Energy conservation — Paper 2 · A Level Physics 9702

8 (a) Explain the concept of work. For Examiner’s … Use … … [2] (b) A table tennis ball falls vertically through air. Fig. 8.1 shows the variation of the kinetic energy EK of the ball with distance h fallen. The ball reaches the ground after falling through a distance h0. energy E K 0 0 h h 0 Fig. 8.1 (i) Describe the motion of the ball. … … … … … [3] (ii) On Fig. 8.1, draw a line to show the variation with h of the gravitational potential energy EP of the ball. At h = h0, the potential energy is zero. [3]

8 marks

Mark scheme: 8 (a) product of force and distance M1 moved in the direction of the force A1 [2] (b) (i) falls from rest B1 decreasing acceleration B1 reaches a constant speed B1 [3] (ii) straight line with negative gradient B1 y-axis intercept above maximum EK B1 reasonable gradient (same magnitude as that for EK initially) B1 [3]

This question in 9702/21 Oct/Nov 2005

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

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

4 Two parallel plates P and Q are separated by a distance of 7.6 mm in a vacuum. There is a For potential difference of 250 V between the plates, as illustrated in Fig. 4.1. Examiner’s Use 7.6 mm plate P plate Q X 250 V Fig. 4.1 Electrons are produced at X on plate P. These electrons accelerate from rest and travel to plate Q. The electric field between the plates may be assumed to be uniform. (a) (i) Determine the force on an electron due to the electric field. force = …………………….. N [3] (ii) Show that the change in kinetic energy of an electron as it moves from plate P to plate Q is 4.0 × 10–17 J. [2] (iii) Determine the speed of an electron as it reaches plate Q. For Examiner’s Use speed = … m s–1 [2] (b) The positions of the plates are adjusted so that the electric field between them is not uniform. The potential difference remains unchanged. State and explain the effect, if any, of this adjustment on the speed of an electron as it reaches plate Q. … … … … [3]

10 marks

Mark scheme: 4 (a) (i) either force = e × (V / d) or E = V/d C1 = 1.6 × 10–19 × (250 / 7.6 × 10–3) C1 = 5.3 × 10–15 N A1 [3] (ii) either ∆EK = eV or ∆EK = Fd C1 = 1.6 × 10–19 × 250 = 5.3 × 10–15 × 7.6 × 10–3 M1 = 4.0 × 10–17 J A0 [2] (allow full credit for correct working via calculation of a and v) GCE A/AS LEVEL – October/November 2008 9702 02 (iii) either ∆EK = ½mv2 4.0 × 10–17 = ½ × 9.1 × 10–31 × v2 C1 v = 9.4 × 106 m s–1 A1 [2] or v2 = 2as and a = F/m v2 = (2 × 5.3 × 10–15 × 7.6 × 10–3)/(9.11 × 10–31) (C1) v = 9.4 × 106 m s–1 (A1) (b) speed depends on (electric) potential difference M2 (If states ∆EK does not depend on uniformity of field, then award 1 mark, treated as an M mark) so speed always the same A1 [3]

This question in 9702/21 Oct/Nov 2008

2 A ball B of mass 1.2 kg travelling at constant velocity collides head-on with a stationary ball S For of mass 3.6 kg, as shown in Fig. 2.1. Examiner’s Use v ball B ball S mass 1.2 kg mass 3.6 kg Fig. 2.1 Frictional forces are negligible. The variation with time t of the velocity v of ball B before, during and after colliding with ball S is shown in Fig. 2.2. +4 +3 v / m s–1 +2 +1 0 0 0.1 0.2 0.3 0.4 0.5 t / s –1 –2 Fig. 2.2 (a) State the significance of positive and negative values for v in Fig. 2.2. … … [1] (b) Use Fig. 2.2 to determine, for ball B during the collision with ball S, For Examiner’s (i) the change in momentum of ball B, Use change in momentum = … N s [3] (ii) the magnitude of the force acting on ball B. force = … N [3] (c) Calculate the speed of ball S after the collision. speed = … m s–1 [2] (d) Using your answer in (c) and information from Fig. 2.2, deduce quantitatively whether For the collision is elastic or inelastic. Examiner’s Use … … [2]

11 marks

Mark scheme: 2 (a) ball moving in opposite direction (after collision) … B1 [1] (b) (i) change in momentum = 1.2 (4.0 + 0.8) … C2 (correct values, 1 mark; correct sign {values added}, 1 mark ) = 5.76 N s …(allow 5.8) … A1 [3] (ii) force = ∆p / ∆t or m∆v / ∆t … C1 = 5.76 / 0.08 or 1.2 × 4.8 / 0.08 … C1 = 72 N … A1 [3] (c) 5.76 = 3.6 × V … C1 V = 1.6 m s-1 … A1 [2] (d) either speed of approach = 4.0 m s-1 and speed of separation = 2.4 m s-1 … M1 not equal and so inelastic … A1 or kinetic energy before = 9.6 J and kinetic energy after collision = 4.99 J … M1 kinetic energy after is less / not conserved so inelastic … A1 [2]

This question in 9702/21 May/June 2009

2 (a) Explain what is meant by work done. For Examiner’s … Use … [1] (b) A car is travelling along a road that has a uniform downhill gradient, as shown in Fig. 2.1. 25 m s–1 7.5° Fig. 2.1 The car has a total mass of 850 kg. The angle of the road to the horizontal is 7.5°. Calculate the component of the weight of the car down the slope. component of weight = … N [2] (c) The car in (b) is travelling at a constant speed of 25 m s–1. The driver then applies the brakes to stop the car. The constant force resisting the motion of the car is 4600 N. (i) Show that the deceleration of the car with the brakes applied is 4.1 m s–2. [2] (ii) Calculate the distance the car travels from when the brakes are applied until the car comes to rest. distance = … m [2] (iii) Calculate For Examiner’s 1. the loss of kinetic energy of the car, Use loss of kinetic energy = … J [2] 2. the work done by the resisting force of 4600 N. work done = … J [1] (iv) The quantities in (iii) part 1 and in (iii) part 2 are not equal. Explain why these two quantities are not equal. … … [1]

11 marks

Mark scheme: 2 (a) work done is the force × the distance moved / displacement in the direction of the force or work is done when a force moves in the direction of the force B1 [1] (b) component of weight = 850 × 9.81 × sin 7.5° C1 = 1090 N A1 [2] (use of incorrect trigonometric function, 0/2) (c) (i) Σ F = 4600 – 1090 = (3510) M1 deceleration = 3510 / 850 A1 = 4.1 m s–2 A0 [2] (ii) v2 = u2 + 2as 0 = 252 + 2 × – 4.1 × s C1 s = 625 / 8.2 = 76 m A1 [2] (allow full credit for calculation of time (6.05 s) & then s) (iii) 1. kinetic energy = ½ mv2 C1 = 0.5 × 850 × 252 = 2.7 × 105 J A1 [2] 2. work done = 4600 × 75.7 = 3.5 × 105 J A1 [1] (iv) difference is the loss in potential energy (owtte) B1 [1] GCE AS/A LEVEL – May/June 2011 9702 21

