Forces and elasticity (RP6) – GCSE Physics exam questions with mark scheme

1Grade 1–3Foundation4 marks

A diver stands still on the end of a diving board. The other end of the board is bolted to a fixed support. The board bends.

(a) The weight of the diver acts downwards on the end of the board.
Explain why the board would not bend if the weight of the diver were the only force acting on it. [2]

(b) When the diver jumps off, the board goes back to its original shape.
What type of deformation did the board undergo? [1]

(c) A child presses a thumb into a ball of modelling clay. A dent is left in the clay after the thumb is removed.
Give the reason why this is an example of inelastic deformation. [1]

(a) the board would just move / accelerate downwards (as a whole) (allow: fall; allow: there would be a resultant force on it) — ignore 'it would break'(1)
(a) a second force is needed in a different direction / at a different point, e.g. the upward force from the support (on the bolted end) (allow: more than one force is needed to change the shape of an object)(1)
(b) elastic (deformation) — do not accept 'plastic'(1)
(c) the clay does not return to its original shape (when the force is removed) (allow: it stays deformed / the dent stays; allow: does not go back to its original shape)(1)
Answer: (a) A single force would just move the board (accelerate it downwards); a second force from the support, acting at a different point/in a different direction, is needed to bend it. (b) Elastic. (c) It does not return to its original shape when the force is removed.
Examiner tip: 'Elastic' means it goes back to its original shape when the forces are removed — write the full idea, not just 'it bends back'.

2Grade 4–5Foundation4 marks

A spring inside a toy launcher has a spring constant of 40 N/m.

(a) Write down the equation that links extension ($e$), force ($F$) and spring constant ($k$). [1]

(b) The spring is stretched by 6.0 cm. The spring does not go past its limit of proportionality.
Calculate the force stretching the spring. [3]

(a) $F = k\,e$ (allow: force = spring constant × extension; allow any correct rearrangement, e.g. $k = F/e$)(1)
(b) conversion of extension: 6.0 cm = 0.060 m(1)
(b) $F = 40 \times 0.060$ (substitution with extension in m)(1)
(b) $F = 2.4$ N — allow 2.4 N with no working for 3 marks; an answer of 240 (N) scores 1 mark (correct substitution with e in cm, i.e. 40 × 6.0)(1)
Answer: (a) $F = ke$ (b) 2.4 N
Examiner tip: Spring constants are in N/m, so change centimetres into metres before you substitute — 6.0 cm is 0.060 m.

3Grade 6–76 marks

A pinball machine uses a spring to launch a ball. The spring is compressed by 4.0 cm and then released.

The spring constant of the spring is 300 N/m. The spring is not compressed past its limit of proportionality.

(a) Calculate the elastic potential energy stored in the compressed spring.
Use the Physics Equations Sheet. [3]

(b) The mass of the ball is 0.080 kg. The ball leaves the spring at a speed of 2.0 m/s.
Calculate the kinetic energy of the ball as it leaves the spring. [2]

(c) Calculate the efficiency of the transfer of energy from the elastic potential energy store of the spring to the kinetic energy store of the ball. [1]

(a) 4.0 cm = 0.040 m(1)
(a) $E_e = \tfrac{1}{2} \times 300 \times 0.040^2$(1)
(a) $E_e = 0.24$ J — allow 0.24 J with no working for 3 marks; 2400 (J) scores 1 mark (correct substitution with e in cm)(1)
(b) $E_k = \tfrac{1}{2} \times 0.080 \times 2.0^2$(1)
(b) $E_k = 0.16$ J — allow 0.16 J with no working for 2 marks; 0.080 (J) (speed not squared) scores 1 mark(1)
(c) 0.67 or 67% (allow 0.667 / 66.7%; allow 2/3; allow ecf: their (b) ÷ their (a)) — a value greater than 1 (100%) does not score(1)
Answer: (a) 0.24 J (b) 0.16 J (c) 0.67 (67%)
Examiner tip: The equation $E_e = \tfrac12 ke^2$ works for compression too — $e$ is how far the spring is squashed, in metres; square it before multiplying by $\tfrac12 k$.

