actual 2009 stpm

1 © Majlis Peperiksaan Malaysia 2009

Answer all questions.

1 Which error is not a systematic error?A The zero error of a micrometer screw gaugeB The error due to the calibration of a thermometerC The end error in the slide wire of a Wheatstone bridgeD The error in reading to the smallest scale of a metre rule

2 A stone is thrown vertically upwards from a height of 10.0 m from the ground. If the initial velocity is 5.0 m s–1, what is the time taken by the stone to reach the ground?[The acceleration of free fall is 10.0 m s–2.]A 0.5 s B 1.0 s C 1.5 s D 2.0 s

3 Three coplanar forces 2.0 N, 4.0 N and 5.0 N acting on a body are shown in the diagram.The resultant force isA 6.9 N at an angle of 19.7° above the x-axisB 6.9 N at an angle of 19.7° under the x-axisC 9.1 N at an angle of 44.2° above the x-axisD 9.1 N at an angle of 44.2° under the x-axis

4 A train has an engine that provides a constant thrust force F when it is accelerating from rest. If the train experiences a dragging force F� when moving at speed v, what is the power required to accelerate it?A Fv B F�v C (F – F�)v D (F + F�)v

5 A coin is placed on a horizontal disc at a distance 1.0 m from the axis of rotation. The disc is rotated with an increasing speed. If the coefficient of static friction between the disc and the coin is 0.40, the coin will slip when its speed exceedsA 1.4 m s–1 B 2.0 m s–1 C 2.2 m s–1 D 3.1 m s–1

6 A ball of mass m and radius R rolls at a linear speed of u along a horizontal plane without

slipping. If the moment of inertia of the ball is 23

mR2, the kinetic energy of the ball is

A 16

mu2 B 13

mu2 C 12

mu2 D 56

mu2

7 The static frictional force influences sliding movement. When an object is placed on a surface, the resistance to slide the object will increase whenA a weight is placed on top of the objectB a layer of oil is smeared on the surfaceC air is allowed between the object and the surfaceD ball bearings are placed between the object and the surface

Time: 1 h 45 min

60°

5.0 N

2.0 N

4.0 N x

y

PAPER 1

2009EXPaperVol1&2 for CD.indd 12009EXPaperVol1&2 for CD.indd 1 2/4/10 2:41:44 PM2/4/10 2:41:44 PM

2© Majlis Peperiksaan Malaysia 2009

8 Which statement about a satellite of a planet is not true?A A satellite with a larger orbit moves at a lower speed.B The period of a satellite depends on its distance from the planet.C A satellite moves at the same angular speed as the rotation of the planet.D The centripetal force is due to the gravitational attraction between the planet and a

satellite.

9 The movement of a spacecraft from X at the surface of the Earth to Y at the surface of the Moon along a straight line XY is shown in the diagram.Which graph represents the variation of the potential energy U of the spacecraft with the distance from X to Y?A B C D

U

X Y0Distance

U

X Y0

Distance

U

X Y0Distance

U

X Y0Distance

10 A light platform supported by two identical springs is shown in the diagram.A mass is placed at the centre of the platform so that the springs are compressed by 3.0 cm. The platform is depressed further by 1.0 cm and then released so that it oscillates up and down vertically. Which graph shows the variation of displacement s of the platform with time t?A B C D s/cm

1.0

0.5

0–0.5

–1.0

t /s

s/cm 1.0

0.5

0 –0.5

–1.0

t /s

s/cm

3.0

2.0

1.0

–1.0

–2.0

–3.0

0 t /s

s/cm

3.0 4.0

2.0

1.0

–1.0

–2.0

–3.0

–4.0

0 t /s

11 An object of mass 2.0 kg undergoes a simple harmonic motion according to the equation x = 5.0 cos 100t, where x is the displacement in metres and t is the time in seconds. What is the frequency and kinetic energy of the object when the displacement is 3.0 m?

Frequency Kinetic energy

A 16 Hz 160 kJ

B 16 Hz 250 kJ

C 100 Hz 160 kJ

D 100 Hz 250 kJ

12 Two bodies P and Q are oscillated with initial displacement x0. The variation of the displacement x with time t is shown in the graph.A coersive force with frequency f is then applied to P and Q. Which graph shows the variation of amplitude a for P and Q with the frequency f ?

x

x0 PQ

t

EarthMoon

X Y

Spring

Light platform

Floor

2009EXPaperVol1&2 for CD.indd 22009EXPaperVol1&2 for CD.indd 2 2/4/10 2:41:47 PM2/4/10 2:41:47 PM

3 © Majlis Peperiksaan Malaysia 2009

A B C D

a

PQ

f

a

PQ

f

a

P

Q

f

aP

Q

f

13 The variation of displacement x with time t for a wave is shown in the graph.If the displacement is 1.0 cm at a particular instant, what is the displacement after 2.0 s?A –1.0 cm C 1.0 cm B 0 cm D 1.5 cm

14 A progressive wave of frequency 50.0 Hz overlaps with a reflected wave to produce a standing wave. If the distance between two consecutive nodes is 10.0 cm, the speed of the progressive wave isA 2.5 m s–1 B 5.0 m s–1 C 10.0 m s–1 D 20.0 m s–1

15 The diagram shows the shape of a beat wave which is produced by two tones of frequencies f1 and f2, where f1 > f2.If the resultant wave completes four cycles within 0.20 s, what are the values of f1 and f2?

f1 f2

A 15.0 Hz 5.0 Hz

B 22.5 Hz 17.5 Hz

C 40.0 Hz 35.0 Hz

D 45.0 Hz 35.0 Hz

16 The variation of the intermolecular force F with the separation r between two molecules is shown in the graph.Which statement about the situation at point P is true?A The potential energy is minimum.B The molecule is in a state of equilibrium.C The force is a repulsive force and in its weakest state.D The force is an attractive force and in its strongest state.

