AQA GCSE Physics Physics Paper 2 (Foundation), November 2021: Question 9

17 marks · Standard Demand difficulty · Extended Answer

Explain elastically deformed, describe an experiment to investigate force and extension of a spring, identify the correct equation linking force, extension and spring constant, determine spring constant from a graph, explain how a graph shows direct proportionality, and calculate elastic potential energy.

Practise this question

Question

A multi-part physics question about springs and elasticity. It includes an illustration of a child on a playground toy with springs, a line graph showing force against extension from 0 to 5 N and 0 to 0.14 m, multiple-choice equation options, and calculations involving spring constant and elastic potential energy.
Question text

09 Figure 20 shows a child on a playground toy.

Figure 20

09.1 The springs have been elastically deformed.

Explain what is meant by ‘elastically deformed’.

[2 marks]

A student investigated the relationship between the force applied to a spring and the

extension of the spring.

Figure 21 shows the results.

Figure 21

09.2 Describe a method the student could use to obtain the results given in Figure 21.

You should include a risk assessment for one hazard in the investigation.

Your answer may include a diagram.

[6 marks]

09.3 Which equation links extension (e), force (F) and spring constant (k).

[1 mark]

Tick ( ) one box.

force = spring constant × (extension)2

force = spring constant × extension

extension

force =

spring constant

spring constant

force =

extension

Figure 21 is repeated below.

Figure 21

09.4 Determine the spring constant of the spring.

Use Figure 21.

[3 marks]

Spring constant = N/m

09.5 The student concluded:

‘The extension of the spring is directly proportional to the force applied to the spring.’

Describe how Figure 21 supports the student’s conclusion.

[2 marks]

09.6 The student repeated the investigation using a different spring with a spring constant

of 13 N/m.

Calculate the elastic potential energy of the spring when the extension of the spring

was 20 cm.

Use the Physics Equations Sheet.

[3 marks]

Elastic potential energy = J

Mark scheme

Show the mark scheme The mark scheme for Question 9 detailing point-by-point marking criteria for explaining elastically deformed, a 6-mark level-based response for the required practical method including risk assessment, selecting the correct equation, calculating spring constant from the gradient of the graph, interpreting direct proportionality from the graph, and calculating elastic potential energy using standard formulae.

Question 9

AO/

Question Answers Extra information Mark

Spec. Ref

09.1 will return to its original 1 AO2

shape/length 4.5.3

when the force is removed allow (when) the child gets off 1

the second mark is dependent

on scoring the first mark

09.2 Level 3: The method would lead to the production of a valid 5–6 AO1

outcome. The key steps are identified and logically sequenced. 4.5.3

Level 2: The method would not necessarily lead to a valid 3–4

outcome. Most steps are identified, but the method is not fully

logically sequenced.

Level 1: The method would not lead to a valid outcome. Some 1–2

relevant steps are identified, but links are not made clear.

No relevant content 0

Indicative content

• set up a clamp stand with a clamp

• hang the spring from the clamp

• use a second clamp and boss to fix a (half) metre rule alongside

the spring

• record the ruler reading that is level with the bottom of the spring

• hang a 1 N / a known weight from the bottom of the spring

• record the new position of the bottom of the spring

• calculate the extension of the spring

• measure the extension of the spring

• add further weights to the spring so the force increases 1 N at a

time up to 5 N

• for each new force record the position of the bottom of the spring

and calculate / measure the extension

Indicative content continues on the next page…

Risk Assessment

Hazard: Clamp (stand, boss and masses) might fall off desk

18 Risk: injury to feet

Precaution: Use clamp to fix apparatus to the bench or

Ensure that the slotted masses hang over the

base/foot of the stand or

Ensure that the boss is screwed tightly into the stand

and clamp or

Put (heavy) masses on the base/foot of the stand

or

Stand up so that you can move out of the way

Hazard: Spring could break / come loose

Risk: damage eye

Precaution: Wear safety goggles

If a risk assessment / hazard is not given, the answer can still reach

level 3, but not full marks.

Full marks may be awarded for alternative feasible methods.

