# 9702_s12_ms_23

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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 9702/23 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 23 1 (a) displacement is a vector, distance is a scalar B1 displacement is straight line between two points / distance is sum of lengths moved / example showing difference B1 [2] (either one of the definitions for the second mark) (b) a body continues at rest or at constant velocity unless acted on by a resultant (external) force B1 [1] (c) (i) sum of T1 and T2 equals frictional force B1 these two forces are in opposite directions B1 [2] (allow for 1/2 for travelling in straight line hence no rotation / no resultant torque) (ii) 1. scale vector triangle with correct orientation / vector triangle with correct orientation both with arrows B1 scale given or mathematical analysis for tensions B1 [2] 2. T1 = 10.1 × 103 (± 0.5 × 103) N A1 T2 = 16.4 × 103 (± 0.5 × 103) N A1 [2] 2 (a) weight = 452 × 9.81 component down the slope = 452 × 9.81 × sin 14° M1 = 1072.7 = 1070 N A0 [1] (b) (i) F = ma C1 T – (1070 + 525) = 452 × 0.13 C1 T = 1650 (1653.76) N any forces missing 1/3 A1 [3] (ii) 1. s = ut + ½at2 hence 10 = 0 + ½ × 0.13t2 C1 t = [(2 × 10) / 0.13]1/2 = 12.4 or 12 s A1 [2] 2. v = (0 + 2 × 0.13 × 10)1/2 = 1.61 or 1.6 m s–1 A1 [1] (c) straight line from the origin B1 line down to zero velocity in short time compared to stage 1 B1 line less steep negative gradient B1 final velocity larger than final velocity in the first part – at least 2× B1 [4] 3 (a) V = h × A m=V×ρ B1 W=h×A×ρ×g B1 P=F/A B1 P = hρg P is proportional to h if ρ is constant (and g) B1 [4] (b) density changes with height B1 hence density is not constant with link to formula B1 [2] © University of Cambridge International Examinations 2012 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 23 4 (a) electric field strength is the force per unit positive charge (acting on a stationary charge) B1 [1] (b) (i) E = V / d C1 = 1200 / 14 × 10–3 = 8.57 × 104 V m–1 A1 [2] (ii) W = QV or W = F × d and therefore W = E × Q × d C1 = 3.2 × 10–19 × 1200 = 3.84 × 10–16 J A1 [2] (iii) ∆U = mgh C1 = 6.6 × 10–27 × 9.8 × 14 × 10–3 = 9.06 × 10–28 J A1 [2] (iv) ∆K = 3.84 × 10–16 – ∆U = 3.84 × 10–16 J A1 [1] (v) K = ½mv2 C1 v = [(2 × 3.8 × 10–16) / 6.6 × 10–27]1/2 = 3.4 × 105 m s–1 A1 [2] 5 (a) (i) sum of currents into a junction = sum of currents out of junction B1 [1] (ii) charge B1 [1] (b) (i) ΣE = ΣIR 20 – 12 = 2.0(0.6 + R) (not used 3 resistors 0/2) C1 R = 3.4 Ω A1 [2] (ii) P = EI C1 = 20 × 2 = 40 W A1 [2] (iii) P = I2R C1 P = (2)2 × (0.1 + 0.5 + 3.4) = 16 W A1 [2] (iv) efficiency = useful power / output power C1 24 / 40 = 0.6 or 12 × 2 / 20 × 2 or 60% A1 [2] © University of Cambridge International Examinations 2012 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 23 6 (a) (i) diffraction bending/spreading of light at edge/slit B1 this occurs at each slit B1 [2] (ii) constant phase difference between each of the waves B1 [1] (iii) (when the waves meet) the resultant displacement is the sum of the displacements of each wave B1 [1] (b) d sinθ = nλ n = d / λ = 1 / 450 × 103 × 630 × 10–9 C1 n = 3.52 M1 hence number of orders = 3 A1 [3] (c) λ blue is less than λ red M1 more orders seen A1 each order is at a smaller angle than for the equivalent red A1 [3] 7 (a) thin paper reduces count rate hence α B1 addition of 1 cm of aluminium causes little more count rate reduction hence only other radiation is γ B1 [2] (b) magnetic field perpendicular to direction of radiation B1 look for a count rate in expected direction / area if there were negatively charged radiation present. If no count rate recorded then β not present. B1 [2] © University of Cambridge International Examinations 2012
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