Momentum and Impulse Calculator

Linear momentum is simply p = mv: mass in kilograms times velocity in m/s, giving units of kg·m/s. A 70 kg cyclist moving at 8 m/s carries p = 560 kg·m/s. Momentum is a vector, so direction matters — a car moving north has different momentum from one moving south, even at the same speed. This basic formula is the entry point to conservation laws. In any closed system without external forces, total momentum stays fixed, which is why collision problems start by writing down p_total before and after.

Impulse equals change in momentum: J = Δp = mΔv. It also equals the average force times the time it acts: J = FΔt. So a 0.15 kg ball whose velocity changes from −20 to +25 m/s after a bat hit experiences Δp = 0.15 × 45 = 6.75 kg·m/s. If the bat was in contact for 5 ms, the average force was 6.75/0.005 = 1350 N. The calculator solves any of these from the others. This theorem is the single most useful tool for analysing short, sharp interactions like collisions, kicks, and explosions.

Impulse J equals force F times the time interval Δt over which it acts: J = F × Δt. SI unit is the newton-second (N·s), which is dimensionally identical to kg·m/s — the unit of momentum, since impulse equals change in momentum. So a 50 N force pushing for 0.4 s delivers J = 20 N·s. This works only when force is approximately constant; if F varies, use the integral J = ∫F dt, which graphically equals the area under a force-time curve. Useful for collision analysis, rocket thrust calculations, and crumple-zone safety design.

Δp = m(v_f − v_i). Subtract initial velocity from final, multiply by mass. If a 0.5 kg ball arrives at 12 m/s and rebounds at −10 m/s, Δp = 0.5 × (−10 − 12) = −11 kg·m/s. The minus sign matters — direction-flipping shows up in the sign convention. This calculation is essential for collision and impact problems, where the change in momentum equals the impulse exerted on the object. Use vectors when motion is two-dimensional, since momentum has both magnitude and direction.

If you know the impulse delivered and the contact time, the average force is F = J/Δt. So an 8 N·s impulse over 0.02 s implies an average force of 400 N. This is exactly why airbags, crumple zones, and helmets work — they extend Δt for the same Δp, so the peak force on the body drops dramatically. Catching a fast cricket ball stings less if you let your hands move backwards on impact, increasing the contact time. Same impulse, smaller force. The calculator just runs the arithmetic, but the physics is genuinely life-saving.

Momentum's SI unit is the kilogram-metre per second (kg·m/s). It's also equivalent to the newton-second (N·s), which is what you'd get from F × Δt — the form the impulse-momentum theorem uses. So whenever a calculator returns N·s for impulse, it's the same as kg·m/s for momentum change. Mixing the two is fine. Just don't drift into mixed units like g·m/s or kg·km/h — convert mass to kilograms and speed to m/s up front. Otherwise your numbers come out scaled by factors of 1000 or 3.6 and conservation calculations fall apart.

Both elastic and inelastic collisions conserve total momentum: p_before = p_after. The difference is kinetic energy. In an elastic collision, KE is also conserved (think billiard balls or atomic-scale scattering). In an inelastic collision, some KE turns into heat, sound, or deformation. In a perfectly inelastic case, the objects stick together and you set v_f common to both, then use mass-weighted velocities. The calculator uses the right equations for whichever you select. So always identify the collision type before plugging in — momentum alone isn't enough to solve elastic problems uniquely.