# Dota

Topics: GCE Advanced Level, General Certificate of Secondary Education, Energy Pages: 8 (1275 words) Published: March 7, 2013
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

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MARK SCHEME for the May/June 2010 question paper for the guidance of teachers

9702 PHYSICS
9702/41 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.

CIE will not enter into discussions or correspondence in connection with these mark schemes.

CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

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Mark Scheme: Teachers’ version GCE AS/A LEVEL – May/June 2010 Section A

Syllabus 9702

Paper 41

1

(a) angle (subtended) at centre of circle (by) arc equal in length to radius (b) (i) point S shown below C (ii) (max) force / tension = weight + centripetal force centripetal force = mrω2 15 = 3.0/9.8 × 0.85 × ω2 ω = 7.6 rad s–1

B1 B1 B1 C1 C1 C1 A1 C1 A1 A1 B1 B1 B1 B1 B1

[2] [1]

[4]

2

(a) (i) 27.2 + 273.15 or 27.2 + 273.2 300.4 K (ii) 11.6 K (b) (i) ( is the) mean / average square speed (ii) ρ = Nm/V with N explained so, pV = 1/3 Nm and pV = NkT with k explained so mean kinetic energy / = ½m = 3/2 kT (c) (i) pV = nRT 2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 n = 68 mol (ii) mean kinetic energy = 3/2 kT = 3/2 × 1.38 × 10–23 × 290 = 6.0 × 10–21 J (iii) realisation that total internal energy is the total kinetic energy energy = 6.0 × 10–21 × 68 × 6.02 × 1023 = 2.46 × 105 J

[2] [1] [1]

[4]

C1 A1

[2]

C1 A1 C1 C1 A1 B1

[2]

[3] [1]

3

(a) (i) to-and-fro / backward and forward motion (between two limits)

(ii) no energy loss or gain / no external force acting / constant energy / constant amplitude B1 [1] (iii) acceleration directed towards a fixed point acceleration proportional to distance from the fixed point / displacement (b) acceleration is constant (magnitude) so cannot be s.h.m. B1 B1 M1 A1 [2]

[2]

Page 3 4

Mark Scheme: Teachers’ version GCE AS/A LEVEL – May/June 2010

Syllabus 9702

Paper 41 B1 B1 [2]

(a) ability to do work as a result of the position/shape, etc. of an object (b) (i) 1 ∆Egpe = GMm / r = (6.67 × 10–11 × {2 × 1.66 × 10–27}2) / (3.8 × 10–15) = 1.93 × 10–49 J = Qq / 4πε0r = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) = 6.06 × 10–14 J

C1 C1 A1 C1 C1 A1 B1 M1 A0 B1 B1 B1 M1 A1 B1 M1 A1 B1

[3]

2

∆Eepe

[3]

(ii) idea that 2EK = ∆Eepe – ∆Egpe EK = 3.03 × 10–14 J = (3.03 × 10–14) / 1.6 × 10–13 = 0.19 MeV (iii) fusion may occur / may break into sub-nuclear particles 5 (a) (i) VH depends on angle between (plane of) probe and B-field either VH max when plane and B-field are normal to each other or VH zero when plane and B-field are parallel or VH depends on sine of angle between plane and B-field (ii) 1 calculates VHr at least three times to 1 s.f. constant so valid or approx constant so valid or to 2 s.f., not constant so invalid straight line passes through origin

[2] [1]

[2]

[2] [1]

2

(b) (i) e.m.f. induced is proportional / equal to rate of change of (magnetic) flux (linkage) constant field in coil / flux (linkage) of coil does not change (ii) e.g. vary current (in wire) / switch current on or off / use a.c. current rotate coil move coil towards / away from wire (1 mark each, max 3) 6 (a) all four diodes correct to give output, regardless of polarity connected for correct polarity (b) NS / NP = VS / VP V0 = √2 × Vrms ratio = 9.0 /...