17 A 99.0 cm long rod PQ of negligible weight which is supported at both of its ends by two equally long wires X and Y is shown in the diagram.The cross-sectional areas of X and Y are 1.50 mm2 and 3.00 mm2 respectively. The Young modulus of wires X and Y are 1.80 × 1011 Pa and 1.20 × 1011 Pa respectively. A weight W is suspended from the rod such that X and Y are stretched. At what distance from P should the weight be suspended along the rod so that it remains horizontal?A 24.6 cm B 42.4 cm C 56.6 cm D 74.4 cm

1.5

–1.51.0 2.0

0

x/cm

t/s

F

0r

P

WP

X Y

Q

Rod

0

0.50

Displacement/m

Time/s0.20 0.40

2009EXPaperVol1&2 for CD.indd 32009EXPaperVol1&2 for CD.indd 3 2/4/10 2:41:49 PM2/4/10 2:41:49 PM

4© Majlis Peperiksaan Malaysia 2009

18 The pressure of an ideal gas in a vessel of fixed volume is p. If the gas in the vessel is heated so that the r.m.s. speed of its molecules is increased three times the initial value, what is the new pressure of the gas?

A 13

p B 3p C 3p D 9p

19 Which one is not an assumption of the kinetic theory of gases in an ideal gas model?A The motion of molecules is random.B The mass of a molecule is negligible as compared to the mass of gas.C The collision between a molecule and the wall of a container is elastic.D The volume of molecules is negligible as compared to the volume of gas.

20 As a gas undergoes an isothermal expansion, 20 J of heat is supplied to it. What is the change in the internal energy ΔU and the work done W by the gas in this process?

ΔU W

A –20 J 0

B 0 –20 J

C 0 +20 J

D +20 J 0

21 The variation of internal energy U of one mole of an ideal gas with temperature T is shown in the graph.The ratio of the molar heat capacity at constant pressure to the molar heat capacity at constant volume isA 1.40 C 1.67B 1.42 D 1.71

22 A silver rod and an iron rod of the same mass have densities of 10 500 kg m–3 and 7860 kg m–3 respectively. Both rods conduct the same amount of heat in one second when the differences in temperature between the two ends of the rods are the same. If the thermal conductivities of silver and iron are 420 J s–1 m–1 K–1 and 79 J s–1 m–1 K–1 respectively, what is the ratio of the radius of the silver rod to the radius of the iron rod?A 0.38 B 0.61 C 1.64 D 2.63

23 If work W is required to move a charge q from point X to point Y, the potential difference between X and Y is

A Wq2 B Wq C Wq

D Wq2

24 When an alpha-particle approaches a 9642Mo nucleus head-on, the closest distance between

the alpha-particle and the nucleus is d. If the same alpha-particle approaches a 20984Po nucleus

head-on, what is the closest distance between the alpha-particle and the 20984Po nucleus?

A 0.5d B 2.0d C 2.2d D 4.0d

25 A battery, a resistor and a capacitor connected in series are shown in the circuit.The charge stored in the capacitor is Q when the current through the circuit is I. The potential difference V in the circuit is

A IR + QC

B IR

C QC

D IR – QC

2805800

6000

6200

6400

6600U/J mol–1

300 320 T/ K

C

I R

V

2009 STPM Physics Papers 1 and 2

2009EXPaperVol1&2 for CD.indd 42009EXPaperVol1&2 for CD.indd 4 2/4/10 2:41:50 PM2/4/10 2:41:50 PM

5 © Majlis Peperiksaan Malaysia 2009

26 The capacitance of a parallel-plate capacitor will increase by increasingA the separation between the plates B the amount of charge on the platesC the potential difference between the platesD the permittivity of a medium between the plates

27 The resistivity of an ohmic material varies withA current B temperature C electric field D potential difference

28 A machine requires a current of 50.0 A to operate. The machine will breakdown due to overheating if the rate of heat generated per metre length of a copper wire exceeds 2.0 W. What should be the minimum radius of the copper wire in order to avoid overheating?[Resistivity of copper = 2.0 × 10–8 Ω m.]A 2.8 mm B 4.0 mm C 5.0 mm D 5.6 mm

29 Two resistors R1 and R3 and a rheostat R2 are shown in the circuit.What happens to the readings of A1, A2 and A3 when the resistance R2 is increased?

A1 A2 A3

A Increases Decreases Increases

B Increases Decreases Decreases

C Decreases Decreases Increases

D Decreases Decreases Decreases

30 Two resistors 3.0 Ω and 2.0 Ω connected to a 6.0 V cell of internal resistance 2.0 Ω are shown in the circuit.What is the current flowing through the 3.0 Ω resistor in the circuit?A 0.75 A B 1.13 A C 1.88 A D 2.00 A

31 In a moving coil galvanometer with a radial magnetic field, a torque τ is produced by a current I flowing through a coil. Which graph represents the relationship between τ and I?A B C D

τ

I

τ

I

τ

I

τ

I

32 A cylindrical magnet placed along the axis of a circular coil is shown in the diagram.Which movement does not produce any induced e.m.f. at the terminals of the coil when the magnet is moved in the direction of the arrow?A B C D

