Cambridge A Level Physics 9702 — 2002 May/June Paper 2 · Variant 1

TIME 1 hour INSTRUCTIONS TO CANDIDATES Write your name, Centre number and candidate number in the spaces at the top of this page. Answer all questions. Write your answers in the spaces provided on the question paper. INFORMATION FOR CANDIDATES 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. CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level PHYSICS 9702/2 PAPER 2 MAY/JUNE SESSION 2002 1 hour Candidates answer on the question paper. No additional materials. This question paper consists of 15 printed pages and 1 blank page. SPA (SM/CG) S21690/3 © CIE 2002 [Turn over Candidate Centre Number Number Candidate Name FOR EXAMINER’S USE

2 9702/2 M/J02 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

3 9702/2 M/J02 [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

4 9702/2 M/J02 Answer all the questions in the spaces provided. 1 Make reasonable estimates of the following quantities. (a) mass of an apple mass = … kg [1] (b) number of joules of energy in 1 kilowatt-hour number = … [1] (c) wavelength of red light in a vacuum wavelength = … m [1] (d) pressure due to a depth of 10 m of water pressure = … Pa [1] 2 A student uses a micrometer screw gauge to measure the diameter of a wire. He fails to notice that, with the gauge fully closed, the reading is not zero. (a) State and explain whether the omission introduces a random error or a systematic error into the readings of the diameter. … …[2] (b) Explain why the readings are precise but not accurate. … … …[2] For Examiner’s Use

5 9702/2 M/J02 [Turn over 3 (a) Explain what is meant by the centre of gravity of an object. … … …[2] (b) A non-uniform plank of wood XY is 2.50 m long and weighs 950 N. Force-meters (spring balances) A and B are attached to the plank at a distance of 0.40 m from each end, as illustrated in Fig. 3.1. Fig. 3.1 When the plank is horizontal, force-meter A records 570 N. (i) Calculate the reading on force-meter B. reading = … N (ii) On Fig. 3.1, mark a likely position for the centre of gravity of the plank. (iii) Determine the distance of the centre of gravity from the end X of the plank. distance = … m [6] force-meter A force-meter B 0.40m 0.40m 2.50m X Y For Examiner’s Use

6 9702/2 M/J02 4 A steel ball of mass 73 g is held 1.6 m above a horizontal steel plate, as illustrated in Fig. 4.1. Fig. 4.1 The ball is dropped from rest and it bounces on the plate, reaching a height h. (a) Calculate the speed of the ball as it reaches the plate. speed = … m s–1 [2] (b) As the ball loses contact with the plate after bouncing, the kinetic energy of the ball is 90% of that just before bouncing. Calculate (i) the height h to which the ball bounces, h = … m steel ball mass 73g steel plate 1.6m h For Examiner’s Use

7 9702/2 M/J02 [Turn over (ii) the speed of the ball as it leaves the plate after bouncing. speed = … m s–1 [4] (c) Using your answers to (a) and (b), determine the change in momentum of the ball during the bounce. change = … N s [3] (d) With reference to the law of conservation of momentum, comment on your answer to (c). … … …[3] For Examiner’s Use

8 9702/2 M/J02 5 Some gas is contained in a cylinder by means of a moveable piston, as illustrated in Fig. 5.1. Fig. 5.1 State how, for this mass of gas, the following changes may be achieved. (a) increase its gravitational potential energy …[1] (b) decrease its internal energy … …[1] (c) increase its elastic potential energy … …[1] gas cylinder moveable piston For Examiner’s Use

9 9702/2 M/J02 [Turn over 6 Two horizontal metal plates are situated 1.2 cm apart, as illustrated in Fig. 6.1. Fig. 6.1 The electric field between the plates is found to be 3.0 104N C–1 in the downward direction. (a) (i) On Fig. 6.1, mark with a + the plate which is at the more positive potential. (ii) Calculate the potential difference between the plates. potential difference = … V [3] (b) Determine the acceleration of an electron between the plates, assuming there is a vacuum between them. acceleration = … m s–2 [3] 1.2cm For Examiner’s Use

