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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level PHYSICS 9702/41 Paper 4 A Level Structured Questions May/June…

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Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level PHYSICS 9702/41 Paper 4 A Level Structured Questions May/June 2016 MARK SCHEME Maximum Mark: 100 Published 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 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components. ® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 7 printed pages. © UCLES 2016 [Turn over Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9702 41 1 (a) (gravitational) potential at infinity defined as/is zero B1 (gravitational) force attractive so work got out/done as object moves from infinity (so potential is negative) B1 [2] (b) (i) ∆E = m∆φ = 180 × (14 – 10) × 108 C1 = 7.2 × 1010 J A1 increase B1 [3] (ii) energy required = 180 × (10 – 4.4) × 108 or energy per unit mass = (10 – 4.4) × 108 C1 ½ × 180 × v2 = 180 × (10 – 4.4) × 108 or ½ × v2 = (10 – 4.4) × 108 C1 v = 3.3 × 104 m s–1 A1 [3] 2 (a) e.g. time of collisions negligible compared to time between collisions no intermolecular forces (except during collisions) random motion (of molecules) large numbers of molecules (total) volume of molecules negligible compared to volume of containing vessel or average/mean separation large compared with size of molecules any two B2 [2] 2 (b) (i) mass = 4.0 / (6.02 × 10 23) = 6.6 × 10–24 g or mass = 4.0 × 1.66 × 10–27 × 103 = 6.6 × 10–24 g B1 [1] 3 1 (ii) kT = m c 2 C1 2 2 3 1 × 1.38 × 10–23 × 300 = × 6.6 × 10–27 × c 2 2 2 c 2 = 1.88 × 106 (m2 s–2) C1 r.m.s. speed = 1.4 × 103 m s–1 A1 [3] © Cambridge International Examinations 2016 Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9702 41 3 (a) acceleration/force proportional to displacement (from fixed point) M1 acceleration/force and displacement in opposite directions A1 [2] (b) maximum displacements/accelerations are different B1 graph is curved/not a straight line B1 [2] (c) (i) ω = 2π / T and T = 0.8 s C1 ω = 7.9 rad s–1 A1 [2] (ii) a = (–)ω2 x = 7.852 × 1.5 × 10–2 C1 = 0.93 m s–2 or 0.94 m s–2 A1 [2] (iii) ∆E = ½ mω2 (x02 – x2) C1 = ½ × 120 × 10–3 × 7.852 × {(1.5 × 10–2)2 – (0.9 × 10–2)2} C1 = 5.3 × 10–4 J A1 [3] 4 (a) (i) product of speed and density M1 reference to speed in medium (and density of medium) A1 [2] (ii) α: ratio of reflected intensity and/to incident intensity B1 Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary) B1 [2] (b) in muscle: IM = I0 e–µx = I0 exp(–23 × 3.4 × 10–2) C1 IM / I0 = 0.457 C1 at boundary: α = (6.3 – 1.7)2 / (6.3 + 1.7)2 = 0.33 C1 IT /IM = [(1 – α) =] 0.67 C1 IT / I0 = 0.457 × 0.67 = 0.31 A1 [5] © Cambridge International Examinations 2016 Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9702 41 5 (a) (i) 1011 A1 [1] (ii) 0 0.25 0.50 0.75 1.00 1.25 1.50 1011 0110 1000 1110 0101 0011 0001 All 6 correct, 2 marks. 