Elastic and plastic behaviour — Paper 2 · A Level Physics 9702

5 Fig. 5.1 shows the variation with force F of the extension x of a spring as the force is increased to F3 and then decreased to zero. F3 F F2 F1 0 0 x1 x2 x Fig. 5.1 (a) State, with a reason, whether the spring is undergoing an elastic change. … … [1] (b) The extension of the spring is increased from x1 to x2. Show that the work W done in extending the spring is given by W = k(x22 – x12), where k is the spring constant. [3] Use (c) A trolley of mass 850 g is held between two fixed points by means of identical springs, as shown in Fig. 5.2. trolley spring Fig. 5.2 When the trolley is in equilibrium, the springs are each extended by 4.5 cm. Each spring has a spring constant 16 N cm–1. The trolley is moved a distance of 1.5 cm along the direction of the springs. This causes the extension of one spring to be increased and the extension of the other spring to be decreased. The trolley is then released. The trolley accelerates and reaches its maximum speed at the equilibrium position. Assuming that the springs obey Hooke’s law, use the expression in (b) to determine the maximum speed of the trolley. speed = …………………………. m s–1 [4]

8 marks

Mark scheme: 5 (a) no hysteresis loop/no permanent deformation M1 (do not allow ‘force proportional to extension’) so elastic change A0 [1] (b) work done = area under graph line OR average force × distance B1 = ½Fx ½(F2 + F1)(x2 – x1) A1 F = kx, so work done = = ½kx2 ½k(x2 + x1)(x2 – x1) A1 work done = ½k(x22 – x12) A0 [3] (c) gain in energy of trolley = ½k(0.0602 – 0.0452) + ½k(0.0302 – 0.0452) C1 = 0.36 J C1 kinetic energy = ½ × 0.85 × v2 = 0.36 C1 v = 0.92 m s–1 A1 [4]

This question in 9702/21 May/June 2006

4 A sample of material in the form of a cylindrical rod has length L and uniform area of cross-section A. The rod undergoes an increasing tensile stress until it breaks. Fig. 4.1 shows the variation with stress of the strain in the rod. 0.02 strain breaking point 0.01 0 0 5 10 stress / 108 Pa Fig. 4.1 (a) State whether the material of the rod is ductile, brittle or polymeric. … [1] (b) Determine the Young modulus of the material of the rod. Young modulus = … Pa [2] Examiner’s Use (c) A second cylindrical rod of the same material has a spherical bubble in it, as illustrated in Fig. 4.2. cylindrical rod bubble 1.9 x 103 N 1.9 x 103 N cross-sectional area 3.2 x 10–6 m2 Fig. 4.2 The rod has an area of cross-section of 3.2 × 10–6 m2 and is stretched by forces of magnitude 1.9 × 103 N. By reference to Fig. 4.1, calculate the maximum area of cross-section of the bubble such that the rod does not break. area = … m2 [3] (d) A straight rod of the same material is bent as shown in Fig. 4.3. Fig. 4.3 Suggest why a thin rod can bend more than a thick rod without breaking. … … … [2]

8 marks

Mark scheme: 4 (a) brittle B1 [1] (b) Young modulus = stress / strain C1 = (9.5 × 108) / 0.013 = 7.3 × 1010 Pa (allow ± 0.1 × 1010 Pa) A1 [2] (c) stress = force / area C1 (minimum) area = (1.9 × 103) / (9.5 × 108) = 2.0 × 10–6 m2 C1 (max) area of cross-section = (3.2 – 2.0) × 10–6 = 1.2 × 10–6 m2 A1 [3] (d) when bent, ‘top’ and ‘bottom’ edges have different extensions M1 with thick rod, difference is greater (than with a thin rod) A1 so breaks with less bending A0 [2]

This question in 9702/21 Oct/Nov 2007

4 A spring having spring constant k hangs vertically from a fixed point. A load of weight L, when For hung from the spring, causes an extension e. The elastic limit of the spring is not exceeded. Examiner’s Use (a) State (i) what is meant by an elastic deformation, … … … [2] (ii) the relation between k, L and e. … [1] Question 4 continues on page 10 (b) Some identical springs, each with spring constant k, are arranged as shown in Fig. 4.1. For Examiner’s Use arrangement total extension spring constant of arrangement L L L Fig. 4.1 The load on each of the arrangements is L. For each arrangement in Fig. 4.1, complete the table by determining (i) the total extension in terms of e, (ii) the spring constant in terms of k. [5]

