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1

[Turn over

NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2

PHYSICS 9646/01 Paper 1 Multiple Choice 25 September 2014

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, class and tutor’s name on the Answer Sheet in the spaces provided unless this has been done for you.

There are forty questions on this paper. 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 booklet. The use of an approved scientific calculator is expected, where appropriate.

This document consists of 18 printed pages.

NYJC 9646/01/PRELIM/2014

2

Data

Formulae

uniformly accelerated motion, s = ut + ½at2

v2 = u2 + 2as

work done on/by a gas, W = pΔV

hydrostatic pressure, p = ρgh

gravitational potential, = /Gm r

displacement of particle in s.h.m. x = xo sin ωt

velocity of particle in s.h.m. v = vo cos ωt

= 22xxo

resistors in series, R = R1 + R2 + …

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

electric potential, V = Q / 4πεor

alternating current/voltage, x = xo sin ωt

transmission coefficient, T = exp(-2kd)

where k = 2

2

8 m U E

h

radioactive decay, x = xo exp (-λt)

decay constant λ =

21

693.0

t

speed of light in free space, c = 3.00 × 108 m s-1

permeability of free space, μo = 4π × 10-7 H m-1

permittivity of free space, εo = 8.85 × 10-12 Fm-1

(1 / (36 π)) × 10-9 Fm-1

elementary charge, e = 1.60 × 10-19 C

the Planck constant, h = 6.63 × 10-34 J s

unified atomic mass constant, u = 1.66 × 10-27 kg

rest mass of electron, me = 9.11 × 10-31 kg

rest mass of proton, mp = 1.67 × 10-27 kg

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

the Avogadro constant, NA = 6.02 × 1023 mol-1

the Boltzmann constant, k = 1.38 × 10-23 J K-1

gravitational constant, G = 6.67 × 10-11 N m2 kg-2

acceleration of free fall, g = 9.81 m s-2

3

NYJC 9646/01/PRELIM/2014 [Turn over

1 Intensity of a wave I at a distance r can be determined using the equation

24

PI

r

where P is the power of the source.

The fractional error in the measurement of power is a and that in the measurement of distance

is b. What is the fractional error in the calculated value of I?

A 2a b B 8a b C 2a b

a b

D 2

a b

a b

2 A steel rule can be read to the nearest millimetre. It is used to measure the length of a bar

whose true length is 103 mm. Repeated measurements give the following readings:

Length / mm 102, 101, 101, 101, 102, 101

Are the readings accurate and precise to within 1 mm?

results are accurate to within

1 mm results are precise to within

1 mm

A No No

B No Yes

C Yes No

D Yes Yes

4


3 A tennis ball is released from rest at the top of a tall building.

Which graph best represents the variation with time t of the acceleration a of the ball as it falls,

assuming that the effects of air resistance are appreciable?

A B

C D

4 A hot air balloon carrying a passenger is descending at a constant velocity of 20.0 m s-1. The

passenger throws a stone horizontally at 15.0 m s-1 and 2.00 s later, the rock strikes the

ground.

What is the speed at which the rock strikes the ground?

A 15.0 m s-1 B 30.0 m s-1 C 42.4 m s-1 D 59.6 m s-1

5


5 A box of weight W is being pushed, by a force, F up a frictionless surface inclined at an angle

α. F is directed at an angle of β below the horizontal. What is the magnitude of the

acceleration of the box?

A cos sinF W g

W

B cos sinF W g

W

C cos cos sinF W g

W

D sin sin sinF W g

W

6 A 80.0 kg mass falls from a 3.00 m tall ledge down onto the ground. It hits a hard surface and

comes to rest in 0.0250 s. What is the average force exerted by the surface on it?

A 19.6 N B 785 N C 24 600 N D 25 300 N

7 A metal disc is acted upon by a number of forces. The forces are all in the plane of the disc

and the weight of the disc is negligible.

In which of the following situations is the disc in static equilibrium?

A B

C D

F

F

F

2F

F

F

F 2F

F

F

2F

2F

6


8 The diagram shows a block of copper suspended in water. The block experiences an upthrust from the water.

Which statement is the basis of an explanation for this upthrust?

A Copper is denser than water.

B The pressure of water increases with depth.

C The density of water increases with depth.

D The area of face R is greater than the area of face S.

9 A particle is dropped from rest at a height h above the surface of a viscous liquid column of

height 0.80 m. It attains a speed of 7.5 m s-1 after it has penetrated 0.60 m of the viscous

liquid column. If the mass of the particle is 0.80 kg and the viscous liquid offers a constant

retarding force of 120 N, what is the height, h, from which the particle is released?

A 2.3 m B 3.5 m C 11.4 m D 15.7 m

10 An object of mass 400 kg is lifted through a vertical height of 1200 m in 2.0 minutes by an

electric motor. What is the amount of electrical power needed if the overall efficiency of the

system is 80%?

A 3.1 kW B 4.9 kW C 49 kW D 2900 kW

11 A particle of mass m moves in a circle of radius r at a uniform speed with frequency f. What is

the kinetic energy of the particle?

A

B

C 2 π2 m f2 r2 D 4 π2 m f2 r2

7


12 A car of mass m moving at a constant speed v passes over a humpback bridge of radius of

curvature r. (A humpback bridge is curves in a semicircle above a river).

Given that the car remains in contact with the road, what is the net force R exerted by the car

on the road when it is at the top of the bridge?

A

B

C D

13 The average density of Planet P is twice that of Planet Q, and the radius of Planet P is half

that of Planet Q.

The gravitational field strength at the surface of P is 13.4 N kg-1. What is the gravitational field

strength at the surface of Q?

A 3.4 N kg-1 B 13.4 N kg-1 C 53.6 N kg-1 D 80.4 N kg-1

14 Which of the following correctly shows the relationships between potential energy U, kinetic

energy,K and the total energy, E, of a satellite?

A U = -2K = 2E

B U = -K = E

C U = - ½ K = ½ E

D U = K = ½ E

15 A sphere attached to a horizontal spring is oscillating on a smooth horizontal surface at a

frequency of 2.0 Hz. The amplitude of oscillation is 2.0 cm.

What is the speed of the sphere when it is at a distance 1.0 cm from equilibrium position?

A 6.3 cm s–1 B 22 cm s–1 C 25 cm s–1 D 31 cm s–1

8


16 The diagram below shows a displacement-time graph of a body performing simple harmonic motion.

At which one of the points, U, V, W, X, Y or Z, is the body travelling and accelerating in the opposite direction?

A U, Y B V, X C W, Z D X, Z

17 The pressure of an ideal gas increases at constant volume. Which of the following shows the

correct changes to the quantities?

heat absorbed

work done by the system

change in internal energy

A

B 0

C 0

D 0

18 A pressure of 1.0 107 mm of Hg is achieved in a vacuum system. How many gas molecules

are present in 1000 cm3 of gas if the temperature is 20 C? (760 mm of Hg 1.01 105 Pa).

A 4.8 1016 B 3.3 1015 C 4.8 1013 D 3.3 1012

9


19 The diagram shows a transverse wave on a rope. The wave is travelling from left to right.

At the instant shown, the points P and Q on the rope have zero displacement and maximum

displacement respectively.

Which of the following describes the direction of motion, if any, of the points P and Q at this instant?

point P point Q

A downwards downwards

B downwards stationary

C upwards downwards

D to the right to the right

20 A stationary sound wave is set up between a loudspeaker and a wall. A microphone is connected to a cathode-ray oscilloscope (c.r.o.) and is moved along a line

directly between the loudspeaker and the wall. The amplitude of the trace on the c.r.o. rises to a maximum at a position X, falls to a minimum and then rises once again to a maximum at a position Y.

The distance between X and Y is 33 cm. The speed of sound in air is 330 m s-1. Which diagram represents the c.r.o. trace of the sound received at X?

A B

C D

10


21 The diagram shows two tubes.

The tubes are identical except tube X is closed at its lower end while tube Y is open at its

lower end. Both tubes have open upper ends.

A tuning fork placed above tube X causes resonance of the air at frequency f. No resonance is

found at any lower frequency than f with tube X.

Which tuning fork will produce resonance when placed just above tube Y?

A a fork of frequency

B a fork of frequency

C a fork of frequency

D a fork of frequency 2f

22 When the light from two lamps falls on a screen, no interference pattern can be obtained. This is because the

A lamps are not point sources.

B lamps emit light of different amplitudes.

C light from the lamps is not coherent.

D light from the lamps is white.

23 In which of the following cases does an electric field do positive work on a charged particle?

A A positive charge is moved to a point of higher electric potential.

B A negative charge moves opposite to the direction of the electric field.

C A positive charge completes one circular path around a stationary positive charge.

D A negative charge is moved perpendicular to the electric field between two parallel and oppositely charged plates.

11


24 Which one of the following statements about the electric potential at a point is correct?

A An alternative unit for electrical potential is the joule.

B The potential is given by the rate of change of electric field strength with distance.

C The potential at a point due to a system of point charges is given by the sum of the potentials due to the individual charges at that point.

D The potential at a point is defined as the work done in moving one proton from infinity to the point.

25 A wire has resistance R. A second wire has twice the length, twice the diameter, and twice the

resistivity of the first wire. What is the resistance of the second wire?

A 8R B R C

D

26 If a certain resistor obeys Ohm's law, its resistance will

A change as the voltage across the resistor changes.

B change as the current through the resistor changes.

C change as the energy given off by the electrons in their collisions changes.

D not change.

27 Three resistors, each of different value, are connected in a circuit with an e.m.f. of 12 V with

negligible internal resistance. For which of the following resistor combinations is the total

power dissipated the greatest?

A All three resistors in series.

B All three resistors in parallel.

C Two of the resistors in parallel with the third resistor in series with the parallel pair.

D Insufficient information to determine.

12


28 Three resistors connected in parallel have individual values of 4.0 , 6.0 and 10.0 ,

respectively. If this combination is connected in series with a 12 V battery and a 2.0 resistor,

what is the current in the 10 resistor?

