ANDERSON JUNIOR COLLEGEscore-in-chemistry.weebly.com/uploads/4/8/7/1/48719755/...The orbital period of the satellite is T. A second satellite, in a different circular orbit, has an

1

9646/01/AJC2015/Prelim [Turn Over

2015 JC2 Preliminary Examination PHYSICS 9646/01 Higher 2 Paper 1 Multiple Choice Monday 31 August 2015

1 hour 15 minutes

Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST

Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, PDG and NRIC/FIN and shade the 7 digits of your NRIC/FIN in soft pencil on the

Answer Sheet.

ANDERSON JUNIOR COLLEGE

There are forty questions in this section. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this question paper. The use of an approved scientific calculator is expected, where appropriate.

This document consists of 19 printed pages and 1 blank page

Candidate Name

PDG

( )

2

9646/01/AJC2015/Prelim

Data

speed of light in free space, c = 3.00 x 108 m s−1

permeability of free space, 0 = 4 x 10−7 H m−1

permittivity of free space, 0 = 8.85 x 10−12 F m−1

(1/(36)) x 10−9 F m−1 elementary charge, e = 1.60 x 10−19 C

the Planck constant, h = 6.63 x 10−34 J s unified atomic mass constant, u = 1.66 x 10−27 kg

rest mass of electron, me = 9.11 x 10−31 kg

rest mass of proton, mp = 1.67 x 10−27 kg molar gas constant, R = 8.31 J K−1 mol−1

the Avogadro constant, NA = 6.02 x 1023 mol−1 the Boltzmann constant, k = 1.38 x 10−23 J K−1

gravitational constant, G = 6.67 x 10−11 N m2 kg−2

acceleration of free fall. g = 9.81 m s−2

3


Formulae

uniformly accelerated motion, s = ut + 1

2 at2

v2 = u2 + 2as

work done on/by a gas, W = pV

hydrostatic pressure, p = gh

gravitational potential, ϕ = − r

Gm

displacement of particle in s.h.m., x = x0 sin t

velocity of particle in s.h.m., tvv cos0

22xxo

mean kinetic energy of a E = 3

2 kT

molecule of an ideal gas, resistors in series, R = R1 + R2 + … resistors in parallel, 1/R = 1/R1 + 1/R2 + …

electric potential, V = Q

4 ε0 r

alternating current/voltage, x = x0 sin t

transmission coefficient, T exp(−2kd)

where k = 2

28

h

EUm

radioactive decay, x = x0exp(− t)

decay constant. = 0.693

21t

4


1 The angular deflection of the needle of an ammeter varies with the current in the ammeter as shown in the graph.

Which diagram could represent the appearance of the scale on this meter?

A B

C D

2 The diagram shows part of a thermometer.

What is the correct reading on the thermometer and the uncertainty in this reading?

reading / °C uncertainty in reading / °C

A 24 ± 1

B 24 ± 0.5

C 24 ± 0.2

D 24.0 ± 0.5

3 The acceleration of free fall on the Moon is one-sixth of that on Earth.

On Earth it takes time t for a stone to fall from rest a distance of 2 m.

What is the time taken for a stone to fall from rest a distance of 2 m on the Moon?

A 6t B

6

t C 6t D

6

t

5


4 A hosepipe is fixed as shown.

The jet of water emerges with a horizontal velocity v. The hosepipe is fixed at a height h above the ground. The water jet hits the floor at a horizontal distance d from the nozzle tip. The gravitational field strength is g.

What is the expression for distance d? (Ignore air resistance.)

A h

vg

2 B

g

vh2 C

h

gv

2 D

g

hv

2

5 Water is pumped through a hosepipe at a rate of 90 kg per minute. It emerges from the

hosepipe horizontally with a speed of 20 m s–1. Which force is required from a person holding the hosepipe to prevent it moving backwards?

A 30 N B 270 N C 1800 N D 10800 N 6 Two railway trucks of masses m and 3m move towards each other in opposite directions

with speeds 2v and v respectively. These trucks collide and stick together.

What is the speed of the trucks after the collision?

A

4

v B

2

v C v D

4

5v

6


7 The solid line on the graph shows how the length of a rubber band varies when an increasing load is applied. The dotted line shows how the length subsequently varies as the load is gradually decreased.

Which statement is correct? A The energy recovered when the load is removed is about 10 J.

B The work done in stretching the rubber band is about 5 J.

C The total work done on the rubber band during one cycle of loading and unloading is

about 14 J. D The energy remaining in the rubber band after one cycle of loading and unloading is

about 3 J.

8 A ball is falling at terminal speed in still air. The forces acting on the ball are upthrust,

viscous drag and weight. What is the order of increasing magnitude of these three forces? A upthrust → viscous drag → weight B viscous drag → upthrust → weight C viscous drag → weight → upthrust D weight → upthrust → viscous drag

7


9 An electrical generator is started at time zero. The total electrical energy generated during the first 5 seconds is shown in the graph.

What is the maximum electrical power generated at any instant during these first 5 seconds?

A 10 W B 13 W C 30 W D 50 W

10 A projectile is launched at 30° to the horizontal with initial kinetic energy E.

Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it reaches its highest point?

A 0.50 E B 0.71 E C 0.75 E D 0.87 E

11 A mass m is situated in a uniform gravitational field.

When the mass moves through a displacement x, from P to Q, it loses an amount of potential energy E. Which row correctly specifies the magnitude and the direction of the acceleration due to gravity in this field?

magnitude direction

A

mx

E

B

mx

E

C

x

E

D

x

E

8


12 A small mass is placed at point P on the inside surface of a smooth hemisphere. It is then

released from rest. When it reaches the lowest point T, its speed is 4.0 m s–1. The diagram (not to scale) shows the speed of the mass at other points Q, R and S as it slides down. Air resistance is negligible.

The mass loses potential energy E in falling from P to T.

At which point has the mass lost potential energy4

E?

A Q B R C S D None of these

13 A satellite is in a circular orbit of radius r around the Earth.

The orbital period of the satellite is T.

A second satellite, in a different circular orbit, has an orbital period 64T.

What is the radius of the orbit of the second satellite? A 8 r B 16 r C 64 r D 512 r

9


14 A satellite above the Earth in a circular orbit of radius r1 is moved to a higher circular orbit of

radius r2. The gravitational force-distance graph is shown for the satellite.

What does the shaded area on the graph represent? A the change in gravitational potential energy of the satellite

B the change in kinetic energy of the satellite C the final gravitational potential energy of the satellite

D the final kinetic energy of the satellite

15 An object of mass 0.60 kg is held in place by two horizontal springs.

It is displaced sideways and undergoes simple harmonic motion of period 5.0 s. In each oscillation, it moves from left to right through a total distance of 0.30 m.

What is the total energy of the simple harmonic motion? A 4.3 × 10–3 J

B 1.1 × 10–2 J C 1.7 × 10–2 J

D 4.3 × 10–2 J

10


16 Two objects P and Q are given the same initial displacement and are then released. The graphs show the variation with time t of their displacement x.

P and Q are then subjected to driving forces of the same constant amplitude and of variable frequency f. Which graph represents the variation with f of the amplitudes A of P and of Q?

A B C D

17 There is one temperature, about 0.01 °C, at which water, water vapour and ice can co-exist

in equilibrium. Which statement about the properties of the molecules at this temperature is correct? A Ice molecules are closer to one another than water molecules. B The mean kinetic energy of water molecules is greater than the mean kinetic energy of

ice molecules. C Water vapour molecules are less massive than water molecules. D Water vapour molecules have the same mean square speed as both ice and water

molecules. 18 Before the invention of the modern refrigerator, ice was manufactured industrially and

delivered to households. One method used is the evaporation of ammonia. Energy was required to make the ammonia evaporate and 75 % of this energy came from liquid water at 0 °C, turning the water into ice. In six hours 8.0 × 104 kg of ice was produced. At what rate did the ammonia need to be evaporated? The specific latent heat of fusion of water is 330 kJ kg–1. The specific latent heat of vaporisation of ammonia is 1370 kJ kg–1.

A 0.67 kg s-1 B 1.2 kg s-1 C 12 kg s-1 D 20 kg s-1

P Q x x

t t

11


19 Two metal spheres of different radii are in thermal contact in a vacuum as shown. The spheres are at the same temperature. Which statement must be correct?

