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Physics 9702/22 May/June 2021: question paper and mark scheme

Cambridge AS & A Level Physics (9702), Paper 2 AS Level Structured Questions, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9702_s21_qp_229702_s21_ms_229702/22/M/J/21

Paper
Paper 2 AS Level Structured Questions
Total marks
60
Time
1 hour 15 minutes
Questions
6
Code on the paper
9702/22/M/J/21
9702_s21_ms_22.pdf

9702/22 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 111 marks

Question

(a) Complete Table 1.1 by stating whether each of the quantities is a vector or a scalar. Table 1.1 quantity vector or scalar acceleration electrical resistance momentum [2] (b) State the conditions for an object to be in equilibrium. …… [2] (c) A floating solid cylinder is attached by a wire to the sea bed, as shown in Fig. 1.1. cylinder, cross-sectional weight 28 N area 0.0230 m2 surface of water 0.190 m water, density wire 1.00 × 103 kg m–3 sea bed Fig. 1.1 (not to scale) The density of the water is 1.00 × 103 kg m–3. The base of the cylinder is at a depth of 0.190 m below the surface of the water. The cylinder has a weight of 28 N and a cross-sectional area of 0.0230 m2. The wire and the central axis of the cylinder are both vertical. The cylinder is in equilibrium. (i) Calculate, to three significant figures, the upthrust acting on the cylinder due to the water. upthrust = …… N [2] (ii) Show that the tension T in the wire is 15 N. [1] (iii) The wire has a cross-sectional area of 3.2 mm2. Calculate the stress in the wire. stress = …… Pa [2] (iv) The surface of the water gradually rises until it is level with the top face of the cylinder. State and explain, qualitatively, the variation of the strain energy stored in the wire as the water surface rises. …… [2] [Total: 11]

Mark scheme

1(a) acceleration: vector B2 electrical resistance: scalar momentum: vector 1 mark for two correct, 2 marks for all three correct 1(b) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) is zero B1 1(c)(i) upthrust = ρ g (∆)h × A C1 = (1.00 × 103 × 9.81 × 0.190) × 0.0230 A1 = 42.9 N 1(c)(ii) (T =) 43 – 28 = 15 (N) A1 or (T =) 42.9 – 28 = 14.9 or 15 (N) 1(c)(iii) σ = F / A or T / A C1 = 15 / (3.2 × 10–6) A1 = 4.7 × 106 Pa 1(c)(iv) upthrust (on cylinder) increases (and weight constant) B1 tension/stress increases and (so) strain energy increases B1

Question 212 marks

Question

A ball is thrown vertically downwards to the ground, as illustrated in Fig. 2.1. ball speed u path of ball 1.5 m speed 8.7 m s–1 ground Fig. 2.1 The ball is thrown with speed u from a height of 1.5 m. The ball then hits the ground with speed 8.7 m s–1. Assume that air resistance is negligible. (a) Calculate speed u. u = …… m s–1 [2] (b) State how Newton’s third law applies to the collision between the ball and the ground. …… [2] (c) The ball is in contact with the ground for a time of 0.091 s. The ball rebounds vertically and leaves the ground with speed 5.4 m s–1. The mass of the ball is 0.059 kg. (i) Calculate the magnitude of the change in momentum of the ball during the collision. change in momentum = …… N s [2] (ii) Determine the magnitude of the average resultant force that acts on the ball during the collision. average resultant force = …… N [1] (iii) Use your answer in (c)(ii) to calculate the magnitude of the average force exerted by the ground on the ball during the collision. average force = …… N [2] (d) The ball was thrown downwards at time t = 0 and hits the ground at time t = T. On Fig. 2.2, sketch a graph to show the variation of the speed of the ball with time t from t = 0 to t = T. Numerical values are not required. speed 0 0 T t Fig. 2.2 [1] (e) In practice, air resistance is not negligible. State and explain the variation, if any, with time t of the gradient of the graph in (d) when air resistance is not negligible. …… [2] [Total: 12]