This question in 9702/21 May/June 2011

3 (a) (i) Explain what is meant by work done. For Examiner’s … Use … [1] (ii) Define power. … … [1] (b) Fig. 3.1 shows part of a fairground ride with a carriage on rails. 4.1 m 9.5 m s–1 30° Fig. 3.1 The carriage and passengers have a total mass of 600 kg. The carriage is travelling at a speed of 9.5 m s–1 towards a slope inclined at 30° to the horizontal. The carriage comes to rest after travelling up the slope to a vertical height of 4.1 m. (i) Calculate the kinetic energy, in kJ, of the carriage and passengers as they travel towards the slope. kinetic energy = … kJ [3] (ii) Show that the gain in potential energy of the carriage and passengers is 24 kJ. [2] (iii) Calculate the work done against the resistive force as the carriage moves up the For slope. Examiner’s Use work done = … kJ [1] (iv) Use your answer in (iii) to calculate the resistive force acting against the carriage as it moves up the slope. resistive force = … N [2]

10 marks

Mark scheme: 3 (a) (i) work done equals force × distance moved / displacement in the direction of the force B1 [1] (ii) power is the rate of doing work / work done per unit time B1 [1] (b) (i) kinetic energy = ½ mv2 C1 = 0.5 × 600 (9.5)2 C1 = 27075 (J) = 27 kJ A1 [3] (ii) potential energy = mgh = 600 × 9.81 × 4.1 M1 = 24132 (J) A1 = 24 kJ A0 [2] (iii) work done = 27 – 24 = 3.0 kJ A1 [1] (iv) resistive force = 3000 / 8.2 (distance along slope = 4.1 / sin 30°) C1 = 366 N A1 [2]

This question in 9702/22 May/June 2011

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

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 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

3 (a) Explain what is meant by work done. … … [1] (b) A boy on a board B slides down a slope, as shown in Fig. 3.1. boy on board B 30° horizontal Fig. 3.1 The angle of the slope to the horizontal is 30°. The total resistive force F acting on B is constant. (i) State a word equation that links the work done by the force F on B to the changes in potential and kinetic energy. … … [1] (ii) The boy on the board B moves with velocity v down the slope. The variation with time t of v is shown in Fig. 3.2. 8.0 6.0 v / m s–1 4.0 2.0 0 0 1.0 2.0 3.0 t / s Fig. 3.2 The total mass of B is 75 kg. For B, from t = 0 to t = 2.5 s, 1. show that the distance moved down the slope is 9.3 m, [2] 2. calculate the gain in kinetic energy, gain in kinetic energy = … J [3] 3. calculate the loss in potential energy, loss in potential energy = … J [3] 4. calculate the resistive force F. F = … N [3]

13 marks

Mark scheme: 3 (a) work done is the product of force and the distance moved in the direction of the force or product of force and displacement in the direction of the force B1 [1] GCE AS/A LEVEL – May/June 2014 9702 23 (b) (i) work done equals the decrease in GPE – gain in KE B1 [1] (ii) 1. distance = area under line C1 = (7.4 × 2.5) / 2 = 9.3 m (9.25 m) M1 [2] or acceleration from graph a = 7.4 / 2.5 (= 2.96) (C1) and equation of motion (7.4)2 = 2 × 2.96 × s gives s = 9.3 (9.25) m (A1) 1 2 2. kinetic energy = m v C1 2 1 = × 75 × (7.4)2 C1 2 = 2100 J A1 [3] 3. potential energy = mgh C1 h = 9.3 sin 30 ° C1 PE = 75 × 9.81 × 9.3 sin 30 ° = 3400 J A1 [3] 4. work done = energy loss C1 R = (3421 – 2054) / 9.3 C1 = 150 (147) N A1 [3]

This question in 9702/23 May/June 2014

4 (a) State the principle of conservation of momentum. … … … [2] (b) A ball X and a ball Y are travelling along the same straight line in the same direction, as shown in Fig. 4.1. X Y 400 g 0.65 m s–1 600 g 0.45 m s–1 Fig. 4.1 Ball X has mass 400 g and horizontal velocity 0.65 m s–1. Ball Y has mass 600 g and horizontal velocity 0.45 m s–1. Ball X catches up and collides with ball Y. After the collision, X has horizontal velocity 0.41 m s–1 and Y has horizontal velocity v, as shown in Fig. 4.2. X Y 400 g 0.41 m s–1 600 g v Fig. 4.2 Calculate (i) the total initial momentum of the two balls, momentum = … N s [3] (ii) the velocity v, v = … m s–1 [2] (iii) the total initial kinetic energy of the two balls. kinetic energy = … J [3] (c) Explain how you would check whether the collision is elastic. … … [1] (d) Use Newton’s third law to explain why, during the collision, the change in momentum of X is equal and opposite to the change in momentum of Y. … … … … [2]

13 marks

Mark scheme: 4 (a) for a system (of interacting bodies) the total momentum remains constant M1 provided there is no resultant force acting (on the system) A1 [2] (b) (i) total momentum = m1v1 + m2v2 C1 = 0.4 × 0.65 + 0.6 × 0.45 C1 = 0.26 + 0.27 = 0.53 N s A1 [3] (ii) 0.53 = 0.4 × 0.41 + 0.6 × v C1 v = 0.366 / 0.6 = 0.61 m s–1 A1 [2] (iii) KE = ½ mv2 C1 total initial KE = ½ × 0.4 × (0.65)2 + ½ × 0.6 × (0.45)2 C1 = 0.0845 + 0.06075 = 0.15 (0.145) J A1 [3] (c) check relative speed of approach equals relative speed of separation or: total final kinetic energy equals the total initial kinetic energy B1 [1] (d) the forces on the two bodies (or on X and Y) are equal and opposite B1 time same for both forces and force is change in momentum / time B1 [2]

This question in 9702/23 Oct/Nov 2014

3 Two balls X and Y are supported by long strings, as shown in Fig. 3.1. X Y 4.5 m s–1 2.8 m s–1 Fig. 3.1 The balls are each pulled back and pushed towards each other. When the balls collide at the position shown in Fig. 3.1, the strings are vertical. The balls rebound in opposite directions. Fig. 3.2 shows data for X and Y during this collision. ball mass velocity just before velocity just after collision / m s–1 collision / m s–1 X 50 g +4.5 –1.8 Y M –2.8 +1.4 Fig. 3.2 The positive direction is horizontal and to the right. (a) Use the conservation of linear momentum to determine the mass M of Y. M = … g [3] (b) State and explain whether the collision is elastic. … … … [1] (c) Use Newton’s second and third laws to explain why the magnitude of the change in momentum of each ball is the same. … … … … … [3]

7 marks

Mark scheme: 3 (a) 4.5 × 50 – 2.8 × M ( = ...) C1 (...) = –1.8 × 50 + 1.4 × M C1 (M = ) 75 g A1 [3] (b) total initial kinetic energy/KE not equal to the total final kinetic energy/KE or relative speed of approach is not equal to relative speed of separation so not elastic or is inelastic B1 [1] (c) force on X is equal and opposite to force on Y (Newton III) M1 force equals/is proportional to rate of change of momentum (Newton II) M1 time of collision same for both balls hence change in momentum is the same A1 [3]