8 marks

Mark scheme: 4 (a) (i) change of shape / size / length / dimension … C1 when (deforming) force is removed, returns to original shape / size A1 [2] (ii) L = ke … B1 [1] (b) 2e … B1 ½k …(allow e.c.f. from extension) … B1 ½e and 2k … B1 3 e …(allow e.c.f. from extension in part 2) … B1 2 2 k …(allow e.c.f. from extension) … B1 [5] 3

This question in 9702/21 May/June 2009

4 A uniform wire has length L and area of cross-section A. For The wire is fixed at one end so that it hangs vertically with a load attached to its free end, as Examiner’s shown in Fig. 4.1. Use wire load W Fig. 4.1 When the load of magnitude W is attached to the wire, it extends by an amount e. The elastic limit of the wire is not exceeded. The material of the wire has resistivity ρ. (a) (i) Explain what is meant by extends elastically. … … … [2] (ii) Write down expressions, in terms of L, A, W, ρ and e for 1. the resistance R of the unstretched wire, R = … [1] 2. the Young modulus E of the wire. E = … [1] (b) A steel wire has resistance 0.44 Ω. Steel has resistivity 9.2 × 10–8 Ω m. For Examiner’s A load of 34 N hung from the end of the wire causes an extension of 7.7 × 10–4 m. Use Using your answers in (a)(ii), calculate the Young modulus E of steel. E = … Pa [3]

7 marks

Mark scheme: 4 (a) (i) returns to original shape / size / length etc. … B1 when load / distorting forces / weight / strain is removed … B1 [2] (ii) 1 R = ρL / A … B1 [1] 2 E = WL / Ae … B1 [1] (b) E = WR / eρ … C1 = (34 × 0.44) / (7.7 × 10-4 × 9.2 × 10-8) … C1 = 2.1 × 1011 Pa … A1 [3] [Total: 7]

This question in 9702/21 Oct/Nov 2009

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

5 (a) Explain what is meant by plastic deformation. For Examiner’s … Use … [1] (b) A copper wire of uniform cross-sectional area 1.54 × 10–6 m2 and length 1.75 m has a breaking stress of 2.20 × 108 Pa. The Young modulus of copper is 1.20 × 1011 Pa. (i) Calculate the breaking force of the wire. breaking force = … N [2] (ii) A stress of 9.0 × 107 Pa is applied to the wire. Calculate the extension. extension = … m [2] (c) Explain why it is not appropriate to use the Young modulus to determine the extension when the breaking force is applied. … … [1]

6 marks

Mark scheme: 5 (a) when the load is removed then the wire / body object does not return to its original shape / length B1 [1] (b) (i) stress = force / area C1 F = 220 × 106 × 1.54 × 10–6 = 340 (338.8) N A1 [2] (ii) E = (F × l) / (A × e) C1 e = (90 × 106) × 1.75 / (1.2 × 1011) = 1.31 × 10–3 m A1 [2] (c) the stress is no longer proportional to the extension B1 [1] GCE AS/A LEVEL – October/November 2012 9702 21

This question in 9702/21 Oct/Nov 2012

3 (a) (i) State the principle of conservation of momentum. For Examiner’s … Use … … [2] (ii) State the difference between an elastic and an inelastic collision. … [1] (b) An object A of mass 4.2 kg and horizontal velocity 3.6 m s–1 moves towards object B as shown in Fig. 3.1. A B 3.6 m s–1 1.2 m s–1 4.2 kg 1.5 kg before collision Fig. 3.1 Object B of mass 1.5 kg is moving with a horizontal velocity of 1.2 m s–1 towards object A. The objects collide and then both move to the right, as shown in Fig. 3.2. A B 3.0 m s–1 v 4.2 kg 1.5 kg after collision Fig. 3.2 Object A has velocity v and object B has velocity 3.0 m s–1. (i) Calculate the velocity v of object A after the collision. velocity = … m s–1 [3] (ii) Determine whether the collision is elastic or inelastic. [3]

