Cambridge A Level Physics 9702 — 2023 Oct/Nov Paper 2 · Variant 2

1 (a) In the following list, underline all quantities that are SI base quantities. charge electric current force time [1] (b) Under certain conditions, the distance s moved in a straight line by an object in time t is given by s = 12at 2 where a is the acceleration of the object. State two conditions under which the above expression applies to the motion of the object. 1 ................................................................................................................................................ 2 ................................................................................................................................................ [2] (c) The variation with time t of the velocity v of a car that is moving in a straight line is shown in Fig. 1.1. 25 20 15 v / m s–1 10 5 0 0 2 4 6 8 10 12 14 16 t / s –5 –10 –15 Fig. 1.1 (i) Compare, qualitatively, the acceleration of the car at time t = 8.0 s and at time t = 14.0 s in terms of: ● magnitude ........................................................................................................................................... ........................................................................................................................................... ● direction. ........................................................................................................................................... ........................................................................................................................................... [2] (ii) Determine the magnitude of the acceleration of the car at time t = 4.0 s. acceleration = ................................................ m s–2 [2] (iii) The car is at point X at time t = 0. Determine the magnitude of the displacement of the car from X at time t = 12.0 s. displacement = ...................................................... m [2] [Total: 9]

Mark scheme: Question Answer Marks 1(a) only electric current and time underlined B1 1(b) initial speed / velocity is zero B1 (non-zero magnitude of) acceleration is constant / uniform (and in a straight line) B1 1(c)(i) • magnitude of acceleration at t = 8.0 s is less than that at t = 14.0 s B1 • direction of acceleration at t = 8.0 s is opposite to that at t = 14.0 s B1 1(c)(ii) a = gradient or a = (v – u) / t or a = ∆v / (∆)t C1 a = e.g. (20 + 10) / 12 or (0 + 10) / 4 or (20 – 0) / (12 – 4) A1 a = 2.5 m s–2 1(c)(iii) displacement = [½  (12 – 4)  20] – [½  4  10] C1 or displacement = (–10  12) + (½  2.5  122) or displacement = (20  12) – (½  2.5  122) or displacement = ½  (20 – 10)  12 displacement = 60 m A1

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2 A high‑altitude balloon is stationary in still air. A solid sphere is suspended from the balloon by a string, as shown in Fig. 2.1. balloon string sphere Fig. 2.1 (not to scale) The volume of the balloon is 7.5 m3. The total weight of the balloon, string and sphere is 65 N. The upthrust acting on the string and sphere is negligible. (a) Calculate the density of the air surrounding the balloon. density = .............................................. kg m–3 [2] (b) The string breaks, releasing the sphere. (i) State the magnitude of the acceleration of the sphere immediately after the string breaks. acceleration = ................................................ m s–2 [1] (ii) State and explain the variation, if any, in the magnitude of the acceleration of the sphere when it is moving downwards before it reaches terminal (constant) velocity. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (c) The sphere has a mass of 4.0 kg. Calculate the total resistive force acting on the sphere at the instant when its acceleration is 1.9 m s–2. resistive force = ...................................................... N [2] [Total: 8]

Mark scheme: 2(a) 65 = gV C1 or 65 = mg and m = V  = 65 / (9.81  7.5) A1 = 0.88 kg m–3 2(b)(i) acceleration = 9.8 m s–2 A1 2(b)(ii) air resistance (acting on sphere) increases B1 resultant force (on sphere) decreases B1 (magnitude of) acceleration decreases B1 2(c) F = ma C1 = 4.0  1.9 ( = 7.6 N) resistive force = (4.0  9.81) – (4.0  1.9) A1 = 32 N

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3 A vertical rod is fixed to the horizontal surface of a table, as shown in Fig. 3.1. rod surface of spring table Fig. 3.1 (not to scale) A spring of mass 7.5 g is able to slide along the full length of the rod. The spring is first pushed against the surface of the table so that it has an initial compression of 2.1 cm. The spring is then suddenly released so that it leaves the surface of the table with a kinetic energy of 0.048 J and then moves up the rod. Assume that the spring obeys Hooke’s law and that the initial elastic potential energy of the compressed spring is equal to the kinetic energy of the spring as it leaves the surface of the table. Air resistance is negligible. (a) By using the initial elastic potential energy of the compressed spring, calculate its spring constant. spring constant = ................................................ N m–1 [2] (b) Calculate the speed of the spring as it leaves the surface of the table. speed = ................................................ m s–1 [2] (c) The spring rises to its maximum height up the rod from the surface of the table. This causes the gravitational potential energy of the spring to increase by 0.039 J. (i) Calculate, for this movement of the spring, the increase in height of the spring after leaving the surface of the table. increase in height = ...................................................... m [2] (ii) Calculate the average frictional force exerted by the rod on the spring as it rises. average frictional force = ...................................................... N [2] (d) The rod is replaced by another rod that exerts negligible frictional force on the moving spring. The initial compression x of the spring is now varied in order to vary the maximum increase in height Δh of the spring after leaving the surface of the table. Assume that the spring obeys Hooke’s law for all compressions. On Fig. 3.2, sketch a graph to show the variation with x of Δh. Numerical values are not required. Δh 0 0 x Fig. 3.2 [2] [Total: 10]

