STPM Physics 2008

Time: 1 h 45 min

PAPER 1

1 A fluid flows through a pipe of diameter d and length l. The volume flow rate R is given by cd 4∆p R=— — — — —, ηl where c is a dimensionless constant, ∆p the pressure difference between the two ends of the pipe and η the viscosity of the fluid. The unit of η in terms of base units is A kg m B kg s–2 C kg m–1 s–1 D kg m–3 s–1 2 The figure shows two tugs P and Q pulling a ship R from a point O to X.

1.1

P

15o R

X

O

Q

P pulls with force 6.0  104 N at an angle of 15° to OX and Q pulls with force 3.6  104 N at angle θ to OX. The value of θ is A 8.9° B 15.0° C 24.1° D 25.6° 3 A gun fires 120 bullets per minute at a speed of 200 m s–1 relative to the gun. If the mass of each bullet is 10 g, the recoiling force acting on the gun is A 2N B 4N C 240 N D 4000 N 4 Two cars X and Y moving at the same speed are acted upon by equal braking forces until they come to a stop. If the mass of X is twice that of Y, the ratio of the distance travelled by X to the distance travelled by Y is A 1:2 B 1:1 C 2:1 D 4:1 5 The figure shows a car travelling at a constant speed along the road JKLMN. N

M

L Car

K

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t1 1

J

Actual 2008 STPM Physics Examination Paper

1

2/11/09 10:47:30 AM

At which part of the journey does the car have the greatest acceleration? A JK B KL C LM D

MN

6 The figure shows a rod pivoted at point P on a smooth horizontal surface. 5.0 N 3.0 m

30o P

30o

2.0 m 5.0 N

Two forces each of magnitude 5.0 N acting in opposite directions are applied at the two ends of the rod. The resultant torque on the rod is A 2.5 N m B 12.5 N m C 21.7 N m D 25.0 N m 7 A tyre is fixed to an alloy rim and another identical tyre is fixed to an iron rim. The two rims are of the same dimensions. When the wheels are rotated from rest, it is found that it is relatively easier to rotate the wheel with the alloy rim. This is because A the mass of the alloy rim is smaller B the mass of the alloy rim is more uniformly distributed C the material of the alloy rim is denser D the material of the alloy rim is stronger 8 The figure shows three point masses each with mass m at vertices of an equilateral triangle with the length of its side equal to x. m

x

m

The potential energy of the system is Gm2 Gm2 A –— — — B –— — — 3x 2x

x

x

C

m

2Gm2 –— — — — x

D

3Gm2 –— — — — x

9 The figure shows a uniform rod of weight W placed on a smooth hemisphere with one end of the rod in contact with the floor.

Which figure shows the correct directions of the forces acting on the rod? A B

W

2

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t2 2

W Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:34 AM

C

D

W

W

10 The graph shows the variation of displacement x against time t produced by the oscillation of a simple pendulum. x/cm 5

0 2

4

6

8

10

t /s

–5

Which statement is true? A The velocity at 6 s is zero. B The period of oscillation is 6 s.

C D

The acceleration at 4 s is maximum. The kinetic energy at 2 s is minimum.

11 The graph shows two progressive waves y1 and y2 propagating in opposite directions at the time t = 0 s. y /cm y1

1

y2

0 1

2

3

4

5

6

7

8

9

10

11

12

x /cm

–1

The speed of both waves is 3 cm s–1. Which of the following graphs shows the resultant wave at the time t = 1 s? y /cm 1

A

0 1

2

3

4

5

6

1

2

3

4

5

6

7

8

9

10

11

12

x /cm

9

10

11

12

x /cm

–1 –2

y /cm 2 1

B

0 7

8

–1 –2 Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t3 3

Actual 2008 STPM Physics Examination Paper

3

2/11/09 10:47:34 AM

y /cm 2 1

C

0 1

2

3

4

1

2

3

4

5

6

7

6

7

8

9

10

11

12

x /cm

9

10

11

12

x /cm

–1 –2 y /cm 1

D

0 5

8

–1 –2

12 The displacement of a particle moving in simple harmonic motion is given by x = 5 sin 2t, where x is in centimetres and the time t in seconds. If the period of the motion is T, the acceleration of the T particle at t = — s is 6 B –10.0 cm s–2 C 10.0 cm s–2 D 17.3 cm s–2 A –17.3 cm s–2 13 The figure shows a standing wave in a string fixed at both ends.

Which statement is not true? A The string vibrates the air around it. B The air vibrates at the same frequency as that of the standing wave. C The sound produced has the same wavelength as that of the standing wave. D The standing wave is the superposition of a propagating wave and its reflection. 14 A cyclist approaches a songbird on a tree at a speed of 10 m s–1. If the bird sings at a frequency of 420 Hz, what is the frequency heard by the cyclist? [The speed of sound in air = 343 m s–1] A 396 Hz B 408 Hz C 432 Hz D 445 Hz 15 The figure shows the positions of particles in a medium at a particular instant when a sound wave of frequency 1500 Hz passes through it.

0

5

10

15

20

25

The speed of sound in the medium is A 75 m s–1 B 150 m s–1

30

35

C

40

45

300 m s–1

50

55

D

x /cm

330 m s–1

16 A metal has molar mass M, density ρ and atomic separation d. The Avogadro’s number is given by Md 3 Mρ d3 M A — — — B — — — C — — — D — — — 3 ρ d Mρ ρd 3 17 Three wires P, Q and R of different materials are stretched. P undergoes plastic deformation before it snaps. Q snaps without undergoing plastic deformation. R shows only a small extension with a large tensile force. Which characteristics describe P, Q and R? 4

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t4 4

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:35 AM

P

Q

R

A

Ductile

Brittle

Rigid

B

Ductile

Rigid

Brittle

C

Rigid

Brittle

Ductile

D

Rigid

Ductile

Brittle

18 The r.m.s. speed of the molecules of an ideal gas which has volume V at pressure p is directly proportional to p p pV — C — — D A pV B V V 19 A box contains an ideal gas at 27 °C and 2.0  10–6 Pa. The number of gas molecules per unit volume is A 1.1  108 m–3 B 1.2  109 m–3 C 4.8  1014 m–3 D 5.4  1015 m–3 20 The graph shows the variation of pressure p with volume V for a mass of gas. p S

