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MINISTRY OF EDUCATION, SINGAPORE in collaboration with UNIVERSITY OF CAMBRIDGE LOCAL EXAMINATIONS SYNDICATE General Certificate of Education Advanced Level Higher 1
PHYSICS
8866/01
Paper 1 Multiple Choice
October/November 2013 1 hour
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, Centre number and index number on the Answer Sheet in the spaces provided unless this has been done for you. DO NOT WRITE IN ANY BARCODES . There are thirty 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 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 ,......,
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© UCLES & MOE 2013
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UNIVERSITY of CAMBRIDGE
International Examinations [Turn over
2 Data speed of light in free space, elementary charge ,
e
= 1.60 x
1O- 19C
the Planck constant,
h
= 6.63 X
10- 34 J s
unified atomic mass constant,
u
= 1.66 X
10- 27 kg
rest mass of electron,
me
= 9.11 x 10- 31 kg
rest mass of proton, acceleration of free fall.
9
= 9.81 rns"
Formulae uniformly accelerated motion,
s
~
2 = ut+ .1at 2 = if + 2as
work done on/by a gas ,
w = pf..V
hydrostatic pressure,
p = pgh
resistors in series, resistors in parallel.
© UCLES & MOE 2013
1/R
= 1 / R 1 + 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?
2
A
3 x 103kgms- 1
B
3 x 104 kg m S -1
C
3 x 105kgms- 1
D
3 x 106kgms - 1
To find the resistivity of a sem iconductor, a student makes the following measurements of a cylindrica l rod of the material. length = 25 ± 1 mm
=5.0 ± 0.1 mm resistance = 68 ± 1 n
diameter
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?
3
4
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
Which pair of physica l 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
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
4 5
A ball is released from rest above a hard, horizontal surface. The graph shows how the velocity of the bouncing ball varies with time. At which point on the graph does the ball reach its maximum height after the first bounce? velocity D
A
c B
6
7
Which quantity has the unit of N? A
moment
B
power
C
rate of change of momentum
D
work done
A large bucket, lifted by a rope attached to a crane, is used on a building site to raise heavy loads . The bucket, of total mass m when fully loaded , rises at a uniform speed uniformly to rest in time t.
v before decelerating
What is the difference of tension in the rope supporting the bucket between the time when the bucket is moving at uniform speed and the time when the bucket is decelerating?
A
8
- mg t
B
- mv t
C
m(g - Y)
t
D
m(Y - g) t
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 N s, 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 fau lty 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 equ ilibrium by a third force R. Wh ich diagram represents P, Q and R in vectorial form? A
B
P
P
Q
c
D
Q
Q
Q
P
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
..
4
~ X
L
4
L
8
~
..
fi 1
L
4
L ~
4
..
~ Y
W
L
~
fi
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
W 16
© UCLES & MOE 2013
B
W 8
c
W 4
8866/01/0/N/13
D
W(...§... ) 16
[Turn over
6 12 Charged particles experience forces in an electric field. Current-carrying wires can experience a force in a magnetic field.
Sketc hes 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.
S
R
Q
P
F
1
r
F
F
F
magnetic fields on currents
electric fields on charges
Which row in the tab le correctly identifies the charged particles and direction of the currents in the wires? 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.
model car
h
/ For the car to go around the loop of radius speed of at least
.fir
r without
leaving the track, it must be trave lling at a
at point Y.
What is the minimum possible value of the height h required for the car to rema in on the track while 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 loses all of its energy to the peg, which is knocked 4 ern into the ground.
brick weight 20 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 kg at a steady speed v.
motor 9.6W
bench
0.50kg
What is the value of v?
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
tt~Ml ~
J2a
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 a ripple tank.
diSP,aCementlc:b
o
I
diSP,acement/cmob
(\ .
1.0
V
2.0
o
3.0 distance / cm
I
(\ .
0.05
V
0.10
0.15 time / s
What is the speed of the water wave?
A
0.1 m s"
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 in space. The graphs have the same scales for the time axes.
wave P
displacement
O+-t--;----'t--i-+--t--+--+-~ --~
o
time /ms
wave Q
displacement
O++--+-+-+-+--j--t--t---t-H--~
o
time /ms
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 frequency difference
© 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 to
o
determine 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 x10 13 S- 1
B
1.0 x10 15 S- 1
C
1.0 x10 17 S- 1
D
1.0 x1 0 19 S - 1
22 An electron travels around the circuit shown in the diagram. The cell has negligible internal res istance. At which point in the circuit does the electron have its maximum electrical potential energy?
