Section A [40 marks]

Answer all the question in this section.

1 (a) Explain what is meant by the escape velocity from a planet.

The speed of an object so that it has enough kinetic energy to

overcome the gravitational potential energy

or minimum velocity of a body to overcome the planet

gravity attraction / to escape to infinity / to leave the gravitational field

(b) (i) A planet has a density of 1.4 103 kg m3 and a radius of 3.5 1011 m. Calculate the escape velocity from this planet.

2

2

1 mv = R

GMm or or kinetic energy=change of potential energy

v = Substitute M =

= 3

104.1)105.3(142.31067.68 321111

= 3.1 108 m s1

(ii) State why a body is unable to escape from this planet.

Escape velocity greater than light velocity

2 A form of progressive wave which has frequency of 800 Hz and speed of 350 m s1 can be stated in the form of the following equation.

or minimum

= 2.51 x 1038 kg

Or the planet is a Black Hole

y = A sin ( t kx).(a) Calculate the value of

(i) , = 2f

= 5030 (rad) s-1

(ii) k.

or v = fλ

=

= 0.438 or 0.4375

k =

= 14.3 (rad) m1 or 14.4 (rad) m-1 (14.36)

or v = ,

350 = or 350=

k = 14.4 (rad) m-1 or k = 14.4 (rad) m-1

(b) What is the distance between the two positions which gives the phase difference rad?

or

x = or

= 0.220 m or 0.218 m

or

= 0.438

= 0.219 m

3 The table below shows the Young modulus and the cross-sectional area of two types of materials.

2

MaterialYoung Modulus/

PaCross-sectional area/

m2

Iron 1.00 1011 1.2 104

Steel 2.00 1011 2.0 104

(a) Steel has a higher Young modulus compared to What can you say about the physical property of steel?

harder / stiffer / more rigid / more stronger / higher melting point

(b) A load of 1500 kg requires a wire of 5.0 m long to be hung freely at one of its end. If the

extension cannot exceed 2.0 mm, show quantitatively which material from the table above is suitable to make the wire.

e = or

= 1.8 mm

e iron = 6.1 mm

4 (a) (i) State the SI unit for thermal conductivity.

W m1 K1 or kg m s-3 K-1

Must be kelvin

(ii) In an experiment to determine the thermal conductivity of a material, the measurement of the temperature gradient can be carried out on the condition that a steady state is achieved. What is meant by the steady state?

The temperature gradient is constantThe temperature at any point along material is constant / change/

stable Rate of heat flow is constant/rate of heat flow in = rate of heat flow out

(b) The diagrams (i) and (ii) below show the side views of two concrete walls of a room which is the same dimensions with width L. The concrete wall in diagram (ii) has a layer of air

3

iron

e steel

Steel (more suitable)

/Less extensible/ more difficult to stretch

no

with width at its middle. The thermal conductivity of air is of the thermal conductivity of

concrete.

Outer Inner Outer Inner30 C 20 C 30 C 20 C

(i) (ii)

If the outer and inner temperatures of the room are 30 C and 20 C respectively, calculate the ratio of the rate of flow of heat through the cross-sectional area of the wall in diagram (i) to the rate of flow of heat through the cross-sectional area of the wall in diagram (ii).

For wall Diagram (i):

For wall Diagram (ii):

=

The ratio of heat flowing rate = =2.33

5 In the circuit shown below, the emf of cells E1 and E2 are 2.0 V and 3.0 V respectively. The capacitance of capacitor C is 10 F. Initialy switch S is open.

4E1

LL

General form will do

(a) What is the initial charge of capacitor C ?

Q = CV

= 10 2.0

= 20 C ( 2 sf ) or 2.0 x 10 – 5 C

(b) Explain qualitatively what will happen to the charge in capacitor C after the switch S is closed.

Charge at C increases

because potential difference across C is increased after the switch S is closed

or the capacitor is charged further by E2

6 The diagram below shows a wire PQ of length . The wire falls freely and slides without any contact with the sides of the wire frame. The wire and wire frame are in uniform magnetic field B which is perpendicular to the plane of the wire frame. The wire frame has a resistor of resistance R.