R

V

What happens to the heat transfer, internal energy and work done when the gas changes from states R to S? Heat transfer

Internal energy

Work done

A

Supplied to gas

Increase

By gas

B

Supplied to gas

Decrease

On gas

C

Released by gas

Increase

On gas

D

Released by gas

Increase

By gas

21 The graph shows the variation of pressure p with volume V for 0.04 mol of a diatomic gas. p /10 5 Pa 20.2

X

Y

1.1

0

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t5 5

1.0

8.0

V /10 –4 m 3

Actual 2008 STPM Physics Examination Paper

5

2/11/09 10:47:35 AM

The gas expands adiabatically from state X to state Y. The internal energy change in the process XY is A –7130 J B –4280 J C –285 J D –171 J 22 A wooden box of surface area 1.0 m2 and wall thickness 2.0 cm contains 20.0 kg of ice at 0 °C. The specific latent heat of fusion of ice is 3.3  105 J kg–1 and the thermal conductivity of wood is 0.50 W m–1 K–1. If the room temperature is 33 °C, the time required for all the ice in the box to melt is A 825 s B 863 s C 4000 s D 8000 s 23 The figure shows a metal rod wrapped with an insulating material. Insulation

P

Q

R

S

Which graph shows the variation of temperature with distance along the rod if both ends of the rod are maintained at constant temperatures? A Temperature B Temperature

P

C

Q

R

S

P

Distance

D

Temperature

P

Q

R

S

R

S

Distance

Q

R

S

Distance

Temperature

P

Distance

Q

24 The figure shows the electric field lines for a positive point charge and a nearby negative point charge that are equal in magnitude.

R + Q

6

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t6 6

– P

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:35 AM

The electric field strengths at P, Q and R are EP , EQ and ER respectively. Which arrangement shows the correct relative electric field strengths? A EP > EQ > ER B EP > ER > EQ C EQ > EP > ER D EQ > ER > EP 25 The figure shows a spherical Gaussian surface of radius 20.0 cm enclosing a charge +2 C. Gaussian surface

+2 C

The electric flux through the Gaussian surface is A 4.43  10–12 N m2 C–1 B 1.77  10–11 N m2 C–1 C 2.26  1011 N m2 C–1 D 4.50  1011 N m2 C–1 26 A charged capacitor is connected in series to a switch and a resistor. A voltmeter of high resistance is connected across the capacitor. The switch is closed, and a set of values for the voltage V and the time t is obtained. Graph ln V against t plotted is as follows: In V

2

1

0 10

The time constant of the circuit is A 0.1 s B 2s

20

C

10 s

t (s)

D

20 s

27 Which arrangement of four identical capacitors produces the biggest capacitance between points X and Y?

A

X

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t7 7

Y

B

X

Y

Actual 2008 STPM Physics Examination Paper

7

2/11/09 10:47:36 AM

C

X

D

Y

X

Y

28 Which statement is not true of the model used to explain the electrical conduction in a metal at the microscopic level? A The kinetic energy for the random motion of the free electrons depends on the temperature of the metal. B The excess energy acquired by the free electrons in the electric field is lost during collisions with the lattice atoms. C The mean free time decreases as the temperature of the metal increases because the lattice atoms vibrate more strongly. D At any instant, all the free electrons are moving with the same drift velocity because the electrons are accelerated by the same electric field. 29 A battery with e.m.f. ε1 and internal resistance r1 is connected to a battery with e.m.f. ε2 and internal resistance r2 in the circuit shown. R1

I1

R2

X

Y

W

R3

I2 Z

Which of the following is true of the closed loop WXYZW? A ε1 + ε2 = I1(R1 + R2 + r1) + I2(R3 + r2) B ε1 – ε2 = I1(R1 + R2 + r1) – I2(R3 + r2) C ε1 + ε2 = I1(R1 + R2 – r1) + I2(R3 – r2) D ε1 – ε2 = I1(R1 + R2 – r1) – I2(R3 – r2) 30 Which of the following is true of the internal resistances of a voltmeter and an ammeter?

8

Voltmeter

Ammeter

A

High

High

B

High

Low

C

Low

High

D

Low

Low

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t8 8

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:36 AM

31 The figure shows a 40 cm straight rod carrying a current of 20 A in a 2.0 T magnetic field. B = 2.0 T 20 A

The magnetic field acts in the horizontal plane at an angle of θ with the direction of the current. If the weight of the rod is 8.0 N and the magnetic force is just enough to balance the weight of the rod, what is the angle θ? A 30° B 60° C 120° D 150° 32 The figure shows the trajectories of two charged particles P and Q directed perpendicularly into a region of uniform magnetic field with the same velocity.

P

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

Q

If the radii of the trajectories are the same, what are P and Q? P

Q

A

α-particle

Electron

B

Electron

α-particle

C

Electron

Positron

D

Positron

Electron

33 The figure shows a uniform magnetic field in a rectangular region and a copper ring moving through positions 1 to 4.

1

2

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

3

4

Which statement is not true of the changes experienced by the ring? A At position 1, there is no change in magnetic flux and no current is induced. B At position 2, there is an increase in magnetic flux and a current is induced. C At position 3, there is a constant magnetic flux and a constant current is induced. D At position 4, there is a decrease in magnetic flux and a current is induced. Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t9 9

Actual 2008 STPM Physics Examination Paper

9

2/11/09 10:47:37 AM

34 The figure shows the side view of a circular coil of N turns and area A in a magnetic field B. Circular coil

B

If the normal to the plane of the coil makes an angle α with the magnetic field, the total magnetic flux through the coil is A BA sin α B BA cos α C NBA sin α D NBA cos α 35 In a circuit, an alternating current of r.m.s. value 1 A passes through a resistor. In another circuit, a steady current of magnitude I passes through an identical resistor. If the two resistors dissipate heat at the same rate, the value of I is A 0.71 A B 1.00 A C 1.41 A D 2.00 A 36 The figure shows an operational amplifier circuit.

