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MINISTRY OF EDUCATION, SINGAPOREin collaboration withUNIVERSITY OF CAMBRIDGE LOCAL EXAMINATIONS SYNDICATEGeneral Certificate of Education Advanced LevelHigher 1

READ THESE INSTRUCTIONS FIRST

Write in soft pencil.Do not use staples , paper clips, highlighters, glue or correction fluid.Write your name, Centre number and index number on the Answer Sheet in the spaces provided unlessthis has been done for you.DO NOT WRITE IN ANY BARCODES .

PHYSICSPaper 1 Multiple Choice

Additional Materials: Multiple Choice Answer Sheet

8866/01October/November 2013

1 hour

There are thirty questions on this paper. Answer all questions . For each question there are four possibleanswers 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 working should be done in this booklet.The use of an approved scientific calculator is expected, where appropriate.

This document consists of 12 printed pages.'!l''''()....P~f'

5MB ,......,

~ C.:Jl) Singapore Examinations and Assessment Board.~~

g:,:, UNIVERSITY of CAMBRIDGE1I\'~~. 'I! 1

::: International Examinations

© UCLES & MOE 2013 [Turn over

Data

speed of light in free space,

elementary charge,

the Planck constant,

unified atomic mass constant,

rest mass of electron,

rest mass of proton,

acceleration of free fall.

Formulae

uniformly accelerated motion,

work done on/by a gas ,

hydrostatic pressure,

resistors in series,

resistors in parallel.

© UCLES & MOE 2013

2

e =1.60 x 1O-19C

h =6.63 X 10-34 J s

u =1.66 X 10-27 kg

m e =9.11 x 10-31 kg

9 =9.81 rns"

s =ut+ .1at 22

~ = if + 2as

w=pf..V

p = pgh

1 / R =1/ R1 + 1 / R2 + ...

8866/01/0/N/1 3

3

1 A car is trave lling at its cruising speed on an expressway. It is then brought to rest.

Which value is the best estimate of the car's change of momentum?

A 3 x 103kgms-1

B 3 x 104 kg m S -1

C 3 x 105kgms- 1

D 3 x 106kgms-1

2 To find the resistivity of a semiconductor, a student makes the following measurements of acylindrica l rod of the material.

length = 25 ± 1 mm

diameter =5.0 ± 0.1 mm

resistance =68 ± 1n

He calculates the resistiv ity to be 5.34 x 10-2 n m.

How shou ld the uncertainty be included in his statement of the resistivity?

A (5.34 ± 0.07) x 10-2 n m

B (5.34 ± 0.09) x 10-2 n m

C (5.3 ± 0.4) x 1Q-2 n m

D (5.3 ± 0.5) x 1Q- 2 n m

3 Which pair of physical quantities has the same base units?

A force and momentum

B moment of a force and work

C energy and power

D density and pressure

4 An object is released from the open door of an aircraft in level flight.

It is observed that it takes three seconds for the object to reach termina l velocity.

Which statement about the motion of the object is correct?

A The horizontal component of its velocity is constant.

B The horizontal component of its acceleration is zero.

C The vertical component of its velocity decreases for three seconds.

D The vertical component of its acceleration is zero after three seconds.

© UCLES &MOE 2013 11-'~~

8866/01/0/N/13 [Turn over

S7625274B

Highlight

4

5 A ball is released from rest above a hard, horizontal surface. The graph shows how the velocity ofthe bouncing ball varies with time.

At which point on the graph does the ball reach its maximum height after the first bounce?

velocity

Ac

D

B

6 Which quantity has the unit of N?

A moment

B power

C rate of change of momentum

D work done

7 A large bucket, lifted by a rope attached to a crane, is used on a building site to raise heavyloads .

The bucket, of total mass m when fully loaded , rises at a uniform speed v before deceleratinguniformly to rest in time t.

What is the difference of tension in the rope supporting the bucket between the time when thebucket is moving at uniform speed and the time when the bucket is decelerating?

A - mgt

B - mvt

C m(g - Y)t

D m(Y - g)t

8 A football of mass 0.42 kg is travelling towards a player at 3.0 m S-1 .

The player kicks the ball with an impulse of 6.3 Ns, returning it in the direction of approach.

What is the new speed of the ball?

A 4.6ms- 1

© UCLES & MOE 2013

B 6.2ms-1 C 12ms-1

8866/01/0/N/13

D 18ms-1

5

9 In which process do viscous forces create a significant resistance to motion?

A the application of paint to the surface of a wall

B the compression of air while pumping up a car tyre

C the fall of water drops from a faulty tap

D the spread ing of petrol on a wet road surface

10 Two forces P (horizontal) and Q (vertica l) act on a body. The body is held in equilibrium by a thirdforce R.

Which diagram represents P, Q and R in vectorial form?

