Gravitational potential energy and kinetic energy — Paper 2 · A Level Physics 9702

1 (a) Derive the SI base unit of force. SI base unit of force = ………………………………… [1] (b) A spherical ball of radius r experiences a resistive force F due to the air as it moves through the air at speed v. The resistive force F is given by the expression F = crv, where c is a constant. Derive the SI base unit of the constant c. SI base unit of c = ………………………………… [1] Use (c) The ball is dropped from rest through a height of 4.5 m. (i) Assuming air resistance to be negligible, calculate the final speed of the ball. speed = …………………………… m s–1 [2] (ii) The ball has mass 15 g and radius 1.2 cm. The numerical value of the constant c in the equation in (b) is equal to 3.2 ×10–4 when measured using the SI system of units. Show quantitatively whether the assumption made in (i) is justified. [3]

7 marks

Mark scheme: 1 (a) kg m s–2 B1 [1] (b) kg m–1 s–1 B1 [1] (c) (i) v2 = 2gs = 2 × 9.8 × 4.5 C1 v = 9.4 m s–1 A1 [2] (ii) either F (= 3.2 × 10–4 × 1.2 × 10–2 × 9.4) = 3.6 × 10–5 N M1 weight of sphere (= mg = 15 × 10–3 × 9.8) = 0.15 N M1 3.6 × 10–5 << 0.15, so justified A1 [3] or mg = crvT (M1) terminal speed = 3.8 × 104 m s–1 (M1) 9.4 << 3.8 × 104, so justified (A1)

This question in 9702/21 May/June 2006

5 Fig. 5.1 shows the variation with force F of the extension x of a spring as the force is increased to F3 and then decreased to zero. F3 F F2 F1 0 0 x1 x2 x Fig. 5.1 (a) State, with a reason, whether the spring is undergoing an elastic change. … … [1] (b) The extension of the spring is increased from x1 to x2. Show that the work W done in extending the spring is given by W = k(x22 – x12), where k is the spring constant. [3] Use (c) A trolley of mass 850 g is held between two fixed points by means of identical springs, as shown in Fig. 5.2. trolley spring Fig. 5.2 When the trolley is in equilibrium, the springs are each extended by 4.5 cm. Each spring has a spring constant 16 N cm–1. The trolley is moved a distance of 1.5 cm along the direction of the springs. This causes the extension of one spring to be increased and the extension of the other spring to be decreased. The trolley is then released. The trolley accelerates and reaches its maximum speed at the equilibrium position. Assuming that the springs obey Hooke’s law, use the expression in (b) to determine the maximum speed of the trolley. speed = …………………………. m s–1 [4]

8 marks

Mark scheme: 5 (a) no hysteresis loop/no permanent deformation M1 (do not allow ‘force proportional to extension’) so elastic change A0 [1] (b) work done = area under graph line OR average force × distance B1 = ½Fx ½(F2 + F1)(x2 – x1) A1 F = kx, so work done = = ½kx2 ½k(x2 + x1)(x2 – x1) A1 work done = ½k(x22 – x12) A0 [3] (c) gain in energy of trolley = ½k(0.0602 – 0.0452) + ½k(0.0302 – 0.0452) C1 = 0.36 J C1 kinetic energy = ½ × 0.85 × v2 = 0.36 C1 v = 0.92 m s–1 A1 [4]

This question in 9702/21 May/June 2006

4 (a) A stone of mass 56 g is thrown horizontally from the top of a cliff with a speed of 18 m s–1, as illustrated in Fig. 4.1. 18 m s–1 16 m sea level Fig. 4.1 The initial height of the stone above the level of the sea is 16 m. Air resistance may be neglected. (i) Calculate the change in gravitational potential energy of the stone as a result of falling through 16 m. change = … J [2] (ii) Calculate the total kinetic energy of the stone as it reaches the sea. kinetic energy = … J [3] Examiner’s Use (b) Use your answer in (a)(ii) to show that the speed of the stone as it hits the water is approximately 25 m s–1. [1] (c) State the horizontal velocity of the stone as it hits the water. horizontal velocity = … m s–1 [1] (d) (i) On the grid of Fig. 4.2, draw a vector diagram to represent the horizontal velocity and the resultant velocity of the stone as it hits the water. [1] Fig. 4.2 (ii) Use your vector diagram to determine the angle with the horizontal at which the stone hits the water. angle = … ° [2]

10 marks

Mark scheme: 4 (a) (i) (change in) potential energy = mgh C1 = 0.056 × 9.8 × 16 = 8.78 J (allow 8.8) A1 [2] (ii) (initial) kinetic energy = ½mv2 C1 = ½ × 0.056 × 182 = 9.07 J (allow 9.1) C1 total kinetic energy = 8.78 + 9.07 = 17.9 J A1 [3] (b) kinetic energy = ½mv2 17.9 = ½ × 0.056 × v2 and v = 25(.3) m s-1 B1 [1] (c) horizontal velocity = 18 m s-1 B1 [1] (d) (i) correct shape of diagram (two sides of right-angled triangle with correct orientation) B1 (ii) angle = 41° → 48° (allow trig. solution based on diagram) A2 [3] (for angle 38°→ 41° or 48°→ 51°, allow 1 mark)

This question in 9702/21 May/June 2007

3 (a) (i) Define potential energy. … … [1] (ii) Distinguish between gravitational potential energy and elastic potential energy. gravitational potential energy … … elastic potential energy … … [2] (b) A small sphere of mass 51 g is suspended by a light inextensible string from a fixed point P. The centre of the sphere is 61 cm vertically below point P, as shown in Fig. 3.1. P 18° 61 cm sphere, mass 51g Fig. 3.1 The sphere is moved to one side, keeping the string taut, so that the string makes an angle of 18° with the vertical. Calculate (i) the gain in gravitational potential energy of the sphere, gain = ……………………….. J [2] Examiner’s Use (ii) the moment of the weight of the sphere about point P. moment = … N m [2]

7 marks

Mark scheme: 3 (a) (i) potential energy: stored energy available to do work B1 [1] (ii) gravitational: due to height/position of mass OR distance from mass OR moving mass from one point to another B1 elastic: due to deformation/stretching/compressing B1 [2] (b) (i) height raised = (61 – {61 cos18} =) 3.0 cm C1 energy = (mgh = 0.051 × 9.8 × 0.030 =) 1.5 × 10–2 J A1 [2] (ii) moment = force × perpendicular distance = 0.051 × 9.8 × 0.61 × sin18 C1 = 0.094 N m A1 [2] GCE A/AS LEVEL – October/November 2007 9702 02

This question in 9702/21 Oct/Nov 2007

3 A small ball is thrown horizontally with a speed of 4.0 m s–1. It falls through a vertical height of For 1.96 m before bouncing off a horizontal plate, as illustrated in Fig. 3.1. Examiner’s Use 4.0 m s–1 1.96 m 0.98 m plate Fig. 3.1 Air resistance is negligible. (a) For the ball, as it hits the horizontal plate, (i) state the magnitude of the horizontal component of its velocity, horizontal velocity = … m s–1 [1] (ii) show that the vertical component of the velocity is 6.2 m s–1. [1] (b) The components of the velocity in (a) are both vectors. For Examiner’s Complete Fig. 3.2 to draw a vector diagram, to scale, to determine the velocity of the Use ball as it hits the horizontal plate. Fig. 3.2 velocity = … m s–1] at … ° to the vertical [3] (c) After bouncing on the plate, the ball rises to a vertical height of 0.98 m. (i) Calculate the vertical component of the velocity of the ball as it leaves the plate. vertical velocity = … m s–1 [2] (ii) The ball of mass 34 g is in contact with the plate for a time of 0.12 s. For Examiner’s Use your answer in (c)(i) and the data in (a)(ii) to calculate, for the ball as it bounces Use on the plate, 1. the change in momentum, change = … kg m s–1 [3] 2. the magnitude of the average force exerted by the plate on the ball due to this momentum change. force = … N [2]

12 marks

Mark scheme: 3 (a) (i) speed = 4.0 m s-1 …(allow 1 s.f.) … A1 [1] (ii) v2 = 2gh = 2 × 9.8 × 1.96 … M1 v = 6.2 m s-1 … A0 [1] (use of g = 10 m s-2 loses the mark) (b) correct basic shape with correct directions for vectors … M1 speed = (7.4 ± 0.2) m s-1 … A1 at (33 ± 2)° to the vertical … A1 [3] (for credit to be awarded, speed and angle must be correct on the diagram – not calculated) GCE A/AS LEVEL – October/November 2009 9702 22 (c) (i) either v2 = 2 × 9.8 × 0.98 or v = 6.2 / √2 … C1 speed = 4.4 m s-1 … A1 [2] (allow calculation of t = 0.447 s, then v = 4.4 m s-1) (ii) 1 momentum = mv … C1 change in momentum = 0.034 (6.2 + 4.4) … C1 = 0.36 kg m s-1 … A1 [3] (use of 0.034 (6.2 - 4.4) loses last two marks) 2 force = ∆p / ∆t …….(however expressed) … C1 0.36 = 0.12 = 3.0 N ……(allow 1 s.f.) … A1 [2] [Total: 12]

This question in 9702/22 Oct/Nov 2009

4 (a) Explain what is meant by strain energy (elastic potential energy). For Examiner’s … Use … … [2] (b) A spring that obeys Hooke’s law has a spring constant k. Show that the energy E stored in the spring when it has been extended elastically by an amount x is given by E = 12kx 2. [3] (c) A light spring of unextended length 14.2 cm is suspended vertically from a fixed point, For as illustrated in Fig. 4.1. Examiner’s Use fixed point fixed point fixed point 14.2 cm 16.3 cm 17.8 cm 3.8 N F 3.8 N Fig. 4.1 Fig. 4.2 Fig. 4.3 A mass of weight 3.8 N is hung from the end of the spring, as shown in Fig. 4.2. The length of the spring is now 16.3 cm. An additional force F then extends the spring so that its length becomes 17.8 cm, as shown in Fig. 4.3. The spring obeys Hooke’s law and the elastic limit of the spring is not exceeded. (i) Show that the spring constant of the spring is 1.8 N cm–1. [1] (ii) For the extension of the spring from a length of 16.3 cm to a length of 17.8 cm, For Examiner’s 1. calculate the change in the gravitational potential energy of the mass on the Use spring, change in energy = … J [2] 2. show that the change in elastic potential energy of the spring is 0.077 J, [1] 3. determine the work done by the force F. work done = … J [1]

10 marks

Mark scheme: 4 (a) ability to do work … B1 as a result of a change of shape of an object/stretched etc … B1 [2] (b) work = average force ×distance moved (in direction of the force) … B1 either work = ½ × F × x or work is area under F/x graph which is ½Fx … B1 F = kx … B1 so work / energy = ½kx2 … A0 [3] 3.8 (c) (i) spring constant = … M1 2.1 = 1.8 N cm-1 … A0 [1] (ii) 1 ∆EP = mg∆h or W∆h … C1 = 3.8 × 1.5 × 10-2 = 0.057 J … A1 [2] 2 ∆ES = ½ × 1.8 × 102 (0.0362 – 0.0212) … M1 = 0.077 J … A0 [1] 3 work done = 0.077 – 0.057 = 0.020 J … A1 [1] (allow e.c.f. if ∆ES > ∆EP) [Total: 10] GCE A/AS LEVEL – October/November 2009 9702 22

This question in 9702/22 Oct/Nov 2009

3 A cyclist is moving up a slope that has a constant gradient. The cyclist takes 8.0 s to climb For the slope. Examiner’s The variation with time t of the speed v of the cyclist is shown in Fig. 3.1. Use 8 v / m s–1 6 4 2 0 0 2 4 6 8 t / s Fig. 3.1 (a) Use Fig. 3.1 to determine the total distance moved up the slope. distance = … m [3] (b) The bicycle and cyclist have a combined mass of 92 kg. For The vertical height through which the cyclist moves is 1.3 m. Examiner’s Use (i) For the movement of the bicycle and cyclist between t = 0 and t = 8.0 s, 1. use Fig. 3.1 to calculate the change in kinetic energy, change = … J [2] 2. calculate the change in gravitational potential energy. change = … J [2] (ii) The cyclist pedals continuously so that the useful power delivered to the bicycle is 75 W. Calculate the useful work done by the cyclist climbing up the slope. work done = … J [2] (c) Some energy is used in overcoming frictional forces. For Examiner’s (i) Use your answers in (b) to show that the total energy converted in overcoming Use frictional forces is approximately 670 J. [1] (ii) Determine the average magnitude of the frictional forces. average force = … N [1] (d) Suggest why the magnitude of the total resistive force would not be constant. … … … [2]

