AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2021: Question 1

10 marks · Standard Demand difficulty · Short Answer

Answer questions about a stretched spring, including elastic deformation, measuring extension, elastic potential energy, Hooke’s law, and calculating spring constant.

Practise this question

Question

The question page shows a labelled diagram of a spring hanging from a clamp stand beside a vertical metre rule, with masses attached to the spring. Below it are five parts: a 1-mark multiple-choice question asking what is meant by 'elastically deformed', a 2-mark question asking for a method to determine extension, a 3-mark calculation of elastic potential energy for a spring with extension 80 mm and spring constant 40 N/m, a 1-mark question asking for the equation linking extension, force and spring constant, and a 3-mark calculation where a force of 300 N causes an extension of 0.40 m and the spring constant must be found. Answer spaces and tick boxes are provided, and the total for the full question is 10 marks.
Question text

01 Figure 1 shows a stretched spring.

The spring is elastically deformed.

Figure 1

01.1 What is meant by ‘elastically deformed’?

[1 mark]

Tick ( ) one box.

As the force on the spring increases the length of the spring increases.

Only a very small force is needed to stretch the spring.

The force on the spring causes it to change shape.

The spring will return to its original length when the force is removed.3

01.2 Describe a method to determine the extension of the spring.

[2 marks]

01.3 The extension of the spring is 80 mm.

spring constant = 40 N/m

Calculate the elastic potential energy of the spring.

Use the Physics Equations Sheet.

[3 marks]

Elastic potential energy =4 J

01.4 Write down the equation which links extension (e), force (F) and spring constant (k).

[1 mark]

01.5 A force of 300 N acts on a different spring.

The force causes the spring to extend by 0.40 m.

Calculate the spring constant of the spring.

[3 marks]

Spring constant = N/m

Mark scheme

Show the mark scheme The mark scheme is a table listing answers, extra information, marks, and AO/specification references for parts 01.1 to 01.5. It gives the correct multiple-choice answer that the spring returns to its original length when the force is removed; for measuring extension it states to measure original and extended lengths with a metre rule and subtract original from extended length; for elastic potential energy it converts 80 mm to 0.080 m and uses Ee = 0.5 × 40 × (0.080)^2 to get 0.128 J; for the equation it gives F = ke; and for the final calculation it shows 300 = k × 0.40 leading to k = 750 N/m. The table shows a total of 10 marks and references spec point 6.5.3 throughout.

AO /

Question Answers Extra information Mark

Spec. Ref.

01.1 the spring will return to its 1 6.5.3

original length when the force is AO1

removed

01.2 measure the original length of 1 6.5.3

the spring and the extended AO1

length of the spring (with the

metre rule)

extension = extended length – 1

original length

01.3 e = 0.080 m 1 6.5.3

AO2

E = 0.5 × 40 × (0.080)2 allow a correct substitution using 1

e

an incorrectly / not converted

value of e

Ee = 0.128 (J) allow a correct calculation using 1

an incorrectly / not converted

value of e

01.4 force = spring constant × 1 6.5.3

extension AO1

or

F = k e

01.5 300 = k × 0.40 1 6.5.3

AO2

300 1

k =

0.40

k = 750 (N/m) 1

Total 10

How to answer it

Spring Extension, Elastic Energy and Hooke’s Law

What this question tests

You need to know what elastic deformation means, how to measure extension using a ruler, and how to use the spring equations E = 1/2 ke² and F = ke . Marks are awarded for correct physics ideas, correct substitution, and correct units.

Question focus

Part (a) to (e): springs, extension, elastic potential energy, and spring constant

💡 Key knowledge

  • A spring is elastically deformed if it returns to its original length when the force is removed.
  • Extension = extended length − original length.
  • Elastic potential energy is stored when an elastic object is stretched or compressed.
  • Hooke’s law: force is proportional to extension, as long as the limit of proportionality is not passed.

🧠 Exam technique

  • Use the exact wording the mark scheme wants for definition questions.
  • Convert mm to m before using equations.
  • Always include units: J for energy, N/m for spring constant.
  • Show your substitution and rearrangement for calculation marks.

Part (a) — What is meant by “elastically deformed”?

1 mark

✅ Correct answer

The spring will return to its original length when the force is removed.

💡 Why this scores the mark

The key idea is reversible change. The object changes shape temporarily, then goes back to normal when the force stops.

❌ Common errors

  • “The force causes it to change shape” is too vague — that describes any deformation, not elastic deformation.
  • “Only a very small force is needed” is not the definition.
  • Do not forget the idea of returning to original length.

Part (b) — Describe a method to determine the extension of the spring

2 marks

✅ Mark-scheme answer

Measure the original length of the spring and the extended length of the spring using the metre rule. Then calculate extension using extension = extended length − original length.

🧠 How to get both marks

  • Mark 1: say you measure the original length and the extended length.
  • Mark 2: state the equation extension = extended length − original length .

💡 Good practical detail

  • Read the ruler at eye level to avoid parallax error.
  • Measure from the same reference point each time.
  • Keep the ruler vertical and the spring still before reading.

❌ Common errors

  • Writing only “measure the extension” without explaining how.
  • Using the stretched length alone instead of subtracting the original length.
  • Forgetting the metre rule or a way to measure the lengths.

Part (c) — Calculate the elastic potential energy of the spring

3 marks

📐 Calculation steps

  1. Use the equation: E = 1/2 ke²
  2. Convert the extension to metres:
    80 mm = 0.080 m
  3. Substitute the values:
    E = 1/2 × 40 × (0.080)²
  4. Calculate:
    E = 0.5 × 40 × 0.0064
    E = 0.128 J

✅ Correct answer

Elastic potential energy = 0.128 J

🧠 Examiner insight

The mark scheme allows credit for the correct method even if the extension was not converted properly, but the final answer must be sensible and in joules. Full marks come from showing the equation, substitution, and answer.

❌ Common calculation traps

  • Using 80 instead of 0.080 in the equation.
  • Forgetting to square the extension.
  • Writing the answer in N/m instead of J.
  • Missing the 1/2 in the formula.

Part (d) — Write the equation linking extension, force and spring constant

1 mark

✅ Correct answer

F = ke

💡 Key knowledge

This is Hooke’s law: force = spring constant × extension.

❌ Common errors

  • Mixing up symbols or writing E = ke .
  • Leaving out the symbols entirely when the question asks for the equation.

Part (e) — Calculate the spring constant

3 marks

📐 Calculation steps

  1. Use Hooke’s law: F = ke
  2. Substitute the values:
    300 = k × 0.40
  3. Rearrange:
    k = 300 ÷ 0.40
  4. Calculate:
    k = 750 N/m

✅ Correct answer

Spring constant = 750 N/m

🧠 Exam technique

  • Show the rearrangement to score method marks.
  • Use the correct unit for spring constant: N/m.
  • If the answer is large, don’t panic — strong springs have large spring constants.

❌ Common errors

  • Dividing the wrong way round: 0.40 ÷ 300 .
  • Forgetting the unit or writing just 750 .
  • Using cm instead of m for extension.

Top-mark answer checklist

💡 What the best answers did

  • Used the exact definition of elastic deformation.
  • Gave a clear measurement method for extension.
  • Converted mm to m before using equations.
  • Wrote equations clearly and showed substitutions.
  • Included the correct units in final answers.

🧠 Final exam reminder

If a question says “calculate”, always write the equation first, then substitute values, then work out the answer. This helps you pick up method marks even if the final number is wrong.

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Higher), 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.