A There is no net transfer of thermal energy between the spheres.

B Each sphere has the same internal energy.

C Both spheres radiate electromagnetic energy at the same rate.

D The larger sphere has a greater mean internal energy per atom than the smaller sphere.

20 Which statement describes a situation when polarisation could not occur?

A Light waves are reflected. B Light waves are scattered. C Sound waves pass through a metal grid. D Microwaves pass through a metal grid.

21 The speed of a transverse wave on a stretched string can be changed by adjusting the

tension of the string. A stationary wave pattern is set up on a stretched string using an oscillator set at a frequency of 650 Hz.

How must the wave be changed to maintain the same stationary wave pattern if the applied frequency is increased to 750 Hz? A Decrease the speed of the wave on the string. B Decrease the wavelength of the wave on the string. C Increase the speed of the wave on the string. D Increase the wavelength of the wave on the string.

12


22 Two identical loudspeakers are connected in series to an a.c. supply, as shown.

Which graph best shows the variation of the intensity of the sound with distance along the line XY?

A

B

C

D

23 A diffraction grating experiment is set up using yellow light of wavelength 600 nm. The

grating has 500 lines per mm.

What is the angular separation (θ2 – θ1) between the first and second order maxima of the yellow light?

A 17.5° B 19.4° C 36.9° D 54.3°

13


24 The diagram shows two parallel horizontal metal plates. There is a potential difference V

between the plates.

A small charged liquid drop, midway between the plates, is held in equilibrium by the combination of its weight and the electric force acting on it. The acceleration of free fall is g and the electric field strength is E.

What is the polarity of the charge on the drop, and the ratio of charge to mass of the drop?

polarity mass

charge

A positive

g

E

B positive

E

g

C negative

g

E

D negative

E

g

25 The diagram shows two metal plates connected to a constant high voltage.

Which graph shows the variation of the electric field strength E midway between the two plates as the distance d between the two plates is increased?

A B C D

14


26 The Large Hadron Collider is designed to accelerate groups of protons around a large

circular ring. At any moment, there will be 3000 groups in the ring and each group will contain about 1011 protons. All the protons go around the ring 104 times per second. What is the best estimate of the current in the ring?

A 50 μA B 160 μA C 500 mA D 160 A

27 The graph shows how the electric current I through a conducting liquid varies with the

potential difference V across it.

At which point on the graph does the liquid have the smallest resistance?

28 In the potentiometer circuit shown, the reading on the ammeter is zero.

The light-dependent resistor (LDR) is then covered up and the ammeter gives a non-zero reading. Which change could return the ammeter reading to zero? A Decrease the supply voltage. B Increase the supply voltage. C Move the sliding contact to the left. D Move the sliding contact to the right.

15


29 A circuit is set up as shown, supplied by a 3 V battery. All resistances are 1 kΩ.

What will be the reading on the voltmeter?

A 0 B 0.5 V C 1.0 V D 1.5 V

30 An electron, travelling in a straight line at a speed of 1.46 × 107 m s–1, enters a region where

there is a uniform magnetic field. The diagram shows the path followed by the electron before it enters the magnetic field and within the field.

In the magnetic field, the electron follows a semi-circular path of diameter 0.0700 m. In which direction is the magnetic field and what is the size of the magnetic flux density?

direction of

magnetic field size of magnetic flux density / T

A into page 31019.1

B into page 31038.2

C out of page 31019.1

D out of page 31038.2

16


31 Regions of unbalanced charge are produced inside a cloud as shown.

For the region X, which diagram correctly represents the direction of the electric field and the initial direction in which electrons would move?

32 Two coils of wire connected as shown in the diagram below. The magnet B is suspended

from a spring above the coil on the right and is free to move. Magnet A is moved downwards into the coil on the left.

What is the direction of the force experienced by magnet B as a result of the downward motion of magnet A? A Towards the left. B Towards the right. C Upwards. D Downwards.

N

N

magnet B

magnet A

17


33 A circular coil of diameter 16.0 cm and resistance 4.00 Ω is placed in a uniform magnetic field of flux density 5.00 T directed perpendicularly into the coil.

If the magnetic flux density is reduced to zero at a constant rate over 10.0 ms, what can be deduced about the current flowing in the coil during this change?

magnitude / A direction

A 2.51 clockwise

B 2.51 anticlockwise

C 10.1 clockwise

D 10.1 anticlockwise

34 An electric boiler, designed for travellers, can be used with different supply voltages. It is

rated at 800 W for a 240 Vr.m.s. alternating supply. What will be its power output if it is connected to a 120 V direct supply? A 100 W

B 200 W

C 400 W D 800 W

35 In 2010 the Japanese launched the world’s first interplanetary solar sail spacecraft, called

IKAROS. This works because photons reflected from the sail, of area A, undergo a change

of momentum and, by Newton’s third Law, exert a forward force on the sail. A beam of light of intensity I is reflected at right angles to a solar sail.

The momentum of a photon is given by the expressionc

hf, where f is the frequency of the

light, h is the Planck constant and c is the speed of light. What is the force exerted on the sail?

A

hf

AI B

c

hf2 C

c

AI2 D

c

I

16.0 cm uniform magnetic field directed into the page

18


36 Two phenomena P and Q are described.

P When ultraviolet light shines on zinc, electrons are emitted from the surface. Q When electrons are passed through graphite, a pattern of rings may be observed on

a screen.

Which different models are used to explain the phenomena?

P Q

A particle particle

B particle wave

C wave particle

D wave wave

37 The resistance of a piece of pure silicon falls as the temperature rises.

Which of the following statements is true? A The ratio of the positive to negative charge carriers increases. B The ratio of the positive to negative charge carriers decreases. C The charge carriers can move more easily at a higher temperature. D The total number of charge carriers increases with temperature.

38 In an experiment to learn more about the structure of the atom, Geiger and Marsden fired

α-particles at a thin sheet of gold foil. They found that most of the α-particles passed through

the gold foil with no significant deviation, although a very tiny minority were deflected through large angles, and some were even back-scattered (deflected by more than 90°). The experiment is repeated with a foil made from a heavier isotope of gold. How would the results be different?

A A much greater proportion of the α-particles would be back-scattered.

B A much greater proportion of the α-particles would deflect through a large angle.

C A greater proportion of the α-particles would pass through with no significant deviation.

D There would be no significant change.

19


39 The isotope Rn22286 decays in a sequence of emissions to form the isotope Pb206

82 . At each

stage of the decay sequence, it emits either an α-particle or a β-particle.

What is the number of stages in the decay sequence?

A 4 B 8 C 16 D 20 40 Alpha, beta and gamma radiations

1 are absorbed to different extents in solids,

2 behave differently in an electric field,

3 behave differently in a magnetic field.

The diagrams illustrate these behaviours. diagram 1 diagram 2

diagram 3

Which three labels on these diagrams refer to the same kind of radiation?

A M, P, Z B L, P, Z C L, P, X D N, Q, X

1

H2 Physics/9646/AJC2015/Prelim [Turn Over

2015 AJC H2 Physics Prelim Paper 1 Mark Scheme (40 marks)

1 2 3 4 5

A D C D A

6 7 8 9 10

A D A C C

11 12 13 14 15

A B B A B

16 17 18 19 20

B D B A C

21 22 23 24 25

C D B D A

26 27 28 29 30

C C C B D

31 32 33 34 35

D C A B C

36 37 38 39 40

B D D B A

1 A

From graph, angular deflection of the needle increases at a decreasing rate as the current increases. Hence, the separation of the current scale will decrease as the current increases.

2 D

Measurement of temperature requires 1 scale reading. Uncertainty of scale reading = ½ of smallest division = 0.5° Decimal place of reading follows d.p. of uncertainty

3 C

2

2

1gts on earth, 2'

62

1t

gs

on moon

22 6' tt and tt 6'

4 D Consider vertical motion:

↓: 2

2

1gth

g

ht

2

Consider horizontal motion:

→: g

hvvtd

2

5 A

90 kg min-1 = 1.5 kg s-1

N 305.120)(

t

mv

t

mvF

6 A

fi pp (3m)v – m(2v) = (4m)v’

v' = v/4

7 D

WD to stretch = ½ × 0.5 × 35 = 9 J Energy recovered = ½ × 0.6 × 20 = 6 J Total WD in 1 cycle = Energy remaining in 1 cycle = 9 – 6 = 3 J

8 A

At terminal velocity, U + D = W For a ball in air, U is very small, hence U < D < W

9 C

Instantaneous Power = gradient of Energy – time graph Pmax occurs during t = 2 s to t = 3 s.