Q

A

P

Q

B

P

Q

c

P

Q

D

P

11 A uniform plank, of we ight Wand length L is supported at points X and Y, each at distances !::.4

from the ends of the plank.

L L L L4 4 4 4.. ... ~ .. ~ .. ~

fi 1~ fi~

X YL L8 W 8

What will be the increase of the force on the plank exerted by support X if both X and Yare

moved a distance!::. to the right from their original posit ions?8

A W16

B W8

c W4

D W(...§... )16

© UCLES & MOE 2013 8866/01/0/N/13 [Turn over

6

12 Charged particles experience forces in an electr ic field. Current-carrying wires can experience aforce in a magnetic field.

Sketches P and Q show the forces F acting on charged partic les in an electric field.Sketches Rand S show the forces F acting on current-carry ing wires in a magnetic field.

P

rF

Q

F

1 F

R S

F

electric fields on charges magnetic fields on currents

Which row in the table correctly identifies the charged particles and direction of the currents in thewires?

charge on P charge on Q current in R current in S

A positive negative into the page out of the page

B positive negative out of the page into the page

C negative positive into the page out of the page

0 negative positive out of the page into the page

13 A mode l car is released from rest at a height h on a frict ionless track.

h model car

/

For the car to go around the loop of radius r without leaving the track, it must be trave lling at a

speed of at least .fir at point Y.

What is the minimum possible value of the height h required for the car to remain on the trackwhile going around the loop?

A 2.5 r

© UCLES &MOE 2013

B 2.75r C 3.0r

8866/01/0/N/13

o 4.0r

7

14 A brick of weight 20 N is dropped from a height of 2 m directly onto a wooden peg. The brick losesall of its energy to the peg, which is knocked 4 ern into the ground.

brick weight20 N

2m

peg

What is the approximate average friction force between the peg and the ground?

A 10N B 100N C 1000N D 10000N

15 A small electric motor is 20% efficient. Its input power is 9.6 W when it is lifting a mass of 0.50 kgat a steady speed v.

motor9.6W

What is the value of v?

bench 0.50kg

A 0.39ms-1 B 2.0ms-1 C 2.8ms-1 D 3.0ms-1

16 The loudness of a constant-frequency sound wave increases so that its intensity I is doubled .

By how much does the original amplitude a increase?

A a

© UCLES & MOE 2013

B J2a

tt~Ml~

C (J2 -1)a

8866/01/0 /N/13

D 3a

[Turn over

8

17 The displacement-d istance and displacement-time graphs are for a water wave produced in aripple tank.

diSP,aCementlc:b (\ .

o 1.0 2.0 3.0I V distance / cm

diSP,acement/cmob (\ .

o 0.05 0.10 0.15I V time/ s

What is the speed of the water wave?

A 0.1 ms" B 0.2ms-1 C 10ms-1 o 20m s-1

18 The diagram shows the displacement-time graphs of two pure sound waves P and Q at a point inspace. The graphs have the same scales for the time axes.

displacement wave P

time /msO+-t--;----'t--i-+--t--+--+-~--~

o

displacement wave Q

time /msO++--+-+-+-+--j--t--t---t-H---~

o

The frequency of Q is 125 Hz. The waves are in phase at time t =O.

At what time are the waves next in phase?

A 32ms B 36ms C 64ms o 72ms

19 Which pair of sources is coherent?

A two identica l light sources

B two loudspeakers emitting sounds of frequencies (and 2(

C two ripple tank dippers oscillat ing at identical frequencies in antiphase

o two tuning forks, each producing a single frequency but having a constant frequencydifference

© UCLES &MOE 2013 8866/01/0/N/13

9

20 The equation A = ax for double-slit interference is an approximation which can be used toodetermine the wavelength of light.

Which of the following conditions is necessary?

A a is equal to x.

B a is much greater than x.

C 0 is much greater than a.

D A is much greater than x.

21 A 1.0 kn resistor has a maximum power rat ing of 0.25 W.

What is the maximum rate of flow of electrons that can pass through this resistor?

A 1.0 x1013 S- 1

B 1.0 x1015 S- 1

C 1.0 x1017 S- 1

D 1.0 x1 019 S - 1

22 An electron travels around the circuit shown in the diagram. The cell has negligible internalres istance.

At which point in the circuit does the electron have its maximum electrical potential energy?