A 0.59 A B 1.0 A C 11 A D 16 A

29 Two long straight wires X and Y are placed perpendicular to each other at a distance d apart. A current flows out of the page in wire X while a current flows from left to right in wire Y. What are the directions of the forces acting on wire Y at points P and Q due to the magnetic field produced by wire X?

force at P force at Q

A out of page into page

B into page out of page

C towards X away from X

D towards X towards X

12 V

2.0

4.0 6.0 10.0

13


30 In the diagram, the shaded area represents a uniform magnetic field directed at right-angles into the plane of the paper.

A horizontal beam of electrons enters the field, travelling from left to right. In which direction is this beam deflected by the field?

A upwards

B downwards

C into the plane of the paper

D out of the plane of the paper

31 A conductor in the shape of a solid square is moving with constant velocity in a region of

magnetic field as shown.

The direction of the field is into the plane of the page. Which of the following diagrams correctly represents the separation of the induced charges?

A B C D

electron beam

velocity

magnetic field

into page

conductor

–

–

–

+

+

+

–

–

–

+

+

+

+ + +

– – –

+ + +

– – –

14


32 A horseshoe magnet is brought near to a metallic disc which is free to rotate about the axis XY

as shown.

When the magnet is rotated along the axis XY, the disc will

A remain stationary.

B rotate in the same direction as the magnet.

C rotate in the opposite direction as the magnet.

D oscillate along the axis XY.

33 The diagram below shows a rectangular waveform with a peak current of 3.0 A.

In order to obtain the same r.m.s. current as the waveform shown, a sinusoidal current of the same frequency should have a peak value of

A 2.1 A B 3.0 A C 4.2 A D 6.0 A

N

S

horseshoe magnet metallic disc

X Y

current / A

3.0

-3.0

time

0

15


34 An a.c. supply is connected across a resistor R.

The voltage across AB varies sinusoidally with a peak voltage of 5.0 V as shown.

Which is the following statements about the setup is not correct?

A The frequency of the supply is 50 Hz.

B A steady d.c. supply of

√ V will transfer the same power as the a.c. supply.

C An electron reverses its direction of travel every 0.02 s.

D A p-n semiconductor diode connected to the circuit will be able to rectify the a.c. source

to obtain a d.c. output at R.

35 An electron beam is directed towards a double-slit arrangement as shown. On the other side of the slits, a Geiger-Muller (GM) tube is placed to detect all the electrons that arrive at P.

When slit B is covered the amplitude of the wave function of the electron beam at P is 10 units

and the GM tube detects 200 electrons arriving per unit time. When slit A is covered, the

amplitude of the wave function at P is 6 units. When both slits are open and it happens that

there is a destructive interference at P, the number of electrons per unit time detected at P is

A 32 B 72 C 128 D 180

R

a.c. source

A B

time /s

Voltage / V

5.0

-5.0

0 0.04 0.02

P

GM tube

B

A

electron beam

16


36 Two close parallel plates of separation d are connected to a potential difference Vo / V. An

electron escapes from the positive plate with a kinetic energy E / J (< eVo where e is the

charge of the electron) by absorbing an energetic photon. Considering the quantum behavior

of electrons, the probability for the electron to reach the negative plate

A is zero.

B depends on d only.

C depends on the ratio

o

E

Vonly.

D depends on the ratio

o

E

V and d.

37 When light of frequency less than threshold is shone onto a metal, no electron is emitted

regardless how long is the light radiated on the metal. However, it was found that when laser

light of frequency lower than the threshold is used, electrons could be emitted. Which of the

following is the correct explanation?

A The photon theory does not apply to laser light.

B Each laser photon has a higher momentum that can knock out the electrons.

C Laser light has very high intensity and is highly coherent so the electrons have a chance

to absorb more than a photon at the same time.

D Laser light exhibits wave behaviour rather than particulate behaviour so that the electron

can absorb the wave energy continuously for a longer time interval.

17


38 A P-N junction is connected to an external d.c. source. Which of the following correctly shows

the direction of the conventional current I, relative width of the depletion layer d, directions as

well as relative strengths of the external electric field EEx, internal electric field ED due to the

depletion layer and the resultant electric field EF acting on the P-N junction?

A B

C D

39 A sample of material initially contains atoms of only one radioactive isotope. Which one of the

following quantities is reduced to one half of its initial value during a time equal to the half-life

of the radioactive isotope?

A total mass of the sample

B total number of atoms in the sample

C total number of nuclei in the sample

D activity of the radioactive isotope in the sample

P N

depletion layer

EEx

ED EF

I=0

d

P N

depletion layer

EEx

ED EF

I

d

P N

depletion layer

EEx

ED EF

I

d

P N

depletion layer

EEx

ED EF

I

d

18


40 The activity of a sample of Iodine-131 is plotted as a function of time as shown below. The

activity scale is logarithmic.

The half-life of Iodine-131 is close to

A 180 days B 55 days C 28 days D 8 days

H2 Physics Prelim Paper 1

1. A 11. C 21. D 31. A

2. B 12. D 22. C 32. B

3. D 13. B 23. B 33. C

4. C 14. A 24. C 34. C

5. A 15. B 25. B 35. A

6. D 16. A 26. D 36. D

7. D 17. C 27. B 37. C

8. B 18. D 28. A 38. B

9. C 19. B 29. B 39. D

10. C 20. B 30. B 40. D


1

NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME ______________________________________________________________________________

PHYSICS 9646/02 Paper 2 Structured Questions 19 September 2014

1 hour 45 minutes Candidates answer on the Question Paper.

No Additional Materials are required. ______________________________________________________________________________

READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. 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

Total

72

Data

Formulae


v2 = u2 + 2as

work done on/by a gas, W = p∆V





= 22 xxo






where k = 2

2

8 m U E

h


decay constant λ =

21

693.0

t




(1 / (36 π)) × 10-9 Fm-1











NYJC

1 Astrofall. (a)

(b)

onauts plan

As part offrom the tbefore it h Upon reacfor the accto scale) srate is 10

n a space e

f the prelimtop of a 12

hits the grou

ching Planeceleration oshows the flashes per

9646/02/PR

expedition to

minary inves2 storey buund.

m

et Newtoniaof free fall omotion of a

r second.

3

RELIM/2014

o Planet Ne

stigations couilding. Esti

momentum

a, a scientison Planet Na free falling

Fig

ewtonia to d

onducted omate the m

…………

st takes meaewtonia. A g tennis ba

g. 1.1

determine i

n Earth, a tmomentum

………………

asurementsstroboscop

all released

its accelera

tennis ball of the tenn

…………….

s to determpic photogra

from rest.

[Turn over

ation of free

is releasednis ball just

kg m s1 [2]

ine a valueaph (shownThe strobe

r

e

d t

]

e n e

4


(i) Draw a graph on Fig. 1.2 showing how the displacement of the tennis ball varies with the square of time.

Fig. 1.2 [2]

(ii) Hence determine the acceleration of free fall on Planet Newtonia.

acceleration …………………………………. m s2 [2]

displacement / m

time2 / s2

NYJC

(c)

During theon a ledgrope attac15 m s1. T

(i) Deteproje

(ii) Henland

e space expe. The rescched to it. The acceler

ermine the ectile to the

ce, calculad on the ledg

3

9646/02/PR

pedition on cuer on theThe projec

ration due t

shortest pe astronaut.

te how far ge.

0 m

5

RELIM/2014

the Moon oe ground wactile is direto free fall o

possible tim

the rescue

dis

distance,

of Planet Neants to shoected at anon the Moon

me the resc

time …

er should st

stance …

1

x

55

ewtonia, anot a projec

n initial angn is 1.2 m s

cuer needs

………………

tand in orde

……………

15 m s1

n astronaut ctile to him gle of 55 2.

s to take to

………………

er for the p

………………

[Turn over

is strandedwith a lightand speed

o send the

………. s [2]

projectile to

………. m [1]

r

d t d

e

]

o

]

6


(iii) Suggest how would your answer in (c)(i) change if the situation occurred on Planet Newtonia instead.

……………………………………………………………………………………………….

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

2 (a) State the principle of conservation of momentum.

……………………………………………………………………………………………………...

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

(b) A 0.150 kg toy helicopter is moving at a constant altitude of 75.0 m with a speed of

1.50 m s1 when a shooter fires vertically up and hits it as shown in Fig. 2.1 Given that the mass of the bullet is 1.00 g and the initial speed of the bullet is 100 m s1. Assume negligible air resistance and ignore height of shooter, determine

Fig. 2.1

(i) the distance of the toy helicopter from the shooter when he shoots

distance …………………………………. m [3]

shooter

toy helicopter

75.0 m

ground

1.50 m s1

NYJC

(c)

3 A un

and cable

(a)

(ii) the that

State and helicopter the ground ……………

…………… ……………

……………

……...……

niform sheetby a cable e on the she

Show that

momentumthe bullet is

ma

explain whr in (b) consd.

………………

………………

………………

………………

………………

t of steel wtied to a p

eet is T.

t T is 489 N

9646/02/PR

of the toy s embedded

gnitude of m

hether princisidering its

………………

………………

………………

………………

………………

weighing 800point on its

.

7

RELIM/2014

helicopter ad in the toy

momentum

iple of consmotion imm

……………

……………

……………

……………

………………

0 N is suppleft-edge a

Fig. 3

and bullet ihelicopter a

…………

servation of mediately a

……………

……………

……………

……………

……………

ported by aas shown in

3.1

mmediatelyafter the hit

………………

momentumafter the hit

……………

……………

……………

……………

……………

bolt at its n Fig. 3.1 b

y after the ht.

…………….

m is violatedand just be

………………

………………

………………

………………

……………

lower-left hbelow. The

[Turn over

hit. Assume

kg m s1 [2]

d for the toyefore it hits

……………..

……………..

……………..

……………..

…………. [2]

hand cornerpull by the

[2]

r

e

]

y s

.

.

.

.

r e

]

8


(b) Determine the magnitude of force acting on the bolt by the sheet. magnitude of force …………………………………. N [3]

4 Fig. 4.1 shows a simple electric motor made up of an armature placed in between 2 permanent magnets. The region of space between the 2 magnets has a uniform magnetic flux density of 40 mT. The armature consists of a single square coil of copper wire with each side of length of 20 cm.

