Equilibrium of forces — Paper 2 · A Level Physics 9702

2 (a) Explain what is meant by the centre of gravity of a body. For Examiner’s … Use … … [2] (b) An irregularly-shaped piece of cardboard is hung freely from one point near its edge, as shown in Fig. 2.1. pivot cardboard Fig. 2.1 Explain why the cardboard will come to rest with its centre of gravity vertically below the pivot. You may draw on Fig. 2.1 if you wish. … … … [2]

4 marks

Mark scheme: 2 (a) point where whole weight of body (allow mass) M1 may be considered to act (do not allow ‘acts’) A1 [2] (b) when CG below pivot, weight acts through the pivot B1 (so) weight has no turning effect about pivot B1 [2]

This question in 9702/21 Oct/Nov 2005

3 A stone on a string is made to travel along a horizontal circular path, as shown in Fig. 3.1. For Examiner’s Use path of stone stone Fig. 3.1 The stone has a constant speed. (a) Define acceleration. … … [1] (b) Use your definition to explain whether the stone is accelerating. … … … [2] (c) The stone has a weight of 5.0 N. When the string makes an angle of 35° to the vertical, For the tension in the string is 6.1 N, as illustrated in Fig. 3.2. Examiner’s Use 6.1 N 35° 5.0 N Fig. 3.2 Determine the resultant force acting on the stone in the position shown. magnitude of force = ……………..……………………. N direction of force….………………..………………….. [4]

7 marks

Mark scheme: 3 (a) change in velocity/time (taken) B1 [1] (b) velocity is a vector/velocity has magnitude & direction B1 direction changing so must be accelerating B1 [2] (c) either 6.1 × cos35 = 4.99 N or scale shown B1 so no resultant vertical force triangle of correct shape B1 6.1 sin35 = 3.5 N resultant = 3.5 ± 0.2 N B1 horizontally horizontal ± 3° B1 [4] allow answer based on centripetal force: resultant is centripetal force (which is horizontal) (B1) resultant is horizontal component of tension (B1) 6.1 sin35 = 3.5 N (B1) horizontally (B1)

This question in 9702/21 Oct/Nov 2005

2 A rod AB is hinged to a wall at A. The rod is held horizontally by means of a cord BD, attached to the rod at end B and to the wall at D, as shown in Fig. 2.1. wall D cord T hinge P F C B A rod W Fig. 2.1 The rod has weight W and the centre of gravity of the rod is at C. The rod is held in equilibrium by a force T in the cord and a force F produced at the hinge. (a) Explain what is meant by (i) the centre of gravity of a body, … … … [2] (ii) the equilibrium of a body. … … … … [2] Use (b) The line of action of the weight W of the rod passes through the cord at point P. Explain why, for the rod to be in equilibrium, the force F produced at the hinge must also pass through point P. … … … … [2] (c) The forces F and T make angles α and β respectively with the rod and AC = AB, as shown in Fig. 2.1. Write down equations, in terms of F, W, T, α and β, to represent (i) the resolution of forces horizontally, … [1] (ii) the resolution of forces vertically, … [1] (iii) the taking of moments about A. … [1]

9 marks

Mark scheme: 2 (a) (i) point at which whole weight of body M1 may be considered to act A1 [2] (ii) sum of forces in any direction is zero B1 sum of moments about any point is zero B1 [2] (b) either: T and W have zero moment about P M1 so F must have zero moment, i.e. pass through P A1 [2] or: if all pass through P, distance from P is zero for all forces (M1) so sum of moments about P is zero (A1) (c) (i) Fcosα = Tcosβ B1 [1] (ii) W = Fsinα + Tsinβ B1 [1] (iii) 2W = 3Tsinβ B1 [1]

