a satellite around a central body is inversely proportional to the square root of the radius of the satellite’s 1 orbit. Thus, vs ∝ . r  s

(b)

Making Connections 6.

Recommended web sites are: www.space.com/spacewatch/space_junk.html http://earthwatch.unep.net/solidwaste/spacejunk.html Students are likely to encounter the following points about the problem of space junk: • a study done in 1999 estimated 4 million pounds of space junk in low-Earth orbit • objects that are baseball-size and bigger may threaten the safety of astronauts in space; collisions involving even the smallest of objects may be damaging due to the high speeds of the objects • the U.S. Space Command agency counted the number of objects in space as of June 21, 2000 and found 2671 satellites, 90 space probes, and 6096 chunks of debr is • some objects re-enter Earth’s atmosphere, but most burn up on re-entry, or land in water or uninhabited land •  NASA calculates that if the amount of debri s equal to or larger than 1 centimeter exceeds 150 000, it could make space flight impossible

356 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

• as technology improves, more and more satellites and space objects are being launched into space, not only by space agencies, but other industries including telecommunications and organizations interested in astronomy •  possible solutions include better tracking of space junk, i mproved methods of bringing down satellites to Earth, and launching satellites to sweep up the space junk that is already in orbit

PRACTICE (Page 283)

Understanding Concepts 7.

Relative to the rest of the solar system, Earth’s frame of reference is accelerating, so the geocentric model is the noninertial frame of reference. The heliocentric model is an inertial frame of reference if the solar system is considered to  be isolated. However, it is a noninertial frame with respect to the Milky W ay Galaxy and the rest of the universe. 8. Tycho Brahe (1546–1601) made precise, comprehensive astronomical measurements of the solar system and more than 700 stars. For 20 years, he made countless naked-eye observations using l arge instruments he made hi mself. He was able to collect data for Mercury, Venus, Earth, Mars, and Saturn, because those were planets that he could see. The other planets were beyond his scope of vision and the telescope was not invented until the early 17th century. 9. Using Kepler’s second law, Earth sweeps out equal areas in equal time intervals. Therefore, when Earth is closest to the Sun, it is moving fastest. Conversely, if the orbit is divided into 180 ° “halves,” the portion closest to the Sun will have a smaller area, and therefore a shorter time. Since there are three fewer days between September 21 and March 21, Earth must be closest to the Sun at that time. 3

10. The ratio

r 

2

T 

is calculated for each planet in Table 1.

Table 1 Object

Mean Radius of Orbit (m) 10

5.79 × 10   11 1.08 × 10   11 1.49 × 10   11 2.28 × 10   11 7.78 × 10   12 1.43 × 10   12 2.87 × 10   12 4.50 × 10   12 5.91 × 10  

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

Period of Revolution of Orbit (s) 7.60 1.94 3.16 5.94 3.75 9.30 2.65 5.20 7.82

6

× 10   × 107  × 107  × 107  × 108  × 108  × 109  × 109  × 109 

3

∝ T 2

r  C  =

3

r 

2

T 

3

2

(m  /s ) 18

3.36 × 10 18 3.35 × 10 18 3.31 × 10 18 3.36 × 10 18 3.35 × 10 18 3.38 × 10 18 3.37 × 10 18 3.37 × 10 18 3.38 × 10

All proportionality constants calculated in Table 1 are within 1.5% of the average value. This verifies Kepler’s third law. 3 2 18 3 2 11. (a) The average value (in SI base units) of the constant of proportionality in r  ∝ T   is 3.36 × 10  m /s . 18 3 2 (b) C S = 3.36 × 10  m /s  (from (a)) C S

=

 M S

= =

GM S

4π 

2

C S 4π 

2

G

(3.36 ×1018 m3 /s 2 )4π  2 6.67 × 10 −11 N ⋅ m 2 / kg 2

= 1.99 × 1030 kg 30 The mass of the Sun is 1.99 × 10 kg.  M S

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  357

12. (a) r Moon = 3.84 × 10  m 6 T Moon = 2.36 × 10 s 8

C E

r Moon

=

T Moon

2

(3.84 × 10 m) 8

= C E

3

3

(2.36 × 106 s) 2

= 1.02 × 1013 m 3 /s 2

Thus, Kepler’s third law constant, C E, is 1.02 × 1013 m3/s2 for objects orbiting Earth. (b) Let the subscript s represent the satellite. 24  M E = 5.98 × 10  kg  3600 s   = 1.4 × 104 s T s = 4.0 h = (4.0 h)     1 h   Since

C E

=

GM E

4π  2

3

=

r  s

, then

2

T s

GM E

4π  2

3

=

r  s

2

T s

.

2

3

r  s

r  s

=

GM ET s

4π  2

=3

(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 1024 kg)(1.4 × 104 s) 2 4π  2

= 1.3 × 107 m or 1.3 × 104 km 4 The satellite must be 1.3 × 10  km above the centre of Earth. r  s

vs

= =

vs

2π  r  T 

2π  (1.26 ×10 7 m

)

1.4 × 10 4 s

= 5.6 × 103 m/s

The speed of the satellite is 5.6 × 10  m/s. 3

Applying Inquiry Skills 13.

In the sample diagram above, the shape of each figure is approximately that of a triangle, so the approximate areas are: 1 1  A1

= =

2 1 2

b1h1

 A2

= =

(5.0 mm)(54 mm)

= 1.3 × 102 mm 2 In the diagram, d 2 > d 1, but t 2 = t 1, so v2 > v1.  A1

 A2

2 1 2

b2 h2

(12 mm)(22 mm)

= 1.3 × 102

mm 2

Making Connections 14. (a) Let the subscript C represent the central body around which another body (subscript B) revolves in an orbit of known  period and average radius. Since

358 Unit 2 Energy and Momentum

GM C

4π 

2

=

r  B

3

T B

2

, the equation for the mass of the central body is  M C

=

4π  2 r B3 GT B

2

.

Copyright © 2003 Nelson

(b) According to the equation derived in (a) above, G is known and if the mass of the star can be estimated, the period of revolution of the planet must be found in order to solve for the radius of the orbit. Astronomers have discovered that the mass of a star can be estimated by determining its luminosity. (This applies to “main-sequence” stars, by far the majority of stars.) The period of revolution is determined by measuring the period of the wobble of the star as the  planet tugs on the star. The radius of the orbit is found using the equation

r 

=3

GM Star (T Planet )

4π 

2

2

.

Section 6.2 Questions (Page 284)

Understanding Concepts 1.

According to Kepler’s first law, a comet travels in an elongated elliptical orbit. Kepler’s second law implies that the  portion of the comet’s orbit cl oset to the Sun will have a much smaller area than the rest of its orbit. Thus, the time spent in this region (where the comet may be visible to observers on Earth) will be far less compared to the total orbital period. 2. Kepler’s second law states that Earth sweeps out equal areas in equal time intervals. Therefore, on January 4 when Earth is closest to the Sun, it is moving most rapidly because the distance travelled is greatest for equal time intervals. The Earth is moving least rapidly on July 5 when it is farthest from the Sun. 3. Although the nonrotating frame of reference is placed at the centre of the Sun, the Sun is in orbit around the centre of the Milky Way Galaxy, so it is constantly accelerating. This means it is a noninertial frame of reference within the galaxy. 11 4. r  = 4.8 × 10  m 30  M S = 1.99 × 10  kg C S

=

GM S

4π 

(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.99 × 1030 kg)

= C S C S 2

T 

5.

2

4π  2

= 3.36 × 1018 m 3 /s 2 = =

3

r 

2

T 

3

r 

C S

(4.8 × 1011 m)3

T 

=

T 

= 1.8 ×108 s

3.36 × 1018 m 3 /s 2

The orbital period of the asteroid is 1.8 × 10  s. Let the subscript P represent the unknown planet. C S

C S rP

=

3

TP

2

TP rP

=

=

3

T E r P

2

3

T P

2

r E

3

T E

2

= 2T E

3

(2TE )2

r E

8

=

r E

3

T E

2

4TE 2 r E3

r  P

=3

rP

= 3 4r E = 1.6r E

T E

2

The small planet would be 1.6 times farther from the Sun than Earth. Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  359

6.

5.98 × 10  kg 24 r E = 6.38 × 10  m r s = 2r E 24

 M E =

GM E

4π  2 GM E

4π 

(2 r E )

=

2

=

T 

=

3

2

T 

8r E 3 2

T 

32π 

2

r E

3

GM E

32π  2 (6.38 × 106 m)3

=

(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(5.98 × 10 24 kg)

1 h   = 1.43 ×10 4 s     3600s   T  = 4.0 h 7.

The period of revolution is 4.0 h. Let the subscript P represent Phobos, the subscript D represent Deimos, and the subscript M represent Mars.  3600 s   + 18 min  60 s   = 1.09 × 105 s T D = 30 h 18 min = 30 h     1 min    1 h      r D =

2.3 × 104 km

T P =

7 h 39 min = 7 h 

 3600 s   + 39 min  60 s   = 2.75 × 104 s    1 min    1 h     

C M

= =

C M

r D

3

T D

2

(2.3 × 10 m) 7

(1.09 × 105 s) 2

= 1.02 × 1012 s C M into

Substitute the value of C M rP

3

r  P r  P

3

=

r  P

the equation for Phobos:

3

T P

2

= CMT P 2 = 3 (1.02 × 1012 s)(2.75 × 10 4 s) 2 = 9.2 × 106 m

Phobos is 9.2 × 10  m from the centre of Mars. 6

Applying Inquiry Skills 8.

GM  r 

=

= =

(N ⋅ m / kg )(kg) 2

2

m (kg ⋅ m/s2 )(m2 / kg 2 )(kg) m m3 /s 2 m

= m 2 /s 2 = m/s Therefore, the SI base units are metres per second.

360 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

9.

Since r 3 = CT 2 (from Kepler’s third law), the line on the graph is straight and the slope is Kepler’s third law constant for the Sun.

Making Connections 10. (a) Galileo was born near Pisa, on February 15, 1564. In 1609, after learning that a telescope had been invented in Holland, he built his own telescope of 20 times magnification. The strength of this magnification allowed Galileo to see mountains and craters on the Moon, and to discover the four largest satellites of Jupiter. Tycho’s work was done between 1581 and 1601, during which he made numerous naked-eye observations with large instruments. Kepler began analyzing Tycho’s data in 1601. In 1609, Kepler published his first two laws. He  published his third law in 1619. (b) Relating Kepler’s third law and the law of universal gravitation, we find that the mass of Jupiter is given by  M J

=

4π  2 r M 2 GT M

2

, where r M and T M are the radius of the orbit and period of revolution, respectively, of any moon around

Jupiter. Thus, Galileo would need to know the values of r M and T M for at least one moon, as well as the universal gravitation constant, G. (c) Calculating the mass of Jupiter was not possible until the value of G was determined, which was not possible until Kepler’s third law was formed. (As mentioned in Section 3.3, it was Cavendish who first determined that value in 1798.)

6.3 GRAVITATIONAL POTENTIAL ENERGY IN GENERAL PRACTICE (Pages 287–288)

Understanding Concepts 1.

5.98 × 10  kg  = 0.0123 m  M E 5 r  = 3.84 × 10  km

 M E =

24

 E g

=− =− =−

 E g

GM E m r  GM E

2

0.0123

r 

(6.67 ×10−

11

N ⋅ m 2 /kg 2

)(5.98 ×10

24

kg

2

)

(0.0123 )

3.84 × 108 m

= −7.64 × 1028 J

The gravitational potential energy of the Earth-Moon system is –7.64 × 10  J. 2. (a) m = 1.0 kg 6 r E = 6.38 × 10  m 2 5 r  = 1.0 × 10  km = 1.0 × 10  m 28

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  361

On Earth’s surface, GM E m  E g1 = − r E

(6.67 ×10−

11

=−

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)

kg (1.0 kg )

6.38 × 10 m 6

 E g1

= −6.25 × 10

 E g2

=−

7

J

In orbit,

=−  E g2

GM E m r + r E

(6.67 ×10−

11

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)

kg (1.0 kg )

1.0 × 10 m + 6.38 × 10 m 5

6

= −6.15 ×107 J

∆ Eg = Eg2 − E g1 = −6.15 × 107 J − (−6.25 × 107 J) ∆ E g = 1.0 × 107 J The change in gravitational potential energy for a 1.0-kg mass lifted 1.0 × 102 km above Earth’s surface is 1.0 × 106 J (b)  g  = 9.80 N/kg ∆ Eg = mg ∆y

∆ E g

= (1.0kg)(9.80 N/kg)(1.0 × 105 m) = 9.8 × 105 J

The percentage error can be calculated as:

 measured value − accepted value    × 100% accepted value     9.8 ×105 J − 1.0 ×106 J   =    6  × 100% × 1.0 10 J    % error = − 2.0% Thus, the answer calculated using the equation ∆ E g mg ∆ y would be 2.0% less than the answer calculated in (a). % error = 

(c) The low percentage error (less than 2.0%) tells us that there is little need for more exact treatment in most normal Earth-bound problems. The approximation assuming a constant value of  g  is quite good, even for an altitude of 2 1.0 × 10  km. 24 3.  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m

∆ E g, E K =  E g1 where E g1 is the gravitational potential energy on Earth’s surface and  E g2 is the Since E K =      E g2 –  gravitational potential energy in orbit. Thus, 1 2

mv

 GM E m   − −   r E 2rE     1 1   v = 2GM E  −    rE 2r E   2

=−

GM E m

=

2 6.67 × 10 −

(

11

N ⋅ m /kg 2

2

)(5.98 ×10

24

kg

 

v = 7.91× 10 m/s 3

The object must be projected with an initial velocity of 7.91 × 103 m/s. Unit 2 Energy and Momentum

1

) 6.38 ×10

6

m

−

    2(6.38 ×10 6 m)   1

4.