(a) State the direction of the induced current in the wire PQ. Use Lenz’s law to explain your answer. The direction of the induced current: P to Q

The flow of the induced current should produce a force which is against the

direction of wire motion/ the induced current will oppose the increase in magnetic

flux linkage

5

E2

S

C2.0 5.0

C

Wire frame

P

Q

Magnetic field B

R

(dependent – given only if 1st mark is right)

(b) Deduce an expression of current I which is induced in terms of B, , R and velocity v of the wire.

Induced emf =

Induced current I =

7 The diagram below shows two pieces of Polaroids, P and Q, which are arranged such that the polarisation axes make an angle between one another. A plane unpolarised incident light beam with intensity I0 is applied to Polaroid P. The amplitude of light wave between Polaroid P

and Q is A, and the amplitude after Polaroid Q is .

(a) State, in terms of I0, the intensity I1 of the light beam between Polaroids P and Q.

(b) Calculate the angle between the polarisation axis of Polaroid P and that of Polaroid Q.

= A cos

= 60

(c) Deduce the intensity I2 of the light beam after the Polaroid Q in terms of I0.

or I amplitude2 or I2 = I1 cos2 ө ( or general formula I = Io cos2 ө )

I2

= I0

8 (a) What is meant by the binding energy of a nucleus?

The minimum/required energy/work done required to fully separate all the nucleon in a nucleus

6

Plane unpolarised incident light

Amplitude A Amplitude

I0 I1 I2

Polarisation axis

P Q

Polarisation axis

or I2 =

or I2 = 0.125 Io

(b) Explain from the aspect of mass-energy why the binding energy has a negative value.

The mass and energy are equivalent according

to the Einstein Equation E = mc2 or E = (Δm)c2

Δm = nucleus mass – total nucleon mass or final mass /initial mass

The nucleus mass = the mass of all nucleons + the binding energy

Known that the nucleus mass is smaller than its nucleon mass/ final mass is

less than the initial mass

Section B [60 marks]

Answer any four questions in this section.

9 (a) Define the term torque. [2 marks]

7

or

Torque = rF r = the perpendicular distance from the

rotation axis to the force F

(b) The diagram below shows a solid cylinder of mass M and radius R rolling without slipping from rest state on an inclined plane at an angle with the horizontal. The moment of

inertia of the cylinder about its axis is MR2.

(i) State the force which enables the cylinder to roll without slipping. [1 mark]

Friction (static)

(ii) Copy the diagram above, mark and label the forces that act on the cylinder.[3 marks]

Normal / reaction N

Mg/w

(iii) By using Newton’s law of motion, write the linear motion equation of the cylinder which moves in the direction parallel with the plane. [2 marks]

Mg sin

Mg sin – F = Ma ................ (1)

(iv) By using Newton’s law of motion for rotation, write the rotational motion equation of the cylinder about its axis. [1 mark]

8

R

Friction F

Mg/weight w

RESTRICTED

Or r is the distance from the rotational axis to F/ is the position vector

= MR2 (2)

(v) Derive the expression of the linear acceleration a of the rolling motion in terms of and g. [2 marks]

From (2) F =

F = Ma into (1)

Mg sin – Ma = Ma

a = sin

(vi) Calculate the nett force along the inclined plane on the cylinder in terms of , g and M. [1 mark]

Nett force = Mg sin – F

= Mg sin – M sin

= Mg sin

or

Nett force = Ma

= M g sin

(vii) If the coefficient of friction is = tan , show with a suitable calculation that

the cylinder can roll without slipping. [3 marks]

Friction F on the cylinder = F

= sin

Maximum friction = N

= tan Mg cos

= Mg sin

9

RF

maximum

Friction on the cylinder < limiting friction

10 (a) The diagram below shows a body of mass m connected to the end of a spring which has spring constant k. k

When the body is displaced to perform an oscillation, its displacement from the equilibrium position at any time t is x.

(i) Derive, from Newton’s law of motion, the equation of oscillatory motion of the body. [2 marks]

or F= ma

- kx

10

m

Wall

x

Smooth surface

- kx = ma

(ii) Deduce an expression for the frequency of oscillation of the body. [2 marks]

Simple harmonic motion

= 2x or 2 =

f =

(iii) Explain why the oscillatory motion of this body will finally stop. [2 marks]

Energy loss because of : Air resistance

The work done to compress and extend the

spring/ internal damping/lost as heat in the spring

(b) A body of mass 2.0 kg moves in simple harmonic motion. The displacement x from the equilibrium position at time t is given by x = 6.0 cos 0.22 t where x is in metres and t is in seconds.