Gradient = W3023

1040

10 C

At launch, E=½ m v2 At highest point, KE = ½ m vx

2 where vx = v cos 30 = √3 v / 2 Hence at highest point, KE = ½ m (√3 v / 2)2 = ¾ × ½ m v2 = ¾ E

11 A GPE decreases from P to Q implies the direction of the G-field is towards Since mgx = E, g = E / mx

2

H2 Physics/9646/AJC2015/Prelim

12 B

Ep = ET mgr = ½ m (42) gr = 8 Let x be the point where GPE has decreased by E/4, Ep = Ex mgr = mg(3r/4) + ½ m v2 v2 = ½ gr v = 2.0 ms-1

13 B

Since 32 rT ,

3

3

2

2'64

r

r

T

T

rr 16'

14 A

Since dr

dUF

2

1

r

r

U

U

FdrdUf

i

graph under area U

15 B

E=½ m v2

E=½ m (ω2x02)

2

0

22

2

1x

Tm

J 0106.015.00.5

260.05.0

22

E

16 B

The degree of damping of Q is greater than that of P. Hence the resonant frequency of Q will be slightly lower and the amplitude of vibration at resonance will be lower compared to P.

17 D

Since ice is less dense than water, ice molecules must be further apart than water molecules. Since TKE , the mean KE at the same temperature is the same mean square speed is same Water vapour molecules are identical to water molecules, hence both have the same mass.

18 B

ammonia water E in Gain 75.0E of Loss

vammoniafwater lmlm 75.0

vammonia

fwater l

t

ml

t

m 75.0

t

mammonia

75.0137000036006

33000080000

1-s kg 1895.1t

mammonia

19 A

Thermal equilibrium same temperature

no net transfer of thermal energy The internal energy of a body depends on its mass, which would be different for the 2 spheres of different radii. The radiation of electromagnetic energy depends on the colour and texture of the surface of the spheres. Since both spheres are at the same temperature, the mean internal energy per atom is the same for both spheres.

20 C Microwaves are polarised when passed through a metal grid. Light is also polarised when reflected or scattered. Sound waves are longitudinal and cannot be polarised.

21 C

fv

f

v

To produce the same wave pattern, the wavelength must remain constant. To maintain the same wavelength at a higher frequency, the speed must be increased.

22 D

The amplitude of the sound waves is a maximum in the straight through position and decreases as it moves away from the central peak. The intensity minima further from central maximum do not cancel out completely as amplitude of waves from the individual sources reaching the minima are not equal.

3r / 4

3

H2 Physics/9646/AJC2015/Prelim [Turn Over

23 B

1sind , & 2sin 2 d ,

63 100.2)10500/(1 d

45.17

100.2

10600sin

6

91

1

86.36

100.2

106002sin

6

91

2

Hence 41.1912

24 D Since charge is stationary, it must be experiencing an upward electric force qE and a downward weight. The electric force being upwards implies the charge must be negatively charged. qE = mg q/m = g / E

25 A

Since E = V / d, dE /1

26 C

ItQ

1

106.110103000 19411

t

neI

= 0.48 A ≈ 500 mA

27 C

Smallest R largest I/V

28 C

When LDR is covered, its R↑ leading to V↓. To restore the balance, the sliding contact has to be shifted left. Changing the supply voltage affects V & Vmetal wire at the same time so it does not restore balance.

29 B

1.5 R R 75.01

3

11

RR

Potential at X = 1.5 V

Potential at Y = V0.1375.05.1

75.0

Hence voltmeter reading is 0.5 V

30 D Using FLHR, B must be out of page as F is ↑ and I is towards left.

r

mvBqv

2

mvBqr

T 1038.2

2

07.0106.1

1046.11011.9 3

19

731

B

31 D

32 C

When magnet A moves down, it induces a N pole in the left coil. This cause the induced I to flow, which results in the right coil inducing a N pole beneath magnet B. This results in magnet B experiencing an upward force.

V

E

e movement

N

N

N induced I N

N N

X

Y

4

H2 Physics/9646/AJC2015/Prelim

33 A

)4(

1010

08.053

2

Idt

dE

I = 2.513 A Since flux in coil is into page and decreasing, by lenz law coil will induce current such that to oppose the decrease in flux. Hence current induced will be clockwise.

34 B

R

VP

2

R

2240800

Using 120 V, W200800240

1202

2

P

35 C

IAP IAt

nhf

hf

IA

t

n

t

c

hf

c

hfn

t

pF

c

IA

c

hf

hf

IA

c

hf

t

nF

222

36 B

P: photoelectric effect, particle nature Q: electron diffraction, wave nature

37 D

For intrinsic semiconductors, the ratio of charge carriers is always 1 since electrons that are excited from the conduction band always leaves a hole in the valance band. At higher temperatures, lattice ions vibrate more vigorously and may collide more often with the charge carriers, representing a rise in the resistance.

38 D Back scattering and large angle scattering is due to the positive charge in the gold nucleus. Since the charge of the nucleus is constant, the scattering will not change. Thus A, B & C is incorrect.

39 B

Considering the change in its nucleon number, 222-206=16 This implies that there must be 4 x Helium-4 particles emitted.

ep 01

42

20682

22286 He4PbRn

To balance the proton number, 86 = 82 + 8 + (-p) p=4 Hence,

e01

42

20682

22286 4He4PbRn

40 A

α is L, R & X β is M, P & Z γ is N, Q &Y

1


2015 JC2 Preliminary Examination

PHYSICS 9646/02 Higher 2 Paper 2 Structured Questions Monday 14 September 2015

1 hour 45 minutes

Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name and PDG in the spaces at the top of this page.

Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

1

2

3

4

5

6

7

8

9

Deduction

Total


Candidate Name

PDG

( )

This document consists of 18 printed pages.

2


Data




(1/(36)) x 10−9 F m−1 elementary charge, e = 1.60 x 10−19 C the Planck constant, h = 6.63 x 10−34 J s

unified atomic mass constant, u = 1.66 x 10−27 kg


rest mass of proton, mp = 1.67 x 10−27 kg

molar gas constant, R = 8.31 J K−1 mol−1

the Avogadro constant, NA = 6.02 x 1023 mol−1

the Boltzmann constant, k = 1.38 x 10−23 J K−1



3


Formulae


2 at2

v2 = u2 + 2as



gravitational potential, ϕ = −G m

r


velocity of particle in s.h.m., v = v0 cos t

v = ± ω )x(x 22

0


2 kT

molecule of an ideal gas, resistors in series, R = R1 + R2 + …

resistors in parallel, 1/R = 1/R1 + 1/R2 + …


4 ε0 r



where k = 2

28

h

EUm



21t

4


1 A ball of mass 50 g is released from rest and falls vertically. The ball hits the ground and rebounds vertically, as shown in Fig. 1.1.

Fig. 1.1

The variation with time t of the velocity v of the ball is shown in Fig. 1.2.

Fig. 1.2

Air resistance is negligible.

(a) Without calculation, use Fig. 1.2 to describe the variation with time t of the velocity v of

the ball from t = 0 to t = 2.1 s. …………………………………………………………………………………………………….. …………………………………………………………………………………………………….. …………………………………………………………………………………………………….. …………………………………………………………………………………………………….. …………………………………………………………………………………………………….. ………………………………………………………………………………………………… [3]

5


(b) Calculate, for the ball, from t = 0 to t = 2.1 s,

(i) the distance moved,

distance = …………………………….m [2]

(ii) the displacement from the initial position.

displacement = …………………………….m [1]

(c) While the ball is in contact with the ground, the ground exerts a force on the ball. Determine

(i) the mean acceleration of the ball as it rebounds while in contact with the ground,

acceleration = …………………………….m s-2 [2]

(ii) the mean contact force exerted by the ground on the ball during the impact.

contact force = …………………………….N [2]

6


2 (a) State the conditions for a system of coplanar forces to be in equilibrium. ………………………………………………………………………………………….………….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [2]

(b) Three identical springs S1, S2 and S3 are attached to a point A as shown in Fig. 2.1.