+10 V + Vi

–10 V

Vo

Which graph represents the variation of the output voltage Vo with the input voltage Vi? A B V /V V /V o

o

+10

–2.5

0

+10

2.5

–2

Vi /V

–10

0

+10

2.5

Vi /V

–10

10

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t10 10

Vi /V

Vo /V

D

+10

–2.5

2

–10

Vo /V

C

0

–2

0

2

Vi /V

–10

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:37 AM

37 Which figure shows an oscillator circuit? A

B

+9 V

+9 V

-

-

+

+ –9 V –9 V

V1

Vo

V2

Vo

Vi

C

C

D

R

C

+9 V +9 V +

R -

–9 V

R Vo

+ Vi

–9 V

R

Vo

C R

38 The figure shows the symbol of an operational amplifier. R

-

P

S +

Q

Which terminal and label do not correspond? Terminal

Label

A

P

Inverting input

B

Q

Non-inverting input

C

R

Feedback output

D

S

Signal output

39 A concave mirror has a radius of curvature r. If an object is placed at a distance 2r from the mirror, the linear magnification of the image is 1 1 A — B — C 1 D 3 5 3 40 The distance between an object and a screen is 100 cm. When a lens is placed in between the object and the screen, an image which is real and of the same size as the object is formed on the screen. Which lens is used? A A concave lens with a focal length of 25 cm B A concave lens with a focal length of 50 cm C A convex lens with a focal length of 25 cm D A convex lens with a focal length of 50 cm Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t11 11

Actual 2008 STPM Physics Examination Paper

11

2/11/09 10:47:37 AM

41 The figure shows monochromatic light of wavelength λ which is illuminated nearly normal onto a thin soap film of thickness t and refractive index n.

Air

t

Soap Air

The optical path difference between the reflected rays from the upper and lower surfaces of the soap film is 2t 2t λ λ A — B 2nt C —+— D 2nt + — n n 2 2

42 In the photoelectric effect, the energy of the photon is used A to release the electron from the lattice only B as the kinetic energy of the electron only C to release the electron from the lattice and as the kinetic energy of the electron D by the electron to produce a new electron called photoelectron 43 If λ is the wavelength of a particle with momentum p, which graph has a gradient equal to the Planck constant? A B p

p

0

0

C

D p

0

12

1

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t12 12

p

0

1

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:38 AM

44 The figure shows some of the energy levels of an atom. E4 E3

E2

E1

The maximum number of spectral lines produced by the transition of electrons from these four energy levels is A 3 B 4 C 5 D 6 45 The graph shows the X-ray spectra I and II produced by an X-ray tube when it is operated at two different voltages. Relative intensity

I

II 0

Wavelength

Which statement explains why the minimum wavelength of spectrum II is longer than that of spectrum I? A A lower voltage is used to produce spectrum II. B A higher voltage is used to produce spectrum II. C A lighter element is used as the target material to produce spectrum II. D A heavier element is used as the target material to produce spectrum II. 46 Which statement is not true of stimulated emission? A Photons emitted by stimulated emission are in phase. B Photons produced by stimulated emission are of different frequencies. C Radiation produced by stimulated emission in a laser is monochromatic. D Radiation produced by stimulated emission in a laser has high intensity. 209 47 The nucleus of iron 56 26 Fe is more stable than the nucleus of bismuth 83 Bi. Which of the following is true of the more stable nucleus? A Higher mass per nucleon B Lower ratio of proton to neutron C Lower binding energy per nucleon D Higher binding energy per nucleon

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t13 13

Actual 2008 STPM Physics Examination Paper

13

2/11/09 10:47:38 AM

48 A radioactive element X decays to a radioactive element Y which then decays to element Z. If the sample initially contains X only, which of the following influences the ratio of the number of Y nuclei to the number of X nuclei after a certain time? A Surrounding pressure C Half-life of element Y B Initial number of X nuclei D Half-life of element Z 114 49 When a slow neutron is captured by a stationary 113 48 Cd nucleus, a 48 Cd nucleus is formed and a γ-ray photon is emitted. What is the wavelength of the γ-ray? 114 [Mass of 01 n = 1.0087 u, mass of 113 48 Cd = 112.9044 u, mass of 48 Cd = 113.9034 u] A 2.05  10–23 m C 1.37  10–13 m –16 B 6.63  10 m D 1.45  10–12 m

50 A strong force is responsible for A alpha decay B beta decay

C D

holding the nucleons together holding the electrons in the nucleus

Time: 2 h 30 min

PAPER 2 Section A [40 marks] Answer all questions in this section.

1 The figure shows a ball kicked from a stage of height h with initial velocity u at angle θ with the horizontal. y

u

x

0

Stage

h

L

(a) Write the equations of motion of the ball to show its horizontal and vertical positions at any time t after the ball is kicked. [2 marks] (b) If h = 1.0 m, u = 8.0 m s–1 and θ = 30°, calculate the horizontal distance L of the ball from the stage when it strikes the ground. [4 marks] 2 A flywheel of radius 0.20 m with moment of inertia 0.15 kg m2 rotates at 180 revolutions per minute. A tangential force is applied on the rim of the flywheel and it stops after 12 revolutions. (a) Determine the initial angular velocity of the flywheel in rad s–1. [1 mark] (b) Calculate the angular deceleration. [3 marks] (c) Calculate the magnitude of the tangential force. [2 marks] 14

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t14 14

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:38 AM

3 A pipe of length 20 cm has one of its ends closed and the other end open. Determine its fundamental frequency and first overtone. [4 marks] [Assume the speed of sound to be 345 m s–1] 4 The table shows two measured quantities for aluminium and copper. Material

Young’s modulus (N m–2)

Maximum tensile strength (N m–2)

Aluminium

69  109

110  106

Copper

110  109

250  106

(a) Based on the quantities in the table, compare the physical properties of aluminium and copper. [2 marks] (b) An aluminium wire of length 2.0 m and diameter 0.10 mm is subjected to a tensile force of 10.0 N. Calculate the extension produced, and state your assumption made in the calculation. [3 marks] 5 In an a.c. circuit, the supply voltage is given by V = 240 sin 5000t and the current in the circuit is π given by I = 0.480 sin 5000t – — . 2 (a) Determine the reactance of the circuit. [2 marks] (b) Sketch the phasor diagram of V and I. [1 mark] (c) State the electrical component in the circuit. [1 mark]

