Cambridge A Level Physics 9702 — 2025 Oct/Nov Paper 2 · Variant 4

1 A child kicks a ball so that it leaves horizontal ground with a velocity of 28 m s–1 at an angle of 34° to the horizontal, as shown in Fig. 1.1. 28 m s–1 34° ground Fig. 1.1 Air resistance is negligible. The ball leaves the ground at time t = 0. (a) (i) Calculate the horizontal component vH and the vertical component vV of the velocity of the ball immediately after it has left the ground. vH = ...................................................... m s–1 vV = ...................................................... m s–1 [2] (ii) Show that the ball reaches its maximum height at time t = 1.6 s. [1] (iii) On Fig. 1.2, sketch the variation of vH with time t between t = 0 and t = 3.2 s. Label your line H. 30 velocity / m s–1 20 10 0 0 0.8 1.6 2.4 3.2 t / s –10 –20 –30 Fig. 1.2 [1] (iv) On Fig. 1.2, sketch the variation of vV with time t between t = 0 and t = 3.2 s. Assume that velocity in the upward direction is positive. Label your line V. [3] (b) The total change in momentum of the ball between leaving the ground at t = 0 and landing on the ground at t = 3.2 s is 13 kg m s–1. (i) Define momentum. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Calculate the force that acts on the ball while it is in the air. force = ....................................................... N [2] (iii) Determine the mass of the ball. mass = .................................................... kg [1] [Total: 11]

Mark scheme: Question Answer Marks 1(a)(i) vH = 28 × cos 34° A1 = 23 m s–1 vV = 28 × sin 34° A1 = 16 m s–1 1(a)(ii) time = 16 / 9.81 = 1.6 s A1 1(a)(iii) horizontal straight line at v = 23 m s–1 from t = 0 to t = 3.2 s B1 1(a)(iv) straight diagonal line starting at a positive velocity from t = 0 to t = 3.2 s, crossing the time axis B1 line starting at v = 16 m s–1 and ending at v = –16 m s–1 B1 line passing through v = 0 at t = 1.6 s B1 1(b)(i) product of mass and velocity B1 1(b)(ii) F = p / t C1 = 13 / 3.2 A1 = 4.1 N 1(b)(iii) m = p / v A1 = 13 / (2  16) = 0.41 kg or m = F / g = 4.06 / 9.81 = 0.41 kg

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2 Fig. 2.1 shows a square metal sheet of non-uniform density, with a thin wooden rod fixed at its centre. One of the corners of the sheet is labelled X. X metal sheet rod Fig. 2.1 The rod has negligible mass. The mass of the metal sheet is 2.8 kg. The rod is supported so that the rod is horizontal and the metal sheet is vertical. (a) Define the torque of a couple. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) When the rod is supported in such a way that it can rotate freely within its support, the sheet hangs in equilibrium with point X vertically above the rod, as shown in Fig. 2.2. X rod metal sheet Fig. 2.2 On Fig. 2.2, draw a line to indicate the range of possible positions for the centre of gravity of the metal sheet. [1] (c) When a torque of 3.3 N m is applied to the rod, the sheet is held in equilibrium with two of its edges horizontal, as shown in Fig. 2.3. Point X is at the top-left corner. X metal sheet rod Fig. 2.3 (i) Explain whether the torque applied to the rod to hold the sheet in equilibrium is clockwise or anticlockwise. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Show that the centre of gravity of the sheet has a horizontal displacement of 0.12 m from the rod. [1] (d) The square metal sheet has an average density of 3000 kg m–3 and a uniform thickness of 4.0 mm. Show that the side length of the sheet is 0.48 m. [3] (e) Use the answer in (b) and the information in (c) and (d) to determine the position of the centre of gravity of the sheet. Indicate this position on Fig. 2.3 with a point labelled Y. [2] [Total: 10]

