Stress and strain — Paper 2 · A Level Physics 9702

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

2 A spring is placed on a flat surface and different weights are placed on it, as shown in For Fig. 2.1. Examiner’s Use weights spring Fig. 2.1 The variation with weight of the compression of the spring is shown in Fig. 2.2. 4 compression / cm 3 2 1 0 0 10 20 30 40 weight / N Fig. 2.2 The elastic limit of the spring has not been exceeded. (a) (i) Determine the spring constant k of the spring. k = … N m–1 [2] (ii) Deduce that the strain energy stored in the spring is 0.49 J for a compression of For 3.5 cm. Examiner’s Use [2] (b) Two trolleys, of masses 800 g and 2400 g, are free to move on a horizontal table. The spring in (a) is placed between the trolleys and the trolleys are tied together using thread so that the compression of the spring is 3.5 cm, as shown in Fig. 2.3. thread spring trolley trolley mass 800g mass 2400g Fig. 2.3 Initially, the trolleys are not moving. The thread is then cut and the trolleys move apart. (i) Deduce that the ratio speed of trolley of mass 800 g speed of trolley of mass 2400 g is equal to 3.0. [2] (ii) Use the answers in (a)(ii) and (b)(i) to calculate the speed of the trolley of mass For 800 g. Examiner’s Use speed = … m s–1 [3]

9 marks

Mark scheme: 2 (a) (i) k is the reciprocal of the gradient of the graph C1 k = {32 / (4 × 10–2) = } 800 N m–1 A1 [2] (ii) either energy = average force × extension or ½kx2 or area under graph line C1 energy = ½ × 800 × (3.5 × 10–2)2 or ½ × 28 × 3.5 × 10–2 M1 energy = 0.49 J A0 [2] (b) (i) momentum before cutting thread = momentum after C1 0 = 2400 × V – 800 × v M1 v / V = 3.0 A0 [2] (ii) energy stored in spring = kinetic energy of trolleys C1 0.49 = ½ × 2.4 × ( 1 v)2 + ½ × 0.8 × v2 C1 3 v = 0.96 m s–1 A1 [3] (if only one trolley considered, or masses combined, allow max 1 mark) 2

This question in 9702/21 May/June 2008

4 (a) (i) Define the terms For Examiner’s 1. tensile stress, Use … … [1] 2. tensile strain, … … [1] 3. the Young modulus. … … [1] (ii) Suggest why the Young modulus is not used to describe the deformation of a liquid or a gas. … … [1] (b) The change ∆V in the volume V of some water when the pressure on the water increases by ∆p is given by the expression ∆V ∆p = 2.2 × 109 , V where ∆p is measured in pascal. In many applications, water is assumed to be incompressible. By reference to the expression, justify this assumption. … … … [2] (c) Normal atmospheric pressure is 1.01 × 105 Pa. For Examiner’s Divers in water of density 1.08 × 103 kg m–3 frequently use an approximation that every Use 10 m increase in depth of water is equivalent to one atmosphere increase in pressure. Determine the percentage error in this approximation. error = … % [3]

9 marks

Mark scheme: 4 (a) (i) 1. stress = force / (cross-sectional) area B1 [1] 2. strain = extension / original length B1 [1] 3. Young modulus = stress / strain B1 [1] (ratios must be clear in each answer) (ii) either fluids cannot be deformed in one direction / cannot be stretched or fluids can only have volume change or no fixed shape B1 [1] (b) either unless ∆p is very large or 2.2 × 109 is a large number M1 ∆V is very small or ∆V/V is very small, (so ‘incompressible’) A1 [2] (c) ∆p = hρg 1.01 × 105 = h × 1.08 × 103 × 9.81 C1 h = 9.53 m C1 ∆h / h = 0.47 / 10 or 0.47 / 9.53 error = 4.7% or 4.9% or 5% A1 [3]

This question in 9702/21 May/June 2008

4 (a) Explain what is meant by strain energy (elastic potential energy). For Examiner’s … Use … … [2] (b) A spring that obeys Hooke’s law has a spring constant k. Show that the energy E stored in the spring when it has been extended elastically by an amount x is given by E = 12kx 2. [3] (c) A light spring of unextended length 14.2 cm is suspended vertically from a fixed point, For as illustrated in Fig. 4.1. Examiner’s Use fixed point fixed point fixed point 14.2 cm 16.3 cm 17.8 cm 3.8 N F 3.8 N Fig. 4.1 Fig. 4.2 Fig. 4.3 A mass of weight 3.8 N is hung from the end of the spring, as shown in Fig. 4.2. The length of the spring is now 16.3 cm. An additional force F then extends the spring so that its length becomes 17.8 cm, as shown in Fig. 4.3. The spring obeys Hooke’s law and the elastic limit of the spring is not exceeded. (i) Show that the spring constant of the spring is 1.8 N cm–1. [1] (ii) For the extension of the spring from a length of 16.3 cm to a length of 17.8 cm, For Examiner’s 1. calculate the change in the gravitational potential energy of the mass on the Use spring, change in energy = … J [2] 2. show that the change in elastic potential energy of the spring is 0.077 J, [1] 3. determine the work done by the force F. work done = … J [1]

10 marks

Mark scheme: 4 (a) ability to do work … B1 as a result of a change of shape of an object/stretched etc … B1 [2] (b) work = average force ×distance moved (in direction of the force) … B1 either work = ½ × F × x or work is area under F/x graph which is ½Fx … B1 F = kx … B1 so work / energy = ½kx2 … A0 [3] 3.8 (c) (i) spring constant = … M1 2.1 = 1.8 N cm-1 … A0 [1] (ii) 1 ∆EP = mg∆h or W∆h … C1 = 3.8 × 1.5 × 10-2 = 0.057 J … A1 [2] 2 ∆ES = ½ × 1.8 × 102 (0.0362 – 0.0212) … M1 = 0.077 J … A0 [1] 3 work done = 0.077 – 0.057 = 0.020 J … A1 [1] (allow e.c.f. if ∆ES > ∆EP) [Total: 10] GCE A/AS LEVEL – October/November 2009 9702 22

This question in 9702/22 Oct/Nov 2009

3 (a) The variation with extension x of the tension F in a spring is shown in Fig. 3.1. For Examiner’s Use 200 F /N 150 100 50 0 0 1.0 2.0 3.0 4.0 x /cm Fig. 3.1 Use Fig. 3.1 to calculate the energy stored in the spring for an extension of 4.0 cm. Explain your working. energy = … J [3] (b) The spring in (a) is used to join together two frictionless trolleys A and B of mass M1 and For Examiner’s M2 respectively, as shown in Fig. 3.2. Use spring trolley A trolley B mass M1 mass M2 Fig. 3.2 The trolleys rest on a horizontal surface and are held apart so that the spring is extended. The trolleys are then released. (i) Explain why, as the extension of the spring is reduced, the momentum of trolley A is equal in magnitude but opposite in direction to the momentum of trolley B. … … … … [2] (ii) At the instant when the extension of the spring is zero, trolley A has speed V1 and trolley B has speed V2. Write down 1. an equation, based on momentum, to relate V1 and V2, … … [1] 2. an equation to relate the initial energy E stored in the spring to the final energies of the trolleys. … … [1] (iii) 1. Show that the kinetic energy EK of an object of mass m is related to its For momentum p by the expression Examiner’s Use p2 EK = . 2m [1] 2. Trolley A has a larger mass than trolley B. Use your answer in (ii) part 1 to deduce which trolley, A or B, has the larger kinetic energy at the instant when the extension of the spring is zero. … … [1]

9 marks

Mark scheme: 3 (a) either energy (stored)/work done represented by area under graph or energy = average force × extension ………………………………………… B1 energy = ½ × 180 × 4.0 × 10–2 ……………………………………..………………… C1 = 3.6 J …………………………………………………………………………. A1 [3] (b) (i) either momentum before release is zero ………………………………………. M1 so sum of momenta (of trolleys) after release is zero …..……………. A1 or force = rate of change of momentum (M1) force on trolleys equal and opposite (A1) or impulse = change in momentum (M1) impulse on each equal and opposite (A1) [2] (ii) 1 M1V1 = M2V2 ……………..……………………………………..………………. B1 [1] 2 E = ½ M1V12+ ½ M2V22 ………………………………………………………… B1 [1] (iii) 1 EK = ½mv 2 and p = mv combined to give …………………………………… M1 EK = p 2 / 2m …………………………………………………………………….. A0 [1] 2 m smaller, EK is larger because p is the same/constant …………………… M1 so trolley B …..………………………………………………………………….. A0 [1] GCE AS/A LEVEL – May/June 2010 9702 21

This question in 9702/21 May/June 2010

5 (a) Tensile forces are applied to opposite ends of a copper rod so that the rod is stretched. For The variation with stress of the strain of the rod is shown in Fig. 5.1. Examiner’s Use 2.5 stress / 108 Pa 2.0 1.5 1.0 0.5 0 0 1.0 2.0 3.0 4.0 5.0 strain / 10–3 Fig. 5.1 (i) Use Fig. 5.1 to determine the Young modulus of copper. Young modulus = … Pa [3] (ii) On Fig. 5.1, sketch a line to show the variation with stress of the strain of the rod as the stress is reduced from 2.5 × 106 Pa to zero. No further calculations are expected. [1] (b) The walls of the tyres on a car are made of a rubber compound. For The variation with stress of the strain of a specimen of this rubber compound is shown Examiner’s in Fig. 5.2. Use stress 0 0 strain Fig. 5.2 As the car moves, the walls of the tyres bend and straighten continuously. Use Fig. 5.2 to explain why the walls of the tyres become warm. … … … … [3]

