Momentum and Newton’s laws of motion — Paper 2 · A Level Physics 9702

3 (a) Define velocity. … … [1] (b) A car travels in a straight line up a slope, as shown in Fig. 3.1. ms–1 9.0 car mass 850 kg slope Fig. 3.1 The car has mass 850 kg and travels with a constant speed of 9.0 m s–1. The car’s engine exerts a force on the car of 2.0 kN up the slope. A resistive force FD, due to friction and air resistance, opposes the motion of the car. The variation of FD with the speed v of the car is shown in Fig. 3.2. 0.70 FD / kN 0.60 0.50 0.40 0.30 7 8 9 10 11 12 13 14 15 16 v / m s–1 Fig. 3.2 (i) State and explain whether the car is in equilibrium as it moves up the slope. … … … [2] (ii) Consider the forces that act along the slope. Use data from Fig. 3.2 to determine the component of the weight of the car that acts down the slope. component of weight = … N [2] (iii) Show that the power output of the car is 1.8 × 104 W. [2] (iv) The car now travels along horizontal ground. The output power of the car is maintained at 1.8 × 104 W. The variation of the resistive force FD acting on the car is given in Fig. 3.2. Calculate the acceleration of the car when its speed is 15 m s–1. acceleration = … m s–2 [3] [Total: 10]

10 marks

Mark scheme: 3(a) change of displacement / time (taken) B1 3(b)(i) constant velocity, so resultant force is zero M1 (so car is) in (dynamic) equilibrium A1 3(b)(ii) FD = 0.40 (kN) or 0.40 × 103 (N) C1 component of weight = 2.0 × 103 – 0.40 × 103 = 1.6 × 103 N A1 3(b)(iii) P = Fv C1 = 2.0 ×103 × 9.0 = 1.8 × 104 W A1 3(b)(iv) (driving) force = 1.8 × 104 / 15 (= 1.2 × 103) C1 FD = 0.66 (kN) or 0.66 × 103 (N) C1 acceleration = (1.2 × 103 – 0.66 × 103) / 850 = 0.64 (0.635) m s–2 A1

This question in 9702/22 Feb/March 2017

1 (a) Two forces, with magnitudes 5.0 N and 12 N, act from the same point on an object. Calculate the magnitude of the resultant force R for the forces acting (i) in opposite directions, R = … N [1] (ii) at right angles to each other. R = … N [1] (b) An object X rests on a smooth horizontal surface. Two horizontal forces act on X as shown in Fig. 1.1. 18 N 115° X 55 N Fig. 1.1 (not to scale) A force of 55 N is applied to the right. A force of 18 N is applied at an angle of 115° to the direction of the 55 N force. (i) Use the resolution of forces or a scale diagram to show that the magnitude of the resultant force acting on X is 65 N. [2] (ii) Determine the angle between the resultant force and the 55 N force. angle = … ° [2] (c) A third force of 80 N is now applied to X in the opposite direction to the resultant force in (b). The mass of X is 2.7 kg. Calculate the magnitude of the acceleration of X. acceleration = … m s–2 [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) R = 7(.0) N B1 1(a)(ii) R = 13 N B1 1(b)(i) forces resolved: 18 sin 65° (vertical) and 55 + 18 cos 65° (horizontal) or scale drawing: correct triangle drawn for forces B1 F = [(18 sin 65°)2 + (55 + 18 cos 65°)2]1/2 = 65 (64.7) N or scale drawing: scale given, length of resultant given correctly, ± 1 N A1 1(b)(ii) angle = tan–1 [18 sin 65° / (55 + 18 cos 65°)] = tan–1 (16.3 / 62.6) or scale drawing: correct angle measured/direction correct on diagram below the 55 N force C1 angle = 15 (14.6)° (below the 55 N force) or scale drawing: angle = 15° ± 1° A1 1(c) (resultant) force = mass × acceleration C1 80 − 65 = 2.7a C1 a = 5.6 m s–2 [5.7 if 64.7 N used from (i)] A1

This question in 9702/23 May/June 2017

2 (a) State Newton’s second law of motion. … … [1] (b) A constant resultant force F acts on an object A. The variation with time t of the velocity v for the motion of A is shown in Fig. 2.1. 9.0 v / m s–1 8.0 7.0 6.0 5.0 4.0 0 1.0 2.0 3.0 4.0 t / s Fig. 2.1 The mass of A is 840 g. Calculate, for the time t = 0 to t = 4.0 s, (i) the change in momentum of A, change in momentum = … kg m s–1 [2] (ii) the force F. F = … N [1] (c) The force F is removed at t = 4.0 s. Object A continues at constant velocity before colliding with an object B, as illustrated in Fig. 2.2. A B 840 g 730 g at rest Fig. 2.2 Object B is initially at rest. The mass of B is 730 g. The objects A and B join together and have a velocity of 4.7 m s–1. (i) By calculation, show that the changes in momentum of A and of B during the collision are equal and opposite. [2] (ii) Explain how the answers obtained in (i) support Newton’s third law. … … … … [2] (iii) By reference to the speeds of A and B, explain whether the collision is elastic. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) (resultant) force is proportional/equal to the rate of change of momentum B1 2(b)(i) change in momentum = m(v2 − v1) = 0.84 × (8.8 − 4.2) C1 = 3.9 (3.86) kg m s–1 A1 2(b)(ii) F = (3.9 / 4.0) = 0.97 (0.965) N A1 2(c)(i) change in momentum for A: 0.84 × (4.7 − 8.8) = −3.4 (3.44) change in momentum for B: 0.73 × (4.7 − 0) = 3.4 (3.43) M1 change in momentum for B is equal and opposite to A A1 2(c)(ii) change in momentum equal (for A and B) M1 force is change in momentum / time and time (of collision) is the same hence force on A and B equal and opposite as for Newton’s third law A1 2(c)(iii) inelastic as relative speed of approach not equal to relative speed of separation B1

This question in 9702/23 May/June 2017

1 (a) The drag force FD acting on a sphere moving through a fluid is given by the expression FD = Kρv 2 where K is a constant, ρ is the density of the fluid and v is the speed of the sphere. Determine the SI base units of K. base units … [3] (b) A ball of weight 1.5 N falls vertically from rest in air. The drag force F D acting on the ball is given by the expression in (a). The ball reaches a constant (terminal) speed of 33 m s–1. Assume that the upthrust acting on the ball is negligible and that the density of the air is uniform. For the instant when the ball is travelling at a speed of 25 m s–1, determine (i) the drag force FD on the ball, FD = … N [2] (ii) the acceleration of the ball. acceleration = … m s–2 [2] (c) Describe the acceleration of the ball in (b) as its speed changes from zero to 33 m s–1. … … … … [3] [Total: 10]

10 marks

Mark scheme: 1(a) C1 units of ρ: kg m–3 and units of v: m s–1 C1 units of K: kg m s–2 / [kg m–3 (m s–1)2] = m2 A1 1(b)(i) Kρ = 1.5 / 332 C1 = 1.38 × 10–3 FD = 1.38 × 10–3 × 252 or FD / 1.5 = 252 / 332 FD = 0.86 N A1 1(b)(ii) a = (1.5 – 0.86) / (1.5 / 9.81) or a = 9.81 – [0.86 / (1.5 / 9.81)] C1 a = 4.2 m s–2 A1 1(c) initial acceleration is g/9.81 (m s–2)/acceleration of free fall B1 acceleration decreases B1 final acceleration is zero B1

This question in 9702/21 Oct/Nov 2017

3 A spring is attached at one end to a fixed point and hangs vertically with a cube attached to the other end. The cube is initially held so that the spring has zero extension, as shown in Fig. 3.1. spring with zero extension cube weight 4.0 N 5.1 cm 5.1 cm water 7.0 cm density 1000 kg m–3 Fig. 3.1 Fig. 3.2 The cube has weight 4.0 N and sides of length 5.1 cm. The cube is released and sinks into water as the spring extends. The cube reaches equilibrium with its base at a depth of 7.0 cm below the water surface, as shown in Fig. 3.2. The density of the water is 1000 kg m–3. (a) Calculate the difference in the pressure exerted by the water on the bottom face and on the top face of the cube. difference in pressure = … Pa [2] (b) Use your answer in (a) to show that the upthrust on the cube is 1.3 N. [2] (c) Calculate the force exerted on the spring by the cube when it is in equilibrium in the water. force = … N [1] (d) The spring obeys Hooke’s law and has a spring constant of 30 N m–1. Determine the initial height above the water surface of the base of the cube before it was released. height above surface = … cm [3] (e) The cube in the water is released from the spring. (i) Determine the initial acceleration of the cube. acceleration = … m s–2 [2] (ii) Describe and explain the variation, if any, of the acceleration of the cube as it sinks in the water. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a) C1 ∆p = 1000 × 9.81 × (7.0 × 10–2 – 1.9 × 10–2) or 686 – 186 = 500 Pa A1 3(b) F = pA or (∆)F = ∆p × A C1 upthrust = 500 × (5.1 × 10–2)2 = 1.3 N or upthrust = (686 – 186) × (5.1 ×10–2)2 = 1.3 N or upthrust = 1000 × 9.81 × 5.1 ×10–2 × (5.1 × 10–2)2 = 1.3 N A1 3(c) force = 4.0 – 1.3 = 2.7 N A1 Question Answer Marks 3(d) extension/x/e = 2.7 / 30 C1 = 0.09 (m) or 9 (cm) C1 height above surface = 9 – 7 = 2 cm A1 3(e)(i) mass = 4.0 / 9.81 C1 acceleration = 2.7 / (4.0 / 9.81) = 6.6 m s–2 A1 3(e)(ii) viscous force increases (and then becomes constant) M1 (weight and upthrust constant so) acceleration decreases (to zero) A1

