Equilibrium of forces — Paper 2 · A Level Physics 9702

2 (a) Define the moment of a force. … … [1] (b) A thin disc of radius r is supported at its centre O by a pin. The disc is supported so that it is vertical. Three forces act in the plane of the disc, as shown in Fig. 2.1. A 1.2 N r r 2 O θ C pin disc 6.0 N r 1.2 N B Fig. 2.1 Two horizontal and opposite forces, each of magnitude 1.2 N, act at points A and B on the edge of the disc. A force of 6.0 N, at an angle θ below the horizontal, acts on the midpoint C of a radial line of the disc, as shown in Fig. 2.1. The disc has negligible weight and is in equilibrium. (i) State an expression, in terms of r, for the torque of the couple due to the forces at A and B acting on the disc. … [1] (ii) Friction between the disc and the pin is negligible. Determine the angle θ. θ = … ° [2] (iii) State the magnitude of the force of the pin on the disc. force = … N [1] [Total: 5]

5 marks

Mark scheme: 2(a) B1 2(b)(i) 2.4r or (1.2 × 2r) or (1.2r + 1.2r) A1 2(b)(ii) (anticlockwise moment =) 6.0 × r / 2 × sinθ C1 6.0 × r / 2 × sinθ = 2.4r θ = 53° A1 2(b)(iii) 6.0 N A1

This question in 9702/22 Oct/Nov 2017

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

12 marks

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

This question in 9702/22 Oct/Nov 2017

2 (a) The kilogram, metre and second are all SI base units. State two other SI base units. 1. … 2. … [2] (b) A uniform beam AB of length 6.0 m is placed on a horizontal surface and then tilted at an angle of 31° to the horizontal, as shown in Fig. 2.1. 90 N A 6.0 m Y W X 31° B Fig. 2.1 (not to scale) The beam is held in equilibrium by four forces that all act in the same plane. A force of 90 N acts perpendicular to the beam at end A. The weight W of the beam acts at its centre of gravity. A vertical force Y and a horizontal force X both act at end B of the beam. (i) State the name of force X. … [1] (ii) By taking moments about end B, calculate the weight W of the beam. W = … N [2] (iii) Determine the magnitude of force X. magnitude of force X = … N [1] [Total: 6]

6 marks

Mark scheme: 2(a) ampere kelvin (allow mole, candela) any two correct answers, 1 mark each B2 2(b)(i) frictional (force)/friction B1 2(b)(ii) W cos 31° × 3.0 or 90 × 6.0 C1 W cos 31° × 3.0 = 90 × 6.0 W = 210 N A1 2(b)(iii) X = 90 sin 31° = 46 N A1

This question in 9702/22 Oct/Nov 2018

1 (a) Mass, length and time are all SI base quantities. State two other SI base quantities. 1. … 2. … [2] (b) A wire hangs between two fixed points, as shown in Fig. 1.1. fixed fixed horizontal point 17° 17° point 150 N 150 N wire hook rope tyre Fig. 1.1 (not to scale) A child’s swing is made by connecting a car tyre to the wire using a rope and a hook. The system is in equilibrium with the wire hanging at an angle of 17° to the horizontal. The tension in the wire is 150 N. Assume that the rope and hook have negligible weight. (i) Determine the weight of the tyre. weight = … N [2] (ii) The wire has a cross-sectional area of 7.5 mm2 and is made of metal of Young modulus 2.1 × 1011 Pa. The wire obeys Hooke’s law. Calculate, for the wire, 1. the stress, stress = … Pa [2] 2. the strain. strain = … [2] [Total: 8]

8 marks

Mark scheme: 1(a) current temperature (allow amount of substance, luminous intensity) any two correct answers, 1 mark each B2 1(b)(i) W = 2 × (150 × sin 17°) or 2 × (150 × cos 73°) C1 W = 88 N A1 1(b)(ii) 1. σ = F / A C1 = 150 / (7.5 × 10–6) = 2.0 × 107 Pa A1 2. ε = σ / E C1 = 2.0 × 107 / (2.1 × 1011) = 9.5 × 10–5 A1

