Momentum is mass in motion — literally mass times velocity, p = mv. It captures how hard something is to stop: a slow lorry and a fast bullet can carry comparable momentum despite wildly different speeds. Changing momentum needs an impulse, a force applied over time (J = F·t = Δp), which is the whole idea behind airbags and crumple zones — stretch the stopping time and the peak force drops. And because momentum is conserved in any closed system, it is the key to predicting what happens in collisions and recoil.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: linear momentum and the impulse-momentum theorem.
The momentum equations
With mass in kilograms and velocity in metres per second, momentum is in kg·m/s (equivalently N·s). The impulse-momentum theorem ties force and time to the change in momentum: the same momentum change can come from a big force over a short time or a small force over a long one. That trade-off is why safety design always tries to lengthen the collision time — it cannot reduce the momentum that must be shed, only soften how the force is delivered.
Worked example — stopping a car
Scenario: A 1,500 kg car travels at 20 m/s. Find its momentum, kinetic energy, and the braking force to stop it in 5 s.
The car carries 30,000 kg·m/s of momentum and 300 kJ of kinetic energy. Bringing it to rest in 5 seconds needs an average 6,000 N of braking force; do it in half the time and the force doubles to 12,000 N. This is exactly why a controlled, gradual stop is gentle while a sudden impact — the same momentum lost in a fraction of a second — produces a violent force.
Frequently Asked Questions
p = m × v. A 1500 kg car at 20 m/s has 30,000 kg·m/s. It's a vector in the direction of motion.
kg·m/s, identical to newton-seconds (N·s), since 1 N = 1 kg·m/s².
The change in momentum: J = F·t = Δp. Force to change momentum F = Δp/t — longer time, gentler force.
Yes, in a closed system with no external force. Total momentum before = after, even in collisions.
p = mv (∝ v); KE = ½mv² (∝ v²). Double speed → 2× momentum, 4× energy. Momentum always conserved; KE only if elastic.