A B

D C

© UCLES & MOE 2013 8866/0 1/0 /N/13 [Turn over

10

23 The graph shows the 1-V characteristics of three electrical components, a diode, a filament lampand a resistor, plotted on the same axes.

2.0 V/V1.0o-F-- - "-l- - - -+--

o

I1A

0.2~-----Y'1----Ji'-

Which statement is correct?

A The resistance of the diode equals that of the filament lamp at about 1.2V.

B The resistance of the diode is constant above 0.8 V.

C The resistance of the filament lamp is twice that of the resistor at 1.0 V.

D The resistance of the resistor equals that of the filament lamp when V = 0.8 V.

24 The resistance of a piece of wire, length 1 m and diameter 0.3 mm, is R.

Another piece of wire, made of the same metal , is 2 m longer than the first wire . The diameter is50 % of that of the first wire .

What is the resistance of the second piece of wire?

A 4R B 6R C 8R D 12R

25 A conductor PQ has a potential difference of 12 V between its ends. There is a current of 3 A inPQ.

+ 12V ('-"-) ---.J) OV

P Q

Which statement is correct?

A The charge flowing each second in the conductor is 12 C.

B Electrons flow from P to Q .

C The power dissipated in the conductor is 36 W.

D The resistance of the conductor is 3o.

© UCLES & MOE 2013 8866/01/0 /N/13

11

26 Four magnetic fields are shown.

1 2 3

@ 0 ,,-A~(

0 /y'\

4

Which is the correct order for objects to match these fields?

1 2 3 4

A flat circu lar coil bar magnet solenoid long straight wire

B flat circular coil solenoid bar mag net long straight wire

C long stra ight wire flat circu lar coil bar magnet solenoid

0 long straight wire solenoid bar magnet flat circu lar coil

27 Three para llel wires P, Q and R are fixed perpendicu lar to the paper at three corners of a square.

The current in each wire has the same magnitude. The curre nt in P is into the paper and in Q andR is out of the paper.

In which direction is the force on wire R?

© UCLES &MOE 2013 8866/01/0 /N/13 [Turn over

12

28 An electromagnetic radiation of constant frequency is incident on a metal surface.

Wh ich statement exp lains why the photoe lectr ic current from the meta l surface is proportional tothe intensity of the incident electromagnetic radiation?

A Radiation of greater intens ity causes the meta l surface to get warm and so emit moreelectrons.

B Radiation of greater intensity consists of more energetic photons.

C Radiation of greater intensity means more photons per second strike the metal surface.

D Radiation of greater intensity overcomes the metal's work function energy allowing moreelectrons to escape.

29 An electron in an atom makes a transition from an energy level of energy E1 to a level of energyE2 , emitting a photon of wavelength A.

Which expression gives this wavelength in terms of the Planck constant h and the speed oflight e?

A B C D he

30 Electromagnetic radiation of frequency f and intensity I is directed onto a meta l electrode Xcausing photoelectrons to be emitted . Some of these electrons reach electrode Y causing acurrent in the circu it.

electromagneticradiation

vacuumX

electrons tV

!A reverse voltage Vmin is applied so that the electrons are just prevented from reaching Y.

Which graph represents the variation of Vmin with intensity I when f is constant?AB C D

I

V .min

I

V min

I

V .min

I

Vmin

Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Everyreasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwillingly been included, thepublisher will be pleased to make amends at the earliest possible opportunity.

Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge LocalExaminations Syndicate (UCLES), which is itself a department of the University of Cambridge.

© UCL ES & MOE 2013 8866/01/0/N/13

II

MINISTRY OF EDUCATION, SINGAPOREin collaboration withUNIVERSITY OF CAMBRIDGE LOCAL EXAMINATIONS SYNDICATEGeneral Certificate of Education Advanced LevelHigher 1

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w- PHYSICSNN Paper 2 Structured QuestionsI-'

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\0_ Candidates answer on the Question Paper.ro_w No Additional Materials are required ..-J~

INDEXNUMBER ITIIJ

8866/02OctoberlNovember 2013

2 hours

READ THESE INSTRUCTIONS FIRST

Write your Centre number, index number and name 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.DO NOT WRITE IN ANY BARCODES.

The use of an approved scientific calculator is expected, where appropriate.

Section AAnswer all questions.

Section 8Answer 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 partquestion.

For Examiner's Use

Section A .>1

2

3

4

5

6

7

Section 8 .>8

9

10

Total

This document consists of 22 printed pages and 2 blank pages.~,+OAPO,,~

SMII~

~ 0l\Singapore Examinations and Assessment Board.~~

~ UNIVERSITY of CAMBRIDGE~ International Examinations

© UCLES & MOE 2013

L ~..",

DC (NF/JG) 61576/6 [Turn over

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• 3740107702 • CI

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Data C~

ac =3.00 X 108 rn s"

C ~speed of light in free space, u'C...

