Cambridge A Level Physics 9702 — 2005 May/June Paper 2 · Variant 1

This document consists of 16 printed pages. SP (CW/AR) S92061/3.1 © UCLES 2005 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level PHYSICS Paper 2 9702/02 May/June 2005 1 hour Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen in the spaces provided on the Question Paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. You may lose marks if you do not show your working or if you do not use appropriate units. DO NOT WRITE IN THE BARCODE. DO NOT WRITE IN THE GREY AREAS BETWEEN THE PAGES. For Examiner’s Use 1 2 3 4 5 6 7 8 Total Candidate Name Centre Number Candidate Number

2 9702/02/M/J/05 Data speed of light in free space, c = 3.00 × 108 m s–1 permeability of free space,
0 = 4 × 10–7 H m–1 permittivity of free space, 0 = 8.85 × 10–12 F m–1 elementary charge, e = 1.60 × 10–19 C the Planck constant, h = 6.63 × 10–34 J s unified atomic mass constant, u = 1.66 × 10–27 kg rest mass of electron, me = 9.11 × 10–31 kg rest mass of proton, mp = 1.67 × 10–27 kg molar gas constant, R = 8.31 J K–1 mol–1 the Avogadro constant, NA = 6.02 × 1023 mol–1 the Boltzmann constant, k = 1.38 × 10–23 J K–1 gravitational constant, G = 6.67 × 10–11 N m2 kg–2 acceleration of free fall, g = 9.81 m s–2 © UCLES 2005

3 9702/02/M/J/05 [Turn over Formulae uniformly accelerated motion, s = ut +
 at 2 v2 = u2 + 2as work done on/by a gas, W = pV gravitational potential, φ = – simple harmonic motion, a = –
2x velocity of particle in s.h.m., v = v0 cos
t v = ±
√(x2 0 – x2) resistors in series, R = R1 + R2 + . . . resistors in parallel, 1/R = 1/R1 + 1/R2 + . . . electric potential, V = capacitors in series, 1/C = 1/C1 + 1/C2 + . . . capacitors in parallel, C = C1 + C2 + . . . energy of charged capacitor, W =
 QV alternating current/voltage, x = x0 sin
t hydrostatic pressure, p = qgh pressure of an ideal gas, p =
 <c2> radioactive decay, x = x0 exp(– t) decay constant,  = critical density of matter in the Universe, q0 = equation of continuity, Av = constant Bernoulli equation (simplified), p1 +
 qv2 1 = p2 +
 qv2 2 Stokes’ law, F = Arv Reynolds’ number, Re = drag force in turbulent flow, F = Br2qv2 qvr  3H0 2 8G 0.693 t
 Nm V Q 40r Gm r © UCLES 2005

4 9702/02/M/J/05 Answer all the questions in the spaces provided. 1 Make estimates of the following quantities. (a) the speed of sound in air speed = … [1] (b) the density of air at room temperature and pressure density = … [1] (c) the mass of a protractor mass = … [1] (d) the volume, in cm3, of the head of an adult person volume = … cm3 [1] For Examiner’s Use © UCLES 2005

5 9702/02/M/J/05 [Turn over 2 The Brownian motion of smoke particles in air may be observed using the apparatus shown in Fig. 2.1. Fig. 2.1 (a) Describe what is seen when viewing a smoke particle through the microscope. … … …[2] (b) Suggest and explain what difference, if any, would be observed in the movement of smoke particles when larger smoke particles than those observed in (a) are viewed through the microscope. … … …[2] light microscope smoke cell For Examiner’s Use © UCLES 2005

6 9702/02/M/J/05 3 A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g. The block is suspended from strings so that it is free to move in a vertical plane. The bullet buries itself in the block. The block and bullet rise together through a vertical distance of 8.6 cm, as shown in Fig. 3.1. Fig. 3.1 (a) (i) Calculate the change in gravitational potential energy of the block and bullet. change = … J [2] (ii) Show that the initial speed of the block and the bullet, after they began to move off together, was 1.3 m s–1. [1] 8.6 cm bullet wood block For Examiner’s Use © UCLES 2005