Mark scheme: 2(a) product of force and distance M1 perpendicular distance between the (line of action of the two) forces A1 2(b) straight vertical line drawn from centre to bottom corner B1 2(c)(i) centre of gravity is to the right of the rod so torque is anticlockwise B1 or moment of weight of sheet about rod is clockwise so torque is anticlockwise 2(c)(ii) (horizontal displacement) = 3.3 / (2.8  9.81) = 0.12 (m) A1 2(d)  = m / V C1 V = 4.0  10–3 × (side length)2 C1 A1 0.48m side length =   2.8 / ( 3000  0.0040 )  = 2(e) point lies on a straight line at 45° to the horizontal from the bottom-right corner to the edge of the rod B1 point lies on a vertical line half-way between rod and right-hand edge B1

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3 A bungee jumper of mass 64 kg secures one end of an elastic rope to a bridge. The other end is attached to the bungee jumper. The jumper falls from rest from the bridge and descends into the valley below, as shown in Fig. 3.1. rope h jumper Fig. 3.1 (not to scale) Fig. 3.2 shows the variation of the tension T in the rope with the vertical distance h of the jumper below the level of the bridge. 2000 T / N 1500 1000 500 0 0 20 40 60 80 100 120 h / m Fig. 3.2 (a) The rope obeys Hooke’s law. State Hooke’s law. ................................................................................................................................................... ............................................................................................................................................. [1] (b) (i) Determine the unstretched length of the rope. length = ..................................................... m [1] (ii) Determine the spring constant k of the rope. k = ............................................... N m–1 [2] (c) For the position of the bungee jumper at a distance of 120 m below the bridge: (i) show that the loss of gravitational potential energy since leaving the bridge is 75 kJ [2] (ii) show that the elastic potential energy in the rope is 75 kJ. [2] (d) Explain what can be deduced from the information in (c) about the speed of the bungee jumper when at a distance of 120 m below the bridge. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 10]

Mark scheme: 3(a) force is proportional to extension B1 3(b)(i) length = 37 m A1 3(b)(ii) spring constant = F / x C1 = e.g. 1800 / (120 – 37) A1 = 22 N m–1 3(c)(i) (()E) = mg()h C1 = 64 × 9.81 × 120 = 75 000 J or 75 kJ A1 3(c)(ii) 1 1 1 C1 (E =) Fx or (E =) kx2 or (E =) F2 / k or (E =) area under graph 2 2 2 1 A1 (E =) × 1800 × (120 – 37) = 75 000 J or 75 kJ 2 or 1 (E =) × 21.7 × (120 – 37)2 = 75 000 J or 75 kJ 2 3(d) (all) gravitational potential energy has been converted / equal to elastic potential energy (so no kinetic energy) B1 kinetic energy is zero so speed is zero B1

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4 (a) State the principle of superposition. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) An electromagnetic wave of wavelength 0.026 m in free space is incident normally on an aluminium sheet, as shown in Fig. 4.1. transmitter aluminium sheet electromagnetic wave Fig. 4.1 The wave reflects at the aluminium sheet and a stationary wave is formed in the region between the transmitter and the sheet. (i) Explain how the stationary wave, including its nodes and antinodes, is formed. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Calculate the frequency of the electromagnetic wave. frequency = .................................................... Hz [2] (iii) State the principal region of the electromagnetic spectrum to which the wave belongs. ..................................................................................................................................... [1] (iv) Determine the distance between a node and an adjacent antinode. distance = ..................................................... m [1] [Total: 9]

Mark scheme: 4(a) when two (or more) waves meet M1 (resultant) displacement equals the sum of the displacements of the (two separate) waves A1 4(b)(i) the incident and reflected waves superpose B1 (the waves superpose so that the resultant) amplitude is maximum at an antinode B1 (the waves superpose so that the resultant) amplitude is minimum / zero at a node B1 4(b)(ii) c = f C1 f = (3.00 × 108) / 0.026 A1 = 1.2 × 1010 Hz 4(b)(iii) microwave B1 4(b)(iv) distance = 0.026 / 4 A1 = 6.5 × 10–3 m

