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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9702 PHYSICS 9702/22 Paper 2 (AS…
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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9702 PHYSICS 9702/22 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 22 1 (a) power = energy / time C1 = (force × distance / time) = kg m2 s–2 / s C1 = kg m2 s–3 A1 [3] (b) (i) units of L2: m2 and units of ρ : kg m–3 and units of v3: m3 s–3 C1 (C = P / L2 ρ v3) hence units of C: kg m2 s–3 m–2 kg–1 m3 m–3 s3 or any correct statement of component units M1 argument /discussion / cancelling leading to C having no units A1 [3] (ii) power available from wind = 3.5 × 105 × 100 / 55 (= 6.36 × 105) C1 v3 = 3.5 × 105 × 100 / (55 × 0.931 × (25)2 × 1.3) C1 v = 9.4 m s–1 A1 [3] (iii) not all kinetic energy of wind converted to kinetic energy of blades B1 generator / conversion to electrical energy not 100% efficient / heat produced in generator / bearings etc B1 [2] (there must be cause of loss and where located) 2 (a) force = rate of change of momentum A1 [1] (b) (i) horizontal line on graph from t = 0 to t about 2.0 s ± ½ square, a 0 M1 horizontal line at 3.5 on graph from 0 to 2 s A1 vertical line at t = 2.0 s to a = 0 or sharp step without a line B1 horizontal line from t = 2 s to t = 4 s with a = 0 B1 [4] (ii) straight line and positive gradient M1 starting at (0,0) A1 finishing at (2,16.8) A1 horizontal line from 16.8 M1 from 2.0 to 4.0 A1 [5] 3 (a) the point where (all) the weight (of the body) M1 is considered / seems to act A1 [2] (b) (i) vertical component of T (= 30 cos 40°) = 23 N A1 [1] (ii) the sum of the clockwise moments about a point equals the sum of the anticlockwise moments (about the same point) B1 [1] (iii) (moments about A): 23 × 1.2 (27.58) M1 = 8.5 × 0.60 + 1.2 × W M1 working to show W = 19 or answer of 18.73 (N) A1 [3] (iv) (M = W / g = 18.73 / 9.81 =) 1.9(09) kg A1 [1] © Cambridge International Examinations 2013 Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 22 (c) (for equilibrium) resultant force (and moment) = 0 B1 upward force does not equal downward force / horizontal component of T not balanced by forces shown B1 [2] 4 (a) apparatus: cell with particles e.g. smoke (container must be closed) B1 diagram showing suitable arrangement with light illumination and microscope B1 [2] (b) specks / flashes of light M1 in random motion A1 [2] (c) cannot see what is causing smoke to move hence molecules smaller than smoke particles (B1) continuous motion of smoke particles implies continuous motion of molecules (B1) random motion of particles implies random motion of molecules (B1) max. 2 [2] 5 (a) (i) v = fλ C1 λ = 40 / 50 = 0.8(0) m A1 [2] (ii) waves (travel along string and) reflect at Q / wall / fixed end B1 incident and reflected waves interfere / superpose B1 [2] (b) (i) nodes labelled at P, Q and the two points at zero displacement B1 antinodes labelled at the three points of maximum displacement B1 [2] (ii) (1.5λ for PQ hence PQ = 0.8 × 1.5) = 1.2 m A1 [1] (iii) T = 1 / f = 1/50 = 20 ms C1 5 ms is ¼ of cycle A1 horizontal line through PQ drawn on Fig. 5.2 B1 [3] 6 (a) charge = current × time B1 [1] (b) (i) P = V 2 / R C1 = (240)2 / 18 = 3200 W A1 [2] (ii) I = V / R = 240 / 18 = 13.3 A A1 [1] (iii) charge = It = 13.3 × 2.6 × 106 C1 = 3.47 × 107 C A1 [2] (iv) number of electrons = 3.47 × 107 / 1.6 × 10–19 (= 2.17 × 1026) C1 number of electrons per second = 2.17 × 1026 / 2.6 × 106 = 8.35 × 1019 A1 [2] © Cambridge International Examinations 2013 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 22 7 (a) (i) W = 206 and X = 82 A1 Y = 4 and Z = 2 A1 [2] (ii) mass-energy is conserved B1 mass on rhs is less because energy is released B1 [2] (b) not affected by external conditions/factors/environment B1 [1] or two examples temperature and pressure © Cambridge International Examinations 2013
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