OCR GCSE Physics A: Deformation, springs and Hooke's law (J249) – 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 statements P2.3a to P2.3f of the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018. All six statements are assessed in both tiers. Where a question needs the weight of a mass, the weight is given in the question.

Links: study guide, revision notes, OCR GCSE Physics hub, printable checklist and the free 10-minute diagnostics.

Questions

1. A spring hangs from a clamp with a mass on its lower end. The spring is stretched. Identify the two forces that stretch it. [2]

2. A paper clip is bent a little and springs back. It is then bent much further and stays bent. Describe the difference between the two kinds of deformation, and say which kind happens in each case. [3]

3. A spring has a natural length of 24.0 cm. A force of 6.0 N makes its length 30.0 cm. The spring is within its linear range. Calculate the spring constant. [3]

4. A spring has a spring constant of 130 N/m. Calculate the force needed to give an extension of 18 mm. Give your answer to 2 significant figures. [3]

5. A student hangs loads on a spring of natural length 30.0 cm.

Force (N)	0	4	8	12	16	20	24
Length (cm)	30.0	32.5	35.0	37.5	40.0	44.0	52.0

(a) Calculate the extension at 24 N. [1] (b) State where the limit of proportionality lies and give evidence from the table. [2] (c) Calculate the spring constant. [2] (d) Calculate the work done in stretching the spring to a force of 12 N. [2]

6. A toy dart launcher has a spring with a spring constant of 800 N/m. It is compressed by 3.5 cm. Calculate the energy stored. [3]

7. A spring has a spring constant of 120 N/m. (a) It stores 0.60 J of energy. Calculate its extension. [3] (b) The energy is doubled. Calculate the new extension. [2]

8. A spring stretched by 0.040 m stores 0.16 J of energy, within its linear range. (a) Calculate the spring constant. [2] (b) Calculate the energy stored at an extension of 0.080 m. [2] (c) Explain your answer to (b) in terms of the first extension. [1]

9. An elastic band is stretched in steps.

Extension (cm)	0	3	6	9	12
Force (N)	0	1.0	1.4	2.0	3.2

(a) Explain why E = ½ke² cannot be used to find the work done. [1] (b) Estimate the work done in stretching the band to 12 cm by treating each 3 cm step as a trapezium. [3] (c) When the load is removed the band returns to its original length. State the type of deformation and the type of relationship the table shows. [2]

10. Spring A extends by 3.0 cm and spring B by 7.5 cm when each is pulled by a force of 4.5 N. Both are within their linear ranges. (a) Calculate the spring constant of each. [3] (b) State which spring is stiffer, with a reason. [1] (c) Calculate the energy stored in each spring. [2]

11. Describe an investigation to find the spring constant of a spring and to decide where it stops obeying Hooke’s law. Take the weight of 100 g as 1.0 N. [6]

12. A pinball plunger contains a spring of spring constant 450 N/m and natural length 12.0 cm. The player pushes the plunger so the spring is 8.0 cm long. (a) State the compression. [1] (b) Calculate the force needed to hold it there. [2] (c) Calculate the energy stored. [2] (d) The spring is compressed further to a length of 6.0 cm. Calculate the new energy stored and the factor by which it has increased. [2] (e) Explain why a force is needed at each end of the spring. [1]

Answers

1. The weight of the mass (or load) pulls the bottom of the spring downwards [1]. The clamp pulls the top of the spring upwards [1]. [2] Examiner insight: “gravity and tension” earns less than naming where each force acts, so say which end of the spring each force pulls.

2. Elastic deformation: the object goes back to its starting shape once the forces stop [1]. Plastic deformation: the object stays permanently deformed [1]. The small bend is elastic and the large bend is plastic [1]. [3] Examiner insight: the test is what happens after the forces are removed, so an answer that only says “elastic bends and plastic breaks” tends to lose the first two marks.

3. Extension = 30.0 − 24.0 = 6.0 cm [1] = 0.060 m [1]. k = F / e = 6.0 / 0.060 = 100 N/m [1]. [3] Examiner insight: using 30.0 cm as the extension is the most common slip, and writing out the substitution lets a reader see which step went wrong.

4. e = 18 mm = 0.018 m [1]. F = ke = 130 × 0.018 [1] = 2.34 N = 2.3 N [1]. [3] Examiner insight: conversion to metres is the step to show. Quoting 2.34 N is fine as a working value, but the question asks for 2 significant figures, so end with 2.3 N.

5. (a) 52.0 − 30.0 = 22.0 cm [1]. (b) Between 16 N and 20 N, close to 16 N [1]. Up to 16 N each extra 4 N adds 2.5 cm, but from 16 N to 20 N it adds 4.0 cm, so the line has stopped being straight [1]. (c) k = F / e with a linear point, 16 N and 0.100 m [1]; k = 160 N/m [1]. (d) e at 12 N = 7.5 cm = 0.075 m [1]. E = ½ × 12 × 0.075 = 0.45 J [1]. [7] Examiner insight: take k from a point before the limit of proportionality. A point such as 20 N and 0.14 m gives a value that does not describe the spring’s linear behaviour.

