AQA A-Level Physics Paper 1, June 2025: Question 7

8 marks · Medium difficulty · Short Answer

Analyze the motion of a racing car around horizontal and banked curved tracks, deriving relationships for radius, calculating slip speed, drawing acting forces, and explaining banked turning.

Practise this question

Question

Question 7 illustrates a car moving on an oval race track. Figure 10 shows an aerial view of a horizontal track with semicircular ends where a car moves at speed v along radius R. The question asks to derive R = 2Ek/F, calculate the slip speed given a centripetal force of 24 kN, mass of 1600 kg, and radius 230 m, and suggest a way to drive faster without slipping. Figure 11 shows the rear view of the car tilted on a banked track inclined to the horizontal, asking students to draw and label all acting forces and explain why banking allows a higher maximum speed.
Question text

07 A racing car of mass m travels around a horizontal oval track. The curved sections of

the track are semicircles.

07.1 At the instant shown in Figure 10 the car is moving with constant speed v in a circle

of radius R.

Figure 10

The car has kinetic energy Ek.

The resultant force acting on the car is F.

2Ek

Show that R =

F

[2 marks]

The maximum centripetal force that can be produced between the car’s tyres and the

track is 24 kN.

The minimum value of R is 230 m.

m = 1600 kg

07.2 The car just starts to slip when it travels on the track in a circular path of radius 230 m.

Deduce the speed of the car.

[1 mark]

speed = m s−1

07.3 The driver wishes to drive this car around the curved section of the track, without

slipping, at a speed greater than your answer to Question 07.2.

*14*Suggest one way in which the driver can achieve this.

[1 mark]

Figure 11 shows the car on the curved section of a different oval track.

The curved section of this track is a slope. This means that cars can travel at greater

speeds than on the track shown in Figure 10.

Figure 11

07.4 The car in Figure 11 stays at the same height on the curved section of the track.

There is no tendency for the car to slip up or down the track and therefore there is no

sideways friction on the tyres.

Draw and label, on Figure 11, arrows to show the direction of each force that acts on

the car as it travels around the curved section of the track.

[2 marks]

07.5 A car has a greater maximum speed on a sloped track than on a horizontal track of

the same radius.

Explain why.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 7 totaling 8 marks. 07.1 gives 2 marks for combining Ek = 1/2 m v^2 and F = m v^2 / R. 07.2 gives 1 mark for calculating v = 58.7 m s^-1. 07.3 gives 1 mark for increasing radius or friction between tyres and track. 07.4 gives 2 marks for drawing downward weight W from centre of mass and normal reaction forces perpendicular to the incline at the wheels. 07.5 gives 2 marks for explaining that the horizontal component of the normal reaction increases the centripetal force, allowing greater speed.

Question Answers Additional comments/Guidance Mark AO

1 mv2

07.1 E = mv2 AND F = In MP1 do not condone use of r for R unless 2 AO2

k

2 R explicitly equated. E.g.r =R

Combines two equations correctly Do not allow use of r for R in MP2.

In MP2 allow r becoming R without

2E explanation.

(To give R = k )

F m E2k

Eg for MP2: F = (need to see m or m)

mR

OR mv2 = FR = 2E

k

07.2 v = 58.7 (m s−1) Calculator value: 58.73670062 1 AO2

Accept 59

Do not accept rounding error.

07.3 Idea that the radius of the path of the car can be increased eg Start at the outside of the track and drive 1 AO3

to the inside of the curve

OR

Condone ‘reduce mass of car’

Idea that friction between tyres and track can be increased Treat banking answer as neutral.

Question Answers Additional comments/Guidance Mark AO 21

07.4 W must be straight and vertically downwards 2 1 × AO1

with a line of action that passess 1 × AO2

approximately through the centre of mass

Normal reaction forces with lines of action

through tyres at right angles to slope

Condone any symbols for W, N1 and N2

Award max 1 if centripetal force appears.

Treat components as neutral

W

N1 and N2

2207.5 (with the slope) there is a (horizontal) component of the 2 AO3

normal reaction force that increases the (maximum)

centripetal force

Idea that v can be greater (for same R) when centripetal

force greater

Total 8

How to answer it

Circular Motion: Forces on Flat and Banked Tracks

AQA A-Level Physics • Further Mechanics • Circular Motion • 8 Marks Total
📌 What this question tests

This question evaluates your ability to model circular motion on both flat and banked surfaces:

  • Algebraic Derivation: Linking kinetic energy (Ek = ½mv²) with centripetal force (F = mv²/R).
  • Quantitative Analysis: Calculating critical slipping speed and converting units (kN to N).
  • Physical Insights: Identifying real-world modifications to increase cornering speed.
  • Free-Body Force Diagrams: Accurately drawing real forces on banked surfaces without including fictitious forces.
  • Resolving Vectors: Explaining how normal contact forces provide centripetal acceleration on inclined tracks.
Question 07.1 • 2 Marks

