Turning effects of forces — Paper 2 · A Level Physics 9702

3 (a) Distinguish between the moment of a force and the torque of a couple. For Examiner’s moment of a force … Use … … torque of a couple … … … [4] (b) One type of weighing machine, known as a steelyard, is illustrated in Fig. 3.1. 4.8 cm pivot 12 N sliding weight metal rod hook 2.5 N sliding weight Fig. 3.1 The two sliding weights can be moved independently along the rod. With no load on the hook and the sliding weights at the zero mark on the metal rod, the metal rod is horizontal. The hook is 4.8 cm from the pivot. A sack of flour is suspended from the hook. In order to return the metal rod to the horizontal position, the 12 N sliding weight is moved 84 cm along the rod and the 2.5 N weight is moved 72 cm. (i) Calculate the weight of the sack of flour. For Examiner’s Use weight = …………………………N [2] (ii) Suggest why this steelyard would be imprecise when weighing objects with a weight of about 25 N. … … [1]

7 marks

Mark scheme: 3 (a) moment: force × perpendicular distance M1 of force from pivot / axis / point A1 couple: (magnitude of) one force × perpendicular distance M1 between the two forces A1 [4] (penalise the ‘perpendicular’ omission once only) (b) (i) W × 4.8 = (12 × 84) + (2.5 × 72) C1 W = 250 N (248 N) A1 [2] (ii) either friction at the pivot or small movement of weights B1 [1]

This question in 9702/21 Oct/Nov 2008

3 (a) Define the torque of a couple. For Examiner’s … Use … … [2] (b) A torque wrench is a type of spanner for tightening a nut and bolt to a particular torque, as illustrated in Fig. 3.1. force F nut torque scale C 45 cm Fig. 3.1 The wrench is put on the nut and a force is applied to the handle. A scale indicates the torque applied. The wheel nuts on a particular car must be tightened to a torque of 130 N m. This is achieved by applying a force F to the wrench at a distance of 45 cm from its centre of rotation C. This force F may be applied at any angle to the axis of the handle, as shown in Fig. 3.1. For the minimum value of F to achieve this torque, (i) state the magnitude of the angle that should be used, = … ° [1] (ii) calculate the magnitude of F. F = … N [2]

5 marks

Mark scheme: 3 (a) product of (magnitude of one) force and distance between forces … M1 reference to either perpendicular distance between forces or line of action of forces and perpendicular distance … A1 [2] (b) (i) 90° … B1 [1] (ii) 130 = F × 0.45 (allow e.c.f. for angle in (i)) … C1 F = 290 N … A1 [2] (allow 1 mark only if angle stated in (i) is not used in (ii)) GCE A/AS LEVEL – May/June 2009 9702 21

This question in 9702/21 May/June 2009

3 (a) State what is meant by the centre of gravity of a body. For Examiner’s … Use … … [2] (b) A uniform rectangular sheet of card of weight W is suspended from a wooden rod. The card is held to one side, as shown in Fig. 3.1. rod card Fig. 3.1 On Fig. 3.1, (i) mark, and label with the letter C, the position of the centre of gravity of the card, [1] (ii) mark with an arrow labelled W the weight of the card. [1] (c) The card in (b) is released. The card swings on the rod and eventually comes to rest. For Examiner’s (i) List the two forces, other than its weight and air resistance, that act on the card Use during the time that it is swinging. State where the forces act. 1. … … 2. … … [3] (ii) By reference to the completed diagram of Fig. 3.1, state the position in which the card comes to rest. Explain why the card comes to rest in this position. … … … [2]