This question in 9702/21 May/June 2015

2 Fig. 2.1 shows an object M on a slope. 3.6 m s–1 Q M P R Fig. 2.1 M moves up the slope, comes to rest at point Q and then moves back down the slope to point R. M has a constant acceleration of 3.0 m s–2 down the slope at all times. At time t = 0, M is at point P and has a velocity of 3.6 m s–1 up the slope. The total distance from P to Q and then to R is 6.0 m. (a) Calculate, for the motion of M from P to Q, (i) the time taken, time = … s [2] (ii) the distance travelled. distance = … m [1] (b) Show that the speed of M at R is 4.8 m s–1. [2] (c) On Fig. 2.2, draw the variation with time t of the velocity v of M for the motion P to Q to R. 6.0 4.0 v / m s–1 2.0 0 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 t / s –2.0 –4.0 –6.0 Fig. 2.2 [3] (d) The mass of M is 450 g. Calculate the difference in the kinetic energy of M at P and at R. difference in kinetic energy = … J [2]

10 marks

Mark scheme: 2 (a) (i) v = u + at C1 0 = 3.6 – 3.0t t (= 3.6 / 3.0) = 1.2 s A1 [2] (ii) (distance to rest from P = (3.6 × 1.2) / 2 =) 2.2 (2.16) m A1 [1] or [0 – (3.6)2] / [2 × (–3.0)] = 2.2 (2.16) m or 3.6 × 1.2 – ½ × 3.0 × (1.2)2 = 2.2 (2.16) m or 0 + ½ × 3.0 × (1.2)2 = 2.2 (2.16) m (b) distance = 6.0 – 2.16 (= 3.84) C1 v2 = u2 + 2as = 2 × 3.0 × 3.84 (= 23.04) M1 or x + 2 × 2.16 = 6.0 gives x = 1.68 (m) (C1) v2 = 3.62 + 2 × 1.68 × 3.0 (= 23.04) (M1) or correct method with intermediate time calculated (t = 1.6 s from Q to R) v = 4.8 m s–1 A0 [2] (c) straight line from v = 3.6 m s–1 to v = 0 at t = 1.2 s B1 straight line continues with the same gradient as v changes sign B1 straight line from v = 0 intercept to v = –4.8 m s–1 B1 [3] (d) difference in KE = ½m(v2 – u2) = 0.5 × 0.45 (4.82 – 3.62) [= 5.184 – 2.916] C1 = 2.3 (2.27) J A1 [2]

This question in 9702/22 Oct/Nov 2015

3 A trolley T moves at speed 1.2 m s–1 along a horizontal frictionless surface. The trolley collides with a stationary block on the end of a fixed spring, as shown in Fig. 3.1. block 1.2 m s–1 fixed end of spring T horizontal frictionless surface Fig. 3.1 The mass of T is 250 g. T compresses the spring by 5.4 cm as it comes to rest. The relationship between the force F applied to the block and the compression x of the spring is shown in Fig. 3.2. 6.0 5.0 x / cm 4.0 3.0 2.0 1.0 0 0 2.0 4.0 6.0 F / N Fig. 3.2 (a) Use Fig. 3.2 to determine (i) the spring constant of the spring, spring constant = … N m–1 [2] (ii) the work done by T compressing the spring by 5.4 cm. work done = … J [2] (b) The spring then expands and causes T to move in a direction opposite to its initial direction. At the time that T loses contact with the block, it is moving at a speed of 0.75 m s–1. From the time that T is in contact with the block, (i) describe the energy changes, … … … … [2] (ii) determine the change in momentum of T. change in momentum = … N s [2]

8 marks

Mark scheme: 3 (a) (i) k = F / x or 1 / gradient C1 (k = 4.4 / (5.4 × 10–2) =) 81 (81.48) N m–1 A1 [2] (ii) work done = area under line or ½Fx or ½kx2 C1 (= 0.5 × 4.4 × 5.4 × 10–2 =) 0.12 (0.119) J A1 [2] (b) (i) kinetic energy/Ek of trolley/T (and block) changes to EPE/strain energy/elastic energy of spring B1 EPE changes to KE of trolley/T and KE of block or to give lower KE to trolley B1 [2] (ii) change in momentum = m(v + u) C1 = 0.25 (0.75 + 1.2) = 0.49 (0.488) N s A1 [2]

This question in 9702/22 Oct/Nov 2015

4 A block is pulled on a horizontal surface by a force P as shown in Fig. 4.1. vertical P = 35 N 60° block horizontal weight = 180 N Fig. 4.1 The weight of the block is 180 N. The force P is 35 N at 60° to the vertical. The block moves a distance of 20 m at constant velocity. (a) Calculate (i) the vertical force that the surface applies to the block (normal reaction force), force = … N [2] (ii) the work done by force P. work done = … J [2] (b) (i) Explain why the block continues to move at constant velocity although work is done on the block by force P. … … … [1] (ii) Explain, in terms of the forces acting, why the block remains in equilibrium. … … … … [2]

7 marks

Mark scheme: 4 (a) (i) reaction/vertical force = weight – P cos 60° C1 = 180 – 35 cos 60° = 160 (163) N A1 [2] (ii) work done = 35 sin 60° × 20 C1 = 610 (606) J A1 [2] (b) (i) work done by force P = work done against frictional force B1 [1] (ii) horizontal component of P is equal and opposite to frictional force B1 vertical component of P + normal reaction force equal and opposite to weight 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

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) 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

5 A solid cylinder is lifted out of oil by a wire attached to a motor. Fig. 5.1 shows two different positions X and Y of the cylinder during the lifting process. beam motor wire cylinder at position Y velocity surface of oil 0.020 m s–1 cylinder at position X oil Fig. 5.1 The motor is fixed to an overhead beam. The cylinder has cross-sectional area 0.018 m2, length 1.2 m and weight 560 N. The density of the oil is 940 kg m–3. Throughout the lifting process, the cylinder moves vertically upwards with a constant velocity of 0.020 m s–1. The viscous force of the oil acting on the cylinder is negligible. (a) Calculate the density of the cylinder. density = … kg m–3 [2] (b) For the cylinder at position X, show that the upthrust due to the oil is 200 N. [2] (c) Calculate, for the moving cylinder at position X, (i) the tension in the wire, tension = … N [1] (ii) the power output of the motor. power = … W [2] (d) The cylinder is raised with constant velocity from position X to position Y. (i) State and explain the variation, if any, of the power output of the motor as the cylinder is raised. Numerical values are not required. … … … … … [3] (ii) The rate of energy output of the motor is less than the rate of increase of gravitational potential energy of the cylinder. Without calculation, explain this difference. … … [1] [Total: 11]

11 marks

Mark scheme: 5(a) C1 = (560 / 9.81) / (1.2 × 0.018) = 2600 kg m–3 A1 5(b) (∆)p = 940 × 9.81 × 1.2 C1 (upthrust =) 940 × 9.81 × 1.2 × 0.018 = 200 N A1 5(c)(i) tension = 560 – 200 = 360 N A1 5(c)(ii) P = Fv C1 = 360 × 0.020 = 7.2 W A1 5(d)(i) upthrust decreases B1 tension (in wire) increases M1 power (output of motor) increases A1 5(d)(ii) there is work done (on the cylinder) by the upthrust or GPE of oil decreases (as it fills the space left by cylinder and so total energy is conserved) B1

