Cambridge A Level Physics 9702 — 2024 May/June Paper 2 · Variant 3

1 The drag force FD acting on a sphere falling through a liquid is given by FD = 6πηr v where r is the radius of the sphere, v is the speed of the sphere in the liquid and η is a property of the liquid called the viscosity. (a) Show that the SI base units of viscosity are kg m–1 s–1. [2] (b) The sphere has a radius of 3.0 cm and is falling vertically downwards at a terminal velocity of 2.0 m s–1 through the liquid. The drag force acting on the sphere is 0.096 N. Calculate the viscosity of the liquid. viscosity = ......................................... kg m–1 s–1 [2] (c) The sphere is shown in Fig. 1.1. sphere liquid Fig. 1.1 On Fig. 1.1, draw and label arrows to represent the directions of the three forces acting on the sphere as it falls at terminal velocity through the liquid. [2] (d) (i) The density of the liquid is 920 kg m–3. Show that the upthrust acting on the sphere is 1.0 N. [2] (ii) Calculate the mass of the sphere. mass = .................................................... kg [2] [Total: 10]

Mark scheme: 1(a) units of F: kg m s–2 C1 units of r: m and units of v: m s–1 units of : kg m s–2 / (m  m s–1) = kg m–1 s–1 A1 1(b) viscosity = 0.096 / (6    0.03  2.0) C1 = 0.085 kg m–1 s–1 A1 1(c) one arrow vertically downwards labelled weight / W B1 arrow(s) vertically upwards labelled U / upthrust and drag / FD/viscous force B1 1(d)(i) V = (4 / 3) r3 C1 upthrust = (4 / 3)    0.033  920  9.81 = 1.0 N A1 1(d)(ii) weight = 1.0 + 0.096 (= 1.096 N) C1 m = 1.096 / 9.81 = 0.11 kg A1

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2 (a) Define displacement from a point. ................................................................................................................................................... ............................................................................................................................................. [1] (b) An object is projected horizontally at a speed of 6.0 m s–1 from a slope, as shown in Fig. 2.1. 6.0 m s–1 object slope θ horizontal Fig. 2.1 (not to scale) The slope is at an angle θ to the horizontal. Air resistance is negligible. The object lands on the slope a time of 0.71 s later and stops without rolling or bouncing. (i) Determine the horizontal distance travelled by the object. distance = ..................................................... m [1] (ii) Determine the vertical distance travelled by the object. distance = ..................................................... m [2] (iii) Use your answers in (b)(i) and (b)(ii) to calculate θ. θ = ....................................................... ° [2] (iv) Determine the magnitude of the displacement of the object from its original position. displacement = ..................................................... m [2] (v) By considering energy, calculate the speed of the object just before it lands. speed = ................................................ m s–1 [3] [Total: 11]

Mark scheme: 2(a) distance (from the point) in a straight line in a given direction B1 2(b)(i) distance = speed  time = 6.0  0.71 = 4.3 m A1 2(b)(ii) s = ut + ½ at2 = ½  9.81  0.712 C1 = 2.5 m A1 2(b)(iii) tan  = 2.5 / 4.3 or hypotenuse = √(4.32 + 2.52) ( = 4.97 m) cos  = 4.3 / 4.97 or sin  = 2.5 / 4.97 C1  = 30° A1 Question Answer Marks 2(b)(iv) displacement = √(4.32 + 2.52) C1 = 4.9 m or 5.0 m A1 or displacement = 2.5 / sin 30° or displacement = 4.3 / cos 30° (C1) = 5.0 m (A1) 2(b)(v) KE = ½mv2 or GPE = mgh C1 initial KE + loss in GPE = final KE (½  m  6.02) + (m  9.81  2.5) = (½  m  v2) C1 v = 9.2 m s–1 A1

