Cambridge A Level Physics 9702 — 2002 Oct/Nov 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 AS Core OCTOBER/NOVEMBER SESSION 2002 1 hour Candidates answer on the question paper. No additional materials. This question paper consists of 14 printed pages and 2 blank pages. SPA (NF/CG) S21714/3 © CIE 2002 [Turn over Candidate Centre Number Number Candidate Name FOR EXAMINER’S USE

2 9702/2/O/N/02 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/O/N/02 [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/O/N/02 Answer all the questions in the spaces provided. 1 (a) (i) Define density. … … (ii) State the base units in which density is measured. … [2] (b) The speed v of sound in a gas is given by the expression v =√ , where p is the pressure of the gas of density ρ. γ is a constant. Given that p has the base units of kg m−1s−2, show that the constant γ has no unit. [3] 2 A student uses a metre rule to measure the length of an elastic band before and after stretching it. The lengths are recorded as length of band before stretching, L0 = 50.0 ± 0.1 cm length of band after stretching, LS = 51.6 ± 0.1 cm. Determine (a) the change in length (LS −L0), quoting your answer with its uncertainty, (LS −L0) = ……………………………………… cm [1] γ ρ p     For Examiner’s Use

5 9702/2/O/N/02 [Turn over (b) the fractional change in length, , fractional change = ………………………………. [1] (c) the uncertainty in your answer in (b). uncertainty = ………………………………… [3] (LS −L0) L0 For Examiner’s Use

6 9702/2/O/N/02 3 A ball falls from rest onto a flat horizontal surface. Fig. 3.1 shows the variation with time t of the velocity v of the ball as it approaches and rebounds from the surface. Fig. 3.1 Use data from Fig. 3.1 to determine (a) the distance travelled by the ball during the first 0.40 s, distance = ……………………………………. m [2] For Examiner’s Use 0.2 0.3 0.4 t/s 0.5 0.6 0.7 0 0 -1 -2 -3 -4 1 2 3 4 5 0.1 v/ms –1

7 9702/2/O/N/02 [Turn over (b) the change in momentum of the ball, of mass 45 g, during contact of the ball with the surface, change = ………………………………….. N s [4] (c) the average force acting on the ball during contact with the surface. force = ……………………………………. N [2] 4 (a) Explain what is meant by the concept of work. … … …[2] (b) Using your answer to (a), derive an expression for the increase in gravitational potential energy ∆Ep when an object of mass m is raised vertically through a distance ∆h near the Earth’s surface. The acceleration of free fall near the Earth’s surface is g. [2] For Examiner’s Use

8 9702/2/O/N/02 5 The variation with time t of the displacement x of a point in a transverse wave T1 is shown in Fig. 5.1. Fig. 5.1 (a) By reference to displacement and direction of travel of wave energy, explain what is meant by a transverse wave. … …[1] (b) A second transverse wave T2, of amplitude A has the same waveform as wave T1 but lags behind T1 by a phase angle of 60°. The two waves T1 and T2 pass through the same point. (i) On Fig. 5.1, draw the variation with time t of the displacement x of the point in wave T2. [2] (ii) Explain what is meant by the principle of superposition of two waves. … … …[2] (iii) For the time t = 1.0 s, use Fig. 5.1 to determine, in terms of A, 1. the displacement due to wave T1 alone, displacement = ……………………………………. 2. the displacement due to wave T2 alone, displacement = ……………………………………. 3. the resultant displacement due to both waves. displacement = ……………………………………. [3] 2 3 4 t/s T1 5 6 0 -A A 1 x For Examiner’s Use

9 9702/2/O/N/02 [Turn over BLANK PAGE Turn over for question 6

10 9702/2/O/N/02 6 An electron travelling horizontally in a vacuum enters the region between two horizontal metal plates, as shown in Fig. 6.1. Fig. 6.1 The lower plate is earthed and the upper plate is at a potential of + 400 V. The separation of the plates is 0.80 cm. The electric field between the plates may be assumed to be uniform and outside the plates to be zero. (a) On Fig. 6.1, (i) draw an arrow at P to show the direction of the force on the electron due to the electric field between the plates, (ii) sketch the path of the electron as it passes between the plates and beyond them. [3] (b) Determine the electric field strength E between the plates. E = ……………………………… V m−1 [2] electron path P + 400V region of electric field For Examiner’s Use