© UCLES & MOE 2013
A
B
D
C
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 lamp and a resistor, plotted on the same axes.
I1A 0 .2~ -----Y'1----Ji'-
o-F-- - "-l- - - -+-- o 1.0 2.0 V/V 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 is 50 % 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 in PQ.
+ 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 3 o.
© UCLES & MOE 2013
8866/01/0 /N/13
11
26 Four magnetic fields are shown. 1
,,-A~
0 0
@
4
3
2
(
/y'\
Which is the correct order for objects to match these fields?
1
2
3
4
A
flat circu lar co il
bar magnet
solenoid
long straight wire
B
flat circu lar coil
solenoid
bar mag net
long straight wire
C
lon g stra ight wire
flat circu lar co il
bar magnet
solenoid
0
long straight wire
solenoid
bar magnet
flat circu lar co il
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 eac h wire has the same mag nitude. The curre nt in P is into the paper and in Q and R is out of the paper. In which direction is the force on wire R?
© UCLES & MOE 2013
8866/0 1/0 /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 lectric current from the meta l surface is proportional to the intensity of the incident electromagnetic radiation? A
Radiation of greater intens ity causes the meta l surface to get warm and so emit more electrons.
B
Radiation of greater intensity consists of more energetic photons.
C
Radiation of greater intensity means more photons per second strike the metal surfac e.
D
Radiation of greater intensity overcomes the metal's work function energy allowing more electrons to esc ape.
29 An electron in an atom makes a transition from an energy level of energy E1 to a level of energy
E2 , emitting a photon of wavelength A. Which expression gives this wavelength in terms of the Planck constant h and the speed of light e?
B
A
D
C
he
30 Electromagnetic rad iation of frequency f and intensity I is directed onto a meta l electrode X causing photoelectrons to be emitted . Some of these electrons reach electrode Y causing a current in the circu it. electromagnetic radiation electrons
t
V
!
X vacuum
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
Vmin.
Vmin
I
D Vmin.
V min
I
I
I
Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwillin gly been included, the publisher 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 Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.
© UCL ES & MOE 2013
8866/01/0/N/13
• 3740107701 •
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MINISTRY OF EDUCATION, SINGAPORE in collaboration with JC UNIVERSITY OF CAMBRIDGE LOCAL EXAMINATIONS SYNDICATE General Certificate of Education Advanced Level Higher 1
CANDIDATE NAME
CENTRE NUMBER
INDEX NUMBER
ITIIJ
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PHYSICS
N N
Paper 2 Structured Questions
8866/02 OctoberlNovember 2013
I-'
2 hours
m \0_
Candidates answer on the Question Paper.
w
No Additional Materials are required .
ro_ .-J ~
READ TH ESE 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.
For Examiner's Use Section A
The use of an approved scientific calculator is expected, where appropriate.
.>
1
Section A Answer all questions.
2
Section 8 Answer any two questions.
3
4
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.
5
6 7 Section 8
.>
8 9 10 Total
This document consists of 22 printed pages and 2 blank pages. ~,+O A PO,,~ SMII ~
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© UCLES & MOE 2013
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DC (NF/JG) 61576/6
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~ UNIVERSITY of CAMBRIDGE
~ International Examinations [Turn over
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• 3740107702 •
2 11111111111111111 11111 1111111111 111111111111111 1111111111111
Data
c = 3.00
speed of light in free space,
X 10
= 1.60 x
8
the Planck constant,
h = 6.63 x 10- 34 J s
unified atom ic mass constant,
u
= 1.66 X 10-27 kg
rest mass of electron,
me
= 9.11 x 10-3 1 kg
rest mass of proton,
mp
= 1.67 x 10- 27 kg
v
W = pl1V
hydrostatic pressure,
p= pgh
resistors in series,
R = R1 + R2 + ...
~UCLES & MOE 2013
a
C~
= ut+ ~at2 = u 2 + 2as
work done on/by a gas,
resistors in parallel,
'"
C~
C~
C
Formulae 5
C~
C~
9 = 9.81 ms- 2
2
C2 C~
C~<
10- 19
e
uniformly accelerated motion,
CI
C...u'
rn s"
elementary charge,
acceleration of free fall,
(;1
1/R = 1/R1 + 1/R2 + . . .
8866/02/0/N/13
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C~ C C C C C C C C C C C C C C C C C C C C C C C C C C C C C CI CI CI CI CI CI CI CI C C C C C C C C C C C C C C C C C C C C C C C
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Sect ion A
JJ-
Answer all the questions in this section.