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3 (a) State Hooke’s law. ................................................................................................................................................... ............................................................................................................................................. [1] (b) The variation of the applied force with the extension for a sample of a material is shown in Fig. 3.1. 10 force / N 8 X 6 4 2 0 0 40 80 120 160 200 extension / mm Fig. 3.1 The sample behaves elastically up to an extension of 80 mm and breaks at point X. (i) On the line in Fig. 3.1, draw a cross (×) to show the limit of proportionality. Label this cross with the letter P. [1] (ii) On the line in Fig. 3.1, draw a cross (×) to show the elastic limit. Label this cross with the letter E. [1] (c) The sample in (b) has a cross-sectional area of 0.40 mm2 and an initial length of 3.2 m. For deformations within the limit of proportionality of the sample, determine: (i) the spring constant of the sample spring constant = ............................................... N m–1 [2] (ii) the Young modulus of the material from which the sample is made. Young modulus = .................................................... Pa [3] (d) Determine an estimate of the work done on the sample as it is extended from zero extension to its breaking point. Explain your reasoning. work done = ...................................................... J [2] (e) A second sample of the same material has a larger cross-sectional area than the original sample but the same initial length. The two samples are each deformed with the limit of proportionality. State and explain qualitatively how the spring constant of the second sample compares with that of the original sample. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 12]

Mark scheme: 3(a) extension is proportional to (applied) force B1 3(b)(i) P at (60, 5.4) A1 3(b)(ii) E at (80, 5.9) A1 3(c)(i) k = F / x or k = gradient of (straight line section of) graph C1 e.g. gradient = 5.4 / 0.060 k = 90 N m–1 A1 3(c)(ii) Young modulus or E =  /  or FL / Ax or kL / A C1 E = (5.4  3.2) / (4.0  10−7  0.06) or 90  3.2 / (4.0  10−7) C1 E = 7.2  108 Pa A1 3(d) work done = area under graph B1 = (1.0  0.2) J A1 3(e) the extension will be smaller (for the same force on the thicker sample) or a greater force is required (to extend the thicker sample by the same amount) or spring constant is proportional to area M1 the spring constant (of the second sample) will be greater A1

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4 A progressive transverse wave travelling from left to right is shown at an instant in time in Fig. 4.1. R wave direction of travel T Fig. 4.1 R and T are points on the wave. (a) State the phase difference between the points R and T. phase difference = ....................................................... ° [1] (b) On Fig. 4.1, draw an arrow at point T to show the direction of movement of point T at the instant shown. [1] (c) The horizontal distance between R and T is 0.62 cm, as shown in Fig. 4.2. 0.62 cm R T Fig. 4.2 (not to scale) The speed of the wave is 0.27 m s–1. Calculate the frequency of the wave. frequency = .................................................... Hz [3] (d) The wave is a water wave produced by a dipper S1 attached to a vibrator in a ripple tank. An identical dipper S2 is attached to the same vibrator. The two dippers produce an interference pattern on the water in the tank, as shown in Fig. 4.3. water P wave crests wave troughs S1 S2 Fig. 4.3 (not to scale) The wave crests from each source are represented by solid lines on Fig. 4.3 and the wave troughs are represented by dashed lines. At point P in Fig. 4.3, the wave from S1 has the same amplitude A as the wave from S2. Describe and explain the amplitude of the resultant wave at point P. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 8]

Mark scheme: 4(a) 270° A1 4(b) arrow pointing vertically downwards at T A1 4(c) v = f or v =  / T and f = 1 / T C1 wavelength = 0.62  10–2  (4 / 3) ( = 0.83  10–2 m) C1 f = 0.27 / (0.83  10–2) = 33 Hz A1 4(d) resultant displacement is the sum of the displacements of the waves (from S1 and S2) or waves (from S1 and S2) superpose (at P) B1 Any one point from:  (at P) the waves (from the two sources) (always) destructively interfere  (at P) the waves have a path difference that is (always) an odd number of half-wavelengths / their path difference is one and a half wavelengths  (at P) the waves have a phase difference that is (always) 180° / they are in antiphase / crest of one wave meets trough of other wave B1 amplitude (of the resultant wave) is zero (at all times) B1