6. e = 3.5 cm = 0.035 m [1]. E = ½ke² = ½ × 800 × 0.035² [1] = 0.49 J, so 0.49 J [1]. [3] Examiner insight: squaring only the 0.035 and then multiplying by ½ and 800 is the sequence to show. Squaring the product ke gives a different and wrong value.

7. (a) E = ½ke², so e² = 2E / k [1] = (2 × 0.60) / 120 = 0.010 [1]; e = 0.10 m [1]. (b) e² = (2 × 1.20) / 120 = 0.020 [1]; e = 0.14 m (about 1.41 × 0.10 m) [1]. [5] Examiner insight: doubling the energy does not double the extension. Answers that state 0.20 m treat the energy as proportional to the extension.

8. (a) k = 2E / e² = (2 × 0.16) / 0.040² [1] = 200 N/m [1]. (b) E = ½ × 200 × 0.080² [1] = 0.64 J [1]. (c) The extension has doubled, so the energy is four times as large, since E depends on e² [1]. [5] Examiner insight: a one-line reason that mentions “squared” is what the explanation needs. “More stretch, more energy” does not use the numbers.

9. (a) The graph is a curve, so force is not proportional to extension and there is no single spring constant [1]. (b) Strip areas in N cm: (0 + 1.0)/2 × 3 = 1.5; (1.0 + 1.4)/2 × 3 = 3.6; (1.4 + 2.0)/2 × 3 = 5.1; (2.0 + 3.2)/2 × 3 = 7.8 [1]. Total = 18.0 N cm [1]. 100 N cm = 1 J, so 0.18 J [1]. (c) Elastic deformation [1]. Non-linear relationship [1]. [6] Examiner insight: many answers leave the work done in N cm. State the conversion to joules, and say “estimate”, since straight-sided strips only approximate a curve.

10. (a) 3.0 cm = 0.030 m and 7.5 cm = 0.075 m [1]. k(A) = 4.5 / 0.030 = 150 N/m [1]. k(B) = 4.5 / 0.075 = 60 N/m [1]. (b) Spring A is stiffer because it has the larger spring constant (it extends less for the same force) [1]. (c) E = ½Fe: A = ½ × 4.5 × 0.030 = 0.0675 J [1]; B = ½ × 4.5 × 0.075 = 0.17 J [1]. [6] Examiner insight: comparing stiffness needs the same force or the values of k, not the extensions alone. Stating a reason, not just “A”, is what earns the mark.

11. Measure the natural length of the spring with a ruler [1]. Add 100 g masses one at a time, each adding 1.0 N, and measure the new length each time with the ruler beside the spring [1]. Work out each extension as length minus natural length [1]. Plot force (vertical) against extension (horizontal) [1]. The gradient of the straight part gives the spring constant [1]. The limit is where the points stop following the straight line, and unloading to check the spring returns to its natural length shows whether it was still elastic [1]. [6] Examiner insight: a plan reads as complete when it says what is measured, what is calculated, what is plotted and how the result is read from the graph. A list of equipment alone is not a method.

12. (a) 12.0 − 8.0 = 4.0 cm (0.040 m) [1]. (b) F = ke = 450 × 0.040 [1] = 18 N [1]. (c) E = ½ × 450 × 0.040² [1] = 0.36 J [1]. (d) Compression 6.0 cm = 0.060 m, E = ½ × 450 × 0.060² = 0.81 J [1]. Factor = 0.81 / 0.36 = 2.25 [1]. (e) The spring must be pushed from both ends, by the player on one end and the frame or ball on the other, because one force alone would just accelerate the spring [1]. [8] Examiner insight: in (d) the compression is 12.0 − 6.0 = 6.0 cm, not the final length of the spring, so check which quantity is squared.

Where marks are usually lost

• Using the stretched length as the extension.
• Substituting centimetres into F = ke or E = ½ke².
• Omitting the square on the extension in E = ½ke².
• Giving a bare answer with no unit, or newtons for energy.
• Reading the spring constant from the curved part of a graph.
• Saying “plastic” when a graph is merely curved.
• Describing an investigation without saying how the result is obtained from the graph.
• Stopping at a number in N cm when the question asks for joules.
• Stating that a single force can stretch an object that has nothing holding the other end.

Next steps

• Re-read the study guide for any question you missed.
• Condense the unit with the revision notes.
• Tick off the unit on the printable checklist and return to the OCR GCSE Physics hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment 2018. Statements P2.3a to P2.3f.