09.3 force = spring constant × 1 AO1

extension 4.5.3

09.4 5.00 0.125 allow any correct pair of values 1 AO2

from the graph 4.5.3

k = 5.00 allow a misread value(s) from 1

0.125 the graph

k = 40 (N/m) allow a correct calculation using 1

their incorrect value(s)

09.5 the line is straight allow the line does not curve 1 AO3

allow a constant gradient 4.5.3

and passes through the origin 1

09.6 e = 0.20 m 1 AO2

4.5.3

E = 0.5 × 13 × 0.202 allow an incorrectly / not 1

e

converted value of e 19

Ee = 0.26 (J) 1

use of two incorrectly/not

converted values scores a

máximum of 1 mark

Total 17

How to answer it

Forces and Elasticity Study Guide

What this question tests

This question assesses your understanding of Hooke's Law and elastic behavior. You will need to recall key definitions, recall and apply the spring constant formula ( F = k × e ) and elastic potential energy formula ( Ee = ½ × k × e² ), interpret linear graph results, describe a core practical method including safety precautions, and analyse direct proportionality.

Question Part 09.1

Defining Elastically Deformed

✅ Correct Answer

  • The spring will return to its original shape/length...
  • ...when the force is removed.
2 marks available (AO2)

💡 Key Knowledge

An elastic deformation means temporary stretching or compressing. If the material does not return to its original length, it has undergone inelastic (or plastic) deformation.

❌ Common Errors

Students often miss the second mark by failing to state that the force must be removed. Note that the second mark is dependent on scoring the first!

Question Part 09.2

Required Practical: Investigating Springs

✅ Correct Answer (Level 3 Response)

  • Setup: Set up a clamp stand with a clamp and hang the spring. Fix a metre rule alongside the spring using a second clamp and boss.
  • Measurement: Record the initial ruler reading level with the bottom of the spring. Add a 1 N (or known mass) weight.
  • Extension: Record the new position and calculate the extension (new reading − original reading).
  • Repetition: Add further weights (e.g., 1 N at a time up to 5 N), recording position and calculating extension for each step.
  • Risk Assessment: Hazard: Clamp/masses falling off desk leading to foot injury. Precaution: Fix apparatus securely to the bench with a G-clamp, or place a heavy mass on the base of the stand.
6 marks available (AO1) - Level marked (Level 3 = 5–6 marks)

🧠 Exam Technique

For 6-mark practical questions, structure your answer chronologically: Setup → Measure initial length → Add load & measure new length → Calculate extension → Repeat. Always include a clear hazard and realistic safety precaution to access top bands.

Question Part 09.3

Identifying the Hooke's Law Equation

✅ Correct Answer

Tick the box for: force = spring constant × extension

1 mark available (AO1)

💡 Key Knowledge

Expressed symbolically: F = k × e (or F = k × x ). Make sure you know how to rearrange this triangle if needed!

Question Part 09.4

Calculating the Spring Constant

📐 Calculation Steps

  1. Rearrange the formula: k = F / e
  2. Choose a point from the graph: E.g., at Force ( F ) = 5.00 N, Extension ( e ) = 0.125 m (Any clear pair from the line can be used).
  3. Substitute values: k = 5.00 / 0.125
  4. Calculate: k = 40 N/m
3 marks available (AO2)

❌ Common Calculation Traps

Watch out for scale reading errors on the axes. Always double-check what each small grid square represents before pulling coordinates from the graph.

Question Part 09.5

Interpreting Direct Proportionality

✅ Correct Answer

  • The line is straight / does not curve.
  • The line passes through the origin (0,0).
2 marks available (AO3)

💡 Key Knowledge

For two variables to be directly proportional, plotted data must form a straight line that goes precisely through the origin.

Question Part 09.6

Calculating Elastic Potential Energy

📐 Calculation Steps

  1. Convert units: Extension e = 20 cm = 0.20 m (Must be in metres!). Spring constant k = 13 N/m .
  2. Recall the equation: Ee = ½ × k × e²
  3. Substitute values: Ee = 0.5 × 13 × (0.20)²
  4. Calculate: Ee = 0.5 × 13 × 0.04 = 0.26 J
3 marks available (AO2)

❌ Common Calculation Traps

Failing to convert centimetres into metres ( 20 cm must become 0.20 m ) or forgetting to square the extension ( e² ) are the most common places students lose marks here.

Topics

Physics · Required Practicals · P5: Forces · Required Practicals

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