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5 (a) (i) State Kirchhoff’s second law. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (ii) State the conservation law that gives rise to Kirchhoff’s second law. ..................................................................................................................................... [1] (b) A circuit contains a cell of internal resistance r and two resistors of resistances R1 and R2, as shown in Fig. 5.1. r R1 I R2 V Fig. 5.1 The potential difference (p.d.) across the two resistors is V. The current in the cell is I. (i) Use Kirchhoff’s laws to show that the total resistance RT of the external circuit is given by 1 1 1 = + . RT R1 R2 [2] (ii) The electromotive force (e.m.f.) of the cell is 1.50 V. When the values of R1 and R2 are 10 Ω and 15 Ω respectively, the p.d. measured by the voltmeter is 1.38 V. Calculate the internal resistance r of the cell. r = ..................................................... Ω [3] (c) A third resistor is added in parallel with R1 and R2 in the circuit in Fig. 5.1. State and explain the effect, if any, of this change on: (i) the current in the cell ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) the p.d. measured by the voltmeter. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 11]

Mark scheme: 5(a)(i) sum of electromotive force(s) = sum of potential difference(s) around a (closed) loop or the (algebraic) sum of the p.d.(s) and e.m.f.(s) is zero around a (closed) loop B1 5(a)(ii) (law of conservation of) energy B1 5(b)(i) (by Kirchhoff’s first law) I = I1 + I2 B1 V / RT = V / R1 + V / R2 therefore 1 / RT = 1 / R1 + 1 / R2 B1 5(b)(ii) resistance of parallel combination = (15  10) / (15 + 10) (= 6.0 ) C1 r = (E – V) / I C1 I= 1.38 / 6.0 = 0.23 A r = (1.50 – 1.38) / 0.23 = 0.52  A1 or (by potential divider principle) r / RT = Ir / V (C1) r / 6.0 = 0.12 / 1.38 r = 0.52  (A1) or (by potential divider equation) V = E  RT / (RT + r) (C1) 1.38 = 1.5  6.0 / (6.0 + r) r = 0.52  (A1) Question Answer Marks 5(c)(i) as the (total) resistance has decreased (and e.m.f. is unchanged) M1 current will (in the cell) increase A1 5(c)(ii) (as greater current means a) bigger drop in p.d. across the internal resistance M1 p.d. (on voltmeter) will decrease A1

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6 Nuclei of an isotope of samarium (Sm) each contain 62 protons and 85 neutrons. (a) State the nuclide notation in the form AZX for this isotope of samarium. [1] (b) This isotope of samarium is radioactive and decays by emitting particles. Gamma-radiation is not emitted. The energy spectrum of the emitted particles is shown in Fig. 6.1. number of particles 0 0 kinetic energy of particle Fig. 6.1 (i) Explain how Fig. 6.1 shows that this isotope of samarium emits α-particles and does not emit β-particles. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) This isotope of samarium decays to an isotope of neodymium (Nd). Give the radioactive decay equation for this decay. Include the nucleon and proton numbers of all the particles involved. [2] (c) A baryon is composed of three quarks which all have different flavours. The baryon has a charge of 0. Two of the quarks in the baryon are an up quark and a bottom quark. (i) Determine, in terms of the elementary charge e, the charge on the third quark in the baryon. charge = .......................................................e [2] (ii) State a possible flavour for the third quark in the baryon. ..................................................................................................................................... [1] [Total: 8]

Mark scheme: 6(a) 147 62Sm 6(b)(i) the (kinetic) energy of the particles is discrete / has only one value (so must be alpha) B1 and beta particles have a (continuous) range of (kinetic) energies (so can’t be beta) B1 6(b)(ii)    147 143 4 62 60 2 Sm Nd values for Sm and Nd correct with no other extra particles on either side of the equation A1 4 2  correct A1 6(c)(i) up quark charge is +(2 / 3) (e) or bottom quark charge is –(1 / 3) (e) C1 0 = +(2 / 3) (e) – (1 / 3) (e) + q (so) charge (on third quark must be) –(1 / 3) (e) A1 6(c)(ii) down or strange A1

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