9 marks

Mark scheme: 3 (a) point at which (whole) weight (of body) (allow mass for weight) M1 appears / seems to act ... (for mass need ‘appears to be concentrated’) A1 [2] (b) (i) point C shown at centre of rectangle ± 5 mm B1 [1] (ii) arrow vertically downwards, from C with arrow starting from the same margin of error as in (b)(i) B1 [1] (c) (i) reaction / upwards / supporting / normal reaction force M1 friction M1 force(s) at the rod A1 [3] (ii) comes to rest with (line of action of) weight acting through rod allow C vertically below the rod B1 so that weight does not have a moment about the pivot / rod B1 [2] GCE A LEVEL – October/November 2010 9702 22

This question in 9702/22 Oct/Nov 2010

3 (a) Explain what is meant by centre of gravity. For Examiner’s … Use … [2] (b) Define moment of a force. … … [1] (c) A student is being weighed. The student, of weight W, stands 0.30 m from end A of a uniform plank AB, as shown in Fig. 3.1. P A B 0.20 m 0.30 m W 80 N 70 N 0.50 m 2.0 m Fig. 3.1 (not to scale) The plank has weight 80 N and length 2.0 m. A pivot P supports the plank and is 0.50 m from end A. A weight of 70 N is moved to balance the weight of the student. The plank is in equilibrium when the weight is 0.20 m from end B. (i) State the two conditions necessary for the plank to be in equilibrium. 1. … … 2. … … [2] (ii) Determine the weight W of the student. For Examiner’s Use W = … N [3] (iii) If only the 70 N weight is moved, there is a maximum weight of student that can be determined using the arrangement shown in Fig. 3.1. State and explain one change that can be made to increase this maximum weight. … … … [2]

10 marks

Mark scheme: 3 (a) point where the weight of an object / gravitational force M1 may be considered to act A1 [2] (b) product of the force and the perpendicular distance (to the pivot) B1 [1] (c) (i) 1. sum / net / resultant force is zero B1 2. net / resultant moment is zero sum of clockwise moments = sum of anticlockwise moments B1 [2] (ii) W × 0.2 = 80 × 0.5 + 70 × 1.3 C1 = 40 + 91 C1 W = 655 N A1 [3] (allow 2/3 for one error in distance but 0/3 if two errors) (iii) move pivot to left (M1) gives greater clockwise moment / smaller anticlockwise moment (A1) or move W to right (M1) gives smaller anticlockwise moment (A1) [2]

This question in 9702/21 May/June 2011

2 (a) Define the torque of a couple. For Examiner’s … Use … [2] (b) A uniform rod of length 1.5 m and weight 2.4 N is shown in Fig. 2.1. 1.5 m rope A 8.0 N pin rod weight 2.4 N rope B 8.0 N Fig. 2.1 The rod is supported on a pin passing through a hole in its centre. Ropes A and B provide equal and opposite forces of 8.0 N. (i) Calculate the torque on the rod produced by ropes A and B. torque = … N m [1] (ii) Discuss, briefly, whether the rod is in equilibrium. … … … … [2] (c) The rod in (b) is removed from the pin and supported by ropes A and B, as shown in For Fig. 2.2. Examiner’s Use 1.5 m rope A rope B 0.30 m P weight 2.4 N Fig. 2.2 Rope A is now at point P 0.30 m from one end of the rod and rope B is at the other end. (i) Calculate the tension in rope B. tension in B = … N [2] (ii) Calculate the tension in rope A. tension in A = … N [1]

8 marks

Mark scheme: 2 (a) torque is the product of one of the forces and the distance between forces M1 the perpendicular distance between the forces A1 [2] (b) (i) torque = 8 × 1.5 = 12 N m A1 [1] (ii) there is a resultant torque / sum of the moments is not zero M1 (the rod rotates) and is not in equilibrium A1 [2] (c) (i) B × 1.2 = 2.4 × 0.45 C1 B = 0.9(0) N A1 [2] (ii) A = 2.4 – 0.9 = 1.5 N / moments calculation A1 [1] 1