Mark scheme: 3(a) E(P) = ½kx2 C1 or E(P) = ½Fx and F = kx 0.048 = ½  k  (2.1  10–2)2 A1 k = 220 N m–1 3(b) E(K) = ½mv2 C1 0.048 = ½  7.5  10–3  v2 A1 v = 3.6 m s–1 3(c)(i) (∆)E = mg(∆)h C1 0.039 = 7.5  10–3  9.81  (∆)h A1 h = 0.53 m 3(c)(ii) F  0.53 = 0.048 – 0.039 C1 F = 0.02 N A1 3(d) sketch: curved line from the origin M1 curved line has increasing gradient A1

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4 (a) A ball Y moves along a horizontal frictionless surface and collides with a ball Z, as illustrated in the views from above in Fig. 4.1 and Fig. 4.2. 0.25 kg 3.7 m s–1 Y 27° P Y Z Q P Q 44° 0.25 kg mZ Z 5.5 m s–1 mZ BEFORE COLLISION AFTER COLLISION Fig. 4.1 (not to scale) Fig. 4.2 (not to scale) Ball Y has a mass of 0.25 kg and initially moves along a line PQ. Ball Z has a mass mZ and is initially stationary. After the collision, ball Y has a final velocity of 3.7 m s–1 at an angle of 27° to line PQ and ball Z has a final velocity of 5.5 m s–1 at an angle of 44° to line PQ. (i) Calculate the component of the final momentum of ball Y in the direction perpendicular to line PQ. component of momentum = .................................................... N s [2] (ii) By considering the component of the final momentum of each ball in the direction perpendicular to line PQ, calculate mZ. mZ = ..................................................... kg [1] (iii) During the collision, the average force exerted on Y by Z is FY and the average force exerted on Z by Y is FZ. Compare the magnitudes and directions of FY and FZ. Numerical values are not required. magnitudes: ....................................................................................................................... directions: .......................................................................................................................... [2] (b) Two blocks, A and B, move directly towards each other along a horizontal frictionless surface, as shown in the view from above in Fig. 4.3. 4 m s–1 6 m s–1 A B Fig. 4.3 The blocks collide perfectly elastically. Before the collision, block A has a speed of 4 m s–1 and block B has a speed of 6 m s–1. After the collision, block B moves back along its original path with a speed of 2 m s–1. Calculate the speed of block A after the collision. speed = ................................................ m s–1 [1] [Total: 6]

Mark scheme: 4(a)(i) component of momentum = 0.25  3.7  sin 27° C1 = 0.42 N s A1 4(a)(ii) m(Z)  5.5  sin 44° = 0.42 A1 or m(Z)  5.5  sin 44° = 0.25  3.7  sin 27° mZ = 0.11 kg 4(a)(iii) magnitudes: equal B1 directions: opposite B1 4(b) 4 + 6 = 2 + v A1 v = 8 m s–1

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5 (a) A beam of vertically polarised light is incident normally on a polarising filter, as shown in Fig. 5.1. vertically polarised transmitted incident light beam light beam transmission filter axis of filter Fig. 5.1 (i) The transmission axis of the filter is initially vertical. The filter is then rotated through an angle of 360° while the plane of the filter remains perpendicular to the beam. On Fig. 5.2, sketch a graph to show the variation of the intensity of the light in the transmitted beam with the angle through which the transmission axis is rotated. maximum value intensity of the light 0 0 90 180 270 360 angle / ° Fig. 5.2 [2] (ii) The intensity of the light in the incident beam is 7.6 W m–2. When the transmission axis of the filter is at angle θ to the vertical, the light intensity of the transmitted beam is 4.2 W m–2. Calculate angle θ. θ = ........................................................° [2] (b) State what is meant by the diffraction of a wave. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) A beam of light of wavelength 4.3 × 10–7 m is incident normally on a diffraction grating in air, as shown in Fig. 5.3. third order beam of light, 68° wavelength 68° 4.3 × 10–7 m diffraction third order grating Fig. 5.3 (not to scale) The third‑order diffraction maximum of the light is at an angle of 68° to the direction of the incident light beam. (i) Calculate the line spacing d of the diffraction grating. d = ...................................................... m [2] (ii) Determine a different wavelength of visible light that will also produce a diffraction maximum at an angle of 68°. wavelength = ...................................................... m [2] [Total: 10]