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5 A student uses a circuit containing an ammeter, a voltmeter and a cell to take measurements to determine the resistance of a length of nichrome wire. (a) (i) Define resistance. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Draw a circuit diagram to show how the components should be connected. Use the symbol for a resistor to represent the nichrome wire. [2] (b) The student also measures the length and the diameter of the wire. Table 5.1 shows the measurements recorded for each quantity. Table 5.1 quantity measurement length (0.864 ± 0.001) m diameter (0.496 ± 0.002) mm voltmeter reading (1.38 ± 0.02) V ammeter reading (0.276 ± 0.001) A (i) Show that the resistance of the wire is 5.00 Ω. [1] (ii) Calculate, to three significant figures, the resistivity ρ of the nichrome. ρ = .................................................. Ω m [3] (iii) Calculate the percentage uncertainty in ρ. percentage uncertainty = ..................................................... % [2] (iv) Determine the absolute uncertainty in ρ. absolute uncertainty = .................................................. Ω m [1] [Total: 10]

Mark scheme: 5(a)(i) potential difference per unit current B1 5(a)(ii) resistor connected to cell in a closed loop and correct circuit symbols used for all components B1 ammeter connected in series with resistor and voltmeter connected in parallel with resistor B1 5(b)(i) (resistance) = 1.38 / 0.276 = 5.00 () A1 5(b)(ii)  = RA / L C1  = 5.00    (0.496  10–3)2 / (4  0.864) C1 = 1.12  10–6  m A1 5(b)(iii) calculation of a fractional or percentage uncertainty in one quantity C1 0.001 / 0.864 or 0.002 / 0.496 or 0.02 / 1.38 or 0.001 / 0.276 percentage uncertainty = [(0.001 / 0.864) + (2  0.002 / 0.496) + (0.02 / 1.38) + (0.001 / 0.276)]  100 A1 = 2.7% 5(b)(iv)  = 0.03  1.12  10–6 A1 = 3  10–8  m

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6 Fig. 6.1 shows four alpha particles W, X, Y and Z moving towards a gold nucleus that is in thin gold foil. gold nucleus W X Y Z Fig. 6.1 (not to scale) (a) (i) The paths of particles X and Z are shown. Complete Fig. 6.1 to show possible paths for particles W and Y. [3] (ii) Describe what may be inferred about the structure of an atom from the path of particle X. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (iii) When a beam containing many alpha particles is incident on thin gold foil, nearly all of the alpha particles follow paths that are similar to the path of particle Z. Describe what may be inferred from this about the structure of an atom. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] (b) State the mass and the charge, in terms of the atomic mass unit u and the elementary charge e, of an alpha particle. mass = ............................................................ u charge = ............................................................ e [2] (c) There are two types of hadron. The hadrons that are in alpha particles are each composed of three quarks. (i) State the name of this type of hadron. ..................................................................................................................................... [1] (ii) Show, by reference to their constituent quarks, that the hadrons in an alpha particle have charges of either zero or +1e. [2] [Total: 10]

Mark scheme: 6(a)(i) W deflects upwards B1 W has deflection that is the mirror image of X B1 Y deflects downwards less than X B1 or path of Y is parallel to particle Z 6(a)(ii) nucleus has most of the mass of the atom B1 or nucleus is charged 6(a)(iii) atom is mostly empty space B1 or nucleus occupies a very small proportion of the space in the atom 6(b) mass = 4 u A1 charge = (+)2e A1 6(c)(i) baryon B1 6(c)(ii) 2 1 C1 charge on up quark = (+) (e) and charge on down quark = − (e) 3 3 (proton is up up down and neutron is up down down) A1  2   1   2   1   2   e −  1   = 1e and  1   e −  2   e = 0  3   3   3   3 

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