This question in 9702/21 Oct/Nov 2011

4 (a) Define the torque of a couple. For Examiner’s … Use … [2] (b) A wheel is supported by a pin P at its centre of gravity, as shown in Fig. 4.1. 25 cm 35 N P 35 N Fig. 4.1 The plane of the wheel is vertical. The wheel has radius 25 cm. Two parallel forces each of 35 N act on the edge of the wheel in the vertical directions shown in Fig. 4.1. Friction between the pin and the wheel is negligible. (i) List two other forces that act on the wheel. State the direction of these forces and where they act. 1. … 2. … [2] (ii) Calculate the torque of the couple acting on the wheel. torque = … N m [2] (iii) The resultant force on the wheel is zero. Explain, by reference to the four forces acting on the wheel, how it is possible that the resultant force is zero. … … [1] (iv) State and explain whether the wheel is in equilibrium. … [1]

8 marks

Mark scheme: 4 (a) torque of a couple = one of the forces / a force × distance M1 multiplied by the perpendicular distance between the forces A1 [2] (b) (i) weight at P (vertically) down B1 normal reaction OR contact force at (point of contact with the pin) P (vertically) up B1 [2] (ii) torque = 35 × 0.25 (or 25) × 2 C1 torque = 18 (17.5) N m A1 [2] (iii) the two 35 N forces are equal and opposite and the weight and the upward / contact / reaction force are equal and opposite B1 [1] (iv) not in equilibrium as the (resultant) torque is not zero B1 [1]

This question in 9702/21 Oct/Nov 2013

4 (a) Define the torque of a couple. For Examiner’s … Use … [2] (b) A wheel is supported by a pin P at its centre of gravity, as shown in Fig. 4.1. 25 cm 35 N P 35 N Fig. 4.1 The plane of the wheel is vertical. The wheel has radius 25 cm. Two parallel forces each of 35 N act on the edge of the wheel in the vertical directions shown in Fig. 4.1. Friction between the pin and the wheel is negligible. (i) List two other forces that act on the wheel. State the direction of these forces and where they act. 1. … 2. … [2] (ii) Calculate the torque of the couple acting on the wheel. torque = … N m [2] (iii) The resultant force on the wheel is zero. Explain, by reference to the four forces acting on the wheel, how it is possible that the resultant force is zero. … … [1] (iv) State and explain whether the wheel is in equilibrium. … [1]

8 marks

Mark scheme: 4 (a) torque of a couple = one of the forces / a force × distance M1 multiplied by the perpendicular distance between the forces A1 [2] (b) (i) weight at P (vertically) down B1 normal reaction OR contact force at (point of contact with the pin) P (vertically) up B1 [2] (ii) torque = 35 × 0.25 (or 25) × 2 C1 torque = 18 (17.5) N m A1 [2] (iii) the two 35 N forces are equal and opposite and the weight and the upward / contact / reaction force are equal and opposite B1 [1] (iv) not in equilibrium as the (resultant) torque is not zero B1 [1]

This question in 9702/22 Oct/Nov 2013

4 (a) Define moment of a force. … … [1] (b) An arrangement for lifting heavy loads is shown in Fig. 4.1. C T 60° B wall beam 2.82.8 mm 30° load 500 N 4000 N A Fig. 4.1 A uniform metal beam AB is pivoted on a vertical wall at A. The beam is supported by a wire joining end B to the wall at C. The beam makes an angle of 30° with the wall and the wire makes an angle of 60° with the wall. The beam has length 2.8 m and weight of 500 N. A load of 4000 N is supported from B. The tension in the wire is T. The beam is in equilibrium. (i) By taking moments about A, show that T is 2.1 kN. [2] (ii) Calculate the vertical component Tv of the tension T. Tv = … N [1] (iii) State and explain why Tv does not equal the sum of the load and the weight of the beam although the beam is in equilibrium. … … [2]