7 marks

Mark scheme: 5 (a) (i) Young modulus = stress/strain ……………………………………………… C1 data chosen using point in linear region of graph ………………………… M1 Young modulus = (2.1 × 108)/(1.9 × 10–3) = 1.1 × 1011 Pa ……………………………………………………………….. A1 [3] (ii) This mark was removed from the assessment, owing to a power-of-ten inconsistency in the printed question paper. (b) area between lines represents energy/area under curve represents energy .. M1 when rubber is stretched and then released/two areas are different …… … A1 this energy seen as thermal energy/heating/difference represents energy released as heat …………………………………………………………………… A1 [3] 2 2

This question in 9702/22 May/June 2010

4 (a) A uniform wire has length L and constant area of cross-section A. For The material of the wire has Young modulus E and resistivity ρ. Examiner’s A tension F in the wire causes its length to increase by DL. Use For this wire, state expressions, in terms of L, A, F, DL and ρ for (i) the stress σ, … [1] (ii) the strain ε, … [1] (iii) the Young modulus E, … [1] (iv) the resistance R. … [1] (b) One end of a metal wire of length 2.6 m and constant area of cross-section 3.8 × 10–7 m2 is attached to a fixed point, as shown in Fig. 4.1. wire 2.6 m load 30 N Fig. 4.1 The Young modulus of the material of the wire is 7.0 × 1010 Pa and its resistivity For is 2.6 × 10–8 Ω m. Examiner’s A load of 30 N is attached to the lower end of the wire. Assume that the area of Use cross-section of the wire does not change. For this load of 30 N, (i) show that the extension of the wire is 2.9 mm, [1] (ii) calculate the change in resistance of the wire. change = … Ω [2] (c) The resistance of the wire changes with the applied load. Comment on the suggestion that this change of resistance could be used to measure the magnitude of the load on the wire. … … … [2]

9 marks

Mark scheme: 4 (a) (i) F / A B1 [1] (ii) ∆L / L B1 [1] (iii) allow FL / A∆L B1 [1] (iv) allow ρL / A or ρ (L + ∆L) / A B1 [1] (b) (i) ∆L = FL / EA = (30 × 2.6) / (7.0 × 1010 × 3.8 × 10–7) M1 = 2.93 × 10–3 m = 2.93 mm A0 [1] (ii) ∆R = ρ∆L / A C1 = (2.6 × 10–8 × 2.93 × 10–3) / (3.8 × 10–7) = 2.0 × 10–4 Ω A1 [2] (c) change in resistance is (very) small M1 so method is not appropriate A1 [2]

This question in 9702/21 Oct/Nov 2010

4 (a) A metal wire has spring constant k. Forces are applied to the ends of the wire to extend For it within the limit of Hooke’s law. Examiner’s Show that, for an extension x, the strain energy E stored in the wire is given by Use E = 12kx 2. [4] (b) The wire in (a) is now extended beyond its elastic limit. The forces causing the extension are then removed. The variation with extension x of the tension F in the wire is shown in Fig. 4.1. 80 60 F / N 40 20 0 0 0.2 0.4 0.6 0.8 1.0 x / mm Fig. 4.1 Energy ES is expended to cause a permanent extension of the wire. (i) On Fig. 4.1, shade the area that represents the energy ES. [1] (ii) Use Fig. 4.1 to calculate the energy ES. For Examiner’s Use ES = … mJ [3] (iii) Suggest the change in the structure of the wire that is caused by the energy ES. … … [1]

9 marks

Mark scheme: 4 (a) energy = average force × extension B1 = ½ × F × x B1 (Hooke’s law) extension proportional to (applied) force B1 hence F = kx B1 so E = ½kx2 A0 [4] (b) (i) correct area shaded B1 [1] (ii) 1.0 cm2 represents 1.0 mJ or correct units used in calculation C1 ES = 6.4 ± 0.2 mJ A2 [3] (for answer > ±0.2 mJ but ≤ ±0.4 mJ, then allow 2/3 marks) (iii) arrangement of atoms / molecules is changed B1 [1]

This question in 9702/22 Oct/Nov 2010

2 Briefly describe the structures of crystalline solids, polymers and amorphous materials. crystalline solids … … … polymers … … … amorphous materials … … … [5]

5 marks

Mark scheme: 2 crystalline: atoms / ions / particles in a regular arrangement / lattice long range order / orderly pattern B1 (lattice) repeats itself (1) polymer: long chain molecules / chains of monomers B1 some cross-linking between chains / tangled chains (1) amorphous: disordered arrangement of molecules / atoms / particles B1 any ordering is short-range (1) (three ‘B’ marks plus any other 2 marks) B2 [5]

This question in 9702/23 Oct/Nov 2010

4 (a) Define, for a wire, For Examiner’s (i) stress, Use … … [1] (ii) strain. … … [1] (b) A wire of length 1.70 m hangs vertically from a fixed point, as shown in Fig. 4.1. wire 25.0 N Fig. 4.1 The wire has cross-sectional area 5.74 × 10–8 m2 and is made of a material that has a Young modulus of 1.60 × 1011 Pa. A load of 25.0 N is hung from the wire. (i) Calculate the extension of the wire. extension = … m [3] (ii) The same load is hung from a second wire of the same material. This wire is twice the length but the same volume as the first wire. State and explain how the extension of the second wire compares with that of the first wire. … … … [3]

8 marks

Mark scheme: 4 (a) (i) stress is force / area B1 [1] (ii) strain is extension / original length B1 [1] (b) (i) E = [F / A] ÷ [e / l] C1 e = (25 × 1.7) / (5.74 × 10–8 × 1.6 × 1011) C1 e = 4.6 × 10–3 m A1 [3] (ii) A becomes A/2 or stress is doubled B1 e ∝ l / A or substitution into full formula B1 total extension increase is 4e A1 [3]

This question in 9702/21 May/June 2011

4 A student measures the Young modulus of a metal in the form of a wire. For Examiner’s (a) Describe, with the aid of a diagram, the apparatus that could be used. Use … … … … [2] (b) Describe the method used to obtain the required measurements. … … … … … … … … … … [4] (c) Describe how the measurements taken can be used to determine the Young modulus. For Examiner’s … Use … … … … … … … … … [4]

10 marks

Mark scheme: 4 (a) clamped horizontal wire over pulley or vertical wire attached to ceiling with mass attached B1 details: reference mark on wire with fixed scale alongside B1 [2] (b) measure original length of wire to reference mark with metre ruler / tape (B1) measure diameter with micrometer / digital calipers (B1) measure initial and final reading (for extension) with metre ruler or other suitable scale (B1) measure / record mass or weight used for the extension (B1) good physics method: measure diameter in several places / remove load and check wire returns to original length / take several readings with different loads (B1) MAX of 4 points B4 [4] (c) determine extension from final and initial readings (B1) plot a graph of force against extension (B1) determine gradient of graph for F / e (B1) calculate area from πd2 / 4 (B1) calculate E from E = F l / e A or gradient × l / A (B1) MAX of 4 points B4 [4] GCE AS/A LEVEL – May/June 2011 9702 22

This question in 9702/22 May/June 2011

4 (a) State Hooke’s Law. For Examiner’s … Use … [1] (b) A spring is compressed by applying a force. The variation with compression x of the force F is shown in Fig. 4.1. 60 F / N 40 20 0 0 10 20 30 40 50 x / mm Fig. 4.1 (i) Calculate the spring constant. spring constant = … N m–1 [1] (ii) Show that the work done in compressing the spring by 36 mm is 0.81 J. [2] (c) A child’s toy uses the spring in (b) to shoot a small ball vertically upwards. The ball has For a mass of 25 g. The toy is shown in Fig. 4.2. Examiner’s Use ball trigger/release for spring spring Fig. 4.2 (i) The spring in the toy is compressed by 36 mm. The spring is released. Assume all the strain energy in the spring is converted to kinetic energy of the ball. Using the result in (b)(ii), calculate the speed with which the ball leaves the spring. speed = … m s–1 [2] (ii) Determine the compression of the spring required for the ball to leave the spring with twice the speed determined in (i). compression = … mm [2] (iii) Determine the ratio maximum possible height for compression in (i) . maximum possible height for compression in (ii) ratio = … [2]

10 marks

Mark scheme: 4 (a) force is proportional to extension B1 [1] (b) (i) gradient of graph determined (e.g. 50 / 40 ×10–3 ) = 1250 N m–1 A1 [1] (ii) W = ½ k x2 or W = ½ final force × extension M1 = 0.5 × 1250 × (36 × 10–3)2 or 0.5 × 45 × 36 × 10–3 M1 = 0.81 J A0 [2] (c) (i) 0.81 = ½ mv2 C1 v = 8.0 (8.0498) m s–1 A1 [2] (ii) 4 × KE / 4 × WD or 3.24 J C1 hence twice the compression = 72 mm A1 [2] (iii) Max height is when all KE or WD or elastic PE is converted to GPE C1 ratio = 1/4 or 0.25 A1 [2]

This question in 9702/23 May/June 2011

6 (a) State Hooke’s law. For Examiner’s … Use … [1] (b) The variation with extension x of the force F for a spring A is shown in Fig. 6.1. 8.0 L 6.0 F / N 4.0 2.0 0 0 2 4 6 8 10 x / 10–2 m Fig. 6.1 The point L on the graph is the elastic limit of the spring. (i) Describe the meaning of elastic limit. … … … [1] (ii) Calculate the spring constant kA for spring A. kA = … N m–1 [1] (iii) Calculate the work done in extending the spring with a force of 6.4 N. For Examiner’s Use work done = … J [2] (c) A second spring B of spring constant 2kA is now joined to spring A, as shown in Fig. 6.2. spring A spring B 6.4 N Fig. 6.2 A force of 6.4 N extends the combination of springs. For the combination of springs, calculate (i) the total extension, extension = … m [1] (ii) the spring constant. spring constant = … N m–1 [1]