4Grade 4–56 marks

A student hung weights on a spring and measured its extension. The graph shows her results.

(a) Describe the relationship between force and extension for forces up to 4.0 N. [1]

(b) Give the name of the point on the graph after which the line is no longer straight. [1]

(d) Use the graph to determine the extension of the spring when the force is 2.5 N. [1]

(a) (directly) proportional (allow: extension doubles when force doubles; allow: linear and passes through the origin) — 'linear' alone or 'positive correlation' alone does not score(1)
(b) limit of proportionality — do not accept elastic limit(1)
(c) correct pair of values read from the straight section, e.g. 4.0 N and 8.0 cm (allow any correct pair, e.g. 2.0 N and 4.0 cm)(1)
(c) $k = \dfrac{4.0}{0.080}$ (extension converted to m)(1)
(c) $k = 50$ N/m — allow 50 with no working for 3 marks; 0.5 (N/cm) scores 2 marks(1)
(d) 5.0 cm (allow 4.9–5.1 cm)(1)
Answer: (a) Force is directly proportional to extension. (b) Limit of proportionality. (c) 50 N/m. (d) 5.0 cm.
Examiner tip: Only use readings from the straight part of the graph to find $k$ — points past the limit of proportionality give the wrong value.

5Grade 6–76 marks

A student wants to investigate the relationship between the force applied to a spring and the extension of the spring.

The diagram shows the equipment she has. Each slotted mass has a mass of 100 g.

Describe a method the student could use to obtain valid results. Include any steps she should take to reduce the risk of harm. [6]

clamp the metre rule vertically beside the spring, with the zero level with the top of the spring (or note the starting reading)(1)
measure the original (unstretched) length of the spring / record the position of the pointer with no masses (or only the hanger) on it(1)
add one 100 g mass at a time(1)
after each mass is added, read the new position of the pointer / new length of the spring(1)
read the ruler at eye level / perpendicular to the scale to avoid parallax error(1)
use a pointer fixed to the bottom of the spring to make readings more precise(1)
calculate extension = new length − original length(1)
calculate the force (weight) from $W = mg$, e.g. 100 g gives about 1 N(1)
keep using the same spring throughout(1)
repeat the readings (e.g. while removing the masses) and calculate a mean(1)
check the spring returns to its original length when masses are removed, i.e. it has not passed its elastic limit(1)
plot a graph of force against extension(1)
safety: wear eye protection in case the spring snaps / flies off(1)
safety: clamp the stand to the bench or place a counterweight on the base so it does not topple(1)
safety: do not stand with feet under the masses / put a soft surface (tray) under the hanger(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered method that could be followed to give valid force and extension readings. Shows how extension is found from the original length, how readings are made accurate, and includes at least one relevant safety step.
Level 2	3–4	A method with most of the key steps, but with gaps or steps out of order; it would produce some useful results. How extension is found or the safety precaution may be missing or unclear.
Level 1	1–2	Some relevant steps or isolated statements (e.g. 'add masses and measure the spring') that are not linked into a method that would give valid results.
Level 0	0	No relevant content.

Answer: Measure the unstretched length with the vertical metre rule and pointer; add 100 g (≈1 N) masses one at a time, reading the pointer at eye level each time; extension = new length − original length; repeat and take means; plot force against extension. Safety: eye protection, clamp the stand to the bench, keep feet clear of the masses.
Examiner tip: Top-band answers say how extension is calculated (new length minus original length) — just 'measure the extension' keeps you in Level 2.

6Grade 4–57 marks

A student measured the length of a spring as she added weights to it. The unstretched length of the spring was 24.0 cm.

