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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

F\Delta t = \Delta p = mv - muImpulse = force × time = change in momentum
SymbolMeaningUnit
FForce (the average force, if it varies)N
ΔtTime the force acts fors
ΔpChange in momentumkg m/s
mMasskg
u, vVelocity before and afterm/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 findUse
ImpulseFΔt, or mv − mu
Average forceF = (mv − mu) ÷ Δt
Time of contactΔt = (mv − mu) ÷ F
Final velocityv = 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.

F\Delta t = 250 \times 0.040 = 10\ \text{N s}

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.

\Delta p = mv - mu = 0.41 \times 22 - 0 = 9.02\ \text{N s}
F = \frac{\Delta p}{\Delta t} = \frac{9.02}{0.009} = 1000\ \text{N}

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:

\Delta p = 0.16 \times (-12) - 0.16 \times 15 = -4.32\ \text{N 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.

Force on a ball during a hit
51015202530354045505560-100100200300400500600700peak 600 N at 25 mscontact ends at 50 ms
Drag to pan, scroll to zoom. x from 0 to 60, y from -120 to 720.

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:

\text{impulse} = \text{area} = \frac{1}{2} \times 0.050 \times 600 = 15\ \text{N s}

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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Momentum and Impulse Calculator

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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.
FΔt = 40 × 0.50 = 20 N s
2. A 2.0 kg trolley speeds up from rest to 5.0 m/s in 0.050 s. Find the average force.
Δp = 2.0 × 5.0 = 10 N s, so F = 10 ÷ 0.050 = 200 N
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.
Δp = 0.060 × (20 + 30) = 3.0 N s
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.
Area = ½ × 0.20 × 100 = 10 N s
5. Why does a crash mat reduce the force on a gymnast who lands on it?
It makes the stop last longer. Δp does not depend on the mat, so the force, Δp ÷ Δt, is smaller.

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.

Papers reset in

Common questions

What is the formula for impulse?
Impulse = FΔt = Δp = mv − mu: force × time equals the change in momentum.
What is the unit of impulse?
The newton second, N s, which is the same as kg m/s.
Is impulse the same as momentum?
No. Impulse is the change in momentum. They share a unit because one equals a change in the other.
How do you find impulse from a force-time graph?
Find the area under the graph. For a straight-line shape use the area formula; for a curve, count squares.
Is impulse a vector?
Yes. It has the direction of the force, which is the direction of the change in momentum.
Why do airbags reduce injuries?
They make the person stop over a longer time. With the same change in momentum, a longer time means a smaller force.

Revision cards

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Where to go next

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Physics · MechanicsImpulse calculator: J = FΔt and change in momentumFree impulse calculator: find impulse from force and time or from a change in velocity, then the average force or contact time, with every step shown.