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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 9702/42 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 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 42 Section A 1 (a) force proportional to product of masses and inversely proportional to square of separation (do not allow square of distance/radius) M1 either point masses or separation @ size of masses A1 [2] (b) (i) ω = 2π / (27.3 × 24 × 3600) or 2π / (2.36 x 106) M1 = 2.66 × 10–6 rad s–1 A0 [1] (ii) GM = r3ω2 or GM = v2r C1 M = (3.84 × 105 × 103)3 × (2.66 × 10–6)2 / (6.67 × 10–11) M1 = 6.0 × 1024 kg A0 [2] (special case: uses g = GM/r2 with g = 9.81, r = 6.4 × 106 scores max 1 mark) (c) (i) grav. force = (6.0 × 1024) × (7.4 × 1022) × (6.67 × 10–11)/(3.84 × 108)2 C1 = 2.0 × 1020 N (allow 1 SF) A1 [2] (ii) either ∆EP = Fx because F constant as x ! radius of orbit B1 ∆EP = 2.0 × 1020 × 4.0 × 10–2 C1 = 8.0 × 1018 J (allow 1 SF) A1 [3] or ∆EP = GMm/r1 – GMm/r2 C1 Correct substitution B1 8.0 × 1018 J A1 (∆EP = GMm/r1 + GMm/r2 is incorrect physics so 0/3) 2 (a) energy = ½mω2a2 and ω = 2πf C1 = ½ × 37 × 10–3 × (2π × 3.5)2 × (2.8 × 10–2)2 M1 = 7.0 × 10–3 J A0 [2] (allow 2π × 3.5 shown as 7π) Energy = ½ mv2 and v = rω (C1) Correct substitution (M1) Energy = 7.0 × 10–3 J (A0) (b) EK = EP ½mω2 (a2 – x2) = ½mω2x2 or EK or EP = 3.5 mJ C1 x = a/√2 = 2.8 /√2 or EK = ½mω2 (a2 – x2) or EP = ½mω x 2 2 C1 = 2.0 cm A1 [3] (EK or EP = 7.0 mJ scores 0/3) Allow: k = 17.9 (C1) E = ½ kx2 (C1) x = 2.0 cm (A1) © University of Cambridge International Examinations 2012 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 42 (c) (i) graph: horizontal line, y-intercept = 7.0 mJ with end-points of line at +2.8 cm and –2.8 cm B1 [1] (ii) graph: reasonable curve B1 with maximum at (0,7.0) end-points of line at (–2.8, 0) and (+2.8, 0) B1 [2] (iii) graph: inverted version of (ii) M1 with intersections at (–2.0, 3.5) and (+2.0, 3.5) A1 [2] (Allow marks in (iii), but not in (ii), if graphs K & P are not labelled) (d) gravitational potential energy B1 [1] 3 (a) sum of potential energy and kinetic energy of atoms/molecules/particles M1 reference to random (distribution) A1 [2] (b) (i) as lattice structure is ‘broken’/bonds broken/forces between molecules reduced (not molecules separate) B1 no change in kinetic energy, potential energy increases M1 internal energy increases A1 [3] (ii) either molecules/atoms/particles move faster/ c2 is increasing or kinetic energy increases with temperature (increases) B1 no change in potential energy, kinetic energy increases M1 internal energy increases A1 [3] 4 (a) (i) as r decreases, energy decreases/work got out (due to) M1 attraction so point mass is negatively charged A1 [2] (ii) electric potential energy = charge × electric potential B1 electric field strength is potential gradient B1 field strength = gradient of potential energy graph/charge A0 [2] (b) tangent drawn at (4.0, 14.5) B1 gradient = 3.6 × 10–24 A2 (for ±0.3 allow 2 marks, for ±0.6 allow 1 mark) field strength = (3.6 × 10–24) / (1.6 × 10–19) = 2.3 × 10–5 V m–1 (allow ecf from gradient value) A1 [4] –5 –1 (one point solution for gradient leading to 2.3 × 10 Vm scores 1 mark only) © University of Cambridge International Examinations 2012 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 42 5 (a) (long) straight conductor carrying current of 1 A M1 current/wire normal to magnetic field M1 (for flux density 1 T,) force per unit length is 1 N m–1 A1 [3] (b) (i) (originally) downward force on magnet (due to current) B1 by Newton’s third law (allow “N3”) M1 upward force on wire A1 [3] (ii) F = BIL 