AQA GCSE Physics Physics Paper 2 (Foundation), November 2021: Question 8
16 marks · Standard Demand difficulty · Short Answer
Explain factors affecting braking distance, identify equations linking force, mass, acceleration, and pressure, and perform related calculations including deceleration and surface area in standard form.
Practise this questionQuestion
Question text
08 The thinking distance and braking distance for a car vary with the speed of the car.
08.1 Explain the effect of two other factors on the braking distance of a car.
Do not refer to speed in your answer.
[4 marks]
08.2 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
08.3 The mean braking force on a car is 7200 N.
The car has a mass of 1600 kg.
Calculate the deceleration of the car.
*35* [3 marks]
37 2
Deceleration = m/s
08.4 Figure 18 shows how the thinking distance and braking distance for a car vary with
the speed of the car.
Figure 18
Determine the stopping distance when the car is travelling at 80 km/h.
[2 marks]
Stopping distance = m
Figure 19 shows part of the braking system for a car.
Figure 19
08.5 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
F
p =
A
A
p = 39
F
08.6 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 8
AO/
Question Answers Mark
Spec. Ref
08.1 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: Points 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
08.2 resultant force = mass × 1 AO1
acceleration 4.5.6.2.2
08.3 7200 = 1600 × a ignore negatives throughout 1 AO2
4.5.6.2.2
a = 7200 1
1600
16 a = 4.5 (m/s2) 1
08.4 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
08.5 F 1 AO1
p = 4.5.5.1.1
A
08.6 120 000 = 60 1 AO2
A
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 and Pressure Study Guide
What this question tests
This exam question assesses your understanding of vehicle stopping distances, Newton's Second Law (F = ma), reading data from line graphs, and hydraulic pressure calculations (p = F / A). You will need to recall equations, substitute values correctly, manipulate formulas, and format answers using standard form and correct units.
Factors Affecting Braking Distance
✅ Correct Answer Structure (Level 2)
To get 3–4 marks, you must identify two distinct factors that affect braking distance (excluding speed) and provide a clear scientific explanation for each linked to friction, kinetic energy, or work done.
- Factor 1: Poor condition of tyres / poor road surface / wet or icy road / worn brakes.
Explanation: Decreases friction between the tyres and the road. - Factor 2: Increased mass of car or passengers.
Explanation: Increases the kinetic energy of the car, meaning more work needs to be done by the brakes to stop it. - Factor 3: Road slopes downhill.
Explanation: A component of gravity opposes the braking force, reducing the overall resultant braking force.
🧠 Exam Technique & Mark Scheme
- Avoid Speed: The question explicitly states: "Do not refer to speed in your answer." Mentioning speed will instantly cap your mark.
- The "Because" Rule: Simply stating a factor (e.g., "icy roads") only scores Level 1 (1–2 marks). You must explain *why* it affects the stopping distance to reach Level 2.
Equation Linking Acceleration, Mass and Resultant Force
✅ Correct Answer
resultant force = mass × acceleration (Tick the first box)
💡 Key Knowledge (Newton's Second Law)
This is standard recall of F = m × a . Make sure you recognise how to rearrange it if needed in future calculations.
Calculating Deceleration
📐 Step-by-Step Calculation [3 marks]
- State the formula: F = m × a (or a = F / m )
- Substitute the values: 7200 = 1600 × a
- Rearrange and solve:
a = 7200 / 1600 = 4.5 m/s²
❌ Common Errors
- Dividing mass by force instead of force by mass ( 1600 / 7200 ). Always check your units: acceleration should be in m/s², so a small number divided by a massive mass makes no physical sense for a standard car braking force!
Determining Stopping Distance from a Graph
📐 Step-by-Step Working [2 marks]
- Locate 80 km/h on the x-axis of Figure 18 and draw a vertical line up to both curves.
- Read the thinking distance: approx. 15 m.
- Read the braking distance: approx. 38 m.
- Add them together for stopping distance: 15 + 38 = 53 m .
🧠 Graph Reading Tips
- Stopping distance is always Thinking Distance + Braking Distance . Examiners award 1 mark for correctly identifying the two values from the graph lines, and 1 mark for their correct sum.
Equation for Pressure, Force and Area
✅ Correct Answer
p = F / A (Tick the third box)
💡 Key Knowledge
Pressure equals force normal to a surface divided by the area of that surface. Units are Pascals (Pa) or N/m².
Calculating Surface Area (Standard Form)
📐 Step-by-Step Calculation [5 marks]
- State equation: p = F / A
- Substitute known values: 120 000 = 60 / A
- Rearrange for area: A = 60 / 120 000
- Calculate decimal value: A = 0.0005 m²
- Convert to standard form: 5 × 10⁻⁴ m²
❌ Common Traps & Errors
- Inverting the fraction: Doing 120 000 / 60 instead of 60 / 120 000 .
- Standard Form formatting: Writing 0.5 × 10⁻³ (not proper standard form because the coefficient is less than 1). Ensure the number is between 1 and 10: 5 × 10⁻⁴ .
- Forgetting units: Always include m² for area!
Topics
Physics · P5: Forces · P1: Energy
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Foundation), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.