Momentum and impulse formulas
Momentum is mass times velocity: p = mv, in kg·m/s. It is a vector, so in one dimension its sign shows direction.
Impulse is the product of the average net force and the time it acts: J = FΔt, in N·s. The impulse-momentum theorem says impulse equals the change in momentum: FΔt = Δp = m(v₂ − v₁).
| Find | Formula |
|---|---|
| Momentum | p = m × v |
| Mass | m = p ÷ v |
| Velocity | v = p ÷ m |
| Impulse | J = F × Δt = m(v₂ − v₁) |
| Average force | F = m(v₂ − v₁) ÷ Δt = J ÷ Δt |
| Time interval | Δt = J ÷ F = m(v₂ − v₁) ÷ F |
| Mass | m = FΔt ÷ (v₂ − v₁) |
Worked example
A 1,500 kg car at 20 m/s has momentum 1,500 × 20 = 30,000 kg·m/s.
Brought to rest in 0.5 s, its impulse is 1,500 × (0 − 20) = −30,000 N·s and the average force is −30,000 ÷ 0.5 = −60,000 N. The minus sign means the force points backward. Stretching the stop to 1.5 s cuts the average force to a third, which is why crumple zones and airbags work.
Solving for force, time or mass
In impulse mode, choose what to find: J = FΔt, F = J ÷ Δt, or Δt = J ÷ F. In change-in-velocity mode, FΔt = m(v₂ − v₁) can be rearranged to m = FΔt ÷ (v₂ − v₁) or Δt = m(v₂ − v₁) ÷ F.
Example: an average braking force of −60,000 N for 0.5 s stops an object moving at 20 m/s. Its mass is (−60,000 × 0.5) ÷ (0 − 20) = 1,500 kg. The same 1,500 kg car stopped by −60,000 N takes Δt = 1,500 × (0 − 20) ÷ (−60,000) = 0.5 s.
Assumptions
The force in FΔt is the average net force over the interval. Peak forces in real impacts can be several times higher. Relativistic effects are ignored, which is accurate for everyday speeds.