AQA GCSE Physics Physics Paper 2 (Foundation), June 2022: Question 4
10 marks · Standard Demand difficulty · Short Answer
Identify the centre of mass of an orange, describe and calculate the relationship between mass and weight, determine the extension and spring constant of a spring under a given force, and state the behavior of the spring when the force is removed.
Practise this questionQuestion
Question text
04 Figure 6 shows the weight of an orange acting from a point labelled X.
Figure 6
04.1 What name is given to point X in Figure 6?
[1 mark]
Tick ( ) one box.
Centre of force
Centre of mass
Centre of balance
Centre of weight
04.2 Weight and mass are not the same.
The relationship between weight and mass for an object can be written as:
weight ∝ mass
Which sentence describes the relationship between weight and mass?
[1 mark]
Tick ( ) one box.
Weight is approximately equal to mass.
Weight is directly proportional to mass.
Weight is less than mass. 17
Figure 7 shows a balance used to measure the mass of 5 oranges.
*16* Figure 7
04.3 All 5 of the oranges have the same mass.
Determine the mass of 1 orange.
[2 marks]
Mass = kg
04.4 Calculate the weight of 1 orange.
gravitational field strength = 9.8 N/kg
Use the equation:
weight = mass × gravitational field strength
[2 marks]
Weight = N
The balance shown in Figure 7 contains a spring.
Figure 8 shows the spring with no force acting on it and with a force of 6.0 N acting
*17* on it.
Figure 8
04.5 What is the extension of the spring when a force of 6.0 N acts on it?
[1 mark]
Tick ( ) one box.
0.015 m
0.035 m
0.050 m
0.085 m
04.6 Calculate the spring constant of the spring.
Use the equation:
force
spring constant =
extension
[2 marks]
Spring constant =19 N/m
04.7 What will happen to the spring when the force is removed?
[1 mark]
Mark scheme
Show the mark scheme
Question 4
AO /
Question Answers Extra information Mark
Spec. Ref.
04.1 centre of mass 1 AO1
4.5.1.3
AO /
Spec. Ref.
04.2 weight is directly proportional to 1 AO1
mass 4.5.1.3
AO /
Spec. Ref.
04.3 reading from balance = 1.1 kg 1 AO2
4.5.1.3
1.1
mass = = 0.22 kg allow correct calculation using 1
5 incorrectly read value from the
balance
AO /
Spec. Ref.
04.4 weight = 0.22 × 9.8 allow ecf from question 04.3 1 AO2
4.5.1.3
2.156 (N) allow correct answer to 2 or 3 1
sig figs
AO /
Spec. Ref.
04.5 0.015 m 1 AO2
4.5.3
AO /
Spec. Ref.
6.0
04.6 spring constant = allow ecf from question 04.5 1 AO2
0.015 4.5.3
400 (N/m) 1
AO /
Spec. Ref.
04.7 returns to its original allow returns to 3.5 cm 1 AO3
length/shape 4.5.3
Total Question 4 10
How to answer it
Forces, Mass, Weight and Hooke's Law Study Guide
What this question tests
This question assesses your understanding of forces and elasticity (AQA GCSE Physics Topic 5). Key skills include identifying the centre of mass, distinguishing between mass and weight, applying the weight equation ( W = m × g ), calculating spring extension, applying Hooke's law to find the spring constant, and recognizing elastic deformation.
Naming the Point of Weight
✅ Correct Answer
Centre of mass (Tick the second box)
💡 Key Knowledge
The centre of mass is the single point through which all the weight of an object can be considered to act.
Relationship Between Weight and Mass
✅ Correct Answer
Weight is directly proportional to mass (Tick the middle box)
💡 Key Knowledge
Because W = m × g and gravitational field strength ( g ) is constant on Earth, multiplying mass by a constant means weight scales directly with mass.
Determining the Mass of 1 Orange
📐 Calculation Steps
- Step 1: Read the total mass of the 5 oranges from the balance scale diagram in Figure 7 = 1.1 kg .
- Step 2: Divide the total mass by the number of oranges: 1.1 ÷ 5 = 0.22 kg .
Mass = 0.22 kg
❌ Common Errors
Students often misread scale increments or forget to divide by 5, giving the total mass of all oranges instead of just one.
Calculating the Weight of 1 Orange
📐 Calculation Steps
- Step 1: State the equation: weight = mass × gravitational field strength
- Step 2: Substitute values: 0.22 kg × 9.8 N/kg
- Step 3: Calculate: 2.156 N (Allow ECF from 04.3). Can round to 2.2 N (2 sig figs) or 2.16 N (3 sig figs).
Weight = 2.156 N
🧠 Exam Technique
Always show your substitution clearly, even for simple one-step calculations. This secures a method mark if your final arithmetic has a typo.
Determining Spring Extension
✅ Correct Answer
0.015 m (Tick the first box)
📐 Working Out
- Original length ( L₀ ) = 3.5 cm
- Stretched length ( L ) = 5.0 cm
- Extension = 5.0 cm - 3.5 cm = 1.5 cm
- Convert cm to m: 1.5 ÷ 100 = 0.015 m
Calculating the Spring Constant
📐 Calculation Steps
- Step 1: Identify force = 6.0 N and extension = 0.015 m (from 04.5).
- Step 2: Substitute into the equation: spring constant = 6.0 ÷ 0.015
- Step 3: Calculate: 400 N/m
Spring constant = 400 N/m
❌ Common Errors
Using 1.5 cm directly instead of converting it to metres ( 0.015 m ) is the single most common reason students lose marks here.
Behaviour of the Spring Upon Force Removal
✅ Correct Answer
It returns to its original length / shape (or returns to 3.5 cm).
💡 Key Knowledge
This demonstrates elastic deformation—the spring will return to its original dimensions as long as the limit of proportionality has not been exceeded.
Topics
Physics · P5: Forces
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Foundation), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.