~:
-: 5:: -
-:::>~ :-: z:::>:
II
3
:1:: -
1
For Examiner's Use
(a) Fig. 1.1 shows two magnetic flux density vectors A and B, drawn to scale.
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Complete the diagram to show the resultant magnetic flux density (A + B) when the two vectors are added.
IE
iE
IE
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!E IE IE IE IE
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IE
IE jlE IE
Fig. 1.1 (b)
[2]
Fig. 1.2 shows a similar situation for magnetic flux density vectors C and D. Draw on the diagram to show the resultant magnetic flux density (C - D) when the two vectors are subtracted.
!E E E E E E E E E E E E E E
n
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rE lE
r~
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Fig. 1.2
[2]
(c) State the unit for magnetic flux density.
E
'E
E
'E
E E
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· 3740107704·
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2
Geologists are able to determine the likelihood of finding oil-bearing rocks beneath the surface of the Earth without drilling . They make very accurate measurements of g, the acceleration of free fall, at the surface.
For Examiner's Use
C~ C~
C~ C~
C~
C~
In one such measurement at a particular place in Singapore , a small marker falls a distance of 0.72056 m from rest in a vacuum. The time it takes is 0.38393s.
CI-
C~
cg C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C
(a) Calculate the value of 9 at that place in Singapore. Give your answer to a suitable number of significant figures .
9=
CI Cz C~
ms-2 [2]
(b) When the measurement was made at a different place in Singapore, an accurate value of 9.7732 m S-2 was obtained for g. Give one deduction that can be made from this different value by considering why 9 varies. .................................. ......... ........................... ................ ........... ......... ..... ...................... ..... .............. ............................................................................................................................
..................................................................................................................................... [2]
C C C C C C C C C C C C C C C C
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· 3740107705·
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3
In a head-on elastic collision between an alpha particle (mass 4 a.m.u.) and a stationary oxygen nucleus (mass 16 a.m.u.), the oxygen nucleus moved off with a velocity of 3.3 x 105 m S -1 after the collision.
For Examiner's Use
Calculate
z"
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(a) the speed of the alpha particle before the collision,
IE E E E E E E E E
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IE
!E
IE IE
iE IE
IE
IE IE IE
IE
IE IE IE IE IE IE IE IE
=
m S-1 [4]
speed after collision =
m S-1 [1 ]
speed before collision (b) the speed of the alpha particle after the collision .
E E E E E E E E E
E E E E
E E
E E
E E E
E
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C~
C~ C~ C~
BLA NK PAGE
C~ C~
C~
C~ CfC~
cg
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4
A typical force -extension graph for metals has the shape shown in Fig. 4.1 .
For Examiner's Use
z:
-wf-: ~~
5: :
force IN
f- :
0: z0:
6.0
D-
E E E E E
E E E
0-+-- - -1--- - - - - - - - - - - o 4.0 extensionlmm
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E E E E E E
E E
E E
Fig. 4.1 In the case shown, the straight line portion of the graph shows elastic deformation. The curved part of the graph shows permanent deformation (plastic deformation) . (a)
Using data from Fig. 4.1, calculate the elastic potential energy stored when the app lied force is 6.0 N.
E E E
E E E E
IE IE E E E E E
E E
IE IE
energy =
J [2]
(b) Plastic deformation is important during the manufacture of a metallic object. In the daily use of the same object, on ly elastic deformation is required. Give an example of such an object, and explain how it illustrates these statements.
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A battery of e.m.f. 15.0V has an internal resistance of 6.0Q. It is connected to a variable external resistance R, as shown in the circuit diagram Fig. 5.1 . 15.0V
(~ (~ (~
(~
(~
6.0Q
(I-
(~
J - -- - +
1
~
(~ (~
For Examiner's Use
.....----------!--; ~ ~T -c= ----i~ I
( z
J
(8 C C C
I
C R tf
C C C C
/ /
C
Fig. 5.1
C
C
The following table gives some of the numerical details for different numerica l values of R.
R IQ
total resistance
current
p.d. across R
power to R
IQ
IA
IV
IW
C C C C C C C C C C ,
efficiency 1%
0
6.0
2.50
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
0
10
16
14
20
24
30
0.50
12
6.0
80
94
100
0.15
14
2.1
94
CI C I C I CI CI CI C I C I CI C I CII CII CII CII CII CII
(a) Insert values in the spaces to complete the table.
~UCLES & MOE 2013
8866/02l0/N/13
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CII CII CII CII CII CI CI CI CI CI CI CI CI CI CI CI CI CI CI CI CI CI CI _
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- • 3740107709 •
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9
:1::: :
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For Examiner's Use
~:
.11 .