This question in 9702/22 Oct/Nov 2017

5 (a) Define the coulomb. … [1] (b) Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 102 V is applied between the plates. Fig. 5.1 shows a side view of this arrangement. 4.0 cm smoke particle weight 3.9 × 10–15 N charge –8.0 × 10–19 C metal plate metal plate +2.0 × 102 V s Fig. 5.1 A smoke particle is in the uniform electric field between the plates. The particle has weight 3.9 × 10–15 N and charge –8.0 × 10–19 C. (i) Show that the electric force acting on the particle is 4.0 × 10–15 N. [2] (ii) On Fig. 5.1, draw labelled arrows to show the directions of the two forces acting on the smoke particle. [1] (iii) The resultant force acting on the particle is F. Determine 1. the magnitude of F, magnitude = … N 2. the angle of F to the horizontal. angle = … ° [3] (c) The electric field in (b) is switched on at time t = 0 when the particle is at a horizontal displacement s = 2.0 cm from the left-hand plate. At time t = 0 the horizontal velocity of the particle is zero. The particle is then moved by the electric field until it hits a plate at time t = T. On Fig. 5.2, sketch the variation with time t of the horizontal displacement s of the particle from the left-hand plate. 4.0 s / cm 2.0 0 0 T t Fig. 5.2 [2] [Total: 9]

9 marks

Mark scheme: 5(a) (coulomb is) ampere second B1 5(b)(i) E = V / d or E = F / Q C1 F = VQ / d F = (2.0 × 102 × 8.0 × 10–19) / 4.0 × 10–2 = 4.0 × 10–15 N A1 5(b)(ii) arrow pointing to the left labelled ‘electric force’ and arrow pointing downwards labelled ‘weight’ B1 5(b)(iii) 1. resultant force = √ [(3.9 × 10–15)2 + (4.0 × 10–15)2] C1 = 5.6 × 10–15 N A1 2. angle = tan–1 (3.9 × 10–15 / 4.0 × 10–15) = 44° A1 5(c) downward sloping line from (0, 2.0) M1 magnitude of gradient of line increases with time and line ends at (T, 0) A1

This question in 9702/22 Oct/Nov 2017

1 (a) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar. quantity vector or scalar acceleration speed power Fig. 1.1 [2] (b) A ball is projected with a horizontal velocity of 1.1 m s–1 from point A at the edge of a table, as shown in Fig. 1.2. table ball 1.1 m s–1 A path of ball B horizontal ground 0.43 m Fig. 1.2 The ball lands on horizontal ground at point B which is a distance of 0.43 m from the base of the table. Air resistance is negligible. (i) Calculate the time taken for the ball to fall from A to B. time = … s [1] (ii) Use your answer in (b)(i) to determine the height of the table. height = … m [2] (iii) The ball leaves the table at time t = 0. For the motion of the ball between A and B, sketch graphs on Fig. 1.3 to show the variation with time t of 1. the acceleration a of the ball, 2. the vertical component sv of the displacement of the ball from A. Numerical values are not required. a sv 0 0 0 t 0 t Fig. 1.3 [2] (c) A ball of greater mass is projected from the table with the same velocity as the ball in (b). Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the time taken for the ball to fall to the ground. … … [1] [Total: 8]

8 marks

Mark scheme: 1(a) acceleration: vector speed: scalar power: scalar All three correct scores 2 marks. Only two correct scores 1 mark. B2 1(b)(i) time = 0.43 / 1.1 = 0.39 s A1 1(b)(ii) s = ut + ½at 2 = ½ × 9.81 × 0.392 C1 = 0.75 m A1 1(b)(iii) 1 horizontal line at a non-zero value of a. B1 2 curved line from origin with increasing gradient. B1 1(c) acceleration (of free fall) is unchanged / not dependent on mass and so no effect (on time taken). A1

This question in 9702/22 Feb/March 2018

2 (a) State Newton’s first law of motion. … … [1] (b) A block of weight 15 N hangs by a wire from a remotely controlled aircraft, as shown in Fig. 2.1. aircraft wire block weight 15 N Fig. 2.1 The aircraft is used to move the block only in a vertical direction. The force on the block due to air resistance is negligible. The variation with time t of the vertical velocity v of the block is shown in Fig. 2.2. The velocity is taken to be positive in the upward direction. 4.0 3.0 v / m s–1 2.0 1.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 t / s –1.0 –2.0 –3.0 –4.0 –5.0 –6.0 –7.0 Fig. 2.2 (i) Determine, for the block, 1. the displacement from time t = 0 to t = 3.0 s, magnitude of displacement = … m direction of displacement … [3] 2. the change in gravitational potential energy from time t = 0 to t = 3.0 s. change in gravitational potential energy = … J [2] (ii) Calculate the magnitude of the acceleration of the block at time t = 2.0 s. acceleration = … m s–2 [2] (iii) Use your answer in (b)(ii) to show that the tension T in the wire at time t = 2.0 s is 20 N. [2] (iv) The wire has a cross-sectional area of 2.8 × 10–5 m2 and is made from metal of Young modulus 1.7 × 1011 Pa. The wire obeys Hooke’s law. Calculate the strain of the wire at time t = 2.0 s. strain = … [3] (v) At some time after t = 3.0 s the tension in the wire has a constant value of 15 N. State and explain whether it is possible to deduce that the block is moving vertically after t = 3.0 s. … … … … [2] [Total: 15]

15 marks

Mark scheme: 2(a) a body continues at (rest or) constant velocity unless acted upon by a resultant force B1 2(b)(i)1. from 0–2 s, distance = ½ × 2 × 6.8 (= 6.8 m) and from 2–3 s, distance = ½ × 1 × 3.4 (= 1.7 m) C1 magnitude of displacement = 5.1 m A1 direction of displacement is down(wards) B1 2(b)(i)2. (∆E) = mg∆h or (E) = mgh or (E) = Wh C1 (∆)E = 15 × 5.1 = (–) 77 J A1 2(b)(ii) a = (v – u) / t or a = gradient or a = dv / dt C1 a = 3.4 m s–2 A1 2(b)(iii) T – W = ma or T – mg = ma C1 T = 15 + (15 / 9.81) × 3.4 = 20 N or 20.2 N A1 2(b)(iv) E = F / Aε or E = σ / ε and σ = F / A C1 ε = 20 / (2.8 × 10–5 × 1.7 × 1011) C1 = 4.2 × 10–6 A1 2(b)(v) block is in equilibrium/has no resultant force B1 block could be stationary (or have constant velocity/speed) (so no, not possible to deduce) B1

This question in 9702/21 May/June 2018

3 (a) State what is meant by the mass of a body. … … [1] (b) Two blocks travel directly towards each other along a horizontal, frictionless surface. The blocks collide, as illustrated in Fig. 3.1. 0.40 m s–1 0.25 m s–1 0.20 m s–1 v block A block B mass mass mass mass 3M M 3M M before after Fig. 3.1 Block A has mass 3M and block B has mass M. Before the collision, block A moves to the right with speed 0.40 m s–1 and block B moves to the left with speed 0.25 m s–1. After the collision, block A moves to the right with speed 0.20 m s–1 and block B moves to the right with speed v. (i) Use Newton’s third law to explain why, during the collision, the change in momentum of block A is equal and opposite to the change in momentum of block B. … … … … [2] (ii) Determine speed v. v = … m s–1 [3] (iii) Calculate, for the blocks, 1. the relative speed of approach, relative speed of approach = … m s–1 2. the relative speed of separation. relative speed of separation = … m s–1 [2] (iv) Use your answers in (b)(iii) to state and explain whether the collision is elastic or inelastic. … … [1] [Total: 9]

9 marks

Mark scheme: 3(a) mass is the property (of a body/object) resisting changes in motion or mass is the quantity of matter (in a body) B1 3(b)(i) force on A (by B) equal and opposite to force on B (by A) or both A and B exert equal and opposite forces on each other B1 force is rate of change of momentum and time (of contact) is same B1 3(b)(ii) p = mv or 3M × 0.40 or M × 0.25 or 3M × 0.2 or Mv C1 (3M × 0.40) – (M × 0.25) = (3M × 0.2) + Mv C1 v = (3 × 0.40) – 0.25 – (3 × 0.2) = 0.35 m s–1 A1 3(b)(iii) 1. relative speed of approach = 0.40 + 0.25 = 0.65 m s–1 A1 2. relative speed of separation = 0.35 – 0.20 = 0.15 m s–1 A1 3(b)(iv) (relative) speed of separation not equal to/less than (relative) speed of approach or answers (to (b)(iii) are) not equal and so inelastic collision B1

This question in 9702/21 May/June 2018

1 (a) Define force. … [1] (b) State the SI base units of force. … [1] (c) The force F between two point charges is given by Q1Q2 F = 4πr 2ε where Q1 and Q2 are the charges, r is the distance between the charges, ε is a constant that depends on the medium between the charges. Use the above expression to determine the base units of ε. base units … [2] [Total: 4]

4 marks

Mark scheme: 1(a) rate of change of momentum B1 1(b) kg m s–2 A1 1(c) units for Q: A s and for r: m C1 units for ε = (A s × A s) / (kg m s–2 × m2) = A2 kg–1 m–3 s4 A1

