← Mechanical
Momentum & Energy

Collisions

Two bodies on a frictionless track. Compare elastic, inelastic, and perfectly inelastic impacts.
Body A Body B Walls are elastic — bodies bounce to stay in view
Final v₁
m/s
Final v₂
m/s
Momentum
kg·m/s
KE lost
%
i About this experiment — click to learn the physics

What you're looking at

Two bodies, A and B, slide toward each other along a frictionless one-dimensional track and collide. You set each body's mass and initial velocity (negative means moving left), choose a collision type, then press Run. The walls are perfectly elastic, so the bodies bounce back into view and can collide again and again. The stat cards show the outcome computed directly from your inputs.

Principle 1 — Conservation of momentum

In every collision — no matter the type — the total momentum of the system is conserved, because the forces the two bodies exert on each other are equal and opposite (Newton's third law) and there is no external horizontal force. Momentum is mass × velocity, so:

m₁·u₁ + m₂·u₂ = m₁·v₁ + m₂·v₂ total momentum before = after

This is why the Momentum readout reads the same before and after impact.

Principle 2 — Kinetic energy & the restitution coefficient

What separates the collision types is how much kinetic energy (KE = ½mv²) survives. This is captured by the coefficient of restitution e, the ratio of the relative speed of separation to the relative speed of approach:

e = (v₂ − v₁) / (u₁ − u₂) 0 = bodies stick · 1 = perfectly bouncy
TypeeWhat happens
Elastic1Kinetic energy is fully conserved; bodies bounce apart cleanly.
Inelastic0–1Some energy is lost to heat, sound and deformation; bodies separate slower.
Perfectly inelastic0Maximum energy loss; the bodies stick together and move as one.

Solving for the final velocities

Combining momentum conservation with the restitution equation gives a closed-form result for the two outgoing velocities:

v₁ = [ m₁u₁ + m₂u₂ + m₂·e·(u₂ − u₁) ] / (m₁ + m₂)
v₂ = [ m₁u₁ + m₂u₂ + m₁·e·(u₁ − u₂) ] / (m₁ + m₂)

For a perfectly inelastic hit (e = 0) both reduce to the same shared velocity (m₁u₁ + m₂u₂)/(m₁ + m₂), the centre-of-mass velocity.

Things to try

  • Give both bodies equal mass and run an elastic collision — they simply swap velocities, a classic result.
  • Send a heavy body into a light stationary one: the light one rockets off at nearly twice the heavy body's speed.
  • Switch to perfectly inelastic and watch the KE lost jump while momentum stays fixed — energy is not conserved, but momentum always is.