This question in 9702/21 May/June 2006

2 (a) State the two conditions that must be satisfied for a body to be in equilibrium. For Examiner’s 1. … Use … 2. … … [2] (b) Three co-planar forces act on a body that is in equilibrium. (i) Describe how to draw a vector triangle to represent these forces. … … … … … … … [3] (ii) State how the triangle confirms that the forces are in equilibrium. … … [1] (c) A weight of 7.0 N hangs vertically by two strings AB and AC, as shown in Fig. 2.1. For Examiner’s Use B C T1 35° T2 50° A 7.0 N Fig. 2.1 For the weight to be in equilibrium, the tension in string AB is T1 and in string AC it is T2. On Fig. 2.1, draw a vector triangle to determine the magnitudes of T1 and T2. T1 = … N T2 = … N [3] (d) By reference to Fig. 2.1, suggest why the weight could not be supported with the strings AB and AC both horizontal. … … [2]

11 marks

Mark scheme: 2 (a) no resultant force/sum of forces zero B1 no resultant moment/torque/sum of moments/torques zero B1 [2] (b) (i) each force is represented by the side of a triangle/by an arrow M1 in magnitude and direction A1 arrows joined, head to tail B1 [3] (could be shown on a sketch diagram) (ii) if the triangle is ‘closed’ (then the forces are in equilibrium) B1 [1] (c) triangle drawn with correct shape (incorrect arrows loses this mark) B1 T1 = 5.4 ± 0.2 N B1 T2 = 4.0 ± 0.2 N B1 [3] (d) forces in strings would be horizontal B1 (so) no vertical force to support the weight B1 [2]

This question in 9702/23 May/June 2010

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

1 (a) Distinguish between scalars and vectors. … … [1] (b) Underline all the vector quantities in the list below. acceleration kinetic energy momentum power weight [2] (c) A force of 7.5 N acts at 40° to the horizontal, as shown in Fig. 1.1. 7.5 N 40° horizontal Fig. 1.1 Calculate the component of the force that acts (i) horizontally, horizontal component = … N [1] (ii) vertically. vertical component = … N [1] (d) Two strings support a load of weight 7.5 N, as shown in Fig. 1.2. For Examiner’s Use T1 T2 50° 40° horizontal 7.5N Fig. 1.2 One string has a tension T1 and is at an angle 50° to the horizontal. The other string has a tension T2 and is at an angle 40° to the horizontal. The object is in equilibrium. Determine the values of T1 and T2 by using a vector triangle or by resolving forces. T1 = … N T2 = … N [4]

9 marks

Mark scheme: 1 (a) scalar has magnitude/size, vector has magnitude/size and direction B1 [1] (b) acceleration, momentum, weight B2 [2] (–1 for each addition or omission but stop at zero) (c) (i) horizontally: 7.5 cos 40° / 7.5 sin 50° = 5.7(45) / 5.75 not 5.8 N A1 [1] (ii) vertically: 7.5 sin 40° / 7.5 cos 50° = 4.8(2) N A1 [1] (d) either correct shaped triangle M1 correct labelling of two forces, three arrows and two angles A1 or correct resolving: T2 cos 40° = T1 cos 50° (B1) T1 sin 50° + T2 sin 40° = 7.5 (B1) T1 = 5.7(45) (N) A1 T2 = 4.8 (N) A1 [4] (allow ± 0.2 N for scale diagram)

This question in 9702/23 Oct/Nov 2011

3 One end of a spring is fixed to a support. A mass is attached to the other end of the spring. For The arrangement is shown in Fig. 3.1. Examiner’s Use mass Fig. 3.1 (a) The mass is in equilibrium. Explain, by reference to the forces acting on the mass, what is meant by equilibrium. … … … [2] (b) The mass is pulled down and then released at time t = 0. The mass oscillates up and down. The variation with t of the displacement of the mass d is shown in Fig. 3.2. 6.0 d / 10–2 m 4.0 2.0 0 0 0.2 0.4 0.6 0.8 1.0 t / s –2.0 –4.0 –6.0 Fig. 3.2 Use Fig. 3.2 to state a time, one in each case, when (i) the mass is at maximum speed, time = … s [1] (ii) the elastic potential energy stored in the spring is a maximum, time = … s [1] (iii) the mass is in equilibrium. time = … s [1] (c) The arrangement shown in Fig. 3.3 is used to determine the length l of a spring when For different masses M are attached to the spring. Examiner’s Use l mass Fig. 3.3 The variation with mass M of l is shown in Fig. 3.4. 35 30 25 l / 10–2 m 20 15 10 5 0 0 0.10 0.20 0.30 0.40 0.50 M / kg Fig. 3.4 (i) State and explain whether the spring obeys Hooke’s law. For Examiner’s … Use … … [2] (ii) Show that the force constant of the spring is 26 N m–1. [2] (iii) A mass of 0.40 kg is attached to the spring. Calculate the energy stored in the spring. energy = … J [3]