This question in 9702/23 Oct/Nov 2018

2 A small charged glass bead of weight 5.4 × 10–5 N is initially at rest at point A in a vacuum. The bead then falls through a uniform horizontal electric field as it moves in a straight line to point B, as illustrated in Fig. 2.1. vertical glass bead weight 5.4 × 10–5 N A horizontal charge –3.7 × 10–9 C uniform horizontal path of the electric field, × 104 V m–1 falling bead field strength 1.3 B side view Fig. 2.1 (not to scale) The electric field strength is 1.3 × 104 V m–1. The charge on the bead is –3.7 × 10–9 C. (a) Describe how two metal plates could be used to produce the electric field. Numerical values are not required. … … … [2] (b) Determine the magnitude of the electric force acting on the bead. electric force = … N [2] (c) Use your answer in (b) and the weight of the bead to show that the resultant force acting on it is 7.2 × 10–5 N. [1] (d) Explain why the resultant force on the bead of 7.2 × 10–5 N is constant as the bead moves along path AB. … … … … [2] (e) (i) Calculate the magnitude of the acceleration of the bead along the path AB. acceleration = … m s–2 [2] (ii) The path AB has length 0.58 m. Use your answer in (i) to determine the speed of the bead at point B. speed = … m s–1 [2] [Total: 11]

11 marks

Mark scheme: 2(a) the (two) plates are vertical (and separated) B1 left plate positively charged and right plate negatively charged/earthed or right plate negatively charged and left plate positively charged/earthed B1 2(b) F = Eq C1 = 1.3 × 104 × 3.7 × 10–9 = 4.8 × 10–5 N A1 2(c) F2 = (4.8 × 10–5)2 + (5.4 × 10–5)2 so F = 7.2 × 10–5 N or F = [(4.8 × 10–5)2 + (5.4 × 10–5)2]0.5 so F = 7.2 × 10–5 N A1 2(d) electric force is constant (because field strength/E is constant) B1 weight is constant (and so resultant force constant) B1 2(e)(i) m = 5.4 × 10–5 / 9.81 (= 5.5 × 10–6) C1 a = 7.2 × 10–5 / (5.5 × 10–6) =13 m s–2 A1 2(e)(ii) v2 = u2 + 2as v2 = 2 × 13 × 0.58 C1 v = 3.9 m s–1 A1

This question in 9702/21 Oct/Nov 2019

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

11 marks

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

This question in 9702/21 Oct/Nov 2019

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

13 marks

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

This question in 9702/21 May/June 2020

3 (a) State the principle of moments. … … … [2] (b) In a bicycle shop, two wheels hang from a horizontal uniform rod AC, as shown in Fig. 3.1. ceiling cord 0.45 m 1.40 m 0.75 m 22 N wall A B C wheel wheel 19 N W W Fig. 3.1 (not to scale) The rod has weight 19 N and is freely hinged to a wall at end A. The other end C of the rod is attached by a vertical elastic cord to the ceiling. The centre of gravity of the rod is at point B. The weight of each wheel is W and the tension in the cord is 22 N. (i) By taking moments about end A, show that the weight W of each wheel is 14 N. [2] (ii) Determine the magnitude and the direction of the force acting on the rod at end A. magnitude = … N direction … [2] (c) The unstretched length of the cord in (b) is 0.25 m. The variation with length L of the tension F in the cord is shown in Fig. 3.2. 60 50 F / N 40 30 20 10 0 0 0.25 0.50 0.75 1.00 L / m Fig. 3.2 (i) State and explain whether Fig. 3.2 suggests that the cord obeys Hooke’s law. … … … [2] (ii) Calculate the spring constant k of the cord. k = … N m–1 [2] (iii) On Fig. 3.2, shade the area that represents the work done to extend the cord when the tension is increased from F = 0 to F = 40 N. [1] [Total: 11]