13 marks

Mark scheme: 3 (a) evidence of use of area below the line B1 distance = 39 m (allow ±0.5 m) A2 [3] (if > ±0.5 m but ≤ 1.0 m, then allow 1 mark) (b) (i) 1 EK = ½mv 2 C1 ∆ EK = ½ × 92 × (62 – 32) = 1240 J A1 [2] 2 EP = mgh C1 ∆ EP = 92 × 9.8 × 1.3 = 1170 J A1 [2] (ii) E = Pt C1 E = 75 × 8 = 600 J A1 [2] (c) (i) energy = (1240 + 600) – 1170 M1 = 670 J A0 [1] (ii) force = 670/39 = 17 N A1 [1] (d) frictional forces include air resistance B1 air resistance decreases with decrease of speed B1 [2] GCE AS/A LEVEL – May/June 2010 9702 23

This question in 9702/23 May/June 2010

5 A spring hangs vertically from a fixed point and a mass of 94 g is suspended from the spring, For stretching the spring as shown in Fig. 5.1. Examiner’s Use mass 94 g 2.6 cm Fig. 5.1 Fig. 5.2 Fig. 5.3 The mass is raised vertically so that the length of the spring is its unextended length. This is illustrated in Fig. 5.2. The mass is then released. The mass moves through a vertical distance of 2.6 cm before temporarily coming to rest. This position is illustrated in Fig. 5.3. (a) State which diagram, Fig. 5.1, Fig. 5.2 or Fig. 5.3, illustrates the position of the mass such that (i) the mass has maximum gravitational potential energy, … [1] (ii) the spring has maximum strain energy. … [1] (b) Briefly describe the variation of the kinetic energy of the mass as the mass falls from its highest position (Fig. 5.2) to its lowest position (Fig. 5.3). … … [1] (c) The strain energy E stored in the spring is given by the expression For Examiner’s 1 Use E = 2kx2 where k is the spring constant and x is the extension of the spring. For the mass moving between the positions shown in Fig. 5.2 and Fig. 5.3, (i) calculate the change in the gravitational potential energy of the mass, change = … J [2] (ii) determine the extension of the spring at which the strain energy is half its maximum value. extension = … cm [3]

8 marks

Mark scheme: 5 (a) (i) Fig. 5.2 B1 [1] (ii) Fig. 5.3 B1 [1] (b) kinetic energy increases from zero then decreases to zero B1 [1] (c) (i) ∆EP = mg∆h / mgh C1 = 94 × 10–3 × 9.8 × 2.6 × 10–2 using g = 10 then –1 = 0.024 J A1 [2] (ii) either 0.024 = ½ k × (2.6 × 10–2)2 or ½ kd2 = ½k × (2.6 × 10–2)2 – ½kd2 C1 0.012 = ½k × d2 kd2 = ½k × (2.6 × 10–2)2 C1 d = 0.018 m d = 0.018 m = 1.8 cm = 1.8 cm A1 [3]

This question in 9702/23 Oct/Nov 2010

4 (a) Distinguish between gravitational potential energy and electric potential energy. For Examiner’s … Use … … [2] (b) A body of mass m moves vertically through a distance h near the Earth’s surface. Use the defining equation for work done to derive an expression for the gravitational potential energy change of the body. [2] (c) Water flows down a stream from a reservoir and then causes a water wheel to rotate, as shown in Fig. 4.1. reservoir stream 120 m water wheel Fig. 4.1 As the water falls through a vertical height of 120 m, gravitational potential energy is converted to different forms of energy, including kinetic energy of the water. At the water wheel, the kinetic energy of the water is only 10% of its gravitational potential energy at the reservoir. (i) Show that the speed of the water as it reaches the wheel is 15 m s–1. [2] (ii) The rotating water wheel is used to produce 110 kW of electrical power. Calculate For the mass of water flowing per second through the wheel, assuming that the Examiner’s production of electric energy from the kinetic energy of the water is 25% efficient. Use mass of water per second = … kg s–1 [3]

9 marks

Mark scheme: 4 (a) electrical potential energy (stored) when charge moved and gravitational potential energy (stored) when mass moved B1 due to work done in electric field and work done in gravitational field B1 [2] (b) work done = force × distance moved (in direction of force) and force = mg M1 mg × h or mg × ∆h A1 [2] (c) (i) 0.1 × mgh = ½ mv2 B1 0.1 × m × 9.81 × 120 = 0.5 × m × v2 B1 v = 15.3 m s–1 A0 [2] (ii) P = 0.5 m v2 / t C1 m / t = 110 × 103 / [0.25 × 0.5 × (15.3)2] C1 = 3740 kg s–1 A1 [3]

This question in 9702/21 Oct/Nov 2011

2 (a) Define For Examiner’s (i) force, Use … … [1] (ii) work done. … … [1] (b) A force F acts on a mass m along a straight line for a distance s. The acceleration of the mass is a and the speed changes from an initial speed u to a final speed v. (i) State the work W done by F. [1] (ii) Use your answer in (i) and an equation of motion to show that kinetic energy of a mass can be given by the expression kinetic energy = ½ × mass × (speed)2. [3] (c) A resultant force of 3800 N causes a car of mass of 1500 kg to accelerate from an initial speed of 15 m s–1 to a final speed of 30 m s–1. (i) Calculate the distance moved by the car during this acceleration. distance = … m [2] (ii) The same force is used to change the speed of the car from 30 m s–1 to 45 m s–1. Explain why the distance moved is not the same as that calculated in (i). … … … [1]

9 marks

Mark scheme: 2 (a) (i) force is rate of change of momentum B1 [1] (ii) work done is the product of the force and the distance moved in the direction of the force B1 [1] (b) (i) W = Fs or W = mas or W = m(v2 – u2) / 2 or W = force × distance s A1 [1] (ii) as = (v2 – u2) / 2 any subject M1 W = mas hence W = m(v2 – u2) / 2 M1 RHS represents terms of energy or with u = 0 KE = ½mv 2 A1 [3] (c) (i) work done = ½ × 1500 × [(30)2 – (15)2] (=506250) C1 distance = WD / F = 506250 / 3800 = 133 m A1 [2] or F = ma a = 2.533 (m s–2) C1 v2 = u2 + 2as s = 133 m A1 (ii) the change in kinetic energy is greater or the work done by the force has to be greater, hence distance is greater (for same force) A1 [1] allow: same acceleration, same time, so greater average speed and greater distance

This question in 9702/22 Oct/Nov 2011

2 (a) A ball is thrown vertically down towards the ground and rebounds as illustrated in For Fig. 2.1. Examiner’s Use ball passing point A A 8.4 m s–1 ball at maximum height after rebound 5.0 m B h Fig. 2.1 As the ball passes A, it has a speed of 8.4 m s–1. The height of A is 5.0 m above the ground. The ball hits the ground and rebounds to B. Assume that air resistance is negligible. (i) Calculate the speed of the ball as it hits the ground. speed = … m s–1 [2] (ii) Show that the time taken for the ball to reach the ground is 0.47 s. [1] (b) The ball rebounds vertically with a speed of 4.2 m s–1 as it leaves the ground. The time For the ball is in contact with the ground is 20 ms. The ball rebounds to a maximum height h. Examiner’s Use The ball passes A at time t = 0. On Fig. 2.2, plot a graph to show the variation with time t of the velocity v of the ball. Continue the graph until the ball has rebounded from the ground and reaches B. v / m s–1 0 0 t / s Fig. 2.2 [3] (c) The ball has a mass of 0.050 kg. It moves from A and reaches B after rebounding. (i) For this motion, calculate the change in 1. kinetic energy, change in kinetic energy = … J [2] 2. gravitational potential energy. change in potential energy = … J [3] (ii) State and explain the total change in energy of the ball for this motion. For Examiner’s … Use … … … [2]

13 marks

Mark scheme: 2 (a) (i) v2 = u2 + 2as = (8.4)2 + 2 × 9.81 × 5 C1 = 12.99 m s–1 (allow 13 to 2 s.f. but not 12.9) A1 [2] (ii) t = (v – u) / a or s = ut + ½at2 = (12.99 – 8.4) / 9.81 or 5 = 8.4t + ½ × 9.81t2 M1 t = 0.468 s A0 [1] (b) reasonable shape M1 suitable scale A1 correctly plotted 1st and last points at (0,8.4) and (0.88 – 0.96,0) with non-vertical line at 0.47 s A1 [3] (c) (i) 1. kinetic energy at end is zero so ∆KE = ½ mv2 or ∆KE = ½ mu2 – ½ mv2 C1 = ½ × 0.05 × (8.4)2 = (–) 1.8 J A1 [2] 2. final maximum height = (4.2)2 / (2 × 9.8) = (0.9 (m)) change in PE = mgh2 – mgh1 C1 = 0.05 × 9.8 × (0.9 – 5) C1 = (–) 2.0 J A1 [3] (ii) change is – 3.8 (J) B1 energy lost to ground (on impact) / energy of deformation of the ball / thermal energy in ball B1 [2]

This question in 9702/22 May/June 2012

3 (a) An object falls vertically from rest through air. State and explain the energy conversions For that occur as the object falls. Examiner’s Use … … … … [3] (b) A ball of mass 150 g is thrown vertically upwards with an initial speed of 25 m s–1. (i) Calculate the initial kinetic energy of the ball. kinetic energy = … J [3] (ii) The ball reaches a height of 21 m above the point of release. For the ball rising to this height, calculate 1. the loss of energy of the ball to air resistance, energy loss = … J [3] 2. the average force due to the air resistance. force = … N [2]

11 marks

Mark scheme: 3 (a) loss in potential energy due to decrease in height (as P.E. = mgh) (B1) gain in kinetic energy due to increase in speed (as K.E. = ½ mv2) (B1) special case ‘as PE decreases KE increases’ (1/2) increase in thermal energy due to work done against air resistance (B1) loss in P.E. equals gain in K.E. and thermal energy (B1) max. 3 [3] GCE AS/A LEVEL – May/June 2013 9702 21 (b) (i) kinetic energy = ½ mv2 C1 = ½ × 0.150 × (25)2 C1 = 46.875 = 47 J A1 [3] (ii) 1. potential energy (= mgh) = 0.150 × 9.81 × 21 C1 loss = KE – mgh = 46.875 – (30.9) C1 = 15.97 = 16 J A1 [3] 2. work done = 16 J work done = force × distance C1 F = 16 / 21 = 0.76 N A1 [2]

This question in 9702/21 May/June 2013

3 (a) State what is meant by work done. For Examiner’s … Use … [1] (b) A trolley of mass 400 g is moving at a constant velocity of 2.5 m s–1 to the right as shown in Fig. 3.1. 2.5 m s–1 trolley 400 g Fig. 3.1 Show that the kinetic energy of the trolley is 1.3 J. [2] (c) The trolley in (b) moves to point P as shown in Fig. 3.2. 2.5 m s–1 trolley F 400 g P Q x Fig. 3.2 At point P the speed of the trolley is 2.5 m s–1. A variable force F acts to the left on the trolley as it moves between points P and Q. The variation of F with displacement x from P is shown in Fig. 3.3. 20 F / N 10 0 0 x P Q Fig. 3.3 The trolley comes to rest at point Q. For Examiner’s (i) Calculate the distance PQ. Use distance PQ = … m [3] (ii) On Fig. 3.4, sketch the variation with x of velocity v for the trolley moving between P and Q. 2.5 v / m s–1 0 P Q x Fig. 3.4 [2]