This question in 9702/22 May/June 2018

3 A child on a sledge slides down a steep hill and then travels in a straight line up an ice-covered slope, as illustrated in Fig. 3.1. ice-covered slope child and sledge total mass 70 kg B 18 m s–1 A Fig. 3.1 (not to scale) The sledge passes point A with speed 18 m s–1 at time t = 0 and then comes to rest at point B. The child applies a brake to the sledge at point B. The brake does not keep the sledge stationary and it immediately slides back down the slope towards A. The variation with time t of the velocity v of the sledge from t = 0 to t = 24 s is shown in Fig. 3.2. 20 v / m s–1 10 0 0 4 8 12 16 20 24 t / s –10 Fig. 3.2 (a) State the time taken for the sledge to travel from A to B. time = … s [1] (b) Determine the displacement of the sledge up the slope from point A at time t = 24 s. displacement = … m [3] (c) Show that the acceleration of the sledge as it moves from B back towards A is 0.50 m s–2. [2] (d) The child and sledge have a total mass of 70 kg. The component of the total weight of the child and sledge that acts down the slope is 80 N. Determine (i) the frictional force on the sledge as it moves from B towards A, frictional force = … N [2] (ii) the angle θ of the slope to the horizontal. θ = … ° [2] (e) The child on the sledge blows a whistle between t = 4.0 s and t = 8.0 s. The whistle emits sound of frequency 900 Hz. The speed of the sound in the air is 340 m s–1. A man standing at point A hears the sound. Use Fig. 3.2 to (i) determine the initial frequency of the sound heard by the man, initial frequency = … Hz [2] (ii) describe and explain qualitatively the variation, if any, in the frequency of the sound heard by the man. … … [1] [Total: 13]

13 marks

Mark scheme: 3(a) time = 12 s A1 3(b) distance (up slope) = ½ × 12 × 18 (= 108) C1 distance (down slope) = ½ × 12 × 6 (= 36) C1 displacement from A = 108 – 36 = 72 m A1 3(c) v = u + at or a = gradient or a = ∆v / (∆)t C1 a = 6 / 12 = 0.50 (m s–2) (other points from the line may be used) A1 or v2 = u2 + 2as and u = 0 or v2 = 2as (C1) a = 6.02 / (2 × 36) = 0.50 (m s–2) (A1) or s = ut + ½at2 and u = 0 or s = ½at2 (C1) a = 2 × 36 / 122 = 0.50 (m s–2) (A1) or s = vt – ½at2 (C1) a = 2 × (6 × 12 – 36) / 122 = 0.50 (m s–2) (A1) Question Answer Marks 3(d)(i) F = 70 × 0.50 (= 35) C1 frictional force = 80 – 35 = 45 N A1 3(d)(ii) sin θ = 80 / (70 × 9.81) C1 θ = 6.7° A1 3(e)(i) f0 = (900 × 340) / (340 + 12) C1 = 870 Hz A1 3(e)(ii) speed/velocity (of sledge) decreases and (so) frequency increases B1

This question in 9702/22 May/June 2018

3 A ball is thrown vertically upwards towards a ceiling and then rebounds, as illustrated in Fig. 3.1. ceiling ball leaving speed 3.8 m s–1 ceiling ball thrown speed 9.6 m s–1 upwards Fig. 3.1 The ball is thrown with speed 9.6 m s–1 and takes a time of 0.37 s to reach the ceiling. The ball is then in contact with the ceiling for a further time of 0.085 s until leaving it with a speed of 3.8 m s–1. The mass of the ball is 0.056 kg. Assume that air resistance is negligible. (a) Show that the ball reaches the ceiling with a speed of 6.0 m s–1. [1] (b) Calculate the height of the ceiling above the point from which the ball was thrown. height = … m [2] (c) Calculate (i) the increase in gravitational potential energy of the ball for its movement from its initial position to the ceiling, increase in gravitational potential energy = … J [2] (ii) the decrease in kinetic energy of the ball while it is in contact with the ceiling. decrease in kinetic energy = … J [2] (d) State how Newton’s third law applies to the collision between the ball and the ceiling. … … … … [2] (e) Calculate the change in momentum of the ball during the collision. change in momentum = … N s [2] (f) Determine the magnitude of the average force exerted by the ceiling on the ball during the collision. average force = … N [2] [Total: 13]

13 marks

Mark scheme: 3(a) v = u + at v = 9.6 – (9.81 × 0.37) = 6.0 m s–1 A1 3(b) s = ½ × (9.6 + 6.0) × 0.37 or 6.02 = 9.62 – (2 × 9.81 × s) or s = (9.6 × 0.37) – (½ × 9.81 × 0.372) or s = (6.0 × 0.37) + (½ × 9.81 × 0.372) C1 s = 2.9 m A1 3(c)(i) (∆)E = mg(∆)h C1 ∆E = 0.056 × 9.81 × 2.9 = 1.6 J A1 3(c)(ii) E = ½mv 2 C1 ∆E = ½ × 0.056 × (6.02 – 3.82) = 0.60 J A1 3(d) force on ball (by ceiling) equal to force on ceiling (by ball) M1 and opposite (in direction) A1 3(e) (p =) mv or 0.056 × 6.0 or 0.056 × 3.8 C1 change in momentum = 0.056 × (6.0 + 3.8) = 0.55 N s A1 Question Answer Mark 3(f) resultant force = 0.55 / 0.085 (= 6.47 N) C1 force by ceiling = 6.47 – (0.056 × 9.81) = 5.9 N A1

This question in 9702/23 May/June 2018

2 A wooden block moves along a horizontal frictionless surface, as shown in Fig. 2.1. 45 m s –1 2.0 m s –1 block steel ball mass 85 g mass 4.0 g horizontal surface Fig. 2.1 The block has mass 85 g and moves to the left with a velocity of 2.0 m s –1. A steel ball of mass 4.0 g is fired to the right. The steel ball, moving horizontally with a speed of 45 m s –1, collides with the block and remains embedded in it. After the collision the block and steel ball both have speed v. (a) Calculate v. v = … m s –1 [2] (b) (i) For the block and ball, state 1. the relative speed of approach before collision, relative speed of approach = … m s–1 2. the relative speed of separation after collision. relative speed of separation = … m s–1 [1] (ii) Use your answers in (i) to state and explain whether the collision is elastic or inelastic. … … [1] (c) Use Newton’s third law to explain the relationship between the rate of change of momentum of the ball and the rate of change of momentum of the block during the collision. … … … … [2]

6 marks

Mark scheme: 2(a) C1 (4.0 × 45) – (2.0 × 85) = 89 v v = 0.11 m s–1 A1 2(b)(i) 1. speed of approach = 47 m s–1 and 2. speed of separation = 0 A1 2(b)(ii) speed of separation less than/not equal to speed of approach and so inelastic collision A1 2(c) force is equal to rate of change of momentum B1 force on ball (by block) equal and opposite to force on block (by ball) so rates of change of momentum are equal and opposite B1 or force on ball (by block) equal and opposite to force on block (by ball) (B1) force is equal to rate of change of momentum so rates of change of momentum are equal and opposite (B1)

This question in 9702/21 Oct/Nov 2018

1 A golfer strikes a ball so that it leaves horizontal ground with a velocity of 6.0 m s–1 at an angle θ to the horizontal, as illustrated in Fig. 1.1. vY 6.0 m s–1 4.8 m s–1 ball θ ground vX Fig. 1.1 (not to scale) The magnitude of the initial vertical component vY of the velocity is 4.8 m s–1. Assume that air resistance is negligible. (a) Show that the magnitude of the initial horizontal component vX of the velocity is 3.6 m s–1. [1] (b) The ball leaves the ground at time t = 0 and reaches its maximum height at t = 0.49 s. On Fig. 1.2, sketch separate lines to show the variation with time t, until the ball returns to the ground, of (i) the vertical component vY of the velocity (label this line Y), [2] (ii) the horizontal component vX of the velocity (label this line X). [2] 5.0 4.0 velocity / m s–1 3.0 2.0 1.0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 t / s –1.0 –2.0 –3.0 –4.0 –5.0 Fig. 1.2 (c) Calculate the maximum height reached by the ball. maximum height = … m [2] (d) For the movement of the ball from the ground to its maximum height, determine the ratio kinetic energy at maximum height . change in gravitational potential energy ratio = … [4] (e) In practice, significant air resistance acts on the ball. Explain why the actual time taken for the ball to reach maximum height is less than the time calculated when air resistance is assumed to be negligible. … … … [1] [Total: 12]

12 marks

Mark scheme: 1(a) or 6.0 sinθ = 4.8 (so θ = 53.1°) and vx = 6.0 cos 53.1° = 3.6 (m s–1) A1 1(b)(i) straight line from (0, 4.8) to (0.49, 0) M1 straight line continues with same slope to (0.98, –4.8) (labelled Y) A1 1(b)(ii) a horizontal line M1 from (0, 3.6) to (0.98, 3.6) (labelled X) A1 1(c) s = ut + ½at2 = (4.8 × 0.49) + (½ × –9.81 × 0.492) or s = ½(u + v)t or area under graph = ½ × (4.8 + 0) × 0.49 or s = vt – ½at2 = ½ × 9.81 × 0.492 or v2 = u2 + 2as s = 4.82 / (2 × 9.81) C1 s = 1.2 m A1 Question Answer Marks 1(d) (∆)E = mg(∆)h C1 E = ½mv2 C1 ratio = (½ × m × 3.62) / (m × 9.81 × 1.2) or ratio = [(½ × m × 6.02) – (m × 9.81 × 1.2)] / (m × 9.81 × 1.2) or ratio = (½ × m × 3.62) / (½ × m × 4.82) C1 ratio = 0.56 A1 1(e) (force due to) air resistance acts in opposite direction to the velocity or (with air resistance, average) resultant force is larger (than weight) B1