7 marks

Mark scheme: 6 (a) extension is proportional to force (for small extensions) B1 [1] (b) (i) point beyond which (the spring) does not return to its original length when the load is removed B1 [1] (ii) gradient of graph = 80 N m–1 A1 [1] (iii) work done is area under graph / ½ Fx / ½ kx2 C1 = 0.5 × 6.4 × 0.08 = 0.256 (allow 0.26) J A1 [2] (c) (i) extension = 0.08 + 0.04 = 0.12 m A1 [1] (ii) spring constant = 6.4 / 0.12 = 53.3 N m–1 A1 [1]

This question in 9702/21 Oct/Nov 2011

3 (a) Define For Examiner’s (i) stress, Use … … [1] (ii) strain. … … [1] (b) Explain the term elastic limit. … … [1] (c) Explain the term ultimate tensile stress. … … … [2] (d) (i) A ductile material in the form of a wire is stretched up to its breaking point. On Fig. 3.1, sketch the variation with extension x of the stretching force F. F 0 0 x Fig. 3.1 [2] (ii) On Fig. 3.2, sketch the variation with x of F for a brittle material up to its breaking For point. Examiner’s Use F 0 0 x Fig. 3.2 [1] (e) (i) Explain the features of the graphs in (d) that show the characteristics of ductile and brittle materials. … … … … … [2] (ii) The force F is removed from the materials in (d) just before the breaking point is reached. Describe the subsequent change in the extension for 1. the ductile material, … … [1] 2. the brittle material. … … [1]

12 marks

Mark scheme: 3 (a) (i) stress = force / (cross-sectional) area B1 [1] (ii) strain = extension / original length or change in length / original length B1 [1] (b) point beyond which material does not return to the original length / shape / size when the load / force is removed B1 [1] GCE AS/A LEVEL – October/November 2011 9702 22 (c) UTS is the maximum force / original cross-sectional area M1 wire is able to support / before it breaks A1 [2] allow one: maximum stress the wire is able to support / before it breaks (d) (i) straight line from (0,0) M1 correct shape in plastic region A1 [2] (ii) only a straight line from (0,0) B1 [1] (e) (i) ductile: initially force proportional to extension then a large extension for small change in force B1 brittle: force proportional to extension until it breaks B1 [2] (ii) 1. does not return to its original length / permanent extension (as entered plastic region) B1 2. returns to original length / no extension (as no plastic region / still in elastic region) B1 [2]

This question in 9702/22 Oct/Nov 2011

3 (a) Define the terms For Examiner’s (i) power, Use … [1] (ii) the Young modulus. … … … [1] (b) A crane is used to lift heavy objects, as shown in Fig. 3.1. motor cable object of mass 1800 kg ground Fig. 3.1 The motor in the crane lifts a total mass of 1800 kg from rest on the ground. The cable supporting the mass is made of steel of Young modulus 2.4 × 1011 Pa. The cross-sectional area of the cable is 1.3 × 10– 4 m2. As the mass leaves the ground, the strain in the cable is 0.0010. Assume the weight of the cable to be negligible. (i) 1. Use the Young Modulus of the steel to show that the tension in the cable is 3.1 × 104 N. [2] 2. Calculate the acceleration of the mass as it is lifted from the ground. acceleration = … m s–2 [3] (ii) The motor now lifts the mass through a height of 15 m at a constant speed. For Examiner’s Calculate Use 1. the tension in the lifting cable, tension = … N [1] 2. the gain in potential energy of the mass. gain in potential energy = … J [2] (iii) The motor of the crane is 30% efficient. Calculate the input power to the motor required to lift the mass at a constant speed of 0.55 m s–1. input power = … W [3]

13 marks

Mark scheme: 3 (a) (i) power = work done per unit time / energy transferred per unit time / rate of work done B1 [1] (ii) Young modulus = stress / strain B1 [1] (b) (i) 1. E = T / (A × strain) (allow strain = ε) C1 T = E × A × strain = 2.4 × 1011 × 1.3 × 10–4 × 0.001 M1 = 3.12 × 104 N A0 [2] 2. T – W = ma C1 [3.12 × 104 – 1800 × 9.81] = 1800a C1 a = 7.52 m s–2 A1 [3] (ii) 1. T = 1800 × 9.81 = 1.8 × 104 N A1 [1] 2. potential energy gain = mgh C1 = 1800 × 9.81 × 15 = 2.7 × 105 J A1 [2] (iii) P = Fv C1 = 1800 × 9.81 × 0.55 C1 input power = 9712 × (100/30) = 32.4 × 103 W A1 [3]

This question in 9702/23 Oct/Nov 2011

5 (a) Define the Young modulus. For Examiner’s … Use … [1] (b) A load F is suspended from a fixed point by a steel wire. The variation with extension x of F for the wire is shown in Fig. 5.1. 6.0 5.0 4.0 F / N 3.0 2.0 1.0 00 0.10 0.20 0.30 x / mm Fig. 5.1 (i) State two quantities, other than the gradient of the graph in Fig. 5.1, that are required in order to determine the Young modulus of steel. 1. … 2. … [1] (ii) Describe how the quantities you listed in (i) may be measured. … … … [2] (iii) A load of 3.0 N is applied to the wire. Use Fig. 5.1 to calculate the energy stored in For the wire. Examiner’s Use energy = … J [2] (c) A copper wire has the same original dimensions as the steel wire. The Young modulus for steel is 2.2 × 1011 N m–2 and for copper is 1.1 × 1011 N m–2. On Fig. 5.1, sketch the variation with x of F for the copper wire for extensions up to 0.25 mm. The copper wire is not extended beyond its limit of proportionality. [2]

8 marks

Mark scheme: 5 (a) E = stress / strain B1 [1] (b) (i) 1. diameter / cross sectional area / radius 2. original length B1 [1] (ii) measure original length with a metre ruler / tape B1 measure the diameter with micrometer (screw gauge) B1 [2] allow digital vernier calipers (iii) energy = ½ Fe or area under graph or ½ kx2 C1 = ½ × 0.25 × 10–3 × 3 = 3.8 × 10–4 J A1 [2] (c) straight line through origin below original line M1 line through (0.25, 1.5) A1 [2]

This question in 9702/22 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) With reference to the arrangement of atoms, distinguish between metals, polymers and For amorphous solids. Examiner’s Use metals: … … polymers: … … amorphous solids: … … [3] (b) On Fig. 3.1, sketch the variation with extension x of force F to distinguish between a metal and a polymer. F F 0 0 0 x 0 x metal polymer Fig. 3.1 [2]

5 marks

Mark scheme: 3 (a) metal: regular / repeated / ordered arrangement / pattern / lattice or long range order (of atoms / molecules / ions) B1 polymer: tangled chains (of atoms / molecules) or long chains (of atoms / molecules / ions) B1 amorphous: disordered / irregular arrangement or short range order (of atoms / molecules / ions) B1 [3] (b) metal: straight line or straight line then curving with less positive gradient B1 polymer: curve with decreasing gradient with steep increasing gradient at end B1 [2]

This question in 9702/22 Oct/Nov 2012

1 Energy is stored in a metal wire that is extended elastically. (a) Explain what is meant by extended elastically. … … [2] (b) Show that the SІ units of energy per unit volume are kg m–1 s–2. [2] (c) For a wire extended elastically, the elastic energy per unit volume X is given by X = Cε2E where C is a constant, ε is the strain of the wire, and E is the Young modulus of the wire. Show that C has no units. [3]

7 marks

Mark scheme: 1 (a) the wire returns to its original length (not ‘shape’) M1 when the load is removed A1 [2] (b) energy: N m / kg m2 s–2 and volume m3 C1 energy / volume: kg m2 s–2 / m3 M1 energy / volume: kg m–1 s–2 A0 [2] (c) ε has no units B1 E: kg m s–2 m–2 M1 units of RHS: kg m–1 s–2 = LHS units / satisfactory conclusion to show C has no units A1 [3]

This question in 9702/21 May/June 2013

4 (a) Define For Examiner’s (i) stress, Use … [1] (ii) strain. … [1] (b) The Young modulus of the metal of a wire is 0.17 TPa. The cross-sectional area of the wire is 0.18 mm2. The wire is extended by a force F. This causes the length of the wire to be increased by 0.095 %. Calculate (i) the stress, stress = … Pa [4] (ii) the force F. F = … N [2]

8 marks

Mark scheme: 4 (a) (i) stress = force / cross-sectional area B1 [1] (ii) strain = extension / original length B1 [1] (b) (i) E = stress / strain C1 E = 0.17 × 1012 C1 stress = 0.17 × 1012 × 0.095 / 100 C1 = 1.6(2) × 108 Pa A1 [4] (ii) force = (stress × area) = 1.615 × 108 × 0.18 × 10–6 C1 = 29(.1) N A1 [2]

This question in 9702/23 May/June 2013

5 (a) Define the Young modulus. … … [1] (b) Two wires P and Q of the same material and same original length l0 are fixed so that they hang vertically, as shown in Fig. 5.1. l 0 l 0 P Q F F Fig. 5.1 (not to scale) The diameter of P is d and the diameter of Q is 2d. The same force F is applied to the lower end of each wire. Show your working and determine the ratio stress in P (i) , stress in Q ratio = … [2] strain in P (ii) . strain in Q ratio = … [2]

5 marks

Mark scheme: 5 (a) (Young modulus / E =) stress / strain B1 [1] (b) (i) stress = F / A or = F / (π d2/4) or = F / (π d2) M1 ratio = 4 (or 4:1) A1 [2] (ii) E is the same for both wires (as same material) [e.g. EP = EQ] M1 strain = stress / E ratio = 4 (or 4:1) [must be same as (i)] A1 [2]

