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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…
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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 9702/21 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 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. Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 21 1 (a) density = mass / volume B1 [1] (b) density of liquids and solids same order as spacing similar / to about 2× B1 density of gases much less as spacing much more or density of gases much lower hence spacing much more B1 [2] (c) (i) density = 68 / [50 × 600 × 900 × 10–9] C1 = 2520 (allow 2500) kg m–3 A1 [2] (ii) P = F / A C1 = 68 × 9.81 / [50 × 600 × 10–6] C1 = 2.2 × 104 Pa A1 [3] 2 (a) torque is the product of one of the forces and the distance between forces M1 the perpendicular distance between the forces A1 [2] (b) (i) torque = 8 × 1.5 = 12 N m A1 [1] (ii) there is a resultant torque / sum of the moments is not zero M1 (the rod rotates) and is not in equilibrium A1 [2] (c) (i) B × 1.2 = 2.4 × 0.45 C1 B = 0.9(0) N A1 [2] (ii) A = 2.4 – 0.9 = 1.5 N / moments calculation A1 [1] 3 (a) (i) horizontal velocity = 15 cos 60° = 7.5 m s–1 A1 [1] (ii) vertical velocity = 15 sin 60° = 13 m s–1 A1 [1] (b) (i) v2 = u2 + 2as s = (13)2 / (2 × 9.81) = 8.6(1) m A1 [1] using g = 10 then max. 1 (ii) t = 13 / 9.81 = 1.326 s or t = 9.95 / 7.5 = 1.327 s A1 [1] (iii) velocity = 6.15 / 1.33 M1 = 4.6 m s–1 A0 [1] (c) (i) change in momentum = 60 × 10–3 [–4.6 – 7.5] C1 = (–)0.73 N s A1 [2] (ii) final velocity / kinetic energy is less after the collision or relative speed of separation relative speed of approach M1 hence inelastic A0 [1] © University of Cambridge International Examinations 2011 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 21 4 (a) electrical potential energy (stored) when charge moved and gravitational potential energy (stored) when mass moved B1 due to work done in electric field and work done in gravitational field B1 [2] (b) work done = force × distance moved (in direction of force) and force = mg M1 mg × h or mg × ∆h A1 [2] (c) (i) 0.1 × mgh = ½ mv2 B1 0.1 × m × 9.81 × 120 = 0.5 × m × v2 B1 v = 15.3 m s–1 A0 [2] (ii) P = 0.5 m v2 / t C1 m / t = 110 × 103 / [0.25 × 0.5 × (15.3)2] C1 = 3740 kg s–1 A1 [3] 5 (a) ohm = volt / ampere B1 [1] (b) ρ = RA / l or unit is Ω m C1 units: V A–1 m2 m–1 = N m C–1 A–1 m2 m–1 C1 = kg m2 s–2 A–1 s–1 A–1 m2 m–1 = kg m3 s–3 A–2 A1 [3] (c) (i) ρ = [3.4 × 1.3 × 10–7] / 0.9 C1 = 4.9 × 10–7 (Ω m) A1 [2] (ii) max = 2.(0) V A1 min = 2 × (3.4 /1503.4) = 4.5 × 10–3 V A1 [2] (iii) P = V2 / R or P = VI and V = IR C1 = (2)2 / 3.4 = 1.18 (allow 1.2) W A1 [2] (d) (i) power in Q is zero when R = 0 B1 [1] (ii) power in Q = 0 / tends to zero as R = infinity B1 [1] © University of Cambridge International Examinations 2011 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 21 6 (a) extension is proportional to force (for small extensions) B1 [1] (b) (i) point beyond which (the spring) does not return to its original length when the load is removed B1 [1] (ii) gradient of graph = 80 N m–1 A1 [1] (iii) work done is area under graph / ½ Fx / ½ kx2 C1 = 0.5 × 6.4 × 0.08 = 0.256 (allow 0.26) J A1 [2] (c) (i) extension = 0.08 + 0.04 = 0.12 m A1 [1] (ii) spring constant = 6.4 / 0.12 = 53.3 N m–1 A1 [1] 7 (a) nuclei with the same number of protons B1 and a different number of neutrons B1 [2] (b) (i) (mass + energy) (taken together) is conserved (B1) momentum is conserved (B1) one point required max. 1 B1 [1] (ii) a = 1 and b = 0 B1 x = 56 B1 y = 92 B1 [3] (c) proton number = 90 B1 nucleon number = 235 B1 [2] © University of Cambridge International Examinations 2011
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