Cambridge A Level Physics 9702 — 2025 May/June Paper 2 · Variant 3

1 (a) Define velocity. ................................................................................................................................................... ............................................................................................................................................. [1] (b) In an experiment, two objects A and B are released from the side of a building, as shown in Fig. 1.1. building A 10.0 m 3.0 m s–1 B h ground Fig. 1.1 (not to scale) Object A is released from rest at a height of 10.0 m above horizontal ground. Object B is released with an initial upward velocity of 3.0 m s−1 at a height h above the ground. Both objects take the same time to reach the ground and they do not collide with each other. Air resistance is negligible. Calculate h. h = ..................................................... m [3] (c) In a second experiment, object B is released from the same height as in (b) but with a speed of 6.0 m s−1 at an angle of 60° to the vertical, as shown in Fig. 1.2. 60° building 6.0 m s–1 B Fig. 1.2 (i) State and explain whether the time taken for object B to reach the ground is less than, the same as, or greater than the time taken in the first experiment. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) By considering energy, state and explain whether the speed at which object B reaches the ground is less than, the same as, or greater than in the first experiment. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: Question Answer Marks 1(a) Rate of change of displacement B1 1(b) For object A: C1 1 2 s = ut + at 2 2  10 so t = 9.81 =1.4 Then for object B: C1 1 2 s = ut + at 2 h = −(3 1.4) + (0.5  9.81  1.42) = 5.7 m A1 OR (C1) h = −(3 1.4) + 10 = 5.7 m (A1) 1(c)(i) time taken (to reach the ground is) same B1 The initial vertical (component of the) velocity is the same (as in part (1b)) B1 1(c)(ii) The (total) initial energy is greater (than in part (1b)) B1 change in gravitational potential energy is same, so speed is greater B1

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2 (a) Define the moment of a force about a pivot. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Three objects A, B and C are placed on a horizontal beam. The beam is in equilibrium, as shown in Fig. 2.1. A B beam C pivot 3.0 m 9.0 m Fig. 2.1 (not to scale) The beam is uniform and has length 9.0 m. A pivot is at the midpoint of the beam. Object A has mass 90 kg and is at one end of the beam. Object B has mass m and is a distance of 3.0 m from the pivot. Object C has mass 150 kg and is at the other end of the beam. (i) Calculate m. m = .................................................... kg [3] (ii) Object A is removed and replaced by a wire fixed to the end of the beam and to the ground, as shown in Fig. 2.2. beam B C wire ground pivot Fig. 2.2 After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged. The wire has a diameter of 1.8 × 10−3 m and has a strain of 1.2 × 10−3. The wire is not extended beyond its limit of proportionality. Calculate the Young modulus of the wire. Young modulus = .................................................... Pa [3] (iii) Object B is now moved to a new position closer to the pivot without passing it. The beam is again horizontal and in equilibrium. State and explain the effect, if any, that this has on the strain in the wire. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 9]

Mark scheme: 2(a) force  perpendicular distance (of line of action of force to / from the point) B1 2(b)(i) 150  9.81  4.5 or 90  9.81  4.5 or 3.0  9.81  m C1 (150  9.81  4.5) = (90  9.81  4.5) + (3.0  9.81  m) C1 m = 90 kg A1 2(b)(ii) Young modulus = σ / ε or F / Aε or FL / Ax C1 Area of wire =   (1.8  10–3 / 2)2 C1 = 2.5  10–6 So Young modulus = ((90  9.81) /   (9.010–4)2 ) / 1.210–3 A1 = 2.9 1011 Pa 2(b)(iii) Moment provided by B will decrease / moment due to wire will increase B1 So force acting on wire will increase (Young’s modulus and area remain constant) and the strain will increase B1