This question in 9702/22 May/June 2014

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

1 (a) The Young modulus of the metal of a wire is 1.8 × 1011 Pa. The wire is extended and the strain produced is 8.2 × 10–4. Calculate the stress in GPa. stress = … GPa [2] (b) An electromagnetic wave has frequency 12 THz. (i) Calculate the wavelength in μm. wavelength = … μm [2] (ii) State the name of the region of the electromagnetic spectrum for this frequency. … [1] (c) An object B is on a horizontal surface. Two forces act on B in this horizontal plane. A vector diagram for these forces is shown to scale in Fig. 1.1. N 2.5 N B 30° W E S 7.5 N Fig. 1.1 A force of 7.5 N towards north and a force of 2.5 N from 30° north of east act on B. The mass of B is 750 g. (i) On Fig. 1.1, draw an arrow to show the approximate direction of the resultant of these two forces. [1] (ii) 1. Show that the magnitude of the resultant force on B is 6.6 N. [1] 2. Calculate the magnitude of the acceleration of B produced by this resultant force. magnitude = … m s–2 [2] (iii) Determine the angle between the direction of the acceleration and the direction of the 7.5 N force. angle = … ° [1]

10 marks

Mark scheme: 1 (a) stress = Young modulus × strain = 1.8 × 1011 × 8.2 × 10–4 or 1.476 × 108 C1 = 0.15 (0.148) GPa A1 [2] (b) (i) wavelength = 3 × 108 / 12 × 1012 C1 = 25 µm A1 [2] (ii) infra-red / IR B1 [1] (c) (i) arrow drawn up to the left of 7.5 N force approximately 5° to 40° to west of north A1 [1] (ii) 1. correct vector triangle or working to show magnitude of resultant force = 6.6 N allow 6.5 to 6.7 N if scale diagram M1 [1] 2. magnitude of acceleration = 6.6 / 0.75 [scale diagram: (6.5 to 6.7) / 0.75] C1 = 8.8 m s–2 [scale diagram: 8.7 – 8.9 m s–2] A1 [2] (iii) 19° [use of scale diagram allow 17° to 21° (a diagram must be seen)] B1 [1]

This question in 9702/22 Oct/Nov 2014

1 (a) The kilogram, metre and second are SI base units. State two other base units. 1. … 2. … [2] (b) Determine the SI base units of (i) stress, SI base units … [2] (ii) the Young modulus. SI base units … [1]

5 marks

Mark scheme: 1 (a) ampere B1 kelvin B1 [2] (allow mole and candela) (b) (i) stress: N m–2 C1 kg m s–2 / m2 = kg m–1 s–2 A1 [2] (ii) Young modulus = stress / strain and strain has no units hence units: kg m–1 s–2 B1 [1]

This question in 9702/23 Oct/Nov 2014

4 Fig. 4.1 shows the values obtained in an experiment to determine the Young modulus E of a metal in the form of a wire. quantity value instrument diameter d 0.48 mm length l 1.768 m 5.0 N to 30.0 N load F in 5.0 N steps extension e 0.25 mm to 1.50 mm Fig. 4.1 (a) (i) Complete Fig. 4.1 with the name of an instrument that could be used to measure each of the quantities. [3] (ii) Explain why a series of values of F, each with corresponding extension e, are measured. … … [1] (b) Explain how a series of readings of the quantities given in Fig. 4.1 is used to determine the Young modulus of the metal. A numerical answer for E is not required. … … … … [2]

6 marks

Mark scheme: 4 (a) (i) diameter and extension: micrometer (screw gauge) or digital calipers B1 length: tape measure or metre rule B1 load: spring balance or Newton meter B1 [3] (ii) to reduce the effect of random errors or to plot a graph to check for zero error in measurement of extension or to see if limit of proportionality is exceeded B1 [1] (b) plot a graph of F against e and determine the gradient B1 E = (gradient × l) / [πd 2 / 4] B1 [2]

This question in 9702/23 May/June 2015

7 A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m–2. (a) Calculate the maximum tension that can be applied to the wire. tension = … N [2] (b) The steel of the wire has density 7800 kg m–3. The wire is hung vertically. Calculate the maximum length of the steel wire that could be hung vertically before the wire breaks under its own weight. length = … m [3] Please turn over for Question 8.

5 marks

Mark scheme: 7 (a) stress or σ = F / A C1 max. tension = UTS × A = 4.5 × 108 × 15 × 10–6 = 6800 (6750) N A1 [2] (b) ρ = m / V C1 weight = mg = ρVg = ρALg 6750 = 7.8 × 103 × 15 × 10–6 × L × 9.81 C1 L = 5.9 (5.88) × 103 m A1 or maximum mass = 6750 / 9.81 = 688 kg (C1) mass per unit length = ρA = 0.117 kg m–1 (C1) L = 688 / 0.117 = 5.9 × 103 m (A1) or maximum mass = 6750 / 9.81 = 688 kg (C1) volume = m / ρ = 0.0882 m3 = LA (C1) L = 0.0882 / 15 × 10–6 = 5.9 × 103 m (A1) [3]

This question in 9702/23 Oct/Nov 2015

4 (a) State what is meant by elastic potential energy. … … [1] (b) A spring is extended by applying a force. The variation with extension x of the force F is shown in Fig. 4.1 for the range of values of x from 20 cm to 40 cm. 11.0 10.0 9.0 ) / N 8.0 7.0 6.0 5.0 20 30 40 [ / cm Fig. 4.1 (i) Use data from Fig. 4.1 to show that the spring obeys Hooke’s law for this range of extensions. … … … [2] (ii) Use Fig. 4.1 to calculate 1. the spring constant, spring constant = … N m–1 [2] 2. the work done extending the spring from x = 20 cm to x = 40 cm. work done = … J [3] (c) A force is applied to the spring in (b) to give an extension of 50 cm. State how you would check that the spring has not exceeded its elastic limit. … … [1] [Total: 9]

9 marks

Mark scheme: 4 (a) the energy (stored) in a body due to its extension/compression/deformation/ change in shape/size B1 [1] (b) (i) two values of F/x are calculated which are the same e.g. 10.4 / 40 = 0.26 and 6.5 / 25 = 0.26 B1 or ratio of two forces and the ratio of the corresponding two extensions are calculated which are the same e.g. 5.2 / 10.4 = 0.5 and 20 / 40 = 0.5 (B1) or gradient of graph line calculated and coordinates of one point on the line used with straight line equation y = mx + c to show c = 0 (B1) (so) force is proportional to extension (and so Hooke’s law obeyed) B1 [2] (b) (ii) 1. k = F / x or k = gradient C1 gradient or values from a single point used e.g. k = 10.4 / (40 × 10–2) k = 26 N m–1 A1 [2] 2. work done = area under graph or ½Fx or ½(F2 + F1)(x2 – x1) or ½kx2 or ½k(x22 – x12) C1 = ½ × 10.4 × 0.4 – ½ × 5.2 × 0.2 C1 or ½ × (5.2 + 10.4) × 20 × 10–2 or ½ × 26 × (0.42 − 0.22) = 1.6 J A1 [3] (c) remove the force and the spring goes back to its original length B1 [1] 3

This question in 9702/21 May/June 2016

3 (a) Define the Young modulus. … … [1] (b) The Young modulus of steel is 1.9 × 1011 Pa. The Young modulus of copper is 1.2 × 1011 Pa. A steel wire and a copper wire each have the same cross-sectional area and length. The two wires are each extended by equal forces. (i) Use the definition of the Young modulus to determine the ratio extension of the copper wire . extension of the steel wire ratio = … [3] (ii) The two wires are each extended by a force. Both wires obey Hooke’s law. On Fig. 3.1, sketch a graph for each wire to show the variation with extension of the force. Label the line for steel with the letter S and the line for copper with the letter C. force 00 extension Fig. 3.1 [1] [Total: 5]

5 marks

Mark scheme: 3 (a) Young modulus = stress / strain B1 [1] (b) (i) E = (F × l) / (A × e) or e = (F × l) / (A × E) B1 e ∝ 1 / E or ratio eC / eS = ES / EC or (1.9 × 1011) / (1.2 × 1011) or 19 / 12 C1 (ratio =) 1.6 (1.58) A1 [3] (ii) two straight lines from (0,0) with S having the steepest gradient B1 [1]

This question in 9702/22 May/June 2016

3 (a) State Hooke’s law. … … [1] (b) The variation with compression x of the force F acting on a spring is shown in Fig. 3.1. 30 F / N 20 10 0 0 1.0 2.0 3.0 4.0 5.0 x / cm Fig. 3.1 The spring is fixed to the closed end of a horizontal tube. A block is pushed into the tube so that the spring is compressed, as shown in Fig. 3.2. block spring tube mass 0.025 kg 4.0 cm BEFORE AFTER Fig. 3.2 (not to scale) The compression of the spring is 4.0 cm. The mass of the block is 0.025 kg. (i) Calculate the spring constant of the spring. spring constant = … N m–1 [2] (ii) Show that the work done to compress the spring by 4.0 cm is 0.48 J. [2] (iii) The block is now released and accelerates along the tube as the spring returns to its original length. The block leaves the end of the tube with a speed of 6.0 m s–1. 1. Calculate the kinetic energy of the block as it leaves the end of the tube. kinetic energy = … J [2] 2. Assume that the spring has negligible kinetic energy as the block leaves the tube. Determine the average resistive force acting against the block as it moves along the tube. resistive force = … N [3] (iv) Determine the efficiency of the transfer of elastic potential energy from the spring to the kinetic energy of the block. efficiency = … [2] [Total: 12]