M S = 1.99 × 10  kg 24  M E = 5.98 × 10  kg 11 r  p = 1.47 × 10  m; the distance from the Sun to Earth at perihelion 11 r a = 1.52 × 10  m; the distance from the Sun to Earth at aphelion  E  p where E a is (a) The maximum change in Earth’s gravitational potential energy during one orbit of the Sun is ∆ E g = E a –  the gravitational potential energy at aphelion and  E  p is the gravitational potential energy at perihelion. 30

 GM S M E    −  −     ra r  p     1 1   = GM S M E  −      r p r a  

∆ E g = −

GM S M E

1 1   = (6.67 ×10−11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg )(1.99 ×10 30 kg ) −   11 11  1.47 × 10 m 1.52 × 10 m   ∆ E g = 1.8 × 1032 J 32 The maximum change in Earth’s gravitational potential energy during one orbit of the Sun is 1.8 × 10  J. (b) Earth is moving the fastest at perihelion, which is its closest approach to the Sun. The maximum change in kinetic 32 energy during one orbit is ∆ E K  = ∆ E g = 1.8 × 10  J.

Making Connections 5.

m = 5.00 × 10  kg r 1= 2r E r 2 = 3r E 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m (a) In the satellite’s initial orbit: GM E m  E g1 = − r 1 2

(6.67 ×10 −

11

=−  E g1

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)(

)

)(

)

kg 5.00 ×10 2 kg

2(6.38 × 10 m) 6

= −1.56 × 10

10

J

In the satellite’s final orbit: GM E m  E g2 = − r 2

(6.67 ×10−

11

=−  E g2

N ⋅ m 2 /kg 2

)(5.98 ×10

24

kg 5.00 ×10 2 kg

3(6.38 × 10 m) 6

= −1.04 × 1010 J

The gravitational potential energy in the satellite’s initial orbit is −1.56 × 1010 J. The gravitational potential energy in 10 the satellite’s final orbit is −1.04 × 10  J. (b) ∆ Eg = Eg2 − E g1

∆ E g

= −1.04 × 1010 J − (− 1.56 × 1010 J) = 5.2 ×109 J

The change in gravitational potential energy from the first orbit to the second orbit is 5.2 × 10  J. (c) W  = ∆ E g 9 W  = 5.2 × 10  J 9 The work done in raising the satellite as it moves from the first orbit to the second orbit is 5.2 × 10  J. 9

Chapter 6 Gravitation and Celestial Mechanics

PRACTICE (Page 293)

Understanding Concepts 6.

The escape speed of a space probe depends on the mass of Earth, not on the mass of the probe, according to the equation v=

2GM E

r  7.  M J = 318 M E r J = 10.9 r E

.

2GM J

=

vJ

 and vE

r J

=

2GM E r E

2G (318 M E ) vJ

10.9r E

=

vE

2GM E r E

 2G (318 M E )   r E   =      2GM E   r 10.9 E      318

= vJ

10.9

= 5.40

vE

The ratio of the escape speed from Jupiter to the escape speed from Earth is 5.40:1. 22 8. mMoon = 7.35 × 10  kg 8 r = 3.84 × 10  m (from Earth’s centre) GM E m (a)  E g = − r 

(6.67 ×10−

11

=−  E g

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)(

kg 7.35 ×10 22 kg

)

3.84 × 108 m

= −7.63 ×10 28 J × 1028 J.

The gravitational potential energy of the Moon with respect to Earth is –7.63 (b)  E K =    E T –  E g 1  EK = Eg − E g 2 1 = −  E g 2  E K  = 3.82 × 1028 J

Since EK  =

1 2

mv

v=

=

2

2 E K  m

2(3.82 × 10 J) 28

7.35 × 10 22 kg

v = 1.02 × 10 m/s 3

The Moon’s kinetic energy is 3.82 × 10  J and its speed in orbit is 1.02 × 10  m/s. 28

(c) The Moon’s binding energy to Earth is E T =

Unit 2 Energy and Momentum

3

1 2

 E g = 3.82 × 10  J. 28

9.

m = 2.0 × 10 2 5 r  = 5.0 × 10  km = 5.0 × 10  m 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m W = ∆E = ET (in orbit) − ET (on Earth)   3

∆ E =

1 2

Eg2

− E g1

 GM E m   − −   2 ( r + rE )  r E   1   1 = GM E m  −    rE 2(r + r E )   =−

1 GM E m

   1 1 = (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg )(2.0 ×10 3 kg )  −   6 5 6  6.38 × 10 m 2(5.0 × 10 m + 6.38 × 10 m)   ∆ E  = 6.7 ×1010 J The total amount of energy needed to place the satellite in circular Earth orbit is 6.7 × 1010 J. 1 GM E m

10. The binding energy of the satellite is

2 r + r E

. Thus, the additional energy that would have to be supplied to the satellite

would be:  E  =

1 GM E m

=

1 GM E m

2 r + r E 2 (r + r E )

(6.67 ×10−

11

=

N ⋅ m /kg 2

2

)(5.98 ×10

24

kg

)(2.0 ×10

3

kg

)

2(5.0 ×105 m + 6.38 × 10 6 m)

 E  = 5.8 ×1010 J

Therefore, 5.8 × 1010 J of additional energy would have to be supplied to the satellite to allow it to escape from Earth’s gravitational field. 4 11. T  = 24 h = 24 h × 3600 s/h = 8.64 × 10  s 24 M E = 5.98 × 10  kg GM E C E = 4π  2

(6.67 ×10−

11

=

r

= 1.01×10

3

= CET 2

r

= 3 CET 2

3

m /s

= 3 (1.01× 1013 r  = 4.23 × 107 m v

= =

2

4π  13

C E

N ⋅ m /kg

2

)(5.98 ×10

24

kg

)

2

2

m 3 /s 2 )(8.64 × 10 4 s) 2

2π r  T 

2π (4.23 × 107 m) 8.64 × 10 s 4

v = 3.1× 10 m/s 3 The satellite’s speed in orbit is 3.1 × 10  m/s. 3

Chapter 6 Gravitation and Celestial Mechanics

1 2

 EK

= ∆E =

mv 2

=−

v2

1 2

Eg2 − E g1

2

r

1 = 2GM E   rE

v=

 GM E m   −−   r E    1   −   2 r  

1 GM E m

(

2 6.67 × 10−11 N ⋅ m2 /kg2

)( 5.98× 10

24

 

1

)  6.38 × 10

kg

6

m

−

1 2(4.23× 10

7

    m)  

v = 1.08 ×10 m/s 4

The satellite must reach a speed of 1.1 12. m = 4.00 M S 30  M S = 1.99 × 10  kg

× 104 m/s during launch to get into a geosynchronous orbit.

8 c = 3.00 × 10 m/s

r  =

=

2GM  c

2

2(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(4.00 × 1.99 × 1030 kg) (3.00 ×10 m/s) 8

2

r  = 1.18 × 10 4 m, or 11.8 km The Schwartzschild radius of a black hole is 11.8 km.

Applying Inquiry Skills 13. As shown in Figure 4 on page 288 of the text, a graph of gravitational potential energy as a function of distance from the centre of the body yields a potential well graph.

Making Connections 14. (a) m = 65.0 kg 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m On Earth’s surface: GM E m  E  = r E

(6.67 ×10−

11

=

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)

kg (65.0 kg )

6.38 × 106 m

 E  = 4.06 ×109 J

The binding energy of a 65.0-kg person on Earth’s surface is 4.06

Unit 2 Energy and Momentum

× 109 J.

(b) To just escape from Earth, E K  = binding energy. Thus, the person would need 4.06 × 10  J of kinetic energy. (c)  g  = 9.80 N/kg ∆ y = 1.00 m W = mg ∆y 9

= (65 kg)(9.80 N/kg)(1.00 m) 2 W  = 6.37 × 10 J To raise the person 1.00 m at Earth’s surface would require 6.37 × 10  J of work. (d) One of NASA’s objectives in designing launches into space is to minimize the mass of the payload. Since the binding 2

 

energy is proportional to the mass of the object  binding energy =

GM E m    , the lower the mass, the lower the binding r E  

energy. Minimizing the mass of the payload would therefore require less energy and work to launch the vehicle into space, reduce the amount of fuel needed, and be more cost effective.

Section 6.3 Questions (Page 294)

Understanding Concepts 1.

2.

Escape energy

=

GM E m r E

. Therefore, the escape energy is proportional to the mass of the object. Since the mass of a

1500-kg rocket is three times the mass of a 500-kg rocket, the escape energy of the 1500-kg rocket will also be three times as great as the escape energy of the 500-kg rocket. The statement, “No satellite can orbit Earth in less than about 80 min” is correct. The minimum period of orbit for a satellite corresponds to the radius of Earth. GM E C E = 4π  2

= C E

(6.67 × 10−11 N ⋅ m2 / kg 2 )(5.98 × 1024 kg) 4π 2

= 1.01×1013 m 3 3

C s T E

2

= =

r s

2

T s r E

3

C E

(6.38 ×106 m)3

T E

=

T E

= (5.07 × 103 s)(1 min/60 s) = 85 min

(1.01× 1013 m 3 / s 2 )

Thus, the minimum period of orbit corresponds to 85 min. 3. (a)  M E = 5.98 × 1024 kg 3 mt = 1.2 × 10  kg (mass of the tank) 6 r E = 6.38 × 10  m 6 6 6 r t = 6.38 × 10  m + 2.0 × 10  m = 8.28 × 10  m W  = ?

Chapter 6 Gravitation and Celestial Mechanics

Applying the law of conservation of energy: W = ∆E g

 GM E mt   − −   rE r t     1 1   = GM E mt  −    rt r E   =−

GM E mt

= (6.67 × 10−11 N ⋅ m 2 /kg2 )(1.2 × 103

1   8.28 ×106

kg) 

m

−

1 6.38 × 10

6

    m  

10 W  = −1.7 × 10 J The work done by gravity on the tank is −1.7 × 1010 J. (The gravitational potential energy at Earth’s surface is more negative than it is at higher altitudes.) (b) v = 0 v′ = ?

Since the loss in gravitational potential energy (found above) equals the gain in kinetic energy, ∆ E K  = −∆ E g = 1.7 × 107 J. 2

m ( v′ )

−

2

m (v)

2

= 1.7 × 107

2 2

(v′) =

J

(

2 1.7 × 107 J

)

m

(

2 1.72 ×10 J

v ′ =

7

)

m

(

2 1.7 × 107 J

=

)

1.2 × 10 kg 3

v ′ = 5.4 × 10 m/s The speed of impact at Earth’s surface is 5.4 × 103 m/s. 9 4.  E K  = 5.0 × 10  J 9  E g = −6.4 × 10  J  E T = E K  + E g 9 9 = (5.0 × 10  J) + (−6.4 × 10  J) 9  E T = −1.4 × 10  J 9 The binding energy of the space vehicle is 1.4 × 10  J. 3 5. m = 2.00 × 10  kg 2 5 r  = 4.00 × 10  km = 4.00 × 10  m 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m GM E m (a)  E g = − r + r E 3

(6.67 ×10−

11

=−  E g

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)(

kg 2.00 × 10 3 kg

)

4.00 × 10 m + 6.38 × 10 m 5

6

= −1.18 × 1011 J

Thus, the average value of gravitational potential energy of the satellite in orbit is (b)  E K  = E T – E g 1  E g =  E g –  2 1 = −  E g 2 10  E K  = 5.88 × 10  J 10 Thus, the average value of orbital kinetic energy is 5.88 × 10  J. Unit 2 Energy and Momentum

−1.18 × 1011 J.

