AQA GCSE Physics Physics Paper 2 (Higher), June 2023: Question 3

13 marks · Standard Demand difficulty · Short Answer

Calculate the mass of a car from its momentum and velocity, determine the total distance traveled from a velocity-time graph, and explain the effects of driver distraction and braking friction.

Practise this question

Question

A series of exam questions about a car's handbrake, velocity-time graph, momentum, stopping distance, and brake temperature. Figure 8 shows a handbrake lever diagram with a pivot and applied force. Figure 9 shows a velocity-time graph with velocity from 0 to 16 m/s on the y-axis and time from 0 to 5 seconds on the x-axis, showing acceleration up to 3 seconds followed by constant velocity.
Question text

03 Some cars have a lever that is used to apply the handbrake.

Figure 8 shows the handbrake lever in a car.

Figure 8

03.1 The driver applies the force shown in Figure 8. The force produces a moment about

the pivot.

How could the driver increase the moment about the pivot without increasing the size

of the force?

[1 mark]

The driver releases the handbrake.

Figure 9 shows how the velocity of the car changes during the first 5 seconds of

a journey.

Figure 9

03.2 After 3 seconds, the momentum of the car is 24 000 kg m/s.

Calculate the mass of the car.

Use the Physics Equations Sheet.

[4 marks]

15 Mass =

kg

03.3 Determine the distance travelled by the car during the first 5 seconds of the journey.

*14* Use Figure 9.

[3 marks]

Distance travelled by the car =16

m

03.4 In an emergency the driver needs to apply the brakes suddenly to stop the

car quickly.

The driver of the car is distracted.

Explain why the distraction will increase the stopping distance.

[3 marks]

03.5 Explain why the temperature of the brakes increases as they are used.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for question 3 detailing answers and mark allocations for five subsections covering lever moments, momentum calculations using the velocity-time graph, distance calculations as area under the graph, reaction time and thinking distance explanations, and work done by friction increasing thermal energy.

Question 3

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 apply the force further away do not allow increase the length 1 AO2

from the pivot of the lever 4.5.4

AO /

Spec. Ref.

03.2 v = 15 (m/s) 1 AO2

4.5.7.1

24 000 = m × 15 allow a value of v = 14.5 (m/s) 1

24 000 1

m =

m = 1600 (kg) 1

AO /

Spec. Ref.

03.3 distance travelled during first 1 AO2

3 seconds = 22.5 (m) 4.5.6.1.5

distance travelled during last 1

2 seconds = 30 (m)

total distance = 52.5 (m) allow 53 (m) 1

allow 1 mark for the correct

addition of their calculated

distances

allow a maximum of 2 marks for

total distance = 50.75 (m) if

velocity used = 14.5 (m/s)–HYSICS – 8463/2H –

AO /

Spec. Ref.

03.4 stopping distance includes allow stopping distance = 1 AO1

thinking distance braking distance + thinking

distance

12 there is an additional time allow the driver’s reaction time 1 AO2

before the driver applies the will increase (due to the

brakes. distraction)

(so) the thinking distance will 1 AO2

increase

4.5.6.1.1

4.5.6.3.1

4.5.6.3.2

AO /

Spec. Ref.

03.5 work is done due to friction (in ignore friction alone 1 AO1

the brakes) 4.5.6.3.4

(causing) an increase in the 1

internal / thermal energy (of the

brakes)

Total Question 3 13

How to answer it

Forces, Moments, and Motion Study Guide

AQA GCSE Physics — Question 3 Revision

What this question tests

This question assesses your understanding of rotational forces (moments), calculating momentum and mass, interpreting velocity-time graphs to find distance travelled, factors affecting stopping distance (thinking vs. braking distance), and energy transfers due to friction in brakes.

Part (0.3.1): Increasing a Moment

How could the driver increase the moment about the pivot without increasing the size of the force? [1 mark]

✅ Correct Answer

Apply the force further away from the pivot.

💡 Key Knowledge

  • Moment = Force × distance (perpendicular distance from the pivot to the line of action of the force).
  • To increase the moment without changing the force, you must increase the distance.

❌ Common Errors

Examiners noted that students often wrote "increase the length of the lever" instead of stating where the force is applied. Make sure you specify applying the force further along the lever!

🎯 Mark Allocation: 1 mark for mentioning distance from the pivot.

Part (0.3.2): Momentum Calculation

After 3 seconds, the momentum of the car is 24 000 kg m/s. Calculate the mass of the car. [4 marks]

📐 Step-by-Step Calculation

  1. Step 1: Find velocity ( v ) at t = 3 s from Figure 9 graph. ( v = 15 m/s )
  2. Step 2: Recall the momentum equation: momentum = mass × velocity ( p = m × v )
  3. Step 3: Substitute values: 24 000 = m × 15
  4. Step 4: Rearrange and solve: m = 24 000 / 15 = 1600 kg

🧠 Exam Technique

Always check the graph first! Data for calculations involving graphs is frequently hidden in preceding or subsequent figures. Show your substitution clearly to pick up method marks even if your final arithmetic slips.

🎯 Mark Allocation: 1 mark for reading velocity = 15 m/s, 1 mark for correct equation/substitution, 1 mark for rearrangement, 1 mark for final answer 1600 kg.

Part (0.3.3): Distance from Velocity-Time Graphs

Determine the distance travelled by the car during the first 5 seconds of the journey. [3 marks]

📐 Step-by-Step Calculation

  1. Step 1: Split the graph into two sections.
    Section 1 (0 to 3 s): A triangle under the line.
    Area = 0.5 × base × height = 0.5 × 3 × 15 = 22.5 m
  2. Step 2: Section 2 (3 to 5 s).
    A rectangle under the constant velocity line.
    Area = width × height = (5 - 3) × 15 = 30 m
  3. Step 3: Add areas together.
    Total distance = 22.5 + 30 = 52.5 m

💡 Key Knowledge

The area under a velocity-time graph represents the total distance travelled. Always split complex shapes into standard geometric shapes (triangles and rectangles).

🎯 Mark Allocation: 1 mark for distance of first section (22.5), 1 mark for distance of second section (30), 1 mark for correct total (52.5 m).

Part (0.3.4): Stopping Distance and Distractions

Explain why the distraction will increase the stopping distance. [3 marks]

✅ Correct Answer

  • Stopping distance is the sum of thinking distance and braking distance.
  • A distraction increases the driver's reaction time.
  • Therefore, the thinking distance (and overall stopping distance) increases.

🧠 Exam Technique

Break your explanation down logically using cause and effect: Distraction → Increased reaction time → Increased thinking distance → Increased stopping distance.

🎯 Mark Allocation: 1 mark for linking stopping distance to thinking distance, 1 mark for noting reaction time increases due to distraction, 1 mark for stating thinking distance increases.

Part (0.3.5): Energy Transfers in Brakes

Explain why the temperature of the brakes increases as they are used. [2 marks]

✅ Correct Answer

  • Work is done by friction between the brake pads and the wheel/disc.
  • This transfers kinetic energy into thermal (internal) energy in the brakes.

❌ Common Errors

Simply writing "friction causes heat" is often penalized or fails to gain full marks. Examiners look for the proper scientific mechanism: work is done by friction, transferring energy to the thermal store.

🎯 Mark Allocation: 1 mark for mentioning work done by friction, 1 mark for an increase in thermal/internal energy.

Topics

Physics · P1: Energy · P5: Forces

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.