9 marks

Mark scheme: 3 (a) (i) the total momentum of a system (of interacting bodies) remains constant M1 provided there are no resultant external forces / isolated system A1 [2] (ii) elastic: total kinetic energy is conserved, inelastic: loss of kinetic energy B1 [1] [allow elastic: relative speed of approach equals relative speed of separation] GCE AS/A LEVEL – May/June 2013 9702 23 (b) (i) initial mom: 4.2 × 3.6 – 1.2 × 1.5 (= 15.12 – 1.8 = 13.3) C1 final mom: 4.2 × v + 1.5 × 3 C1 v = (13.3 – 4.5) / 4.2 = 2.1 m s–1 A1 [3] (ii) initial kinetic energy = ½ mA(vA)2 + ½ mB(vB)2 = 27.21 + 1.08 = 28(.28) M1 final kinetic energy = 9.26 + 6.75 = 16 M1 initial KE is not the same as final KE hence inelastic A1 [3] provided final KE less than initial KE [allow in terms of relative speeds of approach and separation]

This question in 9702/23 May/June 2013

4 A spring hangs vertically from a point P, as shown in Fig. 4.1. P metre rule spring mass M reading x Fig. 4.1 A mass M is attached to the lower end of the spring. The reading x from the metre rule is taken, as shown in Fig. 4.1. Fig. 4.2 shows the relationship between x and M. 0.60 0.40 M / kg 0.20 0 20 22 24 26 28 30 32 x / cm Fig. 4.2 (a) Explain how the apparatus in Fig. 4.1 may be used to determine the load on the spring at the elastic limit. … … … … [2] (b) State and explain whether Fig. 4.2 suggests that the spring obeys Hooke’s law. … … … [2] (c) Use Fig. 4.2 to determine the spring constant, in N m–1, of the spring. spring constant = … N m–1 [3]

7 marks

Mark scheme: 4 (a) add small mass to cause extension then remove mass to see if spring returns to original length M1 repeat for larger masses and note maximum mass for which, when load is removed, the spring does return to original length A1 [2] (b) Hooke’s law requires force proportional to extension B1 graph shows a straight line, hence obeys Hooke’s law M1 [2] (c) k = force / extension C1 = (0.42 × 9.81) / [(30 – 21.2) × 10–2] C1 = 47 (46.8) N m–1 A1 [3]

This question in 9702/23 May/June 2014

4 (a) Compare the molecular motion of a liquid with (i) a solid, … … [2] (ii) a gas. … … [1] (b) (i) A ductile material in the form of a wire is stretched up to its breaking point. On Fig. 4.1, sketch the variation with extension x of the stretching force F. ductile material F 0 0 x Fig. 4.1 [1] (ii) On Fig. 4.2, sketch the variation with extension x of the stretching force F for a brittle material up to its breaking point. brittle material F 0 0 x Fig. 4.2 [1] (c) Describe a similarity and a difference between ductile and brittle materials. similarity: … … difference: … … [2]

7 marks

Mark scheme: 4 (a) (i) solid: (molecules) vibrate B1 no translational motion / fixed position, liquid: translational motion B1 [2] (ii) gas: molecules have random (and translational) motion B1 [1] (b) (i) ductile: straight line through origin then curving towards x-axis B1 [1] (ii) brittle: straight line through origin with no or negligible curved region B1 [1] (c) similarity: obey Hooke’s law / F ∝ x or have elastic regions B1 difference: brittle no or (very) little plastic region ductile has (large(r)) plastic region B1 [2]

This question in 9702/22 Oct/Nov 2014

4 A spring is kept horizontal by attaching it to points A and B, as shown in Fig. 4.1. slider spring cart, mass 1.7 kg v support A B Fig. 4.1 Point A is on a movable slider and point B is on a fixed support. A cart of mass 1.7 kg has horizontal velocity v towards the slider. The cart collides with the slider. The spring is compressed as the cart comes to rest. The variation of compression x of the spring with force F exerted on the spring is shown in Fig. 4.2. 4.5 3.5 F / N 2.5 1.5 0.5 1.0 1.5 2.0 x / cm Fig. 4.2 Fig. 4.2 shows the compression of the spring for F = 1.5 N to F = 4.5 N. The cart comes to rest when F is 4.5 N. (a) Use Fig. 4.2 to (i) show that the compression of the spring obeys Hooke’s law, … … … [2] (ii) determine the spring constant of the spring, spring constant = … N m–1 [2] (iii) determine the elastic potential energy EP stored in the spring due to the cart being brought to rest. EP = … J [3] (b) Calculate the speed v of the cart as it makes contact with the slider. Assume that all the kinetic energy of the cart is converted to the elastic potential energy of the spring. speed = … m s–1 [2]