Fig. 2.1

The springs have extended elastically and the extensions of S1 and S2 are x. Determine,

in terms of x, the extension of S3 such that the system of springs is in equilibrium.

Explain your working.

extension of S3 = ...................................... [2]

point A

S1

S2

S3 150

150

7


(c) Two parallel strings P and Q are attached to a disc of diameter 12 cm, as shown in

Fig. 2.2.

Fig. 2.2

The disc is free to rotate about an axis normal to its plane. The axis passes through the

centre C of the disc.

A lever of length 30 cm is attached to the disc. When a force F is applied at right angles

to the lever at its end, equal forces are produced in P and Q. The disc remains in

equilibrium.

For a force F of magnitude 150 N, determine the force in P.

force = ……………………. N [2]

3 The Moon may be considered to be an isolated sphere of radius 1.74 × 103 km with its mass

of 7.35 × 1022 kg concentrated at its centre.

(a) Show that the gravitational field strength at the surface of the moon is 1.62 N kg -1. [1]

C

string Q

string P

12 cm

30 cm

force F

disc

lever

8


(b) A stone of mass 2.40 kg is situated on the surface of the moon.

(i) The stone is raised through a vertical height of 1800 m. Use the value of field strength given in (a) to determine the change in gravitational potential energy of the

stone. Explain your working.

change in energy = ……………………….. J [2]

(ii) Show that the change in gravitational potential energy of the stone in moving it from the Moon’s surface to infinity is 6.76 × 106 J.

[1]

(iii) The escape speed of the stone is the minimum speed that the stone must be given when it is on the Moon’s surface so that it can escape to infinity. Use the answer in (ii) to determine the escape speed. Explain your working.

speed = …………………. m s-1 [2]

9


(c) The Moon is assumed to be isolated in space. The Moon does, in fact, orbit the Earth. State and explain whether the minimum speed for the stone to reach the Earth from the surface of the Moon is different from the escape speed calculated in (b)(iii).

………………………………………………………………………………………...…………..... ……………………………………………………………………………………..……………….. ………………………………………………………………………………………………....... [2]

4 (a) Fig. 4.1 shows the variation of displacement, xA with time t at point P of sound wave A.

The wave has intensity I.

Fig. 4.1

A second sound wave B of the same frequency as sound wave A also passes

point P. This wave has intensity I9

4. The phase difference between the two

waves is 60°. On Fig. 4.1, sketch the variation with time t of the displacement xB of sound wave

B. [3]

(b) (i) State what is meant by the term polarisation when applied to a wave.

………………………………………………………………………………….................... …………………………………………………………………………………................ [1]

10


(ii) Explain why only transverse waves can be polarised.

………………………………………………………………………………….................... …………………………………………………………………………………................ [1]

(iii) Some films released have enabled viewing in three dimensions (3D). This can be done using two superimposed polarised images on the screen. One of the images is the scene as viewed by a left eye and the other the scene as viewed by a right eye. Explain how the images on the screen need to be polarised and how the spectacles of the cinema-goer also need to be polarised.

………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... …………………………………………………………………………………................[3]

5 (a) Define the term magnetic flux.

……………………………………………………………………………………………..……….. ………………………………………………………………………………………………….…... ……………………………………………………………………..……………......................[1]

(b) State Faraday’s Law of electromagnetic induction.

……………………………………………………………………………………………..……….. ………………………………………………………………………………………………….…... ……………………………………………………………………..……………......................[1]

(c) Fig. 5.1 shows a square flat coil of insulated wire with side x = 0.020 m placed at

position Y. The coil has 1250 turns and the ends of the coil are connected to a

voltmeter. The coil moves sideways steadily through the region of magnetic field of flux density 0.032 T at a speed of 0.10 m s-1 until it reaches position Z. The direction of the field is out of the paper, and the total motion takes 1.0 s.

11


Fig. 5.1

(i) Show that the voltmeter reading as the coil enters the field region, after t = 0.2 s, is

80 mV. Explain your reasoning fully.

[3] (ii) On Fig. 5.2, draw a graph of the voltmeter reading against time for the motion of

the coil from Y to Z. Label the y-axis with a suitable scale.

Fig. 5.2

[2]

A rule measuring distance in metre

t / s voltmeter reading 0 0.2 0.4 0.6 0.8 1.0

t / s

12


6 The variation with time t of the current I in a resistor is shown in Fig. 6.1.

Fig. 6.1

The variation of the current with time is sinusoidal.

(a) Explain why, although the current is not in one direction only, power is converted in the resistor. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [1]

(b) By reference to heating effect, explain what is meant by the root-mean-square (r.m.s.) value of an alternating current. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [2]

(c) Using the relation between root-mean-square (r.m.s.) current and peak current, deduce the value of the ratio

.resistor the in converted pow er maximum

resistor the in converted pow er average

ratio = ……………………... [2]

13


7 (a) Describe how the electronic energy levels of atoms of an element change from sharp distinct levels to broad bands as the phase changes from gas to solid. ……………………………………………………………………………………………..……….. ……………………………………………………………………………………………..……….. ………………………………………………………………………………………………….…... ……………………………………………………………………………………………..……….. …………………………………………………………………………………………………....[3]

(b) A light-dependent resistor (LDR) made of intrinsic semiconductor has a resistance in daylight of less than 1 kΩ, and in the dark about 1 MΩ. Explain in terms of band theory why the resistance is smaller in daylight. …………………………………………………………………………………………................... ……………………………………………………………………………………………..……….. ………………………………………………………………………………………………………. ……………………………………………………………………………………………...........[2]

(i) Explain what is meant by lightly damped oscillations.

………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

8 (a) Fig. 8.1 shows a trolley attached to two identical horizontal springs each connected to a rigid stand. The trolley is at rest at its equilibrium position and both springs remain stretched at all times.

Fig. 8.1

The trolley is pulled to one side and released. Its subsequent oscillations are lightly damped, and the frequency of its oscillations is given by

m

kf

2

1

where k is the spring constant of the system and m is the mass of the oscillating body.

14


(b) The graph in Fig. 8.2 shows how the trolley’s displacement varies with time as it oscillates about its equilibrium position.

Fig 8.2

(i) As time t elapses, the amplitude A of the oscillation changes.

Use data from the dashed curves in Fig. 8.2 to complete the table below for missing values of amplitude A and the natural logarithm of A.

A / cm t / s ln (A / cm)

16.0 0.0 2.8

0.8

10.9 1.6 2.4

9.0 2.4 2.2

7.4 3.2 2.0

[1]

15


(ii) Some data from the table in (b)(i) are used to plot the graph of Fig. 8.3.

Fig. 8.3

On Fig. 8.3,

1. plot the point corresponding to time t = 0.8 s, [1]

2. draw the best-fit line for all the plotted points. [1]

(iii) Determine the gradient of the line drawn in (ii) part 2.

gradient = ……………………………. [1]

(iv) Hence, state the equation for your graph in Fig. 8.3.

……………………………………………………………………………….………….... [1]

(c) Describe how the oscillations in Fig. 8.2 would change if stiffer springs were used in the

setup of Fig. 8.1. …………………………………………………………………………………………................... ……………………………………………………………………………………………..……….. ………………………………………………………………………………………………………. ……………………………………………………………………………………………...........[3]

(d) Suggest a limitation in using the trolley as a simplified model for the motion of an atom

in a crystal. …………………………………………………………………………………………………….. …………………………………………………………………………………………………. [1]

×

×

× 1.0 2.0 3.0

×

16


9 A fairground ride carries passengers in chairs which are attached by metal rods to a rotating

central pole, as shown in Fig. 9.1. When the pole rotates with angular velocity , the rods make an angle θ to the vertical.

Fig. 9.1

It is suggested that cos θ is inversely proportional to ω2.

Design a laboratory experiment, using a small object to represent an occupied chair, to test the relationship between θ and ω.

