Braking a loaded cart

9 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 120 kg cart initially moves at 6.0 m s⁻¹ on a horizontal surface. A brake applies the opposing force shown below. Between listed positions force varies linearly. Other resistive forces are negligible. Treat these model values as exact.

Use the axes and units. The table provides the text alternative.

Braking force against distance
Distance / m	Opposing force / N
0	400
1	400
2	800
3	800

Part (a)

Determine the signed work done by the brake on the cart over the first 3.0 m. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / J

Working

Part (b)

Calculate the speed after travelling 3.0 m. [3 marks]

Give the numerical answer to 2 significant figures.

Final value / m s⁻¹

Working

Part (c)

Explain whether the deceleration is constant over this distance. [2 marks]

Your explanation

Part (d)

Explain where the lost kinetic energy is transferred. [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: -1.8 × 10³ J. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Area magnitude is 400 + 600 + 800 = 1800 J; force opposes displacement. (1 method mark)
Work on the cart is −1.8 × 10³ J; kinetic energy decreases by 1.8 × 10³ J. (1 accuracy mark)

No earlier numerical result is required. Award method credit for valid working and accuracy credit for the stated answer. Accept a defensible equivalent method; wrong physics cannot earn dependent accuracy credit.

Common errors: This uses the maximum braking force throughout. Find each area.

Part (b) marking points

Final answer: 2.4 m s⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Initial kinetic energy is 2160 J. (1 method mark)
Use Kfinal = Kinitial + Wbrake = 2160 − 1800 = 360 J; then use ½mv². (1 method mark)
Obtain 2.4 m s⁻¹ to 2 significant figures. (1 accuracy mark)

Use Kfinal = 2160 J + the signed work from part (a). Carry forward an arithmetic error only if Kfinal is non-negative. A positive braking work contradicts the stated force direction and is not a valid physical model. Credit an equivalent energy-loss magnitude method.

Common errors: This leaves the speed unchanged despite work done by the brake.

Part (c) marking points

The opposing force changes between 1 and 2 m. (1 independent mark)
For constant mass acceleration magnitude follows F/m, so deceleration is not constant throughout. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Part (d) marking points

The brake does work against the motion. (1 independent mark)
Energy is transferred mainly to internal energy of the brake/cart/surroundings. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Common errors: Energy is destroyed: energy is transferred.

Report a problem with this case

Related skills

Scope and reference sources

AQA 3.4.1 (selected); OCR A Module 3 (selected). These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: braking-cart · version 1.1.0 · seed 0. Choose another case