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

8 marks

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

This question in 9702/22 Oct/Nov 2019

1 (a) Determine the SI base units of the moment of a force. SI base units … [1] (b) A uniform square sheet of card ABCD is freely pivoted by a pin at a point P. The card is held in a vertical plane by an external force in the position shown in Fig. 1.1. B 17 cm 45° P A C 4.0 cm G 0.15 N D Fig. 1.1 (not to scale) The card has weight 0.15 N which may be considered to act at the centre of gravity G. Each side of the card has length 17 cm. Point P lies on the horizontal line AC and is 4.0 cm from corner A. Line BD is vertical. The card is released by removing the external force. The card then swings in a vertical plane until it comes to rest. (i) Calculate the magnitude of the resultant moment about point P acting on the card immediately after it is released. moment = … N m [2] (ii) Explain why, when the card has come to rest, its centre of gravity is vertically below point P. … … … … [2] [Total: 5]

5 marks

Mark scheme: 1(a) = kg m2 s–2 A1 1(b)(i) distance of COG from P (= GP) = 17 cos 45° – 4.0 or (144.5)½ – 4.0 (= 8.0 cm) C1 moment = 0.15 × 8.0 × 10–2 = 1.2 × 10–2 N m A1 1(b)(ii) (line of action of) weight acts through pivot/P or distance between (line of action of) weight and pivot/P is zero B1 (so) weight does not have a moment about pivot/P B1

This question in 9702/23 Oct/Nov 2019

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

1 (a) (i) Define the moment of a force about a point. … … [1] (ii) Determine the SI base units of the moment of a force. base units … [1] (b) A uniform rigid rod of length 2.4 m is shown in Fig. 1.1. 2.4 m cross-sectional area A Fig. 1.1 The rod has a weight of 5.2 N and is made of wood of density 790 kg m–3. Calculate the cross-sectional area A, in mm2, of the rod. A = … mm2 [3] (c) A fishing rod AB, made from the rod in (b), is shown in Fig. 1.2. 0.60 m B 0.60 m C T string D 1.20 m 4.6 N 56° stick weight 5.2 N A ground water Fig. 1.2 (not to scale) End A of the rod rests on the ground and a string is attached to the other end B. A support stick exerts a force perpendicular to the rod at point C. The weight of the rod acts at point D. The tension T in the string is in a direction perpendicular to the rod. The rod is in equilibrium and inclined at an angle of 56° to the vertical. The forces and the distances along the rod of points A, B, C and D are shown in Fig. 1.2. (i) Show that the component of the weight that is perpendicular to the rod is 4.3 N. [1] (ii) By taking moments about end A of the rod, calculate the tension T. T = … N [3] [Total: 9]

9 marks

Mark scheme: 1(a)(i) force × perpendicular distance (of line of action of force to the point) B1 1(a)(ii) units: kg m s–2 m = kg m2 s–2 A1 1(b) W = ρVg or W = ρALg C1 A = 5.2 / (790 × 2.4 × 9.81) (= 2.8 × 10–4 (m2)) C1 = 2.8 × 102 mm2 A1 1(c)(i) (component =) 5.2 sin 56° = 4.3 (N) or 5.2 cos 34° = 4.3 (N) A1 1(c)(ii) (T × 2.4) or (4.3 × 1.2) or (4.6 × 1.8) C1 (T × 2.4) + (4.3 × 1.2) = (4.6 × 1.8) C1 T = 1.3 N A1

This question in 9702/21 Oct/Nov 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

3 (a) Define the moment of a force about a point. … … … [2] (b) Fig. 3.1 shows a type of balance that is used for measuring mass. fixed point P mm scale 200 spring 52.6 cm pan 1.8 cm pointer rod pivot 0 6.2 cm Fig. 3.1 (not to scale) A rigid rod is pivoted about a point 6.2 cm from the centre of a pan which is attached to one end. The object being measured is placed on the centre of this pan. A spring, attached to the rod 1.8 cm from the pivot, is attached at its other end to a fixed point P. The spring obeys Hooke’s law over the full range of operation of the balance. A pointer, on the other side of the pivot, is set against a millimetre scale which is a distance 52.6 cm from the pivot. When the system is in equilibrium with no mass on the pan, the rod is horizontal and the pointer indicates a reading on the scale of 86 mm. An object of mass 0.472 kg is now placed on the pan. As a result, the pointer moves to indicate a reading of 123 mm on the scale when the system is again in equilibrium. (i) Show that the increase in the length of the spring is approximately 1.3 mm. [2] (ii) Calculate the magnitude of the moment about the pivot of the weight of the object. moment = … N m [2] (iii) Use your answer in (b)(ii) to determine the increase in the tension in the spring due to the 0.472 kg mass. increase in tension = … N [2] (iv) Use the information in (b)(i) and your answer in (b)(iii) to determine the spring constant k of the spring. Give a unit with your answer. k = … unit … [2] [Total: 10]