12 marks

Mark scheme: 3 (a) force/load is proportional to extension/compression (provided proportionality limit is not exceeded) B1 [1] (b) (i) k = F / x or k = gradient C1 k = 600 N m–1 A1 [2] (ii) (W =) ½kx2 or (W =) ½Fx or (W =) area under graph C1 (W =) 0.5 × 600 × (0.040)2 = 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2] (iii) 1. (EK =) ½mv2 C1 = ½ × 0.025 × 6.02 = 0.45 J A1 [2] 2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0) J] C1 average resistive force = 0.030 / 0.040 C1 = 0.75 N A1 [3] (iv) efficiency = [useful energy out / total energy in] (×100) C1 = [0.45 / 0.48] (×100) = 0.94 or 94% A1 [2]

This question in 9702/21 Oct/Nov 2016

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

11 marks

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

This question in 9702/22 Oct/Nov 2016

3 (a) State Hooke’s law. … … [1] (b) The variation with compression x of the force F acting on a spring is shown in Fig. 3.1. 30 F / N 20 10 0 0 1.0 2.0 3.0 4.0 5.0 x / cm Fig. 3.1 The spring is fixed to the closed end of a horizontal tube. A block is pushed into the tube so that the spring is compressed, as shown in Fig. 3.2. block spring tube mass 0.025 kg 4.0 cm BEFORE AFTER Fig. 3.2 (not to scale) The compression of the spring is 4.0 cm. The mass of the block is 0.025 kg. (i) Calculate the spring constant of the spring. spring constant = … N m–1 [2] (ii) Show that the work done to compress the spring by 4.0 cm is 0.48 J. [2] (iii) The block is now released and accelerates along the tube as the spring returns to its original length. The block leaves the end of the tube with a speed of 6.0 m s–1. 1. Calculate the kinetic energy of the block as it leaves the end of the tube. kinetic energy = … J [2] 2. Assume that the spring has negligible kinetic energy as the block leaves the tube. Determine the average resistive force acting against the block as it moves along the tube. resistive force = … N [3] (iv) Determine the efficiency of the transfer of elastic potential energy from the spring to the kinetic energy of the block. efficiency = … [2] [Total: 12]

12 marks

Mark scheme: 3 (a) force/load is proportional to extension/compression (provided proportionality limit is not exceeded) B1 [1] (b) (i) k = F / x or k = gradient C1 k = 600 N m–1 A1 [2] (ii) (W =) ½kx2 or (W =) ½Fx or (W =) area under graph C1 (W =) 0.5 × 600 × (0.040)2 = 0.48 J or (W =) 0.5 × 24 × 0.040 = 0.48 J A1 [2] (iii) 1. (EK =) ½mv2 C1 = ½ × 0.025 × 6.02 = 0.45 J A1 [2] 2. (work done against resistive force =) 0.48 – 0.45 [= 0.03(0) J] C1 average resistive force = 0.030 / 0.040 C1 = 0.75 N A1 [3] (iv) efficiency = [useful energy out / total energy in] (×100) C1 = [0.45 / 0.48] (×100) = 0.94 or 94% A1 [2]

This question in 9702/23 Oct/Nov 2016

3 The Young modulus of the material of a wire can be determined using the apparatus shown in Fig. 3.1. clamp wire marker on wire C X pulley scale S F bench masses Fig. 3.1 One end of the wire is clamped at C and a marker is attached to the wire above a scale S. A force to extend the wire is applied by attaching masses to the other end of the wire. The reading X of the marker on the scale S is determined for different forces F applied to the end of the wire. The variation with X of F is shown in Fig. 3.2. 40 F / N 30 20 10 0 2.0 4.0 6.0 8.0 10.0 12.0 X / mm Fig. 3.2 (a) The length of the wire from C to the marker for F = 0 is 3.50 m. The diameter of the wire is 0.38 mm. Use the gradient of the line in Fig. 3.2 to determine the Young modulus E of the material of the wire in TPa. E = … TPa [3] (b) The experiment is repeated with a thicker wire of the same material and length. State how the range of the force F must be changed to obtain the same range of scale readings as in Fig. 3.2. … … … [1] [Total: 4]

4 marks

Mark scheme: 3(a) C1 = [gradient × 3.5] / [π × (0.19 × 10–3)2] e.g. E = [{(40 – 5) / ([11.6 – 3.2] × 10–3)} × 3.5] / [π × (0.19 × 10–3)2] or [4170 × 3.5] / [π × (0.19 × 10–3)2] C1 E (= 1.3 × 1011) = 0.13 TPa (allow answers in range 0.120–0.136 TPa) A1 3(b) a larger range of F required or range greater than 35 N B1

This question in 9702/22 May/June 2017

4 (a) Define strain. … … [1] (b) A wire is designed to ensure that its strain does not exceed 4.0 × 10–4 when a force of 8.0 kN is applied. The Young modulus of the metal of the wire is 2.1 × 1011 Pa. It may be assumed that the wire obeys Hooke’s law. For a force of 8.0 kN, calculate, for the wire, (i) the maximum stress, maximum stress = … Pa [2] (ii) the minimum cross-sectional area. minimum cross-sectional area = … m2 [2] [Total: 5]

5 marks

Mark scheme: 4(a) (strain =) extension / original length B1 4(b)(i) E = σ / ε C1 maximum stress = 2.1 × 1011 × 4.0 × 10–4 = 8.4 × 107 Pa A1 4(b)(ii) σ = F / A C1 minimum area = 8.0 × 103 / 8.4 × 107 = 9.5 × 10–5 m2 A1

This question in 9702/21 Oct/Nov 2017

3 (a) For the deformation of a wire under tension, define (i) stress, … … [1] (ii) strain. … … [1] (b) A wire is fixed at one end so that it hangs vertically. The wire is given an extension x by suspending a load F from its free end. The variation of F with x is shown in Fig. 3.1. 8 F / N 7 6 5 4 3 2 1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 x / mm Fig. 3.1 The wire has cross-sectional area 9.4 × 10–8 m2 and original length 2.5 m. (i) Describe how measurements can be taken to determine accurately the cross-sectional area of the wire. … … … … … [3] (ii) Determine the Young modulus E of the material of the wire. E = … Pa [2] (iii) Use Fig. 3.1 to calculate the increase in the energy stored in the wire when the load is increased from 2.0 N to 4.0 N. increase in energy = … J [2] (c) The wire in (b) is replaced by a new wire of the same material. The new wire has twice the length and twice the diameter of the old wire. The new wire also obeys Hooke’s law. On Fig. 3.1, sketch the variation with extension x of the load F for the new wire from x = 0 to x = 0.80 mm. [2] [Total: 11]

11 marks

Mark scheme: 3(a)(i) force / (cross-sectional) area B1 3(a)(ii) extension / original length B1 3(b)(i) measure / determine / find diameter B1 using a micrometer / digital calipers B1 several measurements in different places / along the wire / around the circumference (and average them) B1 3(b)(ii) E = σ / ε or E = FL / Ax or E = gradient × (L / A) E = (4 × 2.5) / (0.8 × 10–3) × (9.4 × 10–8) C1 = 1.3 × 1011 Pa A1 Question Answer Marks 3(b)(iii) E = ½Fx or E = ½kx 2 or E = area under graph E = ½ × (2+4) × 0.4 × 10–3 or E = (½ × 4 × 0.8×10–3) – (½ × 2 × 0.4×10–3) or E = [ ½ × 5000 × (0.8×10–3)2 ] – [ ½ × 5000 × (0.4×10–3)2 ] C1 E = 1.2 × 10–3 J A1 3(c) straight line from the origin and above the original line M1 straight line passes through (0.80, 8.0) A1

This question in 9702/22 Feb/March 2018

4 (a) Define the Young modulus of a material. … … [1] (b) A metal rod is compressed, as shown in Fig. 4.1. rod F F L Fig. 4.1 The variation with compressive force F of the length L of the rod is shown in Fig. 4.2. 151 150 L / mm 149 148 147 146 145 0 10 20 30 40 50 60 70 80 90 F / kN Fig. 4.2 Use Fig. 4.2 to (i) determine the spring constant k of the rod, k = … N m–1 [2] (ii) determine the strain energy stored in the rod for F = 90 kN. strain energy = … J [3] (c) The rod in (b) has cross-sectional area A and is made of metal of Young modulus E. It is now replaced by a new rod of the same original length. The new rod has cross-sectional area A / 3 and is made of metal of Young modulus 2E. The compression of the new rod obeys Hooke’s law. On Fig. 4.2, sketch the variation with F of the length L for the new rod from F = 0 to F = 90 kN. [2] [Total: 8]

8 marks

Mark scheme: 4(a) (Young modulus =) stress / strain B1 4(b)(i) k = F / ∆L or 1 / gradient C1 = 90 × 103 / (2 × 10–3) (or other point on line) = 4.5 × 107 N m–1 A1 4(b)(ii) E = ½F∆L or E = ½k(∆L)2 C1 = ½ × 90 × 103 × 2 × 10–3 or ½ × 4.5 × 107 × (2 × 10–3)2 C1 = 90 J A1 4(c) straight line starting from (0, 150) and below original line M1 line ends at (90, 147) A1