- :
5:x·: ... : 0: z:
a. a• • •
10 -++-t-HH-++l-H-t-l-+1-H+++-H-++-t-HH-++l-H-t-l-+1-H+++-H-++-t-HH-++I-!-+--1
0: 0 ·
E E E E E E
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
E E E
_
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
JmmEm m mEm EEm3IOOmEElilfJi§ o
'I E IE IE IE IE IE IE IE IE IE IE IE E E E E E E E E E E E .I E fi E .I E
o
5
10
15
20
25 R/ Q
Fig. 5.2 (c)
[2]
Explain what is meant by 'efficiency' as used in the last column of the table.
..................................................................................................................................... [2] (d) The battery has efficiency that increases as the external resistance increases. Suggest two consequences of using this battery only with high value external resistors.
1
.
2
.
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..................................................................................................................................... [2]
:E I
: E
:E : E : E
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6
(a)
Explain the phenomenon of photoelectric emission by referring to photon energy and work function energy.
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(b) Explain why the maximum energy of photoelectrons is independent of intensity whereas the photoelectric current is proportional to intensity.
..................................................................................... ....... ......................................... [3]
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7
(a) Briefly describe an experime nt to observe a line emissi on spectrum. You may use a diagram to illustrate your answer.
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..................................................................................................................................... [2]
(b) One line in the Sun's absor ption spectrum has a wavelength of 760 nm. Calculate the energy gap between the discrete electron energy levels that leads to this spectral line. Give your answer in electron volts.
energy gap
=
eV [3]
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Section B
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Answer two of the questions in this section.
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A lamp is marked 240V, 60W.
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Calculate, for the lamp when operating with 240V across it, (i)
the current through it,
=
A [1]
resistance =
Q [1]
current (ii)
(iii)
its resistance ,
the energy supplied to it in 8 hours
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2.
in joules,
energy
=
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energy
=
kW h [2]
in kWh .
(b) The filament of the lamp in (a) has a radius of 0.0043 mm and is made of a material of resistivity 9.2 x 10- 7 Q m.
Calculate the length of the filament.
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Fig. 8.1 illustrates the connection between the copper leads to the filament and the filament itself.
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Describe, giving reasons where appropriate, and without using numerical values, the following features of the current in these three components.
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(i)
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the direction of electron flow
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the rate of electron flow
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the average speed of electrons
........ .... ..................................... ......... ..... ........... ............. ........... ....... ......... ...... ..... [2] (d) Use your answers to (c) to explain why the filament of the lamp gets hot but the copper leads stay relatively cold.
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(e) In practice, the leads to a lamp have some resistance. State how this affects the current in the lamp and its brightness.
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Describe a situation in your home which indicates that the interna l resistance of the mains electrical supply is very low.
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............................................................... ...................................................................... [2] (g) Most electrical extension leads are sold wound on a reel. A label on these reels will have a warning to the user that may read
"Maximum current fully wound 4 A (960W) Maximum current fully unwound 10A (2400 W)" Explain why this warning is necessary.
.......................................... ........................................................................................... [3]
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Question 9 starts on page 16
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9
107
Diesel fuel , when used in a car engine , provides 3.7 x joules of energy per litre of fuel. The car manufacturer's specification for a particular car, travelling at 100 km / hour on a flat road, gives the rate of fuel consumption as 7.8 litres per 100 km travelled . (a) (i)
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Calculate the energy used by the car in travelling 100 km at this speed.
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energy (ii)
=
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Show that the total power being supplied to the car by burning the fuel is 80 kW.
[1]
(b) Fig. 9.1 is a simplified diagram, sometimes called a Sankey diagram , showing how the input power used by this car on a flat road is distributed as various outputs. It is drawn to scale.
%
o powerou tput
20 40
-
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-
60 power wasted as heat in the exhaustgases
80 -
100
--
power used by the . electrical system
power wasted In the transmission system
Fig. 9.1 Using data from the diagram , (i)
determine the efficiency of the car,
efficiency
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=
%
[1]
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(iii)
(iv)
calculate the drivin g force,
calculate the electric current supplied by the car's 15V electrical generator.
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(c) Wh en the car is driving up a hill with a gradient of 1 metre rise for every 25 metres along the road, extra power will be required. The mass of the car is 900 kg. Calculate the percentage increase required in power output for the car. The speed of the car is assumed to be unch anged.
percentage increa se =
% [4]
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(d) Discuss, referring to the law of conservation of energy, ways in which the power output of the car is dissipated to the surroundings of the car when travelling at a constant speed on a flat road.