Mark scheme

2(a) v2 = u2 + 2as C1 u2 = 8.72 – (2 × 9.81 × 1.5) u = 6.8 m s–1 A1 2(b) (magnitude of) force on ball (by ground) equal to force on ground (by ball) B1 (direction of) force on ball (by ground) opposite to force on ground (by ball) B1 2(c)(i) (p = ) 0.059 × 8.7 or 0.059 × 5.4 C1 change in momentum = 0.059 (8.7 + 5.4) A1 = 0.83 N s 2(c)(ii) resultant force = 0.83 / 0.091 or 0.059 [(8.7 + 5.4) / 0.091] A1 = 9.1 N 2(c)(iii) (W =) 0.059 × 9.81 C1 (W =) 0.58 (N) A1 force = 9.1 + 0.58 = 9.7 N 2(d) straight line with a positive gradient and starting from a non-zero value of speed at t = 0 and ending when t = T B1 2(e) air resistance increases B1 resultant force/acceleration decreases so gradient (of curve) decreases B1

Question 38 marks

Question

A child of weight 330 N is at point X at the top of a slide. The slide is at the edge of a swimming pool, as shown in Fig. 3.1. child, X weight 330 N surface of slide 4.0 m surface of water Y water in 1.1 m swimming pool Fig. 3.1 (not to scale) The child moves from rest to the lowest point of the slide that is a vertical distance of 4.0 m below X. The child continues moving towards point Y which is at the end of the slide and a vertical distance of 1.1 m above the lowest point. The kinetic energy of the child at Y is 540 J. (a) Calculate the difference in the gravitational potential energy of the child at points X and Y. difference in gravitational potential energy = …… J [2] (b) An average frictional force of 52 N acts on the child when moving from X to Y. By considering changes of energy, determine the distance moved by the child from X to Y. distance moved = …… m [2] (c) The child leaves the slide at point Y with a velocity that is at an angle of 41° to the horizontal. The path of the child through the air is shown in Fig. 3.2. path of child velocity Z surface of water Y 41° slide water in swimming pool Fig. 3.2 (not to scale) Point Z is the highest point on the path of the child through the air. Assume that air resistance is negligible. Calculate the speed of the child at: (i) point Y speed = …… m s–1 [2] (ii) point Z. speed = …… m s–1 [2] [Total: 8]

Mark scheme

3(a) (Δ)E = mg(Δ)h or W(∆)h C1 = 330 × (4.0 – 1.1) A1 = 960 J 3(b) (work =) 960 – 540 C1 ( = 420 J) distance moved = (960 – 540) / 52 A1 = 8.1 m 3(c)(i) E = ½mv2 C1 540 = ½ × (330 / 9.81) × v2 A1 v = 5.7 m s–1 3(c)(ii) speed = horizontal component of velocity C1 = 5.7 × cos 41° = 4.3 m s–1 A1

Question 411 marks

Question

(a) For a progressive wave, state what is meant by its period. …… [1] (b) State the principle of superposition. …… [2] (c) Electromagnetic waves of wavelength 0.040 m are emitted in phase from two sources X and Y and travel in a vacuum. The arrangement of the sources is shown in Fig. 4.1. X path of detector 1.380 m Z Y 1.240 m Fig. 4.1 (not to scale) A detector moves along a path that is parallel to the line XY. A pattern of intensity maxima and minima is detected. Distance XZ is 1.380 m and distance YZ is 1.240 m. (i) State the name of the region of the electromagnetic spectrum that contains the waves from X and Y. …… [1] (ii) Calculate the period, in ps, of the waves. period = …… ps [3] (iii) Show that the path difference at point Z between the waves from X and Y is 3.5 λ, where λ is the wavelength of the waves. [1] (iv) Calculate the phase difference between the waves at point Z. phase difference = ……° [1] (v) The waves from X alone have the same amplitude at point Z as the waves from Y alone. State the intensity of the waves at point Z. …… [1] (vi) The frequencies of the waves from X and Y are both decreased to the same lower value. The waves stay within the same region of the electromagnetic spectrum. Describe the effect of this change on the pattern of intensity maxima and minima along the path of the detector. …… [1] [Total: 11]

