Assessing grip on a banked road

10 marks · A-level · original practice case

Write your working and explanations below or on paper. Your responses stay in this page; PhysicsHelp does not submit or save them. Copy anything you want to keep before closing it. Check calculations separately from method marks. Use unrounded working between parts.

The situation

A road is banked at 12.0° to horizontal around a bend of radius 60.0 m. A 1000 kg vehicle follows a horizontal circular path at 18.0 m s⁻¹. The limiting static friction coefficient between tyres and road is 0.350. Treat the vehicle as a particle, use g = 9.81 m s⁻² and neglect air resistance. The road rises towards the outside of the bend.

Road and vehicle
Quantity	Value
Bank angle	12.0°
Radius	60.0 m
Speed	18.0 m s⁻¹
Friction coefficient	0.350

Part (a)

Calculate the speed at which no friction is required. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / m s⁻¹

Working

Part (b)

Describe the forces and explain the direction of friction at 18.0 m s⁻¹. Include a labelled force diagram on paper. [3 marks]

Your explanation

Part (c)

Determine the minimum friction coefficient required at this speed. [3 marks]

Give the numerical answer to 3 significant figures.

Final value (no unit)

Working

Part (d)

Assess whether the vehicle can follow the bend without slipping in the stated model. [2 marks]

Your explanation

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 11.2 m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Resolve the normal force vertically and horizontally: N cos θ = mg and N sin θ = mv²/r. (1 method mark)
Divide to obtain v² = rg tan θ and calculate 11.2 m s⁻¹. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (b) marking points

Show weight vertically down and the normal force perpendicular to the road, with labels. (1 independent mark)
Since the speed exceeds the no-friction value, the vehicle tends to slide up the bank; static friction acts down the slope. (1 independent mark)
Static friction has inward and downward components; the resultant force is horizontal towards the circle centre. (1 independent mark)

Credit equivalent correct explanations. Award each distinct point once; a contradiction invalidates the affected point.

Part (c) marking points

Final answer: 0.303 (no unit). Numerical checks assess the final value only; they do not check working or the number of digits typed.

For friction down the slope: N cos θ − f sin θ = mg and N sin θ + f cos θ = mv²/r. (1 method mark)
Solve f/N = (a cos θ − g sin θ)/(g cos θ + a sin θ), where a = v²/r. (1 method mark)
Obtain 0.303. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (d) marking points

The required coefficient 0.303 is below the limiting value 0.350, so static friction can provide the required force. (1 independent mark)
Real road/tyre conditions, speed variation or vehicle tipping may invalidate this particle-model prediction. (1 independent mark)

Accept a consistent threshold comparison after an arithmetic error in part (c), but do not accept friction acting up the slope at this speed.

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Related skills

Scope and reference sources

AQA 3.6.1.1 and 3.4.1.2; OCR A 5.2.2 and 3.2.1. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: banked-road-grip · version 1.1.0 · seed 0. Choose another case