This question in 9702/22 Oct/Nov 2018

3 (a) State the principle of conservation of momentum. … … … [2] (b) The propulsion system of a toy car consists of a propeller attached to an electric motor, as illustrated in Fig. 3.1. propeller moving air 0.045 m speed 1.8 m s–1 electric motor of car body of car 0.045 m ground Fig. 3.1 The car is on horizontal ground and is initially held at rest by its brakes. When the motor is switched on, it rotates the propeller so that air is propelled horizontally to the left. The density of the air is 1.3 kg m–3. Assume that the air moves with a speed of 1.8 m s–1 in a uniform cylinder of radius 0.045 m. Also assume that the air to the right of the propeller is stationary. (i) Show that, in a time interval of 2.0 s, the mass of air propelled to the left is 0.030 kg. [2] (ii) Calculate 1. the increase in the momentum of the mass of air in (b)(i), increase in momentum = … N s 2. the force exerted on this mass of air by the propeller. force = … N [3] (iii) Explain how Newton’s third law applies to the movement of the air by the propeller. … … … [2] (iv) The total mass of the car is 0.20 kg. The brakes of the car are released and the car begins to move with an initial acceleration of 0.075 m s–2. Determine the initial frictional force acting on the car. frictional force = … N [2] [Total: 11]

11 marks

Mark scheme: 3(a) sum/total momentum (of a system of bodies) is constant or sum/total momentum before = sum/total momentum after M1 for an isolated system or no (resultant) external force A1 3(b)(i) m = ρV C1 = 1.3 × π × 0.0452 × 1.8 × 2.0 = 0.030 (kg) A1 3(b)(ii) 1. (∆)p = (∆)mv C1 = 0.030 × 1.8 = 0.054 N s A1 2. F = 0.054 / 2.0 or 0.030 × 1.8 / 2.0 = 0.027 N A1 3(b)(iii) force on air (by propeller) equal to force on propeller (by air) M1 and opposite (in direction) A1 3(b)(iv) resultant force = 0.20 × 0.075 (= 0.015 N) frictional force = 0.027 – 0.015 C1 = 0.012 N A1

This question in 9702/22 Oct/Nov 2018

3 (a) State Newton’s second law of motion. … … [1] (b) A toy rocket consists of a container of water and compressed air, as shown in Fig. 3.1. container compressed air water density 1000 kg m–3 nozzle radius 7.5 mm Fig. 3.1 Water is pushed vertically downwards through a nozzle by the compressed air. The rocket moves vertically upwards. The nozzle has a circular cross-section of radius 7.5 mm. The density of the water is 1000 kg m–3. Assume that the water leaving the nozzle has the shape of a cylinder of radius 7.5 mm and has a constant speed of 13 m s–1 relative to the rocket. (i) Show that the mass of water leaving the nozzle in the first 0.20 s after the rocket launch is 0.46 kg. [2] (ii) Calculate 1. the change in the momentum of the mass of water in (b)(i) due to leaving the nozzle, change in momentum = … N s 2. the force exerted on this mass of water by the rocket. force = … N [3] (iii) State and explain how Newton’s third law applies to the movement of the rocket by the water. … … … [2] (iv) The container has a mass of 0.40 kg. The initial mass of water before the rocket is launched is 0.70 kg. The mass of the compressed air in the rocket is negligible. Assume that the resistive force on the rocket due to its motion is negligible. For the rocket at a time of 0.20 s after launching, 1. show that its total mass is 0.64 kg, 2. calculate its acceleration. acceleration = … m s–2 [3] [Total: 11]

11 marks

Mark scheme: 3(a) (resultant) force proportional/equal to rate of change of momentum B1 3(b)(i) ρ = m / V C1 V = π × (7.5 × 10–3)2 × 13 × 0.2 (= 4.59 × 10–4 m3) m = π × (7.5 × 10–3)2 × 13 × 0.2 × 1000 = 0.46 kg A1 3(b)(ii) 1. (∆)p = (∆m)v C1 (∆)p = 0.46 × 13 = 6.0 N s A1 2. F = 6.0 / 0.20 = 30 N A1 3(b)(iii) force on water (by rocket/nozzle) equal to force on rocket/nozzle (by water) M1 in the opposite direction A1 3(b)(iv) 1. mass = 0.40 + 0.70 – 0.46 = 0.64 kg A1 2. acceleration = [30 – (0.64 × 9.81)] / 0.64 or 30 / 0.64 – 9.81 C1 = 37 m s–2 A1

This question in 9702/23 Oct/Nov 2018

2 (a) Define: (i) displacement … … [1] (ii) acceleration. … … [1] (b) A man wearing a wingsuit glides through the air with a constant velocity of 47 m s–1 at an angle of 24° to the horizontal. The path of the man is shown in Fig. 2.1. 47 m s–1 A man in wingsuit glide path total mass 85 kg h 24° B horizontal Fig. 2.1 (not to scale) The total mass of the man and the wingsuit is 85 kg. The man takes a time of 2.8 minutes to glide from point A to point B. (i) With reference to the motion of the man, state and explain whether he is in equilibrium. … … … … [2] (ii) Show that the difference in height h between points A and B is 3200 m. [1] (iii) For the movement of the man from A to B, determine: 1. the decrease in gravitational potential energy decrease in gravitational potential energy = … J [2] 2. the magnitude of the force on the man due to air resistance. force = … N [2] (iv) The pressure of the still air at A is 63 kPa and at B is 92 kPa. Assume the density of the air is constant between A and B. Determine the density of the air between A and B. density = … kg m–3 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) distance in a specified direction (from a point) B1 2(a)(ii) change in velocity / time (taken) B1 2(b)(i) constant velocity so no resultant force B1 no resultant force so in equilibrium B1 2(b)(ii) (difference in height =) 47 × 2.8 × 60 × sin24° = 3200 m A1 Question Answer Marks 2(b)(iii) 1 (∆)E = mg(∆)h = 85 × 9.81 × 3200 C1 = 2.7 × 106 J A1 2 In terms of energy: work done = 2.7 × 106 J force = 2.7 × 106 / (47 × 2.8 × 60) C1 = 340 N A1 In terms of forces: component of weight along path = force due to air resistance force = 85 × 9.81 × sin24° (C1) = 340 N (A1) 2(b)(iv) (∆)p = ρg(∆)h (92 – 63) × 103 = ρ × 9.81 × 3200 C1 ρ = 0.92 kg m–3 A1

This question in 9702/22 Feb/March 2019

2 (a) State Newton’s second law of motion. … … [1] (b) A car of mass 850 kg tows a trailer in a straight line along a horizontal road, as shown in Fig. 2.1. car trailer tow-bar mass 850 kg horizontal road Fig. 2.1 The car and the trailer are connected by a horizontal tow-bar. The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2. 15 14 v / m s–1 13 12 11 10 9 8 0 5 10 15 20 25 t / s Fig. 2.2 (i) Calculate the distance travelled by the car from time t = 0 to t = 10 s. distance = … m [2] (ii) At time t = 10 s, the resistive force acting on the car due to air resistance and friction is 510 N. The tension in the tow-bar is 440 N. For the car at time t = 10 s: 1. use Fig. 2.2 to calculate the acceleration acceleration = … m s−2 [2] 2. use your answer to calculate the resultant force acting on the car resultant force = … N [1] 3. show that a horizontal force of 1300 N is exerted on the car by its engine [1] 4. determine the useful output power of the engine. output power = … W [2] (c) A short time later, the car in (b) is travelling at a constant speed and the tension in the tow-bar is 480 N. The tow-bar is a solid metal rod that obeys Hooke’s law. Some data for the tow-bar are listed below. Young modulus of metal = 2.2 × 1011 Pa original length of tow-bar = 0.48 m cross-sectional area of tow-bar = 3.0 × 10−4 m2 Determine the extension of the tow-bar. extension = … m [3] (d) The driver of the car in (b) sees a pedestrian standing directly ahead in the distance. The driver operates the horn of the car from time t = 15 s to t = 17 s. The frequency of the sound heard by the pedestrian is 480 Hz. The speed of the sound in the air is 340 m s−1. Use Fig. 2.2 to calculate the frequency of the sound emitted by the horn. frequency = … Hz [2] [Total: 14]

14 marks

Mark scheme: 2(a) (resultant) force proportional/equal to/is rate of change of momentum B1 2(b)(i) distance = area under graph or s = ½ (u + v) t = ½ × (9 + 13) × 10 or s = ut + ½at 2 = (9 × 10) + (½ × 0.40 × 102) or s = vt – ½at 2 = (13 × 10) – (½ × 0.40 × 102) or v 2 = u 2 + 2as 132 = 92 + (2 × 0.40 × s) C1 distance = 110 m A1 Question Answer Marks 2(b)(ii) 1. a = gradient or a = (v – u) / t or a = ∆v / (∆)t e.g. a = (14 – 9) / 12.5 or (13 – 9) / 10 C1 a = 0.40 m s–2 A1 2. resultant force = 850 × 0.40 = 340 N A1 3. (F =) 510 + 440 + 340 = 1300 (N) A1 4. P = Fv C1 = 1300 × 13 = 1.7 × 104 W A1 2(c) E = σ / ε C1 E = (F / A) / (∆L / L) or E = FL / A∆L C1 ∆L = (480 × 0.48) / (3.0 × 10–4 × 2.2 × 1011) = 3.5 × 10–6 m A1 2(d) fo = fs v / (v – vs) 480 = fs × 340 / (340 – 14) C1 fs = 460 Hz A1

