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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the October/November 2011 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 October/November 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
w w w . X t r e m e P a p e r s . c o m
Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 43
© University of Cambridge International Examinations 2011
Section A 1 (a) (i)
weight =
GMm
/
r
2
C1 = (6.67 × 10
–11
× 6.42 × 10
23
× 1.40)/(½ × 6.79 × 10
6
)
2
C1 = 5.20 N A1 [3]
(ii)
potential energy = –
GMm
/
r
C1 = –(6.67 × 10
–11
× 6.42 × 10
23
× 1.40)/(½ × 6.79 × 10
6
) M1 = –1.77 × 10
7
J A0 [2]
(b)
either
½
mv
2
= 1.77 × 10
7
C1
v
2
= (1.77 × 10
7
× 2)/1.40 C1
v
= 5.03 × 10
3
m
s
–1
A1
or
½
mv
2
=
GMm/r
(C1)
v
2
= (2 × 6.67 x 10
–11
× 6.42 × 10
23
)/(6.79 × 10
6
/2) (C1)
v
= 5.02 × 10
3
m
s
–1
(A1) [3]
(c) (i)
½ × 2 × 1.66 × 10
–27
× (5.03 × 10
3
)
2
=
23
× 1.38 × 10
–23
×
T
C1
T
= 2030 K A1 [2]
(ii)
either
because there is a range of speeds M1 some molecules have a higher speed A1
or
some escape from point above planet surface (M1) so initial potential energy is higher (A1) [2]
2
(a)
temperature scale calibrated assuming linear change of property with temperature B1 neither property varies linearly with temperature B1 [2]
(b) (i)
does not depend on the property of a substance B1 [1]
(ii)
temperature at which atoms have minimum/zero energy B1 [1]
(c) (i)
323.15 K A1 [1]
(ii)
30.00 K A1 [1]
Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 43
© University of Cambridge International Examinations 2011
3
(a)
acceleration proportional to displacement/distance from fixed point M1 and in opposite directions/directed towards fixed point A1 [2]
(b)
energy = ½
m
ω
2
x
02
and
ω
= 2
π
f
C1 = ½ × 5.8 × 10
–3
× (2
π
× 4.5)
2
× (3.0 × 10
–3
)
2
C1 = 2.1 × 10
–5
J A1 [3]
(c) (i)
at maximum displacement M1 above rest position A1 [2]
(ii)
acceleration = (–)
ω
2
x
0
and acceleration = 9.81 or
g
C1 9.81 = (2
π
× 4.5)
2
×
x
0
x
0
= 1.2 × 10
–2
m A1 [2]
4
(a)
e.g. storing energy separating charge blocking d.c. producing electrical oscillations tuning circuits smoothing preventing sparks timing circuits
(any two sensible suggestions, 1 each, max 2)
B2 [2]
(b) (i)
–
Q
(induced) on opposite plate of C
1
B1 by charge conservation, charges are –
Q
, +
Q
, –
Q
, +
Q
, –
Q
B1 [2]
(ii)
total p.d.
V
=
V
1
+
V
2
+
V
3
B1
Q
/
C
=
Q
/
C
1
+
Q
/
C
2
+
Q
/
C
3
B1 1/
C
= 1/
C
1
+ 1/
C
2
+ 1/
C
3
A0 [2]
(c) (i)
energy = ½
CV
2
or energy = ½
QV
and
C
=
Q/V
C1 = ½ × 12 × 10
–6
× 9.0
2
= 4.9 × 10
–4
J A1 [2]
(ii)
energy dissipated in (resistance of) wire/as a spark B1 [1]
Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 43
© University of Cambridge International Examinations 2011
5
(a)
supply connected correctly (to left & right) B1 load connected correctly (to top & bottom) B1 [2]
(b)
e.g. power supplied on every half-cycle greater average/mean power
(any sensible suggestion, 1 mark)
B1 [1]
(c) (i)
reduction in the variation of the output voltage/current B1 [1]
(ii)
larger capacitance produces more smoothing M1
either
product
RC
larger
or
for the same load A1 [2]
6 (a)
unit of magnetic flux density B1 field normal to (straight) conductor carrying current of 1 A M1 force per unit length is 1 N
m
–1
A1 [3]
(b) (i)
force on particle always normal to direction of motion M1 (and speed of particle is constant) magnetic force provides the centripetal force A1 [2]
(ii)
mv
2
/
r
=
Bqv
M1
r
=
mv
/
Bq
A0 [1]
(c) (i)
the momentum/speed is becoming less M1 so the radius is becoming smaller A1 [2]
(ii)
1.
spirals are in opposite directions M1 so oppositely charged A1 [2]
2.
equal initial radii M1 so equal (initial) speeds A1 [2]

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