You should draw a labelled diagram to show the arrangement of your apparatus. In your account you should pay particular attention to (a) the equipment you would use, (b) the procedure to be followed, (c) the measurements to be taken, (d) the control of variables, (e) the analysis of the data, (f) any precautions that would be taken to improve the accuracy and safety of the

experiment. Diagram

17


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18


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………………………………………………………………………………………………………. [12]

1

9646/02/AJC2015/PrelimSolutions [Turn Over

2015 Prelim H2 Physics P2 Solutions

1 (a) constant acceleration or linear/uniform increase in velocity until 1.1 s

rebounds or bounces or changes direction decelerates to zero velocity at the same acceleration as initial value

(b) (i) distance = first area above graph + second area below graph = (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 (= 5.94 + 3.96) = 9.9 m

(ii) displacement = first area above graph – second area below graph = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m

(c) (i) Change in velocity = – 8.8 – 10.8 = – 19.6 m s-1 Mean acceleration = change in velocity / time = –19.6 / 0.10 = – 196 m s-2 = – 200 m s-2 (2 s.f.)

(ii) R – mg = ma R = ma + mg = 0.050 (196 + 9.81)

= 10.3 N (accept 10.5 N if acceleration is substituted as 200 m s-2)

2 (a) Resultant/net force on the body must be zero and

Resultant/net torque on the body about any axis must be zero.

(b) Let the force in spring S3 be F3, and the extension in spring S3 be x3. At equilibrium 0F ,

030cos23 FF

30cos233 FkxF

2

323 kxkx

xxx 73.133

(c) Either Moment of F about C = 150 × 0.30

= 45 N m Let the force in P be T 45 = 2T × 0.06 T = 375 = 380 N

OR At eqbm 0 ,

CWM = ACWM 150 × 0.30 = 2T × 0.06

T = 375 = 380 N

3 (a)

26

2211

21074.1

1035.71067.6

R

GMg

-1kg N 62.1

(b) (i) As the change in height is much smaller than the radius of the planet,

hmgGPE = 2.40 × 1.62 × 1800 = 7.00 × 103 J

R

mg

2

9646/02/AJC2015/PrelimSolutions

(ii)

J 106.76

1074.1

40.21035.71067.6

0

6

6

2211

R

GMm

GPEGPEGPE surface

(iii) change in KE = change in GPE

GPEmv 02

1 2

min

62

min 1076.640.22

1v

-13min s m 1037.2 v

(c) Possible reasons could include

Earth would attract the stone

Potential at Earth’s surface is not zero or less than zero

Earth is nearer than infinity, less gain in GPE required to reach earth Thus, the escape velocity would be lower.

4 (a) 2AI

2

41039

4

B

A

B A

I

I

4102 BA cm

Phase difference oo

T

tΔ60360

6

1

103 3

tΔ, so 4100.5 t ms

OR

sinusoidal graph with same frequency as wave A amplitude = 2.0 x 10-4 cm correct phase (ignore lead/lag, look at x-axis only and allow ±1/2 big square)

wave B

wave B

3


(b) (i) Polarisation is where the oscillations in a wave are confined to one direction

only in a plane normal to the direction of transfer of energy of the wave.

(ii) For transverse wave, the displacements of the particles in the wave are at right angles to the direction of transfer of energy of the wave.

(iii) If the left eye image is polarised vertically then the left eye spectacle must also be polarised vertically If the right eye image is polarised horizontally then the right eye spectacle must also be polarised horizontally. Since each image must only be seen by the correct eye, the angle between the planes of polarisation of the two images must be a right angle.

5 (a) Magnetic flux is the product of the magnetic flux density and the area normal to

the field through which the field is passing.

(b) Faraday’s Law states that the emf induced in a conductor is proportional to the

rate of change of magnetic flux linkage.

(c) (i) Area swept out per second by one

side of coil = xv Total flux change per second = NBxv By Faraday’s Law, emf = NBxv emf = 1250 x 0.032 x 0.02 x 0.1 emf = 0.08 V = 80 mV.

ΔMagnetic flux linkage = NBx2 – 0 = 1250 x 0.032 x 0.0202 - 0 = 0.016 Wb-turns Time taken for change = distance / speed = 0.020 / 0.10 = 0.20 s

emf =20.0

016.0

changefortakenTime

linkagefluxMagnetic

emf= 0.08 V = 80 mV.

(ii) appropriate y-scale labelled ‘square pulse’ shape with equal positive and

negative value (+80 mV and -80 mV) value changes within correct time zones, t = 0.2 to 0.4, 0.6 to 0.8 s

voltmeter

reading / mV

50

-50

100

-100

t / s

4


6 (a) power / heating depends on I2, independent of current direction

(b) the value of the steady direct current which

produces heat at the same rate as the alternating current in a given resistor

(c) Since I0 = √2 ×I rms and P = I2R

Pmax = 2 × Paverage

ratio = 0.5

5.0

2ratio

2

2

R

R

rms

rms

I

I

7 (a) From gas to solid, the atoms of the element come closer together.

Electric fields due to their charges / outer electrons overlap/interact, the energy levels begin to split. Solid has large number of atoms so the levels are close together /indistinguishable, thus a band is formed.

(b) Light photons / light energy / thermal energy promote more electrons from valence

band to conduction band, leaving more holes in the valence band. Number of charge carriers increase, so resistance is lower.

8 (a) object undergoing a number of complete to and fro movement about a fixed point

amplitude of vibration decreasing (exponentially) with time

(b) (i)

Point read from graph correctly to appropriate d.p. Value of ln A calculated correctly to appropriate d.p.

A / cm t / s ln (A / cm)

16.0 0.0 2.8

13.0 0.8 2.6

10.9 1.6 2.4

9.0 2.4 2.2

7.4 3.2 2.0

5


(ii)

1. Point corresponding to t = 0.8 s and ln A = 2.6 plotted correctly 2. Best fit line drawn

(iii) gradient = 25.0

2.30

0.28.2

(iv) tA 25.08.2ln

(c) Stiffer springs have higher spring constants.

Since kf , the oscillating frequency will be higher / Period of oscillations will

be shorter Amplitude of oscillation decreases more rapidly.

(d) Trolley oscillates in 1 dimension only, while atom oscillates in 3 dimensions Trolley experiences damping forces, while atom in crystal has total energy conserved (Any relevant point, max 1)

×

×

×

×

× 1.0 2.0 3.0

6


9 Diagram

Procedure 1 Set up the apparatus as shown in the diagram. 2 Clamp a motor using a retort stand. 3 Attach a (metal) pole to a motor. 4 Measure the length of the (rigid) rods l using a metre rule.

5 At the top of the pole, attach a (rigid) rod on each side of the pole. 6 Attach a small object at the end of each rod. 7 Switch on the motor. Measure and record the time taken t for N rotations of the object

using a stopwatch. 8 Repeat step 7 to find the average time tave for N rotations.

9 Find period using T = tave/N and calculate using = 2/T.

10 Measure and record h using the metre rule. Calculate the angle using cos = h/ l.

11 Repeat the experiment by varying the resistance of the resistor connected to motor to

obtain 6 sets of readings of and .

Control of variable: Keep the length of the rod constant by using the same rod. Analysis:

1 Plot a graph of cos against 1/2. 2. The relationship is valid if the graph plotted is a straight line passing through the origin. Safety: 1. Use a protective screen in case the object detaches from the pole. Reliability: 1 Use a set square to ensure that the pole is vertical. 2 Raw time t should be more than 10 seconds (or time at least 10 rotations) to reduce

random error due to human reaction time. 3 Start timing only when the motion is steady so that timing will be accurate.

4 Take preliminary readings to find suitable range of the variables (t and ). 5 Use of a fiducial mark (e.g metre rule) fixed at a point perpendicular to the motion the

object so that the start of the first rotation and the end of last rotation / the counting of rotations can be noted easier.

6 Calculate using cos = h/ l so that measured is more accurate.

motor

object

rod

pole

metre rule

retort stand

resistor dc supply

clamps

bench

h

retort stand

clamps

7


Or 9 Diagram

Procedure 1 Set up the apparatus as shown in the diagram. 2 Clamp a motor using a retort stand. 3 Attach a (metal) pole to a motor. 4 At the top of the pole, attach a (rigid) rod on each side of the pole. 5 Attach a small object at the end of each rod. 6 Switch on the motor. Measure and record the time taken t for N rotations of the object

using a stopwatch. 7 Repeat step 6 to find the average time tave for N rotations.

