Cambridge A Level Physics 9702 — 2022 May/June Paper 2 · Variant 3

1 A solid metal sphere has a diameter of (3.42 ± 0.02) cm and a mass of (67 ± 2) g. (a) Calculate the density, in g cm–3, of the metal. density = .............................................. g cm–3 [3] (b) Determine the percentage uncertainty in the density. percentage uncertainty = ......................................................% [2] [Total: 5]

Mark scheme: 1(a) C1 V = (4 / 3)    r3 = (4 / 3)    (3.42 / 2)3 ( = 20.9 cm3) C1  = 67 / 20.9 = 3.2 g cm−3 A1 1(b) % = %m + 3  %d = [(2 / 67)  100] + [3  (0.02 / 3.42)  100] C1 = 3.0% + 3  0.58% = 4.7% or 5% A1

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2 An archer releases an arrow towards a target at a velocity of 65.0 m s–1 at an angle of 4.30° above the horizontal, as shown in Fig. 2.1. arrow, speed 65.0 m s–1 4.30° centre of target target archer 1.66 m 70.0 m ground Fig. 2.1 (not to scale) When released, the tip of the arrow is a horizontal distance of 70.0 m from the target and 1.66 m above the horizontal ground. The arrow hits the centre of the target. Assume that air resistance is negligible and that all the mass of the arrow is at its tip. (a) Show that the time taken for the arrow to reach the target is 1.08 s. [2] (b) Calculate the height of the centre of the target above the ground. height above ground = ...................................................... m [3] (c) By considering energy changes, state and explain how the final kinetic energy of the arrow as it hits the target compares with its initial kinetic energy immediately after release. A numerical calculation is not required. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 7]

Mark scheme: 2(a) (time =) displacement / velocity C1 (time =) 70.0 / 65.0 cos 4.30° = 1.08 (s) A1 2(b) s = ut + ½at2 = (65  sin 4.30°  1.08) – (0.5  9.81  1.082) C1 s = −0.46 (m) C1 height above ground = 1.66 − 0.46 = 1.2 m A1 2(c) GPE has decreased M1 (total energy conserved so) KE has increased A1

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3 (a) Define velocity. ................................................................................................................................................... ............................................................................................................................................. [1] (b) A constant driving force of 2400 N acts on a car of mass 1200 kg. The car accelerates from rest in a straight line along a horizontal road. Assume that the resistive forces acting on the car are negligible. (i) Calculate the acceleration of the car. acceleration = ................................................ m s–2 [1] (ii) On Fig. 3.1, sketch a graph showing the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line A. 50 40 v / m s–1 30 20 10 0 0 4 8 12 16 20 t / s Fig. 3.1 [2] (c) In reality, a resistive force due to air resistance acts on the car in (b). This resistive force increases with speed until it becomes equal in magnitude to the driving force at time t = 12 s. (i) On Fig. 3.1, sketch a second line to show the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line B. [3] (ii) At time t = 20 s, the driving force is increased to 3000 N and remains constant at this value. Describe how the velocity of the car changes due to this increase in the driving force. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 9]

Mark scheme: 3(a) (velocity =) change in displacement / time B1 3(b)(i) a = 2400 / 1200 = 2.0 m s−2 A1 3(b)(ii) straight line from the origin with positive gradient (labelled A) M1 ending at (20, 40) A1 3(c)(i) line starting at origin (with the same gradient as A) and beneath A at all points B1 gradient decreasing to zero B1 straight horizontal line from t = 12 s and ending at t = 20 s (and labelled B) B1 3(c)(ii) the velocity/speed will increase B1 to a new terminal/constant/maximum velocity/speed B1 or the car has an acceleration (B1) to a new (higher) terminal/constant/maximum velocity/speed (B1)

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4 (a) A mass m moves a vertical distance Δh in a uniform gravitational field and gains gravitational potential energy ΔEP. The acceleration of free fall is g. Use the concept of work done to show that ΔEP = mgΔh. [2] (b) A 0.60 kg mass is attached to a string which is wrapped around the wheel of a generator, as shown in Fig. 4.1. generator wheel string resistor mass, 0.60 kg Fig. 4.1 The mass is held stationary above the floor. When released, the mass initially accelerates and then falls at a steady speed and spins the wheel. The generator causes a current in a resistor. Air resistance is negligible. State the main energy change when the mass is falling at a steady speed. ........................................... energy to ........................................... energy. [1] (c) When falling at a steady speed, the mass in (b) falls through a vertical distance of 1.4 m in a time of 4.0 s. This causes a current of 90 mA in the resistor. The resistance of the resistor is 47 Ω. Calculate: (i) the rate of work done by the falling mass rate of work done = ..................................................... W [2] (ii) the power dissipated in the resistor power = ..................................................... W [2] (iii) the efficiency of the generator. efficiency = ......................................................... [2] [Total: 9]

Mark scheme: 4(a) M1 (force = mg and distance = h) (so) work (done) = mgh and work = E(P) (so E(P) = mgh) A1 4(b) gravitational potential (energy) to heat/thermal (energy) B1 4(c)(i) P = mg()h / ()t or Fv C1 P = (0.60  9.81  1.4) / 4.0 or 0.60  9.81  (1.4 / 4.0) = 2.1 W A1 4(c)(ii) P = I2R or IV or V2 / R C1 = 0.092  47 or 0.09  4.23 or 4.232 / 47 = 0.38 W A1 4(c)(iii) efficiency = Pout / Pin ( 100) or Eout / Ein ( 100) C1 = 0.38 / 2.1 ( 100) or 0.38  4.0 / 2.1  4.0 ( 100) = 0.18 or 18% A1