This question in 9702/22 May/June 2018

2 (a) State what is meant by work done. … … [1] (b) A diver releases a solid sphere of radius 16 cm from the sea bed. The sphere moves vertically upwards towards the surface of the sea. The weight of the sphere is 20 N. The upthrust acting on the sphere is 170 N. The upthrust remains constant as the sphere moves upwards. (i) Calculate the density of the material of the sphere. density = … kg m–3 [2] (ii) Briefly explain the origin of the upthrust acting on the sphere. … … … [1] (iii) Calculate the acceleration of the sphere as it is released from rest. acceleration = … m s–2 [2] (iv) The viscous (drag) force D acting on the sphere is given by D = kr 2v 2 where r is the radius of the sphere and v is its speed. The constant k is equal to 810 kg m–3. Determine the constant (terminal) speed reached by the sphere. speed = … m s–1 [3] (v) The diver releases a different sphere that moves with a constant speed of 6.30 m s–1 directly towards a stationary ship. The sphere emits sound of frequency 4850 Hz. The ship detects sound of frequency 4870 Hz as the sphere moves towards it. Determine, to three significant figures, the speed of the sound in the water. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 2(a)(i) B1 2(b)(i) ρ = m / V C1 = (20 / 9.81) / (4/3 × π × 0.163) = 120 kg m–3 A1 2(b)(ii) the pressure on the lower surface (of sphere) is greater than the pressure on the upper surface (of sphere) B1 2(b)(iii) a = (170 – 20) / (20 / 9.81) C1 = 74 m s–2 A1 2(b)(iv) D = 170 – 20 (= 150) C1 810 × (0.162) × v2 = 150 C1 v = 2.7 m s–1 A1 2(b)(v) 4870 = (4850 × v) / (v – 6.30) C1 v = 1530 m s–1 A1

This question in 9702/23 May/June 2018

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

13 marks

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

This question in 9702/23 May/June 2018

2 (a) State what is meant by kinetic energy. … … [1] (b) A cannon fires a shell vertically upwards. The shell leaves the cannon with a speed of 80 m s–1 and a kinetic energy of 480 J. The shell then rises to a maximum height of 210 m. The effect of air resistance is significant. (i) Show that the mass of the shell is 0.15 kg. [2] (ii) For the movement of the shell from the cannon to its maximum height, calculate 1. the gain in gravitational potential energy, gain in gravitational potential energy = … J [2] 2. the work done against air resistance. work done = … J [1] (iii) Determine the average force due to the air resistance acting on the shell as it moves from the cannon to its maximum height. force = … N [2] (iv) The shell leaves the cannon at time t = 0 and reaches maximum height at time t = T. On Fig. 2.1, sketch the variation with time t of the velocity v of the shell from time t = 0 to time t = T. Numerical values of v and t are not required. v 0 0 t T Fig. 2.1 [2] (v) The force due to the air resistance is a vector quantity. Compare the force due to the air resistance acting on the shell as it rises with the force due to the air resistance as it falls. … … … … … [2] [Total: 12]

12 marks

Mark scheme: 2(a) energy (of a mass/body/object) due to motion/speed/velocity B1 2(b)(i) E = ½mv2 C1 480 = ½ × m × 802 so m = 0.15 kg A1 2(b)(ii) 1. E = mgh or ∆E = mg∆h C1 = 0.15 × 9.81 × 210 = 310 J A1 2. work done = 480 – 310 = 170 J A1 2(b)(iii) work done = Fs C1 force = 170 / 210 = 0.81 N A1 2(b)(iv) curved line from positive value on v-axis to (T, 0) M1 magnitude of gradient decreases A1 2(b)(v) as shell rises force decreases and as shell falls force increases B1 as shell rises force is downward and as shell falls force is upward B1 or as shell rises the force decreases and is downward (B1) as shell falls the force increases and is upward (B1)

This question in 9702/23 Oct/Nov 2018

3 (a) State what is meant by the centre of gravity of a body. … … [1] (b) A uniform square sign with sides of length 0.68 m is fixed at its corner points A and B to a wall. The sign is also supported by a wire CD, as shown in Fig. 3.1. D wire 54 N 35° B C sign E wall 0.68 m W A 0.68 m Fig. 3.1 (not to scale) The sign has weight W and centre of gravity at point E. The sign is held in a vertical plane with side BC horizontal. The wire is at an angle of 35° to side BC. The tension in the wire is 54 N. The force exerted on the sign at B is only in the vertical direction. (i) Calculate the vertical component of the tension in the wire. vertical component of tension = … N [1] (ii) Explain why the force on the sign at B does not have a moment about point A. … … [1] (iii) By taking moments about point A, show that the weight W of the sign is 150 N. [2] (iv) Calculate the total vertical force exerted by the wall on the sign at points A and B. total vertical force = … N [1] (c) The sign in (b) is held together by nuts and bolts. One of the nuts falls vertically from rest through a distance of 4.8 m to the pavement below. The nut lands on the pavement with a speed of 9.2 m s−1. Determine, for the nut falling from the sign to the pavement, the ratio change in gravitational potential energy . final kinetic energy ratio = … [4] [Total: 10]

10 marks

Mark scheme: 3(a) the point where (all) the weight (of the body) is taken to act B1 3(b)(i) vertical component = 54 sin 35° = 31 N A1 3(b)(ii) the (line of action of the) force (at B) passes through (point) A or the (line of action of the) force (at B) has zero (perpendicular) distance from (point) A B1 3(b)(iii) 54 sin 35° × 0.68 or 54 cos 35° × 0.68 or W × 0.34 C1 54 sin 35° × 0.68 + 54 cos 35° × 0.68 = W × 0.34 so W = 150 (N) A1 3(b)(iv) total vertical force = 150 – 31 = 120 N A1 3(c) (∆)E = mg(∆)h C1 E = ½mv 2 C1 ratio = (m × 9.81 × 4.8) / (½ × m × 9.22) or (9.81 × 4.8) / (½ × 9.22) C1 = 1.1 A1

This question in 9702/22 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

2 (a) State what is meant by work done. … … [1] (b) A lift (elevator) of weight 13.0 kN is connected by a cable to a motor, as shown in Fig. 2.1. motor cable lift (elevator) weight 13.0 kN v Fig. 2.1 The lift is pulled up a vertical shaft by the cable. A constant frictional force of 2.0 kN acts on the lift when it is moving. The variation with time t of the speed v of the lift is shown in Fig. 2.2. 3.0 v / m s–1 2.0 1.0 0 0 1 2 3 4 5 6 7 8 t / s Fig. 2.2 (i) Use Fig. 2.2 to determine: 1. the acceleration of the lift between time t = 0 and t = 3.0 s acceleration = … m s–2 [2] 2. the work done by the motor to raise the lift between time t = 3.0 s and t = 6.0 s. work done = … J [2] (ii) The motor has an efficiency of 67%. The tension in the cable is 1.6 × 104 N at time t = 2.5 s. Determine the input power to the motor at this time. input power = … W [3] (iii) State and explain whether the increase in gravitational potential energy of the lift from time t = 0 to t = 7.0 s is less than, the same as, or greater than the work done by the motor. A calculation is not required. … … [1] [Total: 9]

9 marks

Mark scheme: 2(a) B1 2(b)(i) 1. acceleration = gradient or a = (v – u) / t or a = ∆v / t C1 e.g. a = 2.4 / 3.0 = 0.80 m s–2 A1 2. tension in cable = (13.0 + 2.0) × 103 C1 work done = 15 × 103 × (3.0 × 2.4) = 1.1 × 105 J A1 2(b)(ii) power = Fv C1 v = 2.0 (m s–1) C1 input power = (1.6 × 104 × 2.0) / 0.67 = 4.8 × 104 W A1 2(b)(iii) work is done against friction so (increase in) GPE is less (than work done by motor) or energy is lost or transferred or converted to heat/thermal energy due to friction or resistance force or work is done lifting the cable so GPE is less A1