N S

N S

N S

N S

A2

A1

R2 R3

R1

A3

6.0 V, 2.0 Ω

3.0 Ω

2.0 Ω

N S

Magnet

Coil

2009EXPaperVol1&2 for CD.indd 52009EXPaperVol1&2 for CD.indd 5 2/4/10 2:41:51 PM2/4/10 2:41:51 PM

6© Majlis Peperiksaan Malaysia 2009

33 A rectangular coil of width w and length � moving at a constant velocity v through a uniform magnetic field B directed perpendicular to v is shown in the diagram.If the magnetic field is confined to a region of width 3w, which graph shows the variation of induced e.m.f. ε with distance x?A B C D

–w x

ε

w 2w 3w 4w0

–w x

ε

2w 3w 4www0

–w x

ε

2w 3w 4ww0

–w x

ε

2w 3w 4ww0 w

34 In an alternating current circuit, the variation of voltage V and current I with time t for a capacitor can be represented by

V = 100 sin 100t and I = 0.1 sin 100t + π2

,

where V is in volts, I is in amperes and t is in seconds. What is the capacitance of the capacitor?A 1.6 μF B 10 μF C 1.6 F D 10 F

35 Which circuit produces full-wave rectification with terminal X being always positive and terminal Y being always negative?A B C D

X

Y

+

–

X

Y

+

–

X

Y

+

–

X

Y

+

–

36 Which quantity will increase when a voltage supply of an operational amplifier is increased?A Open-loop gain C Saturated voltageB Closed-loop gain D Frequency bandwidth

37 An operational amplifier circuit is shown in the diagram below.

Vi –

+V– V0

Io

Ii

Rf

Ri

V+

Which expression about the circuit is not true?

A V– ≈ V+ B Ii = Io C Vo = – Rf

RiVi D Vo = – 1 + Rf

RiVi

38 The electric field and magnetic field in electromagnetic waves areA 90° out of phase C parallel to each otherB mutually dependent D oscillating in the direction of propagation

39 If the oscillation equation of the electric field in an electromagnetic wave is y = E0 sin (ωt – kx), the oscillation equation of the magnetic field isA x = B0 sin (ωt + kx) C z = B0 sin (ωt – kx)B y = B0 sin (ωt + kx) D z = B0 sin (ωt + kx)

v

w

3w

–w w x2w 3w 4w0

�

2009EXPaperVol1&2 for CD.indd 62009EXPaperVol1&2 for CD.indd 6 2/4/10 2:41:53 PM2/4/10 2:41:53 PM

7 © Majlis Peperiksaan Malaysia 2009

40 Where should an object be placed in front of a convex mirror and a concave mirror so that the image formed is virtual?

Convex mirror Concave mirror

A At a distance less than the focal length At a distance less than the focal length

B At a distance less than the focal length At any position

C At any position At a distance less than the focal length

D At any position At any position

41 Which one is true about the slit separation a, the distance D from slits to the screen and the separation of fringes in a Young’s two-slit experiment?

a D Separation of fringes

A Unchange Increase Increase

B Unchange Decrease Increase

C Increase Unchange Increase

D Decrease Unchange Decrease

42 An air-wedge formed by a strand of hair between two glass slides each of length 5.0 cm is shown in the diagram.The air-wedge is illuminated from above by monochromatic light of wavelength 580 nm. If the separation between two consecutive dark fringes is 0.28 mm as viewed from above, what is the thickness t of the hair?A 2.6 × 10–5 m B 5.2 × 10–5 m C 1.0 × 10–4 m D 1.6 × 10–4 m

43 A photocell consists of a material of threshold frequency f0. If monochromatic light of frequency 3f0 is directed to the photocell, its stopping potential is

A 4hf0e

B 3hf0e

C 2hf0e

D hf0e

44 Proton, electron, deuteron and helium nucleus are particles. If these particles have equal kinetic energy, which particle has the smallest de Broglie wavelength?A Proton B Electron C Deuteron D Helium nucleus

45 Four energy levels of an atom and the line spectrum produced from the electron transitions between the energy levels are shown in the diagrams (a) and (b) respectively.

Which transitions produce P and Q lines?

P Q

A E4 → E3 E2 → E1

B E4 → E2 E4 → E1

C E3 → E2 E4 → E1

D E2 → E1 E4 → E3

46 When a beam of X-rays of wavelength 3.10 × 10–10 m is directed at a glancing angle of 26.5° to a crystal, a second-order diffraction is observed. What is the interplanar distance of the crystal?A 3.88 × 10–10 m B 6.95 × 10–10 m C 7.76 × 10–10 m D 1.39 × 10–19 m

Glass slidesHair

t

5.0 cm

E4

E3

E2

E1

(a)

P

λ increasing

Q

(b)

2009EXPaperVol1&2 for CD.indd 72009EXPaperVol1&2 for CD.indd 7 2/4/10 2:41:55 PM2/4/10 2:41:55 PM

8© Majlis Peperiksaan Malaysia 2009

47 Which statement is not true of stimulated emission in laser production?A The photon emitted is in phase with the incoming photon.B A photon is emitted when an incoming photon is absorbed.C One incoming photon will cause two photons to be emitted.D The photon emitted moves in the same direction as the incoming photon.