 

(c)  E T =

1

 E g 2 10  E T = –5.88 × 10  J 10 Thus, the total energy of the satellite while in orbit is –5.88 × 10  J. (d) Let the subscript p represent the satellite at perigee. 2 5 r  p = 2.80 × 10  km = 2.80 × 10  m GM E m  E gp = − r p + r E

=−  E gp

(6.67 ×10−

11

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)(

kg 2.00 ×10 3 kg

)

2.80 × 105 m + 6.38 × 106 m

= −1.20 × 1011 J

To determine the satellite’s orbital speed at perigee:

1 2

 EK

=−

2

=−

mv

v=

1 2 1 2

E gp

(−1.20 × 1011 J)

1.20 ×1011 J 2.00 ×103 kg

v = 7.74 ×10 m/s 3

The satellite’s orbital speed at perigee is 7.74 × 10  m/s. 2 6. m = 5.00 × 10  kg 2 r  = 2.00 × 10  km 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m GM E m (a)  E g = − r + r E 3

6.67 × 10 − ( =−

11

 E g

N ⋅ m 2 /kg 2

)(5.98 ×10

24

)(

kg 5.00 ×10 2 kg

)

2.00 × 10 m + 6.38 × 10 m 5

6

= −3.03 × 1010 J

The gravitational potential energy of the satellite is −3.03 × 10  J. (b)  E K  = E T – E g 1  E g =  E g –  2 1 10 = −  (−3.03 × 10  J) 2 10  E K  = 1.52 × 10  J 10 Thus, the average value of kinetic energy of the satellite is 1.52 × 10  J. 1 10 (c) binding energy =  E g = –1.52 × 10  J 2 10 Thus, the binding energy of the satellite while in orbit is –1.52 × 10  J. (d) To launch the satellite into orbit: ∆ E  = E T (in orbit) – E T (on Earth) 1 =  Eg2 − E g1 2 10

 GM E m   − −   r E 2 ( r + rE )    1   1 ∆ E = GM E m  −    rE 2(r + r E )   =−

1 GM E m

Chapter 6 Gravitation and Celestial Mechanics

To escape from Earth:

=

 E T

GM E m r E

 GM E m    r     E T  E   = ∆ E  1   1 GM E m  −    rE 2(r + r E )    1    r     E   = 1   1  r − 2(r + r  )    E E   1     6.38 ×106 m      =    1 1  6.38 ×106 m − 2(2.00 × 105 m + 6.38 × 106 m)       E T

∆ E 

= 1.94

The percentage increase in launching energy required for the satellite to escape from Earth is 94%. 30 7. (a)  M S = 1.99 × 10  kg 8 r S = 6.96 × 10  m v = ? (escape speed) v=

=

2GM S r S

2(6.67 × 10

−11

N ⋅ m /kg )(1.99 × 10 2

2

30

kg)

6.96 ×10 m 8

v = 6.18 × 10 m/s 5 The escape speed from our solar system is 6.18 × 10  m/s. (b) The energy needed to escape from the solar system is the addition of the escape energy from Earth and the escape energy from the Sun at the location of Earth’s orbit. The escape energy can be used to determine the escape speed. 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m 11 r E-S = 1.49 × 10  m (radius of Earth’s orbit)  E t = ? (the total escape energy) v = ? GM S m GM E m  E t = + rE-S r   E  5

=  Et

v=

=

N ⋅ m /kg )(1.99 × 10 2

2

30

1.49 × 10

= 9.53 ×10

=

−11

11

8

 Now, E K  = E t = v2

(6.67 × 10

1

( m) J

2

mv

2 2 E K  m

2 E K  m

2(9.53 ×10 ( m) J 8

m

v = 4.37 × 10 m/s The escape speed is 4.37 × 104 m/s. 4

Unit 2 Energy and Momentum

m

kg)(m)

+

−11

(6.67 × 10

N ⋅ m /kg )(5.98 × 10 2

2

6.38 × 10 m 6

24

kg)(m )

8.

mMoon = 7.35 × 10  kg 6 r Moon = 1.74 × 10  m 22

1 2

mv

 GM Moon m   − −   r Moon 2rMoon     1 1   − v = 2GM Moon     rMoon 2r Moon   2

=−

GM Moon m

=

2 6.67 × 10

(

−11

N ⋅ m 2 /kg 2

)(7.35 × 10

22

kg

 

1

) 1.74 ×10

6

m

−

1 2(1.74 × 10

6

    m)  

v = 1.68 × 10 m/s 3

An object projected from the Moon must have a speed of 1.68 × 10  m/s to reach an altitude equal to the Moon’s radius. 30 9.  M S = 1.99 × 10  kg 4 r  = 15.4 km = 1.54 × 10  m 3

c = 3.00 × 10 m/s 8

r  =  M  =

=

2GM  c rc

2

2

2G (1.54 × 10 4 m)(3.00 × 108 m/s) 2 2(6.67 × 10

−11

N ⋅ m /kg ) 2

2

 M  = 1.04 ×10 kg 31

In terms of the Sun’s mass, the mass of the black hole is:  M   M S  M

=

1.04 × 1031 kg 1.99 × 1030 kg

= 5.22 M S

Thus, the mass of the black hole is 5.22  M S.

Applying Inquiry Skills 10. The data for the required graph, generated by using the equation  E g

=

−GM M m r 

, are shown in the table below.

× 109 J) − − − − −

Position

E g (

r  r  r  r  r 

11. (a) M E = 5.98 × 10  kg 24

8 c = 3.00 × 10 m/s

Chapter 6 Gravitation and Celestial Mechanics

r  =

=

2GM E c

2

2(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg) (3.00 ×10 m/s) 8

2

= 8.86 × 10−3 m r  = 8.86mm The theoretical Schwartzschild radius of the black hole is 8.86 mm. (b) The very small radius of the black hole implies that it is extremely dense, according to ρ  ∝ ρ  ρ E

= =

ρ  ρ E

r E

1 3

r 

. Thus,

3

r 3

(6.38 ×106 m)3 (8.86 × 10 −3 m)3

= 3.73 ×10 26

Therefore, the black hole is 3.73 × 1026 times more dense than Earth.

Making Connections 12. The escape energy of an object from the Moon is much less than the escape energy of the same object from Earth. (In fact, calculations show that the ratio of the escape energy from Earth is more than 22 times as great as that from the Moon.) Thus, less fuel is required to send the spacecraft from the Moon to Earth than from Earth to the Moon.

CHAPTER 6 LAB ACTIVITIES Lab Exercise 6.3.1: Graphical Analysis of Energies (Page 295)

Procedure 1.

Changing the numerical values in the given data to megametres and gigajoules is suggested just to make data entry in a software program easier. The results of the graphing and the analysis are not affected by this suggestion. The data are: r  (Mm) E g (GJ)

−

−

−

−

2 – 5. The four lines required on the graph are shown below.

Unit 2 Energy and Momentum

−

−

−

−

−

−

Analysis (a) To analyze the energy data of a spacecraft-moon system, graphs of gravitational potential energy, binding energy, and kinetic energy are useful. All graphs should be as a function of the distance from the centre of the moon, r .

Evaluation (b) Answers will vary. An advantage of a spreadsheet program is that it performs calculations and plots relationships on graphs very quickly. However, a disadvantage is that learning how to use the program for a specific application can be more time consuming.

Synthesis (c) Energy considerations are important in planning space missions in order to determine the amount of fuel required to safely launch and transport the vehicles to and from the destination. The amount of fuel affects the total cost of any mission.

CHAPTER 6 SUMMARY Make a Summary (Page 297)

Details on the diagram will vary. The figure below shows the start of the diagram. Urge students to add enough detail to help them understand and remember as much as possible about the chapter.

CHAPTER 6 SELF QUIZ (Pages 298–299)

True/False 1. 2. 3.

T T T

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  373

4. 5. 6.

T F “Evidence” is to the “Analysis” as Tycho Brahe’s work was to Kepler’s work. F In Figure 1 where the path di stances d 1 and d 2 are equal, the speeds along those path segments are not equal because the area swept out by these paths, and thus the time intervals, are different according to Kepler’s second law. 7. F Kepler’s third law constant for Earth is the same for the Moon as for all satellites in orbit around Earth; the constant depends on Earth’s mass. 8. F The gravitational potential energy of the Earth-Moon system is inversely proportional to the distance between the centres of the two bodies. 9. T 10. F Your escape energy and binding energy are equal because you are at rest, and therefore you have no kinetic energy.

Multiple Choice 11. (a) g ∝ m 1 12. (d) g ∝ r  13. (c) v ∝ 14. (d)

m

v∝

1

r  15. (a) area ∝ ∆t  2 2 16. (c) r ∝ T  or r ∝ T 3 3 2 17. (a) r  ∝ T  1 18. (d) E K  ∝ r  1 19. (e) E g ∝ –  r  3

20. (c) Kepler’s third law allows us to determine Earth’s mass according to the equation C E 21. (c) Since g ∝

1

, then r 2 = 3r 1. Using the equations  g 1

2

r 

 g 2  g1

GM  r 12

 and  g 2

=

GM  3r 12

GM E 4π  2

.

:

 GM     2   2 3r 1    r 1   1    = =    =  GM     3r 1   9  2      r 1   =

 g 2

=

=

1 9

g 1

22. (d) r 1 = 4r 2 v1

=

v1 v2

v1 v2 v1

GM S1 4r S2

 

  =   

GM E

=

r 2

=

4r 2 GM E r 2

and

v2

=

GM 2 r 2

               

4r 2 1 4

= 0.5v2

374 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

23. (c) Let the subscript 1 represent the mass of the Sun at its current value and the subscript 2 represent the mass of the Sun at 1 1 half its current value. Since M S2 =  M S1, C S2 = C S1. Thus, 2 2

=

C S1

C S1 C S2

T 2

3

r 

 and C S2

2

T 

=

3

r 

2

T 

.

 r 3    2   T  =  13    r     2    T 2  

2

=2

T 12 T 2 T 1

=

2

24. (c) A satellite in geosynchronous orbit has a period of revolution of 24 h. 25. (a) The speed of the comet increases as it comes closer to the Sun. Position A is closet to the Sun, and therefore has the greatest speed. Position C is furthest from the Sun, and so has the slowest speed. Positions B and D are equidistant from the Sun and are between positions A and C. Thus, vA > vB = vD > vC. 26. (e) M P = M E 1 r P = r E 4 vE

=

vP vE vP vE vP

2GM E r E

  =    =

  and

8GM E r E 2GM E r E

vP

=

2GM E 1 r E 4

=

8GM E   r E

               

4

= 2vE

CHAPTER 6 REVIEW (Pages 300–301)

Understanding Concepts 1.

The escape energy (and thus the escape speed) from the Sun is much greater than that from Earth, so the rocket given the speed needed to escape from Earth would not have enough speed to escape from the solar system. Space vehicles sent to explore distant planets have a much lower binding energy by the time they reach those distant locations, and could acquire enough energy to escape from the solar system by taking advantage of the force of gravity of the distant planet. 2. Since Earth rotates eastward, an eastward orientation of the rocket as it is being launched means that the rocket already has a component of the required velocity before blasting off. This means that less energy will be needed to launch the rocket eastward than would be required to launch it westward in order to achieve the same speed. 3.  g U = 1.0 N/kg  M U = 8.80× 1025 kg 7 r U = 2.56× 10  m

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  375

 g U

=

r

=

GM U (r + r U ) 2

=

GM U

− r U

 g U

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(8.80 × 10 25 kg)

− 2.56 × 107 m

1.0N/kg

= 5.1× 107 m 4 r  = 5.1× 10 km Uranus has a gravitational field strength of 1.0 N/kg at an elevation of 5.1 23 4.  M  = 1.48 × 10  kg 3 6 r  = 5.55 × 10  km = 5.55 × 10  m GM   g  = 2 r 

=

× 104 km above the surface.

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.48 × 1023 kg) (5.55 × 106 m) 2

 g  = 0.318N/kg The magnitude of Ganymede’s gravitational field strength at a point in space 5.55 5. Use the spacecraft-to-Earth line as the reference for the coordinate system. 24  M E = 5.98 × 10  kg 22  M Moon = 7.35 × 10  kg r E-spacecraft = 3.07 × 108 m 8 r Moon-spacecraft = 2.30 × 10  m

× 103 km from its centre is 0.318 N/kg.

= g E + g Moon  g T, x = g E  g T, y = g Moon 

 g T











 g E

= =



 g E

GM E r 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg) (3.07 × 108 m)2

= 4.23 ×10 −3



 g Moon

= =



 g M 

 gT

= =



 g T

N/kg

GM Moon r 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.35 × 1022 kg) (2.30 × 108 m) 2

= 9.26 × 10−5 g T, x 2

+ g T,y 2

(4.23×10−

= 4.23× 10 −3

376 Unit 2 Energy and Momentum

N/kg

3

N/kg

2

) + (9.26 × 10−

5

N/kg

2

)

N/kg

Copyright © 2003 Nelson

tan θ  =

 g T, y  g T, x

  g    = tan −1  T, y       g T, x   θ  = 1.26° θ 

−3

The total gravitational field strength (magnitude and direction) of the Earth-Moon-spacecraft system is 4.23 × 10  N/kg [1.26° from the spacecraft-to-Earth line]. 6 6. r E = 6.38 × 10  m d M = 0.38d E  g E = 9.8 N/kg  g M = 0.38 g E Since Mercury’s diameter is 0.38 times that of Earth’s, Mercury’s radius is also 0.38 times that of Earth. Therefore, r M = 0.38 r E. GM M  g M = 2 r M  g M

=

GM M 2

(0.38r E )

Substituting g M = 0.38 g E: GM M 0.38 g E = 2 (0.38r E ) 2

 M M

=

(0.38 g E )(0.38r E ) G

2

7.