冢

冣

6 The figure shows an operational amplifier. V1

-

V2

+

Vo

(a) State the typical value of the open-loop gain of an operational amplifier. [1 mark] (b) State the condition for the output voltage Vo to vary linearly with the input voltage difference (V2 – V1). [1 mark] (c) What are the advantages for an operational amplifier having high input impedance and low output impedance? [2 marks] 7 In a photoelectric experiment, light of wavelength 450 nm is incident on a metallic surface with work function 2.3 eV. (a) Determine the velocity of the most energetic electrons ejected from the surface. [4 marks] (b) Calculate the stopping potential. [2 marks] 8 The initial number of atoms in a 2.0 g radioactive element is 6.0  1021. The half-life of the element is 10 hours. (a) Calculate the number of atoms which decay in 24 hours. [4 marks] (b) Determine the mass of the radioactive element left after 24 hours. [1 mark] Section B [60 marks] Answer any four questions in this section. 9 (a) Define simple harmonic motion. [2 marks] (b) The displacement x from the equilibrium position of a mass undergoing simple harmonic motion is given by x = 0.50 cos (2t + φ ), where x is in metres and t in seconds. The initial velocity is – 0.20 m s–1. Calculate (i) the maximum velocity [2 marks] (ii) the maximum acceleration [2 marks] (iii) the phase angle φ . [3 marks] Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t15 15

Actual 2008 STPM Physics Examination Paper

15

2/11/09 10:47:39 AM

(c) The bob of a simple pendulum is displaced and then released. (i) Show that the shortest time when the kinetic energy of the system equals its potential energy is one eighth of the period. [4 marks] (ii) Deduce the subsequent time when the kinetic energy of the system equals its potential energy again. [2 marks] 10 (a)

(i) State two assumptions of an ideal gas. [2 marks] (ii) State two physical conditions under which a gas behaves as an ideal gas. [2 marks] (iii) A 0.35 m3 gas tank contains 7.0 kg of butane gas. Assuming that the gas behaves as an ideal gas, calculate its pressure at 27 °C. [3 marks] [Molecular mass of butane is 58 g mol–1] (iv) Butane gas normally behaves as a real gas. The actual pressure of the butane gas is higher than the calculated value in (a)(iii). Give a reason. [1 mark] (b) (i) What is meant by the degrees of freedom of a gas molecule? [1 mark] (ii) Write an expression relating the total kinetic energy E of a gas molecule to the number of degrees of freedom f. Explain any other symbols used. [2 marks] (iii) The escape velocity of Mars is 5.0  103 m s–1. If the temperature of Mars is 300 K, determine whether oxygen gas can exist on the planet. [4 marks] [Molecular mass of oxygen is 32 g mol–1]

11 (a) Two thin conducting plates have an area of 0.50 m2 each. They are placed parallel to each other and 25 mm apart. One plate is maintained at +75 V while the other at –75 V by a d.c. supply. (i) Determine the amount of charge stored on each plate. [4 marks] (ii) Calculate the energy stored in the electric field between the plates. [2 marks] (b) The figure shows a simple circuit of the photographic flash used in a camera. Switch

Battery

Flash bulb

The capacitance of the capacitor is 40.0 μF, and the resistance of the resistor is 45.0 kΩ. (i) Explain the function of the capacitor in this application. [4 marks] (ii) Calculate the time required to charge the capacitor to 63% so that a good flash can be obtained. [4 marks] (iii) Suggest a way to reduce the charging time of the capacitor. [1 mark] 12 A thin biconvex lens made of glass with refractive index 1.5 has surfaces with equal radius of curvature 15.0 cm. An object is placed 20.0 cm on the principal axis to the left of the lens. (a) Determine (i) the focal length of the lens [2 marks] (ii) the image position [2 marks] (iii) the magnification [2 marks] (iv) the nature of the image. [2 marks] (b) A plane mirror is then placed perpendicularly to the principal axis of the lens with its reflecting surface in contact with the right hand side of the lens. (i) Show, on the same diagram, the image formed by the lens and the subsequent image formed by the mirror. [2 marks] (ii) Determine the final position of the image formed by the combination of the lens and the mirror. [3 marks] (iii) Calculate the overall magnification. [2 marks] 16

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t16 16

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:39 AM

13 (a) Explain briefly the stimulated emission process and the population inversion in the production of a laser. [5 marks] (b) State two differences between the laser and fluorescent light. [2 marks] (c) A laser produces a bright beam of light of wavelength 450 nm. Calculate the energy of a photon in eV. [3 marks] (d) Calculate the wavelength of the light emitted when a hydrogen atom makes a transition from the energy level n = 4 to n = 2 according to the Bohr model. [5 marks] 14 (a) Define the binding energy of a nucleus and the mass defect of a reaction. [2 marks] (b) Determine the binding energy per nucleon of tritium 13 H. [4 marks] [Mass of proton = 1.007276 u, mass of neutron = 1.008665 u, mass of tritium = 3.016049 u, 1 u = 931.5 MeV] (c) The graph shows the variation of the binding energy per nucleon against the mass number. Binding energy 9 per nucleon/MeV 8 62 Ni 28

12 C 6

7

4 2 He

238 U 92

6 5 4 3 2 2

1 0

1

H

50

100

150

200

250 Mass number

(i) Determine the most stable nucleus. (ii) Estimate the energy released for the fusion reaction 2 1

H + 12 H → 24 He

[1 mark] [4 marks]

(d) Explain the working principle of a mass spectrometer to measure the masses of ions. [4 marks]

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t17 17

Actual 2008 STPM Physics Examination Paper

17

2/11/09 10:47:39 AM

SUGGESTED ANSWERS 6. B:

