Cambridge A Level Physics 9702 — 2023 May/June Paper 2 · Variant 2

1 (a) (i) Define pressure. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Use the answer to (a)(i) to show that the SI base units of pressure are kg m–1 s–2. [1] (b) A horizontal pipe has length L and a circular cross‑section of radius R. A liquid of density ρ flows through the pipe. The mass m of liquid flowing through the pipe in time t is given by π(p2 – p1)R 4ρt m = 8kL where p1 and p2 are the pressures at the ends of the pipe and k is a constant. Determine the SI base units of k. SI base units .......................................................... [3] (c) An experiment is performed to determine the value of k by measuring the values of the other quantities in the equation in (b). The values of L and R each have a percentage uncertainty of 2%. State and explain, quantitatively, which of these two quantities contributes more to the percentage uncertainty in the calculated value of k. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [1] [Total: 6]

Mark scheme: 1(a)(i) force / area (normal to the force) B1 1(a)(ii) (p = F / A so units are) kg m s–2 / m2 = kg m–1 s–2 A1 1(b) unit of R: m and unit of t: s and unit of L: m C1 unit of : kg m–3 or  = m / V C1 base units of k: (kg m–1 s–2  m4  kg m–3  s) / (kg  m) = kg m–1 s–1 A1 1(c) R contributes 4  2% or 8% (and L contributes 2%) so R contributes more (to the percentage uncertainty in k) B1

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2 (a) State what is meant by the centre of gravity of an object. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Two blocks are on a horizontal beam that is pivoted at its centre of gravity, as shown in Fig. 2.1. 0.45 m 0.95 m 0.35 m horizontal 30° pivot beam 54 N 2.4 N string support T ground Fig. 2.1 (not to scale) A large block of weight 54 N is a distance of 0.45 m from the pivot. A small block of weight 2.4 N is a distance of 0.95 m from the pivot and a distance of 0.35 m from the right‑hand end of the beam. The right‑hand end of the beam is connected to the ground by a string that is at an angle of 30° to the horizontal. The beam is in equilibrium. (i) By taking moments about the pivot, calculate the tension T in the string. T = ...................................................... N [3] (ii) The string is cut so that the beam is no longer in equilibrium. Calculate the magnitude of the resultant moment about the pivot acting on the beam immediately after the string is cut. resultant moment = ................................................... N m [1] (c) The beam in (b) rotates when the string is cut and the small block of weight 2.4 N is projected through the air. Fig. 2.2 shows the last part of the path of the block before it hits the ground at point Y. path of X block 1.8 m horizontal ground Y Fig. 2.2 (not to scale) At point X on the path, the block has a speed of 3.4 m s–1 and is at a height of 1.8 m above the horizontal ground. Air resistance is negligible. (i) Calculate the decrease in the gravitational potential energy of the block for its movement from X to Y. decrease in gravitational potential energy = ....................................................... J [2] (ii) Use your answer to (c)(i) and conservation of energy to determine the kinetic energy of the block at Y. kinetic energy = ....................................................... J [3] (iii) State the variation, if any, in the direction of the acceleration of the block as it moves from X to Y. ..................................................................................................................................... [1] (iv) The block passes point X at time tX and arrives at point Y at time tY. On Fig. 2.3, sketch a graph to show the variation of the magnitude of the horizontal component of the velocity of the block with time from tX to tY. Numerical values are not required. horizontal component of velocity 0 tX tY time Fig. 2.3 [1] [Total: 12]

Mark scheme: 2(a) the point where (all) the weight (of the object) is taken to act B1 2(b)(i) (54  0.45) or (2.4  0.95) or (T sin 30°  1.3) C1 (54  0.45) = (2.4  0.95) + (T sin 30°  1.3) C1 T = 34 N A1 2(b)(ii) resultant moment = (54  0.45) – (2.4  0.95) or (34 sin 30°  1.3) = 22 N m A1 2(c)(i) (∆)E = mg(∆)h or W(∆)h C1 = 2.4  1.8 = 4.3 J A1 Question Answer Marks 2(c)(ii)  2 1 2 E mv C1 = 1 2  (2.4 / 9.81)  3.42 = 1.4 J (at X) C1 kinetic energy at Y = 4.3  1.4 = 5.7 J A1 or 2 1 2 mv = 2 1 2 mu  mg()h (C1) v2 = 3.42  2  9.81  1.8 v2 = 46.9 so v = 6.85 (m s–1) KE = 1 2  (2.4 / 9.81)  6.852 (C1) = 5.7 J (A1) 2(c)(iii) no variation or acceleration is (always) vertically downwards B1 2(c)(iv) horizontal straight line at a non-zero value of velocity B1