8 marks

Mark scheme: 3 (a) (work =) force × distance moved / displacement in the direction of the force OR when a force moves in the direction of the force work is done B1 [1] (b) kinetic energy = ½ mv2 C1 kinetic energy = ½ 0.4 (2.5)2 = 1.25 / 1.3 J A1 [2] (c) (i) area under graph is work done / work done = ½ Fx C1 1.25 = (14 x) / 2 C1 x = 0.18 (0.179) m [allow x = 0.19 m using kinetic energy = 1.3 J] A1 [3] (ii) smooth curve from v = 2.5 at x = 0 to v = 0 at Q M1 curve with increasing gradient A1 [2] GCE AS/A LEVEL – October/November 2013 9702 21

This question in 9702/21 Oct/Nov 2013

3 (a) State what is meant by work done. For Examiner’s … Use … [1] (b) A trolley of mass 400 g is moving at a constant velocity of 2.5 m s–1 to the right as shown in Fig. 3.1. 2.5 m s–1 trolley 400 g Fig. 3.1 Show that the kinetic energy of the trolley is 1.3 J. [2] (c) The trolley in (b) moves to point P as shown in Fig. 3.2. 2.5 m s–1 trolley F 400 g P Q x Fig. 3.2 At point P the speed of the trolley is 2.5 m s–1. A variable force F acts to the left on the trolley as it moves between points P and Q. The variation of F with displacement x from P is shown in Fig. 3.3. 20 F / N 10 0 0 x P Q Fig. 3.3 The trolley comes to rest at point Q. For Examiner’s (i) Calculate the distance PQ. Use distance PQ = … m [3] (ii) On Fig. 3.4, sketch the variation with x of velocity v for the trolley moving between P and Q. 2.5 v / m s–1 0 P Q x Fig. 3.4 [2]

8 marks

Mark scheme: 3 (a) (work =) force × distance moved / displacement in the direction of the force OR when a force moves in the direction of the force work is done B1 [1] (b) kinetic energy = ½ mv2 C1 kinetic energy = ½ 0.4 (2.5)2 = 1.25 / 1.3 J A1 [2] (c) (i) area under graph is work done / work done = ½ Fx C1 1.25 = (14 x) / 2 C1 x = 0.18 (0.179) m [allow x = 0.19 m using kinetic energy = 1.3 J] A1 [3] (ii) smooth curve from v = 2.5 at x = 0 to v = 0 at Q M1 curve with increasing gradient A1 [2] GCE AS/A LEVEL – October/November 2013 9702 22

This question in 9702/22 Oct/Nov 2013

4 (a) Distinguish between gravitational potential energy and elastic potential energy. For Examiner’s … Use … … [2] (b) A ball of mass 65 g is thrown vertically upwards from ground level with a speed of 16 m s–1. Air resistance is negligible. (i) Calculate, for the ball, 1. the initial kinetic energy, kinetic energy = … J [2] 2. the maximum height reached. maximum height = … m [2] t (ii) The ball takes time t to reach maximum height. For time after the ball has been 2 thrown, calculate the ratio potential energy of ball . kinetic energy of ball ratio = … [3] (iii) State and explain the effect of air resistance on the time taken for the ball to reach maximum height. … … … [1]

10 marks

Mark scheme: 4 (a) gravitational PE is energy of a mass due to its position in a gravitational field B1 elastic PE energy stored (in an object) due to (a force) changing its shape / deformation / being compressed / stretched / strained B1 [2] (b) (i) 1. kinetic energy = ½ mv2 C1 kinetic energy = ½ × 0.065 × 162 = 8.3(2) J A1 [2] 2. v2 = 2gh OR PE = mgh C1 h = 162 / (2 × 9.81) = 13(.05) m A1 [2] GCE AS/A LEVEL – October/November 2013 9702 23 (ii) speed at t = ½ total time = 8 (m s–1) or total t =1.63 or t1/2 = 0.815 s C1 KE is ¼ or h at t1/2 = 9.78 (m) C1 and PE is ¾ of max ratio = 3 or ratio = 9.78 / 3.26 = 3 A1 [3] (iii) time is less because (average) acceleration is greater OR average force is greater B1 [1]

This question in 9702/23 Oct/Nov 2013

4 A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1. metal ball spring spring support Fig. 4.1 (not to scale) The spring is supported and stands vertically. The ball has a speed of 2.8 m s−1 as it makes contact with the spring. The ball is brought to rest as the spring is compressed. (a) Show that the kinetic energy of the ball as it makes contact with the spring is 0.16 J. [2] (b) The variation of the force F acting on the spring with the compression x of the spring is shown in Fig. 4.2. 20 F / N 10 0 0 XB x Fig. 4.2 The ball produces a maximum compression XB when it comes to rest. The spring has a spring constant of 800 N m−1. Use Fig. 4.2 to (i) calculate the compression XB, XB = … m [2] (ii) show that not all the kinetic energy in (a) is converted into elastic potential energy in the spring. [2]

6 marks

Mark scheme: 4 (a) kinetic energy = ½ mv2 C1 = ½ × 0.040 × (2.8)2 = 0.157 J or 0.16 J A1 [2] (b) (i) k = F / x or F = kx C1 XB = 14 / 800 = 0.0175 m A1 [2] (ii) area under graph = elastic potential energy stored C1 or ½ kx2 or ½ Fx (energy stored =) 0.1225 J less than KE (of 0.16 J) A1 [2]

This question in 9702/21 May/June 2014

4 (a) Explain what is meant by gravitational potential energy and kinetic energy. gravitational potential energy: … … .… … … kinetic energy: … … [2] (b) A ball of mass 400 g is thrown with an initial velocity of 30.0 m s–1 at an angle of 45.0° to the horizontal, as shown in Fig. 4.1. path of ball 30.0 m s–1 H ball 45.0° Fig. 4.1 Air resistance is negligible. The ball reaches a maximum height H after a time of 2.16 s. (i) Calculate 1. the initial kinetic energy of the ball, kinetic energy = … J [3] 2. the maximum height H of the ball, H = … m [2] 3. the gravitational potential energy of the ball at height H. potential energy = … J [2] (ii) 1. Determine the kinetic energy of the ball at its maximum height. kinetic energy = … J [1] 2. Explain why the kinetic energy of the ball at maximum height is not zero. … … [1]

11 marks

Mark scheme: 4 (a) GPE: energy of a mass due to its position in a gravitational field B1 KE: energy (a mass has) due to its motion / speed / velocity B1 [2] 1 (b) (i) 1. KE = mv2 C1 2 1 = × 0.4 × (30)2 C1 2 = 180 J A1 [3] 1 2. s = 0 + × 9.81 × (2.16)2 or s = (30 sin 45°)2 / (2 × 9.81) C1 2 = 22.88 (22.9) m = 22.94 (22.9) m A1 [2] 3. GPE = mgh C1 = 0.4 × 9.81 × 22.88 = 89.8 (90) J A1 [2] (ii) 1. KE = initial KE – GPE = 180 – 90 = 90 J A1 [1] 2. (horizontal) velocity is not zero / (object) is still moving / answer explained in terms of conservation of energy B1 [1]

This question in 9702/22 May/June 2014

5 A motor is used to move bricks vertically upwards, as shown in Fig. 5.1. motor bricks container Fig. 5.1 The bricks start from rest and accelerate for 2.0 s. The bricks then travel at a constant speed of 0.64 m s−1 for 25 s. Finally the bricks are brought to rest in a further 3.0 s. The total mass of the bricks is 25 kg. (a) Determine the change in kinetic energy of the bricks (i) in the first 2.0 s, change in kinetic energy = … J [2] (ii) in the next 25 s, change in kinetic energy = … J [1] (iii) in the final 3.0 s. change in kinetic energy = … J [1] (b) The bricks are in a container. The weight of the container and bricks is 350 N. Calculate, for the lifting of the bricks and container when travelling at constant speed, (i) the gain in potential energy, energy gain = … J [3] (ii) the power required. power = … W [2]

9 marks

Mark scheme: 5 (a) (i) change in kinetic energy = ½ mv2 C1 = 0.5 × 25 × (0.64)2 = 5.1(2) J A1 [2] (ii) zero A1 [1] (iii) (–) 5.1(2) J A1 [1] (b) (i) PE = mgh C1 = 350 × 0.64 × 25 C1 = 5600 J A1 [3] (If full length used allow 1/3) (ii) P = Fv or gain in PE / t, EP / t or work done / t, W / t C1 = 350 × 0.64 or 5600 / 25 = 220 (224) W A1 [2]

This question in 9702/21 Oct/Nov 2014

2 The variation with time t of the velocity v of a ball is shown in Fig. 2.1. 5 v / m s–1 0 0 2 4 6 8 10 12 14 16 t / s ï Fig. 2.1 The ball moves in a straight line from a point P at t = 0. The mass of the ball is 400 g. (a) Use Fig. 2.1 to describe, without calculation, the velocity of the ball from t = 0 to t = 16 s. … … … … … [2] (b) Use Fig. 2.1 to calculate, for the ball, (i) the displacement from P at t = 10 s, displacement = … m [2] (ii) the acceleration at t = 10 s, acceleration = … m s–2 [2] (iii) the maximum kinetic energy. kinetic energy = … J [2] (c) Use your answers in (b)(i) and (b)(ii) to determine the time from t = 0 for the ball to return to P. time = … s [2]

10 marks

Mark scheme: 2 (a) constant rate of increase in velocity/acceleration from t = 0 to t = 8 s B1 constant deceleration from t = 8 s to t = 16 s or constant rate of increase in velocity in the opposite direction from t = 10 s to t = 16 s B1 [2] (b) (i) area under lines to 10 s C1 (displacement =) (5.0 × 8.0) / 2 + (5.0 × 2.0) / 2 = 25 m or ½ (10.0 × 5.0) = 25 m A1 [2] (ii) a = (v – u) / t or gradient of line C1 = (–15.0 –5.0) / 8.0 = (–) 2.5 m s–2 A1 [2] (iii) KE = ½ m v2 C1 = 0.5 × 0.4 × (15.0)2 = 45 J A1 [2] (c) (distance =) 25 (m) (= ut + ½ at 2) = 0 + ½ × 2.5 × t 2 C1 (t = 4.5 (4.47) s therefore) time to return = 14.5 s A1 [2]

This question in 9702/23 May/June 2015

3 (a) An object is moved from point P to point R either by a direct path or by the path P to Q to R, as shown in Fig. 3.1. R vertical object horizontal P Q Fig. 3.1 P and Q are on the same horizontal level. R is vertically above Q. Explain whether the work done moving the object against the gravitational field is the same or different along paths PR and PQR. … … … … [2] (b) A ball is thrown with an initial velocity V at an angle θ to the horizontal, as shown in Fig. 3.2. path of ball V e horizontal Fig. 3.2 (not to scale) The variation with time t of the height h of the ball is shown in Fig. 3.3. 12.0 10.0 h / m 8.0 6.0 4.0 2.0 0 0 1.00 2.00 3.00 t / s Fig. 3.3 Air resistance is negligible. (i) Use the time to reach maximum height to determine the vertical component Vv of the velocity of the ball for time t = 0. Vv = … m s–1 [2] (ii) The horizontal displacement of the ball at t = 3.00 s is 25.5 m. On Fig. 3.4, draw the variation with t of the horizontal displacement x of the ball. 30 x / m 20 10 0 0 1.00 2.00 3.00 t / s Fig. 3.4 [1] (iii) For the ball at maximum height, calculate the ratio potential energy of the ball . kinetic energy of the ball ratio = … [3] (iv) In practice, air resistance is not negligible. State and explain the effect of air resistance on the time taken for the ball to reach maximum height. … … … … [2]

10 marks

Mark scheme: 3 (a) work done is force × distance moved in direction of force or no work done along PQ as no displacement/distance moved in direction of force B1 work done is same in vertical direction as same distance moved in direction of force B1 [2] (b) (i) at maximum height t = 1.5 (s) or s = ½(u + v) t, s = 11 m and t = 1.5 s C1 Vv = 0 + 9.81 × 1.5 Vv = (11 × 2) / 1.5 = 15 (14.7) m s–1 A1 [2] (ii) straight line from (0,0) to (3.00, 25.5) B1 [1] (iii) at maximum height Vh = 25.5 / 3 (= 8.5 m s–1) B1 ratio = mgh / ½ mv2 C1 = (2 × 9.81 × 11.0) / (8.5)2 = 3.0 (2.99) A1 [3] (iv) deceleration is greater/resultant force (weight and friction force) is greater M1 time is less A1 [2]

