The impulse formula
The impulse formula is impulse = FΔt = Δp: the force on an object multiplied by the time it acts equals the object's change in momentum. Written out, FΔt = mv − mu, where m is the mass, u the starting velocity and v the final velocity. Impulse is measured in newton seconds, N s, which is the same as kg m/s. A force of 250 N acting for 0.040 s gives an impulse of 250 × 0.040 = 10 N s, so it changes the object's momentum by 10 kg m/s.
The impulse equation
| Symbol | Meaning | Unit |
|---|---|---|
| F | Force (the average force, if it varies) | N |
| Δt | Time the force acts for | s |
| Δp | Change in momentum | kg m/s |
| m | Mass | kg |
| u, v | Velocity before and after | m/s |
It comes straight from Newton's second law written in terms of momentum, F = Δp ÷ Δt: multiply both sides by Δt. AQA's A-level specification states it as impulse = change in momentum, FΔt = Δ(mv), where F is constant, and its data booklet gives FΔt = Δ(mv). At GCSE, AQA's equations sheet gives the same idea for Higher tier as force = change in momentum ÷ time taken. Our guide to momentum covers p = mv and conservation of momentum.
Impulse units
1 N s = 1 kg m/s, because a newton is 1 kg m/s². So impulse and momentum have the same unit, which they must, since one equals a change in the other.
Rearranging the impulse formula
| To find | Use |
|---|---|
| Impulse | FΔt, or mv − mu |
| Average force | F = (mv − mu) ÷ Δt |
| Time of contact | Δt = (mv − mu) ÷ F |
| Final velocity | v = u + FΔt ÷ m |
Impulse formula worked examples
Example 1: from force and time
A force of 250 N acts on a trolley for 0.040 s. Find the impulse.
Example 2: a kick
A footballer kicks a stationary 0.41 kg ball, sending it off at 22 m/s after a 0.009 s contact. Find the impulse and the average force.
The average force is about 1000 N, for less than a hundredth of a second.
Example 3: a rebound
A 0.16 kg ball hits a wall at 15 m/s and bounces straight back at 12 m/s. The contact lasts 0.020 s. Find the impulse and the average force from the wall.
Momentum is a vector, so choose a direction as positive. Taking towards the wall as positive, u = +15 m/s and v = −12 m/s:
The impulse is 4.32 N s away from the wall, and the average force is 4.32 ÷ 0.020 = 216 N. Using 15 − 12 = 3 m/s instead of 15 + 12 = 27 m/s is the most common mistake in rebound questions.
Now try one in the course. Work out the impulse, then press Mark to check it there.
Find a force, time or change in momentum using Ft = Δp.
Press Mark to have your answer marked in the Laws of Motion course.
Impulse and the area under a force-time graph
Real impact forces are not constant: they rise and fall during the contact. AQA's A-level specification asks students to understand the significance of the area under a force-time graph. That area is the impulse, so it gives the change in momentum even when F varies.
Here x is the time in milliseconds and y is the force in newtons. The force rises to 600 N and falls back to zero over 50 ms.
Example 4: impulse from the graph
The shape under the line is a triangle, 50 ms = 0.050 s wide and 600 N tall:
If the ball has a mass of 0.30 kg and starts at rest, its final speed is 15 ÷ 0.30 = 50 m/s. For a curved graph, count squares instead: find the impulse of one square, then multiply by the number of squares under the curve.
Why longer contact times mean smaller forces
Rearranging the impulse formula gives F = Δp ÷ Δt. For a fixed change in momentum, such as a car stopping from a given speed, making the stopping time longer makes the force smaller. AQA's A-level specification relates impact forces to contact times, with examples such as kicking a football, crumple zones and packaging. At GCSE, AQA asks students to explain safety features such as air bags, seat belts, gymnasium crash mats, cycle helmets and cushioned playground surfaces in terms of the rate of change of momentum.
Example 5: crumple zones
A 1000 kg car travelling at 13 m/s stops in a crash. Compare the average force if it stops in 0.10 s with the force if crumple zones stretch the stop to 0.50 s.
The change in momentum is the same either way: 1000 × 13 = 13 000 kg m/s.
- Stopping in 0.10 s: F = 13 000 ÷ 0.10 = 130 000 N
- Stopping in 0.50 s: F = 13 000 ÷ 0.50 = 26 000 N
Five times the stopping time means one fifth of the force. The same idea explains why you bend your knees when you land from a jump, and why a cricket fielder pulls their hands back when catching a ball.
The Momentum and Impulse Calculator works out momentum, impulse and average force, and models collisions, with each step shown.
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Common mistakes
- Ignoring direction in a rebound. Velocities in opposite directions have opposite signs. The change in speed of a bouncing ball is the sum of the two speeds.
- Time in ms. 9 ms is 0.009 s.
- Mass in grams. 160 g is 0.16 kg.
- Using the peak force for the impulse. Impulse uses the average force, or the area under the force-time graph.
- Giving impulse in N or J. The unit is N s (or kg m/s).
Impulse questions with answers
1. A 40 N force acts for 0.50 s. Find the impulse.
2. A 2.0 kg trolley speeds up from rest to 5.0 m/s in 0.050 s. Find the average force.
3. A 60 g ball hits a bat at 30 m/s and leaves at 20 m/s in the opposite direction. Find the size of the impulse.
4. A force rises steadily from 0 to 100 N and falls steadily back to 0, taking 0.20 s in total. Find the impulse.
5. Why does a crash mat reduce the force on a gymnast who lands on it?
Practise momentum in a short timed paper
A-level Physics: Year 1 and GCSE Physics Higher each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so momentum and impulse turn up regularly but not in every paper. The Laws of Motion course has a section on impulse with crash examples.
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Common questions
What is the formula for impulse?
What is the unit of impulse?
Is impulse the same as momentum?
How do you find impulse from a force-time graph?
Is impulse a vector?
Why do airbags reduce injuries?
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