8 Find period using T = tave/N and calculate using = 2/T.

9 Measure and record the angle using a protractor fixed to the pole (using blu-tack). 10 Repeat the experiment by varying the resistance of the resistor connected to motor to

obtain 6 sets of readings of and . Control of variable: Keep the length of the rod constant by using the same rod. Analysis:

1 Plot a graph of cos against 1/2. 2 The relationship is valid if the graph plotted is a straight line passing through the origin. Safety: 1 Use a protective screen in case the object detaches from the pole. Reliability: 1 Use a set square to ensure that the pole is vertical. 2 Raw time t should be more than 10 seconds (or time at least 10 rotations) to reduce

random error due to human reaction time. 3 Start timing only when the motion is steady/stable so that timing will be accurate.

4 Take preliminary readings to find suitable range of the variables (t and ) 5 Use of a fiducial mark (e.g metre rule) fixed at a point perpendicular to the motion the

object so that the start of the first rotation and the end of last rotation (or the counting of rotations) can be noted easier.

6 Use large protractor fixed to pole so that measured is more accurate. [Total: 12]

motor

object

rod

pole

metre rule

retort stand

resistor dc supply

clamps

bench

retort stand

clamps

protractor


1

2015 JC2 Preliminary Examination

PHYSICS 9646/03 Higher 2 Paper 3 Longer Structured Questions Thursday 17 September 2015

2 hours

Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST

Write your name and PDG in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions.

Section B Answer any two questions You are advised to spend about one hour on each section. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.


Candidate Name

PDG

( )

For Examiner’s Use

1

2

3

4

5

6

7

Deduction

Total

This document consists of 23 printed pages and 1 blank page

2


Data




(1/(36)) x 10−9 F m−1 elementary charge, e = 1.60 x 10−19 C the Planck constant, h = 6.63 x 10−34 J s

unified atomic mass constant, u = 1.66 x 10−27 kg


rest mass of proton, mp = 1.67 x 10−27 kg

molar gas constant, R = 8.31 J K−1 mol−1

the Avogadro constant, NA = 6.02 x 1023 mol−1

the Boltzmann constant, k = 1.38 x 10−23 J K−1



3


Formulae


2 at2

v2 = u2 + 2as



gravitational potential, ϕ = −G m

r


velocity of particle in s.h.m., v = v0 cos t

v = ± ω )x(x 22

0


2 kT

molecule of an ideal gas, resistors in series, R = R1 + R2 + …

resistors in parallel, 1/R = 1/R1 + 1/R2 + …


4 ε0 r



where k = 2

28

h

EUm



21t

4


Section A Answer all the questions in this Section.

1 A particle has mass m and charge +q and is travelling with speed v through a vacuum.

The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 1.1.

Fig. 1.1

The uniform electric field between the plates has magnitude 2.8 × 104 V m–1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 1.1.

(a) Explain why the path of the particle in the electric field is not an arc of a circle. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [1]

(b) A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 1.2.

Fig. 1.2

5


(i) State and explain the direction of the magnetic field. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

(ii) The particle has speed 4.7 × 105 m s–1. Calculate the magnitude of the magnetic flux density. Explain your working.

magnetic flux density = ……………………………. T [2]

(c) The particle in (b) has mass m, charge +q and speed v.

Without any further calculation, state the effect, if any, on the path of a particle that has

(i) mass m, charge -q and speed v,

……………………………………………………………………………….………….... [1]

(ii) mass m, charge +q and speed 2v, ……………………………………………………………………………….………….... [1]

(iii) mass 2m, charge +q and speed v. ……………………………………………………………………………….………….... [1]

6


2 A battery of e.m.f. 4.50 V and negligible internal resistance is connected in series with a fixed resistor of resistance 1200 Ω and a thermistor, as shown in Fig. 2.1.

Fig. 2.1

(a) At room temperature, the thermistor has a resistance of 1800 Ω. Deduce that the

potential difference across the thermistor (across AB) is 2.70 V.

[1]

(b) A uniform resistance wire PQ of length 1.00 m is now connected in parallel with the resistor and the thermistor, as shown in Fig. 2.2.

Fig. 2.2

7


A sensitive voltmeter is connected between point B and a moveable contact M on the wire.

(i) Explain why, for constant current in the resistance wire, the potential difference between any two points on the wire is proportional to the distance between the points. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

(ii) The contact M is moved along PQ until the voltmeter shows zero reading.

1. State the potential difference between the contact at M and the point Q. potential difference = ……………………………. V [1]

2. Calculate the length of wire between M and Q. length = ………………………..… cm [2]

(iii) The thermistor is warmed slightly. State and explain the effect on the length of wire between M and Q for the voltmeter to remain at zero deflection. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

8


3 (a) Explain what is meant by a photon.

…………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [2]

(b) An X-ray photon of wavelength 6.50 × 10–12 m is incident on an isolated stationary electron, as illustrated in Fig. 3.1.

Fig. 3.1

The photon is deflected elastically by the electron of mass me. The wavelength of the deflected photon is 6.84 × 10−12 m.

(i) On Fig. 3.1, draw an arrow to indicate a possible initial direction of motion of the electron after the photon has been deflected. [1]

(ii) Calculate

1. the change in energy of the deflected photon. change in photon energy = ……………………………. J [2]

2. the speed of the electron after the photon has been deflected. speed = ………………………..… m s-1 [2]

9


(c) Explain why the magnitude of the final momentum of the electron is not equal to the change in magnitude of the momentum of the photon. …………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [2]

(d) The angle θ through which the photon is deflected is given by the expression

)cos1( cm

h

e

where Δλ is the change in wavelength of the photon, h is the Planck constant and c is

the speed of light in free space.

(i) Calculate the angle θ.

θ = ……………………………. ° [2]

(ii) Use energy considerations to suggest why Δλ must always be positive.

………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

10


4 (a) An equation for one possible nuclear reaction is

H O→ N He 1

1

17

8

14

7

4

2

Data for the masses of the nuclei are given in Fig. 4.1.

mass / u

proton H1

1 1.00728

-particle He4

2 4.00260

nitrogen-14 N147 14.00307

oxygen-17 O178 16.99913

Fig. 4.1

(i) Calculate the energy change, in joules, associated with this reaction.

energy = …………………………….J [2]

(ii) Suggest and explain why, for this reaction to occur, the -particle must have a minimum speed. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

11


(b) One particular fission reaction may be represented by the nuclear equation

n3 Kr Ba n U 1

0

92

36

141

56

1

0

235

92

Barium-141 has a half-life of 18 minutes. The half-life of Krypton-92 is 3.0 s.

In the fission reaction of a mass of Uranium-235, equal numbers of barium and krypton

nuclei are produced.

(i) Estimate the time taken after the fission of the sample of uranium for the ratio

nuclei 92-Krypton of number

nuclei 141-Barium of number

to be approximately equal to 8.

time = …………………………….s [3]

(ii) Suggest why measurement of the mass and activity of a sample of Barium-141 is not appropriate for the determination of its half-life. ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [1]

(c) The isotopes Radium-224 ( Ra224

88 ) and Radium-226 ( Ra226

88 ) both undergo spontaneous

-particle decay. The energy of the -particles emitted from Radium-224 is 5.68 MeV and from Radium-226, 4.78 MeV.

(i) State what is meant by the decay constant of a radioactive nucleus. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [1]

(ii) Suggest, with a reason, which of the two isotopes has the larger decay constant. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

12


Section B Answer two questions from this Section in the spaces provided.

5 (a) State what is meant by a line of force in

(i) a gravitational field,

……………………………………………………………………………….…………........ ……………………………………………………………………………….………….... [1]

(ii) an electric field.

……………………………………………………………………………….…………........ ……………………………………………………………………………….………….... [1]

(b) A charged metal sphere is isolated in space.

State one similarity and one difference between the gravitational force field and the electric force field around the sphere. similarity ……………………………………………………………………................................ …………………………………………………………………………………............................. difference ……………………………………………………………………………….………… …………………………………………………………………………………............................. ……………………………………………………………………………….…………........…. [3]

(c) Two horizontal metal plates are separated by a distance of 1.8 cm in a vacuum.

A potential difference of 270 V is maintained between the plates, as shown in Fig. 5.1. A proton is in the space between the plates.

