Energy stores and changes in energy (KE, GPE, elastic PE) – GCSE Physics exam questions with mark scheme

1Grade 1–3Foundation3 marks

Name the energy store that increases in each situation.

(a) A book is lifted from the floor onto a high shelf. [1]

(b) A catapult's rubber band is pulled back. [1]

(a) gravitational potential (energy store) allow gravitational; 'potential' alone does not score(1)
(b) elastic potential (energy store) allow elastic / strain(1)
(c) thermal (energy store) allow internal; 'heat' alone does not score(1)
Answer: (a) gravitational potential (b) elastic potential (c) thermal
Examiner tip: Write the store name in full: 'potential' alone does not score — say which kind of potential.

2Grade 4–53 marks

A car of mass 1200 kg travels at a speed of 15 m/s.

(a) Write down the equation that links kinetic energy ($E_k$), mass ($m$) and speed ($v$). [1]

(b) Calculate the kinetic energy of the car. [2]

(a) $E_k = \tfrac{1}{2} m v^2$ allow kinetic energy = 0.5 × mass × speed²(1)
(b) $E_k = 0.5 \times 1200 \times 15^2$(1)
(b) 135 000 J allow 135 kJ(1)
Answer: (a) $E_k=\tfrac12 mv^2$ (b) 135 000 J
Examiner tip: Square the speed only — not the mass and the speed together. $0.5 \times 1200 \times 15$ is a common slip.

3Grade 4–55 marks

A climber of mass 65 kg climbs a wall of height 12 m.

gravitational field strength = 9.8 N/kg

(a) Calculate the increase in the climber's gravitational potential energy. [2]

(b) The climb takes 40 s. Calculate the useful power output of the climber.

Give your answer to 2 significant figures. [3]

(a) $\Delta E_p = 65 \times 9.8 \times 12$(1)
(a) 7644 J allow 7600 J(1)
(b) $P = \dfrac{7644}{40}$ allow ecf from (a)(1)
(b) 191(.1) (W)(1)
(b) 190 W (to 2 sf)(1)
Answer: (a) 7644 J (b) 190 W
Examiner tip: The 2 sf mark is separate — write the unrounded value first, then round, so you keep the earlier marks.

4Grade 6–7Higher5 marks

A spring in a toy launcher has a spring constant of 40 N/m. The spring is compressed by 0.15 m.

(a) Calculate the elastic potential energy stored in the spring.

Use the equation: elastic potential energy $= \tfrac{1}{2} \times$ spring constant $\times$ (extension)$^2$ [2]

(b) The spring launches a ball of mass 0.020 kg horizontally. Assume all the elastic potential energy is transferred to the kinetic energy store of the ball.

Calculate the speed of the ball. Give your answer to 2 significant figures. [3]

(a) $E_e = 0.5 \times 40 \times 0.15^2$(1)
(a) 0.45 J(1)
(b) $0.45 = 0.5 \times 0.020 \times v^2$ allow ecf from (a)(1)
(b) $v^2 = 45$ or $v = \sqrt{45}$(1)
(b) 6.7 m/s(1)
Answer: (a) 0.45 J (b) 6.7 m/s
Examiner tip: Rearrange for $v^2$ first, then square-root — students often forget the square root and give 45 m/s.

5Grade 1–3Foundation4 marks

Choose answers from the box.

chemical elastic potential gravitational potential kinetic thermal

(a) A driver presses the brakes and the car slows down.

(i) Which energy store of the car decreases? [1]

(ii) Which energy store of the brakes increases? [1]

(b) An electric kettle is switched on. Which energy store of the water increases? [1]

Which energy store of the ball increases while it is squashed? [1]

(a)(i) kinetic(1)
(a)(ii) thermal (ignore heat; do not accept kinetic)(1)
(b) thermal (allow internal; ignore heat)(1)
(c) elastic potential (do not accept gravitational potential)(1)
Answer: (a)(i) kinetic (ii) thermal; (b) thermal; (c) elastic potential
Examiner tip: Use the store names exactly as in the box — "heat" and "movement" are not energy stores.

6Grade 6–75 marks

A student hangs masses on a spring and measures the extension of the spring. The graph shows her results.

Point P is the limit of proportionality.