Fig. 4.1

(a) Explain what is meant by a magnetic flux density of 40 mT.

……………………………………………………………………………………………………... ……...……………………………………………………………………………………………...

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

(b) On Fig. 4.1, indicate with an arrow the direction of the magnetic field in the region

between the 2 permanent magnets. [1]

12 V R

direction of rotation of armature

A

9


(c) The armature carries a current of 0.55 A just before it starts to move from the instant as shown in Fig 4.1. Determine the magnitude of the torque acting on the armature due to the magnetic force at this instant.

torque …………………………………. N m [3] 5 A metal disc is swinging freely between the poles of an electromagnet, as shown in Fig. 5.1.

Fig. 5.1 When the electromagnet is switched on, the disc comes to rest after 12 s. (a) State Faraday’s law of electromagnetic induction and use the law to explain why an

e.m.f. is induced in the disc. ……………………………………………………………………………………………………... ……...……………………………………………………………………………………………... ……….…………………………………………………………………………………………….. ……...…………………………………………………………………………………………. [2]

metal disc

north pole of electromagnet

south pole of electromagnet

10


(b) An enlarged diagram of the disc as it is leaving the magnetic field is shown in Fig. 5.2. The direction of the magnetic field is shown in the diagram. The dotted circle indicates one possible path of the eddy current generated.

Fig. 5.2

(i) State Lenz’s law of electromagnetic induction. ………………………………………………………………………………………………

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

(ii) Indicate on the dotted path in Fig. 5.2 the direction of the eddy current as the disc is leaving the electromagnet. [1]

(iii) Use Lenz’s law to explain why eddy currents induced in the metal disc are in the

direction as indicated in (b)(ii).

………………………………………………………………………………………………

……………………………………………………………………………………………….

……...……………………………………………………………………………………….

……………………………………………………………………………………………….

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

(c) State and explain how the time taken for the disc to come to rest will change if a metal of higher resistivity is used for the disc.

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

direction of magnetic field is out of the plane

metal disc

path of eddy current

direction of motion

11


6 A helium-neon laser tube consists of a 1:4 mixture of helium and neon gases, neon being the medium in which laser action occurs. Fig. 6.1 shows the few important energy levels involved in the actions.

Fig. 6.1

Helium atoms are excited to a metastable state E3 from ground state by collisions with high

speed electrons. The energy in E3 is then transferred to energy level E

2 by collisions between

the helium and neon atoms. Laser light is then released when the electrons in E2 state fall to

E1 state.

(a) Estimate the order of the time an electron will stay in the following states before falling

to lower states

(i) metastable state E3,

time …………………………………. s

(ii) energy state E2 or E

1.

time …………………………………. s

[1]

(b) Electrons in E3 have energy of 20.61 eV. This is not enough to raise the electrons from

the ground state to E2 which requires 20.66 eV. Suggest why this excitation is possible.

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

12


(c) Lasing occurs when electrons fall from E2 state to E

1 state. Give a brief explanation of

how population inversion is achieved between these two levels.

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

7 A series of data on the performance of one particular modern car are extracted from the

manufacturer’s handbook. The mass of car under test is 1400 kg. Study the following information in Fig. 7.1 to Fig 7.3 and answer the questions that follow.

Speed, v / m s1 13.0 18.0 22.0 27.0 31.0 35.0 36.5 Time to reach the speed from rest, t / s

3.5 5.0 7.0 10.0 13.5 19.5 28.0

Fig. 7.1 Time to reach the speed from rest

Fig. 7.2 Graphs of available force at the wheels (for different gears) and total resistive forces plotted against speed

Total resistive forces on level road, R

Gear 1

Gear 2

Gear 3

Gear 5

Gear 4

13


Fig. 7.3 Climbing resistance of the car on a particular slope

Note: A 10 % gradient means the slope rises 10 metres vertically for every 100 metres of horizontal distance.

(a) (i) On Fig. 7.4, plot a graph of speed v against time t for the car as it accelerates through the gears. [2]

14


Fig. 7.4

(ii) From the above plot, determine the acceleration when the car is travelling at 25 m s1.

acceleration …………………………………. m s2 [2]

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

t /s

v / m s-1

0.0 5.0 10.0 15.0 20.0 25.0 30.0

15


(b) Consider Fig. 7.2, which presents graphs of available force F at the wheels and the resistive forces R against speed v of the car travelling on a level road.

(i) Determine the optimum gear for maximum acceleration at 25 m s1. Justify your

choice. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [2]

(ii) Calculate the maximum theoretical acceleration at 25 m s1. maximum acceleration …………………………………. m s2 [2] (iii) Hence, comment on whether the information provided by the manufacturer is

consistent. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................. [1]

16


(c) The total resistive force FT to the car’s motion on a slope is given by

FT R F

S

where FS is a constant climbing resistance on a particular slope.

By referring to Fig. 7.2 and Fig. 7.3, determine the maximum possible acceleration of the car on a 5 % slope at 15 m s1.

maximum acceleration …………………………………. m s2 [3]

(d) (i) By referring to Fig. 7.2, determine the power required from the engine if this car is to be maintained at a constant speed of 30 m s1 on a level road.

power …………………………………. W [2]

(ii) Determine the fuel consumption in the car’s engine to provide this amount of power if the car travels for 1 hour. Assume that burning one litre of petrol releases 3.5 107 J, and the maximum energy conversion efficiency from the petrol combustion is 20 %.

fuel consumption …………………………………. l [2]

17


(iii) From the answer in (d)(ii), calculate the distance that the car can travel on 1 litre of petrol.

distance travelled per litre of petrol …………………………………. m l1 [2]

(e) Fig. 7.5 shows the hydraulic braking system of the car.

Fig. 7.5

Explain how a braking force is produced when the driver depresses the brake pedal with his foot.

............................................................................................................................................

............................................................................................................................................

............................................................................................................................................

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

brake pads

tyre

wheel

axlebrake fluid

to other brakes

master cylinder piston

brake pedal

disc

18


8 The bow and arrow is a projectile weapon system that predates recorded history and is common to most cultures. A bow is a flexible arc which shoots aerodynamic projectiles called arrows. A string joins the two ends of the bow and when the string is drawn back, the ends of the bow are flexed. (Refer to the diagram on the right; the archer is drawing the string before releasing it.)

When the archer draws its string, elastic potential energy is stored in the bow and string. This stored energy gives the arrow its initial kinetic energy as the string is released. The efficiency of the bow affects arrow flight, bow sound and vibration. Somewhere during the action of drawing, then letting down, energy is lost. In this case, most of the energy is lost to friction in the system; this phenomenon is referred to as “hysteresis” and is common in all mechanical functions that have a return path. The friction comes from the parts of the bow turning on the bearings/axels, bending and relaxing of the limbs as that material shifts and moves, the flex of the riser, losses in the archer’s bones and joints, along with various other minor losses. Design an experiment to investigate how the efficiency of the bow is affected by the distance drawn by the archer. Earlier experiments indicate that x, the length drawn by the archer, is not proportional to F, the force applied by the archer. The following equipment is available: A simple bow and arrows, various weights, a spring balance, light gates with a datalogger, and any other equipment normally available in a school laboratory.

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) how to measure the potential energy of the bow before the arrow is released, (d) how to measure the kinetic energy of the arrow after the string is released, (e) the control of variables. Diagram

19


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1

NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME ______________________________________________________________________________

PHYSICS 9646/02 Paper 2 Structured Questions 19 September 2014

1 hour 45 minutes Candidates answer on the Question Paper.

No Additional Materials are required. ______________________________________________________________________________

READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. 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. ______________________________________________________________________________



1

2

3

4

5

6

7

8

Total

Answer

72

Data

Formulae


v2 = u2 + 2as






= 22 xxo






where k = 2

2

8 m U E

h


decay constant λ =

21

693.0

t




(1 / (36 π)) × 10-9 Fm-1











NYJC

1 Astrofall. (a)

(b)

onauts plan

As part offrom the tbefore it hMass of bFloor-to-ceHeight at tUsing v2 Momentum Upon reacfor the accto scale) srate is 10

t = 0 to 0.1t = 0.1 to 0t = 0.2 to 0t = 0.3 to 0

n a space e

f the prelimtop of a 12

hits the grouall 0.050 eiling heighthe top of b u2 + 2as, vm mv 1.

ching Planeceleration oshows the flashes per

1s: s = 0.000.2s: s = 0.00.3s: s = 0.00.4s: s = 0.1

9646/02/PR

expedition to

minary inves2 storey buund. kg

ht 3 m uilding 12velocity befo.3 Ns

m

et Newtoniaof free fall omotion of a

r second.

07 = ½ a (0.029 = ½ a (065 = ½ a (116 = ½ a (

3

RELIM/2014

o Planet Ne

stigations couilding. Esti

2 3 36 more ball rea

momentum

a, a scientison Planet Na free falling

Fig

.1)2 (0.1)2 (0.1)2

(0.1)2

ewtonia to d

onducted omate the m

m ches groun

…………

st takes meaewtonia. A g tennis ba

g. 1.1

determine i

n Earth, a tmomentum

d 26.6 m/

………………

asurementsstroboscop

all released

its accelera

tennis ball of the tenn

/s

…………….

s to determpic photogra

from rest.

[Turn over

ation of free

is releasednis ball just

kg m s1 [2]

ine a valueaph (shownThe strobe

r

e

d t

]

e n e

4


(i) Draw a graph on Fig. 1.2 showing how the displacement of the tennis ball varies with the square of time.

Fig. 1.2 [2]

All points plotted correctly Best Fit Line drawn

(ii) Hence determine the acceleration of free fall on Planet Newtonia.

Gradient = (0.016 - 0.094) / (0.02 – 0.13 ) = 0.7 m/s2

Acceleration = 2 x gradient = 1.4 m/s2 Coordinates read off correctly Calculation of gradient acceleration …………………………………. m s2 [2]

displacement / m

time2 / s2

NYJC

(c)

During theon a ledgrope attac15 m s1. T

(i) Deteproje

s = u30 = t = 2

(ii) Henlands = u

(iii) SugPlanThe the m……

……

e space expe. The rescched to it. The acceler

ermine the ectile to the

ut + ½ at2

15 sin 55 –2.83 s or 17

ce, calculad on the ledgut = 15 cos

gest how wnet Newtoni

acceleratiomoon. It wo

………………

…...…………

3

9646/02/PR

pedition on cuer on theThe projec

ration due t

shortest pe astronaut.