Fig. 5.1

13


Explain quantitatively why, when predicting the motion of the proton between the plates, the gravitational field is not taken into consideration. [3]

(d) Define electric potential at a point.

…………………………………………………………………………………………….……….. ………………………………………………………………………………………….………….. …………………………………………………………………………………………….......... [1]

(e) Two point charges A and B are separated by a distance of 20 nm in a vacuum, as

illustrated in Fig. 5.2. A point P is a distance x from A along the line AB.

Fig. 5.2

14


The variation with distance x of the electric potential VA due to charge A alone is shown

in Fig. 5.3.

The variation with distance x of the electric potential VB due to charge B alone is also

shown in Fig. 5.3.

(i) State and explain whether the charges A and B are of the same, or opposite, sign.

……………………………………………………………………………….…………........ ……………………………………………………………………………….…………........ ……………………………………………………………………………….………….... [2]

(ii) Use Fig. 5.3 to determine the charge on A.

charge = ……………………………C [2]

(iii) By reference to Fig. 5.3, state how the combined electric potential due to both

charges may be determined. ……………………………………………………………………………….…………........ ……………………………………………………………………………….………….... [1]

Fig. 5.3

15


(iv) Without any calculation, use Fig. 5.3 to estimate the distance x at which the

combined electric potential of the two charges is a minimum. x = ……………………………nm [1]

(v) The point P is a distance x = 10 nm from A.

An -particle has kinetic energy EK when at infinity.

Use Fig. 5.3 to determine the minimum value of EK such that the -particle may travel from infinity to point P. EK = ……………………………J [3]

(vi) On Fig.5.4, sketch the variation with x of the combined electric field strength E due

to the two point charges A and B for values of x from 6 nm to 14 nm.

[2]

Fig. 5.4

x / nm

E

0 2 4 6 8 10 12 14 16 12 14

16


6 (a) A student states, quite wrongly, that temperature measures the amount of thermal energy in a body. State and explain two observations that show why this statement is incorrect. 1.……………………………………………..………………………………………....................

……………………………………………………………….……………………….................... 2…………………………………………………………….……………………….…………....... ……………………………………………………………….……………………….................... [2]

(b) (i) Define specific latent heat.

.………………………………………………………………………………....................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [1]

(ii) A beaker containing a liquid is placed on a balance, as shown in Fig. 6.1.

A heater of power 110 W is immersed in the liquid. The heater is switched on and, when the liquid is boiling, balance readings m are taken at corresponding times t.

Fig. 6.1

17


A graph of the variation with time t of the balance reading m is shown in Fig. 6.2.

1. State the feature of Fig. 6.2 which suggests that the liquid is boiling at a

steady rate.

…………………………………………………………………………….................... ……………………………………………………………………………................[1]

2. Use data from Fig. 6.2 to determine a value for the specific latent heat L of

vaporisation of the liquid.

L = …………………………………. J kg-1 [2] (iii) State, with a reason, whether the value determined in (b)(ii) part 2 is likely to be

an overestimate or an underestimate of the normally accepted value for the specific latent heat of vaporisation of the liquid. ....……………………………………………………………………………….................... ………………………………………………………………………………….................... ...……………………………………………………………………………….…………. [2]

Fig. 6.2

m / g

300 0 2 4 6 8

t / min

320

340

360

380

18


(c) (i) State what is meant by the internal energy of a gas.

..………………………………………………………………………………..................... ………………………………………………………………………………….................... .……………………………………………………………………………….………….. [1]

(ii) The equation

pV = constant T

relates the pressure p and volume V of a gas to its thermodynamic temperature T.

State two conditions for the equation to be valid. 1.………………………………………………………………………………....................

………………………………………………………………………………….................... 2.……………………………………………………………………………….................... ………………………………………………………………………………….................... [2]

(iii) A container has a volume of 2.1 x 10-3 m3. On a day when the temperature is

15°C, the pressure of the air in the container is 280 kPa. Based on the conditions stated in (c)(ii), calculate

1. the number of moles n of air in the container,

n = …………………………….mol [2] 2. the new temperature of the air in the container when the container is heated

until the pressure rises to 290 kPa. Assume that no air has leaked from the container and that the volume is constant.

temperature = …………………………….°C [1]

19


(d) (i) Write down an equation representing the first law of thermodynamics, defining your symbols carefully. ………………………………………………………………………………….................... ….……...……………………………………………………………………….................... …..…………………………………………………………………………….…………. [1]

(ii) The volume occupied by 1.00 mol of liquid water at 100°C is 1.87 x 10-5 m3. When

the water is vaporised at an atmospheric pressure of 1.01 x 105 Pa, the water vapour has a volume of 2.96 x 10-2 m3. The latent heat required to vaporise 1.00 mol of water at 100°C and 1.01 x 105 Pa is 4.05 x 104 J. Determine, for this change of state, the change in internal energy of the system.

change in internal energy = …………………………….J [3] (iii) Using your answer to (d)(ii), estimate the binding energy per molecule in liquid

water.

binding energy per molecule = …………………………….J [2]

20


7 (a) State what is meant by the principle of superposition of waves. ………………………………………………………………………………………….………….. ………………………………………………………………………………………….………….. ……………………………………………………………………………………………......... [2]

(b) Fig. 7.1 shows an arrangement which can be used to determine the speed of sound in air. The loudspeaker emits a sinusoidal sound wave. The electrical signals from the two microphones P and Q are added together in the electronic "signal adder" and the resultant signal is displayed on the cathode-ray oscilloscope (c.r.o.) screen. This process may be regarded as equivalent to the superposition of the waves. Microphone Q is fixed and microphone P is slowly moved back along the edge of the ruler.

(i) Fig. 7.2 shows the appearance of the trace on the c.r.o. when both microphones

are at the left hand end of the ruler i.e. the same distance from the loudspeaker.

Fig. 7.1

Fig. 7.2 1 cm

21


The time-base setting of the c.r.o. is 0.2 ms / cm. Determine the frequency of the sound wave.

frequency = ……………………………. Hz [1] (ii) As P is moved slowly along the edge of the ruler, the amplitude of the trace is seen

to decrease, then increase, then decrease and so on. Explain 1. why the amplitude is a maximum when P and Q are at the left end of the ruler

………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

2. why the amplitude of the trace varies.

………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

(iii) The first minimum of the amplitude occurs when P is at a distance of 6.8 cm from

the left hand end of the ruler. Determine the speed of the sound in air.

speed = ……………………………. m s-1 [2]

22


(c) Fig. 7.3 shows an arrangement for producing stationary waves in a tube that is closed at one end.

Fig. 7.3 (i) Explain how waves from the loudspeaker produce stationary waves in the tube.

………………………………………………………………………………….................... ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

(ii) One of the stationary waves that may be formed in the tube is represented in Fig. 7.4.

Fig. 7.4

1. Describe the motion of the air particles in the tube at

A. point P,

…………………………………………………………………………………..........[1]

B. point S.

…………………………………………………………………………………..........[1]

2. The wavelength of the sound is 0.375 m. Calculate the length of the tube.

length = …………………. m [1]

23


(d) Fig. 7.5 represents light of wavelength 589 nm emitted from two sources. The time axes have the same scales.

(i) Calculate the frequency of the light waves of wavelength 589 nm.

frequency = ……………………………. Hz [1]

(ii) Find the approximate value of time t shown in Fig. 7.5.

t = ……………………………. s [2]

(iii) Explain why the light from these two sources is not coherent.