12 marks

Mark scheme: 3 (a) Resultant force (and resultant torque) is zero B1 Weight (down) = force from/due to spring (up) B1 [2] (b) (i) 0.2, 0.6, 1.0 s (one of these) A1 [1] (ii) 0, 0.8 s (one of these) A1 [1] (iii) 0.2, 0.6, 1.0 s (one of these) A1 [1] GCE AS/A LEVEL – May/June 2012 9702 21 (c) (i) Hooke’s law: extension is proportional to the force (not mass) B1 Linear/straight line graph hence obeys Hooke’s law B1 [2] (ii) Use of the gradient (not just F = kx) C1 K = (0.4 × 9.8) / 15 × 10–2 M1 = 26(.1) N m–1 A0 [2] (iii) either energy = area to left of line or energy = ½ ke2 C1 = ½ × [(0.4 × 9.8) / 15 × 10–2] × (15 × 10–2)2 C1 = 0.294 J (allow 2 s.f.) A1 [3] 2

This question in 9702/21 May/June 2012

1 (a) Explain the differences between the quantities distance and displacement. Use … … … [2] (b) State Newton’s first law. … … … [1] (c) Two tugs pull a tanker at constant velocity in the direction XY, as represented in Fig. 1.1. tug 1 T1 X 25.0° tanker Y 15.0° T2 tug 2 Fig. 1.1 Tug 1 pulls the tanker with a force T1 at 25.0° to XY. Tug 2 pulls the tanker with a force of T2 at 15.0° to XY. The resultant force R due to the two tugs is 25.0 × 103 N in the direction XY. (i) By reference to the forces acting on the tanker, explain how the tanker may be described as being in equilibrium. … … … … [2] (ii) 1. Complete Fig. 1.2 to draw a vector triangle for the forces R, T1 and T2. [2] For Examiner’s Use R 25.0 × 103 N Fig. 1.2 2. Use your vector triangle in Fig. 1.2 to determine the magnitude of T1 and of T2. T1 = … N T2 = … N [2]

9 marks

Mark scheme: 1 (a) displacement is a vector, distance is a scalar B1 displacement is straight line between two points / distance is sum of lengths moved / example showing difference B1 [2] (either one of the definitions for the second mark) (b) a body continues at rest or at constant velocity unless acted on by a resultant (external) force B1 [1] (c) (i) sum of T1 and T2 equals frictional force B1 these two forces are in opposite directions B1 [2] (allow for 1/2 for travelling in straight line hence no rotation / no resultant torque) (ii) 1. scale vector triangle with correct orientation / vector triangle with correct orientation both with arrows B1 scale given or mathematical analysis for tensions B1 [2] 2. T1 = 10.1 × 103 (± 0.5 × 103) N A1 T2 = 16.4 × 103 (± 0.5 × 103) N A1 [2]

This question in 9702/23 May/June 2012

3 (a) Define centre of gravity. For Examiner’s … Use … [2] (b) A uniform rod AB is attached to a vertical wall at A. The rod is held horizontally by a string attached at B and to point C, as shown in Fig. 3.1. C string T wall 1.2 m 50° A B O 8.5 N mass M Fig. 3.1 The angle between the rod and the string at B is 50°. The rod has length 1.2 m and weight 8.5 N. An object O of mass M is hung from the rod at B. The tension T in the string is 30 N. (i) Use the resolution of forces to calculate the vertical component of T. vertical component of T = … N [1] (ii) State the principle of moments. … … [1] (iii) Use the principle of moments and take moments about A to show that the weight of For the object O is 19 N. Examiner’s Use [3] (iv) Hence determine the mass M of the object O. M = … kg [1] (c) Use the concept of equilibrium to explain why a force must act on the rod at A. … … … [2]

