9702_w08_ms_2
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the October/November 2008 question paper
9702 PHYSICS
9702/02 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.
All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given mark s that fairly reflect the relevant k nowledge and sk ills demonstrated.
Mark schemes must be read in conjunction with the question papers and the report on the examination.
?CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the October/November 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
1 (a) (i) Q = It (allow any subject for the equation) B1
[1] (ii)
I B1
t B1 [2]
(allow 1 mark only if all three quoted)
(b)
(i) base unit of I is A
base unit of n is m–3 (not /m–3)
b ase
unit
of
S is m2
b ase
unit
of
q is A s (not C)
b
ase
unit
of
v is m s–1
(–1 for each error or omission) B3 [3]
(ii) A = m–3 m2 A s (m s–1)k M1
e.g. for m: 0 = –3 + 2 + k
k = 1 A1 [2]
2 (a) (i) v2 = 2as
v2 = 2 × 0.85 × 9.8 × 12.8 C1
v = 14.6 m s-1A1 [2]
(ii) time = 29.3 / 14.6 C1
= 2.0 s A1 [2]
(any acceleration scores 0 marks; allow 1 s.f.)
(b) either 60 km h–1 = 16.7 m s–1
or 14.6 m s–1 = 53 km h–1
or 22.1 m s–1 = 79.6 km h–1M1
so driving within speed limit A1
but reaction time is too long / too slow B1 [3]
3 (a) moment: force × perpendicular distance M1
of force from pivot / axis / point A1
couple: (magnitude of) one force × perpendicular distance M1
between the two forces A1 [4]
(penalise the ‘perpendicular’ omission once only)
(b)
(i) W× 4.8 = (12 × 84) + (2.5 × 72) C1
W = 250 N (248 N) A1 [2]
(ii) either friction at the pivot or small movement of weights B1 [1]
4 (a) (i) either force = e × (V / d) or E = V/d C1
= 1.6 × 10–19× (250 / 7.6 × 10–3) C1
=
5.3
× 10–15 N A1 [3]
(ii) either ?E K = e V or ?E K = Fd C1
=
1.6
× 10–19× 250 = 5.3 × 10–15× 7.6 × 10–3M1
=
4.0
× 10–17 J A0 [2]
(allow full credit for correct working via calculation of a and v)
(iii) either ?E K = ?mv2
4.0 × 10–17 = ? × 9.1 × 10–31×v2C1
v = 9.4 × 106 m s–1A1 [2]
or v2 = 2as and a = F/m
v2 = (2 × 5.3 × 10–15× 7.6 × 10–3)/(9.11 × 10–31) (C1)
v = 9.4 × 106 m s–1 (A1)
(b) speed depends on (electric) potential difference M2
(If states ?E K does not depend on uniformity of field, then
award 1 mark, treated as an M mark)
so speed always the same A1 [3]
5 (a) haphazard / random / erratic / zig-zag movement M1
[2]
of (smoke) particles (do not allow molecules / atoms) A1
(b)motion is due to unequal / unbalanced collision rate s (on different faces) B1
(unequal collision rate due to) random motion of (gas) molecules / atoms B1 [2]
(c) either collisions with air molecules average out M1
this prevents haphazard motion A1 [2]
or particle is more massive / heavier / has large inertia (M1)
collision s cause only small movements / accelerations (A1)
6 (a) wave incident at an edge / aperture / slit /(edge of) obstacle M1
bending / spreading of wave (into geometrical shadow) A1 [2]
(award 0/2 for bending at a boundary)
(b) (i) apparatus e.g. laser & slit / point source & slit / lamp and slit & slit
microwave source & slit
water / ripple tank, source & barrier B1
detector e.g. screen
aerial / microwave probe
strobe / lamp B1
what is observed B1 [3]
(ii)apparatus e.g. loudspeaker, and slit / edge B1
detector e.g. microphone & c.r.o. / ear B1
what is observed B1 [3]
7 (a) eith er V = I P B1
current in circuit = E / (P + Q) B1
V = EP / (P + Q) A0[2]
hence
or current is the same throughout the circuit (M1)
V / P = E / (P +Q) (A1)
V = EP / (P + Q) (A0)
hence
(b) (i) (as temperature rises), resistance of (thermistor) decreases M1 either resistance of parallel combination decreases or p.d. across 5 k ? resistor / thermistor decreases M1 p.d. across 2000 ? resistor / voltmeter reading increases A1 [3]
(ii) if R is the resistance of the parallel combination, either 3.6 = (2 × 6) / (2 + R ) or current in 2 k ? resistor = 1.8 mA C1 R = 1.33 k ? current in 5 k ? resistor = 0.48 mA C1
1.331 = 51 + T
1
current in thermistor = 1.32 mA C1
T = 1.82 k ? T = 2.4 / 1.32 = 1.82 k ? A1 [4]
8 (a) nucleus has constant probability of decay
M1 A1 [2] (allow 1 mark for ‘cannot predict which nucleus will decay next’) (b) (i) count rate / activity decreases B1 [1]
(ii) count rate fluctuates / is not smooth B1 [1] (c) eith er the (decay) curves are similar / same or curves indicate same half-life B1 [1]