………………………………………………………………………………….................... ……………………………………………………………………………….………….... [1]

(iv) Explain why sources that are not coherent do not produce a visible interference pattern. ………………………………………………………………………………….................... ………………………………………………………………………………….................... ……………………………………………………………………………….………….... [2]

Fig. 7.5

1


2015 AJC Prelim H2 Phy P3 Suggested solutions Section A 1 (a) either constant speed parallel to plate

or accelerating force is in direction of electric field

so not circular

(b) (i) direction of force due to magnetic field opposite to that due to electric

field magnetic field into plane of page (using Fleming’s left hand rule)

(ii) force due to magnetic field = force due to electric field Bqv = qE B = E / v = (2.8 × 104) / (4.7 × 105) = 6.0 × 10–2 T

(c) (i) no change / not deviated (ii) deviated upwards (iii) no change / not deviated 2 (a)

V70.2

50.43000

1800

21

1

RR

ERV

(b) (i) for a wire, V = IR = I (ρL/A) ρ and A are constant

so V ∝ L

(ii) 1. 2.70 V 2.

cm 0.60

50.4

70.2

100

L

L

(iii) thermistor resistance decreases as temperature rises so QM is shorter

3 (a) a photon is a quantum of electromagnetic energy

which is dependent only on the frequency of the radiation (or mention E=hf)

(b) (i) arrow below axis and pointing to right

2


(ii) 1. change in energy of photon =

if

hchc

=

1212

834

1050.6

1

1084.6

1)100.31063.6(

= J 1052.1 15 (accept +ve value)

2. gain in energy of electron = loss in energy of photon

½ m v2 = 151052.1

v = -17 s m 1078.5

(c) momentum is a vector quantity either must consider momentum in two directions or direction changes so cannot just consider magnitude

(d) (i)

cos1100.31011.9

1063.61050.61084.6

831

341212

θ = 30.7°

(ii) photon loses energy to the deflected electron deflected photon has less energy, longer wavelength (so Δλ always positive)

4 (a) (i) m = (4.00260u + 14.00307u) – (16.99913u + 1.00728u) =-7.4 × 10-4u

Energy = (m)c2

= (7.4 × 10-4 × 1.66 × 10-27) × ( 3.0 × 108 )2 = 1.1 × 10-13 J

(ii) Mass of products greater than mass of reactants

this mass/energy provided from kinetic energy of the -particle

(b) (i) Krypton-92 reduced to 1/8 in 9 s

in 9 s, very little decay of Barium-141(number of nuclei remains almost constant during this period as half-life is 1080 s) so, approximately 9 s OR λKr = 0.231 s-1 or λBa = 6.42 × 10-4 s-1 The initial number of nuclei for Barium-141 and Krypton-92 are the same.

0.231t-

t10×-6.42

e

e=

-4

8

t = 9.0 s (max 2 s.f.)

3


(ii) Sample/activity would decay appreciably whilst measurements are being made OR Barium-141 has large decay constant (short half-life), the activity would change considerably during the time of measurement.

(c) (i) The radioactive decay constant is the probability of decay per unit time

of a nucleus.

(ii) greater energy of -particle means (parent) nucleus less stable,

nucleus more likely to decay hence Radium-224.

Section B

5 (a) (i) (tangent to line gives) direction of force on a (small test) mass (ii) (tangent to line gives) direction of force on a (small test) positive

charge

(b) similarity:

e.g. radial fields or lines normal to surface or greater separation of lines with increased distance from sphere

or field strength ∝ 1 / (distance to centre of sphere)2 difference: e.g. gravitational force (always) towards sphere electric force direction depends on sign of charge on sphere; can be towards or away from sphere or e.g. gravitational field/force is attractive electric field/force is attractive or repulsive

(c) gravitational force = mpg = 1.67 × 10–27 × 9.81

= 1.6 × 10–26 N electric force = qp(E/d) = 1.6 × 10–19 × 270 / (1.8 × 10–2) = 2.4 × 10–15 N electric force very much greater than gravitational force

(d) work done per unit positive charge in moving a point charge from infinity to the

point.

(e) (i) either both potentials are positive or same sign

so same sign or gradients are positive & negative so fields in opposite directions so same sign

(ii) VA = 0.36 V when x = 8 nm (any value from graph of VA)

4


VA = x

Q

o

A

4

QA = 3.2 x 10-19 C (iii) the individual potentials are summed

(iv) x = 11 nm (allow value of x between 10 nm and 13 nm ) (v) V = 0.14 + 0.29 = 0.43 V (allow 0.42 V → 0.44 V)

V at infinity is zero

WD to move particle from infinity to P = qV = 2 × 1.6 × 10–19 × 0.43 = 1.4 × 10–19 J WD = Ek lost Ek at infinity is min if Ek at P is zero. Hence min EK = 1.4 × 10–19 J

(vi)

Or E fields in opposite directions; correct curvature Graph crosses the x-axis at x= 11 nm (same as ans given in b(iv);

allow 0.5 nm for point plotted) and magnitude of E at 6 nm E at 14 nm with correct range (6 nm to 14 nm)

E

x / nm

0

2 4 6 8 10 12 14 16

E

x / nm

0

2 4 6 8 10 12 14 16

5


6 (a) - two objects of different masses of same material require different amount

of thermal energy to raise 1 K. - temperature shows direction of thermal energy transfer, from high to low regardless of objects - when substance melts/boils, thermal energy is supplied but no temperature change (Any two, max 2 marks).

(b) (i) Specific latent heat is the thermal energy required to change the state of unit mass of a substance without any change of temperature.

(ii) 1. constant gradient/straight line (allow constant slope) 2. Pt = mL or power = gradient × L

Use of gradient of graph (or two points separated by at least 3.5 minutes)

110 × (7.0 - 0)× 60 = L × (372 – 325) × 10-3 L = 9.80 × 105 J kg-1 (accept 2 s.f.) (allow 9.8 to 9.9 rounded to 2.s.f.)

(iii) some energy / heat is lost to surroundings or vapour condenses on sides, so value is an overestimate

(c) (i) The internal energy is the sum of the random kinetic

and potential energies of all the molecules of a gas.

(ii)

fixed mass / amount of gas ideal gas

(iii)

1. n =

RT

pV

= 2.8 x 105 x 2.1 x 10-3 / (8.31 x 288) = 0.246 or 0.25 mol

2.

T

p = constant T = (290/280) x 288

= 298 K , i.e. new temperature = 25 °C

Note: Accept using nRTT

pV with n = 0.25 mol,

new temperature = 20 °C

(d) (i) U = q + w

U – increase in internal energy of the system q – energy (heat) supplied to the system w – work done on the system

(ii) work done by the system = pV

= 1.01 × 105 × (2.96 × 10-2 – 1.87 × 10-5) = 2987 J

6


Hence work done on the system = 2987 J energy supplied = 4.05 x 104 J

change in internal energy = 4.05 x 104 + ( 2987) = 3.75 x 104 J

(iii) Number of molecules in 1.00 mol of liquid water = 6.02 x 1023

Binding energy per molecule = 3.75 x 104 / (6.02 x 1023) = 6.23 x 10-20 J

7 (a) The principle of superposition states that when two waves meet at a point,

the resultant displacement is equal to the vector sum of the individual displacements.

(b) (i) 1 cycle represented by 2 cm.

period = 2 x 0.2 ms = 4.0 x 10-4 s frequency = 1/period = 1/(4.0 x 10-4) = 2500 Hz

(ii) 1. P & Q same distance from speaker OR in phase OR zero path

diff. hence constructive interference/superposition (do not allow arguments based on: nodes and antinodes/standing waves OR "microphones closer to loudspeaker”)

2. as P is moved, path difference increases/changes

minima when P moves odd number of ½ s & maxima if P moves

whole number of s OR minima when waves meet out of phase & maxima when waves meet in phase

OR minima when path difference is odd number of ½ s &

maxima when path difference is whole number of s

(iii) First minimum corresponds to ½ path difference

Wavelength, = 2 x 6.8 = 13.6 cm = 0.136 m

v = f

v = 2500 x 0.136 = 340 m s-1

(c) (i) Incident wave superpose with the reflected wave at closed end

stationary wave formed if tube length equivalent to λ / 4, 3λ / 4, etc.

1. A. no motion (as node) / zero amplitude

B. vibration backwards and forwards /

maximum amplitude along length

7


2.

m 281.0

375.075.0

4

3

L

L

(d) (i) frequency, f = (3.00 x 108)/(589 x 10-9) = 5.09 x 1014 Hz (ii)

10.5 waves in 4.3 cm 15 cm will have w aves37153.4

5.10

Accept 32 to 42 waves in t T = 1/f = 1.96 x 10-15 s, so t ≈ 7 x 10-14 s

(iii) from two different sources/not a constant phase difference (iv) Any of the below:

any coherence between one set of pulses/waves and another set cannot last phase difference between the pulses/waves changes with time position of fringes formed on screen varies so any pattern only lasts for a very short time

15 cm

10.5 waves in 4.3 cm

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