9 marks

Mark scheme: 4 (a) (i) two sets of co-ordinates taken to determine a constant value (F / x) M1 F / x constant hence obeys Hooke’s law A1 [2] or gradient calculated and one point on line used (M1) to show no intercept hence obeys Hooke’s law (A1) (ii) gradient or one point on line used e.g. 4.5 / 1.8 × 10–2 C1 (k =) 250 N m–1 A1 [2] (iii) work done or EP = area under graph or ½Fx or ½kx2 C1 = 0.5 × 4.5 × 1.8 × 10–2 or 0.5 × 250 × (1.8 × 10–2)2 C1 = 0.041 (0.0405) J A1 [3] (b) KE = ½mv2 ½mv2 = 0.0405 or KE = 0.0405 (J) C1 (v = [2 × 0.0405 / 1.7]1/2 =) 0.22 (0.218) m s–1 A1 [2]

This question in 9702/21 May/June 2015

4 A spring is supported so that it hangs vertically, as shown in Fig. 4.1. spring mass M Fig. 4.1 Different masses are attached to the lower end of the spring. The extension x of the spring is measured for each mass M. The variation with x of M is shown in Fig. 4.2. 150 M / g 100 50 0 0 40 80 120 160 200 x / mm Fig. 4.2 (a) State and explain whether the spring obeys Hooke’s law. … … [1] (b) State the form of energy stored in the spring due to the addition of the masses. … [1] (c) Describe how to determine whether the extension of the spring is elastic. … … [1] (d) Calculate the work done on the spring as it is extended from x = 40.0 mm to x = 160 mm. work done = … J [3] [Total: 6]

6 marks

Mark scheme: 4(a) the straight line does not go through the origin/the force is not proportional to extension (so does not obey Hooke’s law) A1 4(b) elastic potential energy B1 4(c) remove the force/masses and the spring returns to its original length if elastic B1 4(d) work done is represented by/linked to area under the line (× g) C1 work = ½ (145 + 70) × 10–3 × 9.81 × 120 × 10–3 C1 = 0.13 (0.127) J A1

This question in 9702/23 May/June 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

4 The variation with extension x of the force F applied to a spring is shown in Fig. 4.1. 4.0 3.0 F / N 2.0 1.0 0 0 0.010 0.020 0.030 0.040 0.050 x / m Fig. 4.1 The spring has an unstretched length of 0.080 m and is suspended vertically from a fixed point, as shown in Fig. 4.2. 0.080 m 0.095 m 0.120 m position X position Y block hangs in equilibrium block held before release Fig. 4.2 Fig. 4.3 Fig. 4.4 A block is attached to the lower end of the spring. The block hangs in equilibrium at position X when the length of the spring is 0.095 m, as shown in Fig. 4.3. The block is then pulled vertically downwards and held at position Y so that the length of the spring is 0.120 m, as shown in Fig. 4.4. The block is then released and moves vertically upwards from position Y back towards position X. (a) Use Fig. 4.1 to determine the spring constant of the spring. spring constant = … N m–1 [2] (b) Use Fig. 4.1 to show that the decrease in elastic potential energy of the spring is 0.055 J when the block moves from position Y to position X. [2] (c) The block has a mass of 0.122 kg. Calculate the increase in gravitational potential energy of the block for its movement from position Y to position X. increase in gravitational potential energy = … J [2] (d) Use the decrease in elastic potential energy stated in (b) and your answer in (c) to determine, for the block, as it moves through position X: (i) its kinetic energy kinetic energy = … J [1] (ii) its speed. speed = … m s–1 [2] [Total: 9]