(i) What is the amplitude and the period of the simple harmonic motion? [3 marks]

6.0 m

T =

= 9.1 s

(ii) Calculate the maximum acceleration of the motion. [2 marks]

or rω2

= 6(0.22)2

= 2.9 m s2

(iii) Calculate the kinetic energy of the body at time t = 3 seconds. [4 marks]

11

Note

x = 6.0 cos 0.22(3)

= 2.89 m

Kinetic energy = mv2

= (2)(0.22)2 [62 (2.89)2]

= 13 J

Alternative :

= 13 J

11 (a) (i) What is meant by the degree of freedom of a gas molecules? [1 mark]

The degree of freedom independent ways or modes

for molecules to move or to acquire energy

(ii) Explain why the number of the degrees of freedom of a diatomic gas molecule is different at temperatures 30 K, 300 K and 3000 K. [3 marks]

At 30 K, translational (kinetic) energy only

At 300 K, translational and rotational (kinetic) energy

At 3000 K, translational, rotational and vibration (kinetic) energy

(b) (i) State the law of equipartition of energy. [1 mark]

For one molecule, the mean kinetic energy for every

degree of freedom = 2

1 kT or ½ RT

(No need to explain the symbols except using other symbols.)

12

One mole

Or energy of a molecule is equally shared by each degree of freedom

(ii) A molecule of an ideal gas has f degrees of freedom. By using the law of equipartition of energy, deduce an expression for internal energy U of n moles of ideal gas.

[3 marks]

Mean kinetic energy of 1 molecule = 2f

kT

Mean kinetic energy of 1 mole = NA

kTf2 or or

= 2f

RT

Internal energy of n mole, U = 2f

nRT

(iii) If the pressure of the ideal gas is p, and its volume is V, deduce an expression for internal energy U of the gas in term of f, p and V. [2 marks]

pV = nRT

U = 2f

nRT

= 2f

pV

(c) A container of volume 2.00 103 m3 contains 8.00 g of helium gas at pressure 1.01 105 Pa.

(i) What is the internal energy of the gas? [2 marks]

Internal energy U = 2f

pV f = 3

= )1000.2)(1001.1( 35

2

3

= 3.03 102

= 303 J ( 3 sf)

(ii) What is the mean kinetic energy of one molecule of helium gas? [3 marks]

13

RESTRICTED

For N molecules= N(f/2 kT)

= nNA(f/2 kT)

U = total KEOf themolecules

Number of mole, n = 00.400.8

= 2.0

Mean kinetic energy of one molecule = A2

303N

= 2.52 1022 J(3 sf)

12 (a) An electric field E is applied along a metal rod. Describe the movement of free electrons before the electric field is applied, and when the electric field is applied. [2 marks]

random velocity or random motion without electric field /mean velocity = 0

drift velocity or drift motion with electric field

(b) Explain microscopically why

(i) a metal electric conductor becomes hot when an electric current flows through it, [2 marks]

Free electrons collide with lattice atoms / ion/particles

Kinetic energy of electrons transfer to the lattice atoms

or Make the kinetic energy of lattice atom increases

14

Alternative :n = 2.0

T = = 12.15 K

KE = 3/2 kT = 2.52 x 10 -22 J

(ii) the resistivity of metal increases with a rise in temperature, while the resistivity of a semiconductor decreases with an increase in temperature. [4 marks]

Metal

When the temperature increases, lattice atoms vibrate stronger/

more vigorously

Free electron collision of metal is more rapid / mean free

time decreases / mean free path decreases

Semiconductor

Number of charge carrier increases

The effect of increasing the number of charge carriers

dominates/ more than/overcomes the effect of decreasing of the mean

free time

(c) The diagram below shows a thin strip of silver with a cross-sectional area of 1.00 mm 2.00 mm which carries a current of 14.0 A. A magnetic field of 1.50 T acts on the strip in z-direction. Microvoltmeter connected to PQ show a reading of 2.24 V.

