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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2012 question paper for the guidance of teachers
9702 PHYSICS
9702/43
Paper 4 (A2 Structured Questions), maximum raw mark 100
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination.
ã
Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43
© University of Cambridge International Examinations 2012
Section A 1 (a)
work done in bringing unit mass from infinity (to the point) B1 [1]
(b)
gravitational force is (always) attractive B1
either
as
r
decreases, object/mass/body does work
or
work is done by masses as they come together B1 [2]
(c)
either
force on mass =
mg
(where
g
is the acceleration of free fall /gravitational field strength) B1
g = GM/r
2
B1 if
r
@
h, g
is constant B1
∆
E
P
= force × distance moved M1 =
mgh
A0
or
∆
E
P
=
m
∆
φ
(C1) =
GMm
(1/
r
1
– 1/
r
2
) =
GMm
(
r
2
– r
1
)/
r
1
r
2
(B1) if
r
2
≈
r
1
, then (
r
2
– r
1
) =
h
and
r
1
r
2
=
r
2
(B1)
g = GM/r
2
(B1)
∆
E
P
=
mgh
(A0) [4]
(d)
½
mv
2
=
m
∆
φ
v
2
= 2 ×
GM/r
C1 = (2 × 4.3 × 10
13
) / (3.4 × 10
6
) C1
v
= 5.0 × 10
3
m
s
–1
A1 [3]
(Use of diameter instead of radius to give v = 3.6 × 10
3
m
s
–1
scores 2 marks)
2 (a) (i)
either
random motion
or
constant velocity until hits wall/other molecule B1 [1]
(ii)
(total) volume of molecules is negligible M1 compared to volume of containing vessel A1
or
radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2]
(b)
either
molecule has component of velocity in three directions
or
c
2
= c
X
2
+ c
Y
2
+ c
Z
2
M1 random motion and averaging, so <
c
X
2
> = <
c
Y
2
> = <
c
Z
2
> M1 <
c
2
> = 3<
c
X
2
> A1 so,
pV
= ⅓
Nm
<
c
2
> A0 [3]
(c)
<
c
2
>
∝
T
or
c
rms
∝
C1
temperatures are 300
K and 373
K C1
c
rms
= 580
m
s
–1
A1 [3]
(Do not allow any marks for use of temperature in units of ºC instead of K)
Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43
© University of Cambridge International Examinations 2012
3 (a)
(numerically equal to) quantity of (thermal) energy required to change the state of unit mass of a substance M1 without any change of temperature A1 [2]
(Allow 1 mark for definition of specific latent heat of fusion/vaporisation)
(b)
either
energy supplied = 2400 × 2 × 60 = 288000
J C1 energy required for evaporation = 106 × 2260 = 240000
J C1 difference = 48000
J rate of loss = 48000 / 120 = 400
W A1
or
energy required for evaporation = 106 × 2260 = 240000
J (C1) power required for evaporation = 240000 / (2 × 60) = 2000
W (C1) rate of loss = 2400 – 2000 = 400
W (A1) [3]
4 (a)
a
= (–)
ω
2
x
and
ω
=
2
π
/T
C1
T
= 0.60
s C1
a
= (4
π
2
× 2.0 × 10
–2
) / (0.6)
2
= 2.2
m
s
–2
A1 [3]
(b)
sinusoidal wave with all values positive B1 all values positive, all peaks at
E
K
and energy = 0 at
t
= 0 B1 period = 0.30
s B1 [3]
5 (a)
force per unit positive charge acting on a stationary charge B1 [1]
(b) (i)
E
=
Q
/ 4
πε
0
r
2
C1
Q
= 1.8 × 10
4
× 10
2
× 4
π
× 8.85 × 10
–12
× (25 × 10
–2
)
2
M1
Q
= 1.25 × 10
–5
C = 12.5
µ
C A0 [2]
(ii)
V
=
Q
/ 4
πε
0
r = (1.25 × 10
–5
) / (4
π
× 8.85 × 10
–12
× 25 × 10
–2
) C1 = 4.5 × 10
5
V A1 [2]
(Do not allow use of V = Er unless explained)
Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43
© University of Cambridge International Examinations 2012
6 (a) (i)
peak voltage = 4.0
V A1 [1]
(ii)
r.m.s. voltage (= 4.0/√2) = 2.8
V A1 [1]
(iii)
period
T
= 20
ms M1 frequency = 1 / (20 × 10
–3
) M1 frequency = 50
Hz A0 [2]
(b) (i)
change = 4.0 – 2.4 = 1.6
V A1 [1]
(ii)
∆
Q
=
C
∆
V
or
Q
=
CV
C1 = 5.0 × 10
–6
× 1.6 = 8.0 × 10
–6
C A1 [2]
(iii)
discharge time = 7
ms C1 current = (8.0 × 10
–6
) / (7.0 × 10
–3
) M1 = 1.1(4) × 10
–3
A A0 [2]
(c)
average p.d. = 3.2
V C1 resistance = 3.2 / (1.1 × 10
–3
) = 2900
Ω
(allow 2800
Ω
)
A1 [2]
7 (a)
sketch: concentric circles
(minimum of 3 circles)
M1 separation increasing with distance from wire A1 correct direction B1 [3]
(b) (i)
arrow direction from wire B towards wire A B1 [1]
(ii)
either
reference to Newton’s third law
or
force on each wire proportional to product of the two currents M1 so forces are equal A1 [2]
(c)
force always towards wire A/always in same direction B1 varies from zero (to a maximum value) (1) variation is sinusoidal / sin
2
(1) (at) twice frequency of current (1)
(any two, one each)
B2 [3]
8 (a)
packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1
(allow 1 mark for ‘packet of electromagnetic radiation’)
energy = Planck constant × frequency
(seen here or in
b
)
B1 [3]
(b)
each (coloured) line corresponds to one wavelength/frequency B1 energy = Planck constant × frequency implies specific energy change between energy levels B1 so discrete levels A0 [2]

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