– ½ (1.2t2) .6 s (NA)

te how far ge. 55 (2.83) =

would your a instead.

on of free faould take lon

………………

………………

0 m

5

RELIM/2014

the Moon oe ground wactile is direto free fall o

possible tim

the rescue

= 24.4 m

dis

r answer in

all on Planenger for the

………………

………………

distance,

of Planet Neants to shoected at anon the Moon

me the resc

time …

er should st

stance …

n (c)(i) cha

et Newtoniae projectile t………………

………………

1

x

55

ewtonia, anot a projec

n initial angn is 1.2 m s

cuer needs

………………

tand in orde

……………

nge if the

a (1.4 m/s2)o reach a h

………………

………………

15 m s1

n astronaut ctile to him gle of 55 2.

s to take to

………………

er for the p

………………

situation o

) is larger thheight of 30 ……………

………………

[Turn over

is strandedwith a lightand speed

o send the

………. s [2]

projectile to

………. m [1]

occurred on

han that onm.

……………

……….. [1]

r

d t d

e

]

o

]

n

n

.

]

6


2 (a) State the principle of conservation of momentum. In an isolated system, total momentum is constant. OR When bodies in a system interact the total momentum remains constant provided no resultant external force acts on the system. ……...…………………………………………………………………………………………. [1]

(b) A 0.150 kg toy helicopter is moving at a constant altitude of 75.0 m with a speed of

1.50 m s1 when a shooter fires vertically up and hits it as shown in Fig. 2.1 Given that the mass of the bullet is 1.00 g and the initial speed of the bullet is 100 m s1. Assume negligible air resistance and ignore height of shooter, determine

Fig. 2.1

(i) the distance of the toy helicopter from the shooter when he shoots

Considering bullet, taking up as positive,

2

2

1

21

75.0 100 9.812

0.780 s or 19.6 s

y y ys u t a t

t t

t t

Considering toy, taking right as positive,

1.50 0.780

1.17 m

x xs u t

2 2

2 2

Distance

1.17 75.0

75.0 m

x ys s

distance …………………………………. m [3]

shooter

toy helicopter

75.0 m

ground

1.50 m s1

7


ptoy

pbullet

p

i

(ii) the momentum of the toy helicopter and bullet immediately after the hit. Assume that the bullet is embedded in the toy helicopter after the hit.

Considering bullet, taking up as positive,

1

100 9.81 0.780

92.3 m s

y y yv u a t

By Conservation of Momentum,

f ip p

2 2

2 2

22 3

1

0.150 1.50 1.0 10 92.3

0.243 kg m s

f toy bullet

toy toy bullet bullet

p p p

m u m u

magnitude of momentum …………………………………. kg m s1 [2]

(c) State and explain whether principle of conservation of momentum is violated for the toy helicopter in (b) considering its motion immediately after the hit and just before it hits the ground. No the principle of conservation of momentum is not violated. As there is external force i.e. gravitational force acting on the toy-bullet system after the hit, principle of conservation of momentum is not applicable. Alternatively, Taking the toy-bullet-earth as a system, there is no net external force acting on the system and hence total momentum is constant by principle of conservation of momentum. Gravitational force is exerted by earth on the toy-bullet downwards and by Newton’s Third Law; toy-bullet will exert an equal but opposite force on earth i.e. upwards. By Newton’s Second Law, toy-bullet gains momentum downwards and earth gains an equal but opposite momentum upwards as the time of interaction between the

toy-bullet and earth is the samep

Ft

. Therefore total momentum of the toy-

bullet-earth system is conserved as the sum of the changes of momentum of both toy-bullet and earth is zero. ……...…………………………………………………………………………………………. [2]

ptoy

pbullet

NYJC

3 A unand cable

(a)

(b)

niform sheetby a cable e on the she

Show that

1

Taking m

800(1.5 / 2

489 N

xWd T d

T

Determine

2

2

Using Cos

800

Force on

By Newto

in magnitu

R W

R

Hence, fo

t of steel wtied to a p

eet is T.

t T is 489 N

2

oment abou

2) ( cos 4

N

d

T

e the magni

2

2 2

sine Rule,

2

489 2

sheet by bo

on's 3rd Law

ude but opp

T WT

orce on bolt

964

weighing 800point on its

. ut the bolt,

47)(1.8)

tude of forc

cos

2(800)(489)

otlt, 554

w, force on b

posite in dir

T

R

by sheet

m

8

46/02/PRELIM

0 N is suppleft-edge a

Fig. 3

ce acting on

)cos 43

4 N

bolt by shee

rection to fo

554 N

magnitude o

Redu

Weig

M/2014

ported by aas shown in

3.1

n the bolt by

et is equal

orce on shee

of force …

eaction forcue to bolt, R

ght of sheet

bolt at its n Fig. 3.1 b

y the sheet.

et by bolt.

……………

ce R

, W

800

lower-left hbelow. The

………………

489

0 N

43⁰

hand cornerpull by the

[2]

………. N [3]

N

r e

]

]

9


4 Fig. 4.1 shows a simple electric motor made up of an armature placed in between 2 permanent magnets. The region of space between the 2 magnets has a uniform magnetic flux density of 40 mT. The armature consists of a single square coil of copper wire with each side of length of 20 cm.

Fig. 4.1

(a) Explain what is meant by a magnetic flux density of 40 mT.

40 mT is the magnetic flux density of a magnetic field in which a force per unit length of 40 mili-newton per metre acts on an infinitely long straight conductor carrying a current of 1 ampere which is placed perpendicularly to the magnetic field. [B1] ……...…………………………………………………………………………………………. [1]

(b) On Fig. 4.1, indicate with an arrow the direction of the magnetic field in the region

between the 2 permanent magnets. [1]

(c) The armature carries a current of 0.55 A just before it starts to move from the instant as shown in Fig 4.1. Determine the magnitude of the torque acting on the armature due to the magnetic force at this instant.

340 10 0.55 0.20

0.0044 N

BF BIL

0.0044 0.20

0.00088 N m

torque …………………………………. N m [3]

12 V R

direction of rotation of armature

A

Left to right

10


5 A metal disc is swinging freely between the poles of an electromagnet, as shown in Fig. 5.1.

Fig. 5.1 When the electromagnet is switched on, the disc comes to rest after 12 s. (a) State Faraday’s law of electromagnetic induction and use the law to explain why an

e.m.f. is induced in the disc. Faraday’s Law states that when there is a change in the magnetic flux linkage of a conductor, an e.m.f. is induced in it. The magnetic flux linkage linking the metal disc will increase and decrease respectively as it enters and exits the magnetic field of the electromagnet. Hence, an e.m.f. will be induced in the disc. ……...…………………………………………………………………………………………. [2]

(b) An enlarged diagram of the disc as it is leaving the magnetic field is shown in Fig. 5.2. The direction of the magnetic field is shown in the diagram. The dotted circle indicates one possible path of the eddy current generated.

Fig. 5.2

metal disc

north pole of electromagnet

south pole of electromagnet

direction of magnetic field is out of the plane

metal disc

path of eddy current

direction of motion

11


(i) State Lenz’s law of electromagnetic induction. Lenz’s Law states that the e.m.f. is induced in the direction such as to oppose the change in the flux linkage that is inducing it. ……...…………………………………………………………………………………... [1]

(ii) Indicate on the dotted path in Fig. 5.2 the direction of the eddy current as the disc

is leaving the electromagnet. [1] (iii) Use Lenz’s law to explain why eddy currents induced in the metal disc are in the

direction as indicated in (b)(ii).

As the metal disc is an electrical conductor, charges will be able to flow freely in the disc. As the disc leaves the magnetic field of the electromagnet, to oppose the decrease in magnetic flux linkage, magnetic flux density has to be induced in the same direction as the electromagnet’s magnetic flux density. As such, the charges have to flow in a circular anti-clockwise direction to induce the magnetic flux density out of the plane. ……...…………………………………………………………………………………... [2]

(c) State and explain how the time taken for the disc to come to rest will change if a metal

of higher resistivity is used for the disc.

As the resistivity of the disc is higher, the resistance of the disc will be higher. This will decrease the magnitude of the eddy current induced and hence, the heat energy dissipated. As such, less kinetic energy will need to be converted to heat energy and this will cause the disc to come to rest in a longer time. (Candidates can also explain by commenting on the slower rate of heat dissipation.) ……...…………………………………………………………………………………………. [2]

6 A helium-neon laser tube consists of a 1:4 mixture of helium and neon gases, neon being the medium in which laser action occurs. Fig. 6.1 shows the few important energy levels involved in the actions.

Fig. 6.1

12


Helium atoms are excited to a metastable state E3 from ground state by collisions with high

speed electrons. The energy in E3 is then transferred to energy level E

2 by collisions between

the helium and neon atoms. Laser light is then released when the electrons in E2 state fall to

E1 state.

(a) Estimate the order of the time an electron will stay in the following states before falling

to lower states

(i) metastable state E3,

time ………103…………………………. s

(ii) energy state E2 or E

1.

time ………108…………………………. s

[1]

(b) Electrons in E3 have energy of 20.61 eV. This is not enough to raise the electrons from

the ground state to E2 which requires 20.66 eV. Suggest why this excitation is possible.

The atoms in the mixture possess kinetic energy. Neon atoms absorb kinetic energy during collisions with the helium atoms to make up the difference.

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

(c) Lasing occurs when electrons fall from E2 state to E

1 state. Give a brief explanation of

how population inversion is achieved between these two levels.

Any excited electrons in E2 and E1 in Neon atoms will fall to the ground state in 10-8 s.