11 marks

Mark scheme: 3(a) for a body in (rotational) equilibrium B1 sum/total of clockwise moments about a point = sum/total of anticlockwise moments about the (same) point B1 3(b)(i) (W × 0.45) or (19 × 1.3) or (W × 1.85) or (22 × 2.6) C1 (W × 0.45) + (19 × 1.3) + (W × 1.85) = (22 × 2.6) so W = 14 N A1 3(b)(ii) magnitude = 19 + 14 + 14 – 22 = 25 N A1 direction: vertically upwards A1 3(c)(i) the extension is zero when the force is zero B1 graph is a straight line and (so) Hooke’s law obeyed B1 3(c)(ii) k = F / x or k = gradient C1 e.g. k = 60 / (1.00 – 0.25) k = 80 N m–1 A1 3(c)(iii) area shaded below graph line between L = 0.25 m and L = 0.75 m B1

This question in 9702/23 May/June 2020

2 (a) State what is meant by the centre of gravity of a body. … … … [2] (b) A uniform wooden post AB of weight 45 N stands in equilibrium on hard ground, as shown in Fig. 2.1. B T 0.30 m C horizontal 60° 0.90 m 38 N 45 N A ground Fig. 2.1 (not to scale) End A of the vertical post is supported by the ground. A horizontal wire with tension T is attached to end B of the post. Another wire, attached to the post at point C, is at an angle of 60° to the horizontal and has tension 38 N. The distances along the post of points A, B and C are shown in Fig. 2.1. (i) Calculate the horizontal component of the force exerted on the post by the wire connected to point C. horizontal component of force = … N [1] (ii) By considering moments about end A, determine the tension T. T = … N [2] (iii) Calculate the vertical component of the force exerted on the post at end A. force = … N [1] [Total: 6]

6 marks

Mark scheme: 2(a) point where (all) the weight (of the body) M1 is considered/seems to act A1 2(b)(i) horizontal component of force = 38 cos 60° or 38 sin 30° = 19 N A1 2(b)(ii) (T × 1.2) or (19 × 0.9) or 17 C1 (T × 1.2) = (19 × 0.9) T = 14 N A1 2(b)(iii) F = 45 + 38 sin 60° = 78 N A1

This question in 9702/23 Oct/Nov 2020

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

11 marks

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

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

3 A uniform beam AB is attached by a hinge to a wall at end A, as shown in Fig. 3.1. C 17 N 0.35 m 0.15 m string 50° horizontal A B hinge beam W 12 N Fig. 3.1 (not to scale) The beam has length 0.50 m and weight W. A block of weight 12 N rests on the beam at a distance of 0.15 m from end B. The beam is held horizontal and in equilibrium by a string attached between end B and a fixed point C. The string has a tension of 17 N and is at an angle of 50° to the horizontal. (a) State two conditions for an object to be in equilibrium. 1 … … 2 … … [2] (b) Show that the vertical component of the tension in the string is 13 N. [1] (c) By taking moments about end A, calculate the weight W of the beam. W = … N [2] (d) Calculate the magnitude of the vertical component of the force exerted on the beam by the hinge. force = … N [1] (e) The block is now moved closer to end A of the beam. Assume that the beam remains horizontal. State whether this change will increase, decrease or have no effect on the horizontal component of the force exerted on the beam by the hinge. … [1] [Total: 7]

7 marks

Mark scheme: 3(a) resultant force (in any direction) is zero B1 resultant moment/torque (about any point) is zero B1 3(b) (component =) 17sin50 = 13 (N) A1 or 17cos40 = 13 (N) 3(c) (W  0.25) or (12  0.35) or (13  0.50) C1 (W  0.25) + (12  0.35) = (13  0.50) A1 W = 9.2 N 3(d) F = 9.2 + 12 – 13 A1 = 8 N 3(e) decrease B1

