AQA GCSE Physics Physics Paper 1 (Higher), June 2023: Question 7

8 marks · Standard Demand difficulty · Short Answer

Explain limitations on baby mass in a bouncer, describe the energy transfers as the baby moves downward, and calculate the spring constant given extension and elastic potential energy.

Practise this question

Question

Figure 7 illustrates a baby bouncer consisting of a harness attached to a spring hanging from a door frame. Position A shows the baby suspended higher with the spring unstretched, and Position B shows the baby lower down with the spring stretched. Question 07.1 asks for one reason why babies over 12 kg should not use it (1 mark). Question 07.2 asks to describe the energy transfers as the stationary baby moves from position A to stationary position B (3 marks). Question 07.3 provides an extension of 8.0 cm and elastic potential energy of 4.0 J, asking to calculate the spring constant (4 marks).
Question text

07 A baby bouncer is a harness attached to a spring that hangs from a door frame.

Figure 7 shows a baby in a baby bouncer in two positions.

Figure 7

07.1 The baby bouncer should not be used with babies that have a mass greater

than 12 kg.

Suggest one reason why.

[1 mark]

07.2 In positions A and B the baby is stationary.

Describe the energy transfers as the baby moves from position A to position B.

[3 marks]

07.3 In one position the extension of the spring is 8.0 cm.

The elastic potential energy stored by the spring is 4.0 J.

Calculate the spring constant of the spring.

Use the Physics Equations Sheet.

[4 marks]

Spring constant = N/m

Mark scheme

Show the mark scheme Mark scheme for Question 7: 07.1 awards 1 mark for the spring becoming permanently extended or extension being too great so feet always touch the floor. 07.2 awards 3 marks: 1 for gravitational potential energy in position A, 1 for transfer to kinetic energy and then elastic potential energy, and 1 for all energy being elastic potential energy at position B. 07.3 awards 4 marks: 1 for converting 8.0 cm to 0.080 m, 1 for substitution into Ee = 0.5 * k * e^2 (4.0 = 0.5 * k * 0.080^2), 1 for rearranging for k, and 1 for the final answer k = 1250 N/m.

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 spring may become ignore reference to limit of 1 AO3

permanently extended proportionality 4.1.1.2

allow the harness / spring /

or chain may break

extension of the spring may be

too great (so the baby’s feet are

always on the floor)

ignore baby may be injured /

harmed / may hit doorframe

AO /

Spec. Ref.

07.2 (in position A) the baby has allow Ep for gravitational 1 AO1

gravitational potential energy potential energy 4.1.1.1

(as the baby moves down this) allow Ek for kinetic energy 1

is transferred to kinetic energy

(of the baby) and / then elastic allow Ee for elastic potential

potential energy (of the spring) energy

(in position B) all the energy is ignore energy dissipated to the 1

elastic potential energy surroundings

– HYSICS – –

AO /

Spec. Ref.

AO2

07.3 e = 0.080 (m) 1 4.1.1.2

4.0 = ½ × k × 0.0802 allow a correct substitution using 1

an incorrectly / not converted

value of e

4.0 allow a correct rearrangement 1

k = 2 using an incorrectly / not

(0.5 × 0.080 )

converted value of e

allow an answer consistent with 1

k = 1250 (N/m) their value of e

Total Question 7 8

How to answer it

Energy Transfers & Elastic Potential in a Baby Bouncer

📌 What this question tests

This question assesses core understanding from Topic 1 (Energy) and Topic 5 (Forces):

  • Inelastic deformation & spring safety limits: Explaining physical consequences of exceeding maximum mass.
  • Energy stores and pathways: Describing step-by-step energy changes between gravitational, kinetic, and elastic potential stores.
  • Elastic potential energy calculations: Rearranging Ee = ½ke² , converting units ( cm to m ), and calculating the spring constant.

Question 07.1: Safe Working Limits

Suggest one reason why the baby bouncer should not be used with babies with a mass greater than 12 kg [1 mark]

✅ Acceptable Answers (Any one)

  • The spring may become permanently extended (or inelastically deformed).
  • The extension of the spring may be too great (so the baby's feet are always touching the floor).
  • The spring / harness / chain might break or snap under excessive tension.
awarded: 1 mark (AO3)

❌ Common Errors & Examiner Traps

  • Vague safety answers: Writing "the baby might get hurt" or "it is dangerous" gets 0 marks. You must state the physical mechanical reason.
  • Physics wording: "Exceeding the limit of proportionality" is explicitly ignored in this mark scheme—focus on permanent deformation or snapping.

Question 07.2: Energy Transfers During Oscillation

Describe the energy transfers as the baby moves from position A to position B [3 marks]

💡 Physical Situation Explained

Position A: Baby is at maximum height, stationary before falling.

In Between: Baby falls downwards, accelerating then decelerating, pulling the spring.

Position B: Baby reaches lowest point and is momentarily stationary again.

✅ Mark Scheme Breakdown (3 Marks)

  • Mark 1: At position A, the baby has gravitational potential energy (Ep).
  • Mark 2: As the baby moves downwards, this is transferred to kinetic energy (Ek) and then to elastic potential energy (Ee) of the spring.
  • Mark 3: At position B, all the energy is stored as elastic potential energy (Ee) (since the baby is stationary).

🧠 Exam Technique

Structure your answer chronologically: Start (A) → Middle (falling) → End (B).

Explicitly name both the store of energy and what object has it (e.g., kinetic energy of the baby, elastic potential energy of the spring).

❌ Common Errors

  • Forgetting that the baby is moving between A and B, which means energy must pass through the kinetic energy store.
  • Stating energy is lost: while thermal dissipation occurs, the mark scheme focuses entirely on useful mechanical stores (AQA ignores thermal dissipation here).

Question 07.3: Calculating Spring Constant

Extension = 8.0 cm, Elastic potential energy = 4.0 J. Calculate the spring constant [4 marks]

📐 Step-by-Step Calculation

Step 1: Unit Conversion (Mark 1)
Convert extension from cm to m (divide by 100):
e = 8.0 cm = 8.0 / 100 = 0.080 m
Step 2: Recall and Substitute (Mark 2)
Formula: Ee = ½ × k × e²
Substitute known values:
4.0 = 0.5 × k × (0.080)²
Step 3: Rearrange for Spring Constant (Mark 3)
Square extension: (0.080)² = 0.0064
k = 4.0 / (0.5 × 0.0064) = 4.0 / 0.0032
Step 4: Final Value & Units (Mark 4)
k = 1250 N/m
Final Answer: 1250 N/m (4 marks total)

❌ Calculation Traps to Avoid

  • cm to m failure: Using 8.0 instead of 0.080 . (Gives 0.125 N/m — loses 1 mark, allowed max 3 marks under error carried forward).
  • Forgetting to square: Dividing by 0.080 instead of 0.080² .
  • Dividing by 0.5 incorrectly: Remember dividing by 0.5 is the same as multiplying the top by 2: k = (2 × Ee) / e² = (2 × 4.0) / (0.080)² = 8.0 / 0.0064 = 1250 .

Topics

Physics · P1: Energy · P5: Forces

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.