Inelastic Collision Calculator – Final Velocity and Energy Lost
An inelastic collision is one in which the two bodies conserve momentum but destroy some of their kinetic energy. That covers almost every collision you will ever see: cars crumpling into each other, a bullet burying itself in a block, railway wagons coupling, a lump of clay hitting a cart, a dropped ball returning to only part of its release height. This inelastic collision calculator takes two masses and two initial velocities, applies a coefficient of restitution in the range 0 ≤ e < 1, and returns the velocities after impact, the energy destroyed, and the full momentum bookkeeping.
Why momentum alone is not enough
Conservation of momentum gives you m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ — one equation with two unknowns. In an elastic collision the missing second equation is conservation of kinetic energy. In an inelastic collision energy is not conserved, so that route is closed and many students get stuck. The missing relation is the coefficient of restitution, e = (v₂ − v₁) / (u₁ − u₂), which compares the separation speed after impact with the approach speed before it. Solving the two together gives the closed forms the tool uses:
v₁ = [ m₁u₁ + m₂u₂ + m₂·e·(u₂ − u₁) ] / (m₁ + m₂)v₂ = [ m₁u₁ + m₂u₂ + m₁·e·(u₁ − u₂) ] / (m₁ + m₂)
The perfectly inelastic case
Setting e = 0 makes both expressions collapse to a single common velocity, v = (m₁u₁ + m₂u₂)/(m₁ + m₂) — the centre-of-mass velocity. This is the “they stick together” condition, and it is exactly the second equation the problem was missing. A 1500 kg car at 20 m/s striking a stationary 1000 kg car leaves the scene at 12 m/s: momentum stays at 30 000 kg·m/s, while kinetic energy falls from 300 kJ to 180 kJ. The missing 120 kJ, 40 % of the total, went into bending metal.
Where the energy actually goes
The loss follows ΔKE = ½·μ·(1 − e²)·(u₁ − u₂)², where μ = m₁m₂/(m₁ + m₂) is the reduced mass. It is largest at e = 0 and vanishes at e = 1. The energy becomes permanent deformation, heat inside the crushed material, sound, and fracture. For a stationary target the fraction destroyed can never exceed m₂/(m₁ + m₂), because whatever is left must still carry the original momentum away.
e towards zero and stretches the impact over a longer distance, so the same momentum change is delivered by a much smaller average force.Ballistic pendulums and measured restitution
The ballistic pendulum is the canonical two-stage problem: a perfectly inelastic impact, in which momentum is conserved and energy is not, followed by a swing in which mechanical energy is conserved. Applying energy conservation across the impact instead of momentum is the single most common error in the topic, and it can be wrong by more than a factor of ten — the tool prints that wrong answer beside the right one so the difference is unmissable. The bounce mode handles the other classic laboratory measurement, e = √(h_rebound / h_drop), and projects the whole decaying bounce sequence: heights follow hₙ = h₀e^{2n}, total path length is h₀(1 + e²)/(1 − e²), and the ball comes to rest after √(2h₀/g)·(1 + e)/(1 − e) seconds.
Choosing a coefficient of restitution
Use 0 whenever the bodies stay together. Vehicle-on-vehicle impacts sit around 0.1–0.3, wood on wood near 0.5, steel on steel near 0.6, and a tennis ball on court about 0.75. Nothing macroscopic reaches 1; the elastic case is an idealisation, which is why this tool stops just short of it and hands that limit over to the elastic collision calculator.
Beyond one dimension
In the 2-D mode momentum is conserved component by component. Two vehicles meeting at an intersection — 1200 kg heading east at 15 m/s and 1600 kg heading north at 12 m/s — leave locked together at 9.40 m/s along a bearing of 46.9°, with x-momentum and y-momentum each unchanged. The impact severity panel then converts the result into practical numbers: the impulse exchanged, an average contact force from a contact duration, an average crush force from a crush distance, and the deceleration each body suffers in m/s² and in g.