10 marks

Mark scheme: 3(a) force × distance M1 perpendicular distance of (line of action of) force from the point A1 3(b)(i) distance moved by pointer = 123 – 86 (= 37 mm) C1 (extension =) 37 × (1.8 / 52.6) = 1.3 (mm) or sin or tan θ = 37 / 526 (so θ = 4.0° so extension =) sin or tan θ × 18 = 1.3 (mm) A1 3(b)(ii) moment = 0.472 × 9.81 × 6.2 × 10–2 C1 = 0.29 N m A1 3(b)(iii) (Δ)F × 1.8 × 10–2 = 0.29 C1 ΔF = 16 N A1 3(b)(iv) k = F / x C1 = 16 / (1.3 × 10–3) = 1.2 × 104 N m–1 A1 Question Answer Marks

This question in 9702/23 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) State what is meant by the centre of gravity of an object. … … [1] (b) Two blocks are on a horizontal beam that is pivoted at its centre of gravity, as shown in Fig. 2.1. 0.45 m 0.95 m 0.35 m horizontal 30° pivot beam 54 N 2.4 N string support T ground Fig. 2.1 (not to scale) A large block of weight 54 N is a distance of 0.45 m from the pivot. A small block of weight 2.4 N is a distance of 0.95 m from the pivot and a distance of 0.35 m from the right‑hand end of the beam. The right‑hand end of the beam is connected to the ground by a string that is at an angle of 30° to the horizontal. The beam is in equilibrium. (i) By taking moments about the pivot, calculate the tension T in the string. T = … N [3] (ii) The string is cut so that the beam is no longer in equilibrium. Calculate the magnitude of the resultant moment about the pivot acting on the beam immediately after the string is cut. resultant moment = … N m [1] (c) The beam in (b) rotates when the string is cut and the small block of weight 2.4 N is projected through the air. Fig. 2.2 shows the last part of the path of the block before it hits the ground at point Y. path of X block 1.8 m horizontal ground Y Fig. 2.2 (not to scale) At point X on the path, the block has a speed of 3.4 m s–1 and is at a height of 1.8 m above the horizontal ground. Air resistance is negligible. (i) Calculate the decrease in the gravitational potential energy of the block for its movement from X to Y. decrease in gravitational potential energy = … J [2] (ii) Use your answer to (c)(i) and conservation of energy to determine the kinetic energy of the block at Y. kinetic energy = … J [3] (iii) State the variation, if any, in the direction of the acceleration of the block as it moves from X to Y. … [1] (iv) The block passes point X at time tX and arrives at point Y at time tY. On Fig. 2.3, sketch a graph to show the variation of the magnitude of the horizontal component of the velocity of the block with time from tX to tY. Numerical values are not required. horizontal component of velocity 0 tX tY time Fig. 2.3 [1] [Total: 12]

12 marks

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

This question in 9702/22 May/June 2023

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

11 marks

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

This question in 9702/22 May/June 2024

3 (a) State the principle of moments. … … … [1] (b) A rigid uniform beam rests on a pivot at its centre, as shown in Fig. 3.1. beam x 0.40 m wooden cylinder, weight 4.0 N pivot container h load, weight 2.6 N water Fig. 3.1 (not to scale) A load of weight 2.6 N is suspended from the beam at distance x from the pivot. A wooden cylinder of weight 4.0 N is suspended from the beam at a distance of 0.40 m from the pivot on the opposite side of the pivot to the load. The cylinder rests in a container of water. The lower part of the cylinder is immersed in the water to depth h. Initially, h is equal to 0.10 m and x is equal to 0.40 m. The system is in equilibrium. (i) Use the principle of moments to show that the upthrust U exerted by the water on the cylinder is 1.4 N. [2] (ii) The density of the water is 1.0 × 103 kg m–3. Calculate the area A of the circular cross-section of the cylinder. A = … m2 [3] (c) More water is gradually added to the container in (b), so that depth h in Fig. 3.1 gradually increases. The length x is continuously adjusted so that the system remains in equilibrium. On Fig. 3.2, sketch the variation of x with h. Use the space below for any working. 0.8 0.6 x / m 0.4 0.2 0 0.10 0.15 0.20 0.25 0.30 0.35 0.40 h / m Fig. 3.2 [3] [Total: 9]