(0.38 ) (9.80 N/kg) ((0.38)(6.38 × 106 m) ) = 6.67 ×10−11 N ⋅ m 2 /kg 2  M M = 3.3 × 10 23 kg Therefore, Mercury’s mass is 3.3 × 1023 kg. 3 v = 7.15 × 10  m/s 24  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m (a) v = r  =

GM E r  GM E v

2

(6.67 ×10−

11

= r 

)(5.98 ×10 m/s )

N ⋅ m 2 /kg 2

(7.15 ×10

3

24

kg

)

2

= 7.80 × 106 m

In terms of Earth’s radius, the satellite’s distance from Earth’s centre is r  r E r

=

7.80 × 106 m 6.38 × 106 m

= 1.22

r E

Thus, the satellite is 1.22 r E from Earth’s centre.

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  377

(b) altitude = r − r E

= 1.22 rE − r E altit ude = 0.22 r E The satellite has an altitude of 0.22 r E. 4 8. vTethys = 1.1 × 10  m/s 26  M S = 5.67 × 10  kg (a) v = r  =

GM S r  GM S v

2

(6.67 ×10−

11

= r 

)(5.67 ×10 m/s )

N ⋅ m 2 /kg 2

(1.1×10

4

26

kg

)

2

= 3.1× 108 m, or 3.1× 105 km

The orbital radius of Tethys is 3.1 × 10  km. 2π r  (b) T  = v 5

=

(

2π  3.1× 10 m 8

)

1.1× 10 4 m/s

1 h  1 d    = (1.8 × 105  s)       3600 s  24 h    T  = 2.1 d The orbital period of Tethys is 2.1 d. 30 9.  M S = 1.99 × 10  kg 7 T V = 1.94 × 10  m GM S C S = 2 4π  3

r 

T V 2 r 3

= C S =

TV 2 GM S 4π 2 2

r  =

3

=3

TV GM S 4π  2 (1.94 × 10 s) (6.67 × 10 7

2

−11

N ⋅ m /kg )(1.99 × 10

4π 

2

2

30

kg)

2

r  = 1.08 × 1011 m The average Sun-Venus distance is 1.08 × 1011 m. 24 10.  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m 3 v = 9.00 km/s = 9.00 × 10  m/s mR  = 4.60 kg

378 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

(a) r ′ = ? Applying the law of conservation of energy:

+ EK = Eg′ + E K′

 Eg

−

GM E mR

+

rE

−

1 2

GM E rE

mR v

+

1 2

2

v2

GM E

=−

GM E mR

=−

GM E

+0

r ′

 

 

r ′

=

GM E   1

r ′ =

GM E GM E 1

r′

 

−

r E

2

−

r E

v2

2

v

2

(6.67 ×10−

)(5.98 ×10 kg ) N ⋅ m /s )(5.98 × 10 kg ) 1 − (9.00 × 10 2 6.38 × 10 m 11

=

(6.67 ×10−

11

2

N ⋅ m 2 /s 2

2

24

24

6

3

m/s

2

)

r ′ = 1.85 × 10 m 7

Let the altitude be A.  A = r ′ − r E

= 1.85 × 107 m − 6.38 × 106  A = 1.21× 107 m

m

The altitude above Earth’s surface is 1.21 × 107 m or 1.21 × 104 km. (b) At the altitude found in (a), the gravitational potential energy is negative, the kinetic energy is zero (because the speed is zero), and the binding energy, E B′, is the extra energy needed to give the rocket a total energy of zero.  EB′ + Eg′ + E K ′  EB′

=0 = − E g′ GM E mR    = −  −   r ′    =

GM E mR  r ′

(6.67 ×10−

N ⋅ m 2 /s 2

11

=

)(5.98 ×10

24

)

kg ( 4.60 kg )

1.85 × 10 m 7

 E B′  = 9.92 × 10 J 7

The binding energy is 9.92 × 10  J. 23 11.  M T = 1.35 × 10  kg (mass of Titan) r T = 2.58 × 103 km = 2.58 × 106 m 3 mR  = 2.34 × 10  kg (mass of rocket) (a) vesc = ? (escape speed) 7

vesc

= =

vesc

2GM T r T

(

2 6.67 ×10

−11

N ⋅ m /s 2

2

)(1.35 ×10

23

kg

)

2.58 × 10 m 6

= 2.64 × 10

3

m/s

The escape speed from Titan is 2.64 × 10  m/s. 3

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  379

(b)  E esc = ? (escape speed) At the surface of Titan, the rocket is at rest, so its kinetic energy is zero. Thus, its total energy is  E g and the escape energy is the extra energy needed to give the rocket a total energy of zero.  Eg + E esc = 0  Eesc

= − E g  GM T mR    = −−   r T    =

GM T mR  r T

(6.67 ×10−

11

=

N ⋅ m 2 /s 2

)(1.35 ×10

23

kg

)( 2.34 ×10

3

kg

)

2.58 × 106 m

= 8.17 × 109 J The escape energy is 8.17 × 109 J. This value can also be found by using the escape speed of the rocket in the equation  E esc

 EK

=

1

2

mR ( vesc ) .

2 24 12.  M E = 5.98 × 10  kg 6 r E = 6.38 × 10  m mR  = 1.00 × 104 kg 10 r R = 1.00 × 10  m   (a)  E g = ?  E g

=− =−

 E g

GM E mR  r R 

(6.67 ×10−

11

N ⋅ m /s 2

2

)(5.98 ×10

1.00 × 10

10

= −3.99 × 10

8

24

)(

kg 1.00 × 10 kg 4

)

m

J

The gravitational potential energy is −3.99 × 108 J. (b) Since the total energy, E K  + E g, must be at least zero, the kinetic energy needed to escape is +3.99 (c) vesc = ? vesc

=

2GM E r R 

(6.67 ×10−

11

= vesc

× 108 J.

N ⋅ m 2 /s 2

)(5.98 ×10

24

kg

)

1.00 × 1010 m

= 2.82 × 10 2

m/s

The escape speed from this position is 2.82 × 102 m/s. The escape speed can also be found by applying the e scape energy found in (b) to the equation involving the kinetic energy, vesc

=

2 E K  m

, where E K  = E esc.

13.  M S = 1.99 × 10  kg 24  M E = 5.98 × 10  kg 11 r E = 1.49 × 10  m 30

 E g

=− =−

 E g

GM S M E r E

(6.67 ×10−

11

N ⋅ m 2 /kg 2

)(1.99 ×10

30

)(

kg 5.98 × 10 24 kg

)

1.49 × 10 m 11

= −5.33 × 1033 J

The gravitational potential energy of the Sun-Earth system is −5.33 × 10  J. 33

380 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

14. (a) M M = 3.28 × 10  kg 6 r M = 2.44 × 10  m 23

v

2GM M

=

r M

(

2 6.67 ×10

=

−11

N ⋅ m /kg 2

2

)(3.28 ×10

23

kg

)

22

kg

)

2.44 × 10 m 6

v = 4.23 × 10 m/s, or 4.23 km/s The escape speed from Mercury is 4.23 km/s. (b)  M Moon = 7.35 × 1022 kg 6 r Moon = 1.74 × 10  m 3

v

2GM Moon

=

r Moon 2 6.67 ×10 −11 N ⋅ m 2 /kg 2

(

= v

)(7.35 ×10

1.74 × 10 m 6

= 2.37 × 103 m/s, or 2.37 km/s

The escape speed from Earth’s Moon is 2.37 km/s. 15. (a) M star  = 3.4 × 1030 kg r star  =

1.7 × 104 m 2 v

2GM star 

=

r star 

(

2 6.67 ×10

= v

= 8.5 × 103 m

−11

N ⋅ m /kg 2

2

)(3.4 ×10

30

kg

)

8.5 × 10 m 3

= 2.3 × 108 m/s

The escape speed from a neutron star is 2.3 × 108 m/s. (b) c = 3.00 × 108 m/s v c v c

=

2.3 × 108 m/s 3.00 × 10 m/s 8

= 0.77

Thus, the percentage equals

 v  ×100% = 77% .  c     

The escape speed from a neutron star is 77% the speed of light. 5.06 × 10 m 7

= 2.53 × 10 m 2 4 v = 24 km/s = 2.4 × 10  m/s

16. (a) r  =

v=  M  =

7

2GM  r  rv

2

2G 2

(2.53 ×10 m )(2.4 ×10 m ) = − 2 (6.67 × 10 N ⋅ m /kg ) 7

4

11

 M  = 1.1× 10

26

2

2

kg

The planet’s mass is 1.1 × 1026 kg.

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  381

(b) According to the data in Appendix C, the planet is Neptune. (There is a 6% difference in mass from Appendix C.) 30 17.  M S = 1.99 × 10  kg −27  M P = 1.67 × 10  kg 9 r  = 1.4 × 10  m (initial position) 5 v = 3.5 × 10  m/s (initial speed) 9 (a) r ′ = 2.8 × 10  m (final position) v′ = ? (final speed) Applying the law of conservation of energy:  Eg

−

GM S mP

+

r

−

1 2

GM S r

+ EK = Eg′ + E K′ mP v 2

+

1 2

v2

=− =−

GM S mP r ′ GM S r ′

 

1

+

2

2

mP ( v′ )

1 

+ (v ′)2 2

2    (v ′) = 2GM S  −  + v 2 r ′ r 

1

1



v′ =

=

 

 1 − 1  + v 2    r ′ r  

2GM S 

(

−11

2 6.67 × 10

N ⋅ m /s 2

2

)(1.99 × 10

30

kg

)

1

 2.8 × 108

m

−

1 1.4 × 10

9

 + 3.5 × 105   ( m  

2

)

m/s

v′ = 1.7 × 105 m/s The proton’s speed is 1.7 × 105 m/s. (b) vesc′ = ? (escape speed at the final position) vesc′  =

=

2GM S r ′ 2(6.67 × 10

−11

N ⋅ m 2 /kg 2 )(1.99 × 1030 kg) 2.8 × 109 m

vesc′  = 3.1× 105 m/s The escape speed is 3.1 × 10  m/s at the location indicated; this is greater than the speed found in (a), so the proton will not escape. 18. When light strikes a piece of black paper, a small portion of the light is reflected. However, when light strikes a black hole, the light is absorbed, making the black hole even blacker than black paper. 11 19. m = 1.1 × 10  M S 30  M S = 1.99 × 10  kg 5

c = 3.00 × 10 m/s 8

r  =

=

2GM  c2 2(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(1.1× 1011 (1.99 × 1030 kg)) (3.00 × 108 m/s)2

r  = 3.2 × 1014 m 14 The Schwartzschild radius of the black hole is 3.2 × 10  m.

Applying Inquiry Skills 20.

Table 1 provides

the missing answers concerning some of the moons of Uranus.

(a) Kepler’s third law constant for Uranus (C U) can be calculated using t he ratio

r 3 2

T 

.

(b) The average of the C U values of the calculations in (a) is 1.48 × 10  m /s . 14

382 Unit 2 Energy and Momentum

3

2

Copyright © 2003 Nelson

(c)  M U = 8.80 × 10  kg GM U C U = 4π 2 25

= C U

(6.67 × 10 −11 N ⋅ m 2 / kg 2 )(8.80 × 10 25 kg) 4π  2

= 1.49 × 1014 m 3 /s 2

The values agree (0.7% difference). (d) Table completed using equations T  =

r 3 C U

and r

= 3 CUT 2

.

(e) Students who speculate that only the larger moons can be observed using Earth-based telescopes are right. Thus, only the larger moons were discovered hundreds of years ago. Students who research the physical data of the moons that orbit Uranus will find that Titania and Oberon have diameters greater than 1500 km, whereas all the other moons listed are less than 100 km in diameter.

