Force from a change in momentum calculator

Type Δp [kg·m/s] and Δt [s]. The calculator computes F = Δp/Δt: 10 kg·m/s and 0.1 s is 100 N, and 1000 kg·m/s and 2 s is 500 N. 80 kg·m/s and 0.08 s is 1000 N. Zero seconds does not divide.

Impulse: J = F·Δt. Momentum: p = m·v.

Inputs

Result

Momentum change Δp (kg·m/s) and Time Δt (s). The result shows up here.

How it works

Δp, Δt F F = Δp/Δt N ← kg·m/s, s
From momentum change and time you get mean force. Pair of J = F·Δt.

Mean force from a momentum change is F = Δp / Δt. At 10 kg·m/s and 0.1 s you get 100 N. At 2 kg·m/s and 0.02 s you get 100 N again: the same quotient. At 1000 kg·m/s and 2 s you get 500 N. The same Δp over a longer Δt is a smaller force. 80 kg·m/s in 0.08 s is 1000 N.

The form has two fields: Δp in kg·m/s and Δt in seconds. The result is in newtons. Beside it the result is J = Δp, because impulse equals the momentum change. A comma and a period are the same Δt: 0,1 and 0.1.

Δp and Δt must be positive. Δt = 0 does not divide. Both fields need a number before 100 N can appear. This is the mean force on the Δt interval, not an instant from a graph.

J = F Δt is on the neighbouring page: the pair of this formula. Here F from Δp and Δt. There J from F and Δt. p = m v is momentum at an instant, not a change over time.

F = m a takes acceleration directly. Here you do not type m or a, only Δp and Δt. Same unit: newton. 240 kg·m/s in 0.4 s is 600 N. 5 kg·m/s in 1 s is 5 N.

Type 10 and 0.1, click Calculate, and match 100 N. The header symbol does not brake. Treat the result as a mean on those 0.1 s.

How to use

  1. In the first field enter Δp in kg·m/s, for example 10. That is a momentum change, not p = m v at one instant.
  2. In the second field enter Δt in seconds, for example 0.1. Zero will not run, because you divide by Δt.
  3. Click Calculate. The calculator divides Δp by Δt. 10 and 0.1 give 100 N. J = Δp sits beside it.
  4. Δp and Δt must be positive. Both fields need a number.
  5. For J = F·Δt, open impulse. Instant momentum is on p = m·v. Here it stays F from Δp.

Formula

F = Δp / Δt

Δp > 0, Δt > 0. Unit of F: newton. J = Δp.

F, Δp, and Δt of the impulse

Force from a change in momentum here is F = Δp/Δt. 10 kg·m/s in 0.1 s is 100 N. 1000 kg·m/s in 2 s is 500 N. J = Δp beside it.

F
Mean force [N]. 10 / 0.1 = 100 N. Not instant F from F = m a.
Δp
Momentum change [kg·m/s]. 10 or 1000. Impulse J has the same value.
Δt
Change time [s]. 0.1 s or 2 s. Δt must be positive.

Real-life examples

Example 1

Δp = 10 kg·m/s, Δt = 0.1 s -> F = 100 N.

Example 2

Δp = 50 kg·m/s, Δt = 0.5 s -> F = 100 N.

Example 3

Δp = 2 kg·m/s, Δt = 0.02 s -> F = 100 N.

Example 4

Δp = 1000 kg·m/s, Δt = 2 s -> F = 500 N.

Example 5

Δp = 5 kg·m/s, Δt = 1 s -> F = 5 N.

Example 6

Δp = 80 kg·m/s, Δt = 0.08 s -> F = 1000 N.

Example 7

Δp = 20 kg·m/s, Δt = 0.2 s -> F = 100 N.

Example 8

Δp = 240 kg·m/s, Δt = 0.4 s -> F = 600 N.

Ways to use this calculator

  • You compute 100 N from 10 kg·m/s in 0.1 s.
  • You compare 500 N from 1000 kg·m/s in 2 s.

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Momentum at one instant, not mean force from Δp.
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Engine work calculator W = P·t
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Frequently asked questions

How much F at 10 kg·m/s and 0.1 s?

Mean force is 100 N. That is 10 / 0.1. At 2 kg·m/s and 0.02 s also 100 N: the same Δp/Δt.

Which units do I type?

Momentum change Δp [kg·m/s], time Δt [s]. Result F [N]. Beside it the result is impulse J = Δp.

What if Δt = 0?

The calculator will not divide. Δt must be positive. Both fields need a number.

How is this different from J = F·Δt?

A pair. Here you compute F from Δp and Δt. There you compute J from F and Δt. J = Δp, same number, different fields.

How is this different from p = m·v?

There momentum at an instant, m and v. Here momentum change over time, mean force, with no mass field.

How much at 1000 kg·m/s and 2 s?

Force is 500 N. At 80 kg·m/s and 0.08 s you get 1000 N. A shorter Δt at a similar Δp drives F up.

Does a comma in 0.1 s work?

Yes. 0.1 and 0,1 are the same Δt [s]. A comma and a period mean the same value.

How much at 240 kg·m/s and 0.4 s?

Force is 600 N. At 5 kg·m/s and 1 s you get 5 N. Same formula, another scale.

Is F = m·a the same?

Same unit F [N]. There you type m and a. Here Δp and Δt. Different fields, a mean on a time interval.

Is this an instant force?

No. It is the mean on the Δt from the field. A shorter Δt at the same Δp gives a larger F.

Knowledge sources

The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.

Page updated in 2026.