This question in 9702/22 May/June 2019

3 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]

7 marks

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

This question in 9702/23 May/June 2019

4 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 sand X particle 2.0 m 0 V + 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 m also released from point X. 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]

13 marks

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 or s = ut + ½at 2 and u = 0 C1 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 = 7.5 × 103 N C–1 A1 4(e)(ii) ma = Eq or F = ma and F = Eq C1 q / m = 0.39 / 7.5 × 103 = 5.2 × 10–5 C kg–1 A1 4(f)(i) no effect B1 4(f)(ii) decreases/smaller B1

This question in 9702/23 May/June 2019

3 A small remote-controlled model aircraft has two propellers, each of diameter 16 cm. Fig. 3.1 is a side view of the aircraft when hovering. body of 16 cm 16 cm aircraft propeller propeller air air speed speed 7.6 m s–1 7.6 m s–1 Fig. 3.1 Air is propelled vertically downwards by each propeller so that the aircraft hovers at a fixed position. The density of the air is 1.2 kg m–3. Assume that the air from each propeller moves with a constant speed of 7.6 m s–1 in a uniform cylinder of diameter 16 cm. Also assume that the air above each propeller is stationary. (a) Show that, in a time interval of 3.0 s, the mass of air propelled downwards by one propeller is 0.55 kg. [3] (b) Calculate: (i) the increase in momentum of the mass of air in (a) increase in momentum = … N s [1] (ii) the downward force exerted on this mass of air by the propeller. force = … N [1] (c) State: (i) the upward force acting on one propeller force = … N [1] (ii) the name of the law that explains the relationship between the force in (b)(ii) and the force in (c)(i). … [1] (d) Determine the mass of the aircraft. mass = … kg [1] (e) In order for the aircraft to hover at a very high altitude (height), the propellers must propel the air downwards with a greater speed than when the aircraft hovers at a low altitude. Suggest the reason for this. … … [1] (f) When the aircraft is hovering at a high altitude, an electric fault causes the propellers to stop rotating. The aircraft falls vertically downwards. When the aircraft reaches a constant speed of 22 m s–1, it emits sound of frequency 3.0 kHz from an alarm. The speed of the sound in the air is 340 m s–1. Determine the frequency of the sound heard by a person standing vertically below the falling aircraft. frequency = … Hz [2] [Total: 11]

11 marks

Mark scheme: 3(a) C1 V = π × (0.16 / 2)2 × 7.6 × 3.0 (= 0.458 m3) C1 m = π × (0.16 / 2)2 × 7.6 × 3.0 × 1.2 = 0.55 kg A1 3(b)(i) ∆p = 0.55 × 7.6 = 4.2 N s A1 3(b)(ii) F = 4.2 / 3.0 or 0.55 × 7.6 / 3.0 = 1.4 N A1 3(c)(i) F = 1.4 N A1 3(c)(ii) Newton’s third law (of motion) B1 3(d) 2 × 1.4 = m × 9.81 m = 0.29 kg A1 3(e) the density of air is less at high altitude B1 3(f) fo = fsv / (v – vs) = 3000 × 340 / (340 – 22) C1 = 3200 Hz A1

This question in 9702/21 Oct/Nov 2019

3 (a) State Newton’s third law of motion. … … … [2] (b) A block X of mass mX slides in a straight line along a horizontal frictionless surface, as shown in Fig. 3.1. speed 5v speed v mass mX mass mY X Y X Y Fig. 3.1 Fig. 3.2 The block X, moving with speed 5v, collides head-on with a stationary block Y of mass mY. The two blocks stick together and then move with common speed v, as shown in Fig. 3.2. mY (i) Use conservation of momentum to show that the ratio is equal to 4. mx [2] (ii) Calculate the ratio total kinetic energy of X and Y after collision . total kinetic energy of X and Y before collision ratio = … [3] (iii) State the value of the ratio in (ii) for a perfectly elastic collision. ratio = … [1] (c) The variation with time t of the momentum of block X in (b) is shown in Fig. 3.3. momentum 0 0 10 20 30 40 50 60 t / ms Fig. 3.3 Block X makes contact with block Y at time t = 20 ms. (i) Describe, qualitatively, the magnitude and direction of the resultant force, if any, acting on block X in the time interval: 1. t = 0 to t = 20 ms … 2. t = 20 ms to t = 40 ms. … … [3] (ii) On Fig. 3.3, sketch the variation of the momentum of block Y with time t from t = 0 to t = 60 ms. [3] [Total: 14]

14 marks

Mark scheme: 3(a) force on body A (by body B) is equal (in magnitude) to force on body B (by body A) B1 force on body A (by body B) is opposite (in direction) to force on body B (by body A) B1 3(b)(i) mX × 5v or (mX + mY) × v C1 mX × 5v = (mX + mY) × v (so) mY / mX = 4 A1 3(b)(ii) (E =) ½mv2 C1 ratio = [½ × (mX + mY) × v2] / [½ × mX × (5v)2] C1 ratio = 0.2 A1 3(b)(iii) ratio = 1 A1 3(c)(i) 1. (magnitude of resultant force is) zero B1 2. (magnitude of resultant force is) constant B1 (direction of resultant force is) opposite to the momentum B1 3(c)(ii) horizontal line from (0 ms, 0 squares) ending at (20 ms, 0 squares) B1 straight line from (20 ms, 0 squares) ending at (40 ms, 4.0 squares [= 4.0 cm vertically]) B1 horizontal line from (40 ms, 4.0 squares) ending at (60 ms, 4.0 squares) B1

This question in 9702/22 Oct/Nov 2019

4 (a) A sphere in a liquid accelerates vertically downwards from rest. For the viscous force acting on the moving sphere, state: (i) the direction … [1] (ii) the variation, if any, in the magnitude. … [1] (b) A man of weight 750 N stands a distance of 3.6 m from end D of a horizontal uniform beam AD, as shown in Fig. 4.1. FB FC A B C D 2.0 m 2.0 m 380 N 750 N 3.6 m 9.0 m Fig. 4.1 (not to scale) The beam has a weight of 380 N and a length of 9.0 m. The beam is supported by a vertical force FB at pivot B and a vertical force FC at pivot C. Pivot B is a distance of 2.0 m from end A and pivot C is a distance of 2.0 m from end D. The beam is in equilibrium. (i) State the principle of moments. … … … [2] (ii) By using moments about pivot C, calculate FB. FB = … N [2] (iii) The man walks towards end D. The beam is about to tip when FB becomes zero. Determine the minimum distance x from end D that the man can stand without tipping the beam. x = … m [2] [Total: 8]

8 marks

Mark scheme: 4(a)(i) (vertically) upwards/up B1 4(a)(ii) increases (with time/velocity/depth) B1 4(b)(i) for a body in (rotational) equilibrium B1 sum/total of clockwise moments about a point = sum/total of anticlockwise moments about the (same) point B1 4(b)(ii) (FB × 5.0) or (380 × 2.5) or (750 × 1.6) C1 (FB × 5.0) = (380 × 2.5) + (750 × 1.6) FB = 430 N A1 4(b)(iii) taking moments about C: (380 × 2.5) = 750 × (2.0 – x) C1 (2.0 – x) = 1.3 x = 0.7 m A1 or moments may be taken about other points, e.g. about D: (380 × 4.5) + (750 × x) = 1130 × 2.0 (C1) x = 0.7 m (A1)

This question in 9702/22 Oct/Nov 2019

2 (a) State Newton’s second law of motion. … … [1] (b) A delivery company suggests using a remote-controlled aircraft to drop a parcel into the garden of a customer. When the aircraft is vertically above point P on the ground, it releases the parcel with a velocity that is horizontal and of magnitude 5.4 m s–1. The path of the parcel is shown in Fig. 2.1. 5.4 m s–1 X parcel path of parcel h P Q horizontal ground d Fig. 2.1 (not to scale) The parcel takes a time of 0.81 s after its release to reach point Q on the horizontal ground. Assume air resistance is negligible. (i) On Fig. 2.1, draw an arrow from point X to show the direction of the acceleration of the parcel when it is at that point. [1] (ii) Determine the height h of the parcel above the ground when it is released. h = … m [2] (iii) Calculate the horizontal distance d between points P and Q. d = … m [1] (c) Another parcel is accidentally released from rest by a different aircraft when it is hovering at a great height above the ground. Air resistance is now significant. (i) On Fig. 2.2, draw arrows to show the directions of the forces acting on the parcel as it falls vertically downwards. Label each arrow with the name of the force. parcel velocity Fig. 2.2 [2] (ii) By considering the forces acting on the parcel, state and explain the variation, if any, of the acceleration of the parcel as it moves downwards before it reaches constant (terminal) speed. … … … … … … [3] (iii) Describe the energy conversion that occurs when the parcel is falling through the air at constant (terminal) speed. … … [1] [Total: 11]