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5 (a) Parallel light rays from the Sun are incident normally on a magnifying glass. The magnifying glass directs the light to an area A of radius r, as shown in Fig. 5.1. parallel light rays from Sun r A 5.5 cm magnifying glass Fig. 5.1 (not to scale) The magnifying glass is circular in cross‑section with a radius of 5.5 cm. The intensity of the light from the Sun incident on the magnifying glass is 1.3 kW m–2. Assume that all of the light incident on the magnifying glass is transmitted through it. (i) Calculate the power of the light from the Sun incident on the magnifying glass. power = ..................................................... W [2] (ii) The value of r is 1.5 mm. Calculate the intensity of the light on area A. intensity = ............................................... W m–2 [1] (b) A laser emits a beam of electromagnetic waves of frequency 3.7 × 1015 Hz in a vacuum. (i) Show that the wavelength of the waves is 8.1 × 10–8 m. [2] (ii) State the region of the electromagnetic spectrum to which these waves belong. ..................................................................................................................................... [1] (iii) The beam from the laser now passes through a diffraction grating with 2400 lines per millimetre. A detector sensitive to the waves emitted by the laser is moved through an arc of 180° in order to detect the maxima produced by the waves passing through the grating, as shown in Fig. 5.2. detector diffraction grating laser beam from laser detector moves along this line Fig. 5.2 Calculate the number of maxima detected as the detector moves through 180° along the line shown in Fig. 5.2. Show your working. number of maxima detected = ......................................................... [4] (iv) The laser is now replaced with one that emits electromagnetic waves with a wavelength of 300 nm. Explain, without calculation, what happens to the number of maxima now detected. Assume that the detector is also sensitive to this wavelength of electromagnetic waves. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 12]

Mark scheme: 5(a)(i) C1 = 1.3  103  (  0.0552) = 12 W A1 5(a)(ii) intensity = power / area = 12 / (  0.00152) = 1.7  106 W m−2 A1 5(b)(i) ( =) v / f or c / f C1 ( =) 3.0  108 / 3.7  1015 = 8.1  10−8 (m) A1 5(b)(ii) ultraviolet A1 Question Answer Marks 5(b)(iii) d sin  = n or (1 / N)  sin  = n C1 d = 1 / 2400  103 (m) = 4.2  10–7 (m) or N = 2400  103 (m–1) C1 n = 4.2  10−7  sin 90° / 8.1  10−8 or sin 90° / 2400  103  8.1  10–8 n = 5.2 or 5.1 or when n = 5,  = 76.4° and when n = 6, sin > 1 (so) n = 5 B1 number of maxima = (2  5) + 1 = 11 A1 5(b)(iv) the wavelength has increased M1 (so) number of maxima decreases A1

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6 (a) (i) On Fig. 6.1, sketch the I–V characteristic of a filament lamp. I 0 V 0 Fig. 6.1 [2] (ii) Explain the shape of the line in (a)(i). ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (b) A conducting wire has length 5.8 m and cross‑sectional area 3.4 × 10–8 m2. The resistivity of the metal of the wire is 5.6 × 10–8 Ω m. Calculate the resistance of the wire. resistance = ...................................................... Ω [2] (c) A resistor of resistance R is placed in a circuit with a cell of negligible internal resistance, two switches S1 and S2, a second resistor of resistance 2R and three ammeters X, Y and Z. The circuit is shown in Fig. 6.2. X A S1 R Y A S2 Z 2R A Fig. 6.2 The reading on X is 1.0 A when S1 is open and S2 is closed. Complete Table 6.1. Table 6.1 position of switches ammeter readings S1 S2 reading on X / A reading on Y / A reading on Z / A open open 0 0 0 open closed 1.0 closed open closed closed [4] [Total: 11]

Mark scheme: 6(a)(i) line passes through (0,0) and is in first and third quadrants only M1 gradient of line becoming less steep in both quadrants and roughly symmetrical A1 6(a)(ii) (as I increases) the temperature (of the filament wire/lamp) increases B1 (as I / temperature / V increases) the resistance (of wire/lamp) increases B1 (as I / temperature / V increases the graph curves because) ratio V / I increases or ratio I / V decreases B1 6(b) R = L / A C1 = (5.6  10−8  5.8) / 3.4  10−8 = 9.6  A1 Question Answer Marks 6(c) position of switches ammeter readings S1 S2 X / A Y / A Z / A open open 0 0 0 open closed 1.0 0 1.0 closed open 2.0 2.0 0 closed closed 3.0 2.0 1.0 second row: both values correct (B1) third row: all three values correct (B1) fourth row: X = Y + Z (any values) (B1) all three values correct (B1) B4

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7 (a) Fluorine‑18 (189F) is an isotope that decays to an isotope of oxygen (O) by the emission of a β+ particle. (i) Complete the nuclear equation for the decay, including all the particles involved. 189F [3] (ii) A quark in the fluorine‑18 nucleus changes flavour during the decay. State this change of flavour. ......................... quark to ......................... quark. [1] (b) A hadron has a charge of –2e, where e is the elementary charge. (i) State and explain whether the hadron is a meson or a baryon. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) State a possible quark composition for the hadron. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 7]

Mark scheme: 7(a)(i) 18 9F  18 8O + 0 ( )1 (+) + (0) (0) (e)  or neutrino (B1) 0 ( )1 (+) (B1) 18 8O (B1) 7(a)(ii) up quark to down quark B1 7(b)(i) must be three (anti)quarks as largest (negative) quark charge is (–)2/3 (e) or mesons can only have a charge of 0 or 1(e) M1 (so hadron is) a baryon A1 7(b)(ii) any combination of three from: antiup (quark) / up antiquark and/or anticharm (quark) / charm antiquark and/or antitop (quark) / top antiquark B1

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