OCR A Level Physics A: Springs, stress, strain and the Young modulus (H556) – 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 cover sections 3.4.1 (Springs) and 3.4.2 (Mechanical properties of matter) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Use g = 9.81 N kg⁻¹. Revise first with the study guide or the revision notes. See the OCR A Level Physics hub, the checklist and the free diagnostics.

Questions

1. (a) State Hooke’s law, including the condition under which it applies. [1] (b) A spring of force constant 240 N m⁻¹ is compressed by 3.5 cm. Calculate the force acting on it. [2]

2. The table shows the force F applied to a spring and the extension x produced.

F / N	0	1.0	2.0	3.0	4.0	5.0	6.0
x / mm	0	8	16	24	32	43	58

(a) State, with a reason, the force at which the limit of proportionality is reached. [2] (b) Calculate the force constant of the spring. [2] (c) Calculate the work done in stretching the spring from zero to 6.0 N. [4]

3. A rubber band is loaded in steps and then unloaded in the same steps. (a) Describe the shape of the force-extension graph for loading and for unloading. [3] (b) Explain what the enclosed loop represents. [2]

4. A metal wire of diameter 0.80 mm and original length 1.85 m is stretched by a force of 88 N. The extension is 2.1 mm. (a) Calculate the stress in the wire. [2] (b) Calculate the strain. [1] (c) Calculate the Young modulus of the metal. [1] (d) The diameter is known to ±0.01 mm, the extension to ±0.1 mm and the length to ±5 mm. Treat the force as exact. Estimate the percentage uncertainty in the Young modulus and state the value with its absolute uncertainty. [3]

5. Describe how you would find the Young modulus of the metal in a long thin wire. Include the measurements, how you would reduce uncertainty, how the results are analysed and one safety precaution. [6]

6. A rod made of a ductile metal gives a straight stress-strain line up to a stress of 240 MPa at strain 0.00120. The maximum stress reached is 410 MPa, and the rod breaks at a strain of 0.22. (a) Calculate the Young modulus of the metal. [2] (b) State the ultimate tensile strength. [1] (c) Explain why the data show the metal is ductile and not brittle. [2] (d) A force of 1.8 kN acts on a rod of this metal of cross-sectional area 6.0 mm². Show that Hooke’s law is not obeyed. [2]

7. (a) Describe two ways the stress-strain graph of a brittle material differs from that of a ductile material. [2] (b) State one difference between how a rubber band and a polythene strip behave when loaded well beyond their linear region and then unloaded. [1] (c) Define the term plastic deformation. [1]

8. A spring-loaded launcher has force constant 320 N m⁻¹. It is compressed by 45 mm and releases a ball of mass 25 g along a smooth horizontal surface. Calculate the speed of the ball, assuming all the stored energy becomes kinetic energy. [4]

9. Two wires, A and B, are made from the same metal. B has twice the length of A and half the diameter. The same force is applied to each, and both stay within the limit of proportionality. Show that the extension of B is 8 times that of A. [3]

10. A metal wire is loaded. The force rises linearly from 0 to 30 N as the extension rises from 0 to 0.030 m. It then rises linearly to 42 N at an extension of 0.090 m. The load is then removed. (a) Calculate the work done on the wire during loading. [3] (b) Assuming the unloading line is parallel to the initial loading line, find the permanent extension. [2] (c) Calculate the energy dissipated as thermal energy. [2]

11. A climber of mass 70 kg falls freely 2.5 m before her rope becomes taut. The rope has force constant 1800 N m⁻¹, obeys Hooke’s law, and has negligible mass. (a) Calculate the maximum extension of the rope. [5] (b) Calculate the maximum force in the rope and compare it with the climber’s weight. [2]

Answers

1. (a) Force and extension (or compression) are proportional, so long as the limit of proportionality has not been passed. [1] (b) x = 0.035 m. [1] F = kx = 240 × 0.035 = 8.4 N. [1] Examiner insight: Show the conversion of centimetres to metres; a bare 8400 N comes from leaving x in centimetres.

2. (a) The extension rises by 8 mm for each newton up to 4.0 N. [1] At 5.0 N it is 43 mm, not 40 mm, so the limit is at about 4.0 N. [1] (b) k = F/x = 4.0 / 0.032 [1] = 125 N m⁻¹. [1] (c) Area up to 4.0 N = ½ × 4.0 × 0.032 = 0.064 J. [1] From 4.0 to 5.0 N: ½ × (4.0 + 5.0) × 0.011 = 0.0495 J. [1] From 5.0 to 6.0 N: ½ × (5.0 + 6.0) × 0.015 = 0.0825 J. [1] Total = 0.064 + 0.0495 + 0.0825 = 0.196 J. [1] Examiner insight: Quoting the change in extension per newton is stronger than “the line starts to curve”.

3. (a) The graph is a curve, not a straight line through the origin. [1] The unloading curve lies below the loading curve. [1] Both curves meet at the origin, so the band returns to its original length. [1] (b) The area under loading is the work done on the band and under unloading is the energy recovered. [1] The loop between them is energy dissipated as thermal energy in the rubber. [1] Examiner insight: “Energy lost” is vague. Say where it goes, and which area is which.