E3 is a metastable state which allows electrons to stay for 10-3s. Many helium atoms in E3 state falls to ground state together during collision and pass the energy to the neon atoms and excite them to the E2. So E2 state will have a higher population compared to the E1 state, population inversion is thus achieved. ……...…………………………………………………………………………………………. [2]

13


7 A series of data on the performance of one particular modern car are extracted from the

manufacturer’s handbook. The mass of car under test is 1400 kg. Study the following information in Fig. 7.1 to Fig 7.3 and answer the questions that follow.

Speed, v / m s1 13.0 18.0 22.0 27.0 31.0 35.0 36.5 Time to reach the speed from rest, t / s

3.5 5.0 7.0 10.0 13.5 19.5 28.0

Fig. 7.1 Time to reach the speed from rest

Fig. 7.2 Graphs of available force at the wheels (for different gears) and total resistive forces plotted against speed

Total resistive forces on level road, R

Gear 1

Gear 2

Gear 3

Gear 5

Gear 4

14


Fig. 7.3 Climbing resistance of the car on a particular slope

Note: A 10 % gradient means the slope rises 10 metres vertically for every 100 metres of horizontal distance.

(a) (i) On Fig. 7.4, plot a graph of speed v against time t for the car as it accelerates through the gears. [2]

15


Fig. 7.4

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

t /s

v / m s-1

0.0 5.0 10.0 15.0 20.0 25.0 30.0

16


(ii) From the above plot, determine the acceleration when the car is travelling at

25 m s1.

Acceleration = 0.00.14

5.110.33

= 1.536 = 1.54 ms-2

for drawing tangent at v = 25 ms-1

for correct calculation. (acceptable range: 1.39 – 1.69)

acceleration …………………………………. m s2 [2]

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

0.0 5.0 10.0 15.0 20.0 25.0 30.0

t/s

v/m

s-1

17


(b) Consider Fig. 7.2, which presents graphs of available force F at the wheels and the resistive forces R against speed v of the car travelling on a level road.

(i) Determine the optimum gear for maximum acceleration at 25 m s1. Justify your

choice. Gear 2 Available force for forward motion, F, is the greatest at this gear. ............................................................................................................................. [2]

(ii) Calculate the maximum theoretical acceleration at 25 m s1. From Fig. 7.2, F = 2500 N and R = 375 N Newton’s 2nd law,

2

2500 375 1400

1.52 m s

F R ma

a

a

maximum acceleration …………………………………. m s2 [2] (iii) Hence, comment on whether the information provided by the manufacturer is

consistent. It is almost equal to the acceleration found in (a)(ii), and hence the information provided by the manufacturer had been consistent. ............................................................................................................................. [1]

(c) The total resistive force FT to the car’s motion on a slope is given by

FT R F

S

where FS is a constant climbing resistance on a particular slope.

By referring to Fig. 7.2 and Fig. 7.3, determine the maximum possible acceleration of the car on a 5 % slope at 15 m s1.

From Fig. 7.2, F = 4300 N and R = 250 N From Fig. 7.3, FS = 500 N Newton’s 2nd law,

2

4300 250 500 1400

2.54 m s

TF F ma

a

a

maximum acceleration …………………………………. m s2 [3]

(d) (i) By referring to Fig. 7.2, determine the power required from the engine if this car is

to be maintained at a constant speed of 30 m s1 on a level road. From Fig. 7.2, R 500 N

500 30

15000

P Fv

power …………………………………. W [2]

18


(ii) Determine the fuel consumption in the car’s engine to provide this amount of power if the car travels for 1 hour. Assume that burning one litre of petrol releases 3.5 107 J, and the maximum energy conversion efficiency from the petrol combustion is 20 %.

15000

Power input required0.20

75000 W

7

75000 3600Fuel consumption

3.5 107.71

l

fuel consumption …………………………………. l [2]

(iii) From the answer in (d)(ii), calculate the distance that the car can travel on 1 litre of petrol.

4 1

30 3600Distance travelled on 1 litre

7.71

1.40 10 m l

distance travelled per litre of petrol …………………………………. m l1 [2]

(e) Fig. 7.5 shows the hydraulic braking system of the car.

Fig. 7.5

Explain how a braking force is produced when the driver depresses the brake pedal with his foot.

Force exerted on the piston (through depression of the brake pedal) creates a pressure in the brake fluid. Assuming that the fluid is incompressible, the fluid in turn exerts a force on brake pads, which pushes them towards the disc.

Friction between the disc and brake pads creates the braking force. ....................................................................................................................................... [2]

brake pads

tyre

wheel

axlebrake fluid

to other brakes

master cylinder piston

brake pedal

disc

19


8 The bow and arrow is a projectile weapon system that predates recorded history and is common to most cultures. A bow is a flexible arc which shoots aerodynamic projectiles called arrows. A string joins the two ends of the bow and when the string is drawn back, the ends of the bow are flexed. (Refer to the diagram on the right; the archer is drawing the string before releasing it.)

When the archer draws its string, elastic potential energy is stored in the bow and string. This stored energy gives the arrow its initial kinetic energy as the string is released. The efficiency of the bow affects arrow flight, bow sound and vibration. Somewhere during the action of drawing, then letting down, energy is lost. In this case, most of the energy is lost to friction in the system; this phenomenon is referred to as “hysteresis” and is common in all mechanical functions that have a return path. The friction comes from the parts of the bow turning on the bearings/axels, bending and relaxing of the limbs as that material shifts and moves, the flex of the riser, losses in the archer’s bones and joints, along with various other minor losses. Design an experiment to investigate how the efficiency of the bow is affected by the distance drawn by the archer. Earlier experiments indicate that x, the length drawn by the archer, is not proportional to F, the force applied by the archer. The following equipment is available: A simple bow and arrows, various weights, a spring balance, light gates with a datalogger, and any other equipment normally available in a school laboratory.

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) how to measure the potential energy of the bow before the arrow is released, (d) how to measure the kinetic energy of the arrow after the string is released, (e) the control of variables.

[12] Diagram

20


MARKSCHEME D: Diagram [max 1] Top view of experimental set-up:

- All apparatus drawn in a practical and workable arrangement, labelled clearly B: Basic Procedure [max 2] - Pull the centre of the bow string through different distances x to fire an arrow. Fire arrow through light gates. - Find efficiency for that x where efficiency = kinetic energy of arrow/potential energy of bow. Present final results

as a graph of lg ε vs lg x, assuming the relationship follows the general equation ε = k xn; the constants k and n can be found from the y-intercept and the gradient respectively

M: Measurement [max 5] - Measurement of x using a meter rule - Force on bow string measured using spring balance or slotted masses - Velocity of moving arrow found using photogates and datalogger/timer or high-speed video camera (mention v =

D/t). D must be measured with a ruler or fixed. T is read from datalogger/computer/CRO attached - KE of arrow found by ½ m v2 where m, mass of arrow found by electronic balance - Potential energy of bow found from area under an F-x graph - efficiency = kinetic energy of arrow/potential energy of bow (marks given in B) C: Control of Variables [max 2] - Use the same bow and arrows throughout the experiment - Mark the position on the string where the pulling force is to be applied - Distance between photogates to be ensured constant by measuring in between readings - Ensure the meter rule used to measure x and the bow itself are both clamped securely in position. F: Further accuracy or safety detail [max 2] - The bow must be securely clamped to the work surface. - The F-x graph can be constructed only after all the readings are taken. The area under the graph can be

obtained by counting the squares under the graph from x=0 to the chosen x. The estimation of the area would be better if the graph were to be as large as possible.

- The first photogate should be placed as close to the bow as possible so that a reliable value of the velocity of the arrow as soon as it leaves the bow can be obtained.

- The first photogate should be placed where the arrow is no longer being pushed by the string and bow - Use a thick target made of cork or foam - Surround experimental setup with some form of shielding to catch rebound arrows - Observers cannot be allowed in the area where the experiment is being carried out - Conduct a trial to determine a suitable distance between the photogates

.....................................................................................................................................................

Bow clamped firmly to a stable surface Spring balance attached

to a marked part of the bow string, pulled back to the drawn length x. The pulling force F is read from this spring balance.

Meter rule place under the bow to measure x, the drawn length

photogates connected to datalogger or tmer

target x


CANDIDATE NAME

CLASS

TUTOR’S NAME

PHYSICS 9646/03Paper 3 Longer Structured questions 23 September 2014

2 hours

Candidates answer on the Question Paper.

No Additional Materials are required


Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid.

Section A Answer all questions.

Section B Answer any two 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.


Section A

1

2

3

4

5

Section B

6

7

8

deductions

Total


80


2

Data

Formulae


v2 = u2 + 2as






= 22 xxo






where k = 2

2

8 m U E

h


decay constant λ =

21

693.0

t




(1 / (36 π)) × 10-9 Fm-1












3For

Examiner’s Use

[Turn over

Section A

Answer all the questions in this section.

1 (a) (i) Define gravitational potential. [2] (ii) Explain why gravitational potential has a negative value.

[2] (b) Values for the gravitational potential due to the Earth are given in the table below:

Distance from the Earth’s surface/ m Gravitational potential/ MJ kg-1

0 -62.72

390 000 -59.12

400 000 -59.03

410 000 -58.94

infinity 0 (i) Calculate the loss in the potential energy of a satellite of mass 700 kg if it falls

from a height of 400 000 m to the surface of the Earth.

loss in potential energy = J [2] (ii) Deduce a value for the gravitational field strength of the Earth at a height of

400 000m.

gravitational field strength = N kg-1 [2]


4For

Examiner’s Use

2 (a) State the First Law of Thermodynamics. [1]

Fig. 2.1 shows the variation with volume of the pressure of an ideal gas. The gas which is initially at state X, can be compressed to state Z either directly along the curve path XZ or indirectly from X to Y to Z.

(b) When the gas is compressed from X to Z along the curved path, 9000 J of heat energy

is released to the surrounding.