48 A high energy particle which is emitted when a beryllium foil is bombarded by an alpha-particle is a A proton B neutron C beta-particle D gamma-ray photon

49 A Geiger-Muller tube is placed at a certain distance from a source of radioactive. The count rate R is recorded for a long period of time until the count rate from the source is extremely low as compared to the count rate of the background. Which graph shows the variation of the count rate with time t?A B C D

R

0t

R

0t

R

0t

R

0t

50 Nuclide 12081P decays into AZQ by emitting two α-particles and one positron. What are the

values of A and Z?

A Z

A 201 76

B 202 76

C 202 78

D 206 78

PAPER 2

Time: 2 h 30 min

Section A [40 marks]

Answer all questions in this section.

1 A level circular track of radius 120 m is designed for a car moving at 80.0 km h–1. Calculate the minimum value of the coefficient of static friction between the tyres and the track so that the car does not skid. [4 marks]

2 A 12.0 kg block on a horizontal surface with a horizontal force F applied to it is shown in the diagram.

The coefficient of friction between the block and the surface is 0.85.

(a) Calculate the minimum value of F to slide the block. [3 marks](b) If F is 25.0 N, what is the frictional force acting on the block? [1 mark](c) If F is 125 N, what is the initial acceleration of the block? [2 marks]

3 (a) A wave pulse propagating in a string from left to right is shown in the diagram. Sketch the form of the reflected wave pulse if the end

P of the string is tied

F

Rod

P

2009EXPaperVol1&2 for CD.indd 82009EXPaperVol1&2 for CD.indd 8 2/4/10 2:41:56 PM2/4/10 2:41:56 PM

9 © Majlis Peperiksaan Malaysia 2009

(i) tightly to the rod, [1 mark] (ii) to a bangle which moves freely along the rod. [1 mark]

(b) (i) State the difference between resonant frequency in a closed pipe with that in an open pipe. [1 mark]

(ii) If a standing wave in a pipe is not propagated, how could the sound from the pipe still be heard? [2 marks]

4 (a) State a condition for heat conduction in a medium. [1 mark](b) A good cooking utensil has a high thermal conductivity but a low heat capacity.

Explain the significance of these characteristics. [2 marks](c) A steel cooking pot has an effective surface area of 0.125 m2 and a thickness of

1.0 × 10–3 m. Calculate the rate of heat flow through the bottom of the pot if the temperature difference is 690 °C. [2 marks]

[The thermal conductivity of steel = 66.9 W m–1 K–1.]

5 (a) On the same axes, sketch graphs of resistivity against absolute temperature for pure metal and superconductor. Label your graphs. [2 marks]

(b) A copper wire of radius 1.5 mm and free electron density of 8.4 × 1028 m–3 carries a current of 5.0 A. Calculate the drift velocity of the free electrons in the wire. [3 marks]

6 A combination of electrical components in a circuit is shown in the diagram below.

I

4.0 V

10.0 Ω 20.0 μF

45.0 Ω 15.0 Ω

30.0 Ω

Calculate the final current I. [5 marks]

7 (a) State the principle of the production of X-ray. [1 mark](b) If the potential difference applied to an X-ray tube is 25 kV, calculate

(i) the speed of the electron that strikes the target, [2 marks] (ii) the minimum wavelength of the X-ray produced. [2 marks]

8 The main sections of a mass spectrometer is shown in the diagram below.

BBBBBB111

B2BB

S1 SS22

–P

Ionsource

Detector

+

(a) State the main use of the mass spectrometer. [1 mark](b) Explain the use of (i) parallel plates P and the uniform magnetic field B1, [2 marks] (ii) the uniform magnetic field B2. [2 marks]

Resistivity

Absolute temperature0

2009EXPaperVol1&2 for CD.indd 92009EXPaperVol1&2 for CD.indd 9 2/4/10 2:41:57 PM2/4/10 2:41:57 PM

10© Majlis Peperiksaan Malaysia 2009

Section B [60 marks]

Answer any four questions in this section.

9 (a) (i) Define the work done on a body by a force. [2 marks] (ii) Using an equation of linear motion with constant acceleration, show that the work

done on a body by a constant force is equal to the change in the kinetic energy of the body. [4 marks]

(iii) State the principle of conservation of energy. [1 mark](b) Two blocks X and Y with masses 6.0 kg and 3.0 kg

respectively connected by a string passing over a smooth pulley are shown in the diagram.The base of block X is 4.0 m above the ground while block Y rests on the ground. If block X is released from rest and the mass of the pulley is negligible, calculate (i) the speed of block X just before it strikes the ground, [3 marks] (ii) the distance travelled by block Y to reach maximum height. Explain how your

answer will change if the mass of the pulley is not negligible. [5 marks]

10 (a) One mole of a diatomic ideal gas is heated at a constant pressure of 1.01 × 105 N m–2 from 15.0 °C to 35.0 °C. Calculate (i) the heat absorbed by the gas, [3 marks] (ii) the change in the volume of the gas, [2 marks] (iii) the work done by the gas, [2 marks] (iv) the change in the internal energy of the gas. [2 marks]

(b) A diatomic ideal gas in a container has a pressure of 1.01 × 105 N m–2 at 20.0 °C. The gas undergoes adiabatic compression until its volume becomes a quarter of its original volume. (i) What is meant by an adiabatic compression? [1 mark] (ii) Determine the final temperature of the gas. [3 marks] (iii) Calculate the work done on the gas. [2 marks]

