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CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2015 series 9702 PHYSICS…

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CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2015 series 9702 PHYSICS 9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100 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 October/November 2015 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. Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9702 41 Section A 1 (a) (i) gravitational force provides/is the centripetal force B1 GMmS / x2 = mSv2 / x (allow x or r; allow m or mS) M1 EK = ½mSv2 and clear algebra leading to EK = GMmS / 2x A1 [3] (ii) EP = – GMmS / x (sign essential) B1 [1] (iii) ET = EK + EP = GMmS / 2x – GMmS / x C1 = – GMmS / 2x (allow ECF from (a)(ii)) A1 [2] (b) (i) decreases B1 [1] (ii) decreases B1 [1] (iii) decreases B1 [1] (iv) increases B1 [1] (for answers in (b) allow ECF from (a)(iii)) 2 (a) obeys the equation pV = nRT or pV / T = constant M1 all symbols explained; T in kelvin/thermodynamic temperature A1 [2] (b) (i) temperature rise = 48 K A1 [1] (ii) c2 ∝ T or equivalent C1 c2 = (353 / 305) × 1.9 × 106 C1 cr.m.s. = 1480 m s–1 A1 [3] 3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1 plus work done on the system or minus work done by the system B1 [2] (b) (i) either volume decreases so work done on the system or small volume change so work done on system negligible M1 (thermal) energy absorbed to break lattice structure M1 internal energy increases A1 [3] (ii) gas expands so work done by gas (against atmosphere) M1 no time for thermal energy to enter or leave the gas M1 internal energy decreases A1 [3] 4 (a) free: (body oscillates) without any loss of energy/no resistive forces/no external forces applied B1 forced: continuous energy input (required)/body is made to vibrate by an (external) periodic force/driving oscillator B1 [2] © Cambridge International Examinations 2015 Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9702 41 (b) (i) idea of resonance B1 maximum amplitude at natural frequency B1 frequency = 2.1 Hz (allow 2.08 to 2.12 Hz) B1 [3] (ii) peak not very sharp/amplitude not infinite so frictional forces are present B1 [1] (c) v = ωx0 = 2π × 2.1 × 4.7 × 10–2 (allow ECF from (b)(i)) C1 = 0.62 m s–1 A1 [2] 5 (a) (i) force proportional to the product of the two/point charges B1 and inversely proportional to the square of their separation B1 [2] (ii) 1. force radially away from sphere/to right/to east B1 [1] 2. (maximum) at/on surface of sphere or x = r B1 [1] 3. F ∝ 1 / x2 or F = q1q2 / (4πε 0x2) C1 ratio = 16 A1 [2] (b) E = q / (4πε0x2) or E ∝ q C1 maximum charge = (2.0 / 1.5) × 6.0 × 10–7 C1 = 8.0 × 10–7 C additional charge = 2.0 × 10–7 C A1 [3] 6 (a) (i) force = mg M1 along the direction of the field/of the motion A1 [2] (ii) no force B1 [1] (b) (i) force due to E-field downwards so force due to B-field upwards B1 into the plane of the paper B1 [2] (ii) force due to magnetic field = Bqv B1 force due to electric field = Eq B1 (use of FB and FE not explained, allow 1/2) forces are equal (and opposite) so Bv = E or Eq = Bqv so E = Bv B1 [3] (c) sketch: smooth curved path M1 in ‘upward’ direction A1 [2] 7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1 for emission of electrons from a surface A1 [2] (reference to light/UV rather than e.m. radiation, allow 1/2) © Cambridge International Examinations 2015 Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9702 41 (b) EMAX corresponds to electron emitted from surface B1 electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2] (c) (i) 1/λ0 = 1.85 × 106 (allow 1.82 to 1.88) C1 f0 = c / λ0 = 3.00 × 108 × 1.85 × 106 = 5.55 × 1014 Hz A1 [2] (ii) Φ = hf0 = 6.63 × 10–34 × 5.55 × 1014 (allow ECF from (c)(i)) C1 = 3.68 × 10–19 J A1 [2] (d) sketch: straight line with same gradient M1 intercept between 1.0 and 1.5 A1 [2] 8 (a) nucleus: small central part/core of an atom B1 nucleon: proton or a neutron B1 particle contained within a nucleus B1 [3] (b) (i) 1. decay constant = ln 2 / (3.8 × 24 × 3600) C1 = 2.1 × 10–6 s–1 A1 [2] 2. A = λN 97 = 2.1 × 10–6 × N C1 N = 4.6 × 107 A1 [2] (ii) 1.0 m3 contains (6.02 × 1023) / (2.5 × 10–2) air molecules C1 ratio = (4.6 × 107 × 2.5 × 10–2) / (6.02 × 1023) = 1.9 × 10–18 A1 [2] © Cambridge International Examinations 2015 Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9702 41 Section B 9 (a) (i) (+) 3.0 V B1 [1] (ii) potential = 6.0 × {2.0 / (2.0 + 2.8)} C1 = 2.5 V A1 [2] (iii) potential = 6.0 × {2.0 / (2.0 + 1.8)} = 3.2 V A1 [1] (b) at 10 °C, VA VB M1 VOUT is –9.0 V (allow “negative saturation”) A1 at 20 °C, VOUT is +9.0 V B1 (if 20 °C considered initially, mark as M1,A1,B1) sudden switch (from –9 V to +9 V) when VA = VB B1 [4] 10 (a) sharpness: clarity of edges/resolution (of image) B1 contrast: difference in degree of blackening (of structures) B1 [2] (b) (i) X-rays produced when (high speed) electrons hit target/anode B1 either electrons have been accelerated through 80 kV or electrons have (kinetic) energy of 80 keV B1 [2] (ii) IT / I = e–3.0 × 1.4 C1 = 0.015 A1 [2] (c) for good contrast, µx or eµx or e–µx must be very different B1 µx or eµx or e–µx for bone and muscle will be different than that for muscle M1 so good contrast A1 [3] 11 (a) frequency of carrier wave varies M1 in synchrony with the displacement of the signal/information wave A1 [2] (b) (i) 5.0 V A1 [1] (ii) 720 kHz A1 [1] (iii) 780 kHz A1 [1] (iv) 7500 A1 [1] © Cambridge International Examinations 2015 Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9702 41 12 (a) (i) (gradual) loss of power/intensity/amplitude (not “signal”) B1 [1] (ii) e.g. noise can be eliminated (not “there is no noise”) M1 because pulses can be regenerated A1 e.g. much greater data handling/carrying capacity M1 because many messages can be carried at the same time/greater bandwidth A1 e.g. more secure (M1) because it can be encrypted (A1) e.g. error checking (M1) because extra information/parity bit can be added (A1) [4] (allow any two sensible suggestions with ‘state’ M1 and ‘explain’ A1) (b) attenuation = 10 lg (145 / 29) (= 7.0) C1 attenuation per unit length = 7.0 / 36 = 0.19 dB km–1 A1 [2] © Cambridge International Examinations 2015

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