PAPER 1 cd 4∆p 1. C: R = — — — — — ηl

5.0 N 5.0 sin 30°

cd 4∆p η = ————— Rl

5.0 cos 30° P 2.0 m

1 Unit of η = m4(N m–2)— — — — — — — (m3 s–1)m 1 = m4(kg m s–2 m–2)— — — — — — (m4 s–1) = kg m–1 s–1 2. D:

Y

P

P sin 15o 15o O

X Q sin Q

Since there is no motion in the OY direction, Q sin θ = P sin 15°

冤

冣

θ = 25.6° 3. B: Recoiling force = rate of change of momentum = (number of bullet s–1)  (change of momentum per bullet) 120 = — — — (0.010  200 – 0.010  0) N 60 =4N 4. C: Work done against braking force = loss of kinetic energy 1 Car X: Fs1 = —(2m)(u2) 2

9. C: O

R

F B

(1) A

(2)

(1) s1 — —: — —= 2 (2) s2 v2 5. B: Centripetal acceleration = — — r Since speed v is constant, the centripetal acceleration is the greatest when radius of curvature of the track is the smallest, that is along KL. 18

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t18 18

冥

冣

1 Car Y: Fs2 = —(m)(u2) 2

5.0 sin 30°

5.0 N

Gm2 Gm2 Gm2 = –— — — + (–— — —) + (–— — —) x x x 3Gm2 = –— — — — x

6.0  104 = — — — — — — — — sin 15° 3.6  104

冢

30°

No torque is produced by the horizontal component, (5.0 cos 30°). Resultant torque by the vertical component (5.0 sin 30°) of both the forces about the point P in the clockwise direction = (5.0 sin 30° N)(3.0 m) + (5.0 sin 30° N)(2.0 m) = 12.5 N m 7. A: Moment of inertia I is the resistance to change in rotational motion and depends on the mass and (radius)2 of the wheel. A wheel with a smaller moment of inertia is easier to set into rotation. The moment of inertia of the alloy wheel is smaller. Since both wheels have the same radius, the mass of the alloy wheel must be smaller. 8. D: Gravitational potential energy of the system

P sin θ = — sin 15° Q

冢

3.0 m 5.0 cos 30o

30°

W

Since the forces W, R and F are in equilibrium, their lines of action must pass through a common point, O. 10. C: Acceleration = – ω 2x Acceleration is maximum at t = 4 s when x = –5 cm 11. A: After 1 s, y1 is displaced 3 cm to the right, and y2 is displaced 3 cm to the left as shown in the graph below. Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:39 AM

y /cm

y1

y2

1 0 1

2

3

4

5

6

7

8

9

10 11 12 x /cm

1

2

3

4

5

6

7

8

9

10 11

–1 y = y1 + y 2

1 0 12 x /cm

–2

Apply the principle of superposition of waves to obtain the resultant wave y = y1 + y2 12. A: x = 5 sin 2t dx v=— — = 10 cos 2t dt dv Acceleration = — — = – 20 sin 2t dt 2π ω = 2 = —— T T When t = —, 6 T acceleration = – 20 sin 2 — 6 2π T — — = – 20 sin — T 6 = – 17.3 cm s–2 13. C: Wavelength of sound produced speed of sound =— — — — — — — — — — — — — frequency, f

冢

冣

冢 冣 冢 冣冢 冣

Wavelength of standing wave speed of travelling wave in the string =— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — frequency, f Speed of sound is not equal to the speed of travelling wave in the string. Hence the wavelengths are different. 14. C: Frequency heard by the cyclist, v + uo f′ = — — — — — f v 343 + 10 — — — — — — — (420) Hz = — 343 = 432 Hz 15. C: Wavelength, λ = distance between two successive compressions = (25 – 5) cm Speed = f λ = (1500)(0.20) m s–1 = 300 m s–1 16. D: There are A (Avogadro’s number) of atoms in 1 mole. M Volume of 1 mole, V = — — ρ M A(d 3) = — — ρ M A=— — — ρd3

冢 冢

冣

冣

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t19 19

17. A: Ductile materials undergo plastic deformation before breaking. Brittle materials do not undergo plastic deformation before breaking. Rigid materials require a large force to produce a small extension. 18. B: R.m.s. speed is directly proportional to T , and pV = nRT. Therefore r.m.s. speed is directly proportional to pV . N 19. C: pV = — — RT NA

冢 冣

N pNA — —= — — — V RT (2.0  10–6)(6.02  1023) –3 =— — — — — — — — — — — — — — — — — — — —m (8.31)(273 + 27) = 4.8  1014 m–3 20. A: Since the volume of the gas increases, work is done by the gas. On the graph, the point S is higher than R. Temperature of the gas increases, and its internal energy increases. Heat must be supplied to the gas to increase its internal energy and for the gas to do work. 21. C: Internal energy of a diatomic gas, 5 U = —nRT, pV = nRT 2 Change in the internal energy 5 = —nR(T2 – T1) 2 5 5 = —(pV)2 – —(pV)1 2 2 5 = —[(1.1  105)(8.0  10 –4) 2 – (20.2  105)(1.0  10–4)] = – 285 J 22. D: (Rate of flow of heat)  (time taken) = heat required to melt all the ice

冢 冣

dθ kA — — (t) = ML dx ML Time taken, t = — — — — — — — — (dθ/dx)kA (20.0)(3.3  105 ) =— — — — — — — — — — — — — — — — — — — — —s [(33 – 0)/0.020](0.50)(1.0) = 8000 s 23. D: Along PQ and RS, the rod is insulated, temperature decreases linearly. Along QR, the rod is not insulated, temperature decreases non-linearly. 24. B: Electric field is stronger where the lines of forces are closer. Actual 2008 STPM Physics Examination Paper

19

2/11/09 10:47:40 AM

25. C: Applying Gauss’ law, algebraic sum of electric charge enclosed electric flux = — — — — — — — — — — — — — — — — — — — — —

