OxfordAQA International AS and A-level Physics: Bulk properties of solids and the Young modulus (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow sections 3.2.9 and 3.2.10 of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4 (shared AS and A-level content). Take g as 9.81 N kg⁻¹. Revise first with the study guide if needed. Course pages: OxfordAQA A-level Physics hub and the printable checklist. For more practice, try the free diagnostics.

Questions

1. A metal bar, 8.0 cm by 3.0 cm by 1.5 cm, has mass 0.2844 kg. Calculate its density. [3]

2. A spring is loaded in steps. The extension is 0, 5, 10, 15, 20, 27 and 38 mm for forces of 0, 2, 4, 6, 8, 10 and 12 N.

(a) Show that Hooke’s law is obeyed up to 8 N but not at 10 N. [2] (b) Calculate the spring constant. [2]

3. A cable of diameter 6.0 mm breaks when the tension reaches 21 kN.

(a) Calculate the breaking stress of the material. [2] (b) A cable of the same material has diameter 9.0 mm. Calculate the tension at which it breaks. [2]

4. A spring of stiffness 150 N m⁻¹ obeys Hooke’s law. Calculate the energy stored when its extension is (a) 0.080 m [2], and (b) 0.120 m, and the extra work done in stretching it from 0.080 m to 0.120 m. [2]

5. A wire is loaded so that the force rises in a straight line from 0 to 40 N as the extension increases from 0 to 2.0 mm. The force then rises in a straight line to 60 N at an extension of 6.0 mm. The force is then reduced to zero along a straight line parallel to the first line.

(a) Find the work done in stretching the wire from zero to 6.0 mm. [2] (b) Determine the permanent extension. [2] (c) Calculate the energy recovered on unloading, and the energy dissipated. State what happens to the dissipated energy. [2]

6. Describe the force-extension graph and the behaviour on fracture of (a) a glass rod [2], and (b) a copper wire loaded beyond its elastic limit. [2]

7. A spring of stiffness 420 N m⁻¹ is compressed by 0.045 m and used to fire a 0.025 kg ball vertically upwards.

(a) Calculate the elastic strain energy stored. [1] (b) Calculate the maximum height above the release position, assuming no energy is lost. [2] (c) The ball actually rises 1.40 m above the release position. Calculate the energy dissipated. [2]

8. A copper wire of original length 2.20 m and diameter 0.46 mm extends by 3.4 mm under a load of 28 N. Calculate the Young modulus of copper and the energy stored in the wire. [5]

9. Wire A has length L and diameter d. Wire B is made of the same material, with length 2L and diameter 2d. The same force acts on each and neither passes its elastic limit. Give the ratio B : A for (a) the stress [1], (b) the extension [2], and (c) the energy stored. [2]

10. A student finds the Young modulus of a steel wire of original length 1.85 m. The diameter is 0.28 ± 0.01 mm.

(a) Describe how to measure the diameter to allow for variation along the wire. [2] (b) The gradient of the line of best fit on a graph of extension against force is 1.5 × 10⁻⁴ m N⁻¹. Calculate the Young modulus. [3] (c) Calculate the percentage uncertainty in the cross-sectional area, and say how it affects the uncertainty in the Young modulus. [2] (d) Explain how the student can check that the wire has not been stretched beyond its elastic limit. [1]

11. A wire of length 3.0 m and diameter 1.0 mm has a stress-strain graph that is a straight line from the origin to the elastic limit at strain 0.0018 and stress 290 MPa. The graph then curves to a breaking point at a stress of 380 MPa.

(a) Calculate the Young modulus. [2] (b) Calculate the force at the elastic limit. [2] (c) Calculate the extension at the elastic limit. [1] (d) Calculate the energy stored in the wire at the elastic limit. [2] (e) Calculate the force needed to break the wire. [1]

12. A 0.40 kg mass is attached to a vertical spring of stiffness 65 N m⁻¹ and released from rest with the spring at its natural length.

(a) Calculate the extension when the mass is at equilibrium. [1] (b) Use energy conservation to find the maximum extension. [3] (c) Calculate the speed of the mass as it passes the equilibrium position. [2]

Answers

1. V = 8.0 × 3.0 × 1.5 = 36 cm³ = 3.6 × 10⁻⁵ m³ [1]; ρ = m/V = 0.2844 / 3.6 × 10⁻⁵ [1] = 7.9 × 10³ kg m⁻³ [1]. Examiner insight: Show the cm³ to m³ conversion, so a wrong power of ten can still be traced.

2. (a) F/ΔL = 0.40 N mm⁻¹ for every reading up to 8 N, so force is proportional to extension [1]; at 10 N, F/ΔL = 0.37 N mm⁻¹, so it is no longer proportional [1]. (b) k = F/ΔL = 8 / 0.020 [1] = 400 N m⁻¹ [1]. Examiner insight: Support “proportional” with a calculated ratio, and convert millimetres to metres.

3. (a) A = π × (3.0 × 10⁻³)² = 2.83 × 10⁻⁵ m² [1]; stress = 21 × 10³ / 2.83 × 10⁻⁵ = 7.4 × 10⁸ Pa [1]. (b) A = π × (4.5 × 10⁻³)² = 6.36 × 10⁻⁵ m² [1]; F = 7.43 × 10⁸ × 6.36 × 10⁻⁵ = 4.7 × 10⁴ N [1]. Examiner insight: Breaking stress belongs to the material, so reuse the part (a) value. Using the diameter as the radius affects both parts.

