AQA GCSE Combined Science: Trilogy Physics Paper 2 (Foundation), November 2021: Question 6

10 marks · Low Demand difficulty · Short Answer

Questions on elastic deformation, measuring extension, calculating elastic potential energy, stating Hooke's Law equation, and calculating the spring constant.

Practise this question

Question

Question 6 displays Figure 8, which shows a clamp stand supporting a vertically hanging spring with slotted masses suspended from it, alongside a vertical metre rule. Question 06.1 asks what is meant by 'elastically deformed' via four tick boxes. Question 06.2 asks to describe a method to determine the extension of the spring. Question 06.3 provides an extension of 80 mm and a spring constant of 40 N/m, asking to calculate elastic potential energy. Question 06.4 asks to write down the equation linking extension, force, and spring constant. Question 06.5 states a force of 300 N causes a spring to extend by 0.40 m, asking to calculate its spring constant.
Question text

06 Figure 8 shows a stretched spring.

The spring is elastically deformed.

Figure 8

06.1 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.21

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

[2 marks]

06.3 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 =22 J

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

[1 mark]

06.5 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 Mark scheme for question 6 lists: 06.1 - 'the spring will return to its original length when the force is removed' (1 mark). 06.2 - measure original length and extended length with metre rule (1 mark), extension = extended length minus original length (1 mark). 06.3 - convert e = 0.080 m (1 mark), substitute into E_e = 0.5 * 40 * (0.080)^2 (1 mark), evaluation giving 0.128 J (1 mark). 06.4 - force = spring constant * extension or F = k e (1 mark). 06.5 - substitution 300 = k * 0.40 (1 mark), rearrangement k = 300 / 0.40 (1 mark), calculation gives 750 N/m (1 mark). Total 10 marks.

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 the spring will return to its 1 6.5.3

original length when the force is AO1

removed

06.2 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

06.3 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

06.4 force = spring constant × 1 6.5.3

extension AO1

or

F = k e

06.5 300 = k × 0.40 1 6.5.3

AO2

k = 1

0.40

k = 750 (N/m) 1

Total 10

How to answer it

Forces & Elasticity: Springs, Hooke's Law, and Energy

📌 What this question tests

This question assesses core practical and theoretical understanding of AQA GCSE Physics / Combined Science Specification 6.5.3 (Forces and Elasticity):

  • Understanding the definition of elastic deformation versus inelastic deformation.
  • Describing experimental methods to find spring extension using basic lab apparatus.
  • Calculating elastic potential energy ( Eₑ = ½ k e² ), including millimeter-to-meter unit conversions.
  • Recalling and rearranging Hooke's Law: F = k e .

Question 06.1: Elastic Deformation

Multiple Choice Recall • 1 Mark

✅ Correct Answer

Tick the 4th box:

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

💡 Key Knowledge

  • Elastic deformation: An object returns to its original shape and size when the forces deforming it are removed.
  • Inelastic (plastic) deformation: The object remains permanently stretched and does not return to its original shape.

❌ Common Errors

Choosing "The force on the spring causes it to change shape". While true for any deformation, it does not explain what makes the deformation specifically elastic.

Mark Scheme: 1 mark for selecting the correct statement. No partial credit.

Question 06.2: Determining Extension

Practical Description • 2 Marks

✅ Model Answer

  1. Measure the original (unstretched) length of the spring and the extended length with the metre rule. [1 mark]
  2. Subtract the original length from the extended length:
    extension = extended length − original length [1 mark]

🧠 Exam Technique

To score both marks on "extension" questions, always mention two specific measurements and the subtraction:

  • State what you measure before adding weights.
  • State what you measure after adding weights.
  • Clearly state that extension is the difference between the two.

❌ Common Misconceptions

Writing simply "measure the length of the stretched spring". The total length of the stretched spring is not the extension!

Mark Scheme:
• 1 mark: Measure original length AND extended length (with metre rule).
• 1 mark: extension = extended length − original length.

Question 06.3: Elastic Potential Energy Calculation

Formula Application & Unit Conversion • 3 Marks

📐 Step-by-Step Calculation

Given: Extension e = 80 mm , Spring constant k = 40 N/m

  1. Step 1: Convert units of extension to metres (m) [1 mark]
    80 mm ÷ 1000 = 0.080 m
  2. Step 2: Select formula & substitute correctly [1 mark]
    From the Physics Equations Sheet: Eₑ = 0.5 × k × e²
    Eₑ = 0.5 × 40 × (0.080)²
  3. Step 3: Calculate the final answer [1 mark]
    Eₑ = 20 × 0.0064 = 0.128 J

❌ Calculation Traps

  • Forgetting to convert mm to m: Using 80 instead of 0.080 gives 128 000 J (scores max 2/3 via error carried forward).
  • Forgetting to square e : Entering 0.5 × 40 × 0.080 = 1.6 J loses the final calculation mark.
Mark Breakdown:
• 1 mark for e = 0.080 m
• 1 mark for substitution 0.5 × 40 × (0.080)²
• 1 mark for final answer: 0.128 (J)

Question 06.4: Equation for Hooke's Law

Recall Equation • 1 Mark

✅ Correct Answer

force = spring constant × extension

or in symbols: F = k e

💡 Key Knowledge

This is a required equation that must be memorised. Note the units:

  • F = Force in newtons (N)
  • k = Spring constant in newtons per metre (N/m)
  • e = Extension in metres (m)

❌ Common Errors

Mixing up multiplication and division, such as writing F = k / e or k = F × e . You can remember this with the formula triangle where F is at the top.

Mark Scheme: 1 mark for force = spring constant × extension or F = k e .

Question 06.5: Calculating Spring Constant

Rearranging Equations • 3 Marks

📐 Step-by-Step Calculation

Given: Force F = 300 N , Extension e = 0.40 m

  1. Step 1: Substitute values into the equation [1 mark]
    300 = k × 0.40
  2. Step 2: Rearrange to make k the subject [1 mark]
    k = 300 / 0.40
  3. Step 3: Solve for k [1 mark]
    k = 750 N/m

🧠 Exam Technique: Substitute First!

AQA examiners advise substituting values into the original formula before rearranging. If you rearrange incorrectly at the start (e.g. k = e / F ), you risk losing the substitution mark as well!

Mark Breakdown:
• 1 mark for substitution: 300 = k × 0.40
• 1 mark for rearranging: k = 300 / 0.40
• 1 mark for answer: 750 (N/m)

Topics

Physics · P1: Energy · P5: Forces

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