ε0

2.0 =— — — — — — — — — — — N m2 C–1 8.85  10–12 = 2.26  1011 N m2 C–1 –t/CR

26. C: V = V0e

27. D:

28. B: 29. B:

30. B: 31. A:

32. D:

33. C: 34. D: 35. B:

20

t ln V = ln V0 – — — — which is the equation of CR a straight line. 1 Gradient of graph = – — — — CR 1 Time constant, CR = –— — — — — — — — — — — — — — gradient of graph (20 – 0) = –— — — — — — s = 10 s (0 – 2) C CA = — 4 1 1 1 — — —= — — —+ — — —, CB = 0.4C CB C/2 2C C C CC = — + — = C 2 2 CD = 4C Energy transferred to the lattice atoms is not lost but is dissipated as heat. In the loop WXYZW, ε1 is positive, but ε2 is negative. I1 is positive, but I2 is negative. Applying Kirchhoff’s Second Law, ΣE = Σ(IR) ε1 – ε2 = I1(R1 + R2 + r1) – I2(R3 + r2) A voltmeter has high resistance, and an ammeter has low resistance. Magnetic force, FM = W, weight of rod BIL sin θ = W 8.0 — — — — — — — — — — — sin θ = — (2.0)(20)(0.40) θ = 30° mv2 FM = Bqv = — — — r q v —=— — m rB Since v, r and B for both the particles are the q same, then the magnitude of — must be the m same for the particles. Positive charge is deflected upwards and negative charge downwards. P must be positively charged and Q negatively charged. At position 3, since the magnetic flux is constant, there is no current induced. Magnetic flux through the coil = N(B cos α)A The r.m.s. current of 1 A is the effective value of the a.c. and equals the steady current of 1.00 A.

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t20 20

36. A: Circuit is that of an inverting amplifier. Rf Output voltage, Vo = – — —V Ri i 100 =– — — —V 25 i

冢 冣 冢

冣

= – 4Vi 37. D: Oscillator circuit uses • C and R in series for feedback. • C and R in parallel for generating a.c. 38. C: Terminal R is connected to voltage supply. r 39. B: Focal length, f = —, object distance u = 2r 2 1 1 1 —=—+— f u v 1 1 1 —=— — —– — — v r/2 2r 2r v=— — 3 v 2r 1 Linear magnification, m = — = — — — —= — u 3(2r) 3 40. C: Real image is formed by a convex lens. When image and object are of the same size, u = v = 2f u + v = 4f = 100 cm f = 25 cm λ 41. D: Path difference of — occurs when light is 2 reflected by a denser medium, on the upper surface. The optical path of the ray reflected at the lower surface is greater by 2nt. λ Total path difference = 2nt + — 2 42. C: Apply Einstein’s equation for photoelectric effect. h 43. B: Momentum of photon, p = — λ 1 Gradient of the graph of p against — = h, λ Planck’s constant 44. D: All the possible transitions are as shown in the figure below. E4 E3

E2

E1

45. A: Minimum wavelength λmin is given by hc — — — = eV λmin hc λmin = —— eV λmin for II > λmin for I VI < VII Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:40 AM

ω0 = 180 rpm 180 = — — — (2π) rad s–1 = 18.8 rad s–1 60

46. B: Photons emitted by stimulated emission have the same frequency. 47. D: Stable nuclei have higher binding energy per nucleon. 48. C: The ratio of the number of nuclei of Y to that of X depends on the half-lives of X and Y. The best option is C. 113 49. C: + 01 n → 114 +γ 48 Cd 48 Cd

冢

ω 2 = ω 02 + 2αθ, 0 – (18.8)2 angular acceleration, α = — — — — — — — — 2(2π)(12) = – 2.34 rad s–2 Angular deceleration = 2.34 rad s–2 (c) Torque = Fr = Iα Iα Tangential force, F = — — r (0.15)(2.34) =— — — — — — — — —N 0.20 (b) Using

hc 112.9044 u + 1.0087 u = 113.9034 u + — — λ hc — — = 0.0097 u λ = (0.0097)(1.66  10–27)c2 (6.63  10–34) λ = ————————————————— — — — — — — — — — — —m (0.0097)(1.66  10–27)(3.00  108) = 1.37  10–13 m 50. C: Strong force holds the nucleons together.

= 1.76 N Alternative method Loss of rotational kinetic energy = work done by torque 1 2 —Iω = (Fr)(2πN) 2 (0.15)(18.8)2 F= — — — — — — — — — —N 4π(12)(0.20) = 1.76 N

PAPER 2 Section A 1. u sin

3.

P

u

冣

4 y u cos

Fundamental mode

x

0

0.20 m h = 1.0 m

First overtone Q L

(a) At any time = t, • horizontal displacement, x = (u cos θ)t 1 • vertical displacement, y = (u sin θ)t – —gt 2 2 (g = acceleration due to gravity) (b) When the ball lands at the point Q, vertical displacement, y = – 1.0 m Consider vertical motion 1 Using y = (u sin θ)t – —gt 2, 2 1 –1.0 = (8.0 sin 30°)t – —(9.81)t 2 2 9.81t 2 – 8.0t – 2.0 = 0 – (–8.0)± (8.0)2 – (4)(9.81)(–2.0) t=— — — — — — — — — — — — — — — — — — — — — — — — — — —s 2(9.81) = 1.02 s Horizontal displacement, L = (8.0 cos 30°)(1.02) m = 7.07 m 2. (a) Initial angular velocity, Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t21 21

4

For fundamental mode,

λ0 — — = 0.20 m 4 v — — = 0.20 m 4f0

345 f0 = — — — — — Hz 4(0.20) = 431 Hz Frequency of first overtone, f1 = 3f0 = 1293 Hz stress 4. (a) Young’s modulus = — — — — strain • Young’s modulus of copper is greater than Young’s modulus of aluminium. • A piece of copper wire requires a greater stress to produce the same strain as a piece of aluminium wire. • Therefore copper is more rigid than aluminium. Fundamental frequency,

Actual 2008 STPM Physics Examination Paper

21

2/11/09 10:47:41 AM

Maximum tensile strength of copper is greater than that of aluminium. Hence the elastic limit of copper is higher than that of aluminium. (b) Assuming that the extension is below the limit of proportionality, stress Young’s modulus, Y = — — — — strain F/A =— — — — e/l Extension, Fl e=— — YA (10.0)(2.0) =— — — — — — — — — — — — — — — — — — — — — —m (69  109)π(0.05  10–3)2 = 3.69 cm 5. (a) V = 240 sin 5000t and π I = 0.480 sin (5000t – —) 2 Therefore, V0 = 240 V and I0 = 0.480 A V0 Reactance, X = — — I0 240 =— — — —Ω 0.480 = 500 Ω (b) Phasor diagram V

