A trolley passing a motion sensor

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 0.80 kg trolley moves along a horizontal track. A motion sensor records the values below. Between 0 and 2.0 s, assume constant acceleration. After 2.0 s, the resultant horizontal force is zero. Treat these model values as exact for this exercise.

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

Model velocity readings
t / s	v / m s⁻¹
0	0
1	1.2
2	2.4
3	2.4
4	2.4

Part (a)

Determine the acceleration during the first 2.0 s. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / m s⁻²

Working

Part (b)

Calculate the resultant force during that interval. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / N

Working

Part (c)

Determine the distance travelled in the first 4.0 s. [3 marks]

Give the numerical answer to 2 significant figures.

Final value / m

Working

Part (d)

Explain why zero resultant force after 2.0 s does not mean that the trolley stops. [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: 1.2 m s⁻². Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use Δv/Δt over 0–2.0 s. (1 method mark)
Obtain 1.2 m s⁻². (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 is the final velocity, not Δv/Δt.

Part (b) marking points

Final answer: 0.96 N. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use F = ma, with the acceleration from part (a). (1 method mark)
Obtain 0.96 N. (1 accuracy mark)

Use the acceleration from part (a) in F = 0.80a. A numerical error may be carried forward; mass must not be omitted.

Common errors: This uses acceleration as force. Multiply by mass.

Part (c) marking points

Final answer: 7.2 m. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use the triangular area, ½ × 2.0 × 2.4. (1 method mark)
Add the rectangular area, 2.0 × 2.4. (1 method mark)
Obtain 7.2 m. (1 accuracy mark)

Common errors: This assumes maximum speed throughout; include the accelerating interval.

Part (d) marking points

Zero resultant force means zero acceleration. (1 independent mark)
Velocity therefore remains constant at 2.4 m s⁻¹. (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: A force is needed to maintain velocity: a resultant force is needed to change velocity.

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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: trolley-sensor · version 1.1.0 · seed 0. Choose another case