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CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2013 series
9702 PHYSICS
9702/21
Paper 2 (AS Structured Questions), maximum raw mark 60
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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
w w w . X t r e m e P a p e r s . c o m
Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 21
© Cambridge International Examinations 2013
1 (a)
the wire returns to its srcinal length (
not ‘shape’
) M1
when the load is removed
A1 [2]
(b)
energy: N
m / kg
m
2
s
–2
and volume m
3
C1
energy / volume: kg
m
2
s
–2
/ m
3
M1
energy / volume: kg
m
–1
s
–2
A0 [2]
(c)
ε
has no units B1
E
: kg m
s
–2
m
–2
M1
units of RHS: kg
m
–1
s
–2
= LHS units / satisfactory conclusion to show
C
has no units A1 [3]
2 (a)
mass is the property of a body resisting changes in motion / quantity of
matter in a body / measure of inertia to changes in motion
B1
weight is the force due to the gravitational field/force due to gravity or gravitational force B1 [2]
Allow 1/2 for
‘mass is scalar weight is vector’
(b) (i)
arrow vertically down through O B1
tension forces in correct direction on rope B1 [2]
(ii) 1.
weight =
mg
= 4.9
×
9.81 (= 48.07) C1
69 sin
θ
=
mg
C1
θ
= 44.(1)
°
scale drawing allow ± 2
°
A1 [3]
use of cos or tan 1/3 only
2.
T
= 69 cos
θ
C1
= 49.6 / 50
N
scale drawing
50 ±2 (2/2) 50 ±4 (1/2)
A1 [2]
correct answers obtained using scale diagram or triangle of forces will score full marks
cos in
1.
then sin in
2.
(2/2)
3 (a)
loss in potential energy due to decrease in height (as P.E. =
mgh
) (B1)
gain in kinetic energy due to increase in speed (as K.E. = ½
mv
2
) (B1)
special case ‘as PE decreases KE increases’ (1/2)
increase in thermal energy due to work done against air resistance
(B1)
loss in P.E. equals gain in K.E. and thermal energy (B1)
max. 3 [3]
Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 21
© Cambridge International Examinations 2013
(b) (i)
kinetic energy = ½
mv
2
C1
= ½ × 0.150
×
(25)
2
C1
= 46.875 = 47
J A1 [3]
(ii) 1.
potential energy (=
mgh
) = 0.150
×
9.81
×
21 C1
loss = KE –
mgh
= 46.875 – (30.9) C1
= 15.97 = 16
J A1 [3]
2.
work done = 16
J work done = force
×
distance C1
F
= 16 / 21 = 0.76
N A1 [2]
4 (a)
pressure = force / area (normal to force) A1 [1]
(b)
molecules/atoms/particles in (constant) random/haphazard motion B1
molecules have a change in momentum when they collide with the walls M1
(force exerted on molecules) therefore force on the walls A1
reference to average force from many molecules/many collisions A1 [4]
(c)
elastic collision when kinetic energy conserved B1
temperature constant for gas B1 [2]
5 (a)
waves overlap / meet / superpose (B1)
coherence / constant phase difference (
not constant
λ
or frequency)
(B1
)
path difference = 0,
λ
, 2
λ
or phase difference = 0, 2
π
, 4
π
(B1)
same direction of polarisation/unpolarised (B1)
max. 3 [3]
(b)
λ
=
v
/
f
C1
f
= 12
×
10
9
Hz C1
λ
= 3
×
10
8
/ 12
×
10
9
(
any subject)
M1
= 0.025
m A0 [3]
(c)
maximum at P B1
several
minima or maxima between O and P B1
5 maxima / 6 minima between O and P or 7 maxima / 6 minima including O and P B1 [3]
(d)
slits made narrower B1
slits put closer together B1 [2]
(
not just ‘make slits smaller’)
Allow tilting the slits M1
and explanation of axes of rotation A1
Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 21
© Cambridge International Examinations 2013
6 (a) (i)
chemical to electrical B1 [1]
(ii)
electrical to thermal / heat or heat and light B1 [1]
(b) (i)
(
P
B
=)
E
I
or
I
2
(
R
1
+
R
2
) A1 [1]
(ii)
(
P
R
=)
I
2
R
1
A1 [1]
(c)
R
=
ρ
l
/
A
or clear from the following equation B1
ratio =
I
2
R
1
/
I
2
R
2
=
( ) ( )
2122
of resistance 8hasor
22
R R d d
×
π/ π/
l l
ρ ρ
C1 = 8 or 8:1 A1 [3]
(d)
P
=
V
2
/
R
or
E
2
/
R
C1
(
V
or
E
the same) hence ratio is 1/8 or 1:8 = 0.125 (
allow ecf from
(c)
)
A1 [2]
7 (a)
the majority/most went straight through or were deviated by small angles B1 a very small proportion/a few were deviated by large angles B1 small angles described as < 10
°
and large angles described as >90
°
B1 [3]
(b)
most of the atom is empty space/nucleus very small compared with atom B1
mass and charge concentrated in (very small) nucleus B1
correct links made with statements in
(a)
B1 [3]

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