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Impulse calculator

An impulse calculator works out the impulse on an object: the force on it multiplied by the time the force acts, J = FΔt. Impulse also equals the change in momentum, J = Δp = m(v − u), so you can find it either way. Its unit is the newton second (N s), which is the same as kg m/s. Our Momentum and Impulse Calculator finds the impulse, the average force or the contact time, with the working.

It opens on momentum. To switch it to impulse, choose Impulse in the top row under Calculate, then pick what to find from the second row.

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Momentum and Impulse Calculator

Calculate momentum, impulse and what happens in collisions

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How to use the impulse calculator

  1. Under Calculate, choose Impulse in the top row.
  2. In the second row, pick what you want: Impulse from force and time, From Δv for impulse from mass and the change in velocity, Force for the average force, or Time for the contact time.
  3. Type the values. Each mode starts on an example, such as a 2,000 N force for 0.002 s, so clear it first. Choose units under the inputs: velocity can be in km/h or mph, and time in hours or days.
  4. Read the answer, then the working: the formula, any rearranging, the numbers put in and the result.

Give velocities a sign. Take one direction as positive, and a velocity the other way is negative: a ball bouncing back at 15 m/s has v = −15. The From Δv mode also gives the momentum before and after and draws them, and the Force mode gives the impulse alongside the force. Practice and help sets impulse and collision questions with worked solutions.

The impulse formula

J = F \Delta t = \Delta p = m(v - u)J is impulse in N s, F is force in N, Δt is time in s, m is mass in kg, u and v are the velocities before and after in m/s
To findFormulaMode in the calculator
Impulse from force and timeJ = FΔtImpulse
Impulse from a change in velocityJ = m(v − u)From Δv
Average forceF = m(v − u) ÷ ΔtForce
Contact timeΔt = J ÷ FTime

The link between the two forms is the impulse-momentum theorem. It comes straight from Newton's second law: F = ma = m(v − u)/Δt, so FΔt = m(v − u). Our momentum calculator guide covers p = mv and collisions with the same tool.

Worked examples you can check in the calculator

Example 1: impulse from force and time

A tennis racket pushes on a ball with an average force of 480 N for 0.005 s. What impulse does it give the ball?

J = 480 \times 0.005 = 2.4\ \text{N s}

Choose Impulse, then Impulse, and type 480 and 0.005. The calculator gives 2.4 N s and draws the force against time, with the impulse as the area of the rectangle.

Example 2: impulse from a bounce

A 0.15 kg ball hits a wall at 20 m/s and bounces straight back at 15 m/s. Take towards the wall as positive, so u = 20 and v = −15:

J = 0.15 \times (-15 - 20) = 0.15 \times (-35) = -5.25\ \text{N s}

Choose From Δv and type 0.15, 20 and −15. The calculator gives −5.25 N s, with the momentum before (3 kg m/s) and after (−2.25 kg m/s). The minus sign says the impulse acts away from the wall. If you forget the bounce changes direction and use v = 15, you get −0.75 N s, which is seven times too small.

Example 3: the average force in a collision

The ball in Example 2 is in contact with the wall for 0.012 s. What is the average force on it?

F = \frac{-5.25}{0.012} = -437.5\ \text{N}

Choose Force and type 0.15, 20, −15 and 0.012. The calculator gives −437.5 N and the impulse, −5.25 N s.

Try one in the course, then press Mark:

Find a force, time or change in momentum using Ft = Δp.

Press Mark to have your answer marked in the Laws of Motion course.

Why longer contact times mean smaller forces

For a fixed change in momentum, F = Δp ÷ Δt, so stretching the stopping time shrinks the force. This is how airbags, crumple zones, bike helmets and soft landing mats work.

Example 4: an airbag

A 70 kg passenger moving at 15 m/s is brought to rest in a crash. Their change in momentum is 70 × (0 − 15) = −1,050 N s, whatever stops them. An airbag stops them in 0.15 s; a dashboard would stop them in about 0.015 s:

F_{\text{airbag}} = \frac{-1050}{0.15} = -7000\ \text{N} \qquad F_{\text{dashboard}} = \frac{-1050}{0.015} = -70000\ \text{N}

Ten times the time gives a tenth of the force. Check both in the Force mode. To ask the other way round, how long the stop must last to keep the force below 5,000 N, use the Time mode with J = 1050 and F = 5000: it gives 0.21 s. That mode needs both values positive, so type sizes only.

Impulse from a force-time graph

When the force changes during the impact, find the area under the force-time graph: that area is the impulse. The calculator only takes a single force, which it treats as the average force, so for a changing force find the area by hand.

A kick: force on a ball against time
-22468101214161820222426-100100200300400500600700Peak 600 N
Drag to pan, scroll to zoom. x from -2 to 26, y from -120 to 719.

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 20 ms, which is 0.020 s, so the area is a triangle:

J = \frac{1}{2} \times 0.020 \times 600 = 6.0\ \text{N s}

Example 5: final velocity from an impulse

If that kick acts on a 0.20 kg ball at rest, the final velocity is v = u + J/m = 0 + 6.0 ÷ 0.20 = 30 m/s. The calculator has no mode for the final velocity, but you can check the answer: in From Δv, type 0.2, 0 and 30, and it gives 6 N s, the impulse you started with. The average force over the kick is 6.0 ÷ 0.020 = 300 N, half the peak, as you would expect for a triangle.

Common mistakes

  • Ignoring direction. Velocity is a vector. A bounce reverses it, so the change is v − u with opposite signs, not the difference in speeds.
  • Milliseconds left as they are. Contact times are often given in ms. Divide by 1000 to get seconds before you multiply or divide.
  • Using the peak force. J = FΔt needs the average force. With a graph, use the area instead.
  • Adding the mass when you have the force. J = FΔt does not need the mass at all.
  • Wrong units. Impulse is in N s or kg m/s, the same unit. It is not in newtons.

Practise impulse in a short timed paper

Impulse and momentum come up in AP Physics 1 and College Physics (Calculus Based) practice, and in A-level Physics: Year 1 in the UK. Each daily paper is a short timed paper of 8 questions worth 25 to 35 marks, and the papers mix topics. The Laws of Motion course has a crash simulation that shows impulse at work.

Common questions

How do you calculate impulse?
Multiply the average force by the time it acts: J = FΔt. Or multiply the mass by the change in velocity: J = m(v − u). Both give the answer in N s.
What is the unit of impulse?
The newton second, N s, which is the same as kg m/s, the unit of momentum.
Is impulse the same as change in momentum?
Yes. The impulse-momentum theorem says the impulse on an object equals its change in momentum, J = Δp.
How do you find impulse from a force-time graph?
Find the area under the graph. For a triangle, it is half the base times the height, with the time in seconds.
How do you find the average force from impulse?
Divide the impulse, or the change in momentum, by the contact time: F = J ÷ Δt.
Why do airbags reduce injuries?
They stretch out the stop. Whatever stops you, your momentum falls to zero by the same amount, and spreading that over more time lowers the force, F = Δp ÷ Δt.

Revision cards

Say the answer first, then flip the card to check it.

Where to go next

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