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Physics 9702/23 May/June 2019: question paper and mark scheme

Cambridge AS & A Level Physics (9702), Paper 2 AS Level Structured Questions, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9702_s19_qp_239702_s19_ms_239702/23/M/J/19

Paper
Paper 2 AS Level Structured Questions
Total marks
60
Time
1 hour 15 minutes
Questions
7
Code on the paper
9702/23/M/J/19
9702_s19_ms_23.pdf

9702/23 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 16 marks

Question

(a) (i) Define resistance. ……[1] (ii) A potential difference of 0.60 V is applied across a resistor of resistance 4.0 GΩ. Calculate the current, in pA, in the resistor. current = …… pA [2] (b) The energy E transferred when charge Q moves through an electrical component is given by the equation E = QV where V is the potential difference across the component. Use the equation to determine the SI base units of potential difference. SI base units …… [3] [Total: 6]

Mark scheme

1(a)(i) potential difference / current B1 1(a)(ii) R = 4.0 × 109 (Ω) C1 I = 0.60 / 4.0 × 109 = 1.5 × 10–10 (A) A1 I = 150 pA 1(b) units of energy: kg m2 s–2 C1 units of charge: A s C1 units of potential difference: (kg m2 s–2 / A s =) kg m2 A–1 s–3 A1

Question 211 marks

Question

(a) A resultant force F moves an object of mass m through distance s in a straight line. The force gives the object an acceleration a so that its speed changes from initial speed u to final speed v. (i) State an expression for: 1. the work W done by the force, in terms of a, m and s W = …… [1] 2. the distance s, in terms of a, u and v. s = …… [1] (ii) Use your answers in (i) to show that the kinetic energy of the object is given by 1 kinetic energy = × mass × (speed)2. 2 Explain your working. [2] (b) A ball of mass 0.040 kg is projected into the air from horizontal ground, as illustrated in Fig. 2.1. Y path of ball h ball, mass 0.040 kg X ground Fig. 2.1 The ball is launched from a point X with a kinetic energy of 4.5 J. At point Y, the ball has a speed of 9.5 m s−1. Air resistance is negligible. (i) For the movement of the ball from X to Y, draw a solid line on Fig. 2.1 to show: 1. the distance moved (label this line D) 2. the displacement (label this line S). [2] (ii) By consideration of energy transfer, determine the height h of point Y above the ground. h = …… m [3] (iii) On Fig. 2.2, sketch the variation of the kinetic energy of the ball with its vertical height above the ground for the movement of the ball from X to Y. Numerical values are not required. kinetic energy 0 0 h height Fig. 2.2 [2] [Total: 11]

Mark scheme

2(a)(i) 1. W = mas B1 2. s = (v 2 – u 2) / 2a B1 2(a)(ii) W/work equals energy transferred/gain or change in kinetic energy B1 W (= mas) = ma(v 2 – u 2) / 2a B1 leading to W = m(v 2 – u 2) / 2 (so KE = ½mv 2) 2(b)(i) 1. solid curved line drawn from X to Y along path of ball and labelled D B1 2. solid straight line drawn from X to Y and labelled S B1 2(b)(ii) (∆)E = mg(∆)h C1 4.5 = (0.040 × 9.81 × h) + (½ × 0.040 × 9.52) C1 h = 6.9 m A1 2(b)(iii) line with a negative gradient starting from a non-zero value of kinetic energy when the vertical height is zero M1 straight line ends at a non-zero value of kinetic energy when the vertical height is h A1

