AQA GCSE Physics Physics Paper 2 (Higher), November 2021: Question 1

16 marks · Standard Demand difficulty · Short Answer

Calculate stopping distance, deceleration, and piston surface area using forces, motion, and pressure equations and graphs.

Practise this question

Question

Question text

01 The thinking distance and braking distance for a car vary with the speed of the car.

01.1 Explain the effect of two other factors on the braking distance of a car.

Do not refer to speed in your answer.

[4 marks]

01.2 Which equation links acceleration (a), mass (m) and resultant force (F).

[1 mark]

Tick ( ) one box.

resultant force = mass × acceleration

resultant force = mass × acceleration2

mass

resultant force = 2

acceleration

mass

resultant force =

acceleration

01.3 The mean braking force on a car is 7200 N.

The car has a mass of 1600 kg.

Calculate the deceleration of the car.

[3 marks]

Deceleration = m/s

01.4 Figure 1 shows how the thinking distance and braking distance for a car vary with

the speed of the car.

Figure 1

Determine the stopping distance when the car is travelling at 80 km/h.

[2 marks]

Stopping distance = m

Figure 2 shows part of the braking system for a car.

Figure 2

01.5 Which equation links area of a surface (A), the force normal to that surface (F) and

pressure (p).

[1 mark]

Tick ( ) one box.

p = F × A

p = F × A2

p =

p = 7

01.6 When the brake pedal is pressed, a force of 60 N is applied to the piston.

The pressure in the brake fluid is 120 000 Pa.

Calculate the surface area of the piston.

Give your answer in standard form.

Give the unit.

[5 marks]

Surface area (in standard form) = Unit

Mark scheme

Show the mark scheme

Question 1

AO/

Question Answers Extra information Mark

Spec. Ref

01.1 Level 2: Relevant points (reasons / causes) are identified, given in AO1

3–4

detail and logically linked to form a clear account. 4.5.6.3.3

4.5.6.3.4

Level 1: Point are identified and stated simply, but their relevance 4.1.1.2

1–2

is not clear and there is no attempt at logical linking.

No relevant content 0

Indicative content

Factors

• poor condition of tyres

• poor road surface

• wet or icy road

• poor/worn brakes

Explanation

• because of decreased friction

Factors

• increased mass of car/passengers

Explanation

• increases kinetic energy of car

• more work needs to be done to stop car

• increases momentum of the car

Factor

• road slopes downhill

Explanation

• (a component of) gravity opposes the braking force

• resultant (braking) force is reduced

allow answers in terms of reducing braking distance throughout

A single factor with no related explanation is insufficient to score a

mark

01.2 resultant force = mass × 1 AO1

acceleration 4.5.6.2.2

01.3 7200 = 1600 × a ignore negatives throughout 1 AO2

4.5.6.2.2

a = 7200 1

1600

a = 4.5 (m/s2) 1

01.4 15 (m) 38 (m) two correct values identified 1 AO3

4.5.6.3.1

= 53 (m) allow the correct addition of a 1

misread braking distance and

/or a misread thinking distance

taken from the graph

01.5 F 1 AO1

p =

A 4.5.5.1.1

01.6 120 000 = 60 1 AO2

A = 60 . 1 AO2

120 000

A = 0.0005 1 AO2

A = 5 (.0) × 10–4 allow an answer given to 2 sig 1 AO2

figs from an incorrect calculation

m2 using the given data 1 AO1

4.5.5.1.1

Total 16

How to answer it

Forces, Braking Distance, and Pressure Study Guide

What this question tests

This exam question tests your understanding of factors affecting vehicle stopping distances, Newton's Second Law equations (F = m × a), interpreting graphical data, and pressure calculations involving force and surface area (p = F / A). You will need to demonstrate recall of formulas, rearrange equations, use standard form, and apply correct units.

Part 01.1 (4 Marks)

Factors affecting braking distance

✅ Acceptable Answers (Examples)

Worn tyres or poor road surface: Decreases friction between the tyre and road, increasing braking distance.
Wet or icy roads: Reduces grip/friction, meaning less braking force can be applied without skidding.
Increased mass (e.g., more passengers): Increases the kinetic energy of the car, requiring more work and distance to stop.
Downhill slopes: Gravity opposes the braking force, reducing the total resultant deceleration force.

🧠 Exam Technique & Marking

Marked using a Level of Response scheme (Level 1: 1-2 marks, Level 2: 3-4 marks). To get top marks, you must not just state a factor, but logically link it to a physical reason (e.g., how it affects friction or kinetic energy).

❌ Common Errors

Mentioning speed: The question explicitly states "Do not refer to speed". Mentioning speed instantly limits your score.
Confusing thinking and braking factors: Alcohol, tiredness, and mobile phones affect thinking distance (reaction time), not braking distance.

Part 01.2 (1 Mark)

Equation linking acceleration, mass, and resultant force

✅ Correct Answer

resultant force = mass × acceleration

💡 Key Knowledge

This is Newton's Second Law of Motion ( F = m × a ). Ensure you recall standard physics formulas off-by-heart for multiple-choice sections.

Part 01.3 (3 Marks)

Calculating Deceleration

📐 Step-by-Step Calculation

Identify formula: F = m × a
Substitute known values: 7200 = 1600 × a
Rearrange and solve: a = 7200 / 1600
Final Answer: a = 4.5 m/s²

🧠 Examiner Commentary

1 mark is awarded for correct substitution, 1 mark for rearranging the equation, and 1 mark for the correct final numerical answer. Unit m/s² is expected.

Deceleration = 4.5 m/s²

Part 01.4 (2 Marks)

Determining Stopping Distance from a Graph

✅ Correct Answer

Thinking distance at 80 km/h = 15 m
Braking distance at 80 km/h = 38 m
Total Stopping Distance = 53 m

🧠 Exam Technique

Always read graph axes carefully. Remember that Stopping Distance = Thinking Distance + Braking Distance. You must read both curves on Figure 1 at 80 km/h and add them together.

Stopping distance = 53 m

Part 01.5 (1 Mark)

Equation linking area, force, and pressure

✅ Correct Answer

p = F / A (Pressure equals Force divided by Area)

💡 Key Knowledge

Pressure is defined as force per unit area. Higher force or smaller area results in greater pressure.

Part 01.6 (5 Marks)

Calculating Surface Area in Standard Form

📐 Step-by-Step Calculation

Identify formula: p = F / A rearranged to A = F / p
Substitute values: A = 60 / 120 000
Calculate decimal value: A = 0.0005
Convert to standard form: 5 × 10⁻⁴
Include correct unit: m²

❌ Calculation Traps & Common Errors

Inverted formula: Dividing pressure by force ( p / F ) instead of force by pressure ( F / p ).
Standard form mistakes: Writing 0.5 × 10⁻³ (which is mathematically valid, but standard form requires a coefficient between 1 and 10, i.e., 5 × 10⁻⁴ ).
Missing units: Forgetting to add m² for area, losing the final mark.

Surface area (in standard form) = 5 × 10⁻⁴ Unit: m²

Topics

Physics · P5: Forces · P1: Energy

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.