AQA GCSE Physics Physics Paper 2 (Higher), June 2022: Question 5

9 marks · Standard Demand difficulty · Short Answer

Define centre of mass, calculate the mean weight of a tomato from a balance reading, determine the spring constant of a compression spring, and explain a property of the spring suitable for the balance.

Practise this question

Question

Figure 9 shows a mechanical dial balance holding five tomatoes, with the dial pointer indicating a mass of approximately 425 g. Question 05.1 asks for the definition of 'centre of mass'. Question 05.2 asks to calculate the mean weight of a tomato using gravitational field strength g = 9.8 N/kg. Figure 10 shows a helical spring with an original length of 5.0 cm under no force, compressing to a length of 3.5 cm when subjected to a 6.0 N force. Question 05.3 asks to determine the spring constant in N/m. Question 05.4 asks to explain one property of the spring that makes it suitable for use in the balance.
Question text

05 Figure 9 shows a balance used to measure the mass of five tomatoes.

Figure 9

05.1 What is meant by ‘centre of mass’?

[1 mark]

05.2 Calculate the mean weight of a tomato in Figure 9.

Use the Physics Equations Sheet.

gravitational field strength = 9.8 N/kg

[3 marks]

21 Weight = N

05.3 The balance in Figure 9 contains a spring that compresses when the tomatoes are

placed on the balance.

Figure 10 shows the spring with no force acting and with a 6.0 N force acting.

Figure 10

Determine the spring constant of the spring.

Use the Physics Equations Sheet.

[3 marks]

Spring constant = N/m

05.4 Explain one property of the spring that makes it suitable for use in the balance.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 5 totaling 9 marks. 05.1 gives 1 mark for the point from which weight may be considered to act or mass appears concentrated. 05.2 gives 3 marks: total mass = 0.425 kg (1), mass of 1 tomato = 0.085 kg (1), weight = 0.833 N (1). 05.3 gives 3 marks: substituting into F = ke (6.0 = k × 0.015) (1), rearranging to k = 6.0 / 0.015 (1), and calculating k = 400 N/m (1). 05.4 gives 2 marks for deforming elastically so it returns to original length, or compression directly proportional to force so gives a linear scale.

Question 5

AO /

Question Answers Extra information Mark

Spec. Ref.

05.1 the point from which weight may allow the point through which 1 AO1

be considered to act the line of action of the weight 4.5.1.3

acts

or

the point where the mass allow the point at which the

appears to be concentrated mass is concentrated

AO /

Spec. Ref.

05.2 mass of 5 tomatoes = 0.425 (kg) 1 AO2

4.5.1.3

mass of 1 tomato = 0.085 (kg) allow an incorrect and / or not 1

converted reading correctly

divided by 5

W = (0.085 × 9.8) = 0.833 (N) allow a correct calculation using 1

their value of mass

AO /

Spec. Ref.

AO2

05.3 6.0 = k × 0.015 1 4.5.3

6.0 allow correct rearrangement 1

k = using an incorrectly calculated

0.015

value of e

k = 400 (N/m) allow a correct calculation using 1

an incorrectly calculated value

of e

AO /

Spec. Ref.

18 05.4 deforms elastically 1 AO3

4.5.3

(so) will return to its original 1

length / shape (after force is

removed)

OR

compression is directly

proportional to the force

(applied) (1)

(so) gives a linear scale (1) allow easy to calibrate

Total Question 5 9

How to answer it

Forces, Weight & Hooke's Law

📋 WHAT THIS QUESTION TESTS

This question assesses your ability to recall core definitions of centre of mass, accurately read analog scales, apply the weight equation (W = mg) with unit conversions (g to kg), determine the spring constant using Hooke's Law (F = ke), and explain the physical properties of a spring (elastic deformation and proportionality) required for measuring devices.

QUESTION 05.1 • 1 MARK

Definition of Centre of Mass

What is meant by 'centre of mass'?