2

rad

7. (a) Using Einstein’s equation, maximum kinetic energy, 1 hc 2 —mv max =— —– W 2 λ –34 (6.63  10 )(3.00  108) =— — — — — — — — — — — — — — — — — — — — — — – 2.3(1.60  10–19) 450  10–9 = 7.4  10–20 J 2 — — — — — — — — — — (7.4  10–20 ) m s–1 vmax = — 9.11  10–31

冢

= 4.03  105 m s–1 (b) If Vs is the stopping potential, then 1 2 eVs = —mv max 2 = 7.4  10–20 J 7.4  10–20 Vs = — — — — — — — — — —V 1.60  10–19 = 0.463 V 8. (a) Time, t = 24 h, T1/2 = 10 h 24 x=— — = 2.4 10 After 24 hours, number of atoms, N0 N=— — 2x 6.0  1021 =— — — — — — — — 22.4 = 1.14  1021 Number of atoms which decay in 24 hours = (6.0 – 1.14)  1021 = 4.86  1021 (b) After 24 hours, number of atoms = 1.14  1021 Mass of (6.0  1021) atoms = 2.0 g

I

(c) Since the voltage V leads the current I by π — radians, the component is an inductor. 2 6. (a) Open-loop gain, A0 = 105 (b) Condition for output voltage, Vo = A0(V2 – V1): A0(V2 – V1)  voltage of power supply or voltage of power supply — — — — — — — — — — — — — — — — — — (V2 – V1)  — A0 (c) Advantage of high input impedance is that the op-amp requires a small current for it to operate. The value of the input voltage after connection to the op-amp is the same as before it is connected to the op-amp. Advantage of low output impedance is that the voltage across the component which is connected to the output is the same as the output voltage VO. 22

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t22 22

冣

Mass of (1.14  1021) atoms 1.14  1021 = — — — — — — — — — — (2.0) g 6.0  1021 = 0.38 g

冢

冣

Section B 9. (a) A body is in simple harmonic motion if its acceleration is directly proportional to its displacement x from a fixed point and is constantly directed towards the fixed point. Acceleration = –ω 2x (ω 2 = constant) (b)

(i)

x = 0.50 cos (2t + φ) dx Velocity, v = — — = – 2(0.50) sin (2t + φ) dt = – 1.00 sin (2t + φ) Maximum velocity = 1.00 m s–1 dv (ii) Acceleration = — — dt = – 2.00 cos (2t + φ) Maximum acceleration = 2.00 m s–2 Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:41 AM

(iii) At time t = 0, velocity = – 0.20 m s–1 – 1.00 sin [2(0) + φ] = – 0.20 sin φ = 0.20 φ = 11.5° (c) (i) If the bob is displaced and then released, the displacement of the bob after a time t is given by

冢 冣

x = x0 cos ω t

(iv)

Kinetic energy = potential energy 1 1 —mω 2(x 02 – x2) = —mω 2x2 2 2 x0 x=— — = 0.707x0 2 x0 When x = x0 cos ω t = — —, 2 π ωt = — 4 π π t=— — —= — — — — — — 4ω 4(2π/T) T =— 8 (ii)

(b)

(iii)

x

3kT v= — — — m 3RT — — = — M

0.707x0 3T/8 T/8 T/4

T

t

–0.707x0

From the graph above, the kinetic energy and the potential energy are again equal when the distance of the bob from the equilibrium position is again 0.707x0, which is at time 3T T T t=— — which is — after t = — 8 4 8 10. (a) (i) Assumptions (any two) • Internal energy of the ideal gas consists only of the total kinetic energy of the gas molecules. • There is no force between gas molecules except during collisions. • Collisions between gas molecules are perfectly elastic. • Volume of the gas molecules is negligible compared to volume of the gas. (ii) A gas behaves as an ideal gas when • the pressure is very low, or approaches zero. • the temperature is low. (iii) V = 0.35 m3, mass of gas, m = 7.0 kg, T = 27 °C = 300 K, molecular mass, M = 58 g mol–1 Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t23 23

(i)

(ii)

x0

0

m Using pV = — — RT, M mRT pressure, p = — — — — MV (7.0)(8.31)(300) =— — — — — — — — — — — — — — Pa (58  10–3)(0.35) = 8.60  105 Pa Because of the high pressure, 8.60  105 Pa, butane gas deviates from the behaviour of the ideal gas. Degrees of freedom – independent modes of motion or independent modes of acquiring kinetic energy by the gas molecule. Kinetic energy of a gas molecule, f E = —kT 2 where k is Boltzmann’s constant and T is the temperature of the gas in kelvin. Translational kinetic energy of oxygen molecule at 300 K, 1 3 —mv2 = —kT 2 2 Speed of oxygen molecule,

11. (a)

A

A

冣

=R

3(8.31)(300) = — — — — — — — — — — m s–1 0.032 = 483 m s–1 < 5.0  103 m s–1, the escape speed. Hence, oxygen molecules will not escape and can exist on Mars. ε0A (i) Capacitance of capacitor, C = — — — d Charge on each plate of capacitor, Q = CV ε0A = — — —V d (8.85  10–12)(0.50) =— — — — — — — — — — — — — — — [75 – (–75)] C (25  10–3) = 2.66  10–8 C (ii) Energy stored in the capacitor, 1 U = —QV 2 1 = —(2.66  10–8)(150) J 2 = 2.00  10–6 J (i) The capacitor is charged by the battery. Energy is stored in the electric field between the plates of the capacitor. When the switch is closed, the capacitor discharges. Energy stored in the capacitor is used to produce a flash in the flash bulb.