10 9702/2 M/J02 7 (a) Figs. 7.1(a) and (b) show plane wavefronts approaching a narrow gap and a wide gap respectively. (a) (b) Fig. 7.1 On Figs. 7.1(a) and (b), draw three successive wavefronts to represent the wave after it has passed through each of the gaps. [5] For Examiner’s Use

11 9702/2 M/J02 [Turn over (b) Light from a laser is directed normally at a diffraction grating, as illustrated in Fig. 7.2. Fig. 7.2 The diffraction grating is situated at the centre of a circular scale, marked in degrees. The readings on the scale for the second order diffracted beams are 136° and 162°. The wavelength of the laser light is 630 nm. Calculate the spacing of the slits of the diffraction grating. spacing = … m [4] (c) Suggest one reason why the fringe pattern produced by light passing through a diffraction grating is brighter than that produced from the same source with a double slit. … …[1] diffraction grating laser scale 136° 162° For Examiner’s Use

12 9702/2 M/J02 8 A student has available some resistors, each of resistance 100 Ω. (a) Draw circuit diagrams, one in each case, to show how a number of these resistors may be connected to produce a combined resistance of (i) 200 Ω, (ii) 50 Ω, (iii) 40 Ω. [4] For Examiner’s Use

13 9702/2 M/J02 [Turn over (b) The arrangement of resistors shown in Fig. 8.1 is connected to a battery. Fig. 8.1 The power dissipation in the 100 Ωresistor is 0.81 W. Calculate (i) the current in the circuit, current = … A (ii) the power dissipation in each of the 25 Ωresistors. power = … W [4] 100Ω 25Ω 25Ω For Examiner’s Use

14 9702/2 M/J02 9 The radiation from a radioactive source is detected using the apparatus illustrated in Fig. 9.1. Fig. 9.1 Different thicknesses of aluminium are placed between the source and the detector. The count rate is obtained for each thickness. Fig. 9.2 shows the variation with thickness x of aluminium of the count rate. Fig. 9.2 6 x/mm 5 4 3 2 1 0 0 1000 2000 3000 4000 count rate /s–1 detector shielding radioactive source 6cm aluminium For Examiner’s Use

15 9702/2 M/J02 (a) Suggest why it is not possible to detect the presence of the emission of α-particles from the source. … …[1] (b) State the evidence provided on Fig. 9.2 for the emission from the source of (i) β-particles, … … … (ii) γ-radiation. … … … [4] For Examiner’s Use

CAMBRIDGE INTERNATIONAL EXAMINATIONS JUNE 2002 GCE Advanced Subsidiary Level! MARK SCHEME SYLLABUS/COMPONENT :9702 /2 PHYSICS (STRUCTURED QUESTIONS (AS)) UNIvERsITY of CAMBRIDGE B- ears SP Local Examinations Syndicate

Page T Mark Scheme [Syllabus | Paper ‘AS Level Examinations — June 200Z | Zz Categorisation of marks The marking scheme categorises marks on the MACB scheme. B marks: These are awarded as independent marks, which do not depend on other marks. For a B-mark to be scored, the point to which it refers must be seen specifically in the candidate’s answer. M marks: These are method marks upon which A-marks (accuracy marks) later depend. For an M-mark to be scored, the point to which it refers must be seen in the candidate’s answer. If a candidate fails to score a particular M-mark, then none of the dependent A-marks can be scored. C marks: These are compensatory method marks which can be scored even if the points to which they refer are not written down by the candidate, providing subsequent working gives evidence that they must have known it. For example, if an equation carries a C-mark and the candidate does not write down the actual equation but does correct working which shows he/she knew the equation, then the C-mark is awarded. A marks: These are accuracy or answer marks which either depend on an M-mark, or allow a C-mark to be scored. Conventions within the marking scheme BRACKETS Where brackets are shown in the marking scheme, the candidate is not required to give the bracketed information in order to earn the available marks. UNDERLINING In the marking scheme, underlining indicates information that is essential for marks to be awarded.