This question in 9702/21 Oct/Nov 2015

3 A steel ball falls from a platform on a tower to the ground below, as shown in Fig. 3.1. ball platform path of tower 192 m ball ground Fig. 3.1 The ball falls from rest through a vertical distance of 192 m. The mass of the ball is 270 g. (a) Assume air resistance is negligible. (i) Calculate 1. the time taken for the ball to fall to the ground, time taken = … s [2] 2. the maximum kinetic energy of the ball. maximum kinetic energy = … J [2] (ii) State and explain the variation of the velocity of the ball with time as the ball falls to the ground. … … [1] (iii) Show that the velocity of the ball on reaching the ground is approximately 60 m s–1. [1] (b) In practice, air resistance is not negligible. The variation of the air resistance R with the velocity v of the ball is shown in Fig. 3.2. 4.0 3.0 R / N 2.0 1.0 0 0 20 40 60 80 100 v / m s–1 Fig. 3.2 (i) Use Fig. 3.2 to state and explain qualitatively the variation of the acceleration of the ball with the distance fallen by the ball. … … … … [3] (ii) The speed of the ball reaches 40 m s–1. Calculate its acceleration at this speed. acceleration = … m s–2 [2] (iii) Use information from (a)(iii) and Fig. 3.2 to state and explain whether the ball reaches terminal velocity. … … … [2]

13 marks

Mark scheme: 3 (a) (i) 1. s = ut + ½ at2 192 = ½ × 9.81 × t2 C1 t = 6.3 (6.26) s A1 [2] 2. max Ek (= mgh) = 0.27 × 9.81 × 192 C1 or calculation of v (= 61.4) and use of EK (= ½ mv2) = ½ × 0.27 × (61.4)2 (C1) max Ek = 510 (509) J A1 [2] (ii) velocity is proportional to time or velocity increases at a constant rate as acceleration is constant or resultant force is constant B1 [1] (iii) use of v = at or v2 = 2as or E = ½ mv2 to give v = 61(.4) m s–1 B1 [1] (b) (i) R increases with velocity B1 resultant force is mg – R or resultant force decreases B1 acceleration decreases B1 [3] (ii) at v = 40 m s–1, R = 0.6 (N) C1 0.27 × 9.8 – 0.6 = 0.27 × a a = 7.6 (7.58) m s–2 A1 [2] (iii) R = weight for terminal velocity B1 either weight requires velocity to be about 80 m s–1 or at 60 m s–1, R is less than weight so does not reach terminal velocity B1 [2]

This question in 9702/23 Oct/Nov 2015

3 (a) State what is meant by (i) work done, … … [1] (ii) elastic potential energy. … … [1] (b) A block of mass 0.40 kg slides in a straight line with a constant speed of 0.30 m s−1 along a horizontal surface, as shown in Fig. 3.1. spring 0.30 m s–1 block mass 0.40 kg Fig. 3.1 The block hits a spring and decelerates. The speed of the block becomes zero when the spring is compressed by 8.0 cm. (i) Calculate the initial kinetic energy of the block. kinetic energy = … J [2] (ii) The variation of the compression x of the spring with the force F applied to the spring is shown in Fig. 3.2. 8.0 x / cm 0 0 FMAX F Fig. 3.2 Use your answer in (b)(i) to determine the maximum force FMAX exerted on the spring by the block. Explain your working. FMAX = … N [3] (iii) Calculate the maximum deceleration of the block. deceleration = … m s−2 [1] (iv) State and explain whether the block is in equilibrium 1. before it hits the spring, … … 2. when its speed becomes zero. … … [2] (c) The energy E stored in a spring is given by E = 1 k x 2 2 where k is the spring constant of the spring and x is its compression. The mass m of the block in (b) is now varied. The initial speed of the block remains constant and the spring continues to obey Hooke’s law. On Fig. 3.3, sketch the variation of the maximum compression x0 of the spring with mass m. x0 0 0 m Fig. 3.3 [2] [Total: 12]

12 marks

Mark scheme: 3 (a) (i) (work = ) force × distance moved in the direction of the force. B1 (ii) the energy stored (in an object) due to extension / compression / change of shape B1 (b) (i) EK = ½mv 2 C1 = 0.5 × 0.40 × 0.302 = 1.8 × 10–2 (J) A1 (ii) (change in) kinetic energy = work done on spring / (change in) elastic potential energy C1 1.8 × 10–2 = ½ × F × 0.080 C1 FMAX = 0.45 (N) A1 (iii) a = F / m = 0.45 / 0.40 = 1.1 (m s–2) A1 (iv) 1. constant velocity / resultant force is zero, so in equilibrium B1 2. decelerating / resultant force is not zero, so not in equilibrium B1 (c) curved line from the origin M1 with decreasing gradient A1

This question in 9702/22 Feb/March 2016

3 (a) Explain what is meant by gravitational potential energy and by kinetic energy. gravitational potential energy: … … kinetic energy: … … [2] (b) A motion sensor is used to measure the velocity of a ball falling vertically towards the ground, as illustrated in Fig. 3.1. motion sensor v A B ground Fig. 3.1 The ball passes through points A and B as it falls. The ball has a mass of 1.5 kg. The variation with time t of the velocity v of the ball as it falls from A to B is shown in Fig. 3.2. 8.0 7.0 6.0 v / m s–1 5.0 4.0 3.0 0.40 0.60 0.80 t / s ball at position A ball at position B Fig. 3.2 Use Fig. 3.2 to calculate, for the ball falling from A to B, (i) the displacement, displacement = … m [3] (ii) the acceleration, acceleration = … m s–2 [2] (iii) the change in kinetic energy. change in kinetic energy = … J [3] (c) Show that the work done by the gravitational field on the ball in (b) as it moves from A to B is equal to the change in kinetic energy. [2] [Total: 12]

12 marks

Mark scheme: 3 (a) (gravitational potential energy is) the energy/ability to do work of a mass that it has or is stored due to its position/height in a gravitational field B1 kinetic energy is energy/ability to do work a object/body/mass has due to its speed/velocity/motion/movement B1 [2] (b) (i) s = [(u + v) t] / 2 or acceleration = 9.8/9.75 (using gradient) C1 = [(7.8 + 3.9) × 0.4] / 2 or s = 3.9 × 0.4 + 21 × 9.75 × (0.4)2 C1 s = 2.3(4) m A1 [3] (ii) a = (v – u) / t or gradient of line C1 = (7.8 – 3.9) / 0.4 = 9.8 (9.75) m s–2 (allow ± 21 small square in readings) A1 [2] (iii) KE = 21 mv2 C1 change in kinetic energy = 21 mv2 – 21 mu2 = 21 × 1.5 × (7.82 – 3.92) C1 = 34 (34.22) J A1 [3] (c) work done = force × distance (moved) or Fd or Fx or mgh or mgd or mgx M1 = 1.5 × 9.8 × 2.3 = 34 (33.8) J (equals the change in KE) A1 [2]

This question in 9702/23 May/June 2016

2 A ball of mass 0.030 kg moves along a curved track, as shown in Fig. 2.1. ball mass 0.030 kg speed 1.3 m s–1 A wall 0.31 m B Fig. 2.1 The speed of the ball is 1.3 m s–1 when it is at point A at a height of 0.31 m. The ball moves down the track and collides with a vertical wall at point B. The ball then rebounds back up the track. It may be assumed that frictional forces are negligible. (a) Calculate the change in gravitational potential energy of the ball in moving from point A to point B. change in gravitational potential energy = … J [2] (b) Show that the ball hits the wall at B with a speed of 2.8 m s–1. [3] (c) The change in momentum of the ball due to the collision with the wall is 0.096 kg m s–1. The ball is in contact with the wall for a time of 20 ms. Determine, for the ball colliding with the wall, (i) the speed immediately after the collision, speed = … m s–1 [2] (ii) the magnitude of the average force on the ball. force = … N [2] (d) State and explain whether the collision is elastic or inelastic. … … [1] (e) In practice, frictional effects are significant so that the actual increase in kinetic energy of the ball in moving from A to B is 76 mJ. The length of the track between A and B is 0.60 m. Use your answer in (a) to determine the average frictional force acting on the ball as it moves from A to B. frictional force = … N [2] [Total: 12]

12 marks

Mark scheme: 2 (a) ∆E = mg∆h C1 = 0.030 × 9.81 × (–)0.31 = (–)0.091 J A1 [2] (b) E = ½mv 2 C1 (initial) E = ½ × 0.030 × 1.32 (= 0.0254) C1 0.5 × 0.030 × v 2 = (0.5 × 0.030 × 1.32) + (0.030 × 9.81 × 0.31) so v = 2.8 m s–1 or 0.5 × 0.030 × v 2 = (0.0254) + (0.091) so v = 2.8 m s–1 A1 [3] (c) (i) 0.096 = 0.030 (v + 2.8) C1 v = 0.40 m s–1 A1 [2] (ii) F = ∆p / (∆)t or F = ma = 0.096 / 20 × 10–3 or 0.030 (0.40 + 2.8) / 20 × 10–3 C1 = 4.8 N A1 [2] (d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic or (relative) speed of approach (of ball and wall) not equal to/greater than (relative) speed of separation, so inelastic. B1 [1] (e) force = work done / distance moved = (0.091 – 0.076) / 0.60 C1 = 0.025 N A1 [2]

This question in 9702/22 Oct/Nov 2016

2 (a) Define velocity. … … [1] (b) A ball of mass 0.45 kg leaves the edge of a table with a horizontal velocity v, as shown in Fig. 2.1. ball v path of ball 1.25 m table 1.50 m floor horizontal Fig. 2.1 The height of the table is 1.25 m. The ball travels a distance of 1.50 m horizontally before hitting the floor. Air resistance is negligible. Calculate, for the ball, (i) the horizontal velocity v as it leaves the table, v = … m s–1 [3] (ii) the velocity just as it hits the floor, magnitude of velocity = … m s–1 angle to the horizontal = … ° [4] (iii) the kinetic energy just as it hits the floor, kinetic energy = … J [2] (iv) the loss in gravitational potential energy as it falls from the table to the floor. loss in potential energy = … J [2] (c) Explain why the kinetic energy of the ball in (b)(iii) does not equal the loss of gravitational potential energy in (b)(iv). … … [1] [Total: 13]

13 marks

Mark scheme: 2(a) rate of change of displacement or change in displacement/time taken B1 2(b)(i) s = ut + ½at2 C1 t = [(2 × 1.25) / 9.81]1/2 (= 0.5048 s) C1 or v2 = u2 + 2as vvert = (2 × 9.81 × 1.25)1/2 (= 4.95) (C1) t = [2s / (u + v)] = 2 × 1.25 / 4.95 (= 0.5048 s) (C1) v = d / t = 1.5 / 0.50(48) = 3.0 (2.97) m s–1 A1 2(b)(ii) vertical velocity = at = 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9] C1 velocity = [(vh)2 + (vv)2]1/2 C1 = [(2.97)2 + (4.95)2]1/2 = 5.8 (5.79) [using t = 0.50 leads to 5.7] A1 direction (= tan–1 4.95/2.97) = 59° A1 2(b)(iii) kinetic energy = ½mv2 C1 = ½ × 0.45 × (5.8)2 = 7.6 (7.57) J [using t = 0.50 leads to 7.3 J] A1 Question Answer Marks 2(b)(iv) potential energy = mgh C1 = (0.45 × 9.81 × 1.25) = 5.5 (5.52) J A1 2(c) there is KE of the ball at the start/leaving table or the ball has an initial/constant horizontal velocity or the ball has velocity at start/leaving table B1