C~<elementary charge, e =1.60 x 10- 19 C C~

C~

the Planck constant, h =6.63 x 10- 34 J s C~CC

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

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

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

acceleration of free fall, 9 =9.81 ms- 2 CCCC

Formulae CC

uniformly accelerated motion, 5 =ut+ ~at2 CC

v 2 = u2 + 2as CC

work done on/by a gas, W = pl1V CC

hydrostatic pressure, p= pgh CCC

resistors in series, R = R1 + R2 + ... CCC

resistors in parallel, 1/R =1/R1 + 1/R2 + . . . CCC IC IC IC IC IC IC ICICCCCCCCCCCCCCCCCCC

~UCLES & MOE 2013 -.JC

8866/02/0/N/13 CCCC

Draw on the diagram to show the resultant magnetic flux density (C - D) when the twovectors are subtracted.

Complete the diagram to show the resultant magnetic flux density (A + B) when the twovectors are added.

II

ForExaminer's

Use

[2]

[2]

3

Fig. 1.1

Sect ion A

8866/02/0/N/~3nit = .. ....··.. .. ·· .. .. .. .. ·.. ·..........·.. ··· ·· .. ·[Tu[:~ ov~

Fig. 1.2

n

Answer all the questions in this section.

(b) Fig. 1.2 shows a similar situation for magnetic flux density vectors C and D.

(c) State the unit for magnetic flux density.

1 (a) Fig. 1.1 shows two magnetic flux density vectors A and B, drawn to scale.

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9 = ms-2 [2]

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4

8866/02l0/N/13

Give one deduction that can be made from this different value by considering why 9varies.

In one such measurement at a particular place in Singapore , a small marker falls a distanceof 0.72056 m from rest in a vacuum. The time it takes is 0.38393s.

(a) Calculate the value of 9 at that place in Singapore. Give your answer to a suitablenumber of significant figures .

(b) When the measurement was made at a different place in Singapore, an accurate valueof 9.7732 m S-2 was obtained for g.

2 Geologists are able to determine the likelihood of finding oil-bearing rocks beneath thesurface of the Earth without drilling . They make very accurate measurements of g, theacceleration of free fall, at the surface.

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~UCLES & MOE 2013

speed after collision = m S-1 [1 ]

speed before collision = m S-1 [4]

3 In a head-on elastic collision between an alpha particle (mass 4 a.m.u.) and a stationaryoxygen nucleus (mass 16 a.m.u.), the oxygen nucleus moved off with a velocity of3.3 x 105 m S-1 after the collision.

Calculate

II

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8866/02/0/N/13

(a) the speed of the alpha particle before the collision,

(b) the speed of the alpha particle after the collision .

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Give an example of such an object, and explain how it illustrates these statements.

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

energy = J [2]