This question in 9702/22 Feb/March 2023

2 A hot-air balloon floats just above the ground. The balloon is stationary and is held in place by a vertical rope, as shown in Fig. 2.1. balloon rope ground Fig. 2.1 The balloon has a weight W of 3.39 × 104 N. The tension T in the rope is 4.00 × 102 N. Upthrust U acts on the balloon. The density of the surrounding air is 1.23 kg m–3. (a) (i) On Fig. 2.1, draw labelled arrows to show the directions of the three forces acting on the balloon. [2] (ii) Calculate the volume, to three significant figures, of the balloon. volume = … m3 [3] (iii) The balloon is released from the rope. Calculate the initial acceleration of the balloon. acceleration = … m s–2 [3] (b) The balloon is stationary at a height of 500 m above the ground. A tennis ball is released from rest and falls vertically from the balloon. A passenger in the balloon uses the equation v2 = u2 + 2as to calculate that the ball will be travelling at a speed of approximately 100 m s–1 when it hits the ground. Explain why the actual speed of the ball will be much lower than 100 m s–1 when it hits the ground. … … … … [3] (c) Before the balloon is released, the rope holding the balloon has a strain of 2.4 × 10–5. The rope has an unstretched length of 2.5 m. The rope obeys Hooke’s law. (i) Show that the extension of the rope is 6.0 × 10–5 m. [1] (ii) Calculate the elastic potential energy EP of the rope. EP = … J [2] (iii) The rope holding the balloon is replaced with a new one of the same original length and cross-sectional area. The tension is unchanged and the new rope also obeys Hooke’s law. The new rope is made from a material of a lower Young modulus. State and explain the effect of the lower Young modulus on the elastic potential energy of the rope. … … … [2] [Total: 16]

16 marks

Mark scheme: 2(a)(i) arrow upwards () and labelled upthrust / U B2 arrow downwards () and labelled weight / W / mg arrow downwards () and labelled tension / T 1 mark: One or two correctly labelled arrows 2 marks: Three correctly labelled arrows 2(a)(ii) U = T + W or upthrust = tension + weight C1 Vg = T + W C1 V = [(4.00  102) + (3.39  104)] / (1.23  9.81) V = 2.84  103 m3 A1 2(a)(iii) m = W / g or a = F / m C1 a = (4.00  102) / [(3.39  104) / 9.81)] C1 a = 0.12 m s–2 A1 2(b) there is air resistance (which increases with speed) B1 (average) resultant force is less (than weight) B1 (average) acceleration is less (than g / 9.81, so speed is less than 100 m s–1) B1 2(c)(i) (extension =) 2.5  2.4  10–5 = 6.0  10–5 (m) A1 2(c)(ii) E(P) = ½ Fx C1 or E(P) = ½ kx2 and F = kx E(P) = ½  4.00  102  6.0  10–5 or E(P) = ½  6.7  106  (6.0  10–5)2 A1 E(P) = 0.012 J 2(c)(iii) longer extension M1 or smaller spring constant elastic potential energy is greater A1

This question in 9702/21 Oct/Nov 2023

1 The drag force FD acting on an object falling through air is given by 1 FD = CρAv 2 2 where A is the cross-sectional area of the object, v is the velocity of the object in the air, ρ is the density of the air and C is a constant called the drag coefficient. (a) Use SI base units to show that the drag coefficient has no units. [3] (b) Fig. 1.1 shows a sphere falling at terminal velocity in air. sphere terminal velocity Fig. 1.1 Assume that the upthrust on the sphere is negligible. On Fig. 1.1, draw and label arrows to show the directions of the two forces acting on the sphere. [2] (c) The mass of the sphere is 49 g. Calculate the drag force FD acting on the sphere. FD = … N [2] (d) The sphere is falling in air at a terminal velocity of 25 in SI base units. The density of the air is 1.2 in SI base units. The diameter of the sphere is 0.060 in SI base units. Use your answer in (c) to calculate the drag coefficient C for the sphere. C = … [3] [Total: 10]

10 marks

Mark scheme: 1(a) units of FD: kg m s–2 M1 units of  kg m–3 and units of A: m2 and units of v: m s−1 or units of v2: m2 s–2 M1 kg m s–2 = C kg m s–2 and comment ‘(so) C has no units’ / unit terms cancelled or C = kg m s−2 / (kg m–3 m2 m2 s–2) and comment ‘(so) C has no units’ / unit terms cancelled A1 1(b) one arrow vertically downward labelled weight to within 10° of the vertical B1 one arrow vertically upwards labelled drag / drag force / FD / air resistance / viscous force to within 10° of the vertical B1 1(c) (at terminal velocity) FD = mg C1 FD = 0.049  9.81 = 0.48 N A1 1(d) area =   (0.060 / 2)2 C1 0.48 = ½  C  1.2    (0.060 / 2)2  252 C1 C = 0.45 A1