This question in 9702/22 May/June 2017

7 A stationary nucleus X decays by emitting a β+ particle to form a nucleus of carbon-13 ( 136 C). An incomplete equation to represent this decay is X 136 C + β+. (a) State the name of the class (group) of particles that includes β+. … [1] (b) For nucleus X, state the number of protons, … neutrons. … [1] (c) The carbon-13 nucleus has a mass of 2.2 × 10–26 kg. Its kinetic energy as a result of the decay process is 0.80 MeV. Calculate the speed of this nucleus. speed = … m s–1 [3] (d) Explain why the sum of the kinetic energies of the carbon-13 nucleus and the β+ particle cannot be equal to the total energy released by the decay process. … … [1] [Total: 6]

6 marks

Mark scheme: 7(a) lepton(s) B1 7(b) protons: 7 and neutrons: 6 A1 7(c) E = ½mv2 C1 = 0.80 × 106 × 1.60 × 10–19 C1 = 1.28 × 10–13 (J) v2 = 2 × 1.28 × 10–13 / 2.2 × 10–26 v = 3.4 × 106 m s–1 A1 7(d) an (electron) neutrino/ν(e) is also produced (and this has energy) B1

This question in 9702/22 Oct/Nov 2017

2 (a) Explain what is meant by (i) work done, … … [1] (ii) kinetic energy. … … [1] (b) A leisure-park ride consists of a carriage that moves along a railed track. Part of the track lies in a vertical plane and follows an arc XY of a circle of radius 13 m, as shown in Fig. 2.1. 13 m Y 13 m carriage 22 m s–1 mass 580 kg track X Fig. 2.1 The mass of the carriage is 580 kg. At point X, the carriage has velocity 22 m s–1 in a horizontal direction. The velocity of the carriage then decreases to 12 m s–1 in a vertical direction at point Y. (i) For the carriage moving from X to Y 1. show that the decrease in kinetic energy is 9.9 × 104 J, [2] 2. calculate the gain in gravitational potential energy. gain in gravitational potential energy = … J [2] (ii) Show that the length of the track from X to Y is 20 m. [1] (iii) Use your answers in (b)(i) and (b)(ii) to calculate the average resistive force acting on the carriage as it moves from X to Y. resistive force = … N [2] (iv) Describe the change in the direction of the linear momentum of the carriage as it moves from X to Y. … … [1] (v) Determine the magnitude of the change in linear momentum when the carriage moves from X to Y. change in momentum = … N s [3] [Total: 13]

13 marks

Mark scheme: 2(a)(i) B1 2(a)(ii) energy (of a mass/body) due to motion / speed / velocity B1 2(b)(i) 1 E = ½mv 2 C1 (∆)E = ½ × 580 × (222 – 122) = 9.9 × 104 J A1 2 (∆)E = mg(∆)h ∆E = 580 × 9.81 × 13 C1 = 7.4 × 104 J A1 Question Answer Marks 2(b)(ii) length = (2π×13) / 4 or (π×26) / 4 or (π×13) / 2 = 20 m A1 2(b)(iii) work done against resistive force = 9.9 × 104 – 7.4 × 104 average resistive force = (9.9 × 104 – 7.4 × 104) / 20 C1 = 1300 N A1 2(b)(iv) from horizontal/right to vertical / up or 90° A1 2(b)(v) p = mv or (580 × 22) or (580 × 12) C1 ∆p = [ (580×12)2 + (580×22)2 ]0.5 C1 = 1.5 × 104 N s A1

This question in 9702/22 Feb/March 2018

7 A stationary nucleus X decays to form nucleus Y, as shown by the equation X Y + β– + ν. (a) In the above equation, draw a circle around all symbols that represent a lepton. [1] (b) State the name of the particle represented by the symbol ν. … [1] (c) Energy is released during the decay process. State the form of the energy that is gained by nucleus Y. … [1] (d) By comparing the compositions of X and Y, state and explain whether they are isotopes. … … … [2] (e) The quark composition of one nucleon in X is changed during the emission of a β– particle. Describe this change to the quark composition. … … [1] [Total: 6]

6 marks

Mark scheme: 7(a) B1 7(b) (electron) antineutrino B1 7(c) kinetic (energy) B1 7(d) Y has one more proton (and one less neutron)/X has one less proton (and one more neutron) or Y has more protons (and fewer neutrons)/X has fewer protons (and more neutrons) or a neutron changes to a proton or the number of protons increases M1 (so) not isotopes A1 7(e) up down down changes to up up down or udd → uud or down changes to up or d → u 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

3 (a) (i) Define power. … … [1] (ii) State what is meant by gravitational potential energy. … … [1] (b) An aircraft of mass 1200 kg climbs upwards with a constant velocity of 45 m s–1, as shown in Fig. 3.1. velocity thrust force 45 m s–1 2.0 × 103 N path of aircraft aircraft mass 1200 kg Fig. 3.1 (not to scale) The aircraft’s engine produces a thrust force of 2.0 × 103 N to move the aircraft through the air. The rate of increase in height of the aircraft is 3.3 m s–1. (i) Calculate the power produced by the thrust force. power = … W [2] (ii) Determine, for a time interval of 3.0 minutes, 1. the work done by the thrust force to move the aircraft, work done = … J [2] 2. the increase in gravitational potential energy of the aircraft, increase in gravitational potential energy = … J [2] 3. the work done against air resistance. work done = … J [1] (iii) Use your answer in (b)(ii) part 3 to calculate the force due to air resistance acting on the aircraft. force = … N [1] (iv) With reference to the motion of the aircraft, state and explain whether the aircraft is in equilibrium. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a)(i) work (done) / time (taken) B1 3(a)(ii) energy of a mass due to its position in a gravitational field B1 3(b)(i) P = Fv C1 = 2.0 × 103 × 45 = 9.0 × 104 W A1 3(b)(ii) 1. W = (2.0 × 103) × (45 × 3.0 × 60) or W = 9.0 × 104 × 3.0 × 60 C1 W = 1.6 × 107 J A1 2. (∆)EP = mg(∆)h C1 = 1200 × 9.81 × 3.3 × 3.0 × 60 = 7.0 × 106 J A1 3. W = 1.6 × 107 – 7.0 × 106 = 9.0 × 106 J A1 3(b)(iii) force = (9.0 × 106) / (45 × 3.0 × 60) = 1.1 × 103 N A1 3(b)(iv) constant velocity so no resultant force B1 no resultant force so in equilibrium 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 Two balls, X and Y, move along a horizontal frictionless surface, as illustrated in Fig. 3.1. 60° 3.0 m s–1 X A B 9.6 m s–1 Y 2.5 kg Fig. 3.1 (not to scale) Ball X has an initial velocity of 3.0 m s–1 in a direction along line AB. Ball Y has a mass of 2.5 kg and an initial velocity of 9.6 m s–1 in a direction at an angle of 60° to line AB. The two balls collide at point B. The balls stick together and then travel along the horizontal surface in a direction at right-angles to the line AB, as shown in Fig. 3.2. V X Y A B Fig. 3.2 (a) By considering the components of momentum in the direction from A to B, show that ball X has a mass of 4.0 kg. [2] (b) Calculate the common speed V of the two balls after the collision. V = … m s–1 [2] (c) Determine the difference between the initial kinetic energy of ball X and the initial kinetic energy of ball Y. difference in kinetic energy = … J [2] [Total: 6]

6 marks

Mark scheme: 3(a) C1 (m × 3.0) – (2.5 × 9.6 × cos 60°) = 0 so m = 4.0 (kg) A1 Question Answer Marks 3(b) 2.5 × 9.6 × sin60° = (4.0 + 2.5) × V C1 V = 3.2 m s–1 A1 or use of momentum vector triangle: (4.0 × 3.0)2 + [(4.0 + 2.5) × V]2 = (2.5 × 9.6)2 (C1) V = 3.2 m s–1 (A1) 3(c) E = ½mv 2 difference in EK = ½ × 2.5 × (9.6)2 – ½ × 4.0 × (3.0)2 C1 = 97 J A1

This question in 9702/22 Feb/March 2019

3 The variation with extension x of the force F acting on a spring is shown in Fig. 3.1. 5.0 4.0 F / N 3.0 2.0 1.0 0 0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 x / m Fig. 3.1 The spring of unstretched length 0.40 m has one end attached to a fixed point, as shown in Fig. 3.2. unstretched spring 0.40 m 0.72 m block moving block, downwards weight 2.5 N Fig. 3.2 Fig. 3.3 A block of weight 2.5 N is then attached to the spring. The block is then released and begins to move downwards. At one instant, as the block is continuing to move downwards, the spring has a length of 0.72 m, as shown in Fig. 3.3. Assume that the air resistance and the mass of the spring are both negligible. (a) For the change in length of the spring from 0.40 m to 0.72 m: (i) use Fig. 3.1 to show that the increase in elastic potential energy of the spring is 0.64 J [2] (ii) calculate the decrease in gravitational potential energy of the block of weight 2.5 N. decrease in potential energy = … J [2] (b) Use the information in (a)(i) and your answer in (a)(ii) to determine, for the instant when the length of the spring is 0.72 m: (i) the kinetic energy of the block kinetic energy = … J [1] (ii) the speed of the block. speed = … m s−1 [2] [Total: 7]

7 marks

Mark scheme: 3(a)(i) E = ½Fx or E = ½kx2 or E = area under graph C1 E = ½ × 4.0 × 0.32 = 0.64 J or E = ½ × 12.5 × (0.32)2 = 0.64 J A1 3(a)(ii) E = mgh or E = Wh = 2.5 × 0.32 C1 = 0.80 J A1 3(b)(i) kinetic energy = 0.80 – 0.64 = 0.16 J A1 3(b)(ii) E = ½mv 2 0.16 = ½ × (2.5 / 9.81) × v2 C1 v = 1.1 m s–1 A1

This question in 9702/21 May/June 2019

2 (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]

11 marks

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 leading to W = m(v 2 – u 2) / 2 (so KE = ½mv 2) B1 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

This question in 9702/23 May/June 2019

4 The variation with extension x of the force F applied to a spring is shown in Fig. 4.1. 4.0 3.0 F / N 2.0 1.0 0 0 0.010 0.020 0.030 0.040 0.050 x / m Fig. 4.1 The spring has an unstretched length of 0.080 m and is suspended vertically from a fixed point, as shown in Fig. 4.2. 0.080 m 0.095 m 0.120 m position X position Y block hangs in equilibrium block held before release Fig. 4.2 Fig. 4.3 Fig. 4.4 A block is attached to the lower end of the spring. The block hangs in equilibrium at position X when the length of the spring is 0.095 m, as shown in Fig. 4.3. The block is then pulled vertically downwards and held at position Y so that the length of the spring is 0.120 m, as shown in Fig. 4.4. The block is then released and moves vertically upwards from position Y back towards position X. (a) Use Fig. 4.1 to determine the spring constant of the spring. spring constant = … N m–1 [2] (b) Use Fig. 4.1 to show that the decrease in elastic potential energy of the spring is 0.055 J when the block moves from position Y to position X. [2] (c) The block has a mass of 0.122 kg. Calculate the increase in gravitational potential energy of the block for its movement from position Y to position X. increase in gravitational potential energy = … J [2] (d) Use the decrease in elastic potential energy stated in (b) and your answer in (c) to determine, for the block, as it moves through position X: (i) its kinetic energy kinetic energy = … J [1] (ii) its speed. speed = … m s–1 [2] [Total: 9]

9 marks

Mark scheme: 4(a) C1 e.g. k = 4.0 / 0.050 k = 80 N m–1 A1 4(b) E = ½Fx or E = ½kx2 or E = area under graph C1 (∆)E = (½ × 3.2 × 0.040) – (½ × 1.2 × 0.015) = 0.055 J or (∆)E = (½ × 80 × 0.0402) – (½ × 80 × 0.0152) = 0.055 J or (∆)E = ½ × (1.2 + 3.2) × 0.025 = 0.055 J A1 4(c) (∆)E = mg(∆)h C1 = 0.122 × 9.81 × (0.120 – 0.095) = 0.030 J A1 or (∆)E = W × (∆)h (C1) = 1.2 × 0.025 = 0.030 J (A1) Question Answer Marks 4(d)(i) E = 0.055 – 0.030 = 0.025 J A1 4(d)(ii) E = ½mv2 C1 v = [(2 × 0.025) / 0.122]0.5 = 0.64 m s–1 A1