(i) Using Fig. 2.1, estimate the work done on the gas.

work done on gas = J [2]

(ii) Hence, calculate the change in internal energy of the gas.

change in internal energy = J [1]

1.0

1.1

1.2

1.3

1.4

1.5

0 1 2 3 4 5 6 7 8

Y

X

Z

pressure / MPa

volume / 103 cm3

Fig. 2.1


5For

Examiner’s Use

[Turn over

(c) When the gas is compressed from X to Z along the paths XY and YZ,

(i) determine the quantity of heat supplied to the gas.

heat supplied = J [2]

(ii) using the graph, explain whether the process from Y to Z is isothermal. [2]

3 (a) Define electric field strength at a point in an electric field. [1] (b) Fig. 3.1 shows two isolated point charges X and Y. X carries a charge of +3.2 10-10 C,

while Y carries a charge of -6.2 10-10 C. The two charges are 0.20 cm apart. Sketch on Fig. 3.1, the resultant electric field lines due to the two charges.

Fig. 3.1 [3]

X Y


6For

Examiner’s Use

(c) A test charge T of +1.2 10-10 C is placed 0.10 cm from X such that XTY forms a right angle as shown in Fig. 3.2.

Fig. 3.2

(i) Calculate the magnitude of the net force, F, acting on T and

magnitude of net force, F = N [3]

(ii) indicate its general direction in Fig. 3.2 (Exact calculation for the direction is not

required.) [1]

4 A potential divider circuit consists of two resistors of resistances A and B, as shown in Fig. 4.1.

Fig. 4.1

The battery has e.m.f. E and negligible internal resistance.

E

A B

V

+3.2 10-10 C

+1.2 10-10 C

X Y

-6.2 10-10 C

T

0.20 cm

0.10 cm


7For

Examiner’s Use

[Turn over

(a) Deduce that the potential difference V across the resistor of resistance A is given by the expression

AV E

A B

[2] (b) The resistances A and B are 1500 and 4000 respectively.

A voltmeter is connected in parallel with the 1500 resistor and a thermistor is connected in parallel with the 4000 resistor, as shown in Fig. 4.2.

Fig. 4.2

The battery has e.m.f. 5.0 V. The voltmeter is ideal. (i) State and explain qualitatively the change in the reading of the voltmeter as the

temperature of the thermistor is raised. [3] (ii) The voltmeter reads 2.4 V when the temperature of the thermistor is 20 °C.

Calculate the resistance of the thermistor at 20 °C.

resistance = [3]

5.0 V

1500 4000

V


8For

Examiner’s Use

5 Fig. 5.1 shows the variation with nucleon number of the binding energy per nucleon of a nucleus.

Fig. 5.1

(a) On Fig. 5.1, mark with the letter S the position of the nucleus with the greatest stability. [1]

(b) One possible fission reaction is

235 1 144 90 1

92 0 56 36 0U n Ba Kr 2 n

(i) On Fig. 5.1, mark possible positions for 1 the Uranium-235 nucleus and label this position U, [1] 2 the Krypton-90 nucleus and label this position Kr. [1]

binding energy per nucleon

nucleon number


9For

Examiner’s Use

[Turn over

(ii) The binding energy per nucleon of each nucleus is as follows:

235 1292

144 1256

90 1236

U: 1.2191 10 J

Ba: 1.3341 10 J

Kr: 1.3864 10 J

Use these data to calculate

1 the energy released in this fission reaction, giving your answer to three

significant figures.

energy released = J [3]

2 the mass equivalent of this energy.

mass = kg [1]

(iii) Suggest why the neutrons were not considered in your calculation in (ii). [1]


10For

Examiner’s Use

Section B

Answer two questions in this section.

6 (a) A structure consists of a sphere of mass 0.500 kg, attached firmly to one end of a light rod. The other end of the rod is pivoted freely at point O. The distance between the centre of gravity of the sphere to O is 0.400 m. The structure is held in a position such that the rod is at an angle of 5.00 from the vertical, as shown in Fig. 6.1.

Fig. 6.1 The structure is then released from rest and oscillates in simple harmonic motion. At

one instant during the oscillation, the sphere is directly below O, as shown in Fig. 6.2.

Fig. 6.2

(i) On Fig. 6.2, indicate the forces acting on the structure. [2] (ii) Show that the linear speed of the sphere at the instant in Fig. 6.2 is 0.173 m s-1.

[2]

O

0.400 m5.00

O

0.400 m


11For

Examiner’s Use

[Turn over

(iii) Determine the force exerted on the pivot by the structure in Fig. 6.2.

force = N [3] (iv) With the aid of a diagram, discuss whether the force exerted by the pivot on the

structure is always upward throughout the oscillation.

[3] (b) A horizontal turntable is connected to a motor such that it rotates at exactly 47

revolutions per minute. A peg is fixed vertically on the turntable. A horizontal beam of light casts a shadow of the peg onto a screen in front of which is suspended the structure mentioned in (a), as shown in Fig. 6.3.

Fig. 6.3 The length of the structure is adjusted such that the shadows of the peg and the

structure move in phase on the screen.

parallel beam of light

turntable

structure screen

peg


12For

Examiner’s Use

(i) Explain what is meant by in phase in this context. [1]

The angular speed of the turntable suddenly increases to 1

473

revolutions per minute.

(ii) Define angular speed. [1] (iii) Briefly describe what will be observed on the screen. [2] (iv) Determine the time taken before the two shadows are next in phase.

time taken = s [2] (v) Calculate the number of oscillations made by the peg before the shadows are

next in phase.

number of oscillations = [2] (vi) State two assumptions that were made in the calculations of (b)(iv) and (b)(v). [2]


13For

Examiner’s Use

[Turn over

7 (a) (i) State the principle of superposition. [1] (ii) Explain what is meant when two sources are coherent. [1] (b) Two sources S1 and S2 of sound are situated 80 cm apart in air, as shown in

Fig. 7.1.

Fig. 7.1

The frequency of vibration can be varied. The two sources always vibrate in phase but

have different amplitudes of vibration. A microphone M is situated a distance 100 cm from S1 along a line that is normal to

S1S2. As the frequency of S1 and S2 is gradually increased, the microphone M detects

maxima and minima of intensity of sound. (i) State the two conditions that must be satisfied for the intensity of sound at M to

be zero. 1. 2. [2]


14For

Examiner’s Use

(ii) The speed of sound in air is 330 m s-1. The frequency of sound from S1 and S2 is increased. Determine the number of

minima that will be detected at M as the frequency is increased from 1.0 kHz to 4.0 kHz.

number of minima = [4] (iii) The variation with time of the displacement x of the sound waves arriving at M

from S1 and S2 are as shown in Fig. 7.2a and Fig. 7.2b respectively.

Fig. 7.2a

Fig. 7.2b

Wave from S2

Wave from S1


15For

Examiner’s Use

[Turn over

Determine the ratio of

intensity of sound when a maxima is detected at M

intensity of sound when a minima is detected at M

ratio = [3] (c) Laser beam of red light of wavelength 644 nm is incident normally on a diffraction

grating having 550 lines per millimetre, as illustrated in Fig. 7.3.

Fig. 7.3

Red light of wavelength λ is also incident normally on the grating. The first order

diffracted light of both wavelengths is illustrated in Fig. 7.3. (i) Determine the total number of bright spots of wavelength 644 nm that are visible.

total number of bright spots = [3]


16For

Examiner’s Use

(ii) State and explain 1 whether λ is greater or smaller than 644 nm, [1] 2 in which order of diffracted light there is the greatest separation of the two

wavelengths. [2] (iii) The diffraction grating is now rotated 90⁰ about an axis parallel to the incident

laser beam, as shown in Fig. 7.4.

Fig. 7.4

State what effect, if any, this rotation will have on the diffraction pattern that is

observed. [2]

(iv) In another experiment using the same apparatus, a student notices that the

angular separation between the zero order maxima and the two 1st order maxima are not equal.

Suggest a reason for this difference. [1]


17For

Examiner’s Use

[Turn over

8 (a) ‘X-rays are used to investigate the atomic structure of solids.’ Deduce from this statement the wavelength of the X-rays used.

[1] (b) ‘Sometimes, for example, in the case of rubber, electrons with a de Broglie wavelength

of about 0.11 nm are used instead of X-rays.’ Determine the momentum of each electron.

momentum = N s [2]

(c) An X-ray tube operates with a potential difference of 100 kV between the anode and

cathode. Fig. 8.1 is a sketch of the X-ray spectrum produced by this tube for a particular metal target. Fig. 8.2 shows a sketch of the energy level of target material and how the K line is formed. The tube voltage is 100 kV and the current is 20 mA.

(i) Calculate the maximum energy of an X-ray photon produced, explain your

working.

maximum energy = J [2]

Intensity

Photon energy

0

0

Fig. 8.1

K

K-shell

L-shell

M-shell

N-shell

Fig. 8.2

12

3 4 5


18For

Examiner’s Use

(ii) With reference to Fig. 8.1, write down the numbers that represent the spectrum lines K and L .

K: L: [2] (iii) Sketch on Fig. 8.1 a spectrum for X-ray from the tube if the tube voltage is

reduced to 50 kV, the current at 10 mA. Label this spectrum A. [2] (d) Explain how the characteristic and continuous parts of the spectrum are formed. (i) Formation of characteristic parts of X-ray spectrum: [2] (ii) Formation of continuous parts of X-ray spectrum: [2]

(e) The energy required to remove an electron from the various shells of the nickel atom is: K shell 1.36 × 10-15 J L shell 0.16 × 10-15 J M shell 0.08 × 10-15 J

An X-ray tube with a nickel target emits the X-ray K radiation of nickel. Determine (i) the minimum potential difference across the tube,

potential difference = V [2]


19For

Examiner’s Use

[Turn over

(ii) the energy of the X-ray quantum of longest wavelength in the K-spectrum of nickel.

energy = J [2]

(f) A beam of electrons moving in the x-axis in an X-ray tube with momentum 4 × 10-23 kg m s-1 in the x-axis passes through a 3 mm slit in an anode before it hits the target as shown in Fig. 8.3. The uncertainty of the y-position of the electrons can be considered in the order of the size of the slit. Use uncertainty principle to estimate the possible angular spread of the electron beam after passing through the slit.

= rad [3]

Fig. 8.3

Target Electron beam

anode


CANDIDATE NAME

CLASS

TUTOR’S NAME

Solution

PHYSICS 9646/03Paper 3 Longer Structured questions 23 September 2014

2 hours

Candidates answer on the Question Paper.