Kinetic Energy & Centripetal Force Derivation

Show that R = 2Ek / F

📐 Step-by-Step Derivation

  1. State both fundamental equations using the defined notation:
    Ek = ½mv²  and  F = mv² / R [Mark 1]
  2. Rearrange kinetic energy to isolate the common term mv² :
    mv² = 2Ek
  3. Substitute into the centripetal force equation:
    F = 2Ek / R
  4. Rearrange to make radius R the subject:
    R = 2Ek / F [Mark 2]

❌ Common Errors & Notation Pitfalls

  • Mismatched Symbol Case: Using lowercase r instead of uppercase R . The mark scheme specifically rejects r unless explicitly stated as r = R .
  • Skipping Steps: Jumps straight from the formulas to the final line without showing substitution or equating mv² . "Show that" questions require every step of algebra to be fully visible.
Question 07.2 • 1 Mark

Calculating Maximum Slip Speed

Deduce the speed of the car when it starts to slip (R = 230 m, m = 1600 kg, Fmax = 24 kN)

📐 Calculation

  1. Convert prefix: F = 24 kN = 24 000 N = 2.4 × 10⁴ N
  2. Rearrange formula:
    F = mv² / R &implies; v² = FR / m &implies; v = √(FR / m)
  3. Substitute values:
    v = √( (24 000 × 230) / 1600 )
    v = √(3450) = 58.7367... m s⁻¹
  4. State with appropriate precision:
    Speed = 58.7 m s⁻¹ (or 59 m s⁻¹ to 2 s.f.)
Award 1 mark for 58.7 m s⁻¹ or 59 m s⁻¹. Rounding errors like 58 m s⁻¹ score 0.

🧠 Exam Technique: Alternative Route

You can also use the relationship derived in 07.1:

  • Ek = ½FR = 0.5 × 24 000 × 230 = 2.76 × 10⁶ J
  • v = √(2Ek / m) = √(5.52 × 10⁶ / 1600) = 58.7 m s⁻¹
  • Prefix Alert: Forgetting that 24 kN = 24 000 N is the single most common reason candidates lose this mark.
Question 07.3 • 1 Mark

Increasing Cornering Speed on a Flat Track

Suggest one way the driver can achieve a higher speed without slipping

✅ Accepted Suggestions

  • Increase the radius of the path: Drive a wider arc (e.g., take the "racing line" by starting wide on the outside and cutting through the curve).
  • Increase grip / tyre friction: Fit softer compound tyres, tyres with better tread, or add aerodynamic downforce (spoilers).
  • (Condone: Reduce the mass of the car).

❌ Common Errors & Neutral Answers

  • Banking the track: The question asks what the driver can do to drive this car on this track. The mark scheme treats "bank the track" as neutral because banking is introduced in the next sub-question.
  • Vague answers: Simply writing "go faster" or "drive better" receives zero credit.
Question 07.4 • 2 Marks

Free-Body Diagram on a Banked Track

Draw and label arrows for all forces acting on the car (no sideways friction)

Exact Drawing Specification Required by Examiner:
  • Weight (W or mg): A single straight arrow pointing vertically downwards from near the centre of mass. [1 Mark]
  • Normal Reaction Forces (N, R, or N₁ and N₂): Arrow(s) pointing perpendicular (at 90°) to the slope surface, originating at the contact patches of the tyres (or a single normal contact arrow perpendicular to the slope). [1 Mark]

💡 Physical Reality

There are only two real forces acting on the vehicle:

  1. Gravitational pull of the Earth (Weight, W).
  2. Electrostatic contact push from the track surface (Normal reaction, N).

❌ Fatal Error: "Centripetal Force" Arrow

  • Never draw a separate "centripetal force" arrow! Centripetal force is not an additional physical force; it is the resultant horizontal force.
  • Mark scheme penalty: Award maximum 1 mark if any arrow labeled "centripetal force" or "Fc" is drawn.
Question 07.5 • 2 Marks

Explanation: Advantage of a Banked Track

Explain why a car has a greater maximum speed on a sloped track than on a horizontal track of the same radius

✅ Model Answer (2 Marks)

  • Point 1 (Origin of extra force): On a banked track, there is a horizontal component of the normal reaction force ( N sin θ ) acting towards the centre of the circle, which increases the maximum centripetal force available. [1 Mark]
  • Point 2 (Link to speed): Since F = mv² / R , a greater centripetal force allows a greater speed v for the same radius R. [1 Mark]

🧠 Top-Level Exam Strategy

  • Always specify the direction: Don't just say "the normal reaction helps." You must state the horizontal component of the normal reaction.
  • Explicit equation link: Clearly state that F ∝ v² (for fixed R and m). Greater resultant horizontal force allows larger velocity before the threshold of slipping is reached.

Topics

Physics · 3.6 Further mechanics and thermal physics (A-level only) · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.