11 marks

Mark scheme: 2(a) (resultant) force proportional to rate of change of momentum B1 2(b)(i) arrow drawn vertically downwards from point X B1 2(b)(ii) s = ut + ½at2 h = ½ × 9.81 × 0.812 C1 = 3.2 m A1 2(b)(iii) d = 5.4 × 0.81 = 4.4 m A1 2(c)(i) downward pointing arrow labelled weight B1 upward pointing arrow labelled air resistance B1 2(c)(ii) air resistance increases B1 weight constant or resultant force decreases B1 (so) acceleration decreases B1 2(c)(iii) gravitational potential energy to thermal/internal energy B1

This question in 9702/21 May/June 2020

2 (a) State Newton’s first law of motion. … … [1] (b) A skier is pulled in a straight line along horizontal ground by a wire attached to a kite, as shown in Fig. 2.1. kite wire skier mass 89 kg 28° horizontal ground Fig. 2.1 (not to scale) The mass of the skier is 89 kg. The wire is at an angle of 28° to the horizontal. The variation with time t of the velocity v of the skier is shown in Fig. 2.2. 5.0 4.0 v / m s–1 3.0 2.0 1.0 0 0 1.0 2.0 3.0 4.0 5.0 t / s Fig. 2.2 (i) Use Fig. 2.2 to determine the distance moved by the skier from time t = 0 to t = 5.0 s. distance = … m [2] (ii) Use Fig. 2.2 to show that the acceleration a of the skier is 0.80 m s–2 at time t = 2.0 s. [2] (iii) The tension in the wire at time t = 2.0 s is 240 N. Calculate: 1. the horizontal component of the tension force acting on the skier horizontal component of force = … N [1] 2. the total resistive force R acting on the skier in the horizontal direction. R = … N [2] (iv) The skier is now lifted upwards by a gust of wind. For a few seconds the skier moves horizontally through the air with the wire at an angle of 45° to the horizontal, as shown in Fig. 2.3. 45° horizontal Fig. 2.3 (not to scale) By considering the vertical components of the forces acting on the skier, determine the new tension in the wire when the skier is moving horizontally through the air. tension = … N [2] [Total: 10]

10 marks

Mark scheme: 2(a) a body continues at (rest or) constant velocity unless acted upon by a resultant force B1 2(b)(i) distance = [½ × (2.0 + 4.4) × 3.0] + [4.4 × 2.0] C1 = 9.6 + 8.8 = 18 m A1 2(b)(ii) a = (v – u) / t or gradient or Δv / (Δ)t C1 e.g. a = (4.4 – 2.0) / 3.0 = 0.80 m s–2 A1 2(b)(iii) 1. force = 240 cos 28° or 240 sin 62° = 210 N A1 2. resultant force = 89 × 0.80 (= 71.2 N) C1 R = 210 – 71 = 140 N A1 2(b)(iv) T sin 45° = mg C1 T = (89 × 9.81) / sin 45° = 1200 N A1

This question in 9702/23 May/June 2020

2 A small block is lifted vertically upwards by a toy aircraft, as illustrated in Fig. 2.1. aircraft string velocity block Fig. 2.1 As the block is moving upwards, the string breaks at time t = 0. The block initially continues moving upwards and then falls and hits the ground at time t = 0.90 s. The variation with time t of the velocity v of the block is shown in Fig. 2.2. 1.96 v / m s–1 0 0 0.20 0.900.90 t / s –6.86 Fig. 2.2 Air resistance is negligible. (a) State the feature of the graph in Fig. 2.2 that shows the block has a constant acceleration. … [1] (b) Use Fig. 2.2 to determine the height of the block above the ground when the string breaks at time t = 0. height = … m [3] (c) The block has a weight of 0.86 N. Calculate the difference in gravitational potential energy of the block between time t = 0 and time t = 0.90 s. difference in gravitational potential energy = … J [2] (d) On Fig. 2.3, sketch a line to show the variation of the distance moved by the block with time t from t = 0 to t = 0.20 s. Numerical values of distance are not required. distance moved 0 0 0.20 t / s Fig. 2.3 [2] (e) A block of greater mass is now released from the same height with the same upward velocity. Air resistance is still negligible. State and explain the effect, if any, of the increased mass on the speed with which the block hits the ground. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) constant gradient B1 2(b) (displacement until 0.20 s =) ½ × 1.96 × 0.20 (= 0.196 m) or (displacement after 0.20 s =) ½ × 6.86 × 0.70 (= 2.401 m) C1 height = 2.401 – 0.196 C1 = 2.2 m (alternative methods are possible using equations of uniformly accelerated motion) A1 2(c) (Δ)E = mg(Δ)h or W(Δ)h C1 (Δ)E = 0.86 × 2.2 = 1.9 J A1 2(d) curved line from the origin M1 gradient of curved line decreases and is zero at t = 0.20 s only A1 2(e) acceleration (of free fall) is unchanged/is not dependent on mass and (so) no effect B1

This question in 9702/21 Oct/Nov 2020

3 (a) Define force. … … [1] (b) A ball falls vertically downwards towards a horizontal floor and then rebounds along its original path, as illustrated in Fig. 3.1. ball reaching ball leaving speed the floor speed the floor 3.8 m s–1 1.7 m s–1 Fig. 3.1 The ball reaches the floor with speed 3.8 m s–1. The ball is then in contact with the floor for a time of 0.081 s before leaving it with speed 1.7 m s–1. The mass of the ball is 0.062 kg. (i) Calculate the loss of kinetic energy of the ball during the collision. loss of kinetic energy = … J [2] (ii) Determine the magnitude of the change in momentum of the ball during the collision. change in momentum = … N s [2] (iii) Show that the magnitude of the average resultant force acting on the ball during the collision is 4.2 N. [1] (iv) Use the information in (iii) to calculate the magnitude of: 1. the average force of the floor on the ball during the collision average force = … N 2. the average force of the ball on the floor during the collision. average force = … N [2] [Total: 8]

8 marks

Mark scheme: 3(a) (force =) rate of change of momentum B1 3(b)(i) E = ½mv2 or ½ × 0.062 × 3.82 or ½ × 0.062 × 1.72 C1 loss of KE = ½ × 0.062 × (3.82 – 1.72) = 0.36 J A1 3(b)(ii) p = mv or 0.062 × 3.8 or 0.062 × 1.7 C1 change in momentum = 0.062 × (1.7 + 3.8) = 0.34 N s A1 3(b)(iii) (average resultant force =) 0.34 / 0.081 = 4.2 (N) or (average resultant force =) 0.062 × (1.7 + 3.8) / 0.081 = 4.2 (N) A1 3(b)(iv) 1. average force = 4.2 + (0.062 × 9.81) = 4.8 N A1 2. average force = 4.8 N A1

This question in 9702/21 Oct/Nov 2020

1 (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 power work [2] (b) The variation with time t of the velocity v of an object is shown in Fig. 1.1. 1.50 1.25 1.00 v / m s–1 0.75 0.50 0.25 0 0 2.0 4.0 6.0 8.0 10.0 12.0 t / s Fig. 1.1 (i) Determine the acceleration of the object from time t = 0 to time t = 4.0 s. acceleration = … m s−2 [2] (ii) Determine the distance moved by the object from time t = 0 to time t = 4.0 s. distance = … m [2] (c) (i) Define force. … … [1] (ii) The motion represented in Fig. 1.1 is caused by a resultant force F acting on the object. On Fig. 1.2, sketch the variation of F with time t from t = 0 to t = 12.0 s. Numerical values of F are not required. F 0 00 2.02.0 4.04.0 6.06.0 8.08.0 10.010.0 12.012.0 tt // ss Fig. 1.2 [3] [Total: 10]

10 marks

Mark scheme: 1(a) acceleration: vector work: scalar power: scalar Three correct scores 2 marks. Two correct scores 1 mark. B2 1(b)(i) a = (v – u) / t or a = gradient or a = Δv / (Δ)t e.g. a = (1.40 – 0.70) / 4.0 C1 = 0.18 m s–2 A1 1(b)(ii) distance = 0.5 × (0.70 + 1.40) × 4.0 or (0.70 × 4.0) + (0.5 × 0.70 × 4.0) C1 = 4.2 m A1 1(c)(i) (force equal to) rate of change of momentum B1 1(c)(ii) horizontal line starting from t = 0 and ending at t = 4.0 s at a positive value of F B1 horizontal line starting from t = 4.0 s and ending at t = 8.0 s at F = 0 B1 horizontal line starting from t = 8.0 s and ending at t = 12.0 s at a negative value of F and the magnitude of F is larger than from t = 0 to 4.0 s B1

This question in 9702/22 Feb/March 2021

2 (a) State what is meant by work done. … … [1] (b) A beach ball is released from a balcony at the top of a tall building. The ball falls vertically from rest and reaches a constant (terminal) velocity. The gravitational potential energy of the ball decreases by 60 J as it falls from the balcony to the ground. The ball hits the ground with speed 16 m s−1 and kinetic energy 23 J. (i) Show that the mass of the ball is 0.18 kg. [2] (ii) Calculate the height of the balcony above the ground. height = … m [2] (iii) Determine the average resistive force acting on the ball as it falls from the balcony to the ground. average resistive force = … N [2] (c) State and explain the variation, if any, in the magnitude of the acceleration of the ball in (b) during the time interval when the ball is moving downwards before it reaches constant (terminal) velocity. … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 2(a) force × displacement in the direction of the force B1 2(b)(i) E = ½mv 2 C1 (m =) 23 × 2 / 162 = 0.18 (kg) A1 2(b)(ii) (Δ)E = mg(Δ)h 60 = 0.18 × 9.81 × h C1 h = 34 m A1 2(b)(iii) (work done =) 60 – 23 = 37 (J) C1 average resistive force = 37 / 34 = 1.1 N A1 2(c) air resistance (acting on ball) increases B1 resultant force (on ball) decreases or weight constant and air resistance increases B1 acceleration decreases B1