11 (a) A magnetic field B passing through a horizontal wire loop PQRS of an area A and making an angle θ with the plane of the loop is shown in the diagram. (i) Express, in terms of B, A and θ, the magnetic

flux Φ through the loop. [1 mark] (ii) State two alterations which can produce induced e.m.f. in the loop. [2 marks] (iii) If the magnetic flux through the plane of the loop PQRS is reduced, what is the

direction of the induced current in the loop? State the law used. [2 marks](b) (i) Define self-inductance. [2 marks] (ii) An inductor in the form of solenoid has length �, cross-sectional area A and total

turns N. Show that the self-inductance L is given by

L =

μ0N 2A�

,

where μ0 is the permeability of free space. [2 marks]

S

B

QP

R R Rθ

Y4.0 m

X

2009EXPaperVol1&2 for CD.indd 102009EXPaperVol1&2 for CD.indd 10 2/4/10 2:42:07 PM2/4/10 2:42:07 PM

11 © Majlis Peperiksaan Malaysia 2009

(c) A 50.0 Hz a.c. generator consists of a coil of N turns and area 2.0 × 10–2 m2 which rotates in a uniform magnetic field of 0.15 T. Calculate N so that the root mean square voltage produced is 240 V. [3 marks]

(d) (i) At the moment an electric motor is switched on, a large amount of current flows in the armature and causes damage. Why does this happen? [1 mark]

(ii) The electric motor is usually connected in series to an inductor to avoid the above situation from occurring. How can the inductor help avoid the damage? [2 marks]

12 (a) (i) What is the characteristic of light which enables polarisation to occur? [1 mark]

(ii) A charged particle oscillating in the y-direction is shown in the diagram.If the electromagnetic wave propagates in the positive z-direction, state the directions of the oscillation of the electric field E and magnetic field B components of the electromagnetic wave produced. [2 marks]

(b) Unpolarised light with the electric component of amplitude E0 passing through two polaroids is shown in the diagram. (i) What is meant by unpolarised light? [2 marks] (ii) Explain why, after the light passes through the first polaroid, the intensity of the light

is half of its original value even though the amplitude of the electric component of light before and after passing through the first polaroid is the same. [3 marks]

(iii) Calculate the angle θ so that the original intensity of light can be reduced to 25% after passing through both polaroids. [5 marks]

(c) Light can be polarised by reflection. A light beam is directed to a glass surface with a refractive index of 1.52. Calculate the incident angle so that the reflection of the light beam is fully polarised. [2 marks]

13 In the Bohr model, an electron of mass m moves in a circular orbit around its nucleus at a speed of

v =

nh2πmr

,

where n is an integer, h the Planck constant and r the radius of the orbit. For the hydrogen atom,

(a) show that the radius of the nth orbit is given by

rn =

n2h2ε0

πme2 ,

where ε0 is the permittivity of free space, [3 marks](b) determine the radius of the smallest orbit, [2 marks](c) show that the total energy at the nth level is given by

[3 marks]En = –

1

n2

(e4m

8ε02h2

),

(d) calculate the ionisation energy of the atom, [3 marks](e) calculate the wavelength of the radiation emitted when the electron makes a transition

from the energy level n = 4 to n = 2. Hence, state the type of electromagnetic radiation emitted. [4 marks]

z

xCharge

y

E0 E0 E�

Polaroid Polaroid

Polarisation axis Polarisation axis

θ

2009EXPaperVol1&2 for CD.indd 112009EXPaperVol1&2 for CD.indd 11 2/4/10 2:42:08 PM2/4/10 2:42:08 PM

12© Majlis Peperiksaan Malaysia 2009

14 (a) What is meant by nuclear fusion? [1 mark](b) A fusion reaction is represented by

21H + 31H → X + 10n + 18.2 MeV.

(i) Determine nuclide X. [1 mark] (ii) Calculate the atomic mass of X in u. [4 marks]

[Atomic mass of neutron = 1.008665 u, atomic mass of deuterium = 2.014102 u, atomic mass of tritium = 3.016049 u and 1 u = 931 MeV.]

(c) In the interior of some stars, fusion takes place as two protons experience a head-on collision with a total initial kinetic energy of 1.2 × 10–13 J. (i) Determine the head-on collision energy for each proton. [1 mark] (ii) Show that this fusion could only take place at a very high temperature. [3 marks]

(d) The atomic mass of 2152Mg is 24.985837 u and the atomic mass of 215

3Al is 24.990429 u. (i) What is meant by radioactivity? [1 mark] (ii) Which nucleus will decay to the other nucleus? [1 mark] (iii) What type of decay is involved in (c)(i)? [1 mark] (iv) Calculate the energy, in MeV, released in the decay. [2 marks]

SUGGESTED ANSWERS

Paper 1

1. D: End error, zero error and calibration error are systematic errors.

2. D: u = 5.0 m s–1, a = –10 m s–2, s = –10 m

Using s = ut + 12

at2, –10 = (5.0)t – 5.0 t2

gives t = 2.0 s 3. A: x-component, Rx = 4.0 + 5.0 cos 60° = 6.50 N y-component, Ry = 5.0 sin 60o – 2.0 = 2.33 N

R = 6.502 + 2.332 N = 6.90 N

Angle above x-axis, θ = tan–1 �2.336.50�

= 19.7o

4. D: Total force required = F + F′ Total power = (F + F′)v

5. B: Coin slips when mv2

r = μmg

v = μgr

= (0.40)(9.81)(1.0) m s–1 = 2.0 m s–1

6. D: Total K.E. = 12

mu2 + 12

Iω2

= 12

mu2 + 12�2

3 mR2��u

R�2

= 56

mu2

7. A: Sliding friction ∝ mg 8. C: C is only true for geostationary satellites.

C is not true for other satellites. 9. C: At turning point of curve, resultant

gravitational force = 0

Since MEarth > MMoon, turning point is further from Earth and nearer to Moon.