Question 37 marks

Question

A cylindrical disc of mass 0.24 kg has a circular cross-sectional area A, as shown in Fig. 3.1. cross-sectional force X area A 8.9 N constant 30° speed 0.60 m s–1 disc, disc mass 0.24 kg ground Fig. 3.1 Fig. 3.2 The disc is on horizontal ground, as shown in Fig. 3.2. A force X of magnitude 8.9 N acts on the disc in a direction of 30° to the horizontal. The disc moves at a constant speed of 0.60 m s−1 along the ground. (a) Determine the rate of doing work on the disc by the force X. rate of doing work = …… W [2] (b) The force X and the weight of the disc exert a combined pressure on the ground of 3500 Pa. Calculate the cross-sectional area A of the disc. A = …… m2 [3] (c) Newton’s third law describes how forces exist in pairs. One such pair of forces is the weight of the disc and another force Y. State: (i) the direction of force Y ……[1] (ii) the name of the body on which force Y acts. ……[1] [Total: 7]

Mark scheme

3(a) P = Fv C1 P = 8.9 cos 30° × 0.60 A1 = 4.6 W 3(b) p = F/A C1 F = 8.9 sin 30° + (0.24 × 9.81) C1 ( = 6.80 N) A = 6.80 / 3500 A1 = 1.9 × 10–3 m2 3(c)(i) upwards/up B1 3(c)(ii) the Earth/planet B1

Question 413 marks

Question

Two vertical metal plates in a vacuum are separated by a distance of 0.12 m. Fig. 4.1 shows a side view of this arrangement. 0.080 m X sand particle 2.0 m 0V + 900 V path of particle metal plate metal plate Y 0.12 m Fig. 4.1 (not to scale) Each plate has a length of 2.0 m. The potential difference between the plates is 900 V. The electric field between the plates is uniform. A negatively charged sand particle is released from rest at point X, which is a horizontal distance of 0.080 m from the top of the positively charged plate. The particle then travels in a straight line and collides with the positively charged plate at its lowest point Y, as illustrated in Fig. 4.1. (a) Describe the pattern of the field lines (lines of force) between the plates. ……[2] (b) State the names of the two forces acting on the particle as it moves from X to Y. ……[1] (c) By considering the vertical motion of the sand particle, show that the time taken for the particle to move from X to Y is 0.64 s. [2] (d) Calculate the horizontal component of the acceleration of the particle. horizontal component of acceleration = …… m s−2 [2] (e) (i) Calculate the magnitude of the electric field strength. electric field strength = …… N C−1 [2] (ii) The sand particle has mass m and charge q. Use your answers in (d) and (e)(i) to q determine the ratio . m ratio = …… C kg−1 [2] q (f) Another particle has a smaller magnitude of the ratio than the sand particle. This particle is also released from point X. m For the movement of this particle, state the effect, if any, of the decreased magnitude of the ratio on: (i) the vertical component of the acceleration ……[1] (ii) the horizontal component of the acceleration. ……[1] [Total: 13]

Mark scheme

4(a) straight (horizontal) lines and from the +0.90 kV plate/to the 0 V plate B1 (lines are) equally spaced B1 4(b) weight/gravitational force and electric force B1 4(c) s = ½ at 2 C1 or s = ut + ½ at 2 and u = 0 2.0 = ½ × 9.81 × t 2 so t = 0.64 s A1 4(d) 0.080 = ½ × a × 0.642 C1 a = 0.39 m s–2 A1 4(e)(i) E = (∆)V / (∆)d C1 E = 0.90 × 103 / 0.12 A1 = 7.5 × 103 N C–1 4(e)(ii) ma = Eq C1 or F = ma and F = Eq q / m = 0.39 / 7.5 × 103 A1 = 5.2 × 10–5 C kg–1 4(f)(i) no effect B1 4(f)(ii) decreases/smaller B1