This question in 9702/21 Oct/Nov 2019

3 (a) State two conditions for an object to be in equilibrium. 1. … … 2. … … [2] (b) A sphere of weight 2.4 N is suspended by a wire from a fixed point P. A horizontal string is used to hold the sphere in equilibrium with the wire at an angle of 53° to the horizontal, as shown in Fig. 3.1. P wire string T 53° horizontal F sphere weight 2.4 N Fig. 3.1 (not to scale) (i) Calculate: 1. the tension T in the wire T = … N 2. the force F exerted by the string on the sphere. F = … N [2] (ii) The wire has a circular cross-section of diameter 0.50 mm. Determine the stress σ in the wire. σ = … Pa [3] (c) The string is disconnected from the sphere in (b). The sphere then swings from its initial rest position A, as illustrated in Fig. 3.2. P 75 cm 53° A h B Fig. 3.2 (not to scale) The sphere reaches maximum speed when it is at the bottom of the swing at position B. The distance between P and the centre of the sphere is 75 cm. Air resistance is negligible and energy losses at P are negligible. (i) Show that the vertical distance h between A and B is 15 cm. [1] (ii) Calculate the change in gravitational potential energy of the sphere as it moves from A to B. change in gravitational potential energy = … J [2] (iii) Use your answer in (c)(ii) to determine the speed of the sphere at B. Show your working. speed = … m s–1 [3] [Total: 13]

13 marks

Mark scheme: 3(a) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) is zero B1 3(b)(i) 1. T sin 53° = 2.4 T = 3.0 N A1 2. F = T cos 53° or F 2 = T 2 – 2.42 F = 1.8 N A1 3(b)(ii) σ = T / A or σ = F / A C1 A = πd2 / 4 or A = πr2 C1 σ = 3.0 × 4 / [π × (0.50 × 10–3)2] = 1.5 × 107 Pa A1 3(c)(i) h = 75 – 75 sin 53° = 15 cm A1 3(c)(ii) (Δ)E = mg(Δ)h or (Δ)E = W(Δ)h C1 (Δ)E = 2.4 × 15 × 10–2 = 0.36 J A1 3(c)(iii) E = ½mv2 B1 0.36 = ½ × (2.4 / 9.81) × v2 C1 v = 1.7 m s–1 A1

This question in 9702/21 May/June 2020

3 (a) Explain what is meant by work done. … … [1] (b) A ball of mass 0.42 kg is dropped from the top of a building. The ball falls from rest through a vertical distance of 78 m to the ground. Air resistance is significant so that the ball reaches constant (terminal) velocity before hitting the ground. The ball hits the ground with a speed of 23 m s–1. (i) Calculate, for the ball falling from the top of the building to the ground: 1. the decrease in gravitational potential energy decrease in gravitational potential energy = … J [2] 2. the increase in kinetic energy. increase in kinetic energy = … J [2] (ii) Use your answers in (b)(i) to determine the average resistive force acting on the ball as it falls from the top of the building to the ground. average resistive force = … N [2] (c) The ball in (b) is dropped at time t = 0 and hits the ground at time t = T. The acceleration of free fall is g. On Fig. 3.1, sketch a line to show the variation of the acceleration a of the ball with time t from time t = 0 to t = T. g a 0 0 T t Fig. 3.1 [2] [Total: 9]

9 marks

Mark scheme: 3(a) (work done =) force × displacement in direction of the force B1 3(b)(i) 1. (Δ)E = mg(Δ)h C1 = 0.42 × 9.81 × 78 = 320 J A1 2. E = ½mv2 C1 (Δ)E = ½ × 0.42 × 232 = 110 J A1 3(b)(ii) work done = 320 – 110 (= 210 N) C1 average resistive force = 210 / 78 = 2.7 N A1 3(c) downward sloping line from (0, g) to a non-zero value on the time axis M1 line is curved with a gradient that becomes less negative and the line meets t-axis at time t < T A1

This question in 9702/22 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) A spring is fixed at one end and is compressed by applying a force to the other end. The variation of the force F acting on the spring with its compression x is shown in Fig. 3.1. F1 F 0 0 0.045 x / m Fig. 3.1 A compression of 0.045 m is produced when a force F1 acts on the spring. The spring has a spring constant of 800 N m−1. (i) Determine F1. F1 = … N [2] (ii) Use Fig. 3.1 to show that, for a compression of 0.045 m, the elastic potential energy of the spring is 0.81 J. [2] (b) A child’s toy uses the spring in (a) to launch a ball of mass 0.020 kg vertically into the air. The ball is initially held against one end of the spring which has a compression of 0.045 m. The spring is then released to launch the ball. The kinetic energy of the ball as it leaves the toy is 0.72 J. (i) The toy converts the elastic potential energy of the spring into the kinetic energy of the ball. Use the information in (a)(ii) to calculate the percentage efficiency of this conversion. efficiency = … % [1] (ii) Determine the initial momentum of the ball as it leaves the toy. momentum = … N s [3] (c) The ball in (b) leaves the toy at point A and moves vertically upwards through the air. Point B is the position of the ball when it is at maximum height h above point A, as illustrated in Fig. 3.2. B ball reaches maximum height at point B h ball at point A kinetic energy 0.72 J A mass 0.020 kg Fig. 3.2 (not to scale) The gravitational potential energy of the ball increases by 0.60 J as it moves from A to B. (i) Calculate h. h = … m [2] (ii) Determine the average force due to air resistance acting on the ball for its movement from A to B. average force = … N [2] (iii) When there is air resistance, the ball takes time T to move from A to B. State and explain whether the time taken for the ball to move from A to its maximum height will be more than, less than or equal to time T if there is no air resistance. … … [1] [Total: 13]

13 marks

Mark scheme: 3(a)(i) F = kx C1 F1 = 800 × 0.045 = 36 N A1 3(a)(ii) (E =) ½kx2 or ½Fx or area under graph C1 ½ × 800 × (0.045)2 or ½ × 36 × 0.045 = 0.81 (J) A1 3(b)(i) efficiency = (0.72 / 0.81) × 100 = 89% A1 3(b)(ii) E = ½mv 2 C1 p = mv C1 0.72 = ½ × 0.020 × v 2 and p = 0.020 × v p = 0.17 N s A1 3(c)(i) (Δ)E = mg(Δ)h C1 h = 0.60 / (0.020 × 9.81) = 3.1 m A1 3(c)(ii) F = (0.72 – 0.60) / 3.1 C1 = 0.039 N A1 3(c)(iii) resultant force on ball is less (than that with air resistance) so time (taken) is more (than T) B1

This question in 9702/22 Oct/Nov 2020

4 (a) State Hooke’s law. … … [1] (b) A spring is fixed at one end. A compressive force F is applied to the other end. The variation of the force F with the compression x of the spring is shown in Fig. 4.1. 8 F / N 6 4 2 0 0 4 8 12 16 x / cm Fig. 4.1 Show that the elastic potential energy of the spring is 0.64 J when its compression is 16.0 cm. [2] (c) The spring in (b) is used to project a toy car along a track from point X to point Y, as illustrated in Fig. 4.2. toy car mass 0.076 kg vertical loop compressed 0.12 m of track spring horizontal fixed track block X Y 0.30 m 0.25 m Fig. 4.2 (not to scale) The spring is initially given a compression of 16.0 cm. The car of mass 0.076 kg is held against one end of the compressed spring. When the spring is released it projects the car forward. The car leaves the spring at point X with kinetic energy that is equal to the initial elastic potential energy of the compressed spring. The car follows the track around a vertical loop of radius 0.12 m and then passes point Y. Assume that friction and air resistance are negligible. Calculate: (i) the speed of the car at X speed = … m s–1 [2] (ii) the kinetic energy of the car when it is at the top of the loop kinetic energy = … J [3] (iii) the speed of the car at Y. speed = … m s–1 [1] (d) In practice, a resistive force due to friction and air resistance acts on the car so that its kinetic energy at Y is 0.23 J less than its kinetic energy at X. Determine the average resistive force acting on the car for its movement from X to Y. average resistive force = … N [3] [Total: 12]

12 marks

Mark scheme: 4(a) compression/extension is proportional to force (provided limit of proportionality is not exceeded) B1 4(b) (E) = ½Fx or ½kx2 or area under graph C1 = ½ × 8 × 16 × 10–2 = 0.64 (J) or = ½ × 50 × (16 × 10–2)2 = 0.64 (J) A1 4(c)(i) (E) = ½mv 2 C1 0.64 = ½ × 0.076 × v 2 v = 4.1 m s–1 A1 4(c)(ii) (Δ)(E) = mg(Δ)h C1 = 0.076 × 9.81 × 0.24 (= 0.18 (J)) C1 kinetic energy = 0.64 – 0.18 = 0.46 J A1 4(c)(iii) v = 4.1 m s–1 A1 4(d) W = Fs C1 d = 0.30 + (2π × 0.12) + 0.25 (= 1.3 m) C1 F = 0.23 / 1.3 = 0.18 N A1

This question in 9702/23 Oct/Nov 2020

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

3 A pendulum consists of a solid sphere suspended by a string from a fixed point P, as shown in Fig. 3.1. P θ string 0.93 m Y sphere h X momentum 0.72 N s Fig. 3.1 (not to scale) The sphere swings from side to side. At one instant the sphere is at its lowest position X, where it has kinetic energy 0.86 J and momentum 0.72 N s in a horizontal direction. A short time later the sphere is at position Y, where it is momentarily stationary at a maximum vertical height h above position X. The string has a fixed length and negligible weight. Air resistance is also negligible. (a) On Fig. 3.1, draw a solid line to represent the displacement of the centre of the sphere at position Y from position X. [1] (b) Show that the mass of the sphere is 0.30 kg. [3] (c) Calculate height h. h = … m [2] (d) The distance between point P and the centre of the sphere is 0.93 m. When the sphere is at position Y, the string is at an angle θ to the vertical. Show that θ is 47°. [1] (e) For the sphere at position Y, calculate the moment of its weight about point P. moment = … N m [2] (f) State and explain whether the sphere is in equilibrium when it is stationary at position Y. … … [1] [Total: 10]

10 marks

Mark scheme: 3(a) solid straight line drawn between centre of sphere at X and at Y B1 3(b) p = mv or 0.72 = mv C1 E = ½mv 2 or 0.86 = ½mv 2 C1 (m =) 0.722 / (2 × 0.86) = 0.30 (kg) or v = 2EK / p v = (0.86 × 2) / 0.72 = 2.4 (to 2 s.f.) m = 0.72 / 2.4 = 0.30 (kg) A1 3(c) (Δ)E = mg(Δ)h C1 h = 0.86 / (0.30 × 9.81) = 0.29 m A1 3(d) cosθ = (0.93 – 0.29) / 0.93 so θ = 47° A1 3(e) moment = (0.30 × 9.81) × 0.93 × (sin 47° or cos 43°) or moment = (0.30 × 9.81) × [0.932 – (0.93 – 0.29)2]0.5 C1 = 2.0 N m A1 Question Answer Mark 3(f) there is a resultant force (acting on sphere) or there is a resultant moment (about P acting on pendulum) (so) not in equilibrium B1

This question in 9702/21 May/June 2021

3 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 Z velocity 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]

8 marks

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

This question in 9702/22 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

3 (a) Define power. … … [1] (b) A car of mass 1700 kg moves in a straight line along a slope that is at an angle θ to the horizontal, as shown in Fig. 3.1. B 25 m car, slope A θ mass 1700 kg horizontal Fig. 3.1 (not to scale) The car moves at constant velocity for a distance of 25 m from point A to point B. Air resistance and friction provide a total resistive force of 440 N that opposes the motion of the car. For the movement of the car from A to B: (i) state the change in the kinetic energy change in kinetic energy = … J [1] (ii) calculate the work done against the total resistive force. work done = … J [1] (c) The movement of the car in (b) from A to B causes its gravitational potential energy to increase by 4.8 × 104 J. Calculate: (i) the increase in vertical height h of the car for its movement from A to B h = … m [2] (ii) angle θ. θ = … ° [1] (d) The engine of the car in (b) produces an output power of 1.7 × 104 W to move the car along the slope. Calculate the time taken for the car to move from A to B. time = … s [2] [Total: 8]