Include in your answer an estimate of the fraction of the power wasted as noise. (A really loud sound from a loudspeaker might have a power input of 20W.)
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Question 10 starts on page 20
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(a)
Complete the following diagrams to show (i)
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reflection of a circular wavefront at a plane surface,
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refraction of a plane wavefront at a plane surface,
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(iii)
diffraction of a circul ar wavefront at a narrow gap,
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how an interference pattern is produced when a plane wavefront meets a pair of narrow slits .
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Fig. 10.4 (b) On Fig. 10.4 indicate where maxima and minima of intensity occur.
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(c)
(i)
(ii)
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22
State what is meant by the term coherence of two waves.
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Explain how the coherence condition for interference is obtained in (a)(iv) .
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........................................................................... ...................................... ............ [1] (d) Light of wavelength 5.35 x 10- 7 m is shone on a double slit whose separation is 0.200 mm. Calculate the separation of the fringes on a screen placed 0.875 m from the double slit. The screen and double slit are both placed at right-angles to the beam of light.
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Fig. 10 .5 The intensity of the wave at a distance of 2.0m from its source is 3.0Wm- 2 .
(i)
Using data from the graph, calculate the intensity at a distance of 4.6 m from the source.
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intensity (ii)
=
W m- 2 [3]
Deduce the intensity at a distance of 12.0 m from the source.
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CI University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambr idge.
l..!.UCLES & MOE 2013
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JJC
2013 H1 Physics (8 8866) P1
JURONG JUNIOR CO OLLEGE PHYSIC CS DEPARTM MENT GCE E A levels 20 013 H1 1 Physics (8 8866) Paper 1 solutions
Qn Ans s Suggested solutio on 1
B
Estimate ed mass of car ≈ 1500 kg; Estimate ed speed off car on exp pressway ≈ 80 km h−1 Thereforre, estimate ed change in n car’s mom mentum 3 (15 500)(80 × 10 0 ) ≈0− = −3.3 × 104 kg m s−1 3600
2
D
Using R =
ρL A
=
4 ρL π d2
ven by Error equation is giv
+ρ
⎛ + R ⎞ ⎛ + d ⎞ ⎛ +L ⎞ =⎜ +2 + ρ ⎝ R ⎟⎠ ⎜⎝ d ⎟⎠ ⎜⎝ L ⎟⎠
Thus,
+ρ
1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 0.1 = 0.0947 + = ⎜ ⎟ + 2⎜ ρ ⎝ 668 ⎠ ⎝ 5.00 ⎟⎠ ⎜⎝ 25 ⎟⎠
Î + ρ ≈ 0.5 x 10-2 Ω m (1.S.F F.) 3
B
Both phyysical quanttities can ha ave units ex xpressed as s N m.
4
D
Since terminal veloc city is involvved, there is significant air resistaance acting on the object. The horizontal component of itts velocity decreases d in n the 1st 3 sseconds. The horizontal component of itts accelerattion is non-z zero. The verttical component of its vvelocity incrreases in the 1st 3 secoonds. Stateme ent D is corrrect, since te erminal velo ocity is cons stant.
5
C
At point A, the ball is instantan neously at re est while it is i in its first contact witth the surface. At point B, the ball is still in its first contac ct with the surface, but it has maximum speed and is about to first bou nce from th he surface. At point D, the ball is just in its second con ntact with th he surface aafter falling from its (new reduced) maximum m he eight.
6
C
© Physics Dept
The rate e of change of momenttum of an ob bject is by definition d thee resultant force acting on n it, thus the e physical q quantity has s a unit of newton (sym mbol: N).
Pa age 1 of 6
JJC
2013 H1 Physics (8 8866) P1
JURONG JUNIOR CO OLLEGE PHYSIC CS DEPARTM MENT GCE E A levels 20 013 H1 1 Physics (8 8866) Paper 1 solutions
Qn Ans s Suggested solutio on 7
B
When th he bucket is moving at uniform spe eed, by N2L, the resultan nt force act ing on it is zero z i.e. its tensio on T balanc ces its weigh ht, mg Î Mathe ematically, T = mg When th he bucket is deceleratin ng uniformly y, by N2L, the resultan nt force act ing on it is non-zero n i.e e. its tensio on T’ is less s than its we eight, mg Î Mathe ematically, T '− mg =
Thus, T '− T = 8
C
m (0 − v ) t
− mvv t
Taking the direction n of approacch to be neg gative, Impulse on ball = ch hange in itss momentum m Î 6.3 = 0.42(v − ( −3)) 3 Î v = 12 2 m s -1
9
A
In B, the e air is being g “compresssed” while pumped p into o a tyre; thee viscous fo orces are relattively neglig gible in resissting the mo otion of the additional aamount of air a being pu umped in. In C, the e viscous forces betwee en the wate er lining the interior of tthe “faulty ta ap” and wate er molecule es in the dro op are relatiively small. In D, the e road is “we et”, viscouss forces betw ween the water on the road and petrol p is relativelyy negligible. In A, the ere is almos st no motion n of the pain nt when app plied to the ssurface of a wall.