6 marks

Mark scheme: 4 (a) product of the force and the perpendicular distance to/from a point/pivot B1 [1] (b) (i) 4000 × 2.8 × sin 30° or 500 × 1.4 × sin 30° or T × 2.8 or 4000 × 1.4 or 500 × 0.7 B1 4000 × 2.8 × sin 30° + 500 × 1.4 × sin 30° = T × 2.8 M1 hence T = 2100 (2125) N A0 [2] (ii) (Tv = 2100 cos 60° =) 1100 (1050) N A1 [1] (iii) there is an upward (vertical component of) force at A B1 upward force at A + Tv = sum of downward forces/weight+load/4500 N B1 [2]

This question in 9702/22 Oct/Nov 2015

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

7 marks

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

This question in 9702/23 Oct/Nov 2015

3 (a) State the two conditions for an object to be in equilibrium. 1. … … 2. … … [2] (b) A uniform beam AC is attached to a vertical wall at end A. The beam is held horizontal by a rigid bar BD, as shown in Fig. 3.1. 0.30 m 0.10 m A C 52° B beam : 33 N wire wall bar bucket D 12 N Fig. 3.1 (not to scale) The beam is of length 0.40 m and weight W. An empty bucket of weight 12 N is suspended by a light metal wire from end C. The bar exerts a force on the beam of 33 N at 52° to the horizontal. The beam is in equilibrium. (i) Calculate the vertical component of the force exerted by the bar on the beam. component of the force = … N [1] (ii) By taking moments about A, calculate the weight W of the beam. W = … N [3] (c) The metal of the wire in (b) has a Young modulus of 2.0 × 1011 Pa. Initially the bucket is empty. When the bucket is filled with paint of weight 78 N, the strain of the wire increases by 7.5 × 10–4. The wire obeys Hooke’s law. Calculate, for the wire, (i) the increase in stress due to the addition of the paint, increase in stress = … Pa [2] (ii) its diameter. diameter = … m [3] [Total: 11]

11 marks

Mark scheme: 3 (a) resultant force (in any direction) is zero B1 resultant moment/torque (about any point) is zero B1 [2] (b) (i) force = 33 sin 52° or 33 cos 38° = 26 N A1 [1] (ii) 26 × 0.30 or W × 0.20 or 12 × 0.40 C1 26 × 0.30 = (W × 0.20) + (12 × 0.40) C1 W = 15 N A1 [3] (c) (i) E = ∆σ/ ∆ε or E = σ / ε C1 ∆σ = 2.0 × 1011 × 7.5 × 10–4 = 1.5 × 108 Pa A1 [2] (ii) ∆σ= ∆F / A or σ= F / A C1 A = 78 / 1.5 × 108 (= 5.2 × 10–7 m2) C1 5.2 × 10–7 = πd 2 / 4 d = 8.1 × 10–4 m A1 [3]

This question in 9702/22 Oct/Nov 2016

3 (a) A cylinder is made from a material of density 2.7 g cm–3. The cylinder has diameter 2.4 cm and length 5.0 cm. Show that the cylinder has weight 0.60 N. [3] (b) The cylinder in (a) is hung from the end A of a non-uniform bar AB, as shown in Fig. 3.1. 50 cm 20 cm bar 12 cm A B P X cylinder 0.25 N 0.60 N Fig. 3.1 The bar has length 50 cm and has weight 0.25 N. The centre of gravity of the bar is 20 cm from B. The bar is pivoted at P. The pivot is 12 cm from B. An object X is hung from end B. The weight of X is adjusted until the bar is horizontal and in equilibrium. (i) Explain what is meant by centre of gravity. … … [1] (ii) Calculate the weight of X. weight of X = … N [3] (c) The cylinder is now immersed in water, as illustrated in Fig. 3.2. A B P water X 0.25 N Fig. 3.2 An upthrust acts on the cylinder and the bar is not in equilibrium. (i) Explain the origin of the upthrust. … … … … [2] (ii) Explain why the weight of X must be reduced in order to obtain equilibrium for AB. … … … [1] [Total: 10]

