AQA AS Level Physics Paper 2, June 2022: Question 29

1 mark · Medium difficulty · Multiple Choice

Determine the elastic strain energy stored in a wire of Young modulus E, unstretched length L, cross-sectional area A, and extension e.

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Question

Multiple choice question 29 asking for the elastic strain energy stored in a wire with Young modulus E, length L, cross-sectional area A, and extension e. Four options A, B, C, D are provided with algebraic fractions representing different combinations of these variables.
Question text

29 A wire is made from a material of Young modulus E.

The wire obeys Hooke’s law.

The wire has an unstretched length L and a cross-sectional area A.

When a force is applied to the wire, the extension of the wire is e.

What is the elastic strain energy stored in the wire?

[1 mark]

AEe2

A

2L

L

B

2Ae

Ae2

C

2EL

AEL

D

2e

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is option A, which corresponds to the algebraic expression A E e squared over 2 L.

29 A (AO2)

How to answer it

Deriving Elastic Strain Energy from Young Modulus

What this question tests

This question assesses your ability to combine standard equations of mechanics and materials (Hooke's Law, Young modulus, and elastic potential energy) to derive a new expression involving fundamental parameters ($E, L, A, e$). It tests algebraic manipulation and application of knowledge (AO2).

Question 29 • Multiple Choice • 1 Mark

Derivation of Stored Elastic Strain Energy

✅ Correct Answer: A

The correct option is A ( AEe² / (2L) ).

Mark Awarded: 1 / 1 for selecting option A.

💡 Key Knowledge

  • Young Modulus formula: E = (F × L) / (A × e)
  • Elastic strain energy (W): W = ½ × F × e (area under a force-extension graph)
  • Substitution and rearrangement skills for multi-variable expressions.

🧠 Exam Technique

When faced with algebraic multiple-choice questions, do not guess. Write down the known equations explicitly on your workspace, substitute variables step-by-step, and cross-reference your final algebraic fraction with the given options.

❌ Common Errors

  • Omitting the factor of ½ in the energy equation, leading to incorrect numerator/denominator coefficients.
  • Inverting the Young modulus formula (mixing up length L and extension e ).

📐 Step-by-Step Derivation

  1. Start with the equation for elastic strain energy stored in a stretched wire:
    W = ½ × F × e
  2. Rearrange the Young modulus definition ( E = (F × L) / (A × e) ) to make force F the subject:
    F = (A × E × e) / L
  3. Substitute this expression for F back into the energy equation:
    W = ½ × ((A × E × e) / L) × e
  4. Simplify the algebra to combine terms:
    W = (A × E × e²) / (2 × L) , which matches option A.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.