7 9702/02/M/J/05 [Turn over (b) Using the information in (a)(ii) and the principle of conservation of momentum, determine the speed of the bullet before the impact with the block. speed = … m s–1 [2] (c) (i) Calculate the kinetic energy of the bullet just before impact. kinetic energy = … J [2] (ii) State and explain what can be deduced from your answers to (c)(i) and (a)(i) about the type of collision between the bullet and the block. … … …[2] For Examiner’s Use © UCLES 2005

8 9702/02/M/J/05 4 A glass fibre of length 0.24 m and area of cross-section 7.9 ×10–7m2 is tested until it breaks. The variation with load F of the extension x of the fibre is shown in Fig. 4.1. Fig. 4.1 (a) State whether glass is ductile, brittle or polymeric. …[1] (b) Use Fig. 4.1 to determine, for this sample of glass, (i) the ultimate tensile stress, ultimate tensile stress = … Pa [2] 0 1 2 3 4 0 20 40 60 80 x / 10–4 m F / N For Examiner’s Use © UCLES 2005

9 9702/02/M/J/05 [Turn over (ii) the Young modulus, Young modulus = … Pa [3] (iii) the maximum strain energy stored in the fibre before it breaks. maximum strain energy = … J [2] (c) A hard ball and a soft ball, with equal masses and volumes, are thrown at a glass window. The balls hit the window at the same speed. Suggest why the hard ball is more likely than the soft ball to break the glass window. … … … …[3] For Examiner’s Use © UCLES 2005

10 9702/02/M/J/05 5 (a) Explain what is meant by the diffraction of a wave. … … …[2] (b) Light of wavelength 590 nm is incident normally on a diffraction grating having 750 lines per millimetre. The diffraction grating formula may be expressed in the form d sinθ = nλ. (i) Calculate the value of d, in metres, for this grating. d = … m [2] (ii) Determine the maximum value of n for the light incident normally on the grating. maximum value of n = … [2] For Examiner’s Use © UCLES 2005

11 9702/02/M/J/05 [Turn over (iii) Fig. 5.1 shows incident light that is not normal to the grating. Fig. 5.1 Suggest why the diffraction grating formula, d sinθ = nλ, should not be used in this situation. … …[1] (c) Light of wavelengths 590 nm and 595 nm is now incident normally on the grating. Two lines are observed in the first order spectrum and two lines are observed in the second order spectrum, corresponding to the two wavelengths. State two differences between the first order spectrum and the second order spectrum. 1. … … 2. … …[2] grating incident light diffracted light For Examiner’s Use © UCLES 2005

12 9702/02/M/J/05 6 Two parallel metal plates P and Q are situated 8.0 cm apart in air, as shown in Fig. 6.1. Fig. 6.1 Plate Q is earthed and plate P is maintained at a potential of +160 V. (a) (i) On Fig. 6.1, draw lines to represent the electric field in the region between the plates. [2] (ii) Show that the magnitude of the electric field between the plates is 2.0 ×103V m–1. [1] P Q +160 V 8.0 cm For Examiner’s Use © UCLES 2005

13 9702/02/M/J/05 [Turn over (b) A dust particle is suspended in the air between the plates. The particle has charges of +1.2 ×10–15C and –1.2 ×10–15C near its ends. The charges may be considered to be point charges separated by a distance of 2.5 mm, as shown in Fig. 6.2. Fig. 6.2 The particle makes an angle of 35° with the direction of the electric field. (i) On Fig. 6.2, draw arrows to show the direction of the force on each charge due to the electric field. [1] (ii) Calculate the magnitude of the force on each charge due to the electric field. force = … N [2] (iii) Determine the magnitude of the couple acting on the particle. couple = … N m [2] (iv) Suggest the subsequent motion of the particle in the electric field. … … …[2] 35° +1.2 x 10–15 C 2.5 mm direction of electric field –1.2 x 10–15 C For Examiner’s Use © UCLES 2005