This question in 9702/21 May/June 2024

1 The drag force FD acting on a sphere falling through a liquid is given by FD = 6πηr v where r is the radius of the sphere, v is the speed of the sphere in the liquid and η is a property of the liquid called the viscosity. (a) Show that the SI base units of viscosity are kg m–1 s–1. [2] (b) The sphere has a radius of 3.0 cm and is falling vertically downwards at a terminal velocity of 2.0 m s–1 through the liquid. The drag force acting on the sphere is 0.096 N. Calculate the viscosity of the liquid. viscosity = … kg m–1 s–1 [2] (c) The sphere is shown in Fig. 1.1. sphere liquid Fig. 1.1 On Fig. 1.1, draw and label arrows to represent the directions of the three forces acting on the sphere as it falls at terminal velocity through the liquid. [2] (d) (i) The density of the liquid is 920 kg m–3. Show that the upthrust acting on the sphere is 1.0 N. [2] (ii) Calculate the mass of the sphere. mass = … kg [2] [Total: 10]

10 marks

Mark scheme: 1(a) units of F: kg m s–2 C1 units of r: m and units of v: m s–1 units of : kg m s–2 / (m  m s–1) = kg m–1 s–1 A1 1(b) viscosity = 0.096 / (6    0.03  2.0) C1 = 0.085 kg m–1 s–1 A1 1(c) one arrow vertically downwards labelled weight / W B1 arrow(s) vertically upwards labelled U / upthrust and drag / FD/viscous force B1 1(d)(i) V = (4 / 3) r3 C1 upthrust = (4 / 3)    0.033  920  9.81 = 1.0 N A1 1(d)(ii) weight = 1.0 + 0.096 (= 1.096 N) C1 m = 1.096 / 9.81 = 0.11 kg A1

This question in 9702/23 May/June 2024

2 (a) Define the moment of a force about a pivot. … … [1] (b) Three objects A, B and C are placed on a horizontal beam. The beam is in equilibrium, as shown in Fig. 2.1. A B beam C pivot 3.0 m 9.0 m Fig. 2.1 (not to scale) The beam is uniform and has length 9.0 m. A pivot is at the midpoint of the beam. Object A has mass 90 kg and is at one end of the beam. Object B has mass m and is a distance of 3.0 m from the pivot. Object C has mass 150 kg and is at the other end of the beam. (i) Calculate m. m = … kg [3] (ii) Object A is removed and replaced by a wire fixed to the end of the beam and to the ground, as shown in Fig. 2.2. beam B C wire ground pivot Fig. 2.2 After the change, the beam is again horizontal and in equilibrium. The positions of B and C are unchanged. The wire has a diameter of 1.8 × 10−3 m and has a strain of 1.2 × 10−3. The wire is not extended beyond its limit of proportionality. Calculate the Young modulus of the wire. Young modulus = … Pa [3] (iii) Object B is now moved to a new position closer to the pivot without passing it. The beam is again horizontal and in equilibrium. State and explain the effect, if any, that this has on the strain in the wire. … … … [2] [Total: 9]

9 marks

Mark scheme: 2(a) force  perpendicular distance (of line of action of force to / from the point) B1 2(b)(i) 150  9.81  4.5 or 90  9.81  4.5 or 3.0  9.81  m C1 (150  9.81  4.5) = (90  9.81  4.5) + (3.0  9.81  m) C1 m = 90 kg A1 2(b)(ii) Young modulus = σ / ε or F / Aε or FL / Ax C1 Area of wire =   (1.8  10–3 / 2)2 C1 = 2.5  10–6 So Young modulus = ((90  9.81) /   (9.010–4)2 ) / 1.210–3 A1 = 2.9 1011 Pa 2(b)(iii) Moment provided by B will decrease / moment due to wire will increase B1 So force acting on wire will increase (Young’s modulus and area remain constant) and the strain will increase B1