10 marks

Mark scheme: 3 (a) the point where (all) the weight (of the body) M1 is considered / seems to act A1 [2] (b) (i) vertical component of T (= 30 cos 40°) = 23 N A1 [1] (ii) the sum of the clockwise moments about a point equals the sum of the anticlockwise moments (about the same point) B1 [1] (iii) (moments about A): 23 × 1.2 (27.58) M1 = 8.5 × 0.60 + 1.2 × W M1 working to show W = 19 or answer of 18.73 (N) A1 [3] (iv) (M = W / g = 18.73 / 9.81 =) 1.9(09) kg A1 [1] GCE AS/A LEVEL – May/June 2013 9702 22 (c) (for equilibrium) resultant force (and moment) = 0 B1 upward force does not equal downward force / horizontal component of T not balanced by forces shown B1 [2]

This question in 9702/22 May/June 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 9702/21 Oct/Nov 2013

3 A uniform plank AB of length 5.0 m and weight 200 N is placed across a stream, as shown in Fig. 3.1. FA FB plank A B 880 N 200 N x 5.0 m stream Fig. 3.1 A man of weight 880 N stands a distance x from end A. The ground exerts a vertical force FA on the plank at end A and a vertical force FB on the plank at end B. As the man moves along the plank, the plank is always in equilibrium. (a) (i) Explain why the sum of the forces FA and FB is constant no matter where the man stands on the plank. … … … [2] (ii) The man stands a distance x = 0.50 m from end A. Use the principle of moments to calculate the magnitude of FB. FB = … N [4] (b) The variation with distance x of force FA is shown in Fig. 3.2. 1000 force / N FA 500 0 0 1.0 2.0 3.0 4.0 5.0 x / m Fig. 3.2 On the axes of Fig. 3.2, sketch a graph to show the variation with x of force FB. [3]

9 marks

Mark scheme: 3 (a) (i) resultant force is zero B1 weight of plank + weight of man = FA + FB or 200 (N) + 880 (N) or 1080 = FA + FB B1 [2] (ii) principle of moments used C1 (anticlockwise moments) FB × 5.0 C1 (clockwise moments) 880 × 0.5 + 200 × 2.5 C1 FB = (440 + 500) / 5.0 = 188 N A1 [4] (b) straight line with positive gradient (allow freehand) M1 start point (0, 100) A1 finish point (5, 980) A1 [3] GCE AS/A LEVEL – May/June 2014 9702 21 2

This question in 9702/21 May/June 2014

3 A rod PQ is attached at P to a vertical wall, as shown in Fig. 3.1. R wire wall F 0.64 m 0.96 m 30° Q P rod W Fig. 3.1 The length of the rod is 1.60 m. The weight W of the rod acts 0.64 m from P. The rod is kept horizontal and in equilibrium by a wire attached to Q and to the wall at R. The wire provides a force F on the rod of 44 N at 30° to the horizontal. (a) Determine (i) the vertical component of F, vertical component = … N [1] (ii) the horizontal component of F. horizontal component = … N [1] (b) By taking moments about P, determine the weight W of the rod. W = … N [2] (c) Explain why the wall must exert a force on the rod at P. … … … [1] (d) On Fig. 3.1, draw an arrow to represent the force acting on the rod at P. Label your arrow with the letter S. [1]

6 marks

Mark scheme: 3 (a) (i) (vertical component = 44 sin 30° =) 22 N A1 [1] (ii) (horizontal component = 44 cos 30° =) 38(.1) N A1 [1] (b) W × 0.64 = 22 × 1.60 C1 (W =) 55 N A1 [2] (c) F has a horizontal component (not balanced by W) or F has 38 N acting horizontally or 38 N acts on wall or vertical component of F does not balance W or F and W do not make a closed triangle of forces B1 [1] (d) line from P in direction towards point on wire vertically above W and direction up B1 [1]