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7

8866/02l0/N/13

4.00-+-- - -1--- - - - - - - - - - - -

o

~~~

Fig. 4.1

forceIN 6.0

In the case shown, the straight line portion of the graph shows elastic deformation. Thecurved part of the graph shows permanent deformation (plastic deformation) .

(a) Using data from Fig. 4.1, calculate the elastic potential energy stored when the appliedforce is 6.0 N.

(b) Plastic deformation is important during the manufacture of a metallic object.In the daily use of the same object, only elastic deformation is required.

4 A typical force-extension graph for metals has the shape shown in Fig. 4.1 .

* 3740107707 *

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5 A battery of e.m.f. 15.0V has an internal resistance of 6.0Q. It is connected to a variableexternal resistance R, as shown in the circuit diagram Fig. 5.1 .

R total resistance current p.d. across R power to R efficiencyIQ IQ IA IV IW 1%

0 6.0 2.50 0 0 0

2.0 8.0 1.88 3.8 7.1 25

4.0 10.0

6.0 12.0 1.25 7.5 9.4 50

10 16

14 20

24 30 0.50 12 6.0 80

94 100 0.15 14 2.1 94

(a) Insert values in the spaces to complete the table.

The following table gives some of the numerical details for different numerica l values of R.

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[4]

8

8866/02l0/N/13

R tf/

Fig. 5.1

/

15.0V 6.0Q

.....----------!--; ~ ~T -c=- - - - J--+- - - -i~I 1 I~ J

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1 .

2 .

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

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[2]

25 R/ Q2015

9

105

Fig. 5.2

oJmmEm m mEm EEm3IOOmEElilfJi§o

2-H--t-HH-++l-H-t-l-+1-H+++-H-++-t-HH-++l-H-t-l-+1-H++-t-HH-++l-H-t+l-l-+--1

10 -++-t-HH-++l-H-t-l-+1-H+++-H-++-t-HH-++l-H-t-l-+1-H+++-H-++-t-HH-++I-!-+--1

Suggest two consequences of using this battery only with high value external resistors.

a.a•••power to R

/ W 6 t-t-t-l-+H-t+l-I+++ -I- I-++-H - J-+l-H-l -l-+1-+-1-+ + -H--IH + +l-H-t-l-+H -t++H-t-t-H

4 _

(c) Explain what is meant by 'eff iciency' as used in the last column of the table.

(d) The battery has efficiency that increases as the external resistance increases.

(b) Complete the graph started in Fig. 5.2.

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(b) Explain why the maximum energy of photoelectrons is independent of intensitywhereas the photoelectric current is proportional to intensity.

6 (a) Explain the phenomenon of photoelectric emission by referring to photon energy andwork function energy.

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8866/02l0/N/13

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(b) One line in the Sun's absorption spectrum has a wavelength of 760 nm.

energy gap = eV [3]

Calculate the energy gap between the discrete electron energy levels that leads to thisspectral line. Give your answer in electron volts.

ForExaminer's

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11

8866/02/0/N/13~dfJ~

(a) Briefly describe an experiment to observe a line emission spectrum. You may use adiagram to illustrate your answer.

7

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2. in kWh.

1. in joules,

Calculate the length of the filament.

(ii) its resistance ,

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length = m [3]

12

Section B

8866/02/0/N/13

energy = J [1]

Answer two of the questions in this section.

resistance = Q [1]

current = A [1]

energy = kW h [2]

(i) the current through it,

Calculate, for the lamp when operating with 240V across it,

(iii) the energy supplied to it in 8 hours

(b) The filament of the lamp in (a) has a radius of 0.0043 mm and is made of a material ofresistivity 9.2 x 10-7 Q m.

8 (a) A lamp is marked 240V, 60W.

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(c) Fig. 8.1 illustrates the connection between the copper leads to the filament and thefilament itself.

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Describe, giving reasons where appropriate, and without using numerical values, thefollowing features of the current in these three components.

(i) the direction of electron flow

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(ii) the rate of electron flow

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(iii) the average speed of electrons

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(d) Use your answers to (c) to explain why the filament of the lamp gets hot but the copperleads stay relatively cold.

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

(e) In practice, the leads to a lamp have some resistance.

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Explain why this warning is necessary.

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

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Maximum current fully unwound 10A (2400 W)"

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8866/02/0/N/13

(f) Describe a situation in your home which indicates that the interna l resistance of themains electrical supply is very low.

(g) Most electrical extension leads are sold wound on a reel. A label on these reels willhave a warning to the user that may read

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(a) (i) Calculate the energy used by the car in travelling 100km at this speed.

(ii) Show that the total power being supplied to the car by burning the fuel is 80 kW.

