Impulse and force as rate of change of momentum

Force and momentum

Newton's second law in its general form says the resultant force equals the rate of change of momentum:

F = Δp / Δt

When the mass is constant, Δp = mΔv, so F = mΔv / Δt = ma, the form you met in Unit 1.

Impulse

Rearranging gives impulse: FΔt = Δp

• Impulse is a vector, measured in N s.
• 1 N s = 1 kg m/s, so impulse and momentum have the same unit.
• On a force-time graph, the area under the line is the impulse, even when the force varies.

Using it

For a fixed change in momentum, a longer contact time means a smaller average force. Airbags, crumple zones, crash mats and bending your knees on landing all increase Δt. In sport, 'following through' keeps the force acting for longer, so the ball gets a bigger Δp.

Watch the signs

If an object bounces back, its velocity changes sign. A ball hitting a wall at 20 m/s and rebounding at 15 m/s has Δv = 35 m/s, not 5 m/s.

Worked example

A 0.058 kg tennis ball arrives at 30 m/s and leaves the racket at 40 m/s in the opposite direction. Contact lasts 5.0 ms. Find the average force.

1. Take the outgoing direction as positive: u = -30 m/s, v = +40 m/s.
2. Δp = 0.058 x (40 - (-30)) = 0.058 x 70 = 4.06 kg m/s
3. F = Δp / Δt = 4.06 / 5.0 x 10^-3 = 812 N

Answer: Average force = 812 N (about 810 N)

Check yourself

1. A constant force of 15 N acts on a trolley for 0.40 s. Find the impulse in N s.
   Show answer

   6 N s. Impulse = FΔt = 15 x 0.40 = 6.0 N s.

2. A 0.45 kg football at rest is kicked and leaves the boot at 24 m/s. The contact time is 0.012 s. Find the average force in N.
   Show answer

   900 N. Δp = 0.45 x 24 = 10.8 kg m/s; F = 10.8 / 0.012 = 900 N.

3. A 0.15 kg ball hits a wall at 20 m/s and rebounds at 15 m/s. Contact lasts 0.050 s. Find the average force on the ball in N.
   Show answer

   105 N. Δv = 15 - (-20) = 35 m/s, so Δp = 0.15 x 35 = 5.25 kg m/s; F = 5.25 / 0.050 = 105 N.

4. A force-time graph for a kick is a triangle with a peak of 300 N and a base of 0.020 s. The 0.060 kg ball starts at rest. Find its speed afterwards in m/s.
   Show answer

   50 m/s. Impulse = area = 0.5 x 0.020 x 300 = 3.0 N s, so v = Δp / m = 3.0 / 0.060 = 50 m/s.

5. A 1200 kg car travelling at 15 m/s hits a barrier and stops in 0.60 s. Find the average force on the car in kN.
   Show answer

   30 kN. Δp = 1200 x 15 = 18000 kg m/s; F = 18000 / 0.60 = 30000 N = 30 kN.

6. Why does an airbag reduce the force on a driver in a crash?
   • It reduces the driver's change in momentum
   • It increases the time over which the driver's momentum changes
   • It increases the driver's mass
   • It reduces the driver's speed before the crash
   Show answer

   It increases the time over which the driver's momentum changes. The change in momentum is fixed by the speed; spreading it over a longer time gives a smaller force, F = Δp / Δt.

7. Which unit is the same as the newton second (N s)?
   • kg m/s²
   • J/s
   • N/m
   • kg m/s
   Show answer

   kg m/s. 1 N = 1 kg m/s², so 1 N s = 1 kg m/s, the unit of momentum.

8. What does the area under a force-time graph represent?
   • The work done
   • The acceleration
   • The change in momentum
   • The power
   Show answer

   The change in momentum. Area = force x time = impulse = change in momentum.

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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)
• Pearson Edexcel IAL Physics Unit 4 (WPH14) question paper, June 2024: list of data, formulae and relationships

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