11 9702/2/O/N/02 [Turn over (c) Calculate, for the electron between the plates, the magnitude of (i) the force on the electron, force = …………………………….. N (ii) its acceleration. acceleration = ……………………………… m s−2 [4] (d) State and explain the effect, if any, of this electric field on the horizontal component of the motion of the electron. … … …[2] For Examiner’s Use

12 9702/2/O/N/02 7 A student set up the circuit shown in Fig. 7.1. Fig. 7.1 The resistors are of resistance 15 Ωand 45 Ω. The battery is found to provide 1.6 × 105J of electrical energy when a charge of 1.8 × 104C passes through the ammeter in a time of 1.3 × 105s. (a) Determine (i) the electromotive force (e.m.f.) of the battery, e.m.f. = …………………………………….. V (ii) the average current in the circuit. current = …………………………………….. A [4] A 15Ω 45Ω For Examiner’s Use

13 9702/2/O/N/02 [Turn over (b) During the time for which the charge is moving, 1.1 × 105J of energy is dissipated in the 45 Ωresistor. (i) Determine the energy dissipated in the 15 Ωresistor during the same time. energy = …………………………………. J (ii) Suggest why the total energy provided is greater than that dissipated in the two resistors. … … [4] 8 A nucleus of an atom of francium (Fr) contains 87 protons and 133 neutrons. (a) Write down the notation for this nuclide. ………… Fr ………… [2] (b) The nucleus decays by the emission of an α-particle to become a nucleus of astatine (At). Write down a nuclear equation to represent this decay. [2] For Examiner’s Use

14 9702/2/O/N/02 9 An aluminium wire of length 1.8 m and area of cross-section 1.7 × 10−6m2 has one end fixed to a rigid support. A small weight hangs from the free end, as illustrated in Fig. 9.1. Fig. 9.1 The resistance of the wire is 0.030 Ωand the Young modulus of aluminium is 7.1 × 1010Pa. The load on the wire is increased by 25 N. (a) Calculate (i) the increase in stress, increase = …………………………………… Pa (ii) the change in length of the wire. change = ……………………………………. m [4] wire weight 1.8m For Examiner’s Use

15 9702/2/O/N/02 (b) Assuming that the area of cross-section of the wire does not change when the load is increased, determine the change in resistance of the wire. change = ……………………………… Ω[3] For Examiner’s Use

CAMBRIDGE INTERNATIONAL EXAMINATIONS NOVEMBER 2002 GCE Advanced Subsidiary Level UNIVERSITY of CAMBRIDGE Local Examinations Syndicate

1 (a) (b) 2- (a) (b) © 3 (a) (b) (©) 4 (@) (b) Page 1 of 3 Mark Scheme Syllabus | Pa AS Level Examinations — November 2002 9702 2 per a @ (i) mass/ volume ...(ratio must be clear)... . Bl kgm OR kg/m? vhas unit of m7) o…ecccceeseceeceesceesecececesscendeceseceseeessees P/p has unit of kg m!s? / kgm® (no e.c.f. from (a)) V( p/p) has unit of ms? LHS = RHS soy has no unit oo... eee eeeeceeececeeteeeeeeee eee AO 1.6 + 02cm . BL 1.6/50 = 0.032. ...(ignore any uncertainties)…::..eeeee Bl idea of adding fractional uncertainties …e ee eeeeeeeeeeeeee cl (0.2/1.6) + (0.1/50) = 0.127 OR 12.7% ...(-2 marks if uncertainties not added) … Al actual uncertainty = (+) 0.004 oo... ee eee ceeseceeeeeeeeeeees Al (do not allow more than 2 sig. fig) vw = uy? + 2as OR use of triangle ete ooo. eeeeeeceessessseneeeeee Cl 4.0? = 2x9.8xs OR s=%x4.0x04 s = 0.82m OR 0.0m Ap = m(v -u) OR p=mv speeds are 4.2 ms” and 3.6 ms* Ap = 0.045 (4.2 +3.6) (2/4 only if speeds not added) see = 035Ns Al (1 mark only if only one speed used) any time between 0.14 sand 0.178 ooo eeeeceeeececeeeeeeees Cl force = Ap/ At = 0.35/0.14 (allowe.c.f.) = DSN civcccccccccccacseeseceseccecauseseeeeeeseueeeeeeeeeeeeeanees Al force x distance moved … in the direction of the force weight / force = mg AE, = mg x Ah (no marks for quote of mgAh) se. MI . Al [2] [3] 01] (] [3] [2] [4] [2] [2] [2]