This question in 9702/23 Oct/Nov 2019

3 (a) State what is meant by work done. … … … [1] (b) A skier is pulled along horizontal ground by a wire attached to a kite, as shown in Fig. 3.1. wire kite speed 4.4 m s–1 140 N skier 30° ground horizontal Fig. 3.1 (not to scale) The skier moves in a straight line along the ground with a constant speed of 4.4 m s–1. The wire is at an angle of 30° to the horizontal. The tension in the wire is 140 N. (i) Calculate the work done by the tension to move the skier for a time of 30 s. work done = … J [3] (ii) The weight of the skier is 860 N. The vertical component of the tension in the wire and the weight of the skier combine so that the skier exerts a downward pressure on the ground of 2400 Pa. Determine the total area of the skis in contact with the ground. area = … m2 [3] (iii) The wire attached to the kite is uniform. The stress in the wire is 9.6 × 106 Pa. Calculate the diameter of the wire. diameter = … m [2] (c) The variation with extension x of the tension F in the wire in (b) is shown in Fig. 3.2. 300 F / N 250 200 150 100 50 0 0 0.20 0.40 0.60 0.80 x / mm Fig. 3.2 A gust of wind increases the tension in the wire from 140 N to 210 N. Calculate the change in the strain energy stored in the wire. change in strain energy = … J [3] [Total: 12]

12 marks

Mark scheme: 3(a) force × displacement in the direction of the force B1 3(b)(i) displacement = 4.4 × 30 C1 work done = 140 cos 30° × 4.4 × 30 C1 = 1.6 × 104 J A1 3(b)(ii) p = F / A C1 F = 860 – 140 sin 30° (= 790) C1 A = 790 / 2400 = 0.33 m2 A1 3(b)(iii) σ = F / A or F / πr 2 or 4F / πd 2 C1 9.6 × 106 = 4 × 140 / πd 2 d = 4.3 × 10–3 m A1 Question Answer Marks 3(c) E = ½Fx or ½kx2 or area under graph C1 (Δ)E = ½ × (140 + 210) × 0.20 × 10–3 or (Δ)E = (½ × 210 × 0.60 × 10–3) – (½ × 140 × 0.40 × 10–3) or (Δ)E = (140 × 0.20 × 10–3) + (½ × 0.20 × 10–3 × 70) or (Δ)E = [½×3.5 × 105 × (0.60 × 10–3)2 ] – [½ × 3.5 × 105 × (0.40 × 10–3)2 ] C1 ΔE = 0.035 J A1

This question in 9702/22 Feb/March 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

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

2 A person uses a trolley to move suitcases at an airport. The total mass of the trolley and suitcases is 72 kg. (a) The person pushes the trolley and suitcases along a horizontal surface with a constant speed of 1.4 m s–1 and then releases the trolley. The released trolley moves in a straight line and comes to rest. Assume that a constant total resistive force of 18 N opposes the motion of the trolley and suitcases. (i) Calculate the power required to overcome the total resistive force on the trolley and suitcases when they move with a constant speed of 1.4 m s–1. power = … W [2] (ii) Calculate the time taken for the trolley to come to rest after it is released. time = … s [3] (b) At another place in the airport, the trolley and suitcases are on a slope, as shown in Fig. 2.1. trolley and suitcases 18 N F, 54 N X slope 9.5 m Y Fig. 2.1 (not to scale) The person releases the trolley from rest at point X. The trolley moves down the slope in a straight line towards point Y. The distance along the slope between points X and Y is 9.5 m. The component F of the weight of the trolley and suitcases that acts along the slope is 54 N. Assume that a constant total resistive force of 18 N opposes the motion of the trolley and suitcases. (i) Calculate the speed of the trolley at point Y. speed = … m s–1 [3] (ii) Calculate the work done by F for the movement of the trolley from X to Y. work done = … J [1] (iii) The trolley is released at point X at time t = 0. On Fig. 2.2, sketch a graph to show the variation with time t of the work done by F for the movement of the trolley from X to Y. Numerical values of the work done and t are not required. work done 0 0 t Fig. 2.2 [2] (c) The angle of the slope in (b) is constant. The frictional forces acting on the wheels of the moving trolley are also constant. Explain why, in practice, it is incorrect to assume that the total resistive force opposing the motion of the trolley and suitcases is constant as the trolley moves between X and Y. … … [1] [Total: 12]

12 marks

Mark scheme: 2(a)(i) P = Fv C1 = 18 × 1.4 = 25 W A1 2(a)(ii) a = F / m C1 a = 18 / 72 = 0.25 (m s–2) t = 1.4 / 0.25 C1 = 5.6 s A1 2(b)(i) a = (54 – 18) / 72 or 36 / 72 (= 0.50 m s–2) C1 v 2 = 2 × 0.50 × 9.5 C1 v = 3.1 m s–1 A1 2(b)(ii) W = 54 × 9.5 = 510 J A1 2(b)(iii) curved line from the origin M1 gradient of line increases A1 2(c) (force due to) air resistance increases/changes/not constant or air resistance increases with speed 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) State what is meant by work done. … … [1] (b) Use your answer in (a) to show that the SI base units of energy are kg m2 s–2. [1] (c) A metal rod is heated at one end so that thermal energy flows to the other end. The thermal energy E that flows through the rod in time t is given by cA(T1 – T2)t E = L where A is the cross-sectional area of the rod, T1 and T2 are the temperatures of the ends of the rod, L is the length of the rod, and c is a constant. Determine the SI base units of c. SI base units … [3] [Total: 5]

5 marks

Mark scheme: 2(a) force × displacement in the direction of the force B1 2(b) units: kg m s–2 × m = kg m2 s–2 A1 2(c) T1: K and T2: K C1 A: m2 and t: s and L: m C1 c = (kg m2 s–2 m) / (m2 K s) = kg m s–3 K–1 A1

This question in 9702/23 Oct/Nov 2021

4 A child moves down a long slide, as shown in Fig. 4.1. X child surface of slide Y Fig. 4.1 (not to scale) The child moves from rest at the top end X of the slide. An average resistive force of 76 N opposes the motion of the child as they move to the lower end Y of the slide. The kinetic energy of the child at Y is 300 J. The decrease in gravitational potential energy of the child as it moves from X to Y is 3200 J. (a) Determine the ratio kinetic energy of the child at Y when the resistive force is 76 N . kinetic energy of the child at Y if there is no resistive force ratio = … [1] (b) Use the answer in (a) to calculate the ratio speed of the child at Y when the resistive force is 76 N . speed of the child at Y if there is no resistive force ratio = … [2] (c) Calculate the length of the slide from X to Y. length = … m [2] (d) At end Y of the slide, the child is brought to rest by a board, as shown in Fig. 4.2. child board spring surface of slide Fig. 4.2 (not to scale) A spring connects the board to a fixed point. The spring obeys Hooke’s law and has a spring constant of 63 N m–1. The child hits the board so that it moves to the right and compresses the spring. The speed of the child becomes zero when the elastic potential energy of the spring has increased to its maximum value of 140 J. (i) Calculate the maximum compression of the spring. maximum compression = … m [2] (ii) Calculate the percentage efficiency of the transfer of the kinetic energy of the child to the elastic potential energy of the spring. percentage efficiency = … % [1] (iii) The maximum compression of the spring is x0. On Fig. 4.3, sketch a graph to show the variation of the elastic potential energy of the spring with its compression x from x = 0 to x = x0. Numerical values are not required. elastic potential energy 00 x x0 Fig. 4.3 [2] [Total: 10]