8 marks

Mark scheme: 3(a) work (done) / time (taken) B1 3(b)(i) zero / 0 J A1 3(b)(ii) work done = 440 × 25 = 1.1 × 104 J A1 3(c)(i) (Δ)E(P) = mg(Δ)h C1 h = 4.8 × 104 / (1700 × 9.81) = 2.9 m A1 3(c)(ii) θ = sin–1 (2.9 / 25) = 6.7° A1 3(d) work done = 4.8 × 104 + 1.1 × 104 (= 5.9 × 104 J) C1 time = 5.9 × 104 / 1.7 × 104 = 3.5 s A1

This question in 9702/22 Oct/Nov 2021

7 A stationary nucleus P of mass 243 u decays by emitting an α-particle of mass 4 u to form a different nucleus Q, as illustrated in Fig. 7.1. v 1.6 × 107 m s–1 nucleus P nucleus Q α-particle mass 243 u mass 4 u BEFORE DECAY AFTER DECAY Fig. 7.1 The initial speed of the α-particle is 1.6 × 107 m s–1. (a) Use the principle of conservation of momentum to explain why the initial velocities of nucleus Q and the α-particle must be in opposite directions. … … … … [2] (b) Determine the initial speed v of nucleus Q. v = … m s–1 [2] (c) Calculate the initial kinetic energy, in MeV, of the α-particle. kinetic energy = … MeV [3] (d) A graph of number of neutrons N against proton number Z is shown in Fig. 7.2. 151 150 149 number of P 148 neutrons N 147 146 14592 93 94 95 96 97 98 proton number Z Fig. 7.2 The graph shows a cross that represents nucleus P. A nucleus R has a nucleon number of 242 and is an isotope of nucleus P. Nucleus R decays by emitting a β– particle to form a different nucleus S. (i) On Fig. 7.2, draw a cross to represent: 1. nucleus R (label this cross R) 2. nucleus S (label this cross S). [2] (ii) State the name of the other lepton, in addition to the β– particle, that is emitted during the decay of nucleus R. … [1] [Total: 10]

10 marks

Mark scheme: 7(a) (total) momentum before (decay) is zero or P has zero momentum B1 (total momentum after decay must be zero so) α-particle and Q have momenta in opposite directions (and therefore velocities are in opposite directions) B1 7(b) p = 239 (u) × v or 4 (u) × 1.6 × 107 C1 239 (u) × v = 4 (u) × 1.6 × 107 v = 2.7 × 105 m s–1 A1 7(c) E(K) = ½mv2 C1 = ½ × 4 × 1.66 × 10–27 × (1.6 × 107)2 C1 = 8.5 × 10–13 (J) = 8.5 × 10–13 / 1.60 × 10–13 (MeV) = 5.3 MeV A1 7(d)(i) 1. R plotted at (95,147) B1 2. S plotted at (96,146) B1 7(d)(ii) (electron) antineutrino B1

This question in 9702/22 Oct/Nov 2021

3 (a) Define velocity. … … [1] (b) A remote-controlled toy aircraft is flying horizontally in a wind. Fig. 3.1 shows the velocity vectors, to scale, of the wind and of the aircraft in still air. north wind velocity 54° 23 m s–1 aircraft velocity in still air 42 m s–1 Fig. 3.1 The velocity of the aircraft in still air is 42 m s–1 to the north. The velocity of the wind is 23 m s–1 in a direction of 54° east of south. Determine the magnitude of the resultant velocity of the aircraft. magnitude of velocity = … m s–1 [2] (c) The engine of the aircraft in (b) stops. The aircraft then glides towards the ground with a constant velocity at an angle θ to the horizontal, as illustrated in Fig. 3.2. X aircraft, 280 m weight 46 N glide path θ horizontal of aircraft Y Fig. 3.2 (not to scale) The aircraft has a weight of 46 N and travels a distance of 280 m from point X to point Y. The change in gravitational potential energy of the aircraft for its movement from X to Y is 6100 J. Assume that there is now no wind. (i) Calculate angle θ. θ = … ° [3] (ii) Calculate the magnitude of the force acting on the aircraft due to air resistance. force = … N [2] (d) The aircraft in (c) travels from X to Y in a time of 14 s. Fig. 3.3 shows that, as the aircraft travels from X to Y, it moves directly towards an observer who is standing on the ground. 280 m X aircraft Y observer ground Fig. 3.3 (not to scale) The aircraft emits sound as it travels from X to Y. The observer hears sound of frequency 450 Hz. The speed of the sound in the air is 340 m s–1. Calculate the frequency of the sound that is emitted by the aircraft. frequency = … Hz [3] [Total: 11]

11 marks

Mark scheme: 3(a) change in displacement / time (taken) B1 3(b) by calculation: v 2 = 422 + 232 – (2 × 42 × 23 × cos 54°) or v 2 = (42 – 23 cos 54°)2 + (23 sin 54°)2 or v 2 = (42 – 23 sin 36°)2 + (23 cos 36°)2 C1 v = 34 m s–1 A1 or by scale diagram: triangle of vector velocities drawn (C1) v = 34 m s–1 (allow ± 1 m s–1 if scale diagram used) (A1) 3(c)(i) (Δ)E = mg(Δ)h or (Δ)E = W(Δ)h C1 h = 6100 / 46 (= 133 m) C1 θ = sin–1 (133 / 280) = 28° A1 3(c)(ii) force = 6100 / 280 or 46 sin 28° C1 = 22 N A1 3(d) v(s) = 280 / 14 (= 20 m s–1) C1 fo = fs v / (v – vs) fs = 450 × (340 – 20) / 340 C1 = 420 Hz A1

This question in 9702/23 Oct/Nov 2021

4 A spring is suspended from a fixed point at one end and a vertical force is applied to the other end, as shown in Fig. 4.1. fixed point L spring F Fig. 4.1 The variation of the applied force F with the length L of the spring is shown in Fig. 4.2. 30 F / N 20 10 0 0 20 40 60 L / cm Fig. 4.2 (a) Determine the spring constant k of the spring. k = … N m–1 [2] (b) Determine the elastic potential energy in the spring when the applied force F is 15 N. elastic potential energy = … J [3] [Total: 5]

5 marks

Mark scheme: 4(a) k = F / ∆L or F / x or gradient C1 = e.g. 30 / (0.60 – 0.20) = 75 N m–1 A1 4(b) E =  1 2 F L or 1 2 Fx or    2 1 2 k L or 2 1 2 kx or area under graph C1 = 1 2  15  0.20 or 1 2  75  0.202 C1 = 1.5 J A1

This question in 9702/21 May/June 2022

2 A sphere is attached by a metal wire to the horizontal surface at the bottom of a river, as shown in Fig. 2.1. sphere direction of flow of water water wire horizontal 68° surface Fig. 2.1 (not to scale) The sphere is fully submerged and in equilibrium, with the wire at an angle of 68° to the horizontal surface. The weight of the sphere is 32 N. The upthrust acting on the sphere is 280 N. The density of the water is 1.0 × 103 kg m–3. Assume that the force on the sphere due to the water flow is in a horizontal direction. (a) By considering the components of force in the vertical direction, determine the tension in the wire. tension = … N [2] (b) For the sphere, calculate: (i) the volume volume = … m3 [1] (ii) the density. density = … kg m–3 [2] (c) The centre of the sphere is initially at a height of 6.2 m above the horizontal surface. The speed of the water then increases, causing the sphere to move to a different position. This movement of the sphere causes its gravitational potential energy to decrease by 77 J. Calculate the final height of the centre of the sphere above the horizontal surface. height = … m [3] (d) The extension of the wire increases when the sphere changes position as described in (c). The wire obeys Hooke’s law. (i) State a symbol equation that gives the relationship between the tension T in the wire and its extension x. Identify any other symbol that you use. … … [1] (ii) Before the sphere changed position, the initial elastic potential energy of the wire was 0.65 J. The change in position of the sphere causes the extension of the wire to double. Calculate the final elastic potential energy of the wire after the sphere has changed position. final elastic potential energy = … J [2] [Total: 11]

11 marks

Mark scheme: 2(a) T sin 68° + 32 = 280 C1 T = 270 N A1 2(b)(i) F = gV V = 280 / (1.0  103  9.81) = 0.029 m3 A1 2(b)(ii)  = (32 / 9.81) / 0.029 C1 = 110 kg m–3 A1 Question Answer Marks 2(c) (∆)E = mg(∆)h or (∆)E = W(∆)h C1 (∆)h = (–) 77 / 32 C1 (∆)h = (–) 2.4 final height = 6.2 – 2.4 = 3.8 m A1 2(d)(i) T = kx where k is a constant or T = (EA / L)x where A is (cross-sectional) area, E is Young modulus, L is (original/unstretched) length B1 2(d)(ii)  2 1 2 E kx or 1 2 E Fx and F = kx C1 E = 0.65  22 = 2.6 J A1 or 1 2 E Fx    1 0.65 270 2 x and so x = 4.8  10–3 m k = F / x = 270 / 4.8  10–3 = 5.6  104 (C1) Question Answer Marks 2(d)(ii) final      3 1 540 9.6 10 2 E or E =          3 1 270 2 4.8 10 2 2 or E =        2 4 3 1 5.6 10 9.6 10 2 = 2.6 J (A1)

This question in 9702/22 May/June 2022

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

9 marks

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

This question in 9702/23 May/June 2022

4 A horizontal spring is fixed at one end. A block is pushed against the other end of the spring so that the spring is compressed, as shown in Fig. 4.1. compressed spring block frictionless surface Fig. 4.1 The block is released and accelerates along a horizontal frictionless surface as the spring returns to its original length. The block leaves the end of the spring with a speed of 2.3 m s–1, as shown in Fig. 4.2. speed 2.3 m s–1 spring block leaving the spring Fig. 4.2 The block has a mass of 250 g and the spring has a spring constant of 420 N m–1. Assume that the spring always obeys Hooke’s law and that all the elastic potential energy of the spring is transferred to the kinetic energy of the block. (a) Calculate the kinetic energy of the block as it leaves the spring. kinetic energy = … J [2] (b) Calculate the compression of the spring immediately before the block is released. compression = … m [2] (c) After leaving the spring, the block moves along the surface until it hits a barrier at a speed of 2.3 m s–1. The block then rebounds at a speed of 1.5 m s–1 and moves back along its original path. The block is in contact with the barrier for a time of 0.086 s. Calculate: (i) the change in momentum of the block during the collision change in momentum = … N s [2] (ii) the average resultant force exerted on the block during the collision. average resultant force = … N [1] (d) The maximum compression x of the spring is now varied in order to vary the kinetic energy EK of the block as it leaves the spring. Assume that all the elastic potential energy in the spring is always transferred to the kinetic energy of the block. On Fig. 4.3, sketch a graph to show the variation with x of EK. EK 00 x Fig. 4.3 [1] [Total: 8]

8 marks

Mark scheme: 4(a) E = ½mv2 C1 = ½  0.25  2.32 A1 = 0.66 J 4(b) E = ½kx2 C1 or E = ½Fx and F = kx x = [(2  0.66) / 420]0.5 A1 = 0.056 m or E = ½kx2 (C1) ½mv2 = ½kx2 (A1) x = (0.25  2.32 / 420)0.5 = 0.056 m 4(c)(i) (p =) 0.25  2.3 or 0.25  1.5 C1 change in momentum = 0.25 (2.3 + 1.5) A1 = 0.95 N s 4(c)(ii) resultant force = 0.95 / 0.086 or 0.25  (2.3 + 1.5) / 0.086 A1 = 11 N 4(d) curved line from the origin with an increasing gradient B1