4. (a) F = 150 × 0.080 = 12 N [1]; energy = ½ × 12 × 0.080 = 0.48 J [1]. (b) Energy at 0.120 m = ½ × 150 × 0.120² = 1.08 J [1]; extra work = 1.08 − 0.48 = 0.60 J [1]. Examiner insight: “Extra work” is a difference of two stored energies, not the energy at the new extension.

5. (a) Work = area = ½ × 40 × 2.0 + ½ × (40 + 60) × 4.0 [1] = 40 + 200 = 240 mJ = 0.24 J [1]. (b) The initial gradient is 40 / 2.0 = 20 N mm⁻¹, so unloading from 60 N removes 60 / 20 = 3.0 mm [1]; permanent extension = 6.0 − 3.0 = 3.0 mm [1]. (c) Recovered = ½ × 60 × 3.0 = 90 mJ [1]; dissipated = 240 − 90 = 0.15 J, transferred to internal energy of the wire [1]. Examiner insight: Say where the dissipated energy goes; “lost” alone does not answer the question.

6. (a) Straight line up to fracture [1]; it breaks suddenly at a small extension with no plastic deformation (brittle) [1]. (b) After the straight section the graph curves and a large plastic extension occurs [1]; unloading leaves a permanent extension, with the line parallel to the initial gradient [1]. Examiner insight: Describe the graph. “Glass is brittle” alone is a label, not a description.

7. (a) Energy = ½ × 420 × 0.045² = 0.425 J [1]. (b) mgh = 0.425 [1]; h = 0.425 / (0.025 × 9.81) = 1.73 m [1]. (c) Gravitational potential energy gained = 0.025 × 9.81 × 1.40 = 0.343 J [1]; dissipated = 0.425 − 0.343 = 0.082 J [1]. Examiner insight: Write the energy equation in words first; it shows that the height is measured from the release position.

8. A = π × (0.23 × 10⁻³)² = 1.66 × 10⁻⁷ m² [1]; stress = 28 / 1.66 × 10⁻⁷ = 1.68 × 10⁸ Pa [1]; strain = 3.4 × 10⁻³ / 2.20 = 1.55 × 10⁻³ [1]; E = 1.68 × 10⁸ / 1.55 × 10⁻³ = 1.09 × 10¹¹ Pa [1]; energy = ½ × 28 × 3.4 × 10⁻³ = 0.048 J [1]. Examiner insight: Show stress and strain separately: one long substitution leaves no partial working to credit if a number is entered wrongly.

9. (a) Area of B is four times that of A, so stress ratio B : A = 1 : 4 [1]. (b) ΔL = FL/(AE) [1]; B has twice the length and four times the area, so the extension ratio is 2 / 4 = 1 : 2 [1]. (c) Energy = ½FΔL [1], and F is the same, so the ratio is 1 : 2 [1]. Examiner insight: Start from ΔL = FL/(AE) and compare factors; a bare ratio gives little to credit.

10. (a) Use a micrometer at several positions along the wire [1], and in two perpendicular directions at each position, then take the mean [1]. (b) A = π × (0.14 × 10⁻³)² = 6.16 × 10⁻⁸ m² [1]; E = L / (A × gradient) [1] = 1.85 / (6.16 × 10⁻⁸ × 1.5 × 10⁻⁴) = 2.0 × 10¹¹ Pa [1]. (c) Uncertainty in d = 0.01 / 0.28 = 3.6%, so in A it is 2 × 3.6 = 7.1% [1]; this contributes the same amount to the percentage uncertainty in E, and is likely to be its largest part [1]. (d) Remove the load in steps and check that the reading returns to its original value (or that the graph is linear) [1]. Examiner insight: A calculated percentage, with the squared diameter explained, is more convincing than “the diameter has the largest error”.

11. (a) E = stress / strain = 290 × 10⁶ / 0.0018 [1] = 1.6 × 10¹¹ Pa [1]. (b) A = π × (0.50 × 10⁻³)² = 7.85 × 10⁻⁷ m² [1]; F = 290 × 10⁶ × 7.85 × 10⁻⁷ = 228 N [1]. (c) ΔL = strain × L = 0.0018 × 3.0 = 5.4 mm [1]. (d) Energy = ½FΔL = ½ × 228 × 5.4 × 10⁻³ [1] = 0.615 J [1]. (e) F = 380 × 10⁶ × 7.85 × 10⁻⁷ = 298 N [1]. Examiner insight: Reuse the area from part (b) and keep spare digits until the last line; rounded intermediates can shift the final figure.

12. (a) mg = kx, so x = 0.40 × 9.81 / 65 = 0.0604 m [1]. (b) At maximum extension the mass is momentarily at rest [1]; mgx = ½kx² [1], so x = 2mg/k = 2 × 0.40 × 9.81 / 65 = 0.121 m [1]. (c) ½mv² = mgx − ½kx² at x = 0.0604 m [1]; v = 0.77 m s⁻¹ [1]. Examiner insight: Say that the mass is at rest at maximum extension to justify mgx = ½kx². At the equilibrium position it is still moving, so kinetic energy must appear in the equation.

Where marks are usually lost

• Using the diameter as the radius in A = πd²/4, or leaving the diameter in millimetres.
• Forgetting the ½ in strain energy, or using ½kΔL² for a graph that is not straight.
• Mixing up the limit of proportionality and the elastic limit.
• Treating the gradient of a force-extension graph as E rather than k.
• Measuring the height of a spring-launched object from the wrong position.

Next steps

• Revise from the revision notes and the study guide.
• Tick off topics on the checklist and visit the course hub.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.2.9 and 3.2.10.