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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the…
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 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 May/June 2012 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 – May/June 2012 9702 21 1 (a) (i) V units: m3 (allow metres cubed or cubic metres) A1 [1] (ii) Pressure units: kg m s–2 / m2 (allow use of P = ρgh) M1 Units: kg m–1 s–2 A0 [1] (b) V / t units: m3 s–1 B1 Clear substitution of units for P, r4 and l M1 πP r 4 kg m −1 s −2 m 4 C= = 8V t −1 l m3 s −1 m Units: kg m–1 s–1 A1 [3] (8 or π in final answer –1. Use of dimensions max 2/3) 2 (a) (i) v = u + at C1 = 4.23 + 9.81 × 1.51 M1 = 19.0(4) m s–1 (Allow 2 s.f.) A0 [2] –1 (Use of –g max 1/2. Use of g = 10 max 1/2. Allow use of 9.8. Allow 19 m s ) (ii) either s = ut + ½ at2 (or v2 = u2 + 2as etc.) = 4.23 × 1.51 + 0.5 × 9.81 × (1.51)2 C1 = 17.6 m (or 17.5 m) A1 [2] (Use of –g here wrong physics (0/2)) (b) (i) F = ∆P / ∆t need idea of change in momentum C1 = [0.0465 × (18.6 + 19)] / 12.5 × 10–3 C1 = 140 N A1 (Use of – sign max 2/4. Ignore –ve sign in answer) Direction: upwards B1 [4] (ii) h = ½ × (18.6)2 / 9.81 C1 = 17.6 m (2 s.f. –1) A1 [2] (Use of 19 m s–1, 0/2 wrong physics) (c) either kinetic energy of the ball is not conserved on impact or speed before impact is not equal to speed after hence inelastic B1 [1] 3 (a) Resultant force (and resultant torque) is zero B1 Weight (down) = force from/due to spring (up) B1 [2] (b) (i) 0.2, 0.6, 1.0 s (one of these) A1 [1] (ii) 0, 0.8 s (one of these) A1 [1] (iii) 0.2, 0.6, 1.0 s (one of these) A1 [1] © University of Cambridge International Examinations 2012 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 21 (c) (i) Hooke’s law: extension is proportional to the force (not mass) B1 Linear/straight line graph hence obeys Hooke’s law B1 [2] (ii) Use of the gradient (not just F = kx) C1 K = (0.4 × 9.8) / 15 × 10–2 M1 = 26(.1) N m–1 A0 [2] (iii) either energy = area to left of line or energy = ½ ke2 C1 = ½ × [(0.4 × 9.8) / 15 × 10–2] × (15 × 10–2)2 C1 = 0.294 J (allow 2 s.f.) A1 [3] 4 (a) (i) R = V2 / P or P = IV and V = IR C1 = (220)2 / 2500 = 19.4 Ω (allow 2 s.f.) A1 [2] (ii) R = ρl / A C1 l = [19.4 × 2.0 × 10–7] / 1.1 × 10–6 C1 = 3.53 m (allow 2 s.f.) A1 [3] (b) (i) P = 625, 620 or 630 W A1 [1] (ii) R needs to be reduced C1 Either length ¼ of original length or area 4× greater or diameter 2× greater A1 [2] 5 (a) (i) sum of e.m.f.’s = sum of p.d.’s around a loop/circuit B1 [1] (ii) energy B1 [1] (b) (i) 2.0 = I × (4.0 + 2.5 + 0.5) C1 I = 0.286 A (allow 2 s.f.) A1 [2] (If total resistance is not 7 Ω, 0/2 marks) (ii) R = [0.90 / 1.0] × 4 (= 3.6) C1 V = I R = 0.286 × 3.6 = 1.03 V A1 [2] (If factor of 0.9 not used, then 0/2 marks) (iii) E = 1.03 V A1 [1] (iv) either no current through cell B or p.d. across r is zero B1 [1] 6 (a) (i) coherence: constant phase difference M1 between (two) waves A1 [2] (ii) path difference is either λ or nλ or phase difference is 360° or n × 360° or n2π rad B1 [1] © University of Cambridge International Examinations 2012 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 21 (iii) path difference is either λ/2 or (n + ½) λ or phase difference is odd multiple of either 180° or π rad B1 [1] (iv) w = λD / a C1 = [630 × 10–9 × 1.5] / 0.45 × 10–3 C1 = 2.1 × 10–3 m A1 [3] (b) no change to dark fringes B1 no change to separation/fringe width B1 bright fringes are brighter/lighter/more intense B1 [3] 7 (a) (i) 2 protons and 2 neutrons B1 [1] (ii) e.g. positively charged 2e mass 4u constant energy absorbed by thin paper or few cm of air (3 cm → 8 cm) (not low penetration) highly ionizing deflected in electric/magnetic fields (One mark for each property, max 2) B2 [2] (b) mass-energy is conserved B1 difference in mass ‘changed’ into a form of energy B1 energy in the form of kinetic energy of the products / γ-radiation photons / e.m. radiation B1 [3] © University of Cambridge International Examinations 2012
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