Weight in N	Length 1 in cm	Length 2 in cm	Length 3 in cm	Mean length in cm
1.0	28.1	27.9	28.0	28.0
2.0	32.0	32.2	32.1	32.1
3.0	35.9	36.1	36.0	36.0
4.0	40.2	43.6	39.8	X
5.0	44.1	43.9	44.0	44.0

(a) Identify the anomalous result in the table. [1]

(b) Calculate the value of X. [1]

(d) Use your answer to (c) to calculate the spring constant of the spring, in N/m. [2]

(e) The student concludes that the extension of the spring is directly proportional to the weight.
Use data from the table to support her conclusion. [2]

(a) 43.6 (cm) (allow: Length 2 for 4.0 N)(1)
(b) X = 40.0 (cm) (allow 40) — 41.2 (including the anomaly) does not score(1)
(c) 16.0 cm (allow ecf: their X − 24.0; allow 0.16 m)(1)
(d) $k = \dfrac{4.0}{0.160}$ (allow ecf from (c), extension must be in m)(1)
(d) $k = 25$ N/m — allow 25 with no working for 2 marks; 0.25 (N/cm) scores 1 mark(1)
(e) calculates at least two extensions, e.g. 1.0 N → 4.0 cm, 2.0 N → 8.1 cm, 3.0 N → 12.0 cm, 5.0 N → 20.0 cm(1)
(e) shows that doubling the weight doubles the extension (e.g. 2.0 N → 8.1 cm and 4.0 N → 16.0 cm; allow: 1.0 N → 4.0 cm and 5.0 N → 20.0 cm) or that weight ÷ extension is (about) constant (≈ 0.25 N/cm) — using lengths instead of extensions scores 0 for this mark(1)
Answer: (a) 43.6 cm (b) 40.0 cm (c) 16.0 cm (d) 25 N/m (e) Extensions 4.0, 8.1, 12.0, 16.0, 20.0 cm: doubling the force doubles the extension (F/e ≈ constant).
Examiner tip: The table gives LENGTH, not extension — subtract the unstretched 24.0 cm before you look for proportionality or calculate $k$.

7Grade 8–9Higher6 marks

A spring has a spring constant of 200 N/m. When it is stretched, 0.90 J of energy is stored in its elastic potential energy store. The spring has not gone past its limit of proportionality.

(a) Calculate the extension of the spring. Give your answer in cm. [3]

(b) Calculate the force needed to hold the spring at this extension. [1]

(c) A student says the energy stored should equal force × extension.
Explain why the energy stored is only half of force × extension. [2]

(a) $0.90 = \tfrac{1}{2} \times 200 \times e^2$(1)
(a) $e^2 = 0.0090$ / $e = \sqrt{0.0090} = 0.095$ m (allow 0.0949 m)(1)
(a) $e = 9.5$ cm (allow 9.49 cm) — 9.5 cm with no working scores 3 marks; 0.095 (m) without conversion scores 2 marks(1)
(b) $F = 200 \times 0.095 = 19$ N (allow ecf from (a) with e in m; allow 18.97 N or 18.9–19.0 N)(1)
(c) the force is not constant — it increases from zero (to 19 N) as the spring is stretched (allow: force is proportional to extension)(1)
(c) so the average force is half the final force (allow: work done/energy stored is the area under the force–extension graph, which is a triangle, $\tfrac{1}{2}Fe$)(1)
Answer: (a) 9.5 cm (b) 19 N (c) The force rises from 0 to 19 N as the spring stretches, so the average force (and the triangular area under the F–e graph) is half of $F \times e$.
Examiner tip: When rearranging $E_e = \tfrac12 ke^2$, double the energy and divide by $k$ first, then take the square root — forgetting the square root is the most common slip.

8Grade 6–76 marks

The graph shows how the force applied to two springs, A and B, changes with their extension. Neither spring goes past its limit of proportionality.