This question in 9702/22 Feb/March 2021

2 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]

12 marks

Mark scheme: 2(a) v2 = u2 + 2as u2 = 8.72 – (2 × 9.81 × 1.5) C1 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) = 0.83 N s A1 2(c)(ii) resultant force = 0.83 / 0.091 or 0.059 [(8.7 + 5.4) / 0.091] = 9.1 N A1 2(c)(iii) (W =) 0.059 × 9.81 C1 (W =) 0.58 (N) force = 9.1 + 0.58 = 9.7 N A1 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

This question in 9702/22 May/June 2021

2 (a) Define acceleration. … … [1] (b) A stone falls vertically from the top of a cliff. Fig. 2.1 shows the variation with time t of the velocity v of the stone. 40 v / m s–1 30 20 10 0 0 5 10 15 20 25 30 t / s Fig. 2.1 (i) Explain, with reference to forces acting on the stone, the shape of the curve in Fig. 2.1. … … … … … [3] (ii) Use Fig. 2.1 to determine the speed of the stone when the resultant force on it is zero. speed = … m s–1 [1] (iii) Use Fig. 2.1 to calculate the approximate height through which the stone falls between t = 0 and t = 30 s. height = … m [3] (iv) On Fig. 2.2, sketch the variation with t of the acceleration a of the stone between t = 0 and t = 30 s. 20 a / m s–2 15 10 5 0 0 5 10 15 20 25 30 t / s Fig. 2.2 [3] [Total: 11]

11 marks

Mark scheme: 2(a) change in velocity / time (taken) B1 2(b)(i) air resistance increases (with speed/with time) B1 resultant force decreases (as speed increases/with time) so acceleration decreases (as speed increases/with time) B1 when air resistance equals the weight the speed/velocity/v becomes constant B1 2(b)(ii) speed = 36 m s–1 A1 2(b)(iii) height given by area under the curve C1 height = 950 m Round to two significant figures and award 2 marks for a value in the range 920–980 m and 1 mark for a value in the range 900–910 m or 990–1000 m. A2 2(b)(iv) line starting at (0, 9.8) B1 curve with negative gradient between t = 0 and t = 20 s B1 line showing zero acceleration between t = 20 s and t = 30 s B1 Question Answer Marks

This question in 9702/23 May/June 2021

2 (a) Define momentum. … … [1] (b) Two balls X and Y, of equal diameter but different masses 0.24 kg and 0.12 kg respectively, slide towards each other on a frictionless horizontal surface, as shown in Fig. 2.1. mass 0.24 kg mass 0.12 kg X Y 2.3 m s–1 2.3 m s–1 frictionless surface Fig. 2.1 Both balls have initial speed 2.3 m s–1 before they collide with each other. Fig. 2.2 shows the variation with time t of the force FY exerted on ball Y by ball X during the collision. 400 FY / N 200 0 0 1 2 3 4 5 t / ms –200 – 400 Fig. 2.2 (i) Calculate the kinetic energy of ball X before the collision. kinetic energy = … J [3] (ii) The area enclosed by the lines and the time axis in Fig. 2.2 represents the change in momentum of ball Y during the collision. Determine the magnitude of the change in momentum of ball Y. change in momentum = … N s [2] (iii) Calculate the magnitude of the velocity of ball Y after the collision. velocity = … m s–1 [2] (c) On Fig. 2.3, sketch the variation with time t of the force FX exerted on ball X by ball Y during the collision in (b). 400 FX / N 200 0 0 1 2 3 4 5 t / ms –200 – 400 Fig. 2.3 [3] [Total: 11]

11 marks

Mark scheme: 2(a) mass × velocity B1 2(b)(i) kinetic energy = ½mv2 C1 = ½ × 0.24 × 2.32 C1 = 0.63 J A1 2(b)(ii) change in momentum = ½ × 240 × 5.0 × 10–3 C1 = 0.60 N s A1 2(b)(iii) (change in velocity of Y) = 0.60 / 0.12 ( = 5.0 m s–1) C1 final velocity of Y = 5.0 – 2.3 = 2.7 m s–1 A1 or (final momentum of Y) = 0.60 – 0.12 × 2.3 ( = 0.324 N s) (C1) final velocity of Y = 0.324 / 0.12 = 2.7 m s–1 (A1) 2(c) sloping straight line from (0, 0) to t = 3.0 ms and another straight line continuous with the first from t = 3.0 ms to (5.0, 0) B1 lines showing maximum force of magnitude 240 N B1 lines wholly in the negative F region of the graph B1

This question in 9702/21 Oct/Nov 2021

2 The engine of a toy rocket pushes gases vertically downwards and this results in the rocket accelerating vertically upwards from the ground. The rocket starts to move from rest at time t = 0. The variation with time t of the vertical velocity v of the rocket for the first 0.30 s of the flight is shown in Fig. 2.1. 20 v / m s–1 15 10 5 0 0 0.05 0.10 0.15 0.20 0.25 0.30 t / s Fig. 2.1 As the rocket moves, the thrust force T provided by the rocket engine is 16 N. Assume that the mass of the rocket is constant for this part of its flight. Assume that air resistance is negligible. (a) For this part of the rocket’s flight: (i) show that the acceleration of the rocket is 55 m s–2 [1] (ii) state an expression for the resultant force F experienced by the rocket in terms of the thrust force T and the weight W of the rocket [1] (iii) calculate the mass of the rocket. mass = … kg [2] (b) At time t = 0.30 s, a small piece of metal separates from the rocket. Calculate: (i) the height of the rocket above the ground at t = 0.30 s height = … m [2] (ii) the speed at which the piece of metal strikes the ground. speed = … m s–1 [3] [Total: 9]

9 marks

Mark scheme: 2(a)(i) (e.g. a =) [16.5 – 0] / [0.30 – 0] = 55 (m s–2) A1 2(a)(ii) (F =) T – W A1 2(a)(iii) ma = T – mg C1 m = 16 / (55 + 9.81) m = 0.25 kg A1 2(b)(i) s = ut + ½at 2 or v2 = u2 + 2as or s = vt –½at 2 or s = ½(u + v)t or s = area under graph C1 s = ½  55  0.302 or s = 16.52 / (2  55) or s = 16.5  0.30 – ½  55  0.302 or s = ½  16.5  0.30 s = 2.5 m A1 2(b)(ii) u = 16.5 (m s–1) C1 v2 = u2 + 2as C1 v2 = 16.52 + 2  9.81  2.5 v = 18 m s–1 A1

This question in 9702/23 Oct/Nov 2022

2 A motor uses a wire to raise a block, as illustrated in Fig. 2.1. motor Z wire Y block, weight 1.4 × 104 N X Fig. 2.1 (not to scale) The base of the block takes a time of 0.49 s to move vertically upwards from level X to level Y at a constant speed of 0.64 m s–1. During this time the wire has a strain of 0.0012. The wire is made of metal of Young modulus 2.2 × 1011 Pa and has a uniform cross-section. The block has a weight of 1.4 × 104 N. Assume that the weight of the wire is negligible. (a) Calculate: (i) the cross-sectional area A of the wire A = … m2 [2] (ii) the increase in the gravitational potential energy of the block for the movement of its base from X to Y. increase in gravitational potential energy = … J [3] (b) The motor has an efficiency of 56%. Calculate the input power to the motor as the base of the block moves from X to Y. input power = … W [3] (c) The base of the block now has a uniform deceleration of magnitude 1.3 m s–2 from level Y until the base of the block stops at level Z. Calculate the tension T in the wire as the base of the block moves from Y to Z. T = … N [3] (d) The base of the block is at levels X, Y and Z at times tX, tY and tZ respectively. On Fig. 2.2, sketch a graph to show the variation with time t of the distance d of the base of the block from level X. Numerical values of d and t are not required. d 0 tX tY tZ t Fig. 2.2 [2] [Total: 13]

13 marks

Mark scheme: 2(a)(i) E =  /  or E = F / A C1 A = 1.4  104 / (2.2  1011  0.0012) A1 = 5.3  10–5 m2 2(a)(ii) (∆)h = 0.64  0.49 (= 0.3136) C1 (∆)E = mg(∆)h or W(∆)h C1 = 1.4  104  0.64  0.49 A1 = 4.4  103 J 2(b) P = Fv or W / t C1 = (1.4  104  0.64) / 0.56 or (4.4  103 / 0.49) / 0.56 C1 = 1.6  104 W A1 2(c) m = 1.4  104 / 9.81 C1 ( = 1427 kg) (resultant) F = (1.4  104 / 9.81)  1.3 C1 ( = 1855 N) T = 1.4  104 – 1855 or (1.4104 / 9.81)  (9.81 – 1.3) A1 = 1.2  104 N 2(d) upward sloping straight line from (tX, 0) to tY B1 from tY to tZ: an upward sloping curve with decreasing magnitude of gradient (that is horizontal at tZ) B1