10 marks

Mark scheme: 4(a) ratio = 300 / 3200 = 0.094 A1 4(b) E = ½mv 2 or E ∝ v 2 C1 ratio = (0.094)0.5 = 0.31 A1 4(c) work (done against frictional force) = 3200 – 300 (=2900) C1 length = 2900 / 76 = 38 m A1 4(d)(i) E = ½kx 2 or E = ½Fx and F = kx 140 = ½ × 63 × x 2 or 140 = ½Fx and F = 63x C1 x = 2.1 m A1 4(d)(ii) percentage efficiency = (140 / 300) × 100 = 47% A1 4(d)(iii) curved line from the origin M1 gradient of line increases A1

This question in 9702/22 Feb/March 2022

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

7 marks

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

This question in 9702/23 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

1 (a) State what is meant by work done. … … [1] (b) Use the answer to (a) to determine the SI base units of power. SI base units … [2] (c) The maximum useful output power P of a car travelling on a horizontal road is given by P = v 3 b where v is the maximum speed of the car and b is a constant. For the car, P = 84 kW ± 5% and b = 0.56 ± 7% in SI units. (i) Calculate the value of v. v = … m s–1 [2] (ii) Determine the absolute uncertainty in the value of v. absolute uncertainty = … m s–1 [2] [Total: 7]

7 marks

Mark scheme: Question Answer Marks 1(a) force  displacement in the direction of the force B1 1(b) P = Fs / t C1 = (kg m s–2  m) / s = kg m2 s–3 A1 1(c)(i) 84  103 = v 3  0.56 C1 v = 53 m s–1 A1 1(c)(ii) percentage uncertainty = (5% + 7%) / 3 (= 4%) C1 or fractional uncertainty = (0.05 + 0.07) / 3 (= 0.04) absolute uncertainty = 0.04  53 A1 = () 2 m s–1

This question in 9702/22 Oct/Nov 2022

3 (a) State the principle of moments. … … … [2] (b) A hollow plastic sphere is attached at one end of a bar. The sphere is partially submerged in water and the bar is attached to a fixed vertical support by a pivot P, as shown in Fig. 3.1. 0.29 m P sphere, bar weight 0.30 N fixed support 40° surface of water Fig. 3.1 (not to scale) The sphere has weight 0.30 N. The distance from P to the centre of gravity of the sphere is 0.29 m. Assume that the weight of the bar is negligible. Calculate the moment of the weight of the sphere about P. moment = … N m [2] (c) The system shown in Fig. 3.1 is part of a mechanism that controls the amount of water in a tank. Water enters the tank and causes the sphere to rise. This results in the bar becoming horizontal. Fig. 3.2 shows the system in its new position. 0.29 m R spring 0.017 m P submerged water portion of sphere Fig. 3.2 (not to scale) In this position the rod R exerts a force to compress a horizontal spring that controls the water supply to the tank. R is positioned at a perpendicular distance of 0.017 m above P. The variation of the force F applied to the spring with compression x of the spring is shown in Fig. 3.3. 25 F / N 20 15 10 5 0 0 2 4 6 8 10 x / mm Fig. 3.3 (i) Use Fig. 3.3 to calculate the spring constant k of the spring. k = … N m–1 [2] (ii) At the position shown in Fig. 3.2, the system is stationary and in equilibrium. The radius of the sphere is 0.0480 m and 26.0% of the volume of the sphere is submerged. The density of water is 1.00 × 103 kg m–3. Show that the upthrust on the sphere is 1.18 N. [2] (iii) By taking moments about P, determine the force exerted on the spring by the rod R. force = … N [2] (iv) Calculate the elastic potential energy EP of the compressed spring. EP = … J [2] (d) When the sphere moves from the position shown in Fig. 3.1 to the position shown in Fig. 3.2, the upthrust on the sphere does work. Assume that resistive forces are negligible. Explain why the work done by the upthrust is not equal to the gain in elastic potential energy of the spring. … … [1] [Total: 13]

13 marks

Mark scheme: 3(a) sum of CW moments = sum of ACW moments M1 about the same point for (an object in rotational) equilibrium A1 3(b) moment = 0.3(0)  0.29 cos 40° or 0.3(0)  0.222 C1 = 0.067 N m A1 3(c)(i) k = F / x or k = gradient C1 e.g. k = 21 / 10  10–3 A1 k = 2100 N m–1 3(c)(ii) 4 C1 V(sphere) =    (0.0480)3 3 F = gV A1 4 (upthrust =) 1000  9.81  (    (0.048)3)  0.26(0) = 1.18 (N) 3 3(c)(iii) 1.18  0.29 or 0.30  0.29 or F  0.017 C1 (1.18  0.29) = (0.30  0.29) + (F  0.017) A1 F = 15 N 3(c)(iv) E(P) = ½kx2 or E(P) = ½Fx C1 x = F / k = 15 / 2100 or x determined from graph for F = 15.0 N A1 EP = ½  2100  (15 / 2100)2 or EP = ½  15  (15 / 2100) EP = 0.054 J 3(d) the sphere has gained gravitational potential energy B1

This question in 9702/23 Oct/Nov 2022

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

13 marks

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

This question in 9702/22 Feb/March 2023

3 (a) State the principle of conservation of momentum. … … … [2] (b) A firework is initially stationary. It explodes into three fragments A, B and C that move in a horizontal plane, as shown in the view from above in Fig. 3.1. 6.0 m s–1 fragment B 2m fragment C 4.0 m s–1 3m m θ fragment A v Fig. 3.1 Fragment A has a mass of 3m and moves away from the explosion at a speed of 4.0 m s–1. Fragment B has a mass of 2m and moves away from the explosion at a speed of 6.0 m s−1 at right angles to the direction of A. Fragment C has a mass of m and moves away from the explosion at a speed v and at an angle θ as shown in Fig. 3.1. Calculate: (i) the angle θ θ = … ° [3] (ii) the speed v. v = … m s−1 [2] (c) The firework in (b) contains a chemical that has mass 5.0 g and has chemical energy per unit mass 700 J kg−1. When the firework explodes, all of the chemical energy is transferred to the kinetic energy of fragments A, B and C. (i) Show that the total chemical energy in the firework is 3.5 J. [1] (ii) Calculate the mass m. m = … kg [3] [Total: 11]

11 marks

Mark scheme: 3(a) sum / total momentum before = sum / total momentum after or sum / total momentum (of a system of objects) is constant M1 if no (resultant) external force / for an isolated system A1 3(b)(i) 3m  4 = m  v sin   (v sin  = 12) C1 2m  6 = m  v cos   (v cos = 12) C1 therefore sin = cos or tan = 1 = 45° A1 3(b)(ii) mv  cos 45° = 12m or mv  sin 45° = 12m or (mv)2 = (3m  4)2 + (2m  6)2 C1 v = 17 m s–1 A1 3(c)(i) (chemical energy) = 0.0050  700 = 3.5 (J) or (chemical energy) = 5.0  0.700 = 3.5 (J) A1 Question Answer Marks 3(c)(ii) E = ½mv2 C1 total E = (0.5  3m  42) + (0.5  2m  62) + (0.5  m  172) C1 3.5 = 204m m = 0.017 kg A1