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3 A car of mass 1500 kg is travelling along a straight horizontal road at constant velocity v. The car is subject to a total resistive force F, as shown in Fig. 3.1. v car, mass 1500 kg F horizontal road Fig. 3.1 (a) Show that the power P developed by the engine in overcoming the total resistive force is given by the equation P = Fv . [2] (b) The car now moves up a slope at a constant speed of 30 m s−1. The slope is at an angle to the horizontal of 6.0°, as shown in Fig. 3.2. 30 m s–1 car road 6.0° Fig. 3.2 The total resistive force acting on the car is 1600 N. (i) Show that the increase in gravitational potential energy of the car in a time of 1.0 s is 46 000 J. [2] (ii) Use the information in (b)(i) to determine the power developed by the engine to move the car up the slope. power = ..................................................... W [2] (c) The car picks up a passenger and then continues up the slope at the same speed as in (b). State and explain the effect, if any, that the passenger has on: (i) the air resistance acting on the car ........................................................................................................................................... ..................................................................................................................................... [1] (ii) the power developed by the engine. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 8]

Mark scheme: 3(a) W = Fd B1 P = Fd / t = Fv or P = Fvt / t = Fv B1 3(b)(i) ()E(P) = mg()h C1 increase of gravitational potential energy of car in 1.0 s A1 = 1500  9.81  30  sin 6.0 = 46 000 J 3(b)(ii) (Power to overcome total resistive forces) = 1600  30 C1 = 48 000 W power = 48 000 + 46 000 A1 = 9.4  104 W 3(c)(i) Air resistance is the same, as the speed is the same B1 3(c)(ii) Mass / weight has increased so (power will) increase B1

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4 (a) State the principle of conservation of momentum. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) An object A of mass 4.0 kg travels at a velocity of 6.0 m s−1 to the right on a horizontal frictionless surface. It moves towards a second object B of mass 2.0 kg that is moving at a velocity of 3.0 m s−1 in the same direction as A, as shown in Fig. 4.1. object A, mass 4.0 kg 6.0 m s–1 3.0 m s–1 object B, mass 2.0 kg surface Fig. 4.1 Object A collides with object B. The two objects join and move off together with velocity v. Calculate: (i) velocity v v = ................................................ m s−1 [2] (ii) the percentage of the total initial kinetic energy of the two objects that is transferred to other forms of energy during the collision. percentage = ......................................................% [2] [Total: 6]

Mark scheme: 4(a) sum / total momentum (of a system of bodies) is constant M1 or sum / total momentum before = sum / total momentum after for an isolated system / no (resultant) external force A1 4(b)(i) p = mv or 4.0  6.0 or 2.0  3.0 C1 (4.0  6.0) + (2.0  3.0) = 6.0  v so v = 5.0 m s−1 A1 4(b)(ii) KE before = (0.5  4.0  6.02) + (0.5  2.0  3.02) C1 = 81 J KE after = (0.5  6.0  5.02) (C1) = 75 J percentage transferred = [(81 – 75) / 81]  100 A1 = 7%

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5 (a) State why sound waves cannot be polarised. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A plane-polarised light wave is incident on a polarising filter as shown in Fig. 5.1. polarising filter light wave, intensity I0 Fig. 5.1 The intensity of the light incident on the filter is I0. The light is incident normally on the filter and the transmission axis of the filter is initially perpendicular to the plane of polarisation of the light. The filter is now rotated through 360° about the direction of travel of the light wave. (i) On Fig. 5.2, sketch the variation of the intensity I of the transmitted light with the angle of rotation α as the filter is rotated through 360° from its initial position. I0 I I0 2 0 0 90 180 270 360 α/ ° Fig. 5.2 [3] (ii) The amplitude of the incident light wave is A0 when the intensity of the wave is I0. Use Malus’s law to determine, in terms of A0, the amplitude of the transmitted wave when α = 20°. amplitude = .................................................... A0 [4] [Total: 8]

Mark scheme: 5(a) sound waves are longitudinal / not transverse (and only transverse waves can be polarised) B1 5(b)(i) A periodic curve of at least one period with minimum 0 and maximum I0 B1 Smooth continuous sin2 curve B1 Peaks at 90° and 270° and troughs at 0, 180° and 360° B1 5(b)(ii) I ∝ A2 C1 I = I0 cos2 θ C1 At = 20 the angle between the planes of polarisation of light and filter is 70° (or 110°) C1 I = I0 cos2 70 or I = I0 cos2 110 (I = 0.12 I0) amplitude = 0.34 A0 A1