10. B: In S.H.M., displacement is measured from the equilibrium position. Amplitude of oscillation is 1.0 cm.

11. A: 2πf = 100, f = 16 Hz,

v = ω A2 – x2

= (100) 52 – 32 m s–1 = 400 m s–1

K.E. = 12

mv2 = 12

(2.0)(400)2 J = 160 kJ

12. D: Amplitude of Q decreases faster implies damping of Q is greater. In the graph of a against f, peak of Q is lower and occurs at a lower frequency.

13. C: Period is 1.0 s. 2.0 s later is after 2 cycles, hence displacement is the same at 1.0 cm.

14. C: 10.0 cm = 12

λ, λ = 20.0 cm,

v = f λ = (50)(0.20) m s–1 = 10.0 m s–1

15. B: Period of beats, T = 0.20 s. Beats frequency,

f =

10.20

Hz = 5.0 Hz = f1 – f2 (1)

Frequency of resultant wave

=

40.20

Hz = 20 Hz = 12

( f1 + f2) (2)

Solving (1) and (2), f1 = 22.5 Hz, f2 = 17.5 Hz 16. D: Attractive force is negative. Magnitude of

attraction is maximum at P.

17. C: TX = �EXAX

l � e

= (1.80 × 1011)(1.50 mm2)

l e,

2009EXPaperVol1&2 for CD.indd 122009EXPaperVol1&2 for CD.indd 12 2/4/10 2:42:09 PM2/4/10 2:42:09 PM

13 © Majlis Peperiksaan Malaysia 2009

Current in 3.0 Ω resistor

= � 22.0 + 3.0�(1.875) A = 0.75 A

31. A: Torque is directly proportional to I. 32. A: No change of magnetic flux through the coil. 33. D: No e.m.f. is induced when the coil is outside

the magnetic field, and when it is completely in the field. E.m.f.’s induced as the coil enters and leaves the field are in opposite directions.

34. B: XC = 12πfC

= V0

I0

,

C = 12πf

� I0

V0� = 1100 � 0.1

100� F = 10 μF

35. D: X

Y

I in 1st 1/2-cycle I in 2nd 1/2-cycle 36. C: Saturated voltage = voltage of supply 37. D: The circuit is an inverting amplifier.

D is for non-inverting amplifier. 38. B: EM waves are transverse waves. E and B are

in phase and mutually perpendicular. 39. C: E and B fields are in phase and move in the

x-direction which is the direction of wave propagation.

40. C: Convex mirror – image virtual for all object distance.

Concave mirror – image virtual if u < f.

41. A: Fringe separation, x = λDa

42. B: n = 5.00.028

= 178.6, 2t = (178.6)(580 × 10–9)

t = 5.2 × 10–5 m 43. C: Kmax = eVs = h(3f0) – hf0 = 2hf0

Vs = 2hf0

e

44. D: de Broglie wavelength,

λ = hmv

=

h

2mK, �1

2 mv2 = K�

λ is the shortest if m is the largest. 45. D:

E4

E3

E2

E1

Q

P

46. B: 2d sin θ = nλ, n = 2 2d sin 26.5o = 2(3.10 × 10–10) d = 6.95 × 10–10 m 47. B: The incoming photon is not absorbed, but

stimulates the emission of a photon.

TY = (1.20 × 1011)(3.00 mm2)

l e

Taking moments about the weight,TX(d) = TY (99.0 – d)

�(1.80 × 1011)(1.50 mm2)l �e(d)

= �

(1.20 × 1011)(3.00 mm2)l �e(99.0 – d)

d = 56.6 cm

18. D: p = 13

ρc2, p1 = 13

ρ(3c)2 = 9p

19. B: Mass of the gas = total mass of gas molecules. 20. C: Isothermal expansion, temperature T = constant, ΔU = 0 ΔQ = +20 J = ΔU + W, W = +20 J 21. B: CV,m = gradient of graph

= 6500 – 6100310 – 290

J mol–1 K–1

= 20 J mol–1 K–1

Cp,m = CV + R = (20 + 8.31) J mol–1 K–1

= 28.31 J mol–1 K–1

Cp,m/CV,m = 1.42

22. B: Mass = (πr12)L1ρ1 = (πr2

2)L 2ρ2,

L1

L2 = ρ2

ρ1 � r2

r1�2

dQdt

= k1(πr12) Δθ

L1 = k2(πr 2

2) ΔθL2

� r1

r2�2

= �k2

k1��L1

L2� = �k2

k1��ρ2

ρ1�� r2

r1�2

r1

r2 = � (79)(7860)

(420)(10 500)�1/4

= 0.61

23. C: V = work per unit charge

24. B: 12

mu2 = (2e)(42e)4πε0d

= (2e)(84e)4πε0d1

, d1 = 2.0d

25. A: V = VR + VC = IR + QC

26. D: C = εr ε0Ad

. When εr increases, C increases.

27. B: For ohmic material, VI

= constant if

temperature is constant.

28. A: Power, P = I2R = I2� ρLπr2 �

Radius, r

= � (50.0)2(2.0 × 10–8)(1)π(2.0) �

1/2

m

= 2.8 mm 29. C: When R2 is increased, effective resistance

in the circuit increases. A1 decreases, A2 decreases but A3 increases.