✅ Correct Answer

Either of the following standard AQA definitions:

  • The point from which weight may be considered to act.
  • The point where the mass appears to be concentrated.
Mark allocation: 1 mark for either correct phrasing.

💡 Key Knowledge

  • Every object behaves as if all its mass is focused at one single point.
  • Gravity pulls down on all parts of an object, but we represent the total weight force as an arrow drawn downwards directly from the centre of mass.

🧠 Exam Technique

Be careful not to mix up "mass" and "weight". Notice how the words pair up:

  • "Point where mass is concentrated"
  • "Point from which weight acts"

❌ Common Errors

  • Vague answers like "the exact middle of the tomato" (only true for symmetrical objects of uniform density).
  • Saying "where gravity starts" (unscientific phrasing that scores 0).
QUESTION 05.2 • 3 MARKS

Mean Weight of One Tomato

Calculate the mean weight of a tomato in Figure 9 (g = 9.8 N/kg).

📐 Step-by-Step Calculation

1 Read scale and convert mass to kg:
Pointer indicates 425 g (each major mark is 100 g; the needle points halfway between 400 g and 450 g).
Total mass = 425 ÷ 1000 = 0.425 kg [1 mark]

2 Find mass of one tomato:
There are 5 identical tomatoes:
Mean mass = 0.425 kg ÷ 5 = 0.085 kg [1 mark]

3 Calculate weight using W = mg:
W = 0.085 kg × 9.8 N/kg = 0.833 N [1 mark]

Final Answer: 0.833 N (or 0.83 N)

❌ Calculation Traps

  • Forgetting to divide by 5: Calculating the weight of all 5 tomatoes (4.165 N) instead of the mean weight of a single tomato.
  • Scale misread: Mistaking the pointer reading as 450 g or 420 g. Look closely at the interval divisions.
  • Unit error: Multiplying grams directly by 9.8 (e.g., 85 × 9.8 = 833 N). In physics equations, mass must always be in kilograms (kg)!
QUESTION 05.3 • 3 MARKS

Spring Constant Calculation

Determine the spring constant of the spring compressed from 5.0 cm to 3.5 cm by a 6.0 N force.

📐 Step-by-Step Calculation

1 Calculate compression (extension e):
Compression = Original length - Compressed length
e = 5.0 cm - 3.5 cm = 1.5 cm
Convert to metres: 1.5 ÷ 100 = 0.015 m

2 Substitute into Hooke's Law:
Equation: F = k × e
6.0 = k × 0.015 [1 mark]

3 Rearrange and solve for k:
k = 6.0 ÷ 0.015 [1 mark]
k = 400 N/m [1 mark]

Final Answer: 400 N/m

🧠 Exam Technique & Units

  • Check the printed unit on the answer line: it specifies N/m.
  • If you keep extension as 1.5 cm, your answer would be 4 N/cm, but the given unit line says N/m, which will lose you the final mark! Always convert cm to m first.
  • Error Carried Forward (ECF): If you calculated compression wrong (e.g., 2.0 cm), you can still get 2 marks for correctly rearranging and dividing 6.0 by your value.
QUESTION 05.4 • 2 MARKS

Spring Properties in Measuring Scales

Explain one property of the spring that makes it suitable for use in the balance.

✅ Acceptable Answers (Choose ONE pair)

Option 1 (Elasticity):

  • Property: It deforms elastically [1 mark]
  • Explanation: (so) it will return to its original length / shape once the tomatoes are removed [1 mark]

Option 2 (Linearity / Hooke's Law):

  • Property: Compression is directly proportional to the force applied [1 mark]
  • Explanation: (so) it gives a linear scale / evenly spaced markings / is easy to calibrate [1 mark]

❌ Common Misconceptions

  • Writing everyday descriptions like "it is flexible", "it bounces back", or "it is stretchy". Use the technical physics term: deforms elastically.
  • Only giving the property without the explanation: stating "it obeys Hooke's Law" only earns 1 mark unless you explain that this provides an evenly spaced / linear dial scale.

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.