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GCE Advanced Level
MARK SCHEME for the November 2005 question paper
9702 PHYSICS
9702/04
Core
maximum raw mark 60
This mark scheme is published as an aid to teachers and students, to indicate the
requirements of the examination. It shows the basis on which Examiners were initially
instructed to award marks. It does not indicate the details of the discussions that took place
at an Examiners’ meeting before marking began. Any substantial changes to the mark
scheme that arose from these discussions will be recorded in the published Report on the
Examination.
All Examiners are instructed that alternative correct answers and unexpected approaches in
candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
demonstrated.
Mark schemes must be read in conjunction with the question papers and the Report on the
Examination.
The minimum marks in these components needed for various grades were previously
published with these mark schemes, but are now instead included in the Report on the
Examination for this session.
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
CIE will not enter into discussion or correspondence in connection with these mark
schemes.
CIE is publishing the mark schemes for the November 2005 question papers for most IGCSE
and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary
Level syllabuses.
Page 1
1
(a)
Mark Scheme
A LEVEL – NOVEMBER 2005
GM / R2 = Rω2 …………….…………………...…..…………………..
ω = 2π / (24 × 3600) ………………………………..……..……………
6.67 × 10–11 × 6.0 × 1024 = R3 × ω2
R3 = 7.57 × 1022 …………………………………………………………
R = 4.23 × 107 m ………………………………………………………..
(b)(i) ∆Φ =
=
=
∆EP =
=
2
3
GM/Re – GM/Ro …...………………………………………….…..
(6.67 × 10–11 × 6.0 × 1024) ( 1 / 6.4 × 106 – 1 / 4.2 × 107)
5.31 × 107 J kg–1 ………………………………………………….
5.31 × 107 × 650 ………………………………………………….
3.45 × 1010 J ……………………………………………………..
Paper
4
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A0
[3]
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C1
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[4]
(c)
e.g. satellite will already have some speed in the correct direction
…
B1
[1]
(a)
obeys the law pV = constant × T ………………………..………………..
at all values of p, V and T ……………………………………………….
M1
A1
[2]
(b)
n = (2.9 × 105 × 3.1 × 10–2) / (8.31 × 290) …..………………..………...
= 3.73 mol ……………………………………………………………….
C1
A1
[2]
(c)
at new pressure, nn
= 4.23 mol ….……………………………………….
change = 0.50 mol ……………………………………………………….…
number of strokes = 0.50 / 0.012 = 42 (must round up for mark) …….
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A1
[3]
correct statement, words or symbols
B1
[1]
(a)
(b)(i) w
(c)
3.4
290
×
2.9
300
…..…………………………...…..
A1
[2]
= 4.05 × 104 J
………………………………………………………….
B1
[1]
(iii) ∆U = 4.05 × 104 – 3050 = 37500 J …no e.c.f. from (a)…………………
penalise 2 sig.fig. once only
A1
[1]
number of molecules = NA ……………………………………………….
energy = 37500 / (6.02 × 1023)
= 6.2 × 10–20 J (accept 1 sig.fig.) …………………………..….
C1
(a) (i) ω
(ii) a0
(b)
= 3.73 ×
= p∆V …………………………………………………………………..
= 1.03 × 105 × (2.96 × 10–2 – 1.87 × 10–5)
= (–) 3050 J …………………..……………….…………..……………
(ii) q
4
Syllabus
9702
A1
= 2πf ………………………………………………………....…………..
= 2π × 1400
= 8800 rad s–1 …………………………………………………………..
C1
= (–)ω2x0 ……………………………..…………………………………
= (8800)2 × 0.080 × 10–3
= 6200 m s–2 …………………….…………………………………….
C1
straight line through origin with negative gradient …….……………....
end points of line correctly labelled ……………………………………..
(c) (i) zero displacement
(ii) v
C1
…………………………………………………………
= ωx0 …………………………………………………………………….
= 8800 × 0.080 × 10–3
= 0.70 m s–1 …………………………………………………………….
© University of Cambridge International Examinations 2005
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Page 2
5
(a)
(b)(i)
6
Syllabus
9702
Paper
4
½mv2 = qV ……(or some verbal explanation) …..……………..……
½ × 9.11 × 10-31 × v2 = 1.6 × 10-19 × 1.2 × 104 …………………………
v = 6.49 × 107 m s–1 ………….…………………………………………
B1
B1
A0
[2]
within field:
B1
B1
B1
[3]
circular arc ………………..………………..……………..
in ‘downward’ direction ……………..…………………..
beyond field: straight, with no ‘kink’ on leaving field …………………
(ii) 1. v is smaller ………………………………………………………………….
deflection is larger …………………………………………………………
2. (magnetic) force is larger …………………………………………………
deflection is larger ………………………………………………………..
M1
A1
M1
A1
(numerically equal to) force per unit length …………………….…….…
on straight conductor carrying unit current …………………………….
normal to the field …………………………………………………………
M1
A1
A1
[3]
(b)
flux through coil = BA sinθ ………………………………………………..
flux linkage = BAN sinθ ………………..…………………………………
B1
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[2]
(c) (i)
(induced) e.m.f. proportional to ………………………………..…..…..…
rate of change of flux (linkage) …………………….…………………….
M1
A1
[2]
graph:
B1
B1
B1
[3]
energy required to separate the nucleons in a nucleus ..………..……..
nucleons separated to infinity / completely …………..………………..
M1
A1
[2]
S shown at peak
B1
[1]
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[1]
(a)
(ii)
7
Mark Scheme
A LEVEL – NOVEMBER 2005
(a) (i)
(ii)
(b)(i)
two square sections in correct positions, zero elsewhere …..
pulses in opposite directions ……………………………………
amplitude of second about twice amplitude of first …………..
…..………………………………………………………
4 ……...……………………………………………………………………
(ii) 1. idea of energy as product of A and energy per nucleon ……………
energy = (8.37 × 142 + 8.72 × 90) – 235 × 7.59
= 1189 +785 – 178
= 190 MeV ………(–1 for each a.e.) …………………………
2. energy = mc2 …………………………………………………………….
1 MeV = 1.6 × 10–13 J ………………………………………………….
energy = (190 × 1.6 × 10–13) / (3.0 × 108)2
= 3.4 × 10–28 kg …………………………………………..……
© University of Cambridge International Examinations 2005
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