30. A: 1R1

= 12

+ 13

gives R1 = 1.2 Ω

Current from cell, I = � 6.02.0 + 1.2� A

= 1.875 A

2009EXPaperVol1&2 for CD.indd 132009EXPaperVol1&2 for CD.indd 13 2/4/10 2:42:10 PM2/4/10 2:42:10 PM

14© Majlis Peperiksaan Malaysia 2009

48. B: This reaction led to the discovery of neutrons. 49. A: R decreases exponentially, but not to zero

because of background count. 50. B: 210

81P → 20276Q + 24

2He + 0+1e

Paper 2

Section A

1. μmg = mv2

r

μ = v2

gr =

(80.0 × 103/3600)2

(9.81)(120) = 0.419

2. (a) Minimum force, F = μmg = (0.85)(12.0)(9.81) N = 100 N

(b) F = 25.0 N is not able to move the block.Friction = 25.0 N

(c) 125 – 100 = 12.0a a = 2.08 m s–2

3. (a) (i)

(ii)

(b) (i) For the fundamental note, resonant frequency of a closed pipe is half the resonant frequency of an open pipe.

(ii) Vibrations of the air column produce sound that is propagated through the air around the pipe.

4. (a) There must be a temperature difference (or temperature gradient) across the medium.

(b) High thermal conductivity: Heat flows at a fast rate.Low heat capacity: Small quantity of heat is required to increase its temperature.

(c) Rate of heat flow,dQdt

= kAdθdx

= (66.9)(0.125)� 6901.0 × 10–3� W

= 5.77 × 106 W 5. (a)

Absolute temperature

Superconductor

Pure metal

Resistivity

0

(b) I = nAveDrift velocity,

v = I

nAe =

5.0(8.4 × 1028)π(1.5 × 10–3)2

(1.60 × 10–19)

m s–1

= 5.26 × 10–5 m s–1

6. Finally, the capacitor is fully charged and no current flows in the 10.0 Ω resistor.Effective resistance R in the circuit is given by1R

= 1

30.0 +

1(15.0 + 45.0)

R = 20.0 Ω

Final current, I = VR

= 4.020.0

A = 0.20 A

7. (a) X-ray is produced when fast electrons are retarded on collision with a heavy metal.

(b) (i) 12

mv2 = eV

Speed of electron,

v = 2eVm

= 2(1.60 × 10–19)(25 × 103)9.11 × 10–31

m s–1

= 9.37 × 107 m s–1

(ii) λmin = hceV

= (6.63 × 10–34)(3.00 × 108)(1.60 × 10–19)(25 × 103)

m

= 4.97 × 10–11 m 8. (a) A mass spectrometer is used to determine the

mass of ions of different isotopes. (b) (i) As velocity selector: To select ions with

a certain velocity. (ii) Magnetic field B2 deflects ions of

different isotopes of the same element into different semicircular paths.

Section B 9. (a) (i) Work on a body by a force F,

W = Fswhere s = displacement of the body in the direction of the force

(ii) Mass of body = m, initial velocity = uAfter being displaced a distance s by the force F, velocity = v.Work done = Fs (v2 = u2 + 2as, F = ma)

= (ma)�v2 – u2

2a � = 12

mv2 – 12

mu2

= increase in kinetic energy (iii) Principle of conservation of energy:

Energy can neither be created nor destroyed, but can be transformed from one form into another.

(b) (i) Common acceleration a of the blocksis given by(6.0 + 3.0)a = (6.0 – 3.0)(9.81) a = 3.27 m s–2

v2 = u2 + 2as = 0 + 2(3.27)(4.0) v = 5.11 m s–1

2009EXPaperVol1&2 for CD.indd 142009EXPaperVol1&2 for CD.indd 14 2/4/10 2:42:12 PM2/4/10 2:42:12 PM

15 © Majlis Peperiksaan Malaysia 2009

(ii) For Y: After travelling upwards 4.0 m, u = 5.11 m s–1, a = –9.81 m s–2

At the maximum height, v = 0v2 = u2 + 2as

s = 5.112

2(9.81) m = 1.33 m

Total distance travelled = 5.33 mIf the mass of the pulley is not negligible, part of the energy of the blocks would be transformed into rotational kinetic energy of the pulley. Hence the block Y would rise through a shorter distance.

10. (a) (i) For diatomic gas, CV,m = 52

R,

Cp,m = 52

R + R = 72

R

Heat absorbed, Q

= (1)Cp,m(ΔT) = 72

(8.31)(35.0 – 15.0) J

= 582 J (ii) Use pVm = RT

Initial volume,

Vm = (8.31)(273 + 15.0)

1.01 × 105 m3

= 0.02370 m3

Final volume,

V� = (8.31)(273 + 35.0)

1.01 × 105 m3

= 0.02534 m3

Increase in volume, ΔV = (0.02534 – 0.02370) m3 = 1.64 × 10–3 m3

(iii) Work done by the gas, W = p(ΔV) = (1.01 × 105)( 1.64 × 10–3) J = 166 J

(iv) Use Q = U + WChange in internal energy, ΔU = (582 – 166) J = 416 J

(b) (i) Adiabatic compression: Compression of a gas without heat entering or leaving the gas, Q = 0.