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6 (a) State what is meant by diffraction. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Light of wavelength 720 nm in a vacuum is incident normally on a diffraction grating as shown in Fig. 6.1. second-order maxima light, wavelength 720 nm 52° diffraction grating screen Fig. 6.1 (not to scale) A screen is parallel to the grating. An interference pattern is seen on the screen and the angle between the second-order maxima is 52°. (i) Calculate the frequency of the light. frequency = .................................................... Hz [2] (ii) Calculate the number of lines per unit length in the diffraction grating. number per unit length = .................................................. m−1 [3] (iii) The light in Fig. 6.1 is now replaced with light of a different wavelength λ. It is observed that the third-order maxima of this light are at the same positions as the second-order maxima of the light in Fig. 6.1. Calculate, in nm, the wavelength λ. λ = ................................................... nm [2] [Total: 8]

Mark scheme: 6(a) wave passes (through) an aperture and spreads B1 or wave passes (by / through / around) an edge and spreads 6(b)(i) v = f C1 f = 3.00  108 / 720  10−9 A1 = 4.2  1014 Hz 6(b)(ii) d = nλ / sin θ C1 d = (2  720  10−9) / sin 26 C1 (= 3.3  10−6 m) A1 number of lines per m = 1 / (3.3  10−6) = 3.0  105 m−1 6(b)(iii) 1 C1   3 = 720  2 or sin 26 = 3 3.0  105  = 480 nm A1

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7 A nichrome resistance wire has length 150 cm, cross-sectional area 2.45 × 10−7 m2 and resistivity 1.12 × 10−6 Ω m. (a) Calculate, to three significant figures, the resistance of the wire. resistance = ..................................................... Ω [3] (b) The nichrome wire forms part of a potentiometer circuit together with a cell of electromotive force (e.m.f.) 1.2 V and negligible internal resistance, as shown in Fig. 7.1. 1.2 V 150 cm 64 cm nichrome wire cell X Fig. 7.1 (not to scale) The circuit is used to determine the e.m.f. of cell X. The galvanometer is used in a null method to find the null point 64 cm from the left-hand end of the nichrome wire. (i) Explain what is meant by a null method. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Calculate the e.m.f. of cell X. e.m.f. = ...................................................... V [2] (iii) The cell of e.m.f. 1.2 V is replaced by a new cell with the same e.m.f. but with an internal resistance that is not negligible. State and explain the effect, if any, of the internal resistance of the new cell on the position of the null point. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: 7(a) R = L / A C1 = (1.12  10−6  1.5) / 2.45  10−7 C1 = 6.86  A1 7(b)(i) (A method where the) reading (on the galvanometer) is zero. B1 7(b)(ii) e.m.f. / 1.2 = 64 / 150 C1 e.m.f. = (64 / 150)  1.2 A1 = 0.51 V 7(b)(iii) (the internal resistance will cause a) drop in p.d. across the wire / the terminal p.d. is lower B1 So the null point will move to the right B1

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8 (a) An antiparticle equivalent of the neutron is called the antineutron. The quarks in the antineutron are the antiparticles of the quarks in a neutron. The elementary charge is e. In Table 8.1, state the flavour and charge of the three antiquarks that comprise the antineutron. Table 8.1 flavour charge / e [3] (b) In β− decay, a neutron decays to form a proton. Theory predicts that an antineutron should decay to form an antiproton. A particle and an antiparticle should also be observed. Suggest the names of the particle and the antiparticle. particle: ..................................................................................................................................... antiparticle: ............................................................................................................................... [2] [Total: 5]

Mark scheme: 8(a) flavour charge / e up / u 2 − 3 down / d 1 ( + ) 3 down / d 1 ( + ) 3 3 correct quark flavours B1 Charge on anti-up quark –⅔(e) B1 Charge on anti-down quark (+)⅓(e) B1 8(b) particle: (electron) neutrino B1 antiparticle: positron B1

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