2.4 × 10–3 × 9.8 = B × 5.6 × 6.4 × 10–2 C1 B = 0.066 T (need 2 SF) A1 [2] (g missing scores 0/2, but g = 10 leading to 0.067T scores 1/2) (c) new reading is 2.4√2 g C1 either changes between +3.4 g and –3.4 g or total change is 6.8 g A1 [2] 6 (a) oil drop charged by friction/beta source B1 between parallel metal plates B1 plates are horizontal (1) adjustable potential difference/field between plates B1 until oil drop is stationary B1 mg = q × V/d B1 symbols explained (1) oil drop viewed through microscope (1) m determined from terminal speed of drop (when p.d. is zero) (1) (any two extras, 1 each) B2 [7] (b) 3.2 × 10–19 C A1 [1] 7 (a) minimum energy to remove an electron from the metal/surface B1 [1] (b) gradient = 4.17 × 10–15 (allow 4.1 → 4.3) C1 h = 4.15 × 10–15 × 1.6 × 10–19 or h = 4.1 to 4.3 × 10–15 eV s A1 –34 = 6.6 × 10 J s A0 [2] (c) graph: straight line parallel to given line with intercept at any higher frequency B1 intercept at between 6.9 × 1014 Hz and 7.1 × 1014 Hz B1 [3] © University of Cambridge International Examinations 2012 Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 42 8 (a) nuclei having same number of protons/proton (atomic) number B1 different numbers of neutrons/neutron number B1 [2] (allow second mark for nucleons/nucleon number/mass number/atomic mass if made clear that same number of protons/proton number) (b) probability of decay per unit time is the decay constant C1 λ = ln 2 / t½ = 0.693 / (52 × 24 × 3600) C1 = 1.54 × 10–7 s–1 A1 [3] (c) (i) A = A0 exp(–λt) 7.4 × 106 = A0 exp(–1.54 × 10–7 × 21 × 24 × 3600) C1 A0 = 9.8 × 106 Bq A1 [2] (alternative method uses 21 days as 0.404 half-lives) (ii) A = λN and mass = N × 89 / NA C1 mass = (9.8 × 106 × 89) / (1.54 × 10–7 × 6.02 × 1023) = 9.4 × 10–9 g A1 [2] © University of Cambridge International Examinations 2012 Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 42 Section B 9 (a) e.g. infinite input impedance/resistance zero output impedance/resistance infinite (open loop) gain infinite bandwidth infinite slew rate (any four, one mark each) B4 [4] (b) graph: square wave M1 180° phase change A1 amplitude 5.0 V A1 [3] (c) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 diodes identified correctly A1 [3] (special case: if diode symbol, not LED symbol, allow 2nd and 3rd marks to be scored) 10 (a) e.g. beam is divergent/obeys inverse square law absorption (in block) scattering (of beam in block) reflection (at boundaries) (any two sensible suggestions, 1 each) B2 [2] (b) (i) I = I0 exp(–µx) C1 I0/I = exp(0.27 × 2.4) = 1.9 A1 [2] (ii) I0/I = exp(0.27 × 1.3) × exp(3.0 × 1.1) C1 = 1.42 × 27.1 = 38.5 A1 [2] (c) either much greater absorption in bone than in soft tissue or Io / I much greater for bone than soft tissue B1 [1] 11 (a) (i) loss of (signal) power B1 [1] (ii) unwanted power (on signal) M1 that is random A1 [2] (b) for digital, only the ‘high’ and the ‘low’ / 1 and 0 are necessary M1 variation between ‘highs’ and ‘lows’ caused by noise not required A1 [2] (c) attenuation = 10 lg(P2 / P1) C1 either 195 = 10 lg({2.4 × 103} / P) or –195 = 10 lg(P / 2.4 × 103) C1 P = 7.6 × 10–17 W A1 [3] © University of Cambridge International Examinations 2012 Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 42 12 (a) (i) modulator B1 [1] (ii) serial-to-parallel converter (accept series-to-parallel converter) B1 [1] (b) (i) enables one aerial to be used for transmission and receipt of signals A1 [1] (ii) all bits for one number arrive at one time B1 bits are sent out one after another B1 [2] © University of Cambridge International Examinations 2012

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