AQA A-Level Physics Paper 1, June 2022: Question 27

1 mark · Medium difficulty · Multiple Choice

Calculate the total elastic potential energy stored in a spring when an extra mass of 2M is added to an initial mass M, given that the initial elastic potential energy is E.

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Question

A multiple choice question numbered 27. It states: A mass M is suspended from a spring. When the mass is at rest at the equilibrium position, the elastic potential energy stored is E. An extra mass of 2M is added to the spring and the spring extends while still obeying Hooke's law. What is the total elastic energy stored when the system is at rest at the new equilibrium position? Options are A: 2E, B: 3E, C: 4E, D: 9E.
Question text

27 A mass M is suspended from a spring. When the mass is at rest at the equilibrium

position, the elastic potential energy stored is E.

An extra mass of 2M is added to the spring and the spring extends while still obeying

Hooke’s law.

What is the total elastic energy stored when the system is at rest at the new equilibrium

position?

[1 mark]

A 2E

B 3E

C 4E

D 9E

Mark scheme

Show the mark scheme The mark scheme table shows question number 27 with the correct answer D (AO1), indicating the correct option is 9E.

27 D (AO1) 9E

How to answer it

Question 27: Elastic Potential Energy & Hooke's Law

What this question tests

This question assesses your understanding of Hooke's Law (F = kx) and the equation for elastic potential energy stored in a stretched spring (E = 0.5kx² or E = 0.5Fx). It tests your ability to scale variables proportionally when a system is subjected to an increased load at a new equilibrium position.

Question 27 Breakdown

Multiple Choice Question (1 Mark)

✅ Correct Answer

D (9E)

Mark Scheme: D (AO1) — 1 mark

💡 Key Knowledge

  • Hooke's Law states that extension is directly proportional to force: F = kx .
  • Initial mass is M , so initial weight is Mg and initial extension is x .
  • An extra mass of 2M is added, making the total mass M + 2M = 3M .
  • Total weight becomes 3Mg , meaning the new extension becomes 3x .

📐 Step-by-Step Calculation

  1. Initial State: Force F₁ = Mg , Extension = x . Initial elastic energy E = 0.5 k x² .
  2. New State: Total mass is M + 2M = 3M . Total force F₂ = 3Mg .
  3. New Extension: Since F is proportional to x , the new extension is 3x .
  4. New Energy: Total energy E_total = 0.5 k (3x)² = 0.5 k (9x²) = 9 (0.5 k x²) = 9E .

❌ Common Errors

  • Linear scaling trap: Students often see that the added mass is 2M and incorrectly assume the energy multiplies by 2 ( 2E ) or simply square the added mass ratio without factoring in the total mass.
  • Forgetting total mass: Failing to add the original mass M to the added mass 2M , leading to a force multiplier of 2 instead of 3.

🧠 Exam Technique

For proportional reasoning questions involving squared terms (like energy E = 0.5 k x² ), always write out the algebraic substitution clearly. If the linear variable (extension x ) increases by a factor of 3, the squared term increases by a factor of 3² = 9 .

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.