14 9702/02/M/J/05 7 (a) Define the resistance of a resistor. … …[1] (b) In the circuit of Fig. 7.1, the battery has an e.m.f. of 3.00 V and an internal resistance r. R is a variable resistor. The resistance of the ammeter is negligible and the voltmeter has an infinite resistance. Fig. 7.1 The resistance of R is varied. Fig. 7.2 shows the variation of the power P dissipated in R with the potential difference V across R. Fig. 7.2 0.8 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.9 1.0 1.1 1.2 P / W V / V A V r 3.00 V R For Examiner’s Use © UCLES 2005

15 9702/02/M/J/05 [Turn over (i) Use Fig. 7.2 to determine 1. the maximum power dissipation in R, maximum power = … W 2. the potential difference across R when the maximum power is dissipated. potential difference = … V [1] (ii) Hence calculate the resistance of R when the maximum power is dissipated. resistance = … Ω[2] (iii) Use your answers in (i) and (ii) to determine the internal resistance r of the battery. r = … Ω[3] (c) By reference to Fig. 7.2, it can be seen that there are two values of potential difference V for which the power dissipation is 1.05 W. State, with a reason, which value of V will result in less power being dissipated in the internal resistance. … … … …[3] For Examiner’s Use © UCLES 2005

16 9702/02/M/J/05 8 Fig. 8.1 shows the position of Neptunium-231 (231 93Np) on a diagram in which nucleon number (mass number) A is plotted against proton number (atomic number) Z. Fig. 8.1 (a) Neptunium-231 decays by the emission of an α-particle to form protactinium. On Fig. 8.1, mark with the symbol Pa the position of the isotope of protactinium produced in this decay. [1] (b) Plutonium-243 (243 94Pu) decays by the emission of a β-particle (an electron). On Fig. 8.1, show this decay by labelling the position of Plutonium-243 as Pu and the position of the daughter product as D. [2] 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 224 226 228 230 232 234 236 238 240 242 244 246 248 250 A Z Np For Examiner’s Use © UCLES 2005 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary and Advanced Level MARK SCHEME for the June 2005 question paper 9702 PHYSICS 9702/02 Paper 2 (Structured), maximum raw mark 60 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. This shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Grade thresholds for Syllabus 9702 (Physics) in the June 2005 examination. minimum mark required for grade: maximum mark available A B E Component 2 60 43 39 26 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

June 2005 GCE A AND AS LEVEL MARK SCHEME MAXIMUM MARK: 60 SYLLABUS/COMPONENT: 9702/02 PHYSICS Paper 2 (Structured)