This question in 9702/22 May/June 2015

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

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

2 (a) State the two conditions for a system to be in equilibrium. 1. … … 2. … … [2] (b) A paraglider P of mass 95 kg is pulled by a wire attached to a boat, as shown in Fig. 2.1. parachute paraglider P wire boat horizontal 25° water Fig. 2.1 The wire makes an angle of 25° with the horizontal water surface. P moves in a straight line parallel to the surface of the water. The variation with time t of the velocity v of P is shown in Fig. 2.2. 10.0 8.0 v / m s–1 6.0 4.0 2.0 0 0 2.0 4.0 6.0 8.0 t / s Fig. 2.2 (i) Show that the acceleration of P is 1.4 m s–2 at time t = 5.0 s. [2] (ii) Calculate the total distance moved by P from time t = 0 to t = 7.0 s. distance = … m [2] (iii) Calculate the change in kinetic energy of P from time t = 0 to t = 7.0 s. change in kinetic energy = … J [2] (iv) The tension in the wire at time t = 5.0 s is 280 N. Calculate, for the horizontal motion, 1. the vertical lift force F supporting P, F = … N [3] 2. the force R due to air resistance acting on P in the horizontal direction. R = … N [3] [Total: 14]

14 marks

Mark scheme: 2(a) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) is zero B1 2(b)(i) a = (v − u) / t or gradient or ∆v / (∆)t C1 e.g. a = (8.8 − 4.6) / (7.0 – 4.0) = 1.4 m s–2 A1 2(b)(ii) s = 4.6 × 4 + [(8.8 + 4.6) / 2] × 3 C1 = 18.4 + 20.1 = 39 (38.5) m A1 2(b)(iii) ∆E = ½ × 95 [(8.8)2 − (4.6)2] C1 = 3678 – 1005 = 2700 (2673) J A1 2(b)(iv)1. weight = 95 × 9.81 (= 932 N) C1 vertical tension force = 280 sin 25° or 280 cos 65° (=118.3 N) C1 F = 932 + 118 = 1100 (1050) N A1 2(b)(iv)2. horizontal tension force = 280 cos 25° or 280 sin 65° (= 253.8 N) C1 resultant force = 95 × 1.4 (= 133 N) C1 133 = 253.8 – R R = 120 (120.8) N A1