This question in 9702/23 May/June 2023

4 Fig. 4.1 shows the variation with extension x of the tensile force F for two wires, G and H, made from the same material. 3.5 H 3.0 G F / N 2.5 2.0 1.5 1.0 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 x / mm Fig. 4.1 The elastic limit has not been exceeded for G or H. (a) For the lines in Fig. 4.1: (i) state what is represented by the gradient … [1] (ii) explain why the area under the line represents the elastic potential energy of the wire. … … … … [2] (b) Wires G and H are joined together end‑to‑end to form a composite wire of negligible weight. The composite wire hangs vertically from a fixed support. A block of weight of 2.0 N is attached to the end of the wire, as shown in Fig. 4.2. G fixed support H block, weight 2.0 N Fig. 4.2 (i) Use Fig. 4.1 to determine: ● the extension xG of wire G xG = … mm ● the extension xH of wire H. xH = … mm [1] (ii) Calculate the total elastic potential energy EP of the composite wire due to the weight of the block. EP = … J [2] (iii) The original length of wire G is L and the original length of wire H is 1.5 L. Calculate the ratio cross‑sectional area of wire G cross‑sectional area of wire H. ratio = … [3] [Total: 9]

9 marks

Mark scheme: 4(a)(i) spring constant B1 4(a)(ii) area represents the work done (to extend the wire) B1 work done (to extend the wire) is equal to elastic potential energy B1 4(b)(i) xG = 0.39 mm and xH = 0.29 mm A1 4(b)(ii) E = ½Fx C1 or E = ½kx2 and F = kx or E = area under graph for G: E = ½  2.0  (0.39  10–3) or ½  5.1  103  (0.39  10–3)2 A1 for H: E = ½  2.0  (0.29  10–3) or ½  6.9  103  (0.29  10–3)2 EP = 3.9  10–4 + 2.9  10–4 = 6.8  10–4 J 4(b)(iii) E = FL / Ax or stress / strain = FL / Ax C1 AG / AH = 1  (0.29  10–3) / [1.5  (0.39  10–3)] C1 ratio = 0.50 A1

This question in 9702/23 Oct/Nov 2023

2 (a) Define acceleration. … … [1] (b) An Olympic diver stands on a platform above a pool of water, as shown in Fig. 2.1. 5.9 m s–1 diver 60° horizontal platform 9.0 m surface of water 1.2 m Fig. 2.1 (not to scale) When the diver is on the platform his centre of gravity is a vertical height of 9.0 m above the surface of the water. The diver jumps from the platform with a velocity of 5.9 m s–1 at an angle of 60° to the horizontal. Air resistance is negligible. When the diver hits the surface of the water, his centre of gravity is a vertical height of 1.2 m above the surface of the water. Calculate the speed of the diver at the instant he hits the surface of the water. speed = … m s–1 [3] (c) The diver in (b) enters the water and decelerates. (i) Describe and explain the variation of the viscous drag force acting on the diver in the water as he moves downwards. … … … … [2] (ii) The diver has a volume of 7.5 × 10–2 m3 . The density of the water is 1.0 × 103 kg m–3 . Show that the upthrust acting on the diver when he is entirely underwater is 740 N. [1] (iii) At a particular instant when the diver is entirely underwater his horizontal velocity is zero. The viscous drag force acting on him at this instant is 950 N vertically upwards. The diver has mass 78 kg. Determine the magnitude and direction of the acceleration of the diver. acceleration = … m s–2 direction … [4] [Total: 11]

11 marks

Mark scheme: 2(a) rate of change of velocity B1 2(b) ½ m()v2= mg()h C1 v2 = 5.92 + 2  9.81  7.8 C1 v2 = 188 v = 14 m s–1 A1 or by resolving components (C1) Vertically: v2 = u2 + 2as v2 = (5.9sin60)2 +2  –9.81  (1.2–9.0) vv = 13.4 horizontally: (C1) vh = 5.9cos60 vh = 2.95 resultant velocity = √(13.42 + 2.952) (A1) = 14 m s–1 2(c)(i) (As the diver moves down their) speed decreases B1 (So) viscous force / drag (force) decreases B1 2(c)(ii) (F =) gV A1 = 1000  9.81  7.5  10–2 = 740 (N)

This question in 9702/22 Feb/March 2024

3 (a) State Hooke’s law. … … [1] (b) The variation of the applied force with the extension for a sample of a material is shown in Fig. 3.1. 10 force / N 8 X 6 4 2 0 0 40 80 120 160 200 extension / mm Fig. 3.1 The sample behaves elastically up to an extension of 80 mm and breaks at point X. (i) On the line in Fig. 3.1, draw a cross (×) to show the limit of proportionality. Label this cross with the letter P. [1] (ii) On the line in Fig. 3.1, draw a cross (×) to show the elastic limit. Label this cross with the letter E. [1] (c) The sample in (b) has a cross-sectional area of 0.40 mm2 and an initial length of 3.2 m. For deformations within the limit of proportionality of the sample, determine: (i) the spring constant of the sample spring constant = … N m–1 [2] (ii) the Young modulus of the material from which the sample is made. Young modulus = … Pa [3] (d) Determine an estimate of the work done on the sample as it is extended from zero extension to its breaking point. Explain your reasoning. work done = … J [2] (e) A second sample of the same material has a larger cross-sectional area than the original sample but the same initial length. The two samples are each deformed with the limit of proportionality. State and explain qualitatively how the spring constant of the second sample compares with that of the original sample. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a) extension is proportional to (applied) force B1 3(b)(i) P at (60, 5.4) A1 3(b)(ii) E at (80, 5.9) A1 3(c)(i) k = F / x or k = gradient of (straight line section of) graph C1 e.g. gradient = 5.4 / 0.060 k = 90 N m–1 A1 3(c)(ii) Young modulus or E =  /  or FL / Ax or kL / A C1 E = (5.4  3.2) / (4.0  10−7  0.06) or 90  3.2 / (4.0  10−7) C1 E = 7.2  108 Pa A1 3(d) work done = area under graph B1 = (1.0  0.2) J A1 3(e) the extension will be smaller (for the same force on the thicker sample) or a greater force is required (to extend the thicker sample by the same amount) or spring constant is proportional to area M1 the spring constant (of the second sample) will be greater A1

This question in 9702/23 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

2 (a) Define momentum. … … [1] (b) A child stands on a scooter on horizontal ground. The combined mass of the child and the scooter is 16 kg. The child starts from rest and pushes once on the ground with her foot which causes her to accelerate. The push lasts for a time of 1.1 s. The speed of the child and the scooter after the push is 0.60 m s–1. Determine the average resultant force acting horizontally on the child and the scooter during the push. average force = … N [2] (c) Later, the child in (b) travels down a slope at a constant angle to the horizontal, as shown in Fig. 2.1. A B x Fig. 2.1 (not to scale) At point A her speed is 0.60 m s–1. She has a constant acceleration of 0.85 m s–2 parallel to the slope. After a time of 3.7 s, she reaches point B. Calculate the distance x travelled by the child along the slope from A to B. x = … m [2] (d) At point B, the child in (c) applies the brake with a constant force to maintain a constant velocity. Point C is 18 m from point B, as shown in Fig. 2.2. 18 m A B C Fig. 2.2 (not to scale) The work done by the braking force between B and C is 250 J. (i) Determine the magnitude of the braking force. force = … N [2] (ii) On Fig. 2.3, sketch the variation of the kinetic energy of the child and scooter with distance travelled from point A to point C. Numerical values for kinetic energy are not required. kinetic energy 0 0 x x + 18 m distance from A Fig. 2.3 [3] [Total: 10]