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3 A block is pulled by a force X in a straight line along a rough horizontal surface, as shown in Fig. 3.1. velocity total resistive force 0.80 N X horizontal surface Fig. 3.1 Assume that the total resistive force opposing the motion of the block is 0.80 N at all speeds of the block. The variation with time t of the magnitude of the force X is shown in Fig. 3.2. 2.0 X / N 1.5 1.0 0.5 0 0 1 2 3 4 5 6 t / s Fig. 3.2 (a) (i) Define force. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Determine the change in momentum of the block from time t = 0 to time t = 3.0 s. change in momentum = ............................................ kg m s–1 [2] (b) (i) Describe and explain the motion of the block between time t = 3.0 s and time t = 6.0 s. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Force X produces a total power of 2.0 W when moving the block between time t = 3.0 s and time t = 6.0 s. Calculate the distance moved by the block during this time interval. distance = ...................................................... m [3] (c) The block is at rest at time t = 0. On Fig. 3.3, sketch a graph to show the variation of the momentum of the block with time t from t = 0 to t = 6.0 s. Numerical values of momentum are not required. momentum 0 0 1 2 3 4 5 6 t / s Fig. 3.3 [2] [Total: 10]

Mark scheme: 3(a)(i) rate of change of momentum B1 3(a)(ii) change in momentum = (1.4 – 0.80)  3.0 C1 = 1.8 kg m s–1 A1 3(b)(i) resultant force (on block) is zero B1 (so) velocity is constant B1 3(b)(ii) P = Fv or P = Fs / t C1 v = 2.0 / 0.80 (= 2.5 m s–1) C1 distance = 2.5  3.0 = 7.5 m A1 or P = W / t or P = Fs / t (C1) W = 2.0  3.0 (= 6.0 J) (C1) distance = 6.0 / 0.80 = 7.5 m (A1) 3(c) 0 to 3.0 s: upward sloping straight line from the origin. B1 3.0 to 6.0 s: horizontal line at non-zero value of momentum with no ‘step change’ in momentum at 3.0 s B1

More questions on Momentum and Newton’s laws of motion

4 A spring is suspended from a fixed point at one end. The spring is extended by a vertical force applied to the other end. The variation of the applied force F with the length L of the spring is shown in Fig. 4.1. 12 10 F / N 8 6 4 2 0 0 4 8 12 16 20 24 L / cm Fig. 4.1 For the spring: (a) state the name of the law that gives the relationship between the force and the extension ............................................................................................................................................. [1] (b) determine the spring constant, in N m–1 spring constant = ................................................ N m–1 [2] (c) determine the elastic potential energy when F = 6.0 N. elastic potential energy = ....................................................... J [2] [Total: 5]

Mark scheme: 4(a) Hooke’s (law) B1 4(b) k = F / x or k = gradient C1 = e.g. 12.0 / (0.240 – 0.08) = 75 N m–1 A1 4(c) E = 1 2 Fx or E = 1 2 kx 2 or E = area under graph C1 E = 1 2  6.0  0.080 or 1 2  75  0.082 = 0.24 J A1

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5 (a) A progressive wave travels through a medium. The wave causes a particle of the medium to vibrate along a line P. The energy of the wave propagates along a line Q. Compare the directions of lines P and Q if the wave is: (i) a transverse wave ..................................................................................................................................... [1] (ii) a longitudinal wave. ..................................................................................................................................... [1] (b) A tube is closed at one end. A loudspeaker is placed near the other end of the tube, as shown in Fig. 5.1. tube A A loudspeaker L Fig. 5.1 (not to scale) The loudspeaker emits sound of frequency 1.7 kHz. The speed of sound in the air in the tube is 340 m s–1. A stationary wave is formed with an antinode A at the open end of the tube. There is only one other antinode A inside the tube, as shown in Fig. 5.1. Determine: (i) the wavelength of the sound wavelength = ...................................................... m [2] (ii) the length L of the tube L = ...................................................... m [1] (iii) the maximum wavelength of the sound from the loudspeaker that can produce a stationary wave in the tube. maximum wavelength = ...................................................... m [1] (c) Two polarising filters are arranged so that their planes are vertical and parallel. The first filter has its transmission axis at an angle of 35° to the vertical and the second filter has its transmission axis at angle α to the vertical, as shown in Fig. 5.2. 35° α incident light beam, intensity 8.5 W m–2 intensity 5.2 W m–2 transmission first filter second filter axis of filter Fig. 5.2 Angle α is greater than 35° and less than 90°. A beam of vertically polarised light of intensity 8.5 W m–2 is incident normally on the first filter. (i) Show that the intensity of the light transmitted by the first filter is 5.7 W m–2. [1] (ii) The intensity of the light transmitted by the second filter is 5.2 W m–2. Calculate angle α. α = ........................................................° [2] [Total: 9]