No Additional Materials are required


Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid.

Section A Answer all questions.

Section B Answer any two 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.


Section A

1

2

3

4

5

Section B

6

7

8

deductions

Total


80


2

Data

Formulae


v2 = u2 + 2as






= 22 xxo






where k = 2

2

8 m U E

h


decay constant λ =

21

693.0

t




(1 / (36 π)) × 10-9 Fm-1












3For

Examiner’s Use

[Turn over

Section A

Answer all the questions in this section.

1 (a) (i) Define gravitational potential. [2] The gravitational potential at a point in the gravitational field is defined as the work done per unit mass to move a test mass from infinity to the point by an external agent (without change in kinetic energy).

(ii) Explain why gravitational potential has a negative value. [2] The gravitational potential is always negative as gravitational field is an attractive

field. The work done per unit mass to move a test mass from infinity to the point by an external agent will be negative since the force by the external agent is opposite in direction to the displacement of the test mass as it moves from infinity to a point in the gravitational field.

Since we defined the gravitational potential at infinity to be zero, all gravitational potential in the gravitational field must have a negative value..

(b) Values for the gravitational potential due to the Earth are given in the table below:

Distance from the Earth’s surface/ m Gravitational potential/ MJ kg-1

0 -62.72

390 000 -59.12

400 000 -59.03

410 000 -58.94

infinity 0 (i) Calculate the loss in the potential energy of a satellite of mass 700 kg if it falls

from a height of 400 000 m to the surface of the Earth. [2]

6

9

Loss in potential energy

700[( 62.72) ( 59.03)] 10

2.58 10 J

m

(ii) Deduce a value for the gravitational field strength of the Earth at a height of

400 000m. [2]

6

1

Since , to deduce at 400000 m,

calculate the gradient of it using values at 390000 m and 410000 m

[( 59.12) ( 58.94)] 109.0 N kg

390000 410000

dg g

dr

gr


4For

Examiner’s Use

2 (a) State the First Law of Thermodynamics. [1] The First Law of Thermodynamics states that the increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.

Fig. 2.1 shows the variation with volume of the pressure of an ideal gas. The gas which is initially at state X, can be compressed to state Z either directly along the curve path XZ or indirectly from X to Y to Z.

(b) When the gas is compressed from X to Z along the curved path, 9000 J of heat energy

is released to the surrounding.

(i) Using Fig. 2.1, estimate the work done on the gas. [2] Work done for each square = (0.05 106)(0.5 10-3) = 25 J Estimated number of squares under the curve = 22 + 22 x 12 = 22 +264 = 286 WXZ = 25 286 = 7150 J = 7200 J (accept between 7100 to 7200)

(ii) Hence, calculate the change in internal energy of the gas. [1]

U = W + Q

U = 7200 – 9000

U = –1800 J

(c) When the gas is compressed from X to Z along the paths XY and YZ,

(i) determine the quantity of heat supplied to the gas. [2] WXYZ= Area under XYZ = (1.45 x 106)[(7.0-1.0)x 10-3] = 8700 J UXYZ = QXYZ + WXYZ

QXYZ = –1800 – 8700 = –10500 J

(ii) using the graph, explain whether the process from Y to Z is isothermal. [2] From Y to Z, pressure is constant and volume decreases, hence the product of pV is not constant (decreases). Using pV = nRT, for the same amount of substance, T is not constant (decreases). Therefore process is not isothermal.

1.0

1.1

1.2

1.3

1.4

1.5

0 1 2 3 4 5 6 7 8

Y

X

Z

pressure / MPa

volume / 103 cm3

Fig. 2.1


5For

Examiner’s Use

[Turn over

3 (a) Define electric field strength at a point in an electric field. [1] The electric field strength at a point in an electric field is defined as the electrostatic

force acting per unit positive charge placed at that point. (b) Fig. 3.1 shows two isolated point charges X and Y. X carries a charge of +3.2 10-10 C,

while Y carries a charge of -6.2 10-10 C. The two charges are 0.20 cm apart. Sketch on Fig. 3.1, the resultant electric field lines due to the two charges. [3]

Fig. 3.1

Main features of drawing: 1. Arrows point from positive charge X to negative charge Y. 2. More field lines (nearly double) point towards the negative charge compared to the positive charge. 3. Asymmetry of the field lines between the two charges, with the ‘hump’ closer to the positive charge. Symmetry about the horizontal line joining the two charges.

(c) A test charge T of +1.2 10-10 C is placed 0.10 cm from X such that XTY forms a right

angle as shown in Fig. 3.2.

Fig. 3.2

(i) Calculate the magnitude of the net force, F, acting on T and [3]

Distance TY = 2 20.20 0.10

= 0.03 cm

= 0.173 cm

X Y

+3.2 10-10 C

+1.2 10-10 C

X Y

-6.2 10-10 C

T

0.20 cm

0.10 cm

0.2 cm


6For

Examiner’s Use

Force on T due to Y,

FY = 2

q q 4 r

Y T

o

= 10 10

2 2

(6.2 10 )(1.2 10 )

4 ( 0.03 10 )o

= 2.23 x 10-4 N Force on T due to X,

FX = 2

q q 4 r

X T

o

= 10 10

2 2

(3.2 10 )(1.2 10 )

4 (0.10 10 )o

= 3.45 x 10-4 N Magnitude of the net force acting on T, F

= 2 2 4( 3.45 2.23 ) 10 = 4.11 x 10-4 N

(ii) indicate its general direction in Fig. 3.2 (Exact calculation for the direction is not

required.) [1]

Direction of the net force on T is acting towards the top right hand corner. [This must be indicated on the Figure as indicated in the instructions.]

4 A potential divider circuit consists of two resistors of resistances A and B, as shown in Fig. 4.1.

Fig. 4.1

The battery has e.m.f. E and negligible internal resistance.

E

A B

V

T

FResultant

FX

FY

X Y


7For

Examiner’s Use

[Turn over

(a) Deduce that the potential difference V across the resistor of resistance A is given by the expression

AV E

A B

[2]

Current in series circuit

Potential difference across A

[shown]

tot

E

R

E

A BIA

EA

A BA

EA B

(b) The resistances A and B are 1500 and 4000 respectively.

A voltmeter is connected in parallel with the 1500 resistor and a thermistor is connected in parallel with the 4000 resistor, as shown in Fig. 4.2.

Fig. 4.2

The battery has e.m.f. 5.0 V. The voltmeter is ideal. (i) State and explain qualitatively the change in the reading of the voltmeter as the

temperature of the thermistor is raised. [3] As the temperature of the thermistor is raised, its resistance will decrease. This will result in the combined resistance of the thermistor and the 4000 resistance to decrease due to their parallel circuit arrangement. As a result, based on the potential divider principle, the potential across the 1500 resistance will increase and hence, the voltmeter reading will increase.

5.0 V

1500 4000

V


8For

Examiner’s Use

(ii) The voltmeter reads 2.4 V when the temperature of the thermistor is 20 °C. Calculate the resistance of the thermistor at 20 °C. [3]

3

Potential diff across 1500

15002.4 5.0

1500

1625

1 1 1

1 1 1

1625 4000

2737

2.7 10

ll

ll

ll

ll thermistor

thermistor

thermistor

AE

A R

R

R

R B R

R

R

5 Fig. 5.1 shows the variation with nucleon number of the binding energy per nucleon of a

nucleus.

Fig. 5.1

(a) On Fig. 5.1, mark with the letter S the position of the nucleus with the greatest stability. [1]

S shown at the peak (b) One possible fission reaction is

235 1 144 90 1

92 0 56 36 0U n Ba Kr 2 n

(i) On Fig. 5.1, mark possible positions for 1 the Uranium-235 nucleus and label this position U, [1] 2 the Krypton-90 nucleus and label this position Kr. [1] Kr and U on right of peak in correct relative positions

binding energy per nucleon

nucleon number


9For

Examiner’s Use

[Turn over

(ii) The binding energy per nucleon of each nucleus is as follows:

235 1292

144 1256

90 1236

U: 1.2191 10 J

Ba: 1.3341 10 J

Kr: 1.3864 10 J

Use these data to calculate

1 the energy released in this fission reaction, giving your answer to three

significant figures. [3] binding energy of U-235 = 2.8649 x 10-10

J binding energy of Ba-144 = 1.9211 x 10-10

J binding energy of Kr-90 = 1.2478 x 10-10

J energy release = 3.04 x 10-11

J

2 the mass equivalent of this energy. [1] E = mc2 = 3.38 x 10-28

kg

(iii) Suggest why the neutrons were not considered in your calculation in (ii). [1] Neutrons are single particles which have no binding energy per nucleon

Section B

Answer two questions in this section.

6 (a) A structure consists of a sphere of mass 0.500 kg, attached firmly to one end of a light rod. The other end of the rod is pivoted freely at point O. The distance between the centre of gravity of the sphere to O is 0.400 m. The structure is held in a position such that the rod is at an angle of 5.00 from the vertical, as shown in Fig. 6.1.

Fig. 6.1

O

0.400 m5.00


10For

Examiner’s Use

The structure is then released from rest and oscillates in simple harmonic motion. At one instant during the oscillation, the sphere is directly below O, as shown in Fig. 6.2.

Fig. 6.2

(i) On Fig. 6.2, indicate the forces acting on the structure. [2] (ii) Show that the linear speed of the sphere at the instant in Fig. 6.2 is 0.173 m s-1.

[2]

2

2

1

By conservation of energy,

Loss in GPE Gain in KE

( cos ) 0.5

9.81(0.400)(1 cos5.00) 0.5

0.1728 0.173 m s

mg r r mv

v

v

(iii) Determine the force exerted on the pivot by the structure in Fig. 6.2. [3]

2

2

Since the structure is moving in circular motion,

by considering the structure,

0.17280.500( 9.81)

0.4004.94 N

By Newton's 3rd Law, the force exerted on the structure by the pivot

is

c cF ma

vF W m

r

F

F

equal in magnitude and opposite in direction to the force exerted

on the pivot by the structure. [1] Hence force on pivot is 4.94 N.