5 correct, 1 mark. A2 [2] (b) sketch: 6 horizontal steps of width 0.25 ms shown M1 steps at correct heights and all steps shown A1 steps shown in correct time intervals A1 [3] (c) increase sampling frequency/rate M1 so that step width/depth is reduced A1 increase number of bits (in each number) M1 so that step height is reduced A1 [4] 6 (a) sketch: from x = 0 to x = R, potential is constant at VS B1 smooth curve through (R, VS) and (2R, 0.5VS) B1 smooth curve continues to (3R, 0.33VS) B1 [3] (b) sketch: from x = 0 to x = R, field strength is zero B1 smooth curve through (R, E) and (2R, 0.25E) B1 smooth curve continues to (3R, 0.11E) B1 [3] 7 (a) line has non-zero intercept/line does not pass through origin B1 charge is/should be proportional to potential (difference) or charge is/should be zero when p.d. is zero (therefore there is a systematic error) B1 [2] © Cambridge International Examinations 2016 Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9702 41 (b) reasonable attempt at line of best fit B1 use of gradient of line of best fit clear M1 C = 2800 µF (allow ± 200 µF) A1 [3] (c) energy = ½ CV 2 or energy = ½ QV and C = Q / V C1 ∆ energy = ½ × 2800 × 10–6 × (9.02 – 6.02) C1 = 6.3 × 10–2 J A1 [3] 8 (a) op-amp has infinite/(very) large gain B1 op-amp saturates if V + ≠ V – M1 V+ is at earth potential so P (or V–) must be at earth A1 [3] (b) input resistance to op-amp is very large or current in R2 = current in R1 B1 VIN (– 0) = IR2 and (0) – VOUT = IR1 M1 VOUT / VIN = –R1 / R2 A1 [3] (c) relay coil connected between VOUT and earth M1 correct diode symbol connected between VOUT and coil or between coil and earth M1 correct polarity for diode (‘clockwise’) A1 [3] 9 (a) 0.10 mm B1 [1] (b) VH = (0.13 × 3.8) / (6.0 × 1028 × 0.10 × 10–3 × 1.60 × 10–19) C1 = 5.1 × 10–7 V A1 [2] 10 (a) (non-uniform) magnetic flux in core is changing M1 induces (different) e.m.f. in (different parts of) the core A1 (eddy) currents form in the core M1 which give rise to heating A1 [4] © Cambridge International Examinations 2016 Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9702 41 (b) as magnet falls, tube cuts magnetic flux M1 e.m.f./(eddy) currents induced in metal/aluminium (tube) A1 (eddy) current heating of tube M1 with energy taken from falling magnet A1 or (eddy) currents produce magnetic field (M1) that opposes motion of magnet (A1) so magnet B has acceleration g or magnet B has smaller acceleration/reaches terminal speed A1 [5] 11 (a) period = 15 ms C1 frequency (= 1 / T) = 67 Hz A1 [2] (b) zero A1 [1] (c) Ir.m.s. = I0 / √2 C1 = 0.53 A A1 [2] (d) energy = Ir.m.s.2 × R × t or ½ I02 × R × t or power = Ir.m.s.2 × R and energy = power × t C1 energy = 0.532 × 450 × 30 × 10–3 = 3.8 J A1 [2] 12 (a) (in a solid electrons in) neighbouring atoms are close together (and influence/interact with each other) M1 this changes their electron energy levels M1 (many atoms in lattice) cause a spread of energy levels into a band A1 [3] © Cambridge International Examinations 2016 Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9702 41 (b) photons of light give energy to electrons in valence band B1 electrons move into the conduction band M1 leaving holes in the valence band A1 these electrons and holes are charge carriers B1 increased number/increased current, hence reduced resistance B1 [5] 13 (a) e.g. background count (rate)/radiation multiple possible counts from each decay radiation emitted in all directions dead-time of counter (daughter) product unstable/also emits radiation self-absorption of radiation in sample or absorption in air/detector window three sensible suggestions, 1 each B3 [3] (b) A = A0 exp(– ln 2 × t / T½) 1.21 × 102 = 3.62 × 104 exp(– ln2 × 42.0 / T½) or 1.21 × 102 = 3.62 × 104 exp(–λ × 42.0) C1 T½ = 5.1 minutes (306 s) A1 [2] (c) discrete energy levels (in nuclei) B1 [1] © Cambridge International Examinations 2016

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