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(i) determine the efficiency of the car,

Using data from the diagram ,

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9 Diesel fuel , when used in a car engine , provides 3.7 x 107 joules of energy per litre of fuel.The car manufacturer's specification for a particular car, travelling at 100km/ hour on a flatroad, gives the rate of fuel consumption as 7.8 litres per 100 km travelled.

(b) Fig. 9.1 is a simplified diagram, sometimes called a Sankey diagram , showing how theinput power used by this car on a flat road is distributed as various outputs. It is drawn toscale.

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20

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current = A [3]

(ii ) calculate the power output of the car,

(iii) calculate the drivin g force,

percentage increase = % [4]

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17

8866/02/0/N/13

Calculate the percentage increase required in power output for the car.The speed of thecar is assumed to be unchanged.

power = kW [1]

driving force = N [4]

(iv) calculate the electric current supplied by the car's 15V electrical generator.

(c) When the car is driving up a hill with a gradient of 1 metre rise for every 25 metres alongthe road, extra power will be required. The mass of the car is 900 kg.

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Include in your answer an estimate of the fraction of the power wasted as noise. (Areally loud sound from a loudspeaker might have a power input of 20W.)

(d) Discuss, referring to the law of conservation of energy, ways in which the power outputof the car is dissipated to the surroundings of the car when travelling at a constantspeed on a flat road.

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(ii) refraction of a plane wavefront at a plane surface,

10 (a) Complete the following diagrams to show

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(i) reflection of a circular wavefront at a plane surface,

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(iv) how an interference pattern is produced when a plane wavefront meets a pair ofnarrow slits .

(iii) diffraction of a circular wavefront at a narrow gap,

(b) On Fig. 10.4 indicate where maxima and minima of intensity occur.

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(c) (i) State what is meant by the term coherence of two waves.

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separation = m [3]

8866/02/0/N/13

(ii) Explain how the coherence condition for interference is obtained in (a)(iv) .

(d) Light of wavelength 5.35 x 10- 7 m is shone on a double slit whose separation is0.200 mm. Calculate the separation of the fringes on a screen placed 0.875 m from thedouble slit. The screen and double slit are both placed at right-angles to the beam oflight.

• 3740107722 •

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(ii) Deduce the intensity at a distance of 12.0m from the source.

The intensity of the wave at a distance of 2.0m from its source is 3.0Wm-2.

intensity = W m-2 [3]

II

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8 distance/ m6

intensity = W m-2 [2]

23

4

8866/02/0/N/13

Fig. 10 .5

2

(i) Using data from the graph, calculate the intensity at a distance of 4.6 m from thesource.

(e) Fig. 10.5 shows how the amplitude of a wave is inversely proportional to the distancefrom its source.

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University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University ofCambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambr idge.

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© Physics Dept Page 1 of 10

JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT

GCE A levels 2013 H1 Physics (8866) Paper 2 solutions

General Comments: • In ‘show that’ questions, the full working and clear substitution of all values must be given. A question requiring an

explanation will not gain credit if only a basic statement is given. • Marks were lost for lack of precision of the answers, particularly to definitions, laws or determining information from

diagrams such as graphs.

• Do not spend time writing out part or the entire question before giving any worthwhile answer. The credit available and the space provided for each section gives a guideline to the length and detail required in the answer.

Qn Solution Marks

1(a)

[1] Magnitude [1] Direction

(b)

[1] Magnitude [1] Direction

Common errors : (a),(b) Working lines were left out giving little evidence to justify the correct magnitude and direction of required vector.

(c) tesla [1] Ans

Common errors : Many stated T as the unit. Though T is a legitimate symbol for tesla, it should be spelt in full for this type of question.

A + B

−D

C−D





2(a) s ut at= + 21

2

. g( . )= + 210 72056 0 0 383932

g . −= 29 7768 m s

[1] Sub [1] Ans

Common errors : Insufficient number of significant figures was provided. The question already hinted that more s.f. is required.

(b) Comparing with the new value of g . −= 29 7732 m s

It is deduced that at the new location g is lower. Reasons : - there could be oil-bearing rocks (less dense than the normal rocks) beneath the new location. - measurements were taken at a higher altitude.

[1] Lower

[1] One reason

Common errors : Many mentioned air-resistance but it was clearly stated in the question that the marker falls in a vacuum.