(a) Use the graph to calculate the spring constant of the spring in N/m. [2]

(b) Calculate the elastic potential energy stored in the spring when the extension is 12 cm.

Use the Physics Equations Sheet. [2]

Give the reason why the equation used in (b) cannot be used to calculate the energy stored at this extension. [1]

(a) k = 5.0 ÷ 0.20 (any correct pair of values read from the straight section, with extension converted to m)(1)
(a) k = 25 (N/m) (allow 0.25 for 1 mark only if extension left in cm — i.e. 5.0 ÷ 20 = 0.25 scores 1 mark)(1)
(b) $E_e = \tfrac{1}{2} \times 25 \times 0.12^2$ (allow ecf from (a); extension must be in m)(1)
(b) $E_e$ = 0.18 (J) (allow ecf value from their k)(1)
(c) the extension is beyond the limit of proportionality / force is no longer (directly) proportional to extension (allow the graph is no longer a straight line (through the origin))(1)
Answer: (a) 25 N/m; (b) 0.18 J; (c) the spring is stretched beyond the limit of proportionality, so F is not proportional to e
Examiner tip: Convert the extension from cm to m before using k = F/e or ½ke² — leaving it in cm loses the final mark.

7Grade 6–7Higher5 marks

A sprinter is running at a speed of 6.0 m/s. Her kinetic energy is 1260 J.

(a) Calculate the mass of the sprinter. [3]

(b) The sprinter increases her speed to 9.0 m/s. Her friend says: "Her speed has gone up by 50%, so her kinetic energy will also go up by 50%."

Explain why her friend is wrong. Support your answer with a calculation. [2]

(a) $1260 = \tfrac{1}{2} \times m \times 6.0^2$(1)
(a) $m = \dfrac{2 \times 1260}{6.0^2}$ (rearrangement)(1)
(a) $m$ = 70 (kg) (correct answer with no working scores 3 marks)(1)
(b) kinetic energy depends on speed squared / $E_k \propto v^2$(1)
(b) new kinetic energy = ½ × 70 × 9.0² = 2835 J, which is 2.25 times the original (allow increase of 125%; allow 1575 J increase; allow ecf from (a))(1)
Answer: (a) 70 kg; (b) KE ∝ v², so KE = 2835 J — an increase of 125%, not 50%
Examiner tip: Kinetic energy depends on v², so a 1.5× increase in speed gives a 1.5² = 2.25× increase in kinetic energy.

8Grade 4–56 marks

A skateboarder of mass 50 kg starts from rest at point A at the top of a ramp. She rolls down the ramp to point B.

The diagram shows the ramp.

gravitational field strength = 9.8 N/kg

(a) Calculate the decrease in the gravitational potential energy of the skateboarder between A and B. [2]

(b) Calculate the kinetic energy of the skateboarder at B. [2]

Calculate the energy that is not transferred to the kinetic energy store and name the store it is transferred to. [2]

(a) $E_p = 50 \times 9.8 \times 2.5$(1)
(a) $E_p$ = 1225 (J) (allow 1200 J or 1230 J)(1)
(b) $E_k = \tfrac{1}{2} \times 50 \times 6.0^2$(1)
(b) $E_k$ = 900 (J)(1)
(c) 325 J (allow ecf: their (a) − their (b))(1)
(c) thermal (store of the wheels/ramp/surroundings) (allow internal; ignore heat; ignore sound on its own; do not accept "lost")(1)
Answer: (a) 1225 J; (b) 900 J; (c) 325 J, transferred to the thermal store of the wheels/ramp/surroundings (dissipated)
Examiner tip: Energy is never "lost" — say it is dissipated to the thermal store of the surroundings by friction and air resistance.

9Grade 6–76 marks

A bungee jumper steps off a high platform. She is attached to the platform by an elastic rope.

She falls freely until the rope becomes tight, then the rope stretches until she stops at her lowest point. She then bounces back up and down several times before coming to rest.

Describe the changes in the way energy is stored during the jump, from the moment she steps off the platform until she comes to rest.