9 marks

Mark scheme: 3(a) (for a system in equilibrium,) sum of clockwise moments (about a point) equals sum of anticlockwise moments (about the B1 same point) 3(b)(i) 2.6  0.40 or F  0.40 (any one moment) C1 U = 4.0 – F A1 = 4.0 – (2.6  0.40 / 0.40) = 1.4 N or 2.6  0.4 or (4.0 − U)  0.4 (C1) 2.6  0.4 = (4.0 − U)  0.4 hence U = (4.0 − 2.6) = 1.4 N (A1) 3(b)(ii) U = gV and A = V / h C1 or p = hg and A = U / p or A = U / hg A = 1.4 / (0.10  1.0  103  9.81) C1 = 1.4  10–3 m2 A1 3(c) line starting at (0.10, 0.40) B1 straight line with negative gradient B1 line ending at (0.29, 0) B1

This question in 9702/23 Oct/Nov 2024

2 (a) State the principle of moments. … … … [2] (b) A solid plastic cylinder floats in water. It is used to support one end of a horizontal uniform beam AB as shown in Fig. 2.1. 6.0 m 5.0 m x hinge A B P uniform beam, cylinder weight 1700 N ground water Fig. 2.1 (not to scale) The beam has length 6.0 m and weight 1700 N. The beam is attached to solid ground with a hinge at end A. The cylinder is floating vertically in the water. The top of the cylinder is attached at its centre to the beam at a horizontal distance of 5.0 m from end A. The cylinder applies a vertical force of 1300 N to the beam. A person of weight 660 N stands on the beam at point P. The beam AB is in equilibrium. (i) By taking moments about end A, determine the distance x from A to P. distance = … m [3] (ii) The bottom of the cylinder is submerged in the water to depth y as shown in Fig. 2.2. The beam is still attached to the cylinder but not shown. force from beam, 1300 N cylinder, mass 11 kg y 0.78 m water Fig. 2.2 (not to scale) The cylinder has mass 11 kg and diameter 0.78 m. The beam exerts a vertical force of 1300 N on the cylinder. The cylinder is in equilibrium. Show that the upthrust acting on the cylinder is 1400 N. [1] (iii) The water has density 990 kg m–3. Calculate the depth y. y = … m [2] (iv) The person can stand anywhere between A and B. On Fig. 2.3, sketch the variation of the depth of the bottom of the cylinder with the distance of the person from A, for distances between 0 and 6.0 m. Numerical values are not required. depth / m 0 0 6.0 distance from A / m Fig. 2.3 [2] [Total: 10]

10 marks

Mark scheme: 2(a) in (rotational) equilibrium B1 sum / total of CW moments about a point = sum / total of ACW moments about the (same) point. B1 2(b)(i) The magnitudes of the moments about A are: C1 (1700  3.0) (660 x) (1300  5) Correct magnitude of any one moment about A. Correct magnitudes of a second moment about A. C1 (1700  3.0) + (660 x) = (1300  5) A1 x = 2.1 m 2(b)(ii) (Upthrust =) 11  9.81 + 1300 = 1400 (N) A1 2(b)(iii) y = Upthrust / gA C1 = 1400 / (990  9.81    (0.78 / 2)2) = 0.30 m A1 OR (C1) [P = F / A = 1400 /   (0.78 / 2)2 = 2930] y = P / g = 2930 / (990  9.81) = 0.30 m (A1) 2(b)(iv) line with a non-zero value of depth at distance = 0 B1 A line from distance = 0 to distance = 6.0 m with a gradient that is always positive B1

This question in 9702/22 Feb/March 2025

2 (a) Define the moment of a force about a point. … … [1] (b) A tree of mass 270 kg grows out of sloping ground and is supported by a post, as shown in Fig. 2.1. centre of gravity of tree P F, 1800 N support post θ ground Q 1.2 m 1.6 m Fig. 2.1 (not to scale) The ground applies a total force R on the tree at point Q. The centre of gravity of the tree is a horizontal distance of 1.2 m from Q. The post applies a force F of 1800 N perpendicular to the line PQ. The line of action of F passes through point P at an angle θ to the vertical. P is a horizontal distance of 1.6 m from Q. The tree is in equilibrium and all forces act on the tree in the same plane. (i) By taking moments about point Q, show that θ is 25°. [3] (ii) On Fig. 2.2, draw a labelled scale vector triangle to represent the forces acting on the tree. The weight of the tree has been drawn to scale. weight Fig. 2.2 [2] (iii) The tree exerts a pressure of 150 kPa on the top of the post. Determine the surface area of the tree in contact with the post. area = … m2 [2] [Total: 8]