3(a) Using relative speed of approach = relative speed of separation (Elastic collision) v v u u− = −2 1 1 2

. v uα α× − = −53 3 10 0 Equation 1

By conservation of momentum

u(u ) u(v ) u( . )α α+ = + × 54 0 4 16 3 3 10 where u is the a.m.u

u v ( . )α α= + × 54 3 3 10 Equation 2

Solving Equation 1 and 2 simultaneously

u . −α = × 5 18 25 10 m s

[1] Eqn. 1

[1] Eqn. 2 [1] Working [1] Ans

Common errors : Wrong assumption that alpha particle stopped moving after collision.

(b) Sub u . −α = × 5 18 25 10 m s and solve

v . −α = − × 5 14 95 10 m s

Speed is a scalar quantity therefore is . −× 5 14 95 10 m s

[1] Ans

4 (a) EPE kx= =21 Area under graph (linear portion)

2

( )( )−= × 31 4 10 62

. −= × 21 2 10 J

[1] Sub [1] Ans

Common errors : Forgetting that extension is in mm





(b) In the manufacturing of a steel cooking pan for example, force is applied beyond the elastic limit so that the pan can be permanently in that desired shape. In daily usage any force applied should be within the elastic limit so that the pan can maintain its original shape with no permanent deformation.

[1] Relevant example [1] explanation for manufacturing [1] explanation for daily usage

Common errors : • Not naming a metallic object • Wrongly naming plastic chair as an example

5(a)

[4] 1 mark for every three correct answers

(b)

[1] Use of correct plots

[1] Smooth curve

1.50

0.94

0.75 11

9.4

8.3

8.8

70

63

40 9.0 6.0





(c) Efficiency refers to the percentage of power delivered to the external resistance by the battery.

It can be expressed mathematically as R %×

Power dissipated by 100Power generated by battery

[1] worded explanation [1] expression for efficiency

(d) 1. For high value of R, a small current, I, runs through the circuit. Less power is lost as heat in the internal resistance. Too much heat dissipation may damage the battery. 2. The battery can serve a longer time for an appliance that has a large external resistance since only a small current is drawn each time.

[1] less loss in power

[1] longer battery life

6(a) Photoelectric emission refers to the emission of photoelectrons from a metal surface when an incident radiation of high enough frequency is shone on the surface. The incident radiation of high enough frequency f refers to photons where each photon has an energy of E = hf. This is photon energy. Each photon needs to have a minimum energy known as work function energy before a photoelectron is released from the metal surface. This energy is dependent on the type of metal surface.

[1] general statement

[1] address photon energy [1] address work function energy

Common errors : Students did not distinguish between electrons and photons.

(b) Here intensity refers to the rate of incidence of photons on the metal surface. The maximum energy of photoelectrons on the other hand is referring to the maximum kinetic energy (K.Emax) of the photoelectrons that are released from the metal surface. This K.Emax is only dependent on the frequency, f, of the incident photon and can be related by the equation K.Emax = hf – work function energy. K.Emax not dependent on the rate of incidence of photons. Higher intensity implies higher rate of incidence of these photons which increases the chance of release of photoelectrons proportionally. Hence the photocurrent also increases proportionally.

[1] [1] [1]

7(a) An emission line spectrum of an element consists of a series of separate bright lines of definite frequencies (or wavelengths) on a dark background. It is produced when a stream of photons of different frequencies is passed through a narrow slit and normally through a diffraction grating. These photons are emitted randomly from transitions (from higher to lower energy levels) in the excited atoms of the element in a vapour or gas at low pressure (named hot gas, not necessarily hot).

[1] sub [1] ans

Common errors : Confused with absorption spectrum





(b) GAP

hcE =λ

. ( )−

−

× ×=

×

34 8

9

6 63 10 3 10 J760 10

. ( ) .( . )

−

− −

× ×= =

× ×

34 8

9 19

6 63 10 3 10 1 64 eV760 10 1 6 10

[1] sub to get EGAP in J [1] correct conversion [1] Ans

8(a)(i) P IV=

I .= =60 0 25 A

240

[1] Ans

(ii) V IR=

R.

= = Ω240 960 0 25

[1] Ans

(iii) 1. E Pt=

( ) .= × × = × 660 8 60 60 1 73 10 J

[1] Ans

2. Energy supplied at 60 W for 8 hours

.−= × × =360 10 8 0 48 kWh

[1] Conversion [1] Ans

(b) lRAρ

=

( . )l( . )

−

−

×=π ×

7

3 2

9 2 109600 0043 10

l . −= × 26 06 10 m

[1] Expression [1] Sub [1] Ans

(c)(i) Electrons flow from negative (lower electrical potential) to positive (higher electrical potential) through the filament.

[1]

(ii) Rate of electron flow refers to the conventional current which would flow from positive to negative through the filament. Therefore the rate of electron flow is constant throughout the three components.

[1]

(iii) Since the same current (rate of flow of electrons) flowing in the copper wires have to flow through the very fine filament, the electrons need to speed up at the filament to maintain the constant rate of flow. Therefore the average speed of electrons increases while flowing through the filament.

[1] [1]

Common errors : Many students think that the electrons slowed down in the filament. Some even said electrons travelled at the speed of light.

(d) Since the average speed of electron is higher in the filament, the transfer of kinetic energy to the filament atoms is bigger per collision per second.

[1] [1]