Page 2 Mark Scheme Syllabus | Paper AS Level Examinations - June 2002 9702 2 1 (a) allow50g - 500g {b) allow3 MJ - 4MJ (ce) allow (6.0-8.0) x 107 Mm ieee eeeeeececeeteesenneeneteneteeeneeneesaee neds BI (d) allow (5 x 10°) > (5x 105} Pa sess sesseceeseseteeteeeecenereresereneananens BI (Ignore sig. fig. in (a), (b), (c} and (d). 2 (a) because all readings have same error OR can’t be eliminated by repeating and averaging Bl error is systematic Bl (do not allow ‘systematic’ if argument is fallacious) {b) micrometer measures to fraction of millimetre so is precise OR if repeated, reading is (almost constant) BL but all readings have error so is not accurate BL 3 (a) point at which (whole) weight of body MI may be considered to act... ...seeeceeeeneeeeeeeeeees Al (allow definition based on gravitational force) (b) () BBON ceeecceeececscesescssssceesseeeeeee eeusesassavsvsevasasseavevsessesecseseees BI (ii) position nearer A than B Bl (iii) clear indication about which point moments are taken Bl eg. 950 xx = 380x 1.7 cl KH OB CM ciieeecseeeeeeeee ees te Cl distance = 108cm or 1.08m (accept 2 sig fig) Al 103) p14) 0) 1) [2] (2] 2] (6]

Page 3 Mark Scheme Syllabus | Paper AS Level Examinations — June 2002 9702 2 Cl Al 144m oe. .. Al ii) mgh = Yam’? v = 2x98x 1.44 cl ve Al (eo) Ap = mv =u) OR Pam oo. ccecc ccc cece er eeeereeren ene aeneeeecerteeeeeeees Cc) m = 0.073 kg Ap = 0.073 x (5.6 + 5.3) Cl = 080Ns Al (d) steel plate (and Earth) must gain momentum of 0.80 N s in downward direction (idea of Earth/plate and ball as the system scores 1/3) § (a) increase the height of the cylinder …cccccccceesceseeseeeeseeeseeeseeeeeeen ease Bl (b) take heat out of gas OR expand gas OR Coolit: …-...ccccseeeesceeesseneeeeeeet Bi {c) compress the gas OR increase pressure OR heat at constant volume … Bl 6 {a) (i) top plate positive . Bl (i) E= V/d . Cl V = 3.0x 10x 12x10 BOO V occ ccceccceeeeneeecnseeneecteseneecreeeeeseesanerseeataeeeeeeneae Al {b) F = ma 3.0 x 10° x 1.6 x 107? = 9.1 x 107% @ a = 5.3x10%ms? 2] (4] (3) {3} {1} 03) (1) (3 {3]

[ Page 4 | Mark Scheme L Sytlabus Paper | AS Level Examinations — June 2002 [9702 2 7 (a) Fig. 6.1(a): approximately circular wavefronts ... Centred ON Bap …seeeeseeeeeeeeeereeeeeee constant wavelength (allow this in (a) or (b) Fig. 6.1(b): wavefronts plane at centre Mi curved at edges {5] (D) = Yo (162 - 136) = 13° ieee eeeeeee eee nereeetetetteeeeterae teen ca eee rene reas cl ASING = NA .oeeseseeeeeees cl dsinl3 = 2 x 630 x 10° . Cl d= 56x10%m 2... Al [4] (Use of 6= 162° or 136°, max 2/4) {c) ¢.g. more slits for light to pass through narrow so more diffracted light and ‘off-axis’ fringes clearer … Bl [ij 8 (a) (i) two resistors in series (ii) two resistors in parallel (iii) any correct combination oa {4] (1/2 only in (iii) if connections to externa! circuit not clear) (bk) @ P=PR 0.81 = is (ii) current in 25 Q resistor = 0.045 A power = 0.051 W Al [4] 9 (a) a-particles not able to penetrate air between source and window ..,…506- Bi 1} (b) (i) rapid drop in count rate . Bl for small thicknesses (up to 2 mm) OR most f's stopped by few mm of aluminium … Bl (ii) very slow drop-off in count rate Bl for thicknesses greater than 2 mm OR y much higher penetration than 8 -…::2seesesseseeessseeeeeee retro ees Bt [4} (do not allow ‘ynot stopped by aluminium’)