Table 1 Data of Several Moons of the Planet Uranus for Question 20 r average (km)

Moon

Discovery

Ophelia Desdemona Juliet Portia Rosalind Belinda Titania Oberon

Voyager 2 (1986) Voyager 2 (1986) Voyager 2 (1986) Voyager 2 (1986) Voyager 2 (1986) Voyager 2 (1986) Herschel (1787) Herschel (1787)

T  (Earth

4

5.38 × 10 4 6.27 × 10 4 6.44 × 10 4 6.61 × 10 4 6.99 × 10 4 7.52 × 10 5 4.36 × 10 5 5.85 × 10

days)

0.375 0.475 0.492 0.512 0.556 0.621 8.66 13.46

3

2

C U (m  /s

)

14

1.48 × 10 14 1.46 × 10 14 1.48 × 10 14 1.48 × 10 14 1.48 × 10 14 1.48 × 10 14 1.48 × 10 14 1.48 × 10

21. (a) Some students may think the problem makes sense. However, many students will realize that the (theoretical) radius of an orbit that has a period of 65 min would be less than Earth’s radius. (Students may recall that the typical orbital  period of a satellite in lo w-altitude orbit is about 80 min. For example, see question 22 on page 168 of the text.) 24 (b)  M E = 5.98 × 10  kg 3 T  = 65 min = (65 min)(60 s/min) = 3.90 × 10  s r  = ? r 3 2

T 

r3

=

GM E 2

4π   GM E

=  

r  =

3

4π 

2

  2  T   

GM ET 2 4π 2

(6.67 ×10−

11

=3

N ⋅ m 2 /s 2

)(5.98 ×10

24

)(

kg 3.90 ×10 3 s

)

2

4π  2

r  = 5.36 × 106 m The theoretical radius of the orbit is 5.36 × 106 m. 6 (c) Earth’s radius (6.38 × 10  m) is larger than the theoretical radius found in (b), so the calculated orbit cannot exist. (d) The skill of analyzing a situation is valuable in order to reduce the chances of wasting time on calculations that don’t make sense and to increase the chances of being able to estimate whether or not a solution to a problem is logical.

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  383

22. (a) The rocket’s mass can be calculated from the gravitational potential energy at rest (on Earth’s surface at r E). From the 10 11 graph E g = –10 × 10 J = –1.0× 10 J. −GM E m  E g = r  −rE g m= GM E

=

− (6.38 × 106 m )(−1.0 × 1011 J )

(6.67 ×10−

11

N ⋅ m 2 /kg 2

)(5.98 ×10

24

kg

)

m = 1.6 × 103 kg The rocket’s mass is 1.6 × 10  kg. (b) The escape energy can be determined using the value of gravitational potential energy at rest (1.0 × 1011 J). (c) The launch speed of the rocket can be calculated using the value of the initial kinetic energy E K (on Earth’s surface at   10 11 r E). From the graph E K = –12 × 10 J = –1.2× 10 J.   1  EK  = mv 2 2 3

v=

=

2 E K  m 2(1.2 × 10 J) 11

1.6 × 10 kg 3

v = 1.2 × 10 4 m/s The launch speed is 1.2 × 10  m/s. 10 (d) Extrapolating from the graph, the kinetic energy E K,  approaches 2.0 × 10  J as the distance approaches infinity, where 10 10  E g would approach zero. This can be approximated: at 5r E, the kinetic energy is 4.0 × 10  J and E g is –2.0 × 10  J. 1 2  EK  = mv 2 4

v=

=

2 E K  m 2(2.0 ×10 J) 10

1.6 × 103 kg

v = 5.0 × 103 m/s The speed is 5.0 × 103 m/s.

Making Connections 23. (a) Turning the high-speed craft around would require a fairly large amount of energy, so mission control decided to have the craft continue on toward the Moon. The idea was to take advantage of the Moon’s gravity to act as a sort of slingshot to help the craft accelerate in turning around and begin its return journey at the highest speed possible. (b) One major risk was the chance that there would not be enough electrical power available to guide the craft around the Moon at the most crucial times.

Extension 24. Let L represent the large planet and S represent the small planet. r L = 2r S  DL = DS (densities)

384 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

Thus, using V  for volume, the ratio of the masses is: mS DSV S

=

mL

=

DLV L V S V L

 4  π r 3  3   S =     4  π r  3  3   L     r S3   =  3      r L   3 1    =     2  

mS mL

The centripetal acceleration of the satellite is caused by the force of gravity in each case. Thus, using magnitudes:

Σ F = mac = GMm r2 r3 2

T 

2

T 

r 3

mv 2 r 

=

4π 2 mr

 

2

T 

2

=

4π  mr  

=

GM  4π 

2

=

4π 

2

T 2

GM 

 T L    4π 2    3       r L  =  GmL     T S2   4π 2     3   Gm     r S    S   2

T L

3

2

rL3

×

rS

T S 2 TL 2 TS

2

TL

2

TL 2 TL 2 TL

= =

mS mL r L3 r S 3

×

mS mL

 r  3   mS   = T S2  L3        r S    mL   3 3 22 1   = T S       1 2   = T S2 = T S

The shortest possible period is 40 min. 11 11 11 25. Since the radius of the path is 2.0 × 10  m, the distance between the stars is 2(2.0 × 10  m) = 4.0 × 10  m, 30  M S = 3.0 × 10  kg (mass of each star). The only force acting on each star is the force of gravity of the other star, which causes the circular motion of one star around the other.

Copyright © 2003 Nelson

Chapter 6 Gravitation and Celestial Mechanics  385

Σ F = mac GM S M S 2

( 2r )

GM S 4r 2 GM S 4r  GM S 4r GM S 4r  2

T 

M S v2

=

r  v

=

2

r 

= v2 2

=  

2π r  

   

T 

4π 2 r 2

=

T 2 2 3

16π  r 

=

GM S 2 3

16π  r 

T  =

GM S 16π 

=

(6.67 ×10−

11

3

( 2.0 ×10 m ) N ⋅ m /s )(3.0 × 10

2

11

2

2

30

kg

)

T  = 7.9 × 10 s The period of one complete cycle is 7.9 × 107 s. 7

26.  E  is the amount of energy per unit area, and that area is proportional to

1 r 2

, where r  is the distance from the Sun to the

 planet, so:  E ∝  E  =

1 r 2 k  r 2

From Kepler’s third law: r 3 T 2 r3

= C  = CT 2 2

r

2

where C  is Kepler’s third law constant for the Sun 4

= C 3T 3

Substituting into the first equation: k   E  = 2 4 C 3 T 3 −2

−4

3

T  3

 E

= kC

 E

= (constant)T  3

−4 −4

Thus, E  is proportional to T  3 . (Solving for the “constant” is unnecessary.)

386 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

UNIT 2 PERFORMANCE TASK SAFETY IN TRANSPORTATION AND SPORTS (Pages 302–303)

The three options in this task relate most directly to the principles studied in Chapters 4 and 5. Thus, a good time to discuss the options with the students is at the start of Chapter 5. Share with the students the method of assessment you intend to use with the task. The first two options are research-oriented. Option 1 involves a sport, activity, or piece of equipment of the student’s own choice. Option 2 is somewhat more complex because the students are required to identify the social issues involved before they research and analyze them. Option 3 is a hands-on option that is an alternative to the typical egg-drop device.

Option 1: Protective Equipment in Sports and Recreational Activities In this option, some students will analyze the safety equipment used in a particular sport or activity, such as hockey or  bicycling. Other students will analyze the physics principles involved in the use of a particular piece of equip ment, such as a safety helmet, used in several sports and activities. In both cases, students should relate what they discover to the principles  presented in Chapters 4 and 5, especially collisions in which energy and momentum play an important role. Students will likely find that the Internet is the best resource for this task.

Option 2: Vehicle Safety Features Vehicle safety is an important issue in our society, and students will benefit from analyzing the issue and relating it to the  physics principles in Chapters 4 and 5. Energy, momentum, and collisions are the topics most closely related to vehicle safet y. Students can refer to Appendix A3, text pages 765–766, for suggestions about decision making and defining, researching, and analyzing issues. Students can find information about vehicle safety in reference books, in automobile magazines, and on the Internet.

Option 3: The Bouncing Egg This option applies many of the principles of Chapters 4 and 5, especially the law of conservation of energy, Hooke’s law, and the physics of collisions. The task can be started in conjunction with Section 4.5. As in the case with the model roller coaster (Unit 1 Performance Task, Option 1), it is wise to develop the criteria used to evaluate the success of the device before the students begin their task. Some of the quantitative criteria might be: • height of the drop • number of clear bounces • ability of the device to remain within a horizontal circle of predetermined diameter during its bounces • a maximum allowable mass • maximum allowable outside dimensions (e.g., the device must be able to fit inside a 4-L bucket) Some of the qualitative criteria might be: • originality/creativity • obvious protection of the egg • wise use of materials • aesthetic appeal Students can refer to Appendix A4, text page 767, for a summary of technological problem solving.

Analysis As in the Unit 1 Performance Task, the answers to the Analysis questions depend on various factors, particularly the choice of options. With any of the three options, students are urged to create their own questions and answer them, a feature that helps to make the task more open-ended.

Evaluation Here the students evaluate their own task. Answers will depend on the option chosen.

Copyright © 2003 Nelson

Unit 2 Performance Task  

387

UNIT 2 SELF QUIZ (Pages 304–306)

True/False 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

F One joule is one kilogram metre squared per second squared. T F The thermal energy will be the same if the size of the force of kinetic friction is the same. F The impulse is the change in momentum. T F The only type of collision in which momentum is not conserved is one with a net external force. F The gravitational field strength is inversely proportional to the square of the distance. F The work done on the satellite by Earth is zero. F The geocentric model has Earth at the centre of the universe. F The Sun is located at one focus of a planets orbit. F The speed in orbit depends on the location in the elliptical orbit. F A black hole has an extremely strong gravitational field. T F No form of electromagnetic radi ation can escape.

Multiple Choice 15. (c) Assuming a mass of 75 kg, and a height of 1.8 m: W = ∆E g

= mg ∆y = (75 kg)(9.8 m/s2 )(1.8 m) W  = 1.3 × 103 J 3

The power of 10 is 10 .

Σ F y = ma = 0  F N + FA sin φ − mg cos β  = 0  F N = mg cos β − F A sin φ  17. (a) Σ F x = ma = 0  FA cos φ − mg sin β  − F K  = 0  F + mg sin β   F A = K  16. (c)

cos φ 

388 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

18. (d) W W

= ( F cosθ )∆d  = FA cos φ ( L)

Since sin β  =  L =

h  L h sin β 

Therefore, W  =

 FA h cos φ  sin β 

.

19. (e) The force is always perpendicular to the motion. 1 m   20. (c)  EK  =  m +  v 2 2 5  

= (0.5)(1.2m )v 2  EK  = 0.6 mv2 ′ =  EK 

1

m+ 2

m 2 +   (0.80v)   5 5  

m

= (0.5)(1.4m)(0.64v 2 ) = 0.448mv 2 = 0.75(0.6mv 2 )  EK′ = 0.75 E K  21. (e) All three stones have the same initial total energy, and they all lose the same amount of gravitational energy. Therefore, they will all have the same increase in kinetic energy when they reach the water (even though they will reach the water at different times). 22. (d) 23. (c) To conserve momentum, the same mass must have the same velocity if all of the energy was given to the billiard ball that was stationary. 24. (a) Escape speed is defined as the speed needed to escape from the surface. The current kinetic energies of the rockets are not relevant. 25. (c) The initial kinetic and gravitational energies are the same, giving a total energy of zero. 26. (a) Escape energy is defined as the energy needed to escape from the surface. The current kinetic energies of the rockets are not relevant. 1 27. (d) The gravitational field is proportional to 2 . r  28. (d) r ∝

3

M 

Completion 29. (a) Galileo Galilei (b) Johannes Kepler (c) James Prescott Joule (d) Tycho Brahe (e) Robert Hooke (f) Karl Schwartzschild 30. (a) work (b) the force constant of a spring (c) impulse (d) force (e) thermal energy (f) the mass of Earth 31. completely inelastic; equals; completely inelastic collision 32. zero 33. a singularity; Schwartzschild radius

Copyright © 2003 Nelson

Unit 2 Self Quiz   389

Matching 34. (a) A (b) E 35. impulse law of conservation of momentum kinetic energy thermal energy elastic potential energy escape speed gravitational potential energy Kepler’s third-law constant frequency of a mass spring system in SHM

390 Unit 2 Energy and Momentum

(e) (g) (h) (j) (k) (d) (b) (a) (m)

Copyright © 2003 Nelson

UNIT 2 REVIEW (Pages 307–311)

Understanding Concepts 1. 2. 3.

One situation is a person carrying a book at a constant height across a level floor. A satellite in circular orbit or pushing on a brick wall are two other examples. The work done is transformed into another form of energy. One example is pushing a crate across a level floor. The work done is transformed into thermal energy through friction. Momentum and energy can be related as follows:  E K  =

=  E K  =  p =

4.

5. 6. 7.

8.

9.

mv 2 2 m2 v 2 2m  p 2 2m 2mE K 

If both the baseball and the shot have the same kinetic energy, the shot will have a larger momentum because it has the larger mass. If both objects have the same kinetic energy, they both require the same amount of work done on them to stop them. If the frictional force is the same and the work done is the same, the distance the frictional force acts is the same. Therefore, the two toboggans will have the same stopping distance. During the collision, the initial kinetic energy of the carts is stored as elastic potential energy in the spring. Several common devices that can store elastic potential energy are an elastic band, a bow for shooting arrows, golf balls, and a bungee cord. Bumpers made of springs are impractical because the kinetic energy of the collision would be stored in the spring, and then converted back into kinetic energy again. This last stage could project the car into oncoming traffic or concrete  barriers and cause further collisions. One possibility is:

m = 1.5  103 kg ∆d  = 0.5 m 5  F  = 3.5  10  N h=?