Mark scheme: 5(a)(i) light intensity has maximum value at 0°, 180°, 360° and zero intensity at 90°, 270° M1 ‘sinusoidally-shaped’ curve A1 5(a)(ii) 4.2 = 7.6 cos2 C1 = 42° A1 5(b) wave passes (through) an aperture B1 or wave passes (by / through / around) an edge wave spreads (into geometrical shadow) B1 5(c)(i) n= d sin C1 d = (3  4.3  10–7) / sin 68° A1 = 1.4  10–6 m 5(c)(ii) 1.4  10–6  sin 68° = 2   C1 or 3  4.3  10–7 = 2   = 6.5  10–7 m A1

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6 (a) A metal wire has a resistance per unit length of 0.92 Ω m–1. The wire has a uniform cross‑sectional area of 5.3 × 10–7 m2. Calculate the resistivity of the metal of the wire. resistivity = .................................................. Ω m [2] (b) A battery of electromotive force (e.m.f.) E and negligible internal resistance is connected in series with a fixed resistor and a light‑dependent resistor (LDR), as shown in Fig. 6.1. E 1400 Ω V 1600 Ω Fig. 6.1 The resistance of the fixed resistor is 1400 Ω. The intensity of the light illuminating the LDR causes it to have a resistance of 1600 Ω. A voltmeter connected across the LDR reads 6.4 V. (i) Show that the current in the LDR is 4.0 × 10–3 A. [1] (ii) Calculate the number of free electrons passing through the LDR in a time of 3.2 minutes. number of free electrons = ......................................................... [2] (iii) Calculate the e.m.f. E. E = ...................................................... V [2] (iv) Determine the ratio power dissipated in LDR . power dissipated in fixed resistor ratio = ......................................................... [2] (c) The environmental conditions change causing a decrease in the resistance of the LDR in (b). The temperature of the environment does not change. State whether there is a decrease, increase or no change to: (i) the intensity of the light illuminating the LDR ..................................................................................................................................... [1] (ii) the current in the battery ..................................................................................................................................... [1] (iii) the reading of the voltmeter. ..................................................................................................................................... [1] [Total: 12]

Mark scheme: 6(a) R = L / A C1  = 0.92  5.3  10–7 A1 = 4.9  10–7  m 6(b)(i) (current =) 6.4 / 1600 = 4.0 × 10–3 (A) A1 6(b)(ii) charge = 4.0  10–3  3.2  60 C1 ( = 0.768 C) number = 0.768 / 1.6  10–19 A1 = 4.8  1018 6(b)(iii) 6.4 / E = 1600 / (1400 + 1600) C1 or E = 6.4 + (4.0  10–3  1400) or E = 4.0  10–3 (1600 + 1400) E = 12 V A1 6(b)(iv) P = I2R or P = VI or P = V2 / R C1 ratio = (6.42 / 1600) / [(4.0  10–3)2  1400] A1 = 1.1 6(c)(i) increase B1 6(c)(ii) increase B1 6(c)(iii) decrease B1

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7 (a) In the following list, underline all the particles that are not fundamental. antineutrino baryon nucleon positron [1] (b) A nucleus of thorium‑230 (23090Th) decays in stages, by emitting α‑particles and β– particles, to form a nucleus of lead‑206 (20682Pb). Determine the total number of α‑particles and the total number of β– particles that are emitted during the sequence of decays that form the nucleus of lead‑206 from the nucleus of thorium‑230. number of α‑particles = ............................................................... number of β– particles = ............................................................... [2] (c) A meson has a charge of –1e, where e is the elementary charge. The quark composition of the meson includes a charm antiquark. State and explain a possible flavour (type) of the other quark in the meson. ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 5]

Mark scheme: 7(a) only baryon and nucleon underlined B1 7(b) number of -particles = 6 B1 number of – particles = 4 B1 7(c) charm antiquark (charge) = –⅔(e) B1 or –⅔(e) + q = –1(e) (other quark q = –⅓(e) so) down / strange / bottom B1

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