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

12 marks

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

This question in 9702/22 Oct/Nov 2017

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

6 marks

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

This question in 9702/22 Oct/Nov 2018

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

8 marks

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

This question in 9702/23 Oct/Nov 2018

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

11 marks

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

This question in 9702/21 Oct/Nov 2019

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

11 marks

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

This question in 9702/21 Oct/Nov 2019

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

13 marks

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

This question in 9702/21 May/June 2020

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

11 marks

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

This question in 9702/23 May/June 2020

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

6 marks

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

This question in 9702/23 Oct/Nov 2020

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

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

11 marks

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

This question in 9702/22 May/June 2021

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

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

16 marks

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

This question in 9702/21 Oct/Nov 2023

3 (a) State the conditions for a system to be in equilibrium. … … … [2] (b) Fig. 3.1 shows an airship in flight. The airship is propelled by identical fans that can be angled to control the motion of the airship. airship fans Fig. 3.1 The upthrust on the airship is 93 000 N. The density of the surrounding air is 1.2 kg m–3. (i) Calculate the volume of air displaced by the airship. volume = … m3 [1] (ii) When fully loaded, the weight of the airship is greater than the upthrust. To maintain horizontal flight, the fans provide a total vertical force of 3.0 × 103 N upwards on the airship. Calculate the mass of the airship. mass = … kg [2] (c) At a certain time, the airship in (b) is stationary. The thrust force exerted by a fan on the airship is 2800 N. To produce this force, a mass of 64 kg of air is propelled through the blades of the fan in a time of 0.50 s. Assume that this air is initially stationary at the entrance to the fan. Calculate: (i) the change in momentum ∆p of the air propelled through the fan blades in this time ∆p = … kg m s–1 [2] (ii) the speed of the air as it leaves the fan speed = … m s–1 [2] (iii) the total kinetic energy of this air due to its movement through the fan. kinetic energy = … J [2] [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 3(b)(i) F = Vg A1 V = 93 000 / (1.2  9.81) = 7900 m3 3(b)(ii) weight = 93 000 + 3(.0)  103 C1 m = (93 000 + 3.0  103) / 9.81 A1 = 9800 kg 3(c)(i) (p) = F()t or F = ∆p / ()t or F = rate of change of momentum or F = m (v – u) / t C1 ∆p = 2800  0.50 A1 = 1400 kg m s–1 3(c)(ii) (p =) mv C1 v = 1400 / 64 A1 = 22 m s–1 3(c)(iii) EK = ½mv2 C1 = ½  64  222 A1 = 15 000 J or EK = p2 / 2m (C1) = 14002 / (2  64) (A1) = 15 000 J

This question in 9702/23 Oct/Nov 2023

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

10 marks

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

This question in 9702/21 May/June 2024

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

10 marks

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

This question in 9702/23 May/June 2024

3 (a) The variation of stress with strain for a metal P is shown in Fig. 3.1. 30 stress / 107 Pa E 20 10 0 0 0.5 1.0 1.5 strain (%) Fig. 3.1 Point E is the elastic limit of the metal. (i) Use Fig. 3.1 to determine the Young modulus for P. Young modulus = … Pa [2] (ii) On the line in Fig. 3.1, draw a cross (×) to show the limit of proportionality. Label this point Q. [1] (b) State the conditions necessary for an object to be in equilibrium. … … … [2] (c) A wire is used to hold a uniform shelf AB horizontally in equilibrium as shown in Fig. 3.2. wire wall cup shelf hinge 50° A B 0.12 m 0.65 m Fig. 3.2 (not to scale) The wire is connected to the midpoint of shelf AB at an angle of 50° to the horizontal. The shelf is attached to a wall by a hinge at A. The length of shelf AB is 0.65 m and its weight is 33 N. A cup of weight 1.5 N rests on the shelf with its centre of gravity at a horizontal distance of 0.12 m from B. (i) By taking moments about A, determine the tension in the wire. tension = … N [3] (ii) The stress in the wire is 1.5 × 107 Pa. Determine the radius of the wire. radius = … m [2] (iii) More items are added to the shelf, doubling the stress in the wire. The wire is made of the metal P from (a). Use Fig. 3.1 to state and explain whether the wire will behave plastically or elastically as the stress doubles. … … … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a)(i) E =  /  or E = gradient C1 E = e.g. 12  107 / 0.0050 A1 = 2.4  1010 Pa 3(a)(ii) cross drawn at (1.0%, 24  107 Pa), labelled Q B1 3(b) resultant force (in any direction) is zero B1 resultant moment / torque (about any point) is zero B1 3(c)(i) (moment =) 33  0.65 / 2 C1 or 1.5  (0.65 – 0.12) or T sin 50°  (0.65 / 2) sum of clockwise moments = sum of anticlockwise moments C1 33  (0.65 / 2) + 1.5  (0.65 – 0.12) = T sin 50°  (0.65 / 2) tension = 46 N A1 3(c)(ii) = F / A C1 r2 = 46 / (1.5  107) A1 r = 9.9  10–4 m 3(c)(iii) elastic limit is not reached M1 or (new) stress is less than (stress at) elastic limit or (new) strain is less than (strain at) elastic limit (so the wire behaves) elastically A1

This question in 9702/22 Oct/Nov 2024

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

9 marks

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

This question in 9702/23 May/June 2025

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

9 marks

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

This question in 9702/22 Oct/Nov 2025

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

10 marks

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

This question in 9702/24 Oct/Nov 2025