Copyright © 2003 Nelson

Unit 2 Review   391

Using conservation of energy:

= E T′ mg ∆y = F ∆ d   mg ( h + 0.50 m) = F ∆d    F ∆d  ( h + 0.50 m) =  ET

mg 

h=

=

 F ∆d  mg 

− 0.50 m

(3.5 ×105 N )(0.50 m ) (1.5 × 103 kg)(9.8 N/kg)

− 0.50 m

h = 11 m The pile driver must start from a height of 11 m above the pile. 10. m = 10.0 kg (a) d  = b = 2.00 m W  = ? The work done is equal to the area under the graph, therefore: 1 W = bh 2 1 = (2.00 m)(10 N) 2 W  = 1.0 × 101 J 1 The amount of work done is 1.0  10  J. (b) d  = b = 3.00 m  E K =   ? The change in E K  will be equal to the work done, therefore:  EK  = W 

= =

1 2 1 2

bh + lw (2.00 m)(10 N) + (10 N)(3.00 m − 2.00 m)

 E K  = 2.0 × 101 J The block’s kinetic energy is 2.0 (c) d  = 3.00 m  E K =   20 J v = ? 1  EK  = mv 2 2 v=

=

 101 J, or 20 J.



2 E K  m 2(20 J) (10.0 kg)

v = 2.0 m/s [W] The velocity at the 3.00-m mark is 2.0 m/s [W].

392 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

4

11. W  = E K  = 1.25  10  J v = 50.0 m/s (a) m = ?  EK  = m=

=

1

mv2

2 2 E K  v2

2(1.25 × 104 J) (50.0 m/s)

2

m = 10.0 kg (b)

∆d  =

5.00 m

 F  = ? W

= F ∆d 

 F  =

=

W 

∆d  1.25 × 10 J 4

5.00 m

 F  = 2.50 × 10 N [E] 3 A constant force of 2.50  10  N [E] would give the object the same final velocity. 12. m = 5.0 kg 2  E K  = 5.0  10  J  p = ? 3

 E K  =

=  E K  =

mv 2 2 2 2

m v

2m  p 2 2m

 p =

2mE K 

= 2(5.0 kg)(5.0 × 102  p = 71 kg ⋅ m/s The momentum of the sled is 71 kg ⋅m/s.

J)

13.  x1 = 0.10 m  F  = 5.0 N m = 4.5 kg v = 2.0 m/s  x2 = ? First we must find the force constant of the spring:  F = kx k  =

=

 F   x 5.0 N

0.10 m k  = 50 N/m

Copyright © 2003 Nelson

Unit 2 Review   393

Using the law of conservation of energy:  ET = E T′ 1 2

mv 2

=

1 2

kx 2 mv

 x =

2

k  (4.5 kg)(2.0 m/s)2

=

50 N/m

 x = 0.60 m The maximum compression of the spring will be 0.60 m. 14.  x = 2.00 cm = 0.0200 m (a)  E e = ? 1 2  Ee = kx 2 1 = (50.0 N/m)(0.0200 m)2 2  E e

= 1.00 ×10−2

J  –2

The elastic potential energy is 1.00  10  J. (b)  xi = 0.0200 m  xf  = 6.00 cm = 0.0600 m ∆ E e = ? 1 1 ∆ Ee = kxf2 − kxi2 2 2 1 = k ( xf2 − xi2 ) 2 1 = (50.0 N/m) ( 0.0600 m)2 2

(

∆ E e = 8.00 ×10−2

J

The change in elastic potential energy is 8.00 (c) m = 4.00 kg v = ?  ET = E T′ 1 2

kx 2

=

v

= =

1 2

− ( 0.0200 m)2 )

 –2

 10  J.



mv 2 kx

2

m (50.0 N/m)(0.0600 m)2 0.400 kg

v = 0.671 m/s The cart leaves the spring at a speed of 0.671 m/s. 15. When a small object bounces off a larger stationary object, the change in momentum is greater than if the object sticks to it. If the collision takes the same amount of time, the force applied will therefore have to be much larger (twice as much). When riot police use rubber bullets, the rubber bullet exerts a large force on the person or object it hits. 16. When the net force consists of only one force, impulse is a very useful tool in analyzing the changes in motion produced  by that force.

394 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

17. m = 78 kg ∆t  = 1 min = 60 s v1 = 24 m/s v2 = 0 m/s  F  = ?

Σ F ∆t = ∆p  F  =

=

m(v2

− v1 )

∆t  78 kg(0 m/s − 24 m/s) 60 s

 F  = −31 N The magnitude of the force is 31 N. 18. Let the subscript D represent the dog and W represent the wagon. mD = 9.5 kg vD = 1.2 m/s vW = –3.0 m/s mW = ? Using the conservation of momentum:  p = p ′ 0 = mD vD′ mW

mW

+ mW vW′ −m v ′ = DD ′ vW −(9.5 kg)(1.2 m/s) = −3.0 m/s = 3.8 kg

The mass of the wagon is 3.8 kg. 19. m1 = 2.4 kg v1 = 1.5 m/s [W] m2 = 3.6 kg (a)  E T = ?  ET = E K 

= =

1 2 1

2

m1v1

(2.4 kg)(1.5 m/s)2

2  E T = 2.7 J

The total energy of the system before the system is 2.7 J. (b) v´ = ?  p = p ′ m1v1 + 0 = m1v1′ + m2 v2′ At minimum separation, the two velocities of the carts will be the same, therefore: m1v1 = ( m1 + m2 )v′ v′ =

=

m1v1 m1 + m2 (2.4 kg)(1.5 m/s) 2.4 kg + 3.6 kg

v′ = 0.60 m/s [W] The velocity of each cart will be 0.60 m/s [W].

Copyright © 2003 Nelson

Unit 2 Review   395

(c) ∆ E K  = ?

′ − E KT ∆ EK = EKT 1 1 1 =  m1v1′2 + m1v2′2  −  m1v12     2 2  2    1 1 1 =  (2.4 kg)(0.60 m/s)2 + (3.6 kg)(0.60 m/s)2  −  (2.4 kg)(1.5 m/s)2  2 2  2  ∆ E K  = −1.6 J

     

The change in total kinetic energy of the system is –1.6 J. (d)  x = 12 cm = 0.12 m k  = ? 1  Ee = kx 2 2 2 E  k  = 2e  x 2(1.62 J)

=

2

(0.12 m )

k  = 2.2 × 10 N/m 2 The force constant of the spring is 2.2  10  N/m. 20. m1 = 1.2 kg m2 = 4.8 kg 3 k  = 2.4  10  N/m v2´ = 2.0 m/s  x = ? 2

First we must find the speed of the 1.2-kg trolley:  p = p′ 0 = m1v1′ + m2 v2′

−m2 v2′

v1′ =

m1

−(4.8 kg)(2.0 m/s)

=

(1.2 kg)

v1′ = − 8.0 m/s We can now calculate the force constant:  Ee = EK1 + E K2 1 2

kx

2

=

1 2

+

m1v1′

+ m2 v2′2

2

2

 x =

=

1

m1v1′

2

m2 v2′

2

k  (1.2 kg)(−8.0 m/s)2

+ (4.8 kg)(2.0 m/s)2

2.4 × 103 N/m

 x = 0.20 m The force constant of the spring is 0.20 m. 21. Choose right as positive. m1 = 15 kg v1 = –6.0 m/s m2 = 25 kg v2 = 3.0 m/s v2 ´ = ?

396 Unit 2 Energy and Momentum

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(a) v1´ = –0.30 m/s

 p = p ′

= m1v1′ + m2 v2′ m v + m2 v2 − m1v1′ v2′ = 1 1

m1v1 + m2 v2

m2

=

m2 v2

+ m1 (v1 − v1′ ) m2

= v2 +

m1 (v1 − v1′ ) m2

= 3.0 m/s + v2′

15 kg( −6.0 m/s − ( −0.30 m/s)) 25 kg

= −0.42 m/s [right], or 0.42 m/s [left]

The velocity of the 25-kg object would be 0.42 m/s [left]. (b) v1´ = 0.45 m/s  p = p ′

= m1v1′ + m2 v2′ m v + m2 v2 − m1v1′ v2′ = 1 1

m1v1 + m2 v2

m2

=

m2 v2

= v2 +

+ m1 (v1 − v1′ ) m2

m1 (v1 − v1′ ) m2

= 3.0 m/s + v2′

15 kg (−6.0 m/ s − 0.45 m/s) 25 kg

= −0.87 m/s [right], or 0.87 m/s [left]

The velocity of the 25-kg object would be 0.87 m/s [left]. (c) Since the objects stick together, their final velocities will be equal to each other, therefore:  p = p ′

= m1v1′ + m2 v2′ m1v1 + m2 v2 = ( m1 + m2 ) v′ m v + m2 v2 v ′ = 1 1 m1 + m2 (15 kg)(−6.0 m/s) + (25 kg)(3.0 m/s) = 15 kg + 25 kg v ′ = −0.38 m/s [right], or 0.38 m/s [left] m1v1 + m2 v2

The velocity of the 25-kg object would be 0.38 m/s [left]. 22. m1 = 1.4  104 kg 4 v2´ = 2.5  10  m/s 1 v1´ = 2.0  10  m/s First we must calculate the mass of the sled and the rocket combined:  p = p ′ 0 = m1v1′ + m2 v2′

−m1v1′ v2′ −(1.4 × 104 kg)(50 m/s) = −2.5 × 104 m/s m2 = 28 kg m2

=

The total mass required is 28 kg. At 10 kg/s, it will take 28 ÷ 10 = 2.8 s  to burn that much fuel. Copyright © 2003 Nelson

Unit 2 Review   397

23. m1 = 2.3 kg v1 = 18 m/s [E] v2 = 19 m/s [W] v´ = 3.1 m/s m2 = ? Choose east as positive. Using the conservation of momentum:  p = p ′ m1v1 + m2 v2

= m1v1′ + m2 v2′

′ = v2′ = v′ , therefore:

Since it is a completely inelastic collision v1

− m2 v′ = m1v′ − m1v1 m2 (v2 − v ′) = m1 (v′ − v1 ) m ( v′ − v1 ) m2 = 1 (v2 − v′)

m2 v2

= m2

(2.3 kg)(3.1 m/s − 18 m/s)

−19 m/s − 3.1 m/s

= 1.6 kg

The mass of the second bird is 1.6 kg. 24. v1 = v v2

=0

v1′ = −

v

5 m2 = ?  p = p ′

= m1v1′ + m2 v2′ m1v1 + 0 = m1v1′ + m2 v2′ m1v1 − m1v1′ = m2 v2′ m1 ( v1 − v1′ ) = m2 v2′ m v′ m1 = 2 2 (Equation 1) v1 − v1′

m1v1 + m2 v2

 ET 1 2

m1v12

+

1 2

m2 v22

= E T′ =

1 2

m1v1′2

+

1 2

m2 v2′2

+ 0 = m1v1′ + m2 v2′2 m1v12 − m1v1′2 = m2 v2′2 m1 (v12 − v1′2 ) = m2 v2′2 2 m v′ m1 = 2 2 2 2 v1 − v1′ m2 v2′2 m1 = (v1 + v1′ )(v1 − v1′ ) m1v12

398 Unit 2 Energy and Momentum

2

(Equation 2)

Copyright © 2003 Nelson

Set Equations 1 and 2 equal to each other: m2 v2′ v1 − v1′

=

1=

m2 v2′2 (v1 + v1′ )(v1 − v1′ ) v2′ v1 + v1′

v2′

= v1 + v1′ v = v +  −      5  

v2′

=

4 5

v

(Equation 3)

Substitute Equation 3 into Equation 1: m v′ m1 = 2 2 v1 − v1′

 4 v     5    =  v   v −  −    5    4mv      =  5    6v    5      m

m1

=

2 3

m

The mass of the second nucleus is

2

m. 3 25. The total momentum before and after must be the same. The two initial momentum vectors will be the components of the final momentum vector. The dotted line on the diagram represents the direction “after.” A completely inelastic collision means they will stick together.

m2 = 2.0  103 kg v2  = 2.4  101 m/s [E] 

3

m1 = 3.6  10  kg v1 = 1.0  101 m/s [S] 

v = ?

