GCSE Physics resources

AQA GCSE Physics exam skills

How to Answer GCSE Physics Calculation Questions

Practise connecting two physical stages. Keep intermediate values and units visible so a later valid step remains checkable after an arithmetic slip.

AQA 8463AQA 8464 GCSE PhysicsCombined Science: Trilogy FoundationHigher

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The revision snapshot

Practise connecting two physical stages. Keep intermediate values and units visible so a later valid step remains checkable after an arithmetic slip.

By the end, you should be able to:

  • Connect energy, power and efficiency stages.
  • Cross-check electrical energy by charge or power.
  • Keep linked temperature and state-change calculations separate.

Before you revise

Diagnostic question

Choose an answer from memory. Your result tells you what to watch for in the guide.

A motor’s useful output is 60 W and efficiency 0.75. What input power is needed?

Read the feedback for every option
  1. Multiplying gives an input smaller than useful output.
  2. 60/0.75 = 80 W.
  3. This reverses the useful-power relationship.
  4. Adding a dimensionless ratio to power is invalid.

Core revision guide

Learn the model, then use it

Practise connecting two physical stages. Keep intermediate values and units visible so a later valid step remains checkable after an arithmetic slip.

A lift with efficiency

Scope: 8463 and 8464 · Foundation and Higher

A 4 kg mass rises 3 m in 2 s with g = 9.8 N/kg. Useful energy = mgh = 117.6 J and useful power = 58.8 W. If the motor is 70% efficient, input power = useful/0.70 = 84 W. Dividing by efficiency gives a larger input; multiplying would incorrectly make input smaller than useful output.

An electrical cross-check

Scope: 8463 and 8464 · Foundation and Higher

A 12 V, 0.50 A lamp running for 4 min has P = 6 W and E = 6 × 240 = 1440 J. Alternatively Q = 0.50 × 240 = 120 C and E = QV = 1440 J. Matching independent routes checks the method. 4 minutes is not 4 seconds, and 500 mA is not 500 A.

Separate state-change stages

Scope: 8463 and 8464 · Foundation and Higher

Melting ice at 0 °C uses E = mL. Heating the melted water uses E = mcΔθ. Add energies for sequential stages, divide the total by heater power for time, and account for efficiency only if given. Do not substitute zero Δθ into the state-change equation or round each stage before summing.

A charge calculation converts milliamps and minutes before substituting into Q = It to obtain 120 coulombs.
Convert units before substitution so every value matches the units expected by the equation.Open the full-size exam diagram

Misconception clinic

Replace the tempting answer

Read each belief, say what is wrong with it, then compare your correction.

Tempting idea: 45 W.

Use instead: Multiplying gives an input smaller than useful output. 60/0.75 = 80 W.

Tempting idea: 2000 C.

Use instead: Both units need conversion. 0.500 × 240 = 120.

Retrieval practice

Quick checks

1. A 500 mA current flows for 4 min. What charge passes?

Read the feedback for every option
  1. Both units need conversion.
  2. This treats minutes as seconds.
  3. 0.500 × 240 = 120.
  4. This leaves milliamperes unconverted.

Worked examples

See the method being built

Apply a lift with efficiency in the worked reasoning, then try the changed-context questions below.

Example 1 · Lift plus efficiency

Scope: 8463 and 8464 · Foundation and Higher

A 4 kg mass rises 3 m in 2 s; g = 9.8 N/kg and efficiency is 70%. Find input power.

  1. Euseful = mgh = 117.6 J.
  2. Puseful = 117.6/2 = 58.8 W.
  3. Pinput = 58.8/0.70 = 84 W.
Show the answer

84 W input, with 58.8 W useful output and 25.2 W unwanted transfer.

Exam precision

Use the physics precisely

Do this: Write each intermediate result with its unit and retain unrounded values. Check that input exceeds useful output when efficiency is below one.

Independent practice

Exam-style questions

Write an answer before opening the marking guidance. Then edit the exact phrase or step that would gain the next mark.

1. A motor lifts 5.0 kg through 2.0 m in 4.0 s. g = 10 N/kg and efficiency = 80%. Find useful power and input power.

[4 marks]

Scope: 8463 and 8464 · Foundation and Higher

Show marking guidance and model answer

Original PhysicsUK mark scheme: Point 1 (1 mark): Useful energy = mgh = 5.0 × 10 × 2.0 = 100 J. Point 2 (1 mark): Useful power = 100/4.0 = 25 W. Point 3 (1 mark): Efficiency = 0.80; input power = useful power/efficiency. Point 4 (1 mark): Input power = 25/0.80 = 31.25 W (31 W acceptable). Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.

  1. Useful energy = mgh = 5.0 × 10 × 2.0 = 100 J. (1 mark)
  2. Useful power = 100/4.0 = 25 W. (1 mark)
  3. Efficiency = 0.80; input power = useful power/efficiency. (1 mark)
  4. Input power = 25/0.80 = 31.25 W (31 W acceptable). (1 mark)

Model answer: Useful energy = mgh = 5.0 × 10 × 2.0 = 100 J. Useful power = E/t = 100/4.0 = 25 W. Efficiency = 80/100 = 0.80, so input power = useful power/efficiency = 25/0.80 = 31.25 W (31 W to two significant figures). Input exceeds useful output because efficiency is below one.

Marking notes: Carry a numerical error from energy into a correct E/t method, then carry useful power into correct division by 0.80. Do not carry a physically invalid efficiency inversion.

2. A 6.0 V device carries 300 mA for 5.0 min. Find energy transferred using charge.

[3 marks]

Scope: 8463 and 8464 · Foundation and Higher

Show marking guidance and model answer

Original PhysicsUK mark scheme: Point 1 (1 mark): I = 0.300 A and t = 300 s. Point 2 (1 mark): Q = It = 90 C. Point 3 (1 mark): E = QV = 90 × 6.0 = 540 J. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.

  1. I = 0.300 A and t = 300 s. (1 mark)
  2. Q = It = 90 C. (1 mark)
  3. E = QV = 90 × 6.0 = 540 J. (1 mark)

Model answer: Convert 300 mA to 0.300 A and 5.0 min to 300 s. Q = It = 0.300 × 300 = 90 C. E = QV = 90 × 6.0 = 540 J. Alternatively P = VI = 6.0 × 0.300 = 1.8 W, then E = Pt = 1.8 × 300 = 540 J.

Marking notes: Accept P = VI then E = Pt as an alternative complete method with the same conversions. Later method marks may carry a previous arithmetic error.

Questions pupils ask

How to Answer GCSE Physics Calculation Questions FAQs

Why keep intermediate calculation values unrounded?

Early rounding can change the final result. Round the final answer to an appropriate precision after completing all linked steps.

Useful next steps

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