This question in 9702/22 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

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

12 marks

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

This question in 9702/22 May/June 2023

3 A vertical rod is fixed to the horizontal surface of a table, as shown in Fig. 3.1. rod surface of spring table Fig. 3.1 (not to scale) A spring of mass 7.5 g is able to slide along the full length of the rod. The spring is first pushed against the surface of the table so that it has an initial compression of 2.1 cm. The spring is then suddenly released so that it leaves the surface of the table with a kinetic energy of 0.048 J and then moves up the rod. Assume that the spring obeys Hooke’s law and that the initial elastic potential energy of the compressed spring is equal to the kinetic energy of the spring as it leaves the surface of the table. Air resistance is negligible. (a) By using the initial elastic potential energy of the compressed spring, calculate its spring constant. spring constant = … N m–1 [2] (b) Calculate the speed of the spring as it leaves the surface of the table. speed = … m s–1 [2] (c) The spring rises to its maximum height up the rod from the surface of the table. This causes the gravitational potential energy of the spring to increase by 0.039 J. (i) Calculate, for this movement of the spring, the increase in height of the spring after leaving the surface of the table. increase in height = … m [2] (ii) Calculate the average frictional force exerted by the rod on the spring as it rises. average frictional force = … N [2] (d) The rod is replaced by another rod that exerts negligible frictional force on the moving spring. The initial compression x of the spring is now varied in order to vary the maximum increase in height Δh of the spring after leaving the surface of the table. Assume that the spring obeys Hooke’s law for all compressions. On Fig. 3.2, sketch a graph to show the variation with x of Δh. Numerical values are not required. Δh 0 0 x Fig. 3.2 [2] [Total: 10]

10 marks

Mark scheme: 3(a) E(P) = ½kx2 C1 or E(P) = ½Fx and F = kx 0.048 = ½  k  (2.1  10–2)2 A1 k = 220 N m–1 3(b) E(K) = ½mv2 C1 0.048 = ½  7.5  10–3  v2 A1 v = 3.6 m s–1 3(c)(i) (∆)E = mg(∆)h C1 0.039 = 7.5  10–3  9.81  (∆)h A1 h = 0.53 m 3(c)(ii) F  0.53 = 0.048 – 0.039 C1 F = 0.02 N A1 3(d) sketch: curved line from the origin M1 curved line has increasing gradient A1

This question in 9702/22 Oct/Nov 2023

4 A pinball machine uses a spring to launch a small metal ball of mass 4.5 × 10 –2 kg up a ramp. The spring is compressed by 8.0 × 10 –2 m and held in equilibrium, as shown in Fig. 4.1. original length 8.0 × 10–2 m ramp spring fixed end ball, mass 4.5 × 10–2 kg 15° horizontal Fig. 4.1 (not to scale) The ramp is at an angle of 15° to the horizontal. (a) The spring obeys Hooke’s law and has a spring constant of 29 N m–1. Calculate the elastic potential energy in the compressed spring. elastic potential energy = … J [2] (b) The spring is released and expands quickly back to its original length. (i) Calculate the increase in gravitational potential energy of the ball when the spring returns to its original length. increase in gravitational potential energy = … J [3] (ii) The ball leaves the spring when the spring reaches its original length. Assume that all the elastic potential energy of the spring is transferred to the ball. Calculate the speed of the ball as it leaves the spring. speed = … m s–1 [3] (c) The ball comes to rest on a horizontal trapdoor of negligible mass at a distance d from its pivot. A force F acts vertically downwards at a distance of 2.0 cm from the pivot, as shown in Fig. 4.2. 2.0 cm d ball F pivot trapdoor Fig. 4.2 (not to scale) (i) The trapdoor is in equilibrium when F is 1.7 N. Calculate d. d = … m [2] (ii) Force F is decreased from 1.7 N. State the direction of the resultant moment about the pivot on the trapdoor. … [1] [Total: 11]

11 marks

Mark scheme: 4(a) E = ½kx2 or E= ½Fx and F = kx C1 E = ½  29  (8.0  10–2)2 or E = ½  2.32  8.0  10–2 E = 9.3  10–2 J A1 4(b)(i) ()E(P) = mg()h C1 = 4.5  10–2  9.81  8.0  10–2 sin 15° C1 = 9.1  10–3 J A1 4(b)(ii) E(K) = ½mv2 C1 (9.3  10–2 – 9.1  10–3) = ½  4.5  10–2  v2 C1 v = (2  8.4  10–2 / 4.5  10–2)0.5 = 1.9 m s–1 A1 4(c)(i) 1.7  2.0 ( 10–2) or 4.5  10–2  9.81  d C1 1.7  2.0  10–2 = 4.5  10–2  9.81  d d = 7.7  10–2 m A1 4(c)(ii) clockwise B1

This question in 9702/22 May/June 2024

3 (a) State what is meant by the work done by a force. … … [1] (b) A block of mass m is raised vertically at constant speed. The vertical height gained by the block is Δh, as shown in Fig. 3.1. block, mass m Δh Fig. 3.1 Derive an expression, in terms of m and Δh, for the change in gravitational potential energy ΔEP of the block. State the meaning of any other symbols you use. [2] (c) An electric motor has an input power of 900 W. The motor takes 1.0 minute to lift a load of weight 240 N at constant speed through a vertical height of 150 m. Resistive forces are negligible. (i) Show that the work done by the motor on the load in 1.0 minute is 36 kJ. [1] (ii) Determine the useful output power of the motor. power = … W [2] (iii) Use your answer in (c)(ii) to determine the efficiency of the motor. efficiency = … [2] (iv) Some of the power wasted in the motor is dissipated by the resistance of its coil. This dissipated power is 280 W. The coil of the motor is made from wire of total length 23 m. The wire has a cross- sectional area of 2.6 × 10–8 m2 and is made from metal of resistivity 1.7 × 10–8 Ω m. Calculate the current in the coil. current = … A [3] [Total: 11]

11 marks

Mark scheme: 3(a) product of force and displacement in direction of force B1 3(b) weight/force = mg, and g identified as acceleration of free fall B1 EP identified as work (done by lifting force/weight), and so = mg  h B1 3(c)(i) 240  150 = 36 kJ A1 3(c)(ii) P = W / t C1 = 36 000 / 60 A1 = 600 W 3(c)(iii) efficiency = useful output power / total input power C1 = 600 / 900 A1 = 0.67 3(c)(iv) P = I2R and R = L / A C1 280 = I2  (1.7  10–8  23) / (2.6  10–8) C1 I = 4.3 A A1

This question in 9702/21 Oct/Nov 2024

3 (a) A truck R of mass 9400 kg moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1. R A 180 m B Fig. 3.1 At point A the speed of the truck is 13 m s–1 and at point B the speed of the truck is 22 m s–1. A and B are a distance of 180 m apart. (i) Calculate the acceleration of the truck between A and B. acceleration = … m s–2 [2] (ii) Determine the gain in kinetic energy of the truck between A and B. gain in kinetic energy = … J [3] (b) A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes. Fig. 3.2 shows the variation with time of the momentum of the truck. 25 20 momentum / 104 kg m s–1 15 10 5 0 0 5 10 15 20 25 time / s Fig. 3.2 (i) Define force. … … [1] (ii) Show that the average resultant force F acting on truck R between time t = 0 and t = 15 s is –1.2 × 104 N. [1] (iii) An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force F in (b)(ii). State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R. … … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 3(a)(i) a = (v2 – u2) / 2s C1 = (222 – 132) / (2  180) = 0.88 m s–2 A1 OR (C1) [t = (180  2) / (22 + 13) = 10.3] 180 = 13  10.3 + ½ a  10.32 or 180 = 22  10.3 – ½ a  10.3 2 or 22 = 13 + a  10.3 a = 0.88 m s–2 (A1) 3(a)(ii) ()E = ½m()v2 C1 gain in KE = ½m(v2 – u2) C1 = ½  9400  (222 – 132) = 1.5  106 J A1 OR (C1) W = Fs = ma  d gain in KE = 9400  0.88  180 (C1) = 1.5  106 J (A1) 3(b)(i) rate of change of momentum B1 3(b)(ii) (Force =) (2.5  104 – 21  104) / 15 = – 1.2  104 (N) A1 3(b)(iii) The change of momentum (for S and R to come to rest) is the same B1 (Average) force on R (to come to rest) is B1 (–21  104 / 23 =) 0.91  104 N or (Average) force on R (to come to rest) is less than the force on S / less than F (S will come to rest in) less time (than R). B1 OR (B1) The change of momentum (for S and R to come to rest) is the same Time for S to come to rest is (–21  104 / 1.2  104 =) 17.5 s (and time for R to come to rest is 23 s) (B1) (S comes to rest in) less time (than R). (B1)

This question in 9702/22 Feb/March 2025

4 A small ball is dropped from rest from height h1 above the ground and falls vertically downwards. The ball collides with the ground and bounces back vertically upwards, reaching a maximum height h2. Fig. 4.1 shows the ball just before and just after hitting the ground. ball, mass 0.25 kg speed 3.6 m s–1 ground speed 5.2 m s–1 before hitting ground after hitting ground Fig. 4.1 The ball has mass 0.25 kg and is in contact with the ground for a time of 0.18 s. Just before the ball hits the ground, it has speed 5.2 m s–1. Just after it leaves the ground, it has speed 3.6 m s–1. Air resistance acting on the ball is negligible. (a) State and explain whether the collision is elastic or inelastic. … … … [1] (b) (i) Calculate the change in momentum of the ball during the collision with the ground. change in momentum = … kg m s–1 [2] (ii) Determine the average force on the ball during the collision with the ground. force = … N [2] h2 (c) Calculate the ratio . h1 ratio = … [3] [Total: 8]

8 marks

Mark scheme: 4(a) The (total) kinetic energy changes / decreases so (the collision is) inelastic B1 OR (B1) (relative) speed of approach not equal to / greater than (relative) speed of separation so (collision is) inelastic 4(b)(i) p = mv or 0.25  3.6 or 0.25  5.2 C1 p = 0.25  (3.6 + 5.2) = 2.2 kg m s–1 A1 4(b)(ii) F = p / ()t C1 = 2.2 / 0.18 = 12 N A1 OR (C1) F = ma and a = (v–u) / t = mv / t = 0.25  (3.6 + 5.2) / 0.18 = 12 N (A1) 4(c) ½ mv2 = mg()h C1 ½  0.25  5.22 = 0.25  g  h1 h1 = 5.22 / 2g h1 = 1.38 ½  0.25  3.62 = 0.25  g  h2 C1 h2 = 3.62/2g h2 = 0.66 h2 / h1 = 0.66 / 1.38 ratio = 0.48 A1

This question in 9702/22 May/June 2025

3 A car of mass 1500 kg is travelling along a straight horizontal road at constant velocity v. The car is subject to a total resistive force F, as shown in Fig. 3.1. v car, mass 1500 kg F horizontal road Fig. 3.1 (a) Show that the power P developed by the engine in overcoming the total resistive force is given by the equation P = Fv . [2] (b) The car now moves up a slope at a constant speed of 30 m s−1. The slope is at an angle to the horizontal of 6.0°, as shown in Fig. 3.2. 30 m s–1 car road 6.0° Fig. 3.2 The total resistive force acting on the car is 1600 N. (i) Show that the increase in gravitational potential energy of the car in a time of 1.0 s is 46 000 J. [2] (ii) Use the information in (b)(i) to determine the power developed by the engine to move the car up the slope. power = … W [2] (c) The car picks up a passenger and then continues up the slope at the same speed as in (b). State and explain the effect, if any, that the passenger has on: (i) the air resistance acting on the car … … [1] (ii) the power developed by the engine. … … [1] [Total: 8]

8 marks

Mark scheme: 3(a) W = Fd B1 P = Fd / t = Fv or P = Fvt / t = Fv B1 3(b)(i) ()E(P) = mg()h C1 increase of gravitational potential energy of car in 1.0 s A1 = 1500  9.81  30  sin 6.0 = 46 000 J 3(b)(ii) (Power to overcome total resistive forces) = 1600  30 C1 = 48 000 W power = 48 000 + 46 000 A1 = 9.4  104 W 3(c)(i) Air resistance is the same, as the speed is the same B1 3(c)(ii) Mass / weight has increased so (power will) increase B1

This question in 9702/23 May/June 2025