Mark scheme

4(a) time for one oscillation/vibration/cycle B1 or time between adjacent wavefronts (passing the same point) or shortest time between two wavefronts (passing the same point) 4(b) (when two or more) waves meet/overlap (at a point) B1 (resultant) displacement is sum of the individual displacements B1 4(c)(i) microwave(s) B1 4(c)(ii) v = λ/T C1 or v = fλ and f = 1/T T = 0.040 / 3.00 × 108 C1 = 1.33 × 10–10 (s) A1 = 1.33 × 10–10 / 10–12 (ps) = 130 ps 4(c)(iii) (1.380 – 1.240) / 0.040 = 3.5 A1 or 1.380 / 0.040 – 1.240 / 0.040 = 3.5 4(c)(iv) phase difference = 1260° or 180° A1 4(c)(v) (always) zero A1 4(c)(vi) increase in distance between (adjacent intensity) maxima/minima A1

Question 58 marks

Question

(a) Define the ohm. …… [1] (b) A wire is made of metal of resistivity ρ. The length L of the wire is gradually increased. Assume that the volume V of the wire remains constant as its length is increased. Show that the resistance R of the extending wire is proportional to L2. [2] (c) A battery of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 5.1. E r I A R V Fig. 5.1 An ammeter measures the current I in the circuit. A voltmeter measures the potential difference V across the variable resistor. The resistance R is now varied to change the values of I and V. The variation with I of V is shown in Fig. 5.2. 3 V/V 2 1 0 0 2 4 6 I/A Fig. 5.2 (i) Use Fig. 5.2 to state the e.m.f. E of the battery. E = …… V [1] (ii) Use Fig. 5.2 to determine the power dissipated in the variable resistor when there is a current of 5.0 A. power = …… W [3] (iii) State what is represented by the value of the gradient of the graph. …… [1] [Total: 8]

Mark scheme

5(a) volt / ampere B1 5(b) R = ρL / A B1 (A = V / L) B1 (so) R = ρL2 / V (with ρ and V constant so R ∝ L2) 5(c)(i) E = 2.4 V A1 5(c)(ii) P = VI or I2R or V2 / R C1 = 1.3 × 5.0 or 5.02 × 0.26 or 1.32 / 0.26 C1 = 6.5 W A1 5(c)(iii) (–) internal resistance or (–) r B1

Question 610 marks

Question

(a) One of the results of the α-particle scattering experiment is that a very small minority of the α-particles are scattered through angles greater than 90°. State what may be inferred about the structure of the atom from this result. …… [2] (b) An α-particle is made up of other particles. One of these particles is a proton. State and explain whether a proton is a fundamental particle. …… [1] (c) A radioactive source produces a beam of α-particles in a vacuum. The average current produced by the beam is 6.9 × 10–9 A. Calculate the average number of α-particles passing a fixed point in the beam in a time of 1.0 minute. number = …… [3] (d) The α-particles in the vacuum in (c) enter a uniform electric field. The α-particles enter the field with their velocity in the same direction as the field. State and explain whether the magnitude of the acceleration of an α-particle due to the field decreases, increases or stays constant as the α-particle moves through the field. …… [2] (e) A nucleus X is an isotope of a nucleus Y. The mass of nucleus X is greater than that of Y. Both of the nuclei are in the same uniform electric field. State and explain whether the magnitude of the electric force acting on nucleus X is greater than, less than or the same as that acting on nucleus Y. …… [2] [Total: 10]

Mark scheme

6(a) the nucleus is charged B1 the majority of the mass (of atom) is in the nucleus B1 6(b) made up of quarks (so) not a fundamental particle B1 6(c) (Q =) 6.9 × 10–9 × 60 C1 number = (6.9 × 10–9 × 60) / (2 × 1.60 × 10–19) C1 = 1.3 × 1012 A1 6(d) (magnitude of electric) force is constant B1 (so magnitude of) acceleration is constant B1 6(e) (nuclei have) same charge/same number of protons B1 (so) same (magnitude of) force B1

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Where can I find the 9702/22 May/June 2021 mark scheme?

On this page, free: the mark scheme for AS & A Level Physics 9702/22, May/June 2021 (9702_s21_ms_22), and the question paper (9702_s21_qp_22), to read or download, with each question's marking points written out underneath.

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Physics 9702/22 for May/June 2021 is out of 60 marks, with 1 hour 15 minutes to answer it, across 6 questions.

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