This question in 9702/23 May/June 2018

2 (a) State Newton’s second law of motion. … … [1] (b) A car of mass 850 kg tows a trailer in a straight line along a horizontal road, as shown in Fig. 2.1. car trailer tow-bar mass 850 kg horizontal road Fig. 2.1 The car and the trailer are connected by a horizontal tow-bar. The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2. 15 14 v / m s–1 13 12 11 10 9 8 0 5 10 15 20 25 t / s Fig. 2.2 (i) Calculate the distance travelled by the car from time t = 0 to t = 10 s. distance = … m [2] (ii) At time t = 10 s, the resistive force acting on the car due to air resistance and friction is 510 N. The tension in the tow-bar is 440 N. For the car at time t = 10 s: 1. use Fig. 2.2 to calculate the acceleration acceleration = … m s−2 [2] 2. use your answer to calculate the resultant force acting on the car resultant force = … N [1] 3. show that a horizontal force of 1300 N is exerted on the car by its engine [1] 4. determine the useful output power of the engine. output power = … W [2] (c) A short time later, the car in (b) is travelling at a constant speed and the tension in the tow-bar is 480 N. The tow-bar is a solid metal rod that obeys Hooke’s law. Some data for the tow-bar are listed below. Young modulus of metal = 2.2 × 1011 Pa original length of tow-bar = 0.48 m cross-sectional area of tow-bar = 3.0 × 10−4 m2 Determine the extension of the tow-bar. extension = … m [3] (d) The driver of the car in (b) sees a pedestrian standing directly ahead in the distance. The driver operates the horn of the car from time t = 15 s to t = 17 s. The frequency of the sound heard by the pedestrian is 480 Hz. The speed of the sound in the air is 340 m s−1. Use Fig. 2.2 to calculate the frequency of the sound emitted by the horn. frequency = … Hz [2] [Total: 14]

14 marks

Mark scheme: 2(a) (resultant) force proportional/equal to/is rate of change of momentum B1 2(b)(i) distance = area under graph or s = ½ (u + v) t = ½ × (9 + 13) × 10 or s = ut + ½at 2 = (9 × 10) + (½ × 0.40 × 102) or s = vt – ½at 2 = (13 × 10) – (½ × 0.40 × 102) or v 2 = u 2 + 2as 132 = 92 + (2 × 0.40 × s) C1 distance = 110 m A1 Question Answer Marks 2(b)(ii) 1. a = gradient or a = (v – u) / t or a = ∆v / (∆)t e.g. a = (14 – 9) / 12.5 or (13 – 9) / 10 C1 a = 0.40 m s–2 A1 2. resultant force = 850 × 0.40 = 340 N A1 3. (F =) 510 + 440 + 340 = 1300 (N) A1 4. P = Fv C1 = 1300 × 13 = 1.7 × 104 W A1 2(c) E = σ / ε C1 E = (F / A) / (∆L / L) or E = FL / A∆L C1 ∆L = (480 × 0.48) / (3.0 × 10–4 × 2.2 × 1011) = 3.5 × 10–6 m A1 2(d) fo = fs v / (v – vs) 480 = fs × 340 / (340 – 14) C1 fs = 460 Hz A1

This question in 9702/22 May/June 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

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

4 (a) Define, for a wire: (i) stress … … [1] (ii) strain. … … [1] (b) (i) A school experiment is performed on a metal wire to determine the Young modulus of the metal. A force is applied to one end of the wire which is fixed at the other end. The variation of the force F with extension x of the wire is shown in Fig. 4.1. F1 F 00 x Fig. 4.1 The maximum force applied to the wire is F1. The gradient of the graph line in Fig. 4.1 is G. The wire has initial length L and cross-sectional area A. Determine an expression, in terms of A, G and L, for the Young modulus E of the metal. E = … [2] (ii) A student repeats the experiment in (b)(i) using a new wire that has twice the diameter of the first wire. The initial length of the wire and the metal of the wire are unchanged. On Fig. 4.1, draw the graph line representing the new wire for the force increasing from F = 0 to F = F1. [2] (iii) Another student repeats the original experiment in (b)(i), increasing the force beyond F1 to a new maximum force F2. The new graph obtained is shown in Fig. 4.2. F2 F F1 00 x Fig. 4.2 1. On Fig. 4.2, shade an area that represents the work done to extend the wire when the force is increased from F1 to F2. [1] 2. Explain how the student can check that the elastic limit of the wire was not exceeded when force F 2 was applied. … … … [1] (iv) Each student in the class performs the experiment in (b)(i). The teacher describes the values of the Young modulus calculated by the students as having high accuracy and low precision. Explain what is meant by low precision. … … [1] [Total: 9]

9 marks

Mark scheme: 4(a)(i) (stress =) force / cross-sectional area B1 4(a)(ii) (strain =) extension / original length B1 4(b)(i) E = FL / Ax C1 = GL / A A1 4(b)(ii) straight line from origin above the original line M1 line ends at point (4 small squares, F1). A1 4(b)(iii) 1. shaded area below the graph line and between the two vertical dashed lines B1 2. remove the force/F/F2 and the wire goes back to original length/zero extension B1 4(b)(iv) values have a large range B1

This question in 9702/21 Oct/Nov 2020

3 (a) A uniform metal bar, initially unstretched, has sides of length w, x and y, as shown in Fig. 3.1. w y x Fig. 3.1 The bar is now stretched by a tensile force F applied to the shaded ends. The changes in the lengths x and y are negligible. The bar now has sides of length x, y and z, as shown in Fig. 3.2. F z y x F Fig. 3.2 Determine expressions, in terms of some or all of F, w, x, y and z, for: (i) the stress σ applied to the bar by the tensile force σ = … [1] (ii) the strain ε in the bar due to the tensile force ε = … [1] (iii) the Young modulus E of the metal from which the bar is made. E = … [2] (b) A copper wire is stretched by a tensile force that gradually increases from 0 to 280 N. The variation with extension of the tensile force is shown in Fig. 3.3. 320 force / N 240 160 80 0 0 2 4 6 8 10 12 extension / mm Fig. 3.3 (i) State the maximum extension of the wire for which it obeys Hooke’s law. extension = … mm [1] (ii) Use Fig. 3.3 to determine the strain energy in the wire when the tensile force is 120 N. strain energy = … J [3] (iii) Explain why the work done in stretching the wire to an extension of 12 mm is not equal to the energy recovered when the tensile force is removed. … … … [2] [Total: 10]

10 marks

Mark scheme: 3(a)(i) σ = F / xy B1 3(a)(ii) ε = (z – w) / w B1 3(a)(iii) E = σ / ε C1 = Fw / xy(z – w) A1 3(b)(i) extension = 2.2 mm (allow 2.0–2.4 mm) A1 3(b)(ii) strain energy = area under graph/line or ½Fx or ½kx2 C1 = ½ × 120 × 1.4 × 10–3 or ½ × 8.6 × 104 × (1.4 × 10–3)2 C1 = 0.084 J A1 3(b)(iii) (some of the) deformation of the wire is plastic/permanent/not elastic or wire goes past the elastic limit/enters plastic region B1 energy (that cannot be recovered) is dissipated as thermal energy/becomes internal energy B1

This question in 9702/21 Oct/Nov 2021

6 The ends of a metal resistance wire are connected to a battery of electromotive force (e.m.f.) 8.0 V and negligible internal resistance, as shown in Fig. 6.1. 8.0 V resistance wire Fig. 6.1 The power dissipated by the resistance wire is 36 W. (a) Calculate: (i) the current in the resistance wire current = … A [2] (ii) the number of free electrons that pass through the resistance wire in a time of 50 s number = … [2] (iii) the resistance of the wire. resistance = … Ω [2] (b) The metal of the resistance wire in the circuit has a resistivity of 1.4 × 10–6 Ω m. The cross-sectional area of the wire is 0.25 mm2. Determine the length of the wire. length = … m [2] (c) The circuit shown in Fig. 6.1 is modified by replacing the original resistance wire with a second resistance wire. The second wire has a greater diameter than the original wire. There are no other differences between the second wire and the original wire. By reference to resistance, state and explain whether the power dissipated by the second wire is more than, less than or the same as the power dissipated by the original wire. … … … … [2] (d) The circuit shown in Fig. 6.1 is modified by connecting a second battery, of e.m.f. 8.0 V and negligible internal resistance, in parallel with the original battery and the original resistance wire, as shown in Fig. 6.2. 8.0 V 8.0 V original resistance wire Fig. 6.2 By reference to the current in the resistance wire, state and explain whether the addition of the second battery causes the power in the original resistance wire to decrease, increase or stay the same. … … … … [2] [Total: 12]

12 marks

Mark scheme: 6(a)(i) C1 I = 36 / 8.0 = 4.5 A A1 6(a)(ii) charge = 4.5 × 50 = 225 C1 number = 225 / 1.6 × 10–19 = 1.4 × 1021 A1 6(a)(iii) R = V 2 / P or R = V / I or R = P / I 2 = 8.02 / 36 or = 8.0 / 4.5 or = 36 / 4.52 C1 = 1.8 Ω A1 Question Answer Marks 6(b) R = ρL / A C1 L = (1.8 × 0.25 × 10–6) / 1.4 × 10–6 = 0.32 m A1 6(c) (larger cross-sectional area, same length, same resistivity and so) less resistance M1 (same p.d. and more current so) more power (dissipated) A1 6(d) current (in wire) is the same M1 (same p.d. across wire so) power stays the same A1

This question in 9702/22 Feb/March 2022

2 A sphere is attached by a metal wire to the horizontal surface at the bottom of a river, as shown in Fig. 2.1. sphere direction of flow of water water wire horizontal 68° surface Fig. 2.1 (not to scale) The sphere is fully submerged and in equilibrium, with the wire at an angle of 68° to the horizontal surface. The weight of the sphere is 32 N. The upthrust acting on the sphere is 280 N. The density of the water is 1.0 × 103 kg m–3. Assume that the force on the sphere due to the water flow is in a horizontal direction. (a) By considering the components of force in the vertical direction, determine the tension in the wire. tension = … N [2] (b) For the sphere, calculate: (i) the volume volume = … m3 [1] (ii) the density. density = … kg m–3 [2] (c) The centre of the sphere is initially at a height of 6.2 m above the horizontal surface. The speed of the water then increases, causing the sphere to move to a different position. This movement of the sphere causes its gravitational potential energy to decrease by 77 J. Calculate the final height of the centre of the sphere above the horizontal surface. height = … m [3] (d) The extension of the wire increases when the sphere changes position as described in (c). The wire obeys Hooke’s law. (i) State a symbol equation that gives the relationship between the tension T in the wire and its extension x. Identify any other symbol that you use. … … [1] (ii) Before the sphere changed position, the initial elastic potential energy of the wire was 0.65 J. The change in position of the sphere causes the extension of the wire to double. Calculate the final elastic potential energy of the wire after the sphere has changed position. final elastic potential energy = … J [2] [Total: 11]

