Interactive practical

Force, impulse and change of momentum

Measure trolley speeds and signed force–time traces, then compare impulse with change of momentum.

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A-Level Physics

Force, impulse and change of momentum

Pearson Edexcel 9PH0 — matched to your course.

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At a glance

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Practical focuses1
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Available question marks4

Practical focus

Practical method

Aim

Investigate the relationship between force, impulse and change of momentum using a force sensor and a moving trolley or ball.

Variables

Independent variable: change in momentum (mass or speed of the trolley)
Dependent variable: impulse from the area under the force–time graph
Control variables: the same force sensor and buffer; a level track; data logger sample rate.

Apparatus

Method

  1. Measure the mass of the trolley.
  2. Push the trolley so it passes through a light gate, hits the force sensor and rebounds through the gate.
  3. Record the speed before and after, and the force–time graph during impact.
  4. Calculate the change in momentum m(v − u), taking the rebound velocity as negative.
  5. Find the impulse from the area under the force–time graph and compare.
  6. Repeat with different speeds and masses.

Sample results

Trolley mass 0.80 kg; rebound velocities negative.

u / m s⁻¹v / m s⁻¹Δp / kg m s⁻¹Impulse (area) / N s
0.50−0.40−0.720.71
0.75−0.60−1.081.10
1.00−0.80−1.441.42

Analysis

In each run the impulse from the force–time graph equals the magnitude of the change in momentum to within about 2%, supporting FΔt = Δp. The force on the trolley acts opposite to its initial motion.

Uncertainty

Light-gate speeds are good to about ±0.01 m s⁻¹, giving about 2% uncertainty in Δp. The area under the force–time curve depends on the sample rate; a higher rate gives a smoother curve and a better area.

Errors and improvements

ProblemEffectImprovement
Track not levelSpeed changes between gate and sensorLevel the track so the trolley moves at constant speed
Low sample ratePeak force and area poorly recordedSample at 1 kHz or more
Sign of rebound ignoredΔp far too smallTake velocities after the rebound as negative

Risk assessment

HazardRiskControl measure
Trolley leaving the trackInjuryUse end stops
Pinched fingers at the bufferMinor injuryKeep hands clear on impact

Common mistakes

Worked analysis

A 0.80 kg trolley hits the sensor at 1.00 m s⁻¹ and rebounds at 0.80 m s⁻¹. Δp = 0.80 × (−0.80 − 1.00) = −1.44 kg m s⁻¹. If contact lasts 0.12 s, the average force = \(\dfrac{1.44}{0.12}\) = 12 N.

Common misconception

Impulse is the area under a force–time graph, not the maximum force. A softer buffer gives a smaller peak force over a longer time for the same impulse.

Assessment coverage

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2 questions available, worth 4 marks in total.

Practical 2
Foundation 1Standard 1
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Retrieval cards

What is impulse?

Force × time, equal to the change in momentum: FΔt = Δp.

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