Elastic vs Inelastic Collisions

In an elastic collision both momentum and kinetic energy are conserved, while in an inelastic collision momentum is conserved but some kinetic energy is converted to other forms, with the maximum loss occurring