10
A
© Physics Dept
Since forces P, Q and a R are in n equilibrium m, they will form f a closeed vector triangle, ad-to-tail in either a clocckwise or antia with the direction off the arrowss joined hea clockwisse orientatio on.
Pa age 2 of 6
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2013 H1 Physics (8 8866) P1
JURONG JUNIOR CO OLLEGE PHYSIC CS DEPARTM MENT GCE E A levels 20 013 H1 1 Physics (8 8866) Paper 1 solutions
Qn Ans s Suggested solutio on 11
C
Before th he pivots arre translated d, taking moments m abo out pivot Y,
W ⎛L⎞ ⎛L⎞ ⎜ ⎟ W = ⎜ ⎟ Fx Î Fx = 2 ⎝4⎠ ⎝2⎠ After the e pivots are translated, taking moments m abo out pivot Y,
3W ⎛L L⎞ ⎛L⎞ ⎜ + ⎟ W = ⎜ ⎟ Fx ' Î Fx ' = 4 ⎝2⎠ ⎝4 8⎠ Thus, inccrease in th he contact fo force acting on the plan nk by suppoort X =
Fx '− Fx = 12
D
W 4
For P an nd Q, note that an electrric force actts on a posiitive charge ed particle in n the same direction as s the electric field f that the e charged p particle is in n while an electrric force actts on a nega ative charged particle in the oppoosite directio on as ed particle is in. the electtric field tha at the charge For R an nd S, use Fleming’s Le eft-Hand Rule. wer this ques stion, candiidates probably have to o rely on theeir O-level To answ knowled dge of static electricity ssince the to opic of “electric fields” iss not covere ed in the H1 syllabus. s
13
A
By conse ervation of energy, GPE of model m car at a top of tracck = GPE + KE of mod del car at topp of loop
1 m 2 2.5r Î h=2 Î mgh h=
14
C
(
gr g
)
2
+ mg 2r
By conse ervation of energy, Loss in GPE G of bric ck = average e WD again nst frictional forces Î mgh h = Fave d Î (20)(2 + 0.04) = Fave (0.04) Î Fave ≈ 1000 N
15
A
© Physics Dept
Efficienccy of motor = rate of wo ork done in lifting load / input powe wer Î 0.2 = (0.5)(9.81)v / 9.6 Î v = 0.39 m s -1
Pa age 3 of 6
JJC
2013 H1 Physics (8 8866) P1
JURONG JUNIOR CO OLLEGE PHYSIC CS DEPARTM MENT GCE E A levels 20 013 H1 1 Physics (8 8866) Paper 1 solutions
Qn Ans s Suggested solutio on 16
C
For a co onstant-frequency soun nd wave, I ∝ a 2 . Thus wh hen I is doub bled, the ne ew amplitud de will be a 2 and the amplitude increases by a ( 2 − 1) .
17
B
From the e graphs, th he waveleng gth is 2.0 cm m while the period is 0 .10 s. -1 Thus, sp peed = 0.02 2 / 0.1 = 0.2 2ms
18
D
The period of Q, TQ, is 1/125 s = 8.0 ms. When t = 4.5TQ = 4.0TP , P and d Q are in anti-phase. The period of P, TP, is 4.5(8)/4 4 = 9.0 ms. P and Q will be nex xt in phase, when the e time is eq quals to the Lowe est Commo on Multiple ((L.C.M) of TQ and TP, i.e. (8)(9 9) = 72 ms.
19
C
The ripp ple tank dipp pers have a constant phase p differe ence of π raad, which means m they are coherent. For B an nd D, the so ources have e a constant frequency y difference.. For A, a light source e can gene rate photon ns over a range of frequuency e.g. within w the visib ble and infra ared region of the electtromagnetic c spectrum, thus the ph hase differencce between 2 photons from 2 diffe erent light so ources are uunlikely to have h a constantt phase diffe ference. Fo or monochro omatic sources, each liight source generate es photons at random, thus the ph hase difference betweeen 2 photon ns from 2 different light sources are un nlikely to hav ve a consta ant phase diifference.