(i) Determine the electric pole at P. [1 mark]

positive

(ii) Calculate the drift velocity of free electrons. [3 marks]

15

B

P1.00 mm

2.00 mm

14.0 A

y

z

x

V Microvoltmeter

Q

eEH = evB or FE = FM or VH =

or = vB or

3

6

1000.21024.2

= v 1.50

v = 7.47 104 m s1

(iii) Calculate the density of free electrons. [3 marks]

I = nA ve

14.0 = n 19433 1060.11047.71000.11000.2

n = 5.86 2810 m3 or 5.85 x 1028 m-3

13 (a) (i) Write the lens maker’s formula. Explain the symbols that you use. [1 mark]

or

with correct symbol explanations

(ii) The diagram below shows the lens of a student’s spectacles which is made of glass of refraction index 1.57.

Lens surfaces X and Y have radii of curvature 15.0 cm and 13.0 cm respectively. Determine the focal length of the lens. State whether the lens is the converging or diverging type.

[4 marks]

r2 = 13.0 cm

[ or r2 = 13.0 cm]

16

n / n1 / n2 r1 / r2

Alternative :

VH=

Substituting n = 5.86 2810 m3 or 5.85 x 1028 m-3

Any one of the “n” and any one of the “r”

f = 171 cm

diverging

(iii) The student wears the spectacles and dives in water of refraction index 1.33 to see an object in the water. Explain quantitatively why the student is unable to see the object in the water clearly. [3 marks]

cm

Focal length of lens has changed

(b) A glass rod of 50.0 cm long has convex ends with radii of curvature r1 = 5.00 cm and r2 = 4.00 cm respectively. The refraction index of the glass rod is 1.40. An object O at a distance of 30.0 cm from the X end of the glass rod is shown in the diagram below.

Determine the image position when the object O is observed from the Y end side of the glass rod. Draw a ray diagram to illustrate your answer. [7 marks]

(b)

v1 = 30.0 cm

u2 = (50.0 30.0) = 20.0 cm

v2 = 33.3 cm( 3 sf )

( Alternative

v1 = 30 cm

u2 = -20 cm

v2 = 33.3 cm( 3 sf ) ]

17

14 (a) Explain how X-ray is diffracted at high intensity in certain directions by a crystal.[3 marks]

Atoms as secondary source or

X-ray is diffracted by atoms or 2d sin or diagram

path difference = n / in phase

Constructive interference happen

(b) The diagram below shows atoms in a crystal.

The distance a between atoms is 5.63 1010 m. The distance between atom planes is d.

(i) Calculate the value of d. [2 marks]

(5d)2 = a2 + 4a2

d = 5

a = 2.52 1010 m

18

O

Ray

I1 I2

θ

a

θa

Sin2θ + cos2θ = 1

sin θ =

tan θ = ½

sin θ =

------------------ 1

d = ---------- 1

(ii) A beam of X-ray with wavelength of range 0.800 1010 m to 1. 50 1010 m is directed at a glancing angle of 50.0 to the atomic planes which are shown in the diagram. Determine the wavelengths which are diffracted at high intensity by the atomic planes. [4 marks]

2d sin = n

n

50sin)1052.2(2 10

n = 3 3 = 129 pm = 1.29 1010 m (3 sf)

n = 4 4 = 96.5 pm = 0.965 1010 m

(c) Radiation emission from a radioactive source is random and spontaneous. Explain what is meant by random and spontaneous. [2 marks]

Random: NtN

dd

or same probability

or cannot be predicted which or when a decay can happen

or number of particles emitted is not constant,

or time interval between successive emission is not constant

Spontaneous: is not planned

or cannot control

or is not influence by surrounding state

19

d

½ a

θa

a

d

or happen by itself

or no stimulation required

(d) The diagram below shows the radiation which is emitted from a radioactive source in a strong uniform magnetic field.

(i) Explain why all the lines of radiation P are of nearly the same length. [1 mark]

Is emitted with same ( kinetic) energy / same speed

(ii) Explain why the radius of curvature of radiation R varies. [2 marks]

Different kinetic energy / different speed

Centripetal force / magnetic force depends on the speed

/ r depends on v

(iii) Determine the direction of the magnetic field. [1 mark]

-particle -particle

Or direction: (Enter) into the paper

20

F + F

vv

B B or

21

CONFIDENTIAL