Mark scheme: 5(a)(i) (they are) perpendicular B1 5(a)(ii) (they are) parallel B1 5(b)(i)  = v / f C1 = 340 / 1700 = 0.20 m A1 5(b)(ii) L = 3 4   = 3 4  0.20 = 0.15 m A1 5(b)(iii)  = 4  0.15 or 0.20  3 = 0.60 m A1 5(c)(i) (I =) 8.5  cos2 35° = 5.7 (W m–2) A1 5(c)(ii) 5.2 = 5.7 cos2  ( = 17°) C1  = 35° + 17° = 52° A1

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6 (a) The current in a filament lamp decreases. State and explain how the resistance of the lamp changes. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A cylindrical wire has length L and resistance R. The total number of free electrons (charge carriers) contained in the volume of the wire is N. Each free electron has charge e. The potential difference between the ends of the wire is V. Determine expressions, in terms of some or all of the symbols e, L, N, R and V for: (i) the current in the wire current = ......................................................... [1] (ii) the average drift speed of the free electrons average drift speed = ......................................................... [2] (iii) the average time taken for a free electron to move along the full length of the wire. time taken = ......................................................... [1] [Total: 5]

Mark scheme: 6(a) temperature decreases (so) resistance decreases B1 6(b)(i) current = V / R A1 6(b)(ii) I = Anvq n = N / V or n = N / AL C1 v = (V / R) / [(V / L) (N / V) e] or (V / R) / [A (N / AL) e] = VL / RNe A1 or v = L / t = L / (Q / I) (C1) = LI / Q = L(V / R) / Ne = VL / RNe (A1) 6(b)(iii) time = distance / speed or Q / I = L / (VL / RNe) or Ne / (V / R) time = RNe / V A1

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7 (a) A battery of electromotive force (e.m.f.) 9.0 V and negligible internal resistance is connected to a light‑dependent resistor (LDR) and a fixed resistor, as shown in Fig. 7.1. 9.0 V 1800 Ω 1200 Ω Fig. 7.1 The LDR and fixed resistor have resistances of 1800 Ω and 1200 Ω respectively. Calculate the potential difference across the LDR. potential difference = ...................................................... V [2] (b) The circuit in (a) is now modified by adding a uniform resistance wire XY and a galvanometer, as shown in Fig. 7.2. 9.0 V Z X Y 1800 Ω 1200 Ω Fig. 7.2 (not to scale) The length of the wire XY is 1.2 m. The movable connection Z is positioned on the wire XY so that the galvanometer reading is zero. (i) Calculate the length XZ along the resistance wire. length XZ = ...................................................... m [2] (ii) The environmental conditions change causing a decrease in the resistance of the LDR. The temperature of the LDR remains constant. State whether there is a decrease, increase or no change to: the intensity of the light illuminating the LDR • ........................................................................................................................................... the total power produced by the battery • ........................................................................................................................................... the length XZ so that the galvanometer reads zero. • ........................................................................................................................................... [3] [Total: 7]

Mark scheme: 7(a) V / 9.0 = 1800 / (1800 + 1200) C1 V = 5.4 V A1 or I = 9.0 / (1800  1200) = 3.0  10–3 (A) V = 3.0  10–3  1800 (C1) = 5.4 V (A1) 7(b)(i) L / 1.2 = 5.4 / 9.0 or XZ / 1.2 = 5.4 / 9.0 C1 L = 0.72 m A1 or L / 1.2 = 1800 / (1800  1200) or XZ / 1.2 = 1.8 / (1.8  1.2) (C1) L = 0.72 m (A1) 7(b)(ii)  (intensity) increase B1  (power) increase B1  (length XZ) decrease B1

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8 (a) Nucleus P and nucleus Q are isotopes of the same element. Nucleus Q is unstable and emits a β– particle to form nucleus R. (i) For nuclei P and Q, compare: the number of protons • ........................................................................................................................................... the number of neutrons. • ........................................................................................................................................... [2] (ii) When nucleus Q decays to form nucleus R, the quark composition of a nucleon changes. State the change to the quark composition of the nucleon. ..................................................................................................................................... [1] (iii) State the name of another particle that must be emitted from nucleus Q in addition to the β– particle. ..................................................................................................................................... [1] (b) A hadron consists of two charm quarks and one bottom quark. Determine, in terms of the elementary charge e, the charge of the hadron. charge = ....................................................... e [2] [Total: 6]

Mark scheme: 8(a)(i) number of protons: equal/same B1 number of neutrons: unequal/different B1 8(a)(ii) down (quark) changes to up (quark) or up down down (quarks) change to up up down (quarks) B1 8(a)(iii) (electron) antineutrino B1 8(b) charm (quark charge) is ()2/3(e) or 2 charm (quark charges) is ()4/3(e) or bottom (quark charge) is –1/3(e) C1 charge = 2/3(e) 2/3(e) –1/3(e) = (+)1(e) A1

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