4. (a) A = π × (0.40 × 10⁻³)² = 5.03 × 10⁻⁷ m². [1] Stress = 88 / 5.03 × 10⁻⁷ = 1.75 × 10⁸ Pa. [1] (b) Strain = 2.1 × 10⁻³ / 1.85 = 1.14 × 10⁻³. [1] (c) E = 1.75 × 10⁸ / 1.14 × 10⁻³ = 1.5 × 10¹¹ Pa. [1] (d) Percentage uncertainties: diameter 0.01/0.80 = 1.25%, doubled to 2.5% in area; extension 4.8%; length 0.3%. [1] Total = 2.5 + 4.8 + 0.3 = 7.5%. [1] Absolute uncertainty = 7.5% of 1.54 × 10¹¹ ≈ 1.2 × 10¹⁰, so E = (1.5 ± 0.1) × 10¹¹ Pa. [1] Examiner insight: Doubling the percentage uncertainty in the diameter is the point most often missed.

5. Use a long thin wire clamped firmly at the top, kept straight by a small fixed weight, and measure the original length with a metre rule or tape. [1] Find the diameter with a micrometer at several positions and in perpendicular directions, average it, and calculate A = πd²/4. [1] Add equal masses in steps, recording the position of a marker against a fixed scale each time. [1] Unload and repeat to check the wire is still elastic. [1] Plot extension against force: the gradient is L/(EA), so E = L / (A × gradient). [1] Safety: wear eye protection and put sand or foam under the masses. [1] Examiner insight: Naming the instrument for each measurement, and saying what is plotted and how E comes from the gradient, covers more than a general “measure the extension”.

6. (a) E = stress/strain = 240 × 10⁶ / 0.00120 [1] = 2.0 × 10¹¹ Pa. [1] (b) The maximum stress is 410 MPa. [1] (c) The rod undergoes a large strain (0.22) beyond the linear region before it breaks. [1] A brittle material would fracture soon after the straight section. [1] (d) A = 6.0 × 10⁻⁶ m², so stress = 1800 / 6.0 × 10⁻⁶ = 3.0 × 10⁸ Pa. [1] This exceeds 240 MPa, so it is beyond the limit of proportionality. [1] Examiner insight: A “show that” answer needs the calculated stress and the comparison with 240 MPa. M means 10⁶.

7. (a) A brittle material’s graph is linear almost to the breaking point. [1] It has little or no plastic region and breaks suddenly, with no necking. [1] (b) Rubber returns to its original length on unloading, but polythene keeps a permanent extension. [1] (c) A permanent change in the size or shape remains after the force is removed. [1] Examiner insight: A definition of plastic deformation needs the idea of the force being removed.

8. x = 0.045 m. [1] E = ½kx² = ½ × 320 × 0.045² = 0.324 J. [1] ½mv² = 0.324, so v² = 2 × 0.324 / 0.025 = 25.9. [1] v = 5.1 m s⁻¹. [1] Examiner insight: Mass in kilograms and compression in metres are both needed; quoting the energy first shows the method.

9. x = FL/(EA) with A ∝ d². [1] B has L doubled, giving a factor of 2, and d halved, so A is a quarter, giving a factor of 4. [1] Extension ratio = 2 × 4 = 8. [1] Examiner insight: State the starting relationship, then apply each change separately and show the factors.

10. (a) Triangle: ½ × 30 × 0.030 = 0.45 J. [1] Trapezium: ½ × (30 + 42) × (0.090 - 0.030) = 2.16 J. [1] Total = 2.61 J. [1] (b) Initial slope = 30 / 0.030 = 1000 N m⁻¹. [1] The unloading line reaches zero force at 0.090 - 42/1000 = 0.048 m, so the permanent extension is 0.048 m. [1] (c) Energy recovered = ½ × 42 × (0.090 - 0.048) = 0.882 J. [1] Energy dissipated = 2.61 - 0.882 = 1.7 J. [1] Examiner insight: The recovered energy uses the triangle whose base is the extension recovered, 0.042 m, not the full 0.090 m.

11. (a) Loss of gravitational potential energy = mg(2.5 + x). [1] Gain in elastic energy = ½kx² = 900x². [1] Equating (no kinetic energy at the lowest point): 70 × 9.81 × (2.5 + x) = 900x². [1] So 900x² - 686.7x - 1716.75 = 0. [1] The positive root is x = 1.8 m. [1] (b) F = kx = 1800 × 1.81 = 3.3 kN. [1] The weight is 70 × 9.81 = 687 N, so the force is about 4.8 times the weight. [1] Examiner insight: The gravitational energy lost includes the extra distance x as well as the 2.5 m of free fall.

Where marks are usually lost

• Using the total length as the extension, or leaving mm, cm or kN unconverted.
• Using the diameter in place of the radius in the area, or forgetting that A is proportional to d².
• Treating the gradient of an extension-force graph as k.
• Using ½Fx for a curved or kinked graph.
• Calling a rubber band’s behaviour plastic, or its graph straight.
• Defining plastic deformation without saying what happens when the force is removed.
• Forgetting the percentage uncertainty in A is double that in d.
• Missing the extra extension in a gravitational energy balance.

Next steps

• Read the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick items off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 3.4.1 Springs and 3.4.2 Mechanical properties of matter.