10 marks

Mark scheme: 3(a) C1 V = πd2L / 4 or πr2L C1 weight = 2.7 × 103 × π (1.2 × 10–2)2 × 5.0 × 10–2 × 9.81 = 0.60 N A1 3(b)(i) the point from where (all) the weight (of a body) seems to act B1 3(b)(ii) W × 12 C1 (0.25 × 8) + (0.6 × 38) C1 W = (2 + 22.8) / 12 = 2.1 (2.07) N A1 3(c)(i) pressure changes with depth (in water) or pressure on bottom (of cylinder) different from pressure on top B1 pressure on bottom of cylinder greater than pressure on top or force (up) on bottom of cylinder greater than force (down) on top B1 3(c)(ii) anticlockwise moment reduced and reducing the weight of X reduces clockwise moment or anticlockwise moment reduced so clockwise moment now greater than (total) anticlockwise moment B1

This question in 9702/21 May/June 2017

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

4 A rigid plank is used to make a ramp between two different horizontal levels of ground, as shown in Fig. 4.1. 45 N 1.10 m B 0.30 m D C 1.50 m rope 38° T 96 N A Fig. 4.1 (not to scale) Point A at one end of the plank rests on the lower level of the ground. A force acts on, and is perpendicular to, the plank at point B. The plank is held in equilibrium by a rope that connects point D on the plank to the ground. The plank has a weight that may be considered to act from its centre of gravity C. The rope is perpendicular to the plank and has tension T. The plank is at an angle of 38° to the vertical. The forces and the distances along the plank of points A, B, C and D are shown in Fig. 4.1. (a) Show that the component of the weight that is perpendicular to the plank is 59 N. [1] (b) By taking moments about end A of the plank, calculate the tension T. T = … N [3] [Total: 4]

4 marks

Mark scheme: 4(a) (component =) 96 sin 38° = 59 (N) or 96 cos 52° = 59 (N) 4(b) (45 × 2.9) or (T × 1.8) or (59 × 1.5) C1 (45 × 2.9) = (T × 1.8) + (59 × 1.5) C1 T = 23 N A1

This question in 9702/22 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 pendulum consists of a solid sphere suspended by a string from a fixed point P, as shown in Fig. 3.1. P θ string 0.93 m Y sphere h X momentum 0.72 N s Fig. 3.1 (not to scale) The sphere swings from side to side. At one instant the sphere is at its lowest position X, where it has kinetic energy 0.86 J and momentum 0.72 N s in a horizontal direction. A short time later the sphere is at position Y, where it is momentarily stationary at a maximum vertical height h above position X. The string has a fixed length and negligible weight. Air resistance is also negligible. (a) On Fig. 3.1, draw a solid line to represent the displacement of the centre of the sphere at position Y from position X. [1] (b) Show that the mass of the sphere is 0.30 kg. [3] (c) Calculate height h. h = … m [2] (d) The distance between point P and the centre of the sphere is 0.93 m. When the sphere is at position Y, the string is at an angle θ to the vertical. Show that θ is 47°. [1] (e) For the sphere at position Y, calculate the moment of its weight about point P. moment = … N m [2] (f) State and explain whether the sphere is in equilibrium when it is stationary at position Y. … … [1] [Total: 10]