8 marks

Mark scheme: 2(a) force  perpendicular distance (of line of action of force to / from the point) B1 2(b)(i) (moment due to weight =) 1.2  270  9.81 B1 (moment due to post =) 1800  (1.6 / cos) B1 1.2  270  9.81 = 1800  (1.6 / cos) so = 25(°) or B1 1.2  270  9.81 – 1800  (1.6 / cos) = 0 so = 25(°) 2(b)(ii) A closed tip-to-tail vector triangle M1 vector labelled F at an angle of 25°  3° anticlockwise from vertical and A1 vector labelled R at an angle of 37°  3° clockwise from vertical 2(b)(iii) A = F / p C1 = 1800 / (150  103) = 0.012 m2 A1

This question in 9702/22 May/June 2025

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

2 A spacecraft in deep space uses jets of hot gas from its thrusters to change its velocity. Fig. 2.1 shows a side view of the spacecraft and some of its thrusters. upwards thruster C 0.40 m centre of gravity thruster A thruster B 1.6 m leftwards Fig. 2.1 (not to scale) Thruster A is a distance of 1.6 m leftwards from the centre of gravity of the spacecraft. Thruster C is a distance of 0.40 m upwards from the centre of gravity of the spacecraft. Thrusters A and B can produce forces on the spacecraft in the upwards direction only. Thruster C can produce a force on the spacecraft in the leftwards direction only. All the thrusters shown produce forces entirely in the same plane as the centre of gravity. (a) (i) Thruster A is activated, producing a force of 60 N upwards on the spacecraft. Thruster C is also activated, producing a force of 220 N in the leftwards direction on the spacecraft. Calculate the resultant moment due to these forces about the centre of gravity. resultant moment = … N m [2] (ii) State and explain whether the forces from A and C are a couple. … … … [1] (b) Thrusters A and C are now switched off and the spacecraft is stationary. Thruster B is activated at time t1, producing a constant force on the spacecraft until the fuel runs out at time t2. As the fuel is used, the total mass of the spacecraft decreases. On Fig. 2.2, sketch the variation of speed of the spacecraft with time from t1 to t2. speed 0 t1 t2 time Fig. 2.2 [2] (c) The spacecraft now splits apart into a carrier and a payload as shown in Fig. 2.3. payload upwards carrier Fig. 2.3 During the split, an average force of 5500 N acts on the payload for a time of 0.36 s. The velocity of the payload increases by 8.5 m s–1 in the upwards direction. The combined mass of the carrier and payload is 2.5 × 103 kg. (i) State the principle of conservation of momentum. … … … [2] (ii) Show that the mass of the payload is 230 kg. [2] (iii) Calculate the magnitude of the change in velocity of the carrier. change in velocity = … m s–1 [3] [Total: 12]

12 marks

Mark scheme: 2(a)(i) 60  1.6 (= 96 N m) C1 or 220  0.40 (= 88 N m) resultant moment = 96 – 88 A1 = 8.0 N m 2(a)(ii) the resultant force (of A and C) is not zero B1 or forces (from A and C) are not parallel or forces (from A and C) are not equal (magnitude) or forces (from A and C) are not opposite (direction) or forces (from A and C) act through the same point so (they are) not a couple 2(b) line with a positive gradient from (t1, 0) to t2 B1 line with increasing positive gradient from t1 to t2 B1 2(c)(i) sum / total momentum before = sum / total momentum after M1 or sum / total momentum (of a system of objects) is constant if no (resultant) external force / for an isolated system A1 2(c)(ii) ()p = F()t C1 m = Ft / ()v A1 (m =) 5500 × 0.36 / 8.5 = 230 (kg) or F = ma and a = ()v / ()t (C1) (m =) 5500 / (8.5 / 0.36) = 230 (kg) (A1) 2(c)(iii) (p =) 5500  0.36 (= 1980 N s) C1 or (p =) 230  8.5 (= 1955 N s) 1980 = (2.5  103 – 230)v C1 or 1955 = (2.5  103 – 230)v v = 0.87 m s–1 or 0.86 m s–1 A1 or a = 5500 / (2.5  103 – 230) (C1) (= 2.42 m s–2) v = 2.42  0.36 (C1) v = 0.87 m s–1 or 0.86 m s–1 (A1)

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