O

0.400 m

Weight of structure, W

Force on structure by pivot, F


11For

Examiner’s Use

[Turn over

(iv) With the aid of a diagram, discuss whether the force exerted by the pivot on the structure is always upward throughout the oscillation. [3] Not always vertical. This is because as the structure is oscillating (at any instant between Fig. 6.1 and Fig. 6.2), there is a centripetal acceleration toward O, which is diagonal. This suggests that the resultant force needs to have a horizontal component. As weight of the structure is always vertically downward, the only force that can provide the horizontal component can only be from the pivot.

(b) A horizontal turntable is connected to a motor such that it rotates at exactly 47

revolutions per minute. A peg is fixed vertically on the turntable. A horizontal beam of light casts a shadow of the peg onto a screen in front of which is suspended the structure mentioned in (a), as shown in Fig. 6.3.

Fig. 6.3 The length of the structure is adjusted such that the shadows of the peg and the

structure move in phase on the screen. (i) Explain what is meant by in phase in this context. [1]

In phase means that the shadows of the peg and structure are always synchronized. The shadows always move to the same extreme location at the same time.

The angular speed of the turntable suddenly increases to 1

473

revolutions per minute.

(ii) Define angular speed. [1]

Angular velocity ω is the rate of change in angular displacement of a body. Or The rate of change in phase of an oscillator

(iii) Briefly describe what will be observed on the screen. [2]

The shadow of the peg will oscillate faster than that of the structure, and the two shadows will not be in phase anymore. The phase difference will increase with time.

(iv) Determine the time taken before the two shadows are next in phase. [2]

Since the peg is 1/3 revolution per minute faster than the structure, this means that the peg will cover 1/3 revolution more every one minute. To be in phase again, the peg needs to cover exactly one revolution more than the structure.

Hence 1

time required 1 3 min3

= 180 s

parallel beam of light

turntable

structure screen

peg


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(v) Calculate the number of oscillations made by the peg before the shadows are next in phase. [2]

Number of revolution = 3 × 47 + 1 = 142 revolutions (vi) State two assumptions that were made in the calculations of (b)(iv) and (b)(v). [2] There is no loss in energy of the structure throughout the motion and hence the

period/angular speed of the structure remains constant. The turntable is able to maintain its angular speed/period due to the motor.

7 (a) (i) State the principle of superposition. [1] When two (or more) waves meet at (a point in space), the resultant displacement is the vector sum of the individual displacements

(ii) Explain what is meant when two sources are coherent. [1] Waves emitted from the sources have a constant phase difference. (b) Two sources S1 and S2 of sound are situated 80 cm apart in air, as shown in

Fig. 7.1.

Fig. 7.1

The frequency of vibration can be varied. The two sources always vibrate in phase but

have different amplitudes of vibration. A microphone M is situated a distance 100 cm from S1 along a line that is normal to

S1S2. As the frequency of S1 and S2 is gradually increased, the microphone M detects

maxima and minima of intensity of sound. (i) State the two conditions that must be satisfied for the intensity of sound at M to

be zero. [2] 1. Waves from S1 and S2 arriving at M have a phase difference of 180o (or π rad). or Waves from S1 and S2 arriving at M have a path difference that is an odd integral of half wavelength. 2. Waves from S1 and S2 arriving at M have the same amplitude (or intensity). or Sources S1 and S2 have a ratio of amplitudes of 1.28.


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[Turn over

(ii) The speed of sound in air is 330 m s-1. The frequency of sound from S1 and S2 is increased. Determine the number of

minima that will be detected at M as the frequency is increased from 1.0 kHz to 4.0 kHz. [4] Path difference between waves from S1 and S2 is 28 cm. Wavelength of the sources changes from 33 cm to 8.25 cm.

For f = 1.0 kHz: n = 0.34 For f = 4.0 kHz: n = 2.9

Minima occurs when wavelength of the sources are (56 cm,) 18.7 cm, 11.2 cm, (8.0 cm, 6.22 cm …)

Since n can only be an integer, minima occurs when n = 1 & n = 2

Hence 2 minima observed. (iii) The variation with time of the displacement x of the sound waves arriving at M

from S1 and S2 are as shown in Fig. 7.2a and Fig. 7.2b respectively.

Fig. 7.2a

Fig. 7.2b

Determine the ratio of

intensity of sound when a maxima is detected at M

intensity of sound when a minima is detected at M

Wave from S2

Wave from S1


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Amplitude of resultant wave at minima is 0.6 units and at maxima is 3.4 units

2

2

Intensity amplitude

intensity when maxima is detected 3.432

intensity when minima is detected 0.6

[3] (c) Laser beam of red light of wavelength 644 nm is incident normally on a diffraction

grating having 550 lines per millimetre, as illustrated in Fig. 7.3.

Fig. 7.3

Red light of wavelength λ is also incident normally on the grating. The first order

diffracted light of both wavelengths is illustrated in Fig. 7.3. (i) Determine the total number of bright spots of wavelength 644 nm that are visible.

[3]

39

sin

10sin90 (644 10 )

550

2.8

d n

n

n

Highest order that can be seen is the 2nd order. Hence total number of bright lines observed is 5.

(ii) State and explain 1 whether λ is greater or smaller than 644 nm, [1] Since θ is greater, λ is also greater. 2 in which order of diffracted light there is the greatest separation of the two

wavelengths. [2] When n is larger, ∆θ is larger, thus the greatest separation occurs in the second order.


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(iii) The diffraction grating is now rotated 90⁰ about an axis parallel to the incident laser beam, as shown in Fig. 7.4.

Fig. 7.4

State what effect, if any, this rotation will have on the diffraction pattern that is

observed. [2] 0th order (or central) maxima remains in the same position. Diffraction pattern will rotate through 90o

(iv) In another experiment using the same apparatus, a student notices that the

angular separation between the zero order maxima and the two 1st order maxima are not equal.

Suggest a reason for this difference. [1]

Grating is not normal to incident light or Screen is not parallel to the grating

8 (a) ‘X-rays are used to investigate the atomic structure of solids.’ Deduce from this statement the wavelength of the X-rays used.

The wavelength is in the order of the separation of the atoms. 10-10m [1] (b) ‘Sometimes, for example, in the case of rubber, electrons with a de Broglie wavelength

of about 0.11 nm are used instead of X-rays.’ Determine the momentum of each electron.

34

19

6.63 x 10 . kg m s

0.11 10

hp

x

246 0 x 10

momentum = N s [2]


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(c) An X-ray tube operates with a potential difference of 100 kV between the anode and cathode. Fig. 8.1 is a sketch of the X-ray spectrum produced by this tube for a particular metal target. Fig. 8.2 shows a sketch of the energy level of target material and how the K line is formed. The tube voltage is 100 kV and the current is 20 mA.

(i) Calculate the maximum energy of an X-ray photon produced, explain your

working.

An accelerated electron loses all its kinetic energy in one single collision when hitting the target and the energy is converted to only one photon. This most energetic photon is thus obtained.

hfmax = ½ mv2 = eV = 1.60 x 10-19 x 100 x103 = 1.60 x 10-14 J

maximum energy = J [2]

(ii) With reference to Fig. 8.1, write down the numbers that represent the spectrum lines K and L .

K: L: [2] (iii) Sketch on Fig. 8.1 a spectrum for X-ray from the tube if the tube voltage is

reduced to 50 kV, the current at 10 mA. Label this spectrum A. [2] (d) Explain how the characteristic and continuous parts of the spectrum are formed. (i) Formation of characteristic parts of X-ray spectrum: [2]

K

K-shell

L-shell

M-shell

N-shell

Fig. 8.2

3 2

Line spectrum: When a high speed electron knocks out an orbiting electron in the inner shells of a target atom, a ‘hole’ is formed. When an electron from an outer shell of this atom fall into this ‘hole’, an X-ray photon of a particular frequency will be emitted.

Fig. 8.1

12

3 4 5

Intensity

Photon energy

0

0

A


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(ii) Formation of continuous parts of X-ray spectrum: [2]

(e) The energy required to remove an electron from the various shells of the nickel atom is: K shell 1.36 × 10-15 J L shell 0.16 × 10-15 J M shell 0.08 × 10-15 J

An X-ray tube with a nickel target emits the X-ray K radiation of nickel. Determine (i) the minimum potential difference across the tube, To have a K line formed, the incident electrons must have enough energy

to knock out an atom from K-shell to infinity. eVmin = 1.36 × 10-15 Vmin = 1.36 × 10-15/1.60 x 10-19 = 8490 V

potential difference = V [2] (ii) the energy of the X-ray quantum of longest wavelength in the K-spectrum of

nickel.

The least energetic line is given by K, when electron from L-shell falls to fill the hole in the K-shell.

h fmin = 1.36 × 10-15 - 0.16 × 10-15 = 1.20 x 10-15 J

energy = J [2]

According to classical physics, when the electrons are accelerating towards the target and when they are decelerating when hitting the target, electromagnetic waves will be emitted. In this case, the EM waves emitted is of high energy and is in the X-rays region. Because the acceleration and deceleration are continuous processes, the energy converted to X-rays can be any values. Hence continuous spectrum is formed.


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(f) A beam of electrons moving in the x-axis in an X-ray tube with momentum 4 × 10-23 kg m s-1 in the x-axis passes through a 3 mm slit in an anode before it hits the target as shown in Fig. 8.3. The uncertainty of the y-position of the electrons can be considered in the order of the size of the slit. Use uncertainty principle to estimate the possible angular spread of the electron beam after passing through the slit.

343

3 35

32 1

3210

23

p2

6.63 10p x 3 x 10 [1]

4

p x 3 x 10 5.27 x 10

p 1.8 x 10 kg m s [1]

pθ 1.8 x 10tan = 4.5 10

2 p 4 x 10

θ = 4.5

y

y

y

y

y

x

y

x

x

-10 -10x 10 x 2 = 9 x 10 rad [1]

= rad [3]

Fig. 8.3

Target Electron beam

anode

px

py

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MergedFile - score-in-chemistry.weebly.comscore-in-chemistry.weebly.com/uploads/4/8/7/1/48719755/nyjc_2014...NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 PHYSICS 9646/01