Include in your answer why each bounce is lower than the one before. [6]

as she falls, her gravitational potential energy store decreases(1)
her kinetic energy store increases while she falls freely (before the rope is tight)(1)
once the rope is tight, energy is transferred to the elastic potential energy store of the rope(1)
her kinetic energy store decreases as the rope stretches (she slows down) (allow kinetic energy decreases once tension in rope exceeds her weight)(1)
at the lowest point her kinetic energy is zero, gravitational potential energy is at a minimum and elastic potential energy is at a maximum(1)
on the way back up, elastic potential energy is transferred to kinetic and then gravitational potential energy stores(1)
some energy is dissipated each time / during every transfer, e.g. by air resistance and as the rope stretches(1)
dissipated energy goes to the thermal store of the surroundings / of the rope (allow internal energy of the rope)(1)
so less energy is available to transfer back to the gravitational potential energy store, so each bounce reaches a lower height(1)
when she comes to rest, all the original decrease in gravitational potential energy (apart from that stored elastically in the rope) has been dissipated to the thermal store of the surroundings(1)
total energy is conserved throughout (energy is not lost / destroyed)(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically sequenced description of the store changes through the fall, the stretch and the rebound, using correct store names, with a coherent explanation that dissipation to thermal stores reduces the energy returned to the gravitational potential energy store so each bounce is lower.
Level 2	3–4	Describes most of the store changes in a mainly correct order, including the elastic potential store of the rope, and gives some reference to energy being dissipated or transferred to the surroundings, though the link to the lower bounces may be incomplete.
Level 1	1–2	Identifies one or two relevant store changes (e.g. gravitational potential to kinetic) or a simple statement that energy is wasted, with little or no sequencing.
No relevant content	0	Nothing worthy of credit.

Answer: GPE → KE while falling; once the rope is tight, KE and GPE → elastic PE of the rope; at the lowest point KE = 0 and elastic PE is a maximum; on rebound elastic PE → KE → GPE; energy is dissipated to the thermal store of the surroundings and rope at each stage, so each bounce is lower until she comes to rest.
Examiner tip: Top-band answers follow the jump in order AND explain that dissipation to thermal stores leaves less energy for the GPE store each bounce — naming stores is not enough.

10Grade 8–9Higher8 marks

A student throws a ball of mass 0.15 kg vertically upwards. The ball leaves her hand at a speed of 12 m/s.

gravitational field strength = 9.8 N/kg

(a) Show that the ball could rise to a maximum height of about 7.3 m above her hand if air resistance is ignored. [3]

(b) A heavier ball is thrown upwards at the same speed. Air resistance is still ignored.

Explain why the heavier ball would reach the same maximum height. [2]

Calculate the percentage of the ball's initial kinetic energy that is dissipated on the way up.

Give your answer to 2 significant figures. [3]

(a) $E_k = \tfrac{1}{2} \times 0.15 \times 12^2 = 10.8$ (J)(1)
(a) (all kinetic energy transferred to gravitational potential energy so) $10.8 = 0.15 \times 9.8 \times h$(1)
(a) $h$ = 7.35 (m) (must see answer to at least 2 decimal places, or 7.34…, for a "show that"; allow $h = v^2 / 2g$ = 144 ÷ 19.6 = 7.35 m for full marks)(1)
(b) both kinetic energy and gravitational potential energy are (directly) proportional to mass (allow both stores depend on mass in the same way)(1)
(b) so mass cancels / $\tfrac{1}{2}mv^2 = mgh$ gives $h = v^2/2g$, which does not depend on mass(1)
(c) $E_p$ at top = 0.15 × 9.8 × 6.1 = 8.967 (J) (allow 8.97 J or 9.0 J)(1)
(c) energy dissipated = 10.8 − 8.967 = 1.833 (J) (allow ecf from (a); allow height route: (7.35 − 6.1) ÷ 7.35 for the first two marks)(1)
(c) percentage = 1.833 ÷ 10.8 × 100 = 17 (%) (2 s.f. required; an unrounded answer such as 16.97 or 17.0 scores 2 marks; allow 16 (%) from using the given height of 7.3 m in either route)(1)
Answer: (a) Ek = 10.8 J = mgh → h = 7.35 m; (b) both Ek and Ep are proportional to m, so h = v²/2g is independent of mass; (c) 17%
Examiner tip: In a "show that" question give your calculated value to more significant figures than the value given (7.35 m, not 7.3 m).

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