Page 2 of 3 Mark Scheme Syllabus Paper AS Level Examinations — November 2002 9702 2 5 (a) displacement & direction of energy travel normal to one another ...B1 [1] (b) (@) _ phase angle of 60° correct .. (need to see 1% wavelengths) … Bl lags behind Ty …sccceceeeecceescceeecceeeeeuseseeeeeeeesaeeeeeeeees Bl [2] (ii) waves must be in same place (at same time) … se ceeeeee seer eee Bl resultant displacement = sum of individual displacements … Bl [2] (ii) 1. -“4A «+ BI 2. YA … (allow e.c.f.) . Bl 3. zero … (allowe.c.f.) ... . Bl £3] 6 (a) (i) arrow in upward direction, foot near Poo... eeeeeeee eee eee Bl (ii) _ curved path consistent with (i) between plates …0…eee Bl then straight (with no kink at change-over) …:::eceeeeeeeeee Bl [3] (b) EH VI ciccsccscccececccescnenenencseceeeeenseeceececesaeesseeeeeeeneeses Cl = 400/(0.8 x 107) =5.0x10°Vm! oo. (allow 1 sig fig) …ceeeeeeeeee Al [2] (COR) Rn tt 2 eae Cl « = 5.0 x 10*x 1.6 x 10°? =8.0x10°N_ … (allow 1 sig fig ande.c.£) 0.0.0.0... Al (ii) DH FIM cacccccccccceecsceeccccenceceeecececeeeececeseccesseeceneecaeeeees Cl = (8.0 x 107") /(9.1 x 107!) = 88.x10%ms? … (allow | sig fig ande.c.f.) … Al [4] (d) because Fy is normal to horizontal motion no effect (2]

Page 3 of 3 Mark Scheme Syllabus | Paper | AS Level Examinations — November 2002 9702 2 7 (a) @ Gi) - &) @ 8 (a) (b) 9@ (b) @ @ (ii) emf. = energy / charge = (1.6 x 10°) /(1.8 x 10*) = BDV ceecececccceeccecceeeeseeeeceeeeeeceeeeeseceeeseeeesseeeees Al current = AQ/Ar ... Cl = (1.80 x 10%) / (1.3 x 10°) = OFA ccccceecsessceseescsssssescessscescesscsnecseearstseens Al energy « R (or formula) see energy = (15/45)x 1.14 x 10° = 3.7x10J energy dissipated in internal resistance (of battery) …06 Bl OR in extra resistance in circuit shows nucleon number as 220 ... shows proton number as 87 shows products as$He OR ‘a and {At …(allow e.c.f. from (a)) … stress = F/ A viccsceccssscceeseseeeseeeeceeeseueesesssesseeeesseeees cl” = 25 / (1.7x 10 = 147x107Pa … (do not allow 1 sig fig) … Al stress = E x strain …eecceeeeecceeeccceeeececececceeeeceneeseeeeeees Cl 1.47 x 107 = 7.1 x 10! x (AI/1.8) Al = O.37 MM oe ee eeeedecce ee cee cee ceeseseeeeeceeseeseeeeaeeeseeneeeenees Al R = pl/A OR RL *°. AR/R = AI/1 eeeceece eee = (3.7 x 10“/ 1.8) x 0.03 = 62x10°Q May calculate p = 2.833... x 10%°Q m giving new R as 3.0006167 x 107Q hence AR - full credit possible However, if rounds off p as 2.83 x 10*Q m, then Raew < Roi! Allow 1 mark only for R < L 4 (4) [2] [2] [4] B)