Page 1 Mark Scheme Syllabus Paper A and AS LEVEL – June 2005 9702 2 © University of Cambridge International Examinations 2005 1 (a) allow 100 m s-1 → 900 m s-1 B1 [1] (b) allow 0.5 kg m-3 → 1.5 kg m-3 B1 [1] (c) allow 5 g → 50 g B1 [1] (d) allow 2 × 103 cm3 → 9 × 103 cm3 B1 [1] 2 (a) speck of light B1 that moves haphazardly/randomly/jerkily/etc. B1 [2] (b) randomness of collisions would be ‘averaged out’ B1 so less (haphazard) movement B1 [2] (do not allow ‘more massive so less movement’) 3 (a) (i) ∆Ep = mg∆h C1 = 0.602 × 9.8 × 0.086 = 0.51 J A1 [2] (do not allow g = 10, m = 0.600 or answer 0.50 J) (ii) v2 = (2gh =) 2 × 9.8 × 0.086 or (2 x 0.51)/0.602 M1 v = 1.3 (m s-1) A0 [1] (b) 2 × V = 602 × 1.3 (allow 600) C1 V = 390 m s-1 A1 [2] (c) (i) Ek = ½mv2 C1 = ½ × 0.002 × 3902 = 152 J or 153 J or 150 J A1 [2] (ii) Ek not the same/changes or Ek before impact>Ek after/Ep after M1 so must be inelastic collision A1 [2] (allow 1 mark for ‘bullet embeds itself in block’ etc.) 4 (a) brittle B1 [1] (b) (i) stress = force/area C1 = 60/(7.9 × 10-7) = 7.6 × 107 Pa A1 [2] (ii) Young modulus = stress/strain C1 limiting strain = 0.03/24 (= 1.25 × 10-3) C1 Young modulus = (7.6 × 107)/(1.25 × 10-3) = 6.1 × 1010 Pa A1 [3] (iii) energy = ½ × 60 × 3.0 × 10-4 C1 = 9.0 × 10-3 J A1 [2] (c) If hard, ball does not deform (much) B1 and either (all) kinetic energy converted to strain energy B1 If soft, Ek becomes strain energy of ball and window B1 (no mention of strain energy, max 2 marks) or impulse for hard ball takes place over shorter time (B1) larger force/greater stress (B1) [3]

Page 2 Mark Scheme Syllabus Paper A and AS LEVEL – June 2005 9702 2 © University of Cambridge International Examinations 2005 5 (a) When a wave (front) is incident on an edge or an obstacle/slit/gap M1 Wave ‘bends’ into the geometrical shadow/changes direction/spreads A1 [2] (b) (i) d = 1/(750 × 103) C1 = 1.33 × 10-6 m A1 [2] (ii) 1.33 × 10-6 × sin90° = n × 590 × 10-9 C1 n = 2 (must be an integer) A1 [2] (iii) formula assumes no path difference of light before entering grating or there is a path difference before the grating B1 [1] (c) e.g. lines further apart in second order lines fainter in second order (allow any sensible difference: 1 each, max 2) B2 [2] (if differences stated but without reference to the orders, max 1 mark) 6 (a) (i) lines normal to plate and equal spacing (at least 4 lines) B1 direction from (+) to earthed plate B1 [2] (ii) E = 160/0.08 M1 = 2.0 × 103 V m-1 A0 [1] (b) (i) correct directions with line of action of arrows passing through charges B1 [1] (ii) force = Eq C1 = 2.0 × 103 × 1.2 × 10-15 = 2.4 × 10-12 N A1 [2] (iii) couple = force × perpendicular separation M1 = 2.4 × 10-12 × 2.5 × 10-3 × sin35° = 3.4(4) × 10-15 N m A1 [2] (iv) either rotates to align with the field or oscillates (about a position) M1 with the positive charge nearer to the earthed plate/clockwise A1 [2] 7 (a) potential difference/current B1 [1] (b) (i) 1) 1.13 W 2) 1.50 V B1 [1] (ii) power = V2 / R or power = VI and V = IR C1 R = 1.502/1.13 = 1.99 Ω A1 [2]

Page 3 Mark Scheme Syllabus Paper A and AS LEVEL – June 2005 9702 2 © University of Cambridge International Examinations 2005 (iii) either E = IR + Ir or voltage divided between R and r C1 I = 1.5 / 2.0 (=0.75 A) p.d. across R = p.d. Across r = 1.5 C1 3.0 = 1.5 + 0.75r r = 2.0 Ω so R = r = 1.99 Ω A1 [3] (c) larger p.d. across R means smaller p.d. across r M1 smaller power dissipation at larger value of V A1 since power is VI and I is same for R and r A1 [3] 8 (a) position shown as A = 227, Z = 91 B1 [1] (b) Pu shown as A = 243, Z = 94 B1 D shown with A = APu and with Z = (ZPu + 1) B1 [2]