AQA GCSE Physics Physics Paper 2 (Higher), November 2021: Question 2
17 marks · Standard Demand difficulty · Short Answer
Investigate the relationship between force and extension for a spring, including definitions, method description, equation selection, graph analysis, and elastic potential energy calculations.
Practise this questionQuestion
Question text
02 Figure 3 shows a child on a playground toy.
Figure 3
02.1 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 4 shows the results.
Figure 4
02.2 Describe a method the student could use to obtain the results given in Figure 4.
You should include a risk assessment for one hazard in the investigation.
Your answer may include a diagram.
[6 marks]
02.3 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 4 is repeated below.
Figure 4
02.4 Determine the spring constant of the spring.
Use Figure 4.
[3 marks]
Spring constant = N/m
02.5 The student concluded:
‘The extension of the spring is directly proportional to the force applied to the spring.’
Describe how Figure 4 supports the student’s conclusion.
[2 marks]
02.6 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
Question 2
AO/
Question Answers Extra information Mark
Spec. Ref
02.1 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
02.2 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
Risk: injury to feet 9
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.
02.3 force = spring constant × 1 AO1
extension 4.5.3
02.4 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)
02.5 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
02.6 e = 0.20 m 1 AO2
4.5.3
E = 0.5 × 13 × 0.202 allow an incorrectly / not 1
e
10 converted value of e
Ee = 0.26 (J) 1
use of two incorrectly/not
converted values scores a
maximum of 1 mark
Total 17
How to answer it
Forces and Springs Study Guide
What this question tests
This question assesses your understanding of Hooke's Law, elastic deformation, planning a practical investigation involving springs, interpreting force-extension graphs, and calculating elastic potential energy.
Defining Elastically Deformed
✅ Correct Answer
- The spring will return to its original shape/length...
- ...when the force is removed.
💡 Key Knowledge
Elastic deformation contrasts with inelastic (plastic) deformation, where an object is permanently stretched and does not return to its original dimensions.
Describing a Practical Investigation
🧠 Exam Technique (Level 3 Response)
To achieve 5–6 marks, your method must be logically sequenced and reproducible by another student. Always include apparatus setup, how variables are measured, and a clear risk assessment.
💡 Method Outline & Risk Assessment
- Setup: Clamp a spring to a clamp stand; hang a metre rule alongside it vertically.
- Measurement: Record the initial ruler reading at the bottom of the spring without weights.
- Loading: Add known weights (e.g., 1 N masses) incrementally up to 5 N.
- Extension: Record the new position and calculate extension (new length minus original length).
- Hazard & Precaution: Clamp stand might fall off the desk causing foot injury. Precaution: Use a G-clamp to secure the base to the desk, or place a heavy mass on the base.
Hooke's Law Equation
✅ Correct Answer
Tick the box for: force = spring constant × extension
💡 Key Knowledge
Expressed as the formula: F = k × e (found on your Physics Equations Sheet).
Calculating the Spring Constant from a Graph
📐 Step-by-Step Calculation
- Rearrange the formula: k = F / e (Spring constant = Force ÷ Extension).
- Choose coordinate points from the graph: At Force = 5.00 N, Extension = 0.125 m (or any other correct pair on the straight line).
- Substitute values: k = 5.00 / 0.125
- Calculate final answer: k = 40 N/m
❌ Common Errors
- Failing to read values accurately from the grid intersections.
- Inverting the fraction (calculating extension divided by force instead).
Interpreting Direct Proportionality
✅ Correct Answer
- The line is straight / does not curve.
- The line passes through the origin (0,0).
🧠 Exam Technique
When asked how a graph supports direct proportionality, always reference two geometric features: it must be a straight line AND it must pass through the origin.
Calculating Elastic Potential Energy
📐 Step-by-Step Calculation
- Convert units: Extension must be in metres. e = 20 cm = 0.20 m .
- Recall formula from Equations Sheet: E_e = 0.5 × k × e²
- Substitute values: E_e = 0.5 × 13 × (0.20)²
- Calculate squared term: 0.20² = 0.04
- Final calculation: E_e = 0.5 × 13 × 0.04 = 0.26 J
❌ Common Errors & Traps
- Unit Trap: Forgetting to convert centimetres to metres (using 20 instead of 0.20). Using un-converted values limits you to a maximum of 1 mark!
- Square Trap: Forgetting to square the extension value in the formula.
Topics
Physics · Required Practicals · P5: Forces · Required Practicals
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.