Copyright © 2003 Nelson

Unit 2 Review   399

First we must calculate the initial momentum of each vehicle. For the truck:  p1 = m1 v1 



= (3.6 × 103 kg)(1.0 × 101 m/s) = 3.6 × 104 kg ⋅ m/s

 p1 

For the car:

= m2 v2 = (2.0 ×103 kg)(2.4 × 101 m/s) = 4.8 × 104 kg ⋅ m/s

 p2 



 p2 

The momentum of the car and truck coupled together after the collision:

′  p12 

=

p1 

2

+

p2 

2

= (3.6 × 104 kg ⋅ m/s) 2 + (4.8 × 104 kg ⋅ m/s)2 ′ = 6.0 ×104 kg ⋅ m/s  p12 

To calculate the final velocity of the car and truck coupled together:

′  p12

= m12 v12′  p ′ ′ = 12 v12





m12

=

6.0 × 104 kg ⋅ m/s 3.6 ×10 kg + 2.0 ×10 kg 3

3

= 10.714 m/s ′ = 11 m/s v12 

tan θ  =

 p1  p2 

−1   p1

         p2   4 −1  3.6 × 10 kg ⋅ m/s   = tan    4  4.8 ×10 kg ⋅ m/s   θ  = 37° 

θ  = tan



The final speed of the cars is 11 m/s [37º S of E]. 26. (a) m1 = 2.3  104 kg v1 = 15 m/s [51° S of W] m2 = 1.2  104 kg v´ = 11 m/s [35° S of W]

400 Unit 2 Energy and Momentum

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90° = 35° + θ  + 39°

θ  = 16° Units have been omitted until the final step for clarity. Using the cosine law: 2

 p2

m22 v22 v2

= p12 + p12′2 − 2 p1 p12′ cos θ  = m12 v12 + m122 v12′2 − 2(m1v1 )(m12 v12′ ) cosθ  = =

m12 v12

+ m122 v12′2 − 2( m1v1 )(m12 v12′ ) cosθ  m22

(2.3 × 10 4 ) 2 (15)2

+ (2.3× 104 + 1.2 × 104 ) 2 (11) 2 − 2(2.3× 104 )(15)(2.3× 104 + 1.2× 104 )(11) cos16° (1.2 × 104 ) 2

= 9.0871 m/s v2 = 9.1 m/s sin φ  p1

=

sin θ  p2

  p1 sin θ        p2    m v sin θ    = sin −1  1 1    m2 v2   4 −1  (2.3× 10 kg)(15 m/s)sin16°   = sin    4  (1.2 × 10 kg)(9.0871 m/s)   φ  = 61° φ  = sin −1 

Since 61º – 35º = 26º, the initial velocity of the second truck was 9.1 m/s [26º N of W]. (b) % lost = ?  E − E T′ × 100% % lost = T  E T

 1 m v 2 + 1 m v 2  −  1 m v ′2 + 1 m v ′ 2    2 1 1 2 2 2  2 1 1 2 2 2       × 100%   = 1 1 m1v12 + m2 v22 2

2

− m1v1′ − m2 v2′2 = × 100% + m2 v22 m1v12 − m1v1′2 + m2 v22 − m2 v2′ 2 = × 100% m1v12 + m2 v22 m ( v2 − v′2 ) + m ( v2 − v′2 ) = 1 1 1 2 2 22 2 m1v1 + m2 v2 2 m1v1

+

2 m2 v2 m1v12

(

2

(2.3 × 10 kg) (15 m/s ) 4

=

2

− (11 m/s )2 ) + (1.2 × 104 kg) ((9.0871 m/s )2 − (11 m/s )2 )

( 2.3 × 10 kg )(15 m/s) 4

2

+ (1.2 × 104 kg )(9.0871 m/s)2

× 100%

% lost = 31% The percentage of the initial kinetic energy lost is 31%.

Copyright © 2003 Nelson

Unit 2 Review   401

27. θ  = 180º – 25º = 155º m1 = 82 kg v1 = 8.3 m/s [N] m2 = 95 kg v2 = 6.7 m/s [25° W of N] m3 = 85 kg v´ = ?

 p ′

= p12 + p22 − 2 p1 p2 cos θ  m2 v′2 = m12 v12 + m22 v22 − 2( m1v1 )( m2 v2 ) cosθ  2

v ′ =

=

m12 v12

+ m22 v22 − 2( m1v1 )(m2 v2 ) cosθ  m 2

2 2

(82 kg) (8.3 m/s)

+ (95 kg)2 (6.7 m/s)2 − 2(82 kg)(8.3 m/s)(95 kg)(6.7 m/s)cos155° (82 kg + 95 kg + 85 kg) 2

v ′ = 4.9 m/s sin φ  p2

=

sin θ  p′

  p2 sin θ        p′   m v sin θ    = sin −1  2 2    mv′    (95 kg)(6.7 m/s)sin155°   = sin −1     (85 kg + 95 kg + 82 kg )( 4.9 m/s)   φ  = 12° φ  = sin −1 

The final velocity of the combined players is 4.9 m/s [12º W of N] 28. We are most likely to see a comet when it is moving at its fastest speed. When a comet is within the solar system, it is at its closest approach to the sun which means most of the comet’s energy is in the form of kinetic energy.

402 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

29. The farther an orbiting vehicle is from Earth, the slower its speed. To increase the radius of orbit from a particular point, the satellite would first need an increase in speed to increase its total energy. As it moved to a higher orbit, this additional kinetic energy would be converted into gravitational potential energy. At the higher orbit, the speed of the vehicle would  be less than at the lower orbit. v

=

GM 

r  30. The speed required to escape from the surface of Earth is always 11.2 km/s. If a space craft is not on the surface of Earth, it would not need that much speed. The term escape speed is defined as the launch speed necessary to just escape Earth’s gravitational pull.  –11 2 2 31. G = 6.67  10  N·m /kg 24  M  = 5.98  10  kg 6 r E = 6.38  10  m  g  = ? First we must calculate the radius of the satellite’s orbit:

= r E + 6.55 × 10 2 km = 6.38 × 106 m + 6.55 × 105 m 6 r 2 = 7.035 × 10 m r2

To calculate the gravitational acceleration: GM   g  = 2 r 

=

(6.67 × 10

−11

N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg)

(7.035 × 106 m) 2

 g  = 8.06 m/s 2 2

The magnitude of the gravitational acceleration is 8.06 m/s . 32. Let the subscript X represent the unknown planet and E represent Earth. mX = 0.25 M E r X = 0.60r E  g X = ? GM E  g E = r E2  g X

=

GM X

=

G (0.25M E )

= =

r X2 (0.60r E )

2

0.25  GM E

 0.36 

r E2

       

0.25

( g E ) 0.36  g X = 0.69 g E The surface gravitational acceleration of the unknown planet is 0.69 g E.

Copyright © 2003 Nelson

Unit 2 Review   403

 –11

2

33. G = 6.67  10  N·m /kg  M S = 1.99  1030 kg r S = 6.96  108 m 11 r E = 1.49  10  m  g  = ?

2

GM 

 g  =

r 2 (6.67 × 10

=

−11

N ⋅ m /kg )(1.99 × 10 kg) 2

2

30

(6.96 ×108 m + 1.49 × 1011 m) 2

 g  = 5.92× 10−3 N/kg [toward the Sun]  –3

The gravitational field of the Sun at the position of Earth is 5.92  10  N/kg [toward the Sun]. 34. Let the subscript M represent Mars, J represent Jupiter, and S represent the Sun. 11 (a) r M = 2.28  10  m 11 r J = 7.78  10  m  M S = 1.99  1030 kg T  = ? The radius of orbit will be: r + r  r  = M J 2 2.28 ×1011 m + 7.78 ×1011 m

=

2

r  = 5.03 × 10 m 11

To calculate the period: GM S 2

4π 

=

r 3 2

T 

2 3 4π  r 

T  =

=

GM S 2 11 3 4π  (5.03 × 10 m)

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)

T  = 1.9456 × 10 s 8

1.9456 ×10 s × 8

1 hr

×

1d

×

1a

3600 s 24 h 365.26 d The period of the asteroid’s orbit in Earth years is 6.16 a. (b) v = ? v=

=

= 6.16 a

GM  r  (6.67 × 10

−11

N ⋅ m /kg )(1.99 × 10 2

2

5.03 × 10

11

30

kg)

m

v = 1.62 ×10 m/s 4 The speed of the asteroid is 1.62  10  m/s. 4

404 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

35. T  = 24 h = 24 h ×

60 min

1h (a)  M  N = 1.03  1026 kg C  = ?

60 s

×

C  =

4

1 min

 = 8.64  10  s

GM  4π 2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.03 × 1026 kg)

=

4π 2

C  = 1.74 ×1014 m 3 /s 2 14 3 2 Kepler’s third-law constant is 1.74  10  m /s . (b) r  = ? C  = r

3

r 

T 2

= 3 CT 2

= 3 (1.74 × 1014 m3 /s2 )(8.64 × 104 s) 2 r  = 1.09 ×108 m The satellite must be 1.09

8

 10  m from Neptune to maintain its circular orbit.



8.4 ×10 m 7

8

7

(c) 1.09 × 10  m – 2.48 × 10  m =

4

 = 8.42 × 10  km 1000 4 The altitude of the orbit is 8.42 × 10  km. 24 h 60 min 60 s 5 36. T  = 1.77 d = 1.77 d ×  = 1.529  10  s × × 1d 1h 1 min 8 r  = 4.22  10  m  M J = ? GM J 2 4π 

 M J

= = =

 M J The mass of Jupiter is 1.90  –26 37. (a) m = 5.32  10  kg v = ?

r 3 2

T 

4π 2 r 3 2

GT 

2 8 3 4π  (4.22 × 10 m)

(6.67 × 10

−11

= 1.90 × 1027

N ⋅ m 2 /kg 2 )(1.529 × 105 s) 2

kg

27

 10  kg.



=0  EK + E g = 0  EK = − E g 1  GMm   mv 2 = −  −   2  r     E T

v=

=

2GM  r  2(6.67 × 10

−11

N ⋅ m /kg )(3.28 × 10 kg) 2

2

23

2.44 ×10 m 6

v = 4.23 × 103 m/s In order to escape, the minimum speed of the molecule must have been 4.23

Copyright © 2003 Nelson

3

 10  m/s.



Unit 2 Review   405

(b) v = ?  ET  EK

1 2

=

+ Eg =

1 2 1 2

E g E g 1

 EK

=−

mv 2

= − − 2

v=

=

E g 2 1  GMm  

   

r 

GM  r  (6.67 × 10 −11 N ⋅ m 2 /kg 2 )(3.28 × 10 23 kg) 2(2.44 ×10 m) 6

v = 2.12 × 103 m/s The speed of the molecule in an orbit around Mercury would be 2.12 (c) Let the subscript M represent Mercury and O represent the orbit. v = ?  ET = E T′  EK 1 2

mv 2

+ EgL =

1 2



3

 10  m/s.

E gO

 GMm   1  GMm   +−  =  −    rM  2  r O    1 2

v

2

=

GM rM

−

GM   2r O

 2 1   −   r  M r O    2 1   = GM  −    rM 2r M  

v = GM 

= =

3GM  2r M 3(6.67 × 10

−11

N ⋅ m 2 /kg 2 )(3.28 × 1023 kg)

2(2.44 × 10 m) 6

v = 3.67 × 103 m/s The molecule would need to launch from the surface at a speed of 3.67 (d) binding energy = ? 1  binding energy = − ET = − E g 2 1  GMm  

= − − 2 = =

   

r 

GMm 2r  (6.67 × 10

 binding energy = 2.39 ×10

−11

N ⋅ m /kg )(3.28 × 10 kg)(5.32 × 10 2

2

23

−26

kg)

2(2.44 × 10 m) 6

−19

J

In this orbit, the molecule’s binding energy would be 2.39

406 Unit 2 Energy and Momentum

3

 10  m/s.





 –19

 10

 J.

Copyright © 2003 Nelson

38. (a) Let the subscript J represent Jupiter, and B represent the black hole.  M J = 1.90  1027 kg  M B = 85 M J = 85(1.90  1027 kg) = 1.615  1029 kg r  = ? For a black hole, the escape speed is the speed of light, c.  E T = 0

+ E g = 0  EK = − E g 1  GMm   2 mv = −  −   2  r   

 EK

r  =

=

2GM  v

2

2(6.67 × 10

−11

N ⋅ m /kg )(1.615 × 10 kg) 2

2

29

(3.00 × 108 m/s) 2

r  = 2.4 ×102 m The Schwartzschild radius of the black hole is 2.4  102 m. (b) For a black hole, the escape speed is the speed of light, c = 3.00 × 108 m/s.

Applying Inquiry Skills 39.

40.