This question in 9702/22 Feb/March 2023

4 Two blocks slide directly towards each other along a frictionless horizontal surface, as shown in Fig. 4.1. The blocks collide and then move as shown in Fig. 4.2. 0.37 kg m s–1 0.65 kg m s–1 0.13 kg m s–1 X Y X Y BEFORE COLLISION AFTER COLLISION Fig. 4.1 Fig. 4.2 Block X initially moves to the right with a momentum of 0.37 kg m s–1. Block Y initially moves to the left with a momentum of 0.65 kg m s–1. After the blocks collide, block X moves to the left back along its original path with a momentum of 0.13 kg m s–1. Block Y also moves to the left after the collision. (a) Block X has an initial kinetic energy of 0.30 J. Calculate the mass of block X. mass = … kg [3] (b) Determine the magnitude of the momentum of block Y after the collision. momentum = … kg m s–1 [1] (c) Block X exerts an average force of 7.7 N on block Y during the collision. Calculate the time that the blocks are in contact with each other. time = … s [2] [Total: 6]

6 marks

Mark scheme: 4(a) E = ½mv2 C1 p = mv C1 m = 0.372 / (2  0.30) or 0.37 / 1.6 or (0.30  2) / 1.62 A1 = 0.23 kg 4(b) 0.37 – 0.65 = –0.13 – p A1 p = 0.15 kg m s–1 4(c) 7.7 = (0.13 + 0.37) / (∆)t C1 or 7.7 = (0.65 – 0.15) / (∆)t time = 0.065 s A1

This question in 9702/22 Feb/March 2023

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]

10 marks

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

This question in 9702/22 May/June 2023

3 A trolley A moves along a horizontal surface at a constant velocity towards another trolley B which is moving at a lower constant speed in the same direction. Fig. 3.1 shows the trolleys at time t = 0. A B horizontal surface Fig. 3.1 Table 3.1 shows data for the trolleys. Table 3.1 trolley mass / kg initial speed / m s–1 A 0.25 0.48 B 0.75 0.12 The two trolleys collide elastically and then separate. Resistive forces are negligible. Fig. 3.2 shows the variation with time t of the velocity v for trolley B. 0.5 v / m s–1 0.4 0.3 B 0.2 0.1 0 / s 0 0.1 0.2 0.3 0.4 0.5t –0.1 –0.2 –0.3 –0.4 –0.5 Fig. 3.2 (a) State what is represented by the area under a velocity–time graph. … [1] (b) Use Table 3.1 and Fig. 3.2 to determine: (i) the acceleration of trolley B during the collision acceleration of B = … m s–2 [2] (ii) the magnitude and direction of the final velocity of trolley A. magnitude = … m s–1 direction … [3] (c) On Fig. 3.2, sketch the variation of the velocity of trolley A with time t from t = 0 to t = 0.50 s. [3] [Total: 9]

9 marks

Mark scheme: 3(a) displacement A1 3(b)(i) a = gradient or a = v / ()t or a = (v – u) / t C1 e.g. a = (0.30 – 0.12) / (0.35 – 0.15) A1 a = 0.90 m s–2 3(b)(ii) (0.25  0.48) + (0.75  0.12) = (0.25 v) + (0.75  0.30) C1 or (0.48 – 0.12) = (0.30 – v) or (½  0.25  0.482) + (½  0.75  0.122) = (½  0.25 × v 2) + (½ × 0.75  0.302) v = (–)0.060 m s–1 A1 direction: to the left / from the right / opposite to (its) initial velocity / opposite to (initial / final) velocity of B B1 3(c) sketch: horizontal line from (0, 0.48) to (0.15, 0.48) B1 horizontal line from (0.35, –0.06) to (0.5, –0.06) B1 straight line between (0.15, 0.48) and (0.35, –0.06) B1

This question in 9702/21 Oct/Nov 2023

3 (a) State the principle of conservation of momentum. … … … [2] (b) An object of mass 2m is travelling at a speed of 5.0 m s–1 in a straight line. It collides with an object of mass 3m which is initially stationary, as shown in Fig. 3.1. 5.0 m s–1 object, mass 2m object, mass 3m Fig. 3.1 After the collision, the object of mass 2m moves with velocity v at an angle of 30° to its original direction of motion. The object of mass 3m moves with velocity w also at an angle of 30°, as shown in Fig. 3.2. object, mass 2m v original path 30° 30° w object, mass 3m Fig. 3.2 By considering the conservation of momentum in two dimensions, calculate the magnitudes of v and w. v = … m s–1 w = … m s–1 [4] (c) An object of mass 4.2 kg is travelling in a straight line at a speed of 6.0 m s–1. The object is brought to rest in a distance of 0.050 m by a constant force. Calculate the magnitude of this force. force = … N [3] [Total: 9]

9 marks

Mark scheme: 3(a) sum / total momentum before (a collision) = sum / total momentum after (a collision) or sum / total momentum (of a system) is constant M1 if no (resultant) external force (acts) / for an isolated system A1 3(b) along direction of motion: 10m = 2mv cos 30° + 3mw cos 30° C1 perpendicular to direction of motion: 2mv cos 60° = 3mw cos 60° (v = 3w / 2) C1 v = 2.9 m s–1 A1 w = 1.9 m s–1 A1 3(c) EK = ½  m  v2 ( = ½  4.2  6.02) ( = 76 J) C1 force = work done / distance C1 force = 76 / 0.050 = 1500 N A1 Question Answer Marks 3(c) or a = (–)u2 / 2s = (–)6.02 / (2  0.050) ( = (–)360 m s–2) (C1) F = ma (C1) F = 4.2  360 = 1500 N (A1) or a = (–)u2 / 2s = (–)6.02 / (2  0.050) ( = (–)360 m s–2) (C1) t = –u / a = –6.0 / –360 (= 0.017 s) F = p / t (C1) F = (0 – 4.2  6) / 0.017 = 1500 N (A1)

This question in 9702/21 May/June 2024

2 A skydiver jumps from an aircraft at time t = 0 and falls vertically downwards. The variation with t of her velocity v is shown in Fig. 2.1. 45 40 v / m s–1 35 30 25 20 15 10 5 0 0 5 10 15 20 25 30 t / s Fig. 2.1 (a) (i) Using Fig. 2.1, state the terminal velocity of the skydiver. terminal velocity = … m s–1 [1] (ii) By drawing a suitable line on Fig. 2.1, determine the acceleration of the skydiver at time t = 9.0 s. acceleration = … m s–2 [2] (b) The mass of the skydiver and her equipment is 68 kg. The upthrust on the skydiver is negligible. After reaching terminal velocity, the skydiver opens her parachute at time t1. A total drag force of 1800 N acts on the skydiver. Determine the magnitude and direction of the acceleration of the skydiver at time t1. acceleration = … m s–2 direction = … [3] (c) The parachute is fully open at time t2. At a later time t3 the skydiver reaches a constant velocity of 5.7 m s–1. (i) Describe and explain the variation with time of the magnitude of her acceleration between time t2 and time t3. … … … … … [2] (ii) Calculate the change in momentum of the skydiver between time t1 and time t3. change in momentum = … N s [2] [Total: 10]

10 marks

Mark scheme: 2(a)(i) 39 m s–1 A1 2(a)(ii) tangent line to curve drawn on Fig. 2.1 C1 a = gradient of tangent line = v / t e.g. = (44 – 26) / (18 – 0) 0.9 ⩽ a ⩽ 1.1 m s–2 A1 2(b) ()F = 68  9.81 – 1800 ( = –1133 N) C1 a = ()F / m = – 1133 / 68 = (–)17 m s–2 A1 upwards B1 2c(i) drag force decreases (as speed decreases) B1 (as speed decreases) resultant force decreases so (magnitude of) acceleration decreases (to zero) B1 2(c)(ii) ()p = mv or ()p = m(v – u) C1 = 68  (5.7 – 39) = (–)2300 N s A1

This question in 9702/22 May/June 2024

2 (a) State the principle of conservation of momentum. … … … [2] (b) A ball X has mass 240 g and moves in a straight line on a horizontal frictionless surface with an initial speed of 16 m s–1. The ball collides with a stationary ball Y that has mass 480 g. After the collision, ball X is stationary, as shown in Fig. 2.1. ball X, ball Y, ball X, ball Y, mass 240 g mass 480 g mass 240 g mass 480 g 16 m s–1 v surface surface BEFORE AFTER Fig. 2.1 (i) Show that the speed v of ball Y after the collision is 8.0 m s–1. [1] (ii) Calculate the change in the total kinetic energy ∆EK of the balls due to the collision. ∆EK = … J [3] (c) The collision in (b) lasts for a time of 2.0 ms. Assume that the contact force between the balls is constant during this time. (i) Determine the magnitude and direction of the force exerted on ball X by ball Y during the collision. magnitude = … N direction … [3] (ii) Compare the magnitude and direction of the force exerted on ball Y by ball X during the collision with the answers in (c)(i). No further calculations are required. … … … [2] [Total: 11]

11 marks

Mark scheme: 2(a) sum / total momentum (of a system of bodies) is constant M1 or sum / total momentum before = sum / total momentum after for an isolated system / no (resultant) external force A1 2(b)(i) 240  16 = 480v and so v = 8.0 m s–1 A1 or (initial momentum =) 240  16 (= 3840 g m s–1) and v = 3840 / 480 = 8.0 m s–1 2(b)(ii) (EK =) ½ mv2 C1 EK = ½ [(0.24  162) – (0.48  8.02)] C1 = 15 J A1 2(c)(i) F = (0.24  16) / (2.0  10–3) or F = (0.48  8) / (2.0  10–3) C1 = 1900 N A1 direction: to the left B1 2(c)(ii) equal (magnitude) B1 opposite (direction) B1

This question in 9702/23 Oct/Nov 2024