Momentum and impulse calculator

The formula

$$p = m v$$ $$J = \Delta p = m(v - u)$$ $$\bar{F} = \frac{J}{\Delta t}$$

$m$ is the mass, $u$ and $v$ the velocities before and after, $p$ the momentum, $J$ the impulse (the change in momentum), $\Delta t$ the contact time and $\bar{F}$ the average force during contact.

Where it comes from

Newton’s second law says force equals mass times acceleration, $F = ma$. Acceleration is the change in velocity divided by the time it takes:

$$F = m\,\frac{v - u}{\Delta t}$$

Multiply both sides by $\Delta t$:

$$F\,\Delta t = mv - mu$$

The right side is the change in the quantity $mv$, which we call momentum. The left side, force times the time it acts, is called impulse. So a force acting for a time changes momentum by exactly the impulse.

During a real collision the force isn’t steady. It rises from zero to a peak and falls back. The equation still holds if $\bar{F}$ is the average force over the contact.

Worked example

A 0.16 kg cricket ball arrives at 30 m/s and leaves the bat at 40 m/s in the opposite direction. Take the batted direction as positive, so $u = -30$ m/s. Contact lasts 1.5 ms.

$$J = 0.16 \times (40 - (-30)) = 11.2\ \text{N·s}$$ $$\bar{F} = \frac{11.2}{0.0015} = 7470\ \text{N}$$

That’s the weight of about 760 kg, for just over a thousandth of a second.

Watch out for

• Direction matters. Momentum is a vector. A bounce reverses the velocity, so the change is the sum of the two speeds, not the difference. Forgetting the minus sign makes the force far too small.
• Longer contact means less force. For the same change in momentum, stretching the time lowers the force. That’s how airbags, crumple zones and bent knees on landing protect you.
• Peak force is higher than average. The peak during a collision is often twice the average or more.

Sources

• OpenStax, University Physics Volume 1, section 9.2 "Impulse and Collisions"

Last checked November 11, 2026