AQA GCSE Combined Science: Trilogy Physics Paper 2 (Foundation), 2024: Question 7

14 marks · Standard Demand difficulty · Short Answer

Analyze energy transfers in a gymnast's jump and perform calculations related to spring constant and elastic potential energy from experimental data and graphs.

Practise this question

Question

The image shows a series of physics exam questions. Figure 12 shows a gymnast in mid-air over a floor containing springs. Figure 13 shows experimental equipment: a spring on a scale and several large masses. Figure 14 is a graph of Force in newtons (0 to 400) against Compression in metres (0 to 0.06). A linear line of best fit is plotted with a gradient triangle showing vertical change delta y and horizontal change delta x.
Question text

07 Figure 12 shows an Olympic gymnast performing a floor routine.

Figure 12

The floor contains springs.

When the gymnast lands on the floor, a force compresses the springs in the floor.

07.1 When a spring is compressed, the elastic potential energy of the spring increases.

Explain why compressing the springs in the floor helps the gymnast to jump higher.

Use ideas about energy in your answer.

[2 marks]

07.2 When the gymnast lands on the floor, one of the springs compresses by 1.2 cm.

spring constant = 8500 N/m

Calculate the elastic potential energy stored in the spring.

Use the Physics Equations Sheet.

Give the unit.

[4 marks]

Elastic potential energy = Unit

A student investigated a spring with a different spring constant.

When masses are placed on the spring it compresses.

The student measured the compression of the spring for different masses.

Figure 13 shows some of the equipment used.

Figure 13

07.3 Describe how the compression of the spring could be determined.

[2 marks]

07.4 Explain why the investigation should be done on the laboratory floor rather than on

a table.

[2 marks]

Figure 14 shows the results.

Figure 14

The spring constant is the gradient of the line of best fit shown on Figure 14.

07.5 Determine the value Δy on Figure 14.

[1 mark]

Δy = N

07.6 Determine the value Δx on Figure 14.

[1 mark]

Δx = m

07.7 Determine the spring constant of the spring.

Use your answers to Question 07.5 and Question 07.6.

Give your answer to 3 significant figures.

[2 marks]

Spring constant (3 significant figures) = N/m

Mark scheme

Show the mark scheme The mark scheme table lists answers for questions 07.1 through 07.7. For 07.1, it explains energy transfers from elastic potential to kinetic or gravitational potential. For 07.2, it shows the calculation for elastic potential energy: 0.5 times 8500 times 0.012 squared equals 0.612 Joules. For 07.7, it calculates the spring constant as the gradient: 250 divided by 0.042, resulting in 5950 N/m to three significant figures.

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 (as a spring decompresses) the 1 AO1

elastic potential energy of the 6.5.3

spring decreases

(as the gymnast leaves the 1

floor) the kinetic energy of the

gymnast increases

or

the gravitational potential energy

of the gymnast increases

OR

(as a spring decompresses) the

spring exerts a force on the

gymnast (1)

(so) work is done on the

gymnast (1)

AO /

Spec. Ref.

07.2 e = 0.012 m 1 AO2

E = 0.5 × 8500 × 0.0122 allow a correct substitution using 1 AO2

e

an incorrectly / not converted

value of e

Ee = 0.612 allow 0.61 1 AO2

allow a correct calculation using

an incorrectly / not converted

value of e

J or joule 1 AO1

– – – 6.5.3

AO /

Spec. Ref.

07.3 measure the original length of 1 AO3

the spring and the compressed 6.5.3

length of the spring (using a RPA18

metre rule)

compression = original length − 1

compressed length

OR

calculate / measure the weight

of the mass on the spring (1)

use the equation

F

e = (1)

k

AO /

Spec. Ref.

07.4 masses could fall (off the spring) 1 AO1

6.5.3

(so) less likely to cause injury / allow less hazardous 1 RPA18

damage (if done on the floor) allow lower risk (of injury)

allow specific cause of injury

– eg landing on foot – –

AO /

Spec. Ref.

07.5 Δy = (330 – 80 =) 250 (N) 1 AO3

6.5.3

RPA18

AO /

Spec. Ref.

07.6 Δx = (0.052 – 0.010 =) 1 AO3

0.042 (m) 6.5.3

RPA18

AO /

Spec. Ref.

allow ecf from question 07.5

07.7 AO3

and question 07.6

6.5.3

RPA18

k = allow k = 5952.38 1

0.042

5950 (N/m) allow their calculated gradient to 1

3 significant figures

Total Question 7 14

How to answer it

Springs, Energy, and Hooke's Law

What this question tests

This question assesses your ability to explain energy transfers in a system, calculate elastic potential energy using the correct units, describe practical methods for measuring spring compression, and use graph data to determine a spring constant (gradient).

Part 07.1

Energy Transfers in Jumping

Explaining how springs help a gymnast jump higher

Correct Answer

  • As the spring decompress, the elastic potential energy of the spring decreases. [1 mark]
  • This energy is transferred to the gymnast, increasing their kinetic energy or gravitational potential energy. [1 mark]

Exam Technique

When an "Explain" question asks about energy, always describe what happens to the stores. If one store decreases, another must increase. Use the phrase "transferred to" to link them.

Part 07.2

Calculating Elastic Potential Energy

Using the equation Eₑ = 0.5 × k × e²

Step-by-Step Calculation

  1. Convert units: The compression is 1.2 cm. You must convert this to metres.
    1.2 ÷ 100 = 0.012 m [1 mark]
  2. Substitute values: Use the formula from the sheet.
    Eₑ = 0.5 × 8500 × (0.012)² [1 mark]
  3. Calculate: Eₑ = 0.5 × 8500 × 0.000144 = 0.612 [1 mark]
  4. State the unit: Energy is measured in Joules (J). [1 mark]

Common Errors

  • Forgetting to square: Many students forget to square the extension (e²).
  • Unit conversion: Using 1.2 instead of 0.012 will lead to a massive error.

Key Knowledge

k = spring constant (N/m)
e = extension or compression (m)

Part 07.3 & 07.4

Practical Skills (RPA 18)

07.3 Determining Compression

Measure the original length of the spring and the compressed length (using a ruler). [1 mark]

Compression = original length − compressed length. [1 mark]

07.4 Laboratory Safety

Why use the floor?

The masses used are large. If the investigation is done on a table, the masses could fall off, potentially causing injury to feet or damage to equipment. Doing it on the floor reduces this risk. [2 marks]

Part 07.5, 07.6 & 07.7

Graph Analysis & Spring Constant

07.5 Determine Δy

Look at the vertical triangle on the graph:
330 N - 80 N = 250 N [1 mark]

07.6 Determine Δx

Look at the horizontal triangle on the graph:
0.052 m - 0.010 m = 0.042 m [1 mark]

07.7 Calculate Spring Constant (k)

The spring constant is the gradient of the line.

  1. Formula: k = Δy ÷ Δx
  2. Substitution: k = 250 ÷ 0.042
  3. Raw Result: 5952.38...
  4. Final Answer (3 sig figs): 5950 N/m [2 marks]

Examiner Tip: Significant Figures

The question explicitly asks for 3 significant figures. 5952.38 rounds to 5950. If you wrote 5952 or 6000, you would lose the final mark even if your calculation was perfect!

Topics

Physics · P1: Energy · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Foundation), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.