Practise 33 exam-style questions on Energy stores and systems 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.
33 questions
1–6 marks each
Instant marking
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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 pendulum bob is pulled to one side and released. It swings down to its lowest point. Air resistance can be ignored.
Describe how the energy stored by the bob changes as it swings from its release point to its lowest point.
Model answer
As the bob falls, its gravitational potential energy store decreases and its kinetic energy store increases by the same amount. At the lowest point the bob has its maximum kinetic energy.
Explanation
As the bob swings down it loses height, so its gravitational potential store decreases, and it speeds up, so its kinetic store increases. With no air resistance no energy is dissipated, so the decrease in one store equals the increase in the other.
Mark scheme
1 mark
the gravitational potential energy store (of the bob) decreases
allow energy is transferred from the gravitational potential store to the kinetic store for 2 marks
1 mark
the kinetic energy store (of the bob) increases
allow by the same amount
Question 2
3 marks
A boy throws a ball vertically upwards. Air resistance can be ignored.
Describe the changes in the way energy is stored from the moment the ball leaves his hand until it reaches its highest point.
Model answer
As the ball rises it slows down, so its kinetic energy store decreases. It gains height, so its gravitational potential energy store increases by the same amount. At the highest point the ball is momentarily at rest, so all the energy that was in its kinetic store is now in its gravitational potential store.
Explanation
With no air resistance, no energy is dissipated, so energy simply moves from one store to the other. A ball thrown straight up stops for an instant at the top, so its kinetic store is empty there and its gravitational potential store is at its maximum.
Mark scheme
1 mark
the kinetic energy store of the ball decreases (as it slows down)
1 mark
the gravitational potential energy store of the ball increases (as it gains height)
1 mark
at the highest point the kinetic store is zero / all the energy from the kinetic store is now in the gravitational potential store
allow the decrease in kinetic energy equals the increase in gravitational potential energy
Question 3
3 marks
A lift of mass 650 kg moves up 4.5 m from one floor of a building to the next.
gravitational field strength = 9.8 N/kg
Show that the energy transferred to the gravitational potential energy store of the lift is about 29 kJ.
Model answer
gravitational potential energy = mass × gravitational field strength × height Eₚ = 650 × 9.8 × 4.5 Eₚ = 28 665 J = 28.7 kJ, which is about 29 kJ
Explanation
In a show that question the answer is given, so the marks are for the working and for a more precise value than the one quoted. Remember to compare like with like: 28 665 J is 28.7 kJ.
Mark scheme
1 mark
Eₚ = m g h
allow the equation in words; allow a correct substitution for this mark
1 mark
Eₚ = 650 × 9.8 × 4.5
1 mark
Eₚ = 28 665 (J)
allow 28.7 kJ or 28.665 kJ; must see a value more precise than 29 kJ
Question 4
4 marks
A pendulum bob of mass 0.20 kg is pulled to one side so that it is raised 5.0 cm above its lowest position. It is then released. Air resistance can be ignored.
gravitational field strength = 9.8 N/kg
Calculate the maximum speed of the bob.
Model answer
h = 5.0 cm = 0.050 m Eₚ = m g h = 0.20 × 9.8 × 0.050 = 0.098 J At the lowest point all of this is in the kinetic store: 0.098 = 0.5 × 0.20 × v² v² = 0.98 v = 0.99 m/s
Explanation
The bob is fastest at its lowest point, where all the energy that was in its gravitational potential store has been transferred to its kinetic store. The height must be converted from centimetres to metres before substituting.
Mark scheme
1 mark
5.0 cm = 0.050 m
an answer of 0.99 (m/s) scores 4 marks
1 mark
Eₚ = 0.20 × 9.8 × 0.050 = 0.098 (J)
allow ecf from an incorrect height conversion
1 mark
0.098 = 0.5 × 0.20 × v²
allow gravitational potential energy lost = kinetic energy gained; allow ecf
1 mark
v = 0.99 (m/s)
allow 0.98995 or 1.0 (m/s); allow ecf; an answer of 9.9 (m/s) shows the height was not converted and scores 3 marks
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