Energy conservation — Paper 2 · A Level Physics 9702

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

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

12 marks

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

This question in 9702/23 Oct/Nov 2020

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

10 marks

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

This question in 9702/22 Feb/March 2021

3 A 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

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

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

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