Question 59 marks

Question

A vertical tube of length 0.60 m is open at both ends, as shown in Fig. 5.1. A tube N 0.60 m A direction of incident sound wave Fig. 5.1 An incident sinusoidal sound wave of a single frequency travels up the tube. A stationary wave is then formed in the air column in the tube with antinodes A at both ends and a node N at the midpoint. (a) Explain how the stationary wave is formed from the incident sound wave. ……[2] (b) On Fig. 5.2, sketch a graph to show the variation of the amplitude of the stationary wave with height h above the bottom of the tube. amplitude 0 0 0.20 0.40 0.60 h/m Fig. 5.2 [2] (c) For the stationary wave, state: (i) the direction of the oscillations of an air particle at a height of 0.15 m above the bottom of the tube ……[1] (ii) the phase difference between the oscillations of a particle at a height of 0.10 m and a particle at a height of 0.20 m above the bottom of the tube. phase difference = …… ° [1] (d) The speed of the sound wave is 340 m s−1. Calculate the frequency of the sound wave. frequency = …… Hz [2] (e) The frequency of the sound wave is gradually increased. Determine the frequency of the wave when a stationary wave is next formed. frequency = …… Hz [1] [Total: 9]

Mark scheme

5(a) (incident) wave reflects at end/top of tube B1 (incident) wave and reflected wave interfere/superpose B1 5(b) line has maximum value of amplitude at h = 0 and h = 0.60 m only B1 line has minimum/zero value of amplitude at h = 0.30 m only B1 5(c)(i) vertical/along length of tube/along axis of tube B1 5(c)(ii) phase difference = 0 A1 5(d) v = fλ C1 v = 340 / (2 × 0.60) A1 = 280 Hz 5(e) f = 340 / 0.60 A1 = 570 Hz

Question 610 marks

Question

(a) Define the ohm. ……[1] (b) A battery of electromotive force (e.m.f.) E and internal resistance 1.5 Ω is connected to a network of resistors, as shown in Fig. 6.1. E 1.5 I 2.0 RZ 1.8 A Y Z 8.0 0.60 A X Fig. 6.1 Resistor X has a resistance of 8.0 Ω. Resistor Y has a resistance of 2.0 Ω. Resistor Z has a resistance of RZ. The current in X is 0.60 A and the current in Y is 1.8 A. (i) Calculate: 1. the current I in the battery I = …… A [1] 2. resistance RZ RZ = …… Ω [2] 3. e.m.f. E. E = …… V [2] (ii) Resistors X and Y are each made of wire. The two wires have the same length and are made of the same metal. Determine the ratio: cross-sectional area of wire X 1. cross-sectional area of wire Y ratio = …… [2] average drift speed of free electrons in X 2. . average drift speed of free electrons in Y ratio = …… [2] [Total: 10] Please turn over for Question 7.

Mark scheme

6(a) volt / ampere B1 6(b)(i) 1. I = 1.8 + 0.60 A1 = 2.4 A 2. (8.0 × 0.60) = 1.8 × (2.0 + RZ) C1 RZ = 0.67 Ω A1 3. E – (2.4 × 1.5) = (0.60 × 8.0) C1 or E – (2.4 × 1.5) = 1.8 × (2.0 + 0.67) or E = 2.4 × [1.5 + (8.0 × 2.67) / (8.0 + 2.67)] E = 8.4 V A1 6(b)(ii) 1. R = ρL / A or R ∝ 1 / A C1 ratio = RY / RX = 2.0 / 8.0 A1 = 0.25 2. I ∝ Av or IX / IY = AXvX / AYvY C1 ratio = (0.60 / 1.8) × (1 / 0.25) A1 = 1.3

Question 74 marks

Question

A sample of a radioactive substance may decay by the emission of either α-radiation or β-radiation and/or γ-radiation. State the type of radiation, one in each case, that: (a) consists of leptons ……[1] (b) contains quarks ……[1] (c) cannot be deflected by an electric field ……[1] (d) has a continuous range of energies, rather than discrete values of energy. ……[1] [Total: 4]

Mark scheme

7(a) beta/β B1 7(b) alpha/α B1 7(c) gamma/γ B1 7(d) beta/β B1

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Where can I find the 9702/23 May/June 2019 mark scheme?

On this page, free: the mark scheme for AS & A Level Physics 9702/23, May/June 2019 (9702_s19_ms_23), and the question paper (9702_s19_qp_23), to read or download, with each question's marking points written out underneath.

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Physics 9702/23 for May/June 2019 is out of 60 marks, with 1 hour 15 minutes to answer it, across 7 questions.

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