10 marks

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

This question in 9702/21 May/June 2021

3 (a) 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

2 (a) State what is meant by the centre of gravity of an object. … … [1] (b) A non-uniform rod XY is pivoted at point P, as shown in Fig. 2.1. 4.00 m 1.70 m 1.10 m X Y wire P sphere F 44.0 N 3.0 N Fig. 2.1 (not to scale) The rod has length 4.00 m and weight 44.0 N. The centre of gravity of the rod is 1.70 m from end X of the rod. Point P is 1.10 m from end X. A sphere hangs by a wire from end Y of the rod. The weight of the sphere is 3.0 N. The weight of the wire is negligible. A force F is applied vertically downwards at end X so that the horizontal rod is in equilibrium. (i) By taking moments about P, calculate F. F = … N [3] (ii) Calculate the force exerted on the rod by the pivot. force = … N [1] (c) The sphere in (b) is now immersed in a liquid in a container, as shown in Fig. 2.2. P liquid, density 1100 kg m–3 F 44.0 N container Fig. 2.2 The density of the liquid is 1100 kg m–3. The upthrust acting on the sphere due to the liquid is 2.5 N. The magnitude of F is unchanged so that the horizontal rod is not in equilibrium. (i) Use Archimedes’ principle to determine the radius r of the sphere. r = … m [3] (ii) Calculate the magnitude and direction of the resultant moment of the forces on the rod about P. magnitude of resultant moment = … N m direction of resultant moment … [2] [Total: 10]

10 marks

Mark scheme: 2(a) point where (all) the weight (of the object) is taken to act B1 2(b)(i) moments about P are: (F  1.10), (44.0  0.60) and (3.0  2.90) 1 mark for any one correct moment and 2 marks for two correct moments C2 (F  1.10) = (44.0  0.60) + (3.0  2.90) F = 32 N A1 2(b)(ii) force = 32 + 44 + 3 = 79 N A1 2(c)(i) (F = gV) V = 2.5 / (1100  9.81) (= 2.32  10–4) C1 V = (4 / 3)r3 C1 r = [(3  2.32  10–4) / 4]1/3 r = 0.038 m A1 2(c)(ii) resultant moment = 2.5  2.9 = 7.3 N m or resultant moment = F  1.1  (44  0.6 + 0.5  2.9) = 7.3 N m (allow 7.2 N m or 7.4 N m from rounded values of F) A1 direction: anticlockwise A1

This question in 9702/21 May/June 2022

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

5 marks

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

This question in 9702/21 May/June 2022

3 (a) State what is meant by the centre of gravity of an object. … … [1] (b) A uniform beam AB is attached by a frictionless hinge to a vertical wall at end A. The beam is held so that it is horizontal by a metal wire CD, as shown in Fig. 3.1. C wire wall 45 N block D 37° A B 23 N W hinge 0.56 m 0.20 m 0.20 m Fig. 3.1 (not to scale) The beam is of length 0.96 m and weight 23 N. A block of weight W rests on the beam at a distance of 0.20 m from end B. The wire is attached to the beam at point D which is a distance of 0.40 m from end B. The wire exerts a force on the beam of 45 N at an angle of 37° to the horizontal. The beam is in equilibrium. (i) Calculate the vertical component of the force exerted by the wire on the beam. vertical component of the force = … N [1] (ii) By taking moments about A, calculate the weight W of the block. W = … N [3] (iii) The hinge exerts a force on the beam at end A. Calculate the horizontal component of this force. horizontal component of force = … N [1] (iv) The block is now placed closer to point D on the beam. State whether this change will increase, decrease or have no effect on the tension in the wire. … [1] (v) The stress in the wire is 5.3 × 107 Pa. The wire is now replaced by a second wire that has a radius which is three times greater than that of the original wire. The tension in the wire is unchanged. Calculate the stress in the second wire. stress = … Pa [2] [Total: 9]

9 marks

Mark scheme: 3(a) point where (all) the weight (of an object) is taken to act B1 3(b)(i) vertical component = 45 sin 37° A1 = 27 N 3(b)(ii) the magnitudes of the three moments about A are (23  0.48), (27  0.56) and (W  0.76) C1 correct magnitude of any one moment about A correct magnitudes of any two moments about A C1 (23  0.48) + (W  0.76) = 27  0.56 A1 W = 5.4 N 3(b)(iii) horizontal component = 45 cos 37° A1 = 36 N 3(b)(iv) decrease B1 3(b)(v) = F / A C1 = F / r 2 or 4F / d 2 A1 so  = 5.3  107  r 2 / (3r)2 = 5.3  107 / 9 = 5.9  106 Pa