11 marks

Mark scheme: 2(a) T sin 68° + 32 = 280 C1 T = 270 N A1 2(b)(i) F = gV V = 280 / (1.0  103  9.81) = 0.029 m3 A1 2(b)(ii)  = (32 / 9.81) / 0.029 C1 = 110 kg m–3 A1 Question Answer Marks 2(c) (∆)E = mg(∆)h or (∆)E = W(∆)h C1 (∆)h = (–) 77 / 32 C1 (∆)h = (–) 2.4 final height = 6.2 – 2.4 = 3.8 m A1 2(d)(i) T = kx where k is a constant or T = (EA / L)x where A is (cross-sectional) area, E is Young modulus, L is (original/unstretched) length B1 2(d)(ii)  2 1 2 E kx or 1 2 E Fx and F = kx C1 E = 0.65  22 = 2.6 J A1 or 1 2 E Fx    1 0.65 270 2 x and so x = 4.8  10–3 m k = F / x = 270 / 4.8  10–3 = 5.6  104 (C1) Question Answer Marks 2(d)(ii) final      3 1 540 9.6 10 2 E or E =          3 1 270 2 4.8 10 2 2 or E =        2 4 3 1 5.6 10 9.6 10 2 = 2.6 J (A1)

This question in 9702/22 May/June 2022

2 A motor uses a wire to raise a block, as illustrated in Fig. 2.1. motor Z wire Y block, weight 1.4 × 104 N X Fig. 2.1 (not to scale) The base of the block takes a time of 0.49 s to move vertically upwards from level X to level Y at a constant speed of 0.64 m s–1. During this time the wire has a strain of 0.0012. The wire is made of metal of Young modulus 2.2 × 1011 Pa and has a uniform cross-section. The block has a weight of 1.4 × 104 N. Assume that the weight of the wire is negligible. (a) Calculate: (i) the cross-sectional area A of the wire A = … m2 [2] (ii) the increase in the gravitational potential energy of the block for the movement of its base from X to Y. increase in gravitational potential energy = … J [3] (b) The motor has an efficiency of 56%. Calculate the input power to the motor as the base of the block moves from X to Y. input power = … W [3] (c) The base of the block now has a uniform deceleration of magnitude 1.3 m s–2 from level Y until the base of the block stops at level Z. Calculate the tension T in the wire as the base of the block moves from Y to Z. T = … N [3] (d) The base of the block is at levels X, Y and Z at times tX, tY and tZ respectively. On Fig. 2.2, sketch a graph to show the variation with time t of the distance d of the base of the block from level X. Numerical values of d and t are not required. d 0 tX tY tZ t Fig. 2.2 [2] [Total: 13]

13 marks

Mark scheme: 2(a)(i) E =  /  or E = F / A C1 A = 1.4  104 / (2.2  1011  0.0012) A1 = 5.3  10–5 m2 2(a)(ii) (∆)h = 0.64  0.49 (= 0.3136) C1 (∆)E = mg(∆)h or W(∆)h C1 = 1.4  104  0.64  0.49 A1 = 4.4  103 J 2(b) P = Fv or W / t C1 = (1.4  104  0.64) / 0.56 or (4.4  103 / 0.49) / 0.56 C1 = 1.6  104 W A1 2(c) m = 1.4  104 / 9.81 C1 ( = 1427 kg) (resultant) F = (1.4  104 / 9.81)  1.3 C1 ( = 1855 N) T = 1.4  104 – 1855 or (1.4104 / 9.81)  (9.81 – 1.3) A1 = 1.2  104 N 2(d) upward sloping straight line from (tX, 0) to tY B1 from tY to tZ: an upward sloping curve with decreasing magnitude of gradient (that is horizontal at tZ) B1

This question in 9702/22 Feb/March 2023

6 (a) Define the Young modulus. … … [1] (b) A uniform wire is suspended from a fixed support. Masses are added to the other end of the wire, as shown in Fig. 6.1. fixed support wire masses Fig. 6.1 (not to scale) The variation of the length l of the wire with the force F applied to the wire by the masses is shown in Fig. 6.2. 2.003 l / m 2.002 2.001 2.000 1.999 1.998 0 10 20 30 F / N Fig. 6.2 The cross-sectional area of the wire is 0.95 mm2. (i) Determine the unstretched length of the wire. unstretched length = … m [1] (ii) For an applied force F of 30 N, determine: ● the stress in the wire stress = … Pa ● the strain of the wire. strain = … [3] [Total: 5]

5 marks

Mark scheme: 6(a) (Young modulus =) stress / strain B1 6(b)(i) unstretched length = 1.9980 m A1 6(b)(ii) stress = F / A C1 = 30 / 9.5  10–7 = 3.2  107 Pa A1 strain = 0.0050 / 1.9980 = 2.5  10–3 A1

This question in 9702/23 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

4 Fig. 4.1 shows the variation with extension x of the tensile force F for two wires, G and H, made from the same material. 3.5 H 3.0 G F / N 2.5 2.0 1.5 1.0 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 x / mm Fig. 4.1 The elastic limit has not been exceeded for G or H. (a) For the lines in Fig. 4.1: (i) state what is represented by the gradient … [1] (ii) explain why the area under the line represents the elastic potential energy of the wire. … … … … [2] (b) Wires G and H are joined together end‑to‑end to form a composite wire of negligible weight. The composite wire hangs vertically from a fixed support. A block of weight of 2.0 N is attached to the end of the wire, as shown in Fig. 4.2. G fixed support H block, weight 2.0 N Fig. 4.2 (i) Use Fig. 4.1 to determine: ● the extension xG of wire G xG = … mm ● the extension xH of wire H. xH = … mm [1] (ii) Calculate the total elastic potential energy EP of the composite wire due to the weight of the block. EP = … J [2] (iii) The original length of wire G is L and the original length of wire H is 1.5 L. Calculate the ratio cross‑sectional area of wire G cross‑sectional area of wire H. ratio = … [3] [Total: 9]

9 marks

Mark scheme: 4(a)(i) spring constant B1 4(a)(ii) area represents the work done (to extend the wire) B1 work done (to extend the wire) is equal to elastic potential energy B1 4(b)(i) xG = 0.39 mm and xH = 0.29 mm A1 4(b)(ii) E = ½Fx C1 or E = ½kx2 and F = kx or E = area under graph for G: E = ½  2.0  (0.39  10–3) or ½  5.1  103  (0.39  10–3)2 A1 for H: E = ½  2.0  (0.29  10–3) or ½  6.9  103  (0.29  10–3)2 EP = 3.9  10–4 + 2.9  10–4 = 6.8  10–4 J 4(b)(iii) E = FL / Ax or stress / strain = FL / Ax C1 AG / AH = 1  (0.29  10–3) / [1.5  (0.39  10–3)] C1 ratio = 0.50 A1

This question in 9702/23 Oct/Nov 2023

3 A thin metal wire X, of diameter 1.2 × 10–3 m, is used to suspend a model planet, as shown in Fig. 3.1. wire X model planet Fig. 3.1 (not to scale) The variation with strain of the stress for wire X is shown in Fig. 3.2. 1.0 0.8 stress / GPa 0.6 0.4 0.2 0 0 2 4 6 8 10 strain / 10–3 Fig. 3.2 (a) The strain in X is 5.4 × 10–3. (i) Use Fig. 3.2 to calculate the force exerted on the wire by the model planet. force = … N [3] (ii) The elastic potential energy of X is 0.31 J. Calculate the original length of the wire before the model planet was attached. original length = … m [3] (b) Wire X is replaced by a new wire, Y, with the same original length and diameter but double the Young modulus of X. Wire Y also obeys Hooke’s law. On Fig. 3.2, draw a line representing the variation with strain of the stress for Y. [2] [Total: 8]

8 marks

Mark scheme: 3(a)(i) = 0.72  109 C1 force =  A C1 = 0.72  109    (1.2  10–3 / 2)2 = 810 N A1 or (C1) Young modulus = gradient of graph e.g. = 0.80  109 / 6.0  10–3 = 1.33  1011 force = Young modulus  strain  A (C1) = 1.33  1011  5.4 10–3    (1.2  10–3 / 2)2 = 810 N (A1) 3(a)(ii) E(P) = ½ Fx C1 or E(P) = ½ kx2 and F = kx x = 2EP / F C1 x = 2  0.31 / 810 x = 7.7  10–4 L = x /  A1 L = 7.7  10–4 / 5.4  10–3 L = 0.14 m or (C1) E(P) = ½ Fx+ or E(P) = ½ kx2 and k = EA/L x = 2EP / EA  (C1) x = 2  0.31 / (1.33  1011   (1.2  10–3 / 2)2  5.4  10–3) x = 7.6  10–4 L = x /  (A1) L = 7.6  10–4 / 5.4  10–3 L = 0.14 m 3(b) A straight line, passing through the origin with a larger gradient than wire X. M1 Gradient of the line is twice the gradient of wire X. A1