This question in 9702/23 May/June 2025

3 A spring is fixed at one end and attached to the frame of a pulley at the other end. A cable is passed around the wheel of the pulley. The spring is stretched to a fixed length using the cable and pulley. Fig. 3.1 shows the view from above of the spring, cable and pulley. spring fixed end pulley frame pulley wheel cable Fig. 3.1 The spring obeys Hooke’s law and has a spring constant k of 250 N m–1. A force F acts on the spring. The tension in the cable is T. The pulley is in equilibrium. (a) On Fig. 3.2, draw labelled arrows to show the directions of the forces acting on the pulley. Fig. 3.2 [2] (b) The force F is 110 N. (i) Determine T. T = … N [1] (ii) Calculate the extension of the spring. extension = … m [2] (c) A second identical spring with the same spring constant of 250 N m–1 is now also connected to the pulley, as shown in Fig. 3.3. springs fixed end Fig. 3.3 The tension in the cable is kept the same. The pulley is again in equilibrium. (i) Determine the extension of the springs. extension = … m [2] (ii) The elastic potential energy stored in the spring in Fig. 3.1 is E1. The total elastic potential energy stored in the two springs in Fig. 3.3 is E2. E1 Calculate the ratio . E2 ratio = … [2] [Total: 9]

9 marks

Mark scheme: 3(a) an arrow horizontally on the page to the left labelled F B1 two arrows horizontally on the page to the right each labelled T B1 3(b)(i) T = 110 / 2 A1 = 55 N 3(b)(ii) x = F / k C1 = 110 / 250 A1 = 0.44 m 3(c)(i) extension = 55 / 250 or 110 / (2  250) C1 extension = 0.22 m A1 3(c)(ii) 1 1 1 C1 E = Fx or E = kx2 or E = F2 / k 2 2 2 1 1 A1 E1 / E2 = (  110  0.44) / (2  (  55  10.22)) 2 2 or 1 1 E1 / E2 = ( k  0.442) / (2  ( k  0.222)) 2 2 or 1 1 E1 / E2 = (  1102 / k) / (2  (  552 / k)) 2 2 E1 / E2 = 2.0 (no ECF from 3(b)(ii) and 3(c)(i))

This question in 9702/22 Oct/Nov 2025

2 Fig. 2.1 shows a square metal sheet of non-uniform density, with a thin wooden rod fixed at its centre. One of the corners of the sheet is labelled X. X metal sheet rod Fig. 2.1 The rod has negligible mass. The mass of the metal sheet is 2.8 kg. The rod is supported so that the rod is horizontal and the metal sheet is vertical. (a) Define the torque of a couple. … … … [2] (b) When the rod is supported in such a way that it can rotate freely within its support, the sheet hangs in equilibrium with point X vertically above the rod, as shown in Fig. 2.2. X rod metal sheet Fig. 2.2 On Fig. 2.2, draw a line to indicate the range of possible positions for the centre of gravity of the metal sheet. [1] (c) When a torque of 3.3 N m is applied to the rod, the sheet is held in equilibrium with two of its edges horizontal, as shown in Fig. 2.3. Point X is at the top-left corner. X metal sheet rod Fig. 2.3 (i) Explain whether the torque applied to the rod to hold the sheet in equilibrium is clockwise or anticlockwise. … … [1] (ii) Show that the centre of gravity of the sheet has a horizontal displacement of 0.12 m from the rod. [1] (d) The square metal sheet has an average density of 3000 kg m–3 and a uniform thickness of 4.0 mm. Show that the side length of the sheet is 0.48 m. [3] (e) Use the answer in (b) and the information in (c) and (d) to determine the position of the centre of gravity of the sheet. Indicate this position on Fig. 2.3 with a point labelled Y. [2] [Total: 10]

10 marks

Mark scheme: 2(a) product of force and distance M1 perpendicular distance between the (line of action of the two) forces A1 2(b) straight vertical line drawn from centre to bottom corner B1 2(c)(i) centre of gravity is to the right of the rod so torque is anticlockwise B1 or moment of weight of sheet about rod is clockwise so torque is anticlockwise 2(c)(ii) (horizontal displacement) = 3.3 / (2.8  9.81) = 0.12 (m) A1 2(d)  = m / V C1 V = 4.0  10–3 × (side length)2 C1 A1 0.48m side length =   2.8 / ( 3000  0.0040 )  = 2(e) point lies on a straight line at 45° to the horizontal from the bottom-right corner to the edge of the rod B1 point lies on a vertical line half-way between rod and right-hand edge B1

This question in 9702/24 Oct/Nov 2025