9 marks

Mark scheme: 4(a) C1 e.g. k = 4.0 / 0.050 k = 80 N m–1 A1 4(b) E = ½Fx or E = ½kx2 or E = area under graph C1 (∆)E = (½ × 3.2 × 0.040) – (½ × 1.2 × 0.015) = 0.055 J or (∆)E = (½ × 80 × 0.0402) – (½ × 80 × 0.0152) = 0.055 J or (∆)E = ½ × (1.2 + 3.2) × 0.025 = 0.055 J A1 4(c) (∆)E = mg(∆)h C1 = 0.122 × 9.81 × (0.120 – 0.095) = 0.030 J A1 or (∆)E = W × (∆)h (C1) = 1.2 × 0.025 = 0.030 J (A1) Question Answer Marks 4(d)(i) E = 0.055 – 0.030 = 0.025 J A1 4(d)(ii) E = ½mv2 C1 v = [(2 × 0.025) / 0.122]0.5 = 0.64 m s–1 A1

This question in 9702/21 Oct/Nov 2019

3 (a) State what is meant by work done. … … … [1] (b) A skier is pulled along horizontal ground by a wire attached to a kite, as shown in Fig. 3.1. wire kite speed 4.4 m s–1 140 N skier 30° ground horizontal Fig. 3.1 (not to scale) The skier moves in a straight line along the ground with a constant speed of 4.4 m s–1. The wire is at an angle of 30° to the horizontal. The tension in the wire is 140 N. (i) Calculate the work done by the tension to move the skier for a time of 30 s. work done = … J [3] (ii) The weight of the skier is 860 N. The vertical component of the tension in the wire and the weight of the skier combine so that the skier exerts a downward pressure on the ground of 2400 Pa. Determine the total area of the skis in contact with the ground. area = … m2 [3] (iii) The wire attached to the kite is uniform. The stress in the wire is 9.6 × 106 Pa. Calculate the diameter of the wire. diameter = … m [2] (c) The variation with extension x of the tension F in the wire in (b) is shown in Fig. 3.2. 300 F / N 250 200 150 100 50 0 0 0.20 0.40 0.60 0.80 x / mm Fig. 3.2 A gust of wind increases the tension in the wire from 140 N to 210 N. Calculate the change in the strain energy stored in the wire. change in strain energy = … J [3] [Total: 12]

12 marks

Mark scheme: 3(a) force × displacement in the direction of the force B1 3(b)(i) displacement = 4.4 × 30 C1 work done = 140 cos 30° × 4.4 × 30 C1 = 1.6 × 104 J A1 3(b)(ii) p = F / A C1 F = 860 – 140 sin 30° (= 790) C1 A = 790 / 2400 = 0.33 m2 A1 3(b)(iii) σ = F / A or F / πr 2 or 4F / πd 2 C1 9.6 × 106 = 4 × 140 / πd 2 d = 4.3 × 10–3 m A1 Question Answer Marks 3(c) E = ½Fx or ½kx2 or area under graph C1 (Δ)E = ½ × (140 + 210) × 0.20 × 10–3 or (Δ)E = (½ × 210 × 0.60 × 10–3) – (½ × 140 × 0.40 × 10–3) or (Δ)E = (140 × 0.20 × 10–3) + (½ × 0.20 × 10–3 × 70) or (Δ)E = [½×3.5 × 105 × (0.60 × 10–3)2 ] – [½ × 3.5 × 105 × (0.40 × 10–3)2 ] C1 ΔE = 0.035 J A1

This question in 9702/22 Feb/March 2020

5 One end of a wire is attached to a fixed point. A force F is applied to the wire to cause extension x. The variation with F of x is shown in Fig. 5.1. 0.6 0.5 x / mm 0.4 0.3 0.2 0.1 0 0 5 10 15 20 25 30 35 40 45 F / N Fig. 5.1 The wire has a cross-sectional area of 4.1 × 10–7 m2 and is made of metal of Young modulus 1.7 × 1011 Pa. Assume that the cross-sectional area of the wire remains constant as the wire extends. (a) State the name of the law that describes the relationship between F and x shown in Fig. 5.1. … [1] (b) The wire has an extension of 0.48 mm. Determine: (i) the stress stress = … Pa [2] (ii) the strain. strain = … [2] (c) The resistivity of the metal of the wire is 3.7 × 10–7 Ω m. Determine the change in resistance of the wire when the extension x of the wire changes from x = 0.48 mm to x = 0.60 mm. change in resistance = … Ω [3] (d) A force of greater than 45 N is now applied to the wire. Describe how it may be checked that the elastic limit of the wire has not been exceeded. … … [1] [Total: 9]