(ii) Diatomic gas,

γ = 75

, use TV γ – 1 = constant

T2 = (273 + 20.0)� V1

V1/4�7/5–1

= 510 K

(iii) Work done,

W = –ΔU �U = 52

nRT� = – 5

2 nR(T2 – T1)

= – 52

n(8.31)(510 – 293) J

= – 4508n J where n = number of moles

Work done on the gas = 4508n J

If n = 1, work done = 4508 J 11. (a) (i) Magnetic flux, Φ = (B sin θ)A (ii) Rotate the coil about an axis parallel to

the plane of the coil.Remove the coil from the magnetic field.Reduce or increase the magnetic flux density B.(Any two)

(iii) Induced current in the direction PQRSP.Lenz’s law states that the direction of the induced current is such as to oppose the change in the magnetic flux linked with the coil.When the magnetic field is reduced, the induced current flows in the direction PQRSP so that the magnetic field in the coil produced by the current is in the original direction of B.

(b) (i) From E = – L dIdt

,

self-inductance,

L = – E(dI/dt)

= e.m.f. induced in the inductor

rate of change of current in the inductor

(ii) Magnetic flux,

Φ = LI = NBA, and B = μ0 �Nl � I

LI = N�μ0Nl � IA

L = μ0N2A

l (c) Maximum e.m.f. induced, E0 = NBAω

R.m.s. voltage produced = NBAω2

= 240 V

N = 2 (240)

BA(2π f )

= 2 (240)

(0.15)(2.0 × 10–2)2π(50) = 360 turns

(d) (i) No back-e.m.f. is induced yet.

Current in armature, I0 = VR

, where V

is the applied voltage, and R is the resistance of the coil. Since R is small, the current I is large and causes the coil to be heated up.

(ii) When the motor is switched on, the change of current in the inductor produces a back-e.m.f. that reduces the current.

12. (a) (i) Light waves are transverse waves. (ii) Direction of EM wave propagation is in

the direction of the vector product, E × B.

2009EXPaperVol1&2 for CD.indd 152009EXPaperVol1&2 for CD.indd 15 2/4/10 2:42:14 PM2/4/10 2:42:14 PM

16© Majlis Peperiksaan Malaysia 2009

Direction of E-field: +y direction, direction of B-field: –x direction, orDirection of E-field: –y direction, direction of B-field: +x direction

y

E

EB

B

–x

–y

x

z

z

v

v

or

(b) (i) Unpolarised light: Oscillations of the E-field (and B-field) are in all possible planes.

(ii) Components of the electric field oscillations parallel to the polarisation axis of the Polaroid are transmitted, but components perpendicular to the axis are absorbed. Therefore, only half of the energy is transmitted.

(iii) I0 = intensity of unpolarised lightIntensity of light transmitted by

first Polaroid, I1 = 12

I0

Intensity after second Polaroid,

I2 = 14

I0 = I1 cos2 θ = �12

I0� cos2 θ

cos θ = 12

θ = 45o

(c) Using Brewster’s law, tan i = n, refractive index i = tan–11.52 = 56.7o

13. (a) mv2

rn =

e2

4πε0r2n

, v = nh

2πmrn

mrn

� nh2πmrn

�2

= e2

4πε0r2n

rn = n2h2ε0

πme2

(b) Radius is the smallest when n = 1,

r1 = (1)2h2ε0

πme2

= (6.63 × 10–34)2(8.85 × 10–12)

π(9.11 × 10–31)(1.60 × 10–19)2 m

= 5.31 × 10–11 m (c) Total energy, En

= kinetic energy + electric potential energy

= 12

mv2 + (e)(–e)4πε0rn

�From mv2

rn =

e2

4πε0r2n

, 12

mv2 = e2

8πε0rn�

= –e2

8πε0rn = –� e2

8πε0��πme2

n2h2ε0�

= –1n2� e4m

8h2ε20�

(d) Ionisation energy = –E1

= 1

(1)2 � e4m8h2ε2

0�

= (1.60 × 10–19)4(9.11 × 10–31)

8(6.63 × 10–34)2(8.85 × 10–12)2 J

= 2.17 × 10–18 J

(e) E1 = – 1

(1)2 � e4m8h2ε2

0� = –2.17 × 10–18 J

E4 = – 1

(4)2 � e4m8h2ε2

0� = –1.356 × 10–19 J

E2 = – 1

(2)2 � e4m8h2ε2

0� = –5.425 × 10–19 J

hcλ

= E4 – E2 = (5.425 – 1.356) × 10–19 J

λ = (6.63 × 10–34)(3.00 × 108)

4.069 × 10–19 m

= 4.89 × 10–7 m which is visible light. 14. (a) Nuclear fusion: Two small nuclei combine to

form a larger nucleus. (b) (i) 2

1H + 31H → 42X + 10n + 18.2 MeV X is 42He. (ii) 2

1H + 31H → 42X + 10n + 18.2 MeV(2.014102 + 3.016049) u

= mX + �1.008665 + 18.2931� u

mX = 4.001937 u (c) (i) Energy of each proton

= 12

(1.2 × 10–13) J = 6.0 × 10–14 J

(ii) 32

kT = 6.0 × 10–14 J

Temperature,

T = 2(6.0 × 10–14)3(1.38 × 10–23)

K

= 2.90 × 109 K which is very high. (d) (i) Radioactivity is the spontaneous and

random emission of radiations from unstable nuclei.

(ii) 2513Al decays to 25

12Mg, because mass of Al > mass of Mg

(iii) Positron decay (iv) 25

13Al → 2512Mg + +1

0 eΔm = 24.990429 u –

�24.985837 + 9.11 × 10–31

1.66 × 10–27� u

= 0.004043 uEnergy released= (0.004043)(931) MeV = 3.76 MeV

2009EXPaperVol1&2 for CD.indd 162009EXPaperVol1&2 for CD.indd 16 2/4/10 2:42:15 PM2/4/10 2:42:15 PM