(a) Which spring is stiffer? Give a reason for your answer using the graph. [2]

(b) A force of 2.0 N is applied to each spring.
Show that spring A stores twice as much elastic potential energy as spring B.
Use the Physics Equations Sheet. [4]

(a) (spring) B(1)
(a) B needs a larger force for the same extension / B stretches less for the same force (allow: B has the steeper line / larger gradient, so a larger spring constant) — reason only scores if B chosen(1)
(b) extensions read from graph at 2.0 N: A = 5.0 cm (0.050 m) and B = 2.5 cm (0.025 m)(1)
(b) spring constants: $k_A = \dfrac{2.0}{0.050} = 40$ N/m and $k_B = \dfrac{2.0}{0.025} = 80$ N/m (allow from any correct pair of graph values)(1)
(b) $E_A = \tfrac{1}{2} \times 40 \times 0.050^2 = 0.050$ J (allow ecf)(1)
(b) $E_B = \tfrac{1}{2} \times 80 \times 0.025^2 = 0.025$ J and $0.050 = 2 \times 0.025$ (allow comparison shown as a ratio) — alternative method: energy = area under graph = $\tfrac12 \times 2.0 \times 0.050$ and $\tfrac12 \times 2.0 \times 0.025$ scores full marks(1)
Answer: (a) B — it needs a larger force for the same extension (steeper line). (b) $k_A = 40$ N/m, $k_B = 80$ N/m; $E_A = 0.050$ J, $E_B = 0.025$ J, so A stores twice as much.
Examiner tip: In a 'show that' question every step must be written down — including the extensions you read off the graph, converted to metres.

9Grade 1–3Foundation3 marks

(a) Complete the sentence.
When a spring is stretched, work is done on it and energy is transferred to the ____________ energy store of the spring. [1]

(b) A spring has been stretched past its elastic limit and then the force is removed.
Tick (✓) one box.

☐ The spring returns to its original length.
☐ The spring stays longer than its original length.
☐ The spring becomes shorter than its original length. [1]

(c) A force of 2.0 N stretches a spring by 3.0 cm. The spring obeys Hooke's law (force is proportional to extension).
What is the extension when the force is 6.0 N? [1]

(a) elastic potential (allow: elastic; do not accept 'potential' alone or 'gravitational potential')(1)
(b) 'The spring stays longer than its original length' only — more than one box ticked scores 0(1)
(c) 9.0 cm (allow 9 cm; allow 0.090 m)(1)
Answer: (a) elastic potential (b) The spring stays longer than its original length. (c) 9.0 cm
Examiner tip: If the force is 3 times bigger, the extension is 3 times bigger — but only while the spring is below its limit of proportionality.

10Grade 8–9Higher6 marks

A car of mass 1200 kg rests on four identical suspension springs. The weight of the car is shared equally between the four springs. Each spring is compressed by 3.0 cm.

Gravitational field strength = 9.8 N/kg

(a) Calculate the spring constant of one spring. [4]

(b) A passenger of mass 75 kg gets into the car. Assume the passenger's weight is also shared equally between the four springs and the springs do not go past their limit of proportionality.
Calculate the extra compression of each spring. Give your answer in mm. [2]

(a) $W = 1200 \times 9.8 = 11\,760$ N(1)
(a) force on one spring $= 11\,760 \div 4 = 2940$ N(1)
(a) $k = \dfrac{2940}{0.030}$(1)
(a) $k = 98\,000$ N/m (allow 9.8 × 10⁴ N/m) — correct answer with no working scores 4 marks; 392 000 N/m (weight not shared) scores 2 marks; 980 (N/m) (e not converted to m) scores 3 marks(1)
(b) extra force per spring $= 75 \times 9.8 \div 4 = 183.75$ N and $e = \dfrac{183.75}{98\,000}$ (allow ecf from (a); allow proportion method: extra compression $= 30 \text{ mm} \times \dfrac{75}{1200}$)(1)
(b) $e = 1.9$ mm (allow 1.875 mm / 1.88 mm; allow ecf) — 0.0019 (m) without conversion does not score this mark(1)
Answer: (a) 98 000 N/m (b) 1.9 mm
Examiner tip: $F = ke$ applies to compression as well as stretching — but use the force on ONE spring, not the whole weight of the car.

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