Changes in energy - AQA GCSE Physics Exam Questions

Practise 39 exam-style questions on Changes in energy for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

39 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A runner of mass 60 kg is running at a speed of 5.0 m/s.

Calculate the kinetic energy of the runner.

Model answer

Eₖ = 0.5 × m × v²
Eₖ = 0.5 × 60 × 5.0²
Eₖ = 0.5 × 60 × 25
Eₖ = 750 J

Explanation

Square the speed first (5.0² = 25), then multiply by the mass and by 0.5. A common mistake is to square the whole of 0.5 × 60 × 5.0, or to forget to square the speed at all.

Mark scheme

1 mark
Eₖ = 0.5 × 60 × 5.0²
an answer of 750 (J) scores 2 marks
1 mark
750 (J)
an answer of 150 (J) (speed not squared) scores 0 marks

Question 2

2 marks

Estimate the kinetic energy of a typical adult cyclist, including the bicycle, travelling at a typical cycling speed.

Show how you get your answer.

Model answer

Mass of cyclist and bicycle ≈ 80 kg; typical cycling speed ≈ 6 m/s.
Eₖ = 0.5 × m × v² = 0.5 × 80 × 6² = 1440 J
So the kinetic energy is about 1000-1500 J.

Explanation

An estimate uses sensible typical values: an adult is about 60-80 kg, a bicycle about 10-15 kg, and a typical cycling speed is about 6 m/s. The answer only needs to be the right size (about a thousand joules), not an exact value.

Mark scheme

1 mark
sensible values chosen: total mass 60-120 kg and speed 4-8 m/s
allow a typical cycling speed of about 6 m/s
1 mark
correct use of Eₖ = 0.5 × m × v² with their values giving an answer of about 1000 J (allow 500-4000 J)
e.g. 0.5 × 80 × 6² = 1440 J

Question 3

3 marks

A student drops a 0.10 kg ball and a 1.0 kg ball from the same height of 5.0 m. Air resistance can be ignored.

The student predicts that the heavier ball will hit the ground at a higher speed.

Use the equations for gravitational potential energy and kinetic energy to explain why the student is wrong.

Model answer

As each ball falls, the gravitational potential energy it loses is transferred to its kinetic store, so m g h = ½ m v². The mass appears on both sides and cancels, giving v² = 2 g h. The final speed depends only on g and the height, so both balls hit the ground at the same speed (√(2 × 9.8 × 5.0) = 9.9 m/s).

Explanation

The heavier ball has ten times the gravitational potential energy, but it also needs ten times the energy to reach the same speed. Without air resistance all objects fall together, which tackles the common misconception that heavier objects fall faster.

Mark scheme

1 mark
gravitational potential energy lost = kinetic energy gained: m g h = ½ m v²
1 mark
the mass cancels / appears on both sides, so v² = 2 g h
allow the heavier ball has more energy but also more mass, in the same ratio
1 mark
the speed depends only on g and h, so both balls hit the ground at the same speed
allow both hit the ground at 9.9 m/s

Question 4

4 marks

A car of mass 1200 kg accelerates from 20 m/s to 30 m/s.

Calculate the increase in the kinetic energy of the car.

Model answer

Eₖ at 20 m/s = 0.5 × 1200 × 20² = 240 000 J
Eₖ at 30 m/s = 0.5 × 1200 × 30² = 540 000 J
Increase = 540 000 − 240 000 = 300 000 J

Explanation

Work out the kinetic energy at each speed and subtract. Using the change in speed (10 m/s) in the equation gives 60 000 J, which is wrong because kinetic energy depends on speed squared, and 30² − 20² is not the same as 10².

Mark scheme

1 mark
Eₖ at 20 m/s = 0.5 × 1200 × 20² = 240 000 (J)
an answer of 300 000 (J) scores 4 marks
1 mark
Eₖ at 30 m/s = 0.5 × 1200 × 30²
1 mark
= 540 000 (J)
1 mark
increase = 540 000 − 240 000 = 300 000 (J)
allow 300 kJ; allow ecf; an answer of 60 000 (J) (using a change in speed of 10 m/s) scores 0 marks

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