Elastic & Inelastic Collision Calculator

Collision Physics Guide

What do I need to know about Conservation of Momentum?

In all collisions: total momentum is conserved. p_before = p_after. m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Momentum is a vector — directions matter. In 1D problems: use positive for one direction, negative for the opposite. Example: 2kg moving at 5 m/s and 3kg moving at −2 m/s. Total momentum = 2×5 + 3×(−2) = 10 − 6 = +4 kg·m/s. After collision, the total must still equal +4. Conservation of momentum follows from Newton's 3rd law — the forces between colliding objects are equal and opposite, so their changes in momentum are always equal in size and opposite in direction.

What should I know about Elastic Collisions?

Elastic: kinetic energy is also conserved (collisions of hard balls, gas molecules, billiards approximate). Two equations to solve: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (momentum). ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² (KE). Simplification — relative velocity reverses: u₁ − u₂ = −(v₁ − v₂). Special cases: equal masses, one stationary: m₁ stops, m₂ moves with u₁. Mass m₁ hits much heavier m₂: m₁ bounces back at nearly u₁; m₂ barely moves (like a ball off a wall). Tiny mass hits stationary heavy mass: the tiny mass rebounds at close to its original speed while the heavy mass barely moves at all, just like a ball bouncing off a wall.

What's the key thing to understand about Inelastic Collisions?

Inelastic: kinetic energy is NOT conserved — some converts to heat, sound, deformation. Momentum is still conserved. Perfectly inelastic: objects stick together. Common in: car crashes (real cars do not bounce off each other elastically). Lump of clay hitting another lump. Bullet embedding in a wooden block. Formula for perfectly inelastic: v = (m₁u₁ + m₂u₂) / (m₁ + m₂). Both objects end with the same final velocity. Example: 2kg at 10 m/s hits stationary 3kg. v = (2×10 + 0)/(2+3) = 4 m/s. Kinetic energy before the collision is greater than kinetic energy after, since the difference is converted into heat, sound, and deformation of the objects.

What do I need to know about Coefficient of Restitution?

Coefficient of restitution (e) measures how much energy is preserved: e = (v₂ − v₁) / (u₁ − u₂). For elastic: e = 1. For perfectly inelastic: e = 0. For real collisions: 0 < e < 1. Examples: superball: e ≈ 0.9. Basketball off wood floor: e ≈ 0.75. Steel ball off steel anvil: e ≈ 0.6. Tennis ball on tennis racket: e ≈ 0.45-0.55. Cricket ball off bat: e ≈ 0.5. Higher e: more bounce, less energy lost. Sports engineering: balls and bats have regulated e values. ICC regulations require a cricket ball to have a coefficient of restitution within a specific tested range, ensuring consistent bounce behaviour across matches.

What do I need to know about the 1D Elastic Collision Formulas?

For a head-on elastic collision between masses m₁ (velocity u₁) and m₂ (velocity u₂), solving conservation of momentum and conservation of kinetic energy together gives v₁ = [(m₁−m₂)u₁ + 2m₂u₂]/(m₁+m₂) and v₂ = [(m₂−m₁)u₂ + 2m₁u₁]/(m₁+m₂). Two special cases are worth remembering: if the masses are equal, the velocities simply swap; and if a light object strikes a very heavy stationary one (m₁ ≪ m₂), the light object bounces back at nearly its original speed while the heavy one barely moves — like a ball bouncing off a wall.

What should I know about Checking Energy Conservation?

A collision only qualifies as 'elastic' if total kinetic energy is the same before and after: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂². This is a useful check on any elastic-collision calculation — plugging the computed final velocities back into the kinetic-energy formula should reproduce the initial total KE to within rounding error. True elastic collisions are rare at the macroscopic scale (billiard balls and gas molecule collisions come close); most everyday collisions lose some energy to heat, sound, and deformation, which is what makes them inelastic instead.

What should I know about Elastic Collisions in Two Dimensions?

The familiar 1D elastic collision formulas assume both objects move along a single line before and after impact, but real collisions — billiard balls, pucks, molecules — often occur at an angle. Two-dimensional elastic collisions still conserve both momentum and kinetic energy, but now momentum must be conserved separately along two perpendicular axes (typically chosen along and perpendicular to the line joining the centres at the moment of impact, called the line of impact). For a glancing collision between a moving ball and an identical stationary ball, a useful special-case result emerges: if the masses are equal and the collision is perfectly elastic, the two balls always move off at right angles to each other afterward (unless the collision is perfectly head-on). This is a direct consequence of combining conservation of momentum and conservation of kinetic energy for equal masses, and it is exactly why pool and snooker players can predict the cue ball's deflection angle from the geometry of contact alone. For unequal masses or oblique angles generally, the component of velocity along the line of impact behaves like a 1D elastic collision, while the perpendicular component is unchanged for each object, since no force acts in that direction during an idealised instantaneous collision.

Elastic Collision Outcomes by Mass Ratio

The post-collision velocities in a 1D elastic collision depend entirely on the ratio of the two masses, independent of the actual mass values:

The "equal masses" case is the classic Newton's-cradle behavior — it's why, in that desk toy, only one ball on the far end swings out when one ball on the near end is released: equal-mass elastic collisions transfer 100% of the velocity, not a blend of it.

Worked Example: 1D Elastic Collision Between Two Carts

Cart 1 (mass 2 kg, velocity 3 m/s) collides elastically with Cart 2 (mass 4 kg, initially at rest). Find both carts' velocities after the collision. Using the standard 1D elastic collision formulas: v1' = [(m1 − m2)/(m1 + m2)]v1 = [(2 − 4)/6](3) = (−2/6)(3) = −1 m/s. v2' = [2m1/(m1 + m2)]v1 = [4/6](3) = 2 m/s. So Cart 1 bounces backward at 1 m/s, and Cart 2 moves forward at 2 m/s. Check momentum conservation: before = m1v1 = 2(3) = 6 kg·m/s. After = m1v1' + m2v2' = 2(−1) + 4(2) = −2 + 8 = 6 kg·m/s ✓. Check kinetic energy conservation (the defining feature of an elastic collision): before = ½(2)(3)² = 9 J. After = ½(2)(−1)² + ½(4)(2)² = 1 + 8 = 9 J ✓. Both momentum and kinetic energy are exactly conserved, confirming this is a genuine elastic collision — if the after-collision KE had come out lower than 9 J, that would indicate energy was lost to heat or deformation, making it an inelastic collision instead.

Perfectly Inelastic Collisions for Comparison

In a perfectly inelastic collision, the two objects stick together after impact and move with a shared final velocity — momentum is still conserved, but kinetic energy is not (some is lost to heat, sound, and deformation). Using the same carts as the elastic example — Cart 1 (2 kg, 3 m/s) and Cart 2 (4 kg, at rest) — but now assume they collide and stick together. Conservation of momentum gives: m1v1 = (m1+m2)v_f → 2(3) = 6(v_f) → v_f = 6/6 = 1 m/s. Kinetic energy before: ½(2)(3)² = 9 J. Kinetic energy after: ½(6)(1)² = 3 J. So 9 − 3 = 6 J of kinetic energy (two-thirds of the original) was lost to deformation and heat during the sticking collision — compared to the elastic case, where all 9 J was conserved and redistributed between the two separate carts afterward. This stark contrast (0% energy loss for elastic vs. 67% loss for perfectly inelastic, given the same masses and initial velocity) is exactly why elastic and inelastic collisions sit at opposite ends of a spectrum, with most real-world collisions (like car crashes or billiard balls) falling somewhere between the two extremes.