10 marks

Mark scheme: 2(a) product of mass and velocity B1 2(b) F = mv / t C1 = 16  0.60 / 1.1 = 8.7 N A1 2(c) x = ut + ½at2 C1 x = 0.60  3.7 + ½  0.85  3.72 or v = 0.60 + 0.85  3.7 (= 3.75 m s–1) (C1) x = 3.75  3.7 – 0.5  0.85  3.72 or x = ½  (0.60 + 3.75)  3.7 or x = (3.752 – 0.602) / (2  0.85) x = 8.0 m A1 2(d)(i) F = W / s C1 = 250 / 18 A1 = 14 N 2(d)(ii) any line starting at distance = 0 and a positive non-zero value of kinetic energy B1 a straight line from distance = 0 to distance = x with positive gradient B1 a straight horizontal line at a non-zero value of kinetic energy starting at distance = x and ending at distance = x + 18 m that B1 is continuous with the previous line

This question in 9702/22 Oct/Nov 2024

1 (a) Table 1.1 lists some physical quantities. Identify with ticks (✓) which quantities are vectors and which are scalars. Table 1.1 quantity scalar vector acceleration displacement gravitational potential energy speed temperature [2] (b) A constant resultant force F acts on a car of mass m. The car moves from rest with constant acceleration a along horizontal ground. When the car has displacement s, the speed of the car is v. (i) Using the concept of work done on the car, show that the kinetic energy EK of the car is given by the equation 1 EK = mv2. 2 [3] (ii) The mass of the car is 920 kg. At time t = 0, the car is at rest. At time t = 5.8 s, its velocity is 17 m s–1. Calculate the kinetic energy of the car at time t = 5.8 s. kinetic energy = … J [1] (iii) Between time t = 0 and time t = 5.8 s, the work done against resistive forces is 4.7 × 104 J. Determine the average output power of the car during this time. power = … W [3] (iv) At time t = 5.8 s, the speed of the car becomes constant. State and explain whether the output power of the car is greater than, less than or the same as the output power just before t = 5.8 s. … … [1] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a) acceleration and displacement identified as vectors (and no others) B1 speed, temperature and gravitational potential energy identified as scalars (and no others) B1 1(b)(i) W = Fs or W = mas B1 s = v2 / 2a or a = v2 / 2s or as = v2 / 2 B1 W = ma(v2 / 2a) or W = m(v2 / 2s)s or W = m(v2 / 2) B1 and (so EK )= ½mv2 OR (B1) W = Fs or W = mas F = mv / t and s = ½vt (B1) W = mv / t  ½vt and (so EK )= ½mv 2 (B1) OR (B1) W = Fs or W = mas a = v / t and s = ½vt (B1) W = m(v / t)(½vt) and (so EK )= ½mv 2 (B1) OR (B1) W = Fs or W = mas a = v / t and s = ½at2 (B1) W = m(v / t)(½  (v / t)  t2) and (so EK )= ½mv 2 (B1) 1(b)(ii) kinetic energy = ½ mv2 A1 = ½  920  172 = 1.3  105 J 1(b)(iii) P = W / t C1 = (4.7  104 + 1.3  105) / 5.8 C1 = 3.1  104 W A1 1(b)(iv) (at/after t = 5.8 s) B1 the kinetic energy (of the car) does not change / work is done only against resistive forces / no work is done to accelerate (the car) so (power output is) less

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

1 (a) Define acceleration. … … [1] (b) A rocket is launched vertically from the surface of the Earth. Fig. 1.1 shows the variation of the velocity of the rocket with time for the first 20 s after its launch. 400 velocity / m s–1 200 0 0 5 10 15 20 time / s Fig. 1.1 (i) Determine the acceleration of the rocket. acceleration = … m s–2 [1] (ii) Show that the height of the rocket above the surface of the Earth at a time of 20 s after launch is 3.2 km. [2] (c) The mass of the rocket in (b) is 2.9 × 106 kg. Assume that this mass remains constant. For this rocket, from launch to its height at a time of 20 s after launch: (i) calculate the gain in gravitational potential energy ΔEP ΔEP = … J [2] (ii) calculate the gain in kinetic energy ΔEK ΔEK = … J [2] (iii) determine the average power output of the rocket engines. Assume that resistive forces are negligible. power = … W [2] [Total: 10]

10 marks

Mark scheme: Question Answer Marks 1(a) rate of change of velocity B1 1(b)(i) acceleration = 320 / 20 A1 = 16 m s–2 1(b)(ii) 1 C1 s = ((u) + v)t or distance = area under graph 2 1 A1 (height) =  20  320 = 3200 m or 3.2 km 2 1(c)(i) (EP) = mgh C1 = 2.9  106  9.81  3200 A1 = 9.1  1010 J 1(c)(ii) 1 C1 (EK) = m(v2) 2 1 A1 =  2.9  106  3202 2 = 1.5  1011 J 1(c)(iii) (power =) work done / time C1 power = ((1.5 + 0.91)  1011) / 20 A1 = 1.2  1010 W

This question in 9702/23 Oct/Nov 2025

3 A bungee jumper of mass 64 kg secures one end of an elastic rope to a bridge. The other end is attached to the bungee jumper. The jumper falls from rest from the bridge and descends into the valley below, as shown in Fig. 3.1. rope h jumper Fig. 3.1 (not to scale) Fig. 3.2 shows the variation of the tension T in the rope with the vertical distance h of the jumper below the level of the bridge. 2000 T / N 1500 1000 500 0 0 20 40 60 80 100 120 h / m Fig. 3.2 (a) The rope obeys Hooke’s law. State Hooke’s law. … … [1] (b) (i) Determine the unstretched length of the rope. length = … m [1] (ii) Determine the spring constant k of the rope. k = … N m–1 [2] (c) For the position of the bungee jumper at a distance of 120 m below the bridge: (i) show that the loss of gravitational potential energy since leaving the bridge is 75 kJ [2] (ii) show that the elastic potential energy in the rope is 75 kJ. [2] (d) Explain what can be deduced from the information in (c) about the speed of the bungee jumper when at a distance of 120 m below the bridge. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a) force is proportional to extension B1 3(b)(i) length = 37 m A1 3(b)(ii) spring constant = F / x C1 = e.g. 1800 / (120 – 37) A1 = 22 N m–1 3(c)(i) (()E) = mg()h C1 = 64 × 9.81 × 120 = 75 000 J or 75 kJ A1 3(c)(ii) 1 1 1 C1 (E =) Fx or (E =) kx2 or (E =) F2 / k or (E =) area under graph 2 2 2 1 A1 (E =) × 1800 × (120 – 37) = 75 000 J or 75 kJ 2 or 1 (E =) × 21.7 × (120 – 37)2 = 75 000 J or 75 kJ 2 3(d) (all) gravitational potential energy has been converted / equal to elastic potential energy (so no kinetic energy) B1 kinetic energy is zero so speed is zero B1

This question in 9702/24 Oct/Nov 2025