Core practical 9: force and change in momentum

A typical method

• A trolley on a runway is pulled by a constant force, for example from a falling mass over a pulley, or is pushed against a force sensor.
• Tilt the runway slightly so the trolley rolls at constant velocity when given a push: this compensates for friction.
• Two light gates measure the velocity at two points, using v = card length / time the card blocks the gate. A data logger records the time between the gates.
• Measure the moving mass on a balance. Calculate Δp = mΔv.

Analysis

From FΔt = Δp, a graph of Δp (y-axis) against Δt (x-axis) for a constant force is a straight line through the origin with gradient = F. With a force sensor, the data logger plots a force-time graph; the area under it should equal mΔv measured with the light gates.

Errors

• Uncompensated friction makes the measured Δp smaller than FΔt.
• In a pulley set-up, the hanging mass also accelerates, so the moving mass is trolley plus hanging mass.
• Light gates and data loggers remove reaction time and give precise times for short events.

Worked example

A 0.80 kg trolley passes two light gates at 0.30 m/s and 0.75 m/s, 1.2 s apart. Find the average resultant force.

1. Δp = 0.80 x (0.75 - 0.30) = 0.36 kg m/s
2. F = Δp / Δt = 0.36 / 1.2

Answer: F = 0.30 N

Check yourself

1. A 5.0 cm card blocks a light gate for 0.025 s. Find the trolley's speed in m/s.
   Show answer

   2 m/s. v = 0.050 / 0.025 = 2.0 m/s.

2. A 0.50 kg trolley carries a 5.0 cm card. It blocks gate 1 for 0.125 s and gate 2 for 0.0625 s. Find its change in momentum in kg m/s.
   Show answer

   0.2 kg m/s. v1 = 0.40 m/s and v2 = 0.80 m/s, so Δp = 0.50 x 0.40 = 0.20 kg m/s.

3. A force sensor gives an area under the force-time graph of 0.84 N s. The 0.70 kg trolley started at rest. Predict its final speed in m/s.
   Show answer

   1.2 m/s. v = impulse / m = 0.84 / 0.70 = 1.2 m/s.

4. On a graph of Δp against Δt, Δp rises by 0.36 kg m/s over 0.90 s. Find the resultant force in N.
   Show answer

   0.4 N. F = gradient = 0.36 / 0.90 = 0.40 N.

5. In core practical 9, what does the gradient of a graph of change in momentum against time give?
   • The mass of the trolley
   • The acceleration
   • The kinetic energy
   • The resultant force
   Show answer

   The resultant force. FΔt = Δp, so Δp / Δt = F.

6. How is friction compensated for on the runway?
   • Make the runway perfectly level
   • Tilt the runway until a pushed trolley moves at constant velocity
   • Use a heavier trolley
   • Oil the light gates
   Show answer

   Tilt the runway until a pushed trolley moves at constant velocity. When a component of the weight balances friction, the only resultant force is the applied force.

7. Why is a data logger with a force sensor used for a collision?
   • It makes the force constant
   • It removes friction
   • It measures the mass of the trolley
   • The force changes rapidly over a very short time
   Show answer

   The force changes rapidly over a very short time. Collisions last a fraction of a second, so many readings per second are needed to draw the force-time graph.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Cognito: Edexcel International A level Physics specification and core practicals

Last updated 7 November 2026. Spotted a mistake? Tell us.