冢

(b)

M —, kN 冢m = —— N

冣

Actual 2008 STPM Physics Examination Paper

23

2/11/09 10:47:42 AM

Image distance, v = 12.0 cm Final image I2 is 12.0 cm from the lens as shown in the above figure. (iii) Overall magnification 60.0 12.0 = — — — — — — — — 20.0 60.0 = 0.60

(ii) Voltage across the capacitor during charging, V = V0(1 – e–t / CR) When the capacitor is 63% charged, V = 0.63V0 0.63V0 = V0(1 – e–t / CR) 1 et / CR = — — — — 0.37 Time taken, 1 t = (CR) ln — — — — 0.37 1 = (40.0  10–6 )(45.0  103) ln — — — s 0.37 = 1.79 s The charging time can be reduced by using a resistor of resistance smaller than 45.0 kΩ. r1 = r2 = 15.0 cm Focal length f of the lens is given by 1 1 1 — = (n – 1) — + — f r1 r2 2 = (1.5 – 1) — — — — 15.0 Focal length, f = 15.0 cm f = 15.0 cm, u = 20.0 cm 1 1 1 Using — = — + —, f u v 1 1 1 —= — – — v f u 1 1 =— — —– — — — 15.0 20.0 Image distance, v = 60.0 cm v Magnification, m = — u 60.0 =— — — = 3.0 20.0 Image is real, inverted and magnified.

冢

12. (a)

(i)

冢

(ii)

(iii)

(b)

(iv) (i) Lens

O

冣

Mirror

I1

I2

First image formed by lens Final image

(ii) With the mirror behind the lens, light reflected by the mirror passes through the lens. The image I1 acts as a virtual object for the lens. u = – 60.0 cm, f = 15.0 cm 1 1 1 —=—–— v f u 1 1 =— — —– — — — — — 15.0 – 60.0 24

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t24 24

冣

Electron in excited state

Photon

冣

冣

冢

冣冢

13. (a)

冣

冢

(iii)

冢

Electron transition

Photon

Stimulated photon

Stimulated emission occurs when • a photon passes by an excited atom. • an electron in the excited state is stimulated to transit to a lower energy level. • a photon is produced by the transition of the electron. • Condition: frequency of the photon that stimulates the transition should be the same as the frequency of the emitted photon. Population inversion occurs when • excited atoms of the pumping agent collide with atoms of the lasing agent. • atoms of the lasing agent are raised to a metastable state. • this results in more atoms in the excited state than atoms in the ground state. (b) Differences between laser and fluorescent light (any two) Laser

Fluorescent light

Monochromatic

Not monochromatic

Coherent or in phase

Not coherent

Beam does not diverge

Beam diverges

High intensity

Low intensity

(c) Energy of photon hc =— — λ (6.63  10–34 )(3.00  108 ) =— — — — — — — — — — — — — — — — — — — — — — eV (450  10–9 )(1.60  10–19 ) = 2.76 eV (d) In the hydrogen atom, for an electron to orbit the nucleus, mv2 e2 centripetal force, — — —= — — — — — r 4πε0r2 e2 v2 = — — — — — — 4πε0mr Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:42 AM

According to Bohr’s model, angular nh momentum, (mv)r = — — 2π nh 2 m 2v 2r 2 = — — — 2π 2 e nh 2 m2 — — — — — — — r2 = — — — 4πε0mr 2π h2ε0 Radius of electron orbit, r = — — — — — n2 πme 2 Energy of electron, En = kinetic energy + electric potential energy 1 –e2 e2 = —mv2 + — — — — — v2 = — — — — — — 2 4πε0r 4πε0mr

冢

冣

冢

冣

冢

冣

冢

冢

冣

Wavelength, (6.63  10–34)(3.00  108) λ = ——————————————— — — — — — —m 4.064  10–19 = 4.89  10–7 m 14. (a) Binding energy of a nucleus is the energy required to completely separate the nucleons of the nucleus. Mass defect of a nucleus = (total mass before reaction) – (total mass after reaction)

冣

(b) Mass defect = (mp + 2mn) – mN

冥

冤

= [1.007276 + 2(1.008665)] u – 3.016049 u = 0.008557 u Binding energy = (0.008557)(931.5) MeV = 7.97 MeV 7.97 Binding energy per nucleon = — — — MeV 3 = 2.66 MeV (c) (i) The most stable nucleus is the nucleus with the highest binding energy per nucleon, that is 62 28Ni. 2 2 (ii) + → 24 He 1H 1H Energy difference in the = released binding energy = 7 – 2(2  1) MeV = 3 MeV

hε — — —冣n 冥 冤r = 冢—πme

– e2 =— — — — — 8πε0r

2

0 2

2

– me4 =— — — — — — 8ε 02 h2n2 For the transition from n = 4 to n = 2, energy of photon emitted, hc — — = E4 – E2 λ – me4 1 1 =— — — — — — —– — — 8ε 02 h2 42 22 (9.11  10–31)(1.60  10–19)4 =— — — — — — — — — — — — — — — — — — — — — — — — — 8(8.85  10–12)2(6.63  10–34)2 1 1 — — –— — J 2 42 2 = 4.064  10–19 J

冢

冢

冣

冣

(d) Source of positive ions S1

Collimator slits Velocity selector – combined electric and magnetic fields

S2

x -

x

FE

FM

x

Photographic plate d3

+

x

d2

d1

S3

x

v

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

Uniform magnetic field B

x

x

x

x

x

x

x

x

x

x

Bainbridge mass spectrometer

• Ions from the source are collimated by the narrow slits S1 and S2. • In the velocity selector, mutually perpendicular electric field E and magnetic field B act. The selected velocity v is given by

Majlis Peperiksaan Malaysia 2007

Ace Ahead Physics (Actual 08) 5t25 25

Actual 2008 STPM Physics Examination Paper

25

2/11/09 10:47:43 AM

FM = FE qBv = qE E v=— — B • Ions with the selected velocity emerge from the slit S3 and are deflected by the magnetic field B. Ions of different masses follow different semicircular paths to strike a photographic plate.

26

Actual 2008 STPM Physics Examination Paper

Ace Ahead Physics (Actual 08) 5t26 26

mv2 — — — = qBv r qB 2r Mass of ion, m = — — — — E The radii of the paths are obtained by measuring the diameters d1, d2 and d3.

Majlis Peperiksaan Malaysia 2007

2/11/09 10:47:43 AM