20
C
Necessa ary condition for approxximated formula to be applicable.
21
C
Using P = I 2 R , 2
Q⎞ ⎛ nQ P=⎜ ⎟ R ⎝ t ⎠ 2
⎛n ⎞ Î 0.25 = ⎜ x 1.6 x 10-19 ⎟ (1 000) ⎠ ⎝t n Î ≈1 1.0 x 1017 s -1 t 22
B
© Physics Dept
An electron moves from the ne egative term minal of the cell throughh the circuit compone ents to the positive ter minal of the e cell, conve erting its eleectrical pote ential energy to non-electtrical forms of energy in n the circuitt componennts along its journey.
Pa age 4 of 6
JJC
2013 H1 Physics (8 8866) P1
JURONG JUNIOR CO OLLEGE PHYSIC CS DEPARTM MENT GCE E A levels 20 013 H1 1 Physics (8 8866) Paper 1 solutions
Qn Ans s Suggested solutio on 23
A
The resistance of an electrical componentt R can be obtained o froom the grap ph by the ratio of the pote ential differe ence across s the electric cal compon ent V to the e current flowing f through it I i.e. R =
V I
Thus, A is correct because b the e graph of th he diode an nd filament llamp interse ects at about 1.2 V. B and D are false, because b R≠
1 . gradientt
For C, th he resistanc ce of the fila ament lamp (5.0 Ω ) is half that off the resistor (10 Ω ) at 1..0 V. 24
D
R=
ρL A
=
4ρ L L ∝ 2 2 d πd
The seco ond piece of o wire has a length 3 times that off the first pieece of wire and a diameter ½ of that of o the first p piece of wire e, so proporrtionally, thee second piiece of wire hass a resistanc ce (3)(22) = 12 times off that of the first piece of wire. 25
C
The corrrect stateme ent for A is: The charge flowing each secon nd in the co onductor is 3 C. The corrrect stateme ent for B is: Electrons flow from Q to P. The corrrect stateme ent for D is: The resistance of th he conducto or is (12 / 3 =) 4 Ω .
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Pa age 5 of 6
JJC
2013 H1 Physics (8 8866) P1
JURONG JUNIOR CO OLLEGE PHYSIC CS DEPARTM MENT GCE E A levels 20 013 H1 1 Physics (8 8866) Paper 1 solutions
Qn Ans s Suggested solutio on 26
C Illustratio ons with inc clusion of th he objects, where w possible: 4
S
N
In 3 dimensions
27
B
Wires Q and R attra act each oth her while wiires P and R repel eacch other. Thus, the direction of the resulltant force acting a on wire R is indiccated by op ption B.
28
C
Factual Question. Q
29
D
By conse ervation of energy,
E1 − E2 = Î λ=
30
A
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hc
λ
hc E1 − E2
The stop pping potential Vm is in dependent of the inten nsity of the eelectromagn netic radiation n I.
Pa age 6 of 6
JJC
2013 H1 Physics (8866) P2
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
A+B
(b)
−D
[1] Magnitude [1] Direction
C−D
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.
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JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions 2(a)
s = ut +
1 2 at 2
0.72056 = 0 +
1 g( 0.38393 )2 2
g = 9.7768 m s−2
[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 = 9.7732 m s−2 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 2 − v1 = u1 − u2 3.3 × 105 − v α = uα − 0
Equation 1
By conservation of momentum
4u(uα ) + 0 = 4u(v α ) + 16u( 3.3 × 105 ) where u is the a.m.u
uα = v α + 4( 3.3 × 105 )
Equation 2
Solving Equation 1 and 2 simultaneously
uα = 8.25 × 105 m s−1
[1] Eqn. 1 [1] Eqn. 2 [1] Working [1] Ans
Common errors : Wrong assumption that alpha particle stopped moving after collision. (b)
Sub uα = 8.25 × 105 m s−1 and solve
v α = −4.95 × 105 m s−1 Speed is a scalar quantity therefore is 4.95 × 105 m s−1
4 (a)
EPE =
=
[1] Ans
1 2 kx = Area under graph (linear portion) 2
1 ( 4 × 10−3 )( 6 ) 2
= 1.2 × 10−2 J
[1] Sub [1] Ans
Common errors : Forgetting that extension is in mm
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2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (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)
1.50
0.94 0.75
6.0
9.4 11
9.0
40
8.8
63
8.3
70
[4] 1 mark for every three correct answers
(b)
[1] Use of correct plots [1] Smooth curve
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2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (c)
Efficiency refers to the percentage of power delivered to the external resistance by the battery. It can be expressed mathematically as
(d)
6(a)
Power dissipated by R × 100% Power generated by battery
[1] worded explanation [1] expression for efficiency
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
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.