9 marks

Mark scheme: 5(a) Hooke’s (law) B1 5(b)(i) σ = F / A C1 = 36 / (4.1 × 10–7) = 8.8 × 107 Pa A1 5(b)(ii) Young modulus = σ / ε or F / Aε C1 ε = 8.8 × 107 / (1.7 × 1011) = 5.2 × 10–4 A1 5(c) R = ρL / A C1 ΔR = ρΔx / A = 3.7 × 10–7 × 0.12 × 10–3 / (4.1 × 10–7) C1 = 1.1 × 10–4 Ω A1 5(d) remove the force/F and wire returns to original length B1

This question in 9702/22 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

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

12 marks

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

This question in 9702/23 Oct/Nov 2020

3 A spring is extended by a force. The variation with extension x of the force F is shown in Fig. 3.1. 8.0 6.0 F / N 4.0 2.0 0 0 1.0 2.0 3.0 4.0 5.0 x / cm Fig. 3.1 (a) State the name of the law that relates the force and extension of the spring shown in Fig. 3.1. … [1] (b) Determine: (i) the spring constant, in N m−1, of the spring spring constant = … N m−1 [2] (ii) the strain energy (elastic potential energy) in the spring when the extension is 4.0 cm. strain energy = … J [2] (c) One end of the spring is attached to a fixed point. A cylinder that is submerged in a liquid is now suspended from the other end of the spring, as shown in Fig. 3.2. fixed point spring, extension 4.0 cm cylinder, cross-sectional area 1.2 × 10–3 m2 cylinder, cylinder, weight 6.20 N length 5.8 cm liquid Fig. 3.2 The cylinder has length 5.8 cm, cross-sectional area 1.2 × 10−3 m2 and weight 6.20 N. The cylinder is in equilibrium when the extension of the spring is 4.0 cm. (i) Show that the upthrust acting on the cylinder is 0.60 N. [1] (ii) Calculate the difference in pressure between the bottom face and the top face of the cylinder. difference in pressure = … Pa [2] (iii) Calculate the density of the liquid. density = … kg m−3 [2] (d) The liquid in (c) is replaced by another liquid of greater density. State the effect, if any, of this change on: (i) the upthrust acting on the cylinder … [1] (ii) the extension of the spring. … [1] [Total: 12]

12 marks

Mark scheme: 3(a) Hooke’s (law) B1 3(b)(i) k = F / x or k = gradient e.g. k = 7.0 / 5.0 × 10–2 C1 = 140 N m–1 A1 3(b)(ii) E = ½ F x or E = ½ k x 2 or E = area under graph = ½ × 5.6 × 4.0 × 10–2 or ½ × 140 × (4.0 × 10–2)2 C1 = 0.11 J A1 3(c)(i) (upthrust =) 6.20 – 5.60 = 0.60 (N) A1 3(c)(ii) Δp = ΔF / A = 0.60 / 1.2 × 10–3 C1 = 500 Pa A1 3(c)(iii) (Δ)p = ρg(Δ)h ρ = 500 / (9.81 × 5.8 × 10–2) C1 = 880 kg m–3 A1 3(d)(i) (upthrust) increases B1 3(d)(ii) (extension) decreases B1

This question in 9702/22 Feb/March 2021

4 A spring is suspended from a fixed point at one end. The spring is extended by a vertical force applied to the other end. The variation of the applied force F with the length L of the spring is shown in Fig. 4.1. 12 10 F / N 8 6 4 2 0 0 4 8 12 16 20 24 L / cm Fig. 4.1 For the spring: (a) state the name of the law that gives the relationship between the force and the extension … [1] (b) determine the spring constant, in N m–1 spring constant = … N m–1 [2] (c) determine the elastic potential energy when F = 6.0 N. elastic potential energy = … J [2] [Total: 5]