This question in 9702/22 Feb/March 2024

4 (a) Define strain. … … [1] (b) A copper wire of length 4.0 m has a uniform cross-sectional area of 4.5 × 10–7 m2. A tensile force of 18 N is applied to the wire. This causes the wire to extend by 1.4 mm up to its limit of proportionality. (i) Calculate the Young modulus of the wire. Young modulus = … Pa [3] (ii) On Fig. 4.1, draw a line to show how the stress varies with the strain for the wire up to its limit of proportionality. 5 stress / 107 Pa 4 3 2 1 0 0 1 2 3 4 5 strain / 10 – 4 Fig. 4.1 [2] (c) A second copper wire has the same length as the wire in (b) but a larger diameter. Both wires are subjected to a tensile force of 18 N. By placing a tick (3) in each row, complete Table 4.1 to compare the stress and strain of the two wires. Table 4.1 greater in less in the same in second wire second wire both wires stress strain [2] [Total: 8]

8 marks

Mark scheme: 4(a) extension / original length B1 4(b)(i) Young modulus = stress / strain C1 = (18 / 4.5  10–7) / (1.4  10–3 / 4.0) C1 = (4.0  107) / (3.5  10–4) = 1.1  1011 Pa A1 4(b)(ii) straight line through the origin B1 ending at the point (3.5, 4.0) B1 4(c) greater in second wire less in second wire the same in both wires stress  strain  B2

This question in 9702/21 May/June 2024

3 Lightning occurs when charge builds up in the atmosphere, creating a potential difference between the ground and the atmosphere. During a lightning strike there is an average current of 3.3 × 10 4 A for a time of 2.6 × 10 –5 s. (a) Calculate the charge transferred during the lightning strike. charge = … C [2] (b) The potential difference between the ground and the atmosphere is 3.0 × 107 V. Calculate the average power, in GW, transferred during the lightning strike. power = … GW [2] (c) A lightning rod is attached to a tall building to conduct charge safely to the ground. The lightning rod is modelled as a uniform cylindrical copper cable of total length 95 m that runs from the ground to the top of the building, as shown in Fig. 3.1. lightning rod building ground Fig. 3.1 (i) The resistance of the lightning rod is 9.6 Ω. The resistivity of copper is 1.7 × 10 –8 Ω m. Determine the radius of the lightning rod. radius = … m [3] (ii) The radius of the copper lightning rod is doubled with no change to its length. State the effect of this change on the resistance of the lightning rod. … [1] (d) A section of the lightning rod of length 0.12 m is removed for testing. A tensile stress of 1.9 × 106 Pa is applied, as shown in Fig. 3.2. lightning rod fixed support tensile stress 1.9 × 106 Pa 0.12 m Fig. 3.2 (not to scale) The section of the rod obeys Hooke’s law. The Young modulus of copper is 1.3 × 1011 Pa. Calculate the extension of the section. extension = … m [3] [Total: 11]

11 marks

Mark scheme: 3(a) C1 = 3.3  104  2.6  10–5 = 0.86 C A1 3(b) P = IV or P = VQ / t or V = IR and P = V 2/R or P = I 2R C1 P = 3.3  104  3.0  107 or P = (3.0  107  0.86) / (2.6  10–5) or P = (3.0  107)2 / 910 or P = (3.3  104)2  910 P = 9.9  1011 (W) = 990 GW A1 3(c)(i) R = L / A C1 9.6 = 1.7  10–8  95 / r 2 C1 r = 2.3  10–4 m A1 3(c)(ii) (resistance) decreases by a factor of four A1 Question Answer Marks 3(d) E =  /  C1 x = L / E = 1.9  106  0.12 / (1.3  1011) C1 = 1.8  10–6 m A1

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

4 (a) Define the Young modulus of a material. … … [1] (b) A metal wire P that obeys Hooke’s law is stretched within its limit of proportionality. (i) On Fig. 4.1, sketch the variation of tensile force F in the wire with its extension x. F 0 0 x Fig. 4.1 [1] (ii) State the name of the quantity represented by the gradient of the line in Fig. 4.1. … [1] (iii) State the name of the quantity represented by the area under the line in Fig. 4.1. … [1] (c) Another wire Q is made from a metal that has twice the Young modulus of the metal of wire P in (b). Wire Q has the same volume as wire P but has double the cross-sectional area of wire P. The two wires are extended by equal tensile forces within their limits of proportionality. State and explain how the extension of wire Q compares with the extension of wire P. … … … … … [3] [Total: 7]

7 marks

Mark scheme: 4(a) stress per unit strain B1 4(b)(i) straight line with positive gradient passing through origin B1 4(b)(ii) spring constant B1 4(b)(iii) elastic potential energy (stored in wire) B1 4(c) length (of Q) is half (the length of P) B1 extension is proportional to length, and inversely proportional to area and Young modulus B1 extension of Q is = ½ / (2  2) B1 = ⅛ times the extension of P

This question in 9702/21 Oct/Nov 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

4 (a) Define: (i) stress … … [1] (ii) strain. … … [1] (b) Two wires X and Y, with equal unstretched lengths of 0.84 m, are suspended from fixed points that are at the same horizontal level. The lower ends of the wires are attached to a beam of negligible mass. The beam is horizontal and in equilibrium, as shown in Fig. 4.1. wire X wire Y beam load, 18 N Fig. 4.1 Wire X is made from a metal that has a Young modulus of 1.9 × 109 Pa. Wire Y is made from a different metal. A load of weight 18 N is suspended from the beam at a point that is equidistant from the two wires. This load causes both wires to extend by 0.47 mm. (i) Determine the cross-sectional area of wire X. cross-sectional area = … m2 [3] (ii) Wire Y has a greater diameter than wire X. Explain, without calculation, whether the Young modulus of the metal from which wire Y is made is less than, the same as or greater than 1.9 × 109 Pa. … … … … [2] [Total: 7]

7 marks

Mark scheme: 4(a)(i) (normal) force per unit cross-sectional area B1 4(a)(ii) extension per unit unstretched length B1 4(b)(i) E = FL / Ax C1 A = (9.0  0.84) / (1.9  109  0.47  10–3) C1 = 8.5  10–6 m2 A1 4(b)(ii) F, L and x are all the same (for both wires / as in X) B1 or F and strain are the same A is greater (for Y), so the Young modulus (for Y) is less than 1.9  109 Pa or less than that of wire X B1

This question in 9702/23 Oct/Nov 2024

5 (a) Define the Young modulus. … … [1] (b) A wire of unstretched length 0.81 m is made of a metal with Young modulus 95 GPa. The wire obeys Hooke’s law and has a constant cross-sectional area. Fig. 5.1 shows the force–extension graph for the wire. 500 400 force / N 300 200 100 0 0 1 2 3 4 5 extension / 10–3 m Fig. 5.1 (i) Determine the cross-sectional area of the wire. area = … m2 [3] (ii) The extension of the wire is initially 2.0 × 10–3 m. Determine the work done to increase the extension of the wire to 3.0 × 10–3 m. work done = … J [3] [Total: 7]

7 marks

Mark scheme: 5(a) the ratio of stress to strain B1 5(b)(i) A = FL / Ex C1 A = e.g. (500  0.81) / (95  109  4.0  10–3) C1 A = 1.1  10–6 m2 A1 OR (C1) A = kL / E or A = gradient  L / E k = e.g. 500 / 4.0  10–3 k = 1.25  105 A = 1.25  105  0.81 / 95  109 (C1) A = 1.1  10–6 m2 (A1) 5(b)(ii) E = ½ kx2 or E = ½ Fx or E = area (under graph) C1 ()E = ½  1.25  105  ((3.0  10–3)2 – (2.0  10–3)2) C1 or ()E =(½  375  3.0  10–3) – (½  250  2.0  10–3) or ()E = ½  (375 + 250)  1.0  10–3 work done = 0.31 J A1

This question in 9702/22 May/June 2025

3 A wire has length L and cross-sectional area A. The wire is made from a metal that has Young modulus E and resistivity ρ. (a) Define the Young modulus of a material. … … [1] (b) (i) State an expression, in terms of some or all of L, A, E and ρ, for the resistance R0 of the wire. R0 = … [1] (ii) Show that the spring constant k0 of the wire is given by EA k0 = L . [2] (c) The wire is stretched, within the limit of proportionality, by a tensile force F. Assume that any changes in the cross-sectional area of the wire are negligible. (i) On Fig. 3.1, sketch the variation with F of the resistance R of the wire. R R0 0 0 F Fig. 3.1 [1] (ii) On Fig. 3.2, sketch the variation with F of the spring constant k of the wire. k k0 0 0 F Fig. 3.2 [1] (d) Copper has a resistivity of 1.8 × 10–8 Ω m and a Young modulus of 1.3 × 1011 Pa. A copper wire of diameter 1.6 mm has a resistance of 0.034 Ω. (i) Show that the length of the wire is 3.8 m. [1] (ii) Use the equation in (b)(ii) to determine the spring constant of the wire. spring constant = … N m–1 [2] [Total: 9]

9 marks

Mark scheme: 3(a) ratio of stress to strain B1 3(b)(i) R0 = L / A A1 3(b)(ii) k = F / x C1 E = FL / Ax = ((F / x)  (L / A)) = kL / A leading to k0 = EA / L A1 3(c)(i) straight line with positive gradient starting at (0, R0) B1 3(c)(ii) straight horizontal line starting at (0, k0) B1 3(d)(i) L = [0.034    (0.80  10–3)2] / 1.8  10–8 = 3.8 (m) A1 3(d)(ii) k = [1.3  1011    (0.80  10–3)2] / 3.8 C1 or k = EA / [RA / ] = E / R = [1.3  1011  1.8  10–8] / 0.034 k = 6.9  104 N m–1 A1

This question in 9702/23 Oct/Nov 2025