(e) The resistance in the leads will cause the current flowing through the filament to be smaller. Less power (I2Rfilament) is dissipated in the filament and therefore the lamp gets dimmer.

[1]





(f) When there is a spike in the number of appliances used, the current drawn would be large. This would cause a drop in electrical potential across the internal resistance. If the internal resistance of the mains is high, the drop in potential becomes substantial. This may cause some appliance to stop functioning e.g. lightings dim due to insufficient power. But this is not the case, suggesting that the internal resistance is very low.

[1] large potential drop [1] insufficient power

(g) If a power of 2400 W is dissipated while the leads are fully wound, the heat dissipated will be high enough to melt the insulating material covering the leads causing short-circuit. If the same power is dissipated while the leads are fully unwound, the heat could be dissipated fast enough into the surrounding air.

[1] insulation melt [1] short-circuit [1] heat escape fast into air

9 (a)(i) 1 litre of fuel provides . × 73 7 10 J

7.8 litres of fuel provides . . .× × = ×7 87 8 3 7 10 2 89 10 J

[1] working [1] ans

(ii) In 1 hour, for the conversion of . × 82 89 10 J

. .×= = ×

842 89 10Power required 8 03 10 W 80 kW

3600

[1] working

(b)(i) From the graph, looking at power output percentage, efficiency = 32% [1] ans

(ii) PEfficiency %= ×Power output of car, 100Power supplied to car

P..

=× 40 32

8 03 10

P . .= × =42 57 10 W 25 7 kW

[1] ans

(iii) Driving force, F, is produced by the engine to enable the car to cruise at a uniform speed of 100 km /hour on a flat road. P Fv=

.F×

×= 3

4

100 103600

2 57 10

.= × 29 25 10 N

[1] consideration of parameters [1] conversion of 100 km/hr [1] sub [1] ans

(iv) From graph, power used by electrical system = 4% 32% correspond to 2.57 x 104 W

4% correspond to . .×

× = ×4

32 57 10 4 3 21 10 W32

Using P IV=

.I ×=

33 21 1015

.= × 22 14 10 A

[1] electrical power value [1] sub [1] ans





(c) Based on the car speed (100 km /hour), in 1 second, the car travels 27.78 m up-hill, along the road.

the vertical height gained . .= × =1 27 78 1 11 m

25

the gain in GPE mgh=

. . .= × × = × 3900 9 81 1 11 9 80 10 J

Increase in power .= × 39 80 10 W

Percentage increase in power . % . %.

×= × =

×

3

4

9 80 10 100 38 12 57 10

OR Increase in force required to drive up the hill,

aF mg sin ( . )( )= θ = =1900 9 81 353 N

25

Percentage increase in force % . %.

= × =× 2

353 100 38 29 25 10

Since P Fv= and speed is unchanged, power has the same percentage increase of 38.2%

[1] vertical height [1] GPE gained [1] sub [1] ans [2] Fa [1] %increase in F [1] %increase in P

(d) Some of the power required to help maintain the kinetic energy of the car is dissipated at the wheels while overcoming friction. Some of the power output of the car is dissipated while overcoming air-resistance. Power provided to the car to keep it at constant speed on a flat road is typically in the order of 104. So even if the car produces a very loud noise (20 W), it is only one-thousandth of the power used to maintain the kinetic energy of the car. This is very small power dissipation. Alternatively When travelling at a constant speed on a flat road, the car moves through air causing the surrounding air to move. Some of the power output is dissipated as heat and sound. Also, the car tyres flex when parts of the tyre make and break contact with the road. Some output power is also dissipated as heat and sound due to this. Thirdly, vibration of the car parts also dissipates heat and sound. Assuming the power dissipated as sound is 20 W, fraction of power wasted as

noise is 20 1

25700 1300

[1] friction [1] air-resistance [1] fraction due to noise [1] comment [1] moves air [1] tyre flex [1] car parts moving [1] fraction

Common errors : Many students mentioned heat losses from exhaust gases, power losses to electrical system and power wasted in transmission system. The question is looking at the specific output used to maintain constant speed on a level ground and possible ways of dissipation to the surrounding.





10(a)(i)

[2]

(ii)

[2]





(iii)

[2]

(iv)

[2]

Common errors : For (i) and (ii), many students were confused between reflection and refraction.

(b) See Fig. 10.4 [1] maximum intensity [1] minimum intensity

(c)(i) It means that the two waves have a constant phase difference when they superpose at a certain position.

[1]

(ii) When a wave front meets the pair of slits, the wave front emerges from the slits as divergent wave fronts which are in phase with each other, i.e. the two slits act as two coherent sources of divergent waves which are in phase. Hence the coherence condition for interference is obtained.

[1]

minima

maxima





(d) Using

axD

λ =

. x..

−− ×

× =3

7 0 2 105 35 100 875

x . −= × 32 34 10 m

[1] equation [1] sub [1] ans

(e)(i) Given at a distance d = 2.0 m , intensity I = 3.0 W m-2

From the graph, the amplitude A = 0.86 x 10-3 m At d = 4.6 m , A = 0.37 x 10-3 m

Using I kA= 2

k( . )−= × 3 23 0 86 10 ----- (1)

I k( . )−= × 3 20 37 10 ----- (2)

I . −= 20 56 W m

[1] reading A from the graph [1] sub [1] ans

(ii) Using I

d∝ 2

1

kId

= 2

k( )

= 232

k =12 When d = 12.0 m

I . − −= = × 2 22

12 8 33 10 W m12

[1] sub [1] ans

ri ' i i' university of cambridge local examinations...

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