Copyright © 2003 Nelson

Unit 2 Review   407

41. m = 0.55 kg t  = 0.020 s (a) k  = ? From the diagram, there are 14 dots one cycle, therefore: T  = 14(0.020 s) T  = 0.28 s To calculate the force constant: m

T  = 2π  m k 

=

k  =

k 

T 2 2 4π 

4π 2 m T 2 2

=

( 0.55 kg ) 2 (0.28 s )

4π 

k  = 2.8 × 102 N/m 2 The force constant of the springs is 2.8  10  N/m. (b) The results would be the same whether the paper was pulled quickly or slowly. The only value needed from the paper is how many dots there are in each cycle. This can be determined at almost any speed. The only restrictions on speed are that it must be slow enough to get at least one full cycle on the paper, and fast enough that the dots don’t overlap one another. (c) The most likely source of error would be the friction between the paper and the puck. (d) Safety considerations include exercising caution when using an electrical spark timer, handling the vibrating spring coils carefully so as not to be pinched, and being careful not to receive a paper cut from the sliding paper. 42. (a) With no mass hanging from the spring, mark that point as zero. Suspend one weight from the accelerometer and mark that point as 1 g . Suspend a second weight and mark the bottom of the spring as 2 g . Measure the distances between 0 g  and 1 g , and between 1 g  and 2 g . They should be the same. Measure out the average of the two values beyond 2 g  and mark 3 g  and 4 g . (b) Some possibilities are: • spinning in a circle with the accelerometer held horizontally • landing after jumping off a stool while holding the accelerometer • quickly lifting the accelerometer into the air (c) Some possible answers are: • the bottom of a roller coaster ride • during a large swing • during a spin ride (if held horizontal) 43. m = 0.88 kg k  = 36 N/m v = 0.22 m/s ∆d  = 2.5 m  x = 3.6 cm = 0.036 m (a)  E e = ?  E K =   ? To calculate the elastic potential energy: 1  Ee = kx 2 2 1 = (36 N/m)(0.036 m)2 2 = 0.02333 J  E e

= 0.023 J, or 2.3 × 10−2 J

408 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

To calculate the kinetic energy: 1  EK  = mv 2 2 1 = (0.88 kg)(0.22 m/s)2 2 = 0.02130 J  E K  = 0.021 J, or 2.1× 10−2 J The elastic potential energy was 0.023 J and the kinetic energy was 0.021 J. The discrepancy in values was due to energy lost in the form of sound and thermal energy. (b)  F K =   ?  ET = E T′

∆ EK = E th 0−

1 2

mv 2

= FK ∆ d  

 F K  = −

=−

mv 2 2∆d  (0.88 kg)(0.22 m/s) 2 2(2.5 m)

 F K  = −8.5 × 10−3 N  –3

The average kinetic friction is –8.5  10  N. (c) Possible sources of error are: • a desk or track that is not levelled properly • the cart is not initially in contact with the wall • incorrect release of the trigger

Making Connections 44. (a) The jarring forces on your legs can be reduced when jogging by keeping your knees bent and allowing them to flex with each step. (b) By reducing the jarring, you reduce the forces applied internally in your legs. This helps prevent painful damage to  bones and tissue, such as cartilage, in the knee. 45. (a) An air bag is used to reduce injury in two primary ways. Because it is a bag, the force that stops a passenger in the car is spread out over the entire surface of the body, not just a small area such as the points the hands contact on the steering wheel. This reduces the pressure on the person’s body. The second way the airbag reduces injury is by slowing the person down over a longer period of time. This increase in stopping time significantly reduces the total force required to bring the passenger to rest. (b) Other safety devices that used to help are: • crumple zones in a car • snug fit and padding of any type of helmet • design of protective equipment (such as shin pads) 20 46. r  = 2.7  10  m 365.26 d 24 h 3600 s 8 15 T  = 2.0  10  a = 2.0 × 108 a × = 6.312  10  s × × 1a 1d 1h (a)  M  = ? GM 4π 2

=

 M  =

=

3

r 

T 2 4π 2 r 3 2

GT 

2 20 3 4π  (2.7 × 10 m)

(6.67 × 10−11 N ⋅ m 2 /kg 2 )(6.312 × 1015 s) 2

 M  = 2.9 × 1041 kg The total mass of the stars at the hub of our galaxy is 2.9 Copyright © 2003 Nelson

 1041 kg.



Unit 2 Review   409

2.9 × 10 kg 41

(b)

= 1.5 ×1011

2.0 × 10 kg The approximate number of stars that are the size of our Sun is 1.5 30



11

 10 .

Extension 47. m1 = 2.5 kg v1 = 2.3 m/s m2 = 2.0 kg v1 ´ = ? v2 ´ = ? Choose right as positive. Using the conservation of energy (units are omitted): 1 1 1 m1v12 = m1v1′2 + m2 v2′2 2 2 2

= m1v1′2 + m2 v2′2 2.5(2.3) 2 = 2.5v1′2 + 2.0v2′2 2 2 5.29 = v1′ + 0.8v2′ (Equation 1) 2

m1v1

Using the conservation of momentum (units are omitted): m1v1 = m1v1′ + m2 v2′ (2.5)(2.3) = 2.5v1′ + 2.0v2′ 2.3 = v1′ + 0.8v2′ v1′

= 2.3 − 0.8v2′

(Equation 2)

Substitute Equation 2 into Equation 1: 5.29 = (2.3 − 0.8v2′ )

+ 0.8v2′2 2 2 5.29 = (5.29 − 3.68v2′ + 0.64v2′ ) + 0.8v2′ 0 = −3.68v2′ + 1.44v2′2 0 = v2′ (−3.68 + 1.44v2′ ) v2′ = 0, or − 3.68 + 1.44v2′ = 0 Since v2′

2

= 0  is not valid because it represents no change in speed, we use: −3.68 + 1.44 v2′ = 0 1.44v2′  = 3.68 v2′  = 2.5556 v2′  = 2.6 m/s [right]

Substitute back into Equation 2: v1′ = 2.3 − 0.8(2.5556) v1′ = 0.26 m/s [right] The velocity of ball 1 is 0.26 m/s [right], and the velocity of ball 2 is 2.6 m/s [right]. 48. Let the subscript b represent the bullet and w represent the block of wood. m b = 4.0 g = 4.0  10 –3 kg 2 v b = 5.0  10  m/s [N] 2 v b′  = 1.0  10  m/s [N] mw = 2.0 kg ∆d  = 0.40 m

410 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

(a) vw′  = ? Choose north as positive.  p = p ′ m b vb

+ 0 = mb vb′ + mw vw′ m v − mb vb′ vw′ =  b b mw

= = vw′

m b ( vb

− vb′ )

mw (4.0 × 10 −3 kg)(5.0 × 102 m/s − 1.0 × 10 2 m/s) 2.0 kg

= 0.80 m/s [N]

The wooden block moves at a velocity of 0.80 m/s [N] just after the bullet exits. (b)  E K =   ? 1 2  EK  = mv 2 1 = (2.0 kg)(0.80 m/s)2 2  E K  = 0.64 J The maximum kinetic energy of the block is 0.64 J. (c)  F K =   ?  ET = E T′

∆ EK = E th 0−

1 2

mv 2

= FK ∆ d   mv 2

 F K  = −

2∆d  (2.0 kg)(0.80 m/s) 2

=−

2(0.40 m)

 F K  = −1.6 N [N], or 1.6 N [S] The average frictional force stopping the block is 1.6 N [S]. (d) ∆ E K =   ? ∆ EK = EK2 − E K1

= = =

1 2 1 2 1 2

mv22

−

m( v22

1 2

mv12

− v12 )

((

(4.0 × 10 −3 kg) 1.0 × 10 2 m/s

∆ E K  = −4.8 × 102

2

2

) − (5.0 × 102 m/s )

)

J

The decrease in kinetic energy of the bullet is –4.8  102 J. (e) The collision between the bullet and the block is not elastic, so most of the energy is lost to thermal energy. 49. Let the subscript P represent the plane, and B represent the barge. mP = 1.0 Mg = 1.0  103 kg mB = 2.0 Mg = 2.0  103 kg 1 vP = 5.0  10  m/s = 50 m/s vB = 0.0 m/s

Copyright © 2003 Nelson

Unit 2 Review   411

For the landing of the plane: mP vP + mB vB

= (mP + mB )vPB m v + mB vB vPB = P P mP + mB (1.0 × 103 kg)(50 m/s) + 0 = 1.0 × 103 kg + 2.0 × 103 kg vPB = 16.67 m/s

To calculate the frictional force on the plane, and on the barge: 1  FK  = mg  4 1 = (1.0 × 103 kg)(9.8 N/kg) 4  F K  = 2450 N [backward] By Newton’s third law, F K = –2450 N [forward] To calculate the acceleration of the plane: Σ F = mP aP aP

=

Σ F  mP

−2450 N 1.0 × 103 kg = −2.45 m/s 2 =

aP

To calculate the acceleration of the barge: Σ F = mB aB aB

= =

aB

Σ F  mB 2450 N 2.0 × 10 kg 3

= 1.225 m/s2

To calculate the distance the plane will travel during the landing:

= vi2 + 2aP ∆d v 2 − v2 ∆d  = f i vf2

 

2 aP

=

(16.67 m/s)2

− (50 m/s)2

2( −2.45 m/s ) 2

∆d  = 453.5 m To calculate the distance the barge will travel during the landing:

= vi2 + 2aB∆d v 2 − v2 ∆d  = f i vf2

 

2aB

=

(16.67 m/s) 2

− ( 0 m/s)2 2

2(1.225 m/s )

∆d  = 113.4 m

Therefore, the required length of the barge is 453.5 m – 113.4 m = 3.4 × 10 2 m. 412 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

50. Let the subscript C represent the chair. (a) mC = ? k  = ? For the empty chair: m

T  = 2π  m k 

=

k  =

= k  =

k 

2

T 

2 4π 

4π 2 m T 2

+ 0)

4π 2 ( mC

2

(0.901 s ) 4π 2 mC 2

(0.901 s )

For masses on the chair: m

T  = 2π  m k 

=

k  = k  =

k 

T 2 4π 2 4π 2 m 2

T 

+ m)

4π 2 (mC 2

T 

To calculate the mass of the chair: 4π 2 mC 2

2

= mC + m

(0.901 s ) T 2 mC

(0.901 s ) T 2 mC 2

( 0.901 s )

4π 2 ( mC

=

+ m)

T 2

− mC = m

 T 2     m = m − 1  ( 0.901 s )2   C    mC

=

=

mC

Copyright © 2003 Nelson

m

 T 2      − 1  ( 0.901 s )2      14.1 kg

 (1.25 s )2    − 1   ( 0.901 s )2     

= 15.2 kg

Unit 2 Review   413

Similarly, mC

=

mC

=

23.9 kg

 (1.44 s )    − 1  (0.901 s )2    2

56.1 kg

 (1.94s )2    − 1   (0.901s )2     

= 15.4 kg

mC

=

= 15.4 kg

mC

=

33.8 kg

 (1.61 s )    − 1  (0.901 s )2    2

67.1 kg

 ( 2.09 s )2    − 1   ( 0.901s )2     

= 15.4 kg

mC

=

45.0 kg

 (1.79 s )2   − 1  (0.901 s )2   

= 15.3 kg        

= 15.3 kg

To calculate the average value for the mass of the chair: 15.2 kg + 15.4 kg + 15.4 kg + 15.3 kg + 15.4 kg + 15.3 kg mC = 6 mC = 15.3 kg To calculate the spring constant: m

T  = 2π  m k 

=

k  =

=

k 

2

T 

4π 2 4π 2 mC T 2 4π 2 (15.3 kg) 2

(0.901 s )

k  = 744 N/m The mass of the chair is 15.3 kg, and the spring constant is 744 N/m. (b)  f  = ? Assuming a mass of 94 kg, we first calculate the period: T  = 2π 

= 2π 

m k  94 kg + 15.3 kg 744 N/m

T  = 2.4 s Since frequency is inversely proportional to period: 1  f  = T  1

=

2.4 s  f  = 0.42 Hz The frequency is 0.42 Hz.

414 Unit 2 Energy and Momentum

Copyright © 2003 Nelson

(c) T 1 = 2.01 s T 2 = 1.98 s ∆m = ? m

T  = 2π  m k 

=

m=

k 

2

T 

4π 2 2

kT 

4π 2 ∆m = m2 − m1

= = =

kT22

kT12 

4π k 

4π 2

− 2 2

(T22

− T 12 )

4π  744 N/m 2

4π  ∆m = −2.3 kg

((1.98 s )

2

− ( 2.01 s )2 )

The astronaut loses 2.3 kg 51. (a) Venus has the most circular orbit because its eccentricity is closest to zero. Pluto has the most elongated orbit because its eccentricity is the highest. (b)

(c)

(d) The typical elliptical orbits shown for most planets do not represent the true shape of the orbit. Most texts exaggerate the length of the orbit for effect. Even the most elongated orbits have the Sun close to the centre of the ellipse. 52. Let the subscript X represent the planet and E represent Earth. 1 mX =  M E 10 r X =

1

r E 2 2  F gE = 6.1  10  N  F gX = ?

Copyright © 2003 Nelson

Unit 2 Review   415

Half of the diameter also means half of the radius, therefore:  FgE = mg E m=

 F gE  g E

Also,  g E

=

GM E

 g X

=

GM X

r E2 2

r X

 1 M    E   =  10 2    1 r     2 E      G

1

 GM    = 10  2 E   1  r E     4 4

=

( g E ) 10  g X = 0.4 g E Substituting the value of m of the value of g X:  FgX = mg X

 F    =  gE  (0.4 g E )   g E   = 0.4 F gE  F gX The astronaut weighs 2.4

= 0.4(6.1× 102 N) = 2.4 × 102 N

416 Unit 2 Energy and Momentum

2

 10  N on the new planet.



Copyright © 2003 Nelson