This question in 9702/22 Oct/Nov 2022

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

13 marks

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

This question in 9702/23 Oct/Nov 2022

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 rigid uniform beam of weight W is connected to a fixed support by a hinge, as shown in Fig. 2.1. 0.50 m 0.10 m 0.20 m 0.40 m hinge horizontal fixed 30° beam support W 0.30 N 4.8 N string spring ground Fig. 2.1 (not to scale) A compressed spring exerts a total force of 8.2 N vertically upwards on the horizontal beam. A block of weight 0.30 N rests on the beam. The right‑hand end of the beam is connected to the ground by a string at an angle of 30° to the horizontal. The tension in the string is 4.8 N. The distances along the beam are shown in Fig. 2.1. The beam is in equilibrium. Assume that the hinge is frictionless. (a) (i) Show that the vertical component of the tension in the string is 2.4 N. [1] (ii) By taking moments about the hinge, determine the weight W of the beam. W = … N [3] (iii) Calculate the horizontal component of the force exerted on the beam by the hinge. force = … N [1] (b) The spring obeys Hooke’s law and has an elastic potential energy of 0.32 J. Calculate the compression of the spring. compression = … m [2] (c) The string is cut so that the spring extends upwards. This causes the beam to rotate and launch the block into the air. The block reaches its maximum height and then falls back to the ground. Fig. 2.2 shows part of the path of the block in the air shortly before it hits the horizontal ground. A path of block 0.090 m B horizontal ground Fig. 2.2 (not to scale) The block is at a height of 0.090 m above the ground when it passes through point A. The block has a kinetic energy of 0.044 J when it hits the ground at point B. Air resistance is negligible. (i) Calculate the decrease in the gravitational potential energy of the block for its movement from A to B. decrease in gravitational potential energy = … J [2] (ii) Use your answer in (c)(i) and conservation of energy to determine the speed of the block at point A. speed = … m s–1 [3] (iii) By reference to the force on the block, explain why the horizontal component of the velocity of the block remains constant as it moves from A to B. … … [1] (iv) The block passes through point A at time tA and arrives at point B at time tB. On Fig. 2.3, sketch a graph to show the variation of the magnitude of the vertical component vY of the velocity of the block with time t from t = tA to t = tB. Numerical values of vY are not required. vY 0 tA tB t Fig. 2.3 [1] [Total: 14]

14 marks

Mark scheme: 2(a)(i) (component = ) 4.8 sin 30° = 2.4 (N) A1 2(a)(ii) (8.2  0.50) or (W  0.60) or (0.30  0.80) or (2.4  1.2) C1 (8.2  0.50) = (W  0.60) + (0.30  0.80) + (2.4  1.2) C1 W = 1.6 N A1 2(a)(iii) force = 4.8 cos 30° = 4.2 N A1 2(b) E = ½Fx C1 0.32 = ½  8.2  x x = 0.078 m A1 Question Answer Marks 2(c)(i) ((∆)E) = mg(∆)h or W(∆)h C1 = 0.30  0.090 = 0.027 J A1 2(c)(ii) E = ½mv2 C1 E = 0.044 – 0.027 (= 0.017) C1 v2 = (2  0.017) / (0.30 / 9.81) (where v = speed at A) v = 1.1 m s–1 A1 or E = ½mv2 (C1) 0.044 = ½  (0.30 / 9.81)  v2 (where v = speed at B) (v2 = 2.88) v2 = u2 + 2as (where u = speed at A) 2.88 = u2 + 2  9.81  0.090 (C1) u = 1.1 m s–1 (A1) 2(c)(iii) (gravitational / resultant) force / weight is vertical B1 2(c)(iv) straight line with positive gradient starting from non-zero value of vY at time tA to a time tB B1

This question in 9702/21 May/June 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