5 marks

Mark scheme: 4(a) Hooke’s (law) B1 4(b) k = F / x or k = gradient C1 = e.g. 12.0 / (0.240 – 0.08) = 75 N m–1 A1 4(c) E = 1 2 Fx or E = 1 2 kx 2 or E = area under graph C1 E = 1 2  6.0  0.080 or 1 2  75  0.082 = 0.24 J A1

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 Hooke’s law. … … [1] (b) The variation of the applied force with the extension for a sample of a material is shown in Fig. 3.1. 10 force / N 8 X 6 4 2 0 0 40 80 120 160 200 extension / mm Fig. 3.1 The sample behaves elastically up to an extension of 80 mm and breaks at point X. (i) On the line in Fig. 3.1, draw a cross (×) to show the limit of proportionality. Label this cross with the letter P. [1] (ii) On the line in Fig. 3.1, draw a cross (×) to show the elastic limit. Label this cross with the letter E. [1] (c) The sample in (b) has a cross-sectional area of 0.40 mm2 and an initial length of 3.2 m. For deformations within the limit of proportionality of the sample, determine: (i) the spring constant of the sample spring constant = … N m–1 [2] (ii) the Young modulus of the material from which the sample is made. Young modulus = … Pa [3] (d) Determine an estimate of the work done on the sample as it is extended from zero extension to its breaking point. Explain your reasoning. work done = … J [2] (e) A second sample of the same material has a larger cross-sectional area than the original sample but the same initial length. The two samples are each deformed with the limit of proportionality. State and explain qualitatively how the spring constant of the second sample compares with that of the original sample. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a) extension is proportional to (applied) force B1 3(b)(i) P at (60, 5.4) A1 3(b)(ii) E at (80, 5.9) A1 3(c)(i) k = F / x or k = gradient of (straight line section of) graph C1 e.g. gradient = 5.4 / 0.060 k = 90 N m–1 A1 3(c)(ii) Young modulus or E =  /  or FL / Ax or kL / A C1 E = (5.4  3.2) / (4.0  10−7  0.06) or 90  3.2 / (4.0  10−7) C1 E = 7.2  108 Pa A1 3(d) work done = area under graph B1 = (1.0  0.2) J A1 3(e) the extension will be smaller (for the same force on the thicker sample) or a greater force is required (to extend the thicker sample by the same amount) or spring constant is proportional to area M1 the spring constant (of the second sample) will be greater A1

This question in 9702/23 May/June 2024

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

3 A bungee jumper of mass 64 kg secures one end of an elastic rope to a bridge. The other end is attached to the bungee jumper. The jumper falls from rest from the bridge and descends into the valley below, as shown in Fig. 3.1. rope h jumper Fig. 3.1 (not to scale) Fig. 3.2 shows the variation of the tension T in the rope with the vertical distance h of the jumper below the level of the bridge. 2000 T / N 1500 1000 500 0 0 20 40 60 80 100 120 h / m Fig. 3.2 (a) The rope obeys Hooke’s law. State Hooke’s law. … … [1] (b) (i) Determine the unstretched length of the rope. length = … m [1] (ii) Determine the spring constant k of the rope. k = … N m–1 [2] (c) For the position of the bungee jumper at a distance of 120 m below the bridge: (i) show that the loss of gravitational potential energy since leaving the bridge is 75 kJ [2] (ii) show that the elastic potential energy in the rope is 75 kJ. [2] (d) Explain what can be deduced from the information in (c) about the speed of the bungee jumper when at a distance of 120 m below the bridge. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a) force is proportional to extension B1 3(b)(i) length = 37 m A1 3(b)(ii) spring constant = F / x C1 = e.g. 1800 / (120 – 37) A1 = 22 N m–1 3(c)(i) (()E) = mg()h C1 = 64 × 9.81 × 120 = 75 000 J or 75 kJ A1 3(c)(ii) 1 1 1 C1 (E =) Fx or (E =) kx2 or (E =) F2 / k or (E =) area under graph 2 2 2 1 A1 (E =) × 1800 × (120 – 37) = 75 000 J or 75 kJ 2 or 1 (E =) × 21.7 × (120 – 37)2 = 75 000 J or 75 kJ 2 3(d) (all) gravitational potential energy has been converted / equal to elastic potential energy (so no kinetic energy) B1 kinetic energy is zero so speed is zero B1

This question in 9702/24 Oct/Nov 2025