[1] general statement
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] address work function energy
[1] longer battery life
[1] address photon 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.
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]
[1]
[1]
[1] sub
[1] ans
Common errors : Confused with absorption spectrum
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JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (b)
8(a)(i)
(ii)
(iii)
EGAP =
[1] sub to get EGAP in J
hc λ
=
6.63 × 10−34 ( 3 × 108 ) J 760 × 10 −9
[1] correct conversion
=
6.63 × 10−34 ( 3 × 108 ) = 1.64 eV 760 × 10−9 ( 1.6 × 10−19 )
[1] Ans
P = IV 60 I= = 0.25 A 240
[1] Ans
V = IR 240 R= = 960 Ω 0.25
[1] Ans
1. E = Pt
= 60( 8 × 60 × 60 ) = 1.73 × 106 J 2. Energy supplied at 60 W for 8 hours
= 60 × 10 × 8 = 0.48 kWh
[1] Conversion [1] Ans
ρl A
[1] Expression
−3
(b)
R=
[1] Ans
960 =
−7
( 9.2 × 10 )l π( 0.0043 × 10−3 )2
l = 6.06 × 10−2 m
[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]
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JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (f)
(g)
9 (a)(i)
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
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
1 litre of fuel provides 3.7 × 107 J
[1] working [1] ans
7.8 litres of fuel provides 7.8 × 3.7 × 107 = 2.89 × 108 J (ii)
In 1 hour, for the conversion of 2.89 × 108 J
Power required = (b)(i) (ii)
2.89 × 108 = 8.03 × 104 W 80 kW 3600
From the graph, looking at power output percentage, efficiency = 32%
Efficiency = 0.32 =
P 8.03 × 104
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=
2.57 × 104 100×103 3600
[1] ans
[1] ans [1] consideration of parameters [1] conversion of 100 km/hr [1] sub [1] ans
= 9.25 × 102 N (iv)
[1] working
Power output of car, P × 100% Power supplied to car
P = 2.57 × 104 W = 25.7 kW (iii)
[1] insufficient power
From graph, power used by electrical system = 4% 32% correspond to 2.57 x 104 W 4% correspond to
2.57 × 10 4 × 4 = 3.21× 103 W 32
[1] electrical power value
Using P = IV
I=
3.21× 103 15
= 2.14 × 102 A
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[1] sub [1] ans
Page 6 of 10
JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (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
= 900 × 9.81× 1.11 = 9.80 × 103 J
[1] vertical height [1] GPE gained
Increase in power = 9.80 × 103 W Percentage increase in power =
9.80 × 103 × 100% = 38.1% 2.57 × 10 4
OR Increase in force required to drive up the hill,
1 Fa = mg sin θ = 900( 9.81)( ) = 353 N 25
[1] sub [1] ans
[2] Fa [1] %increase in
353 F Percentage increase in force = × 100% = 38.2% 2 9.25 × 10 Since P = Fv and speed is unchanged, power has the same percentage increase [1] %increase in (d)
of 38.2%
P
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 airresistance. 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.
[1] friction [1] air-resistance [1] fraction due to noise [1] comment
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.
[1] moves air [1] tyre flex [1] car parts moving
Assuming the power dissipated as sound is 20 W, fraction of power wasted as noise is
20 1 25700 1300
[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.
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JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions 10(a)(i)
[2]
(ii)
[2]
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JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (iii)
[2]
(iv)
[2]
minima maxima
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]
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JJC
2013 H1 Physics (8866) P2
JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT GCE A levels 2013 H1 Physics (8866) Paper 2 solutions (d)
Using λ =
[1] equation
ax D
5.35 × 10−7 =
0.2 × 10−3 x 0.875
x = 2.34 × 10−3 m (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 = kA2
3 = k( 0.86 × 10 −3 )2 ----- (1)
[1] sub [1] ans [1] reading A from the graph
[1] sub
I = k( 0.37 × 10 −3 )2 ----- (2)
I = 0.56 W m−2 (ii)
Using I ∝
[1] ans
1 d2
k d2 k 3= ( 2 )2
I=
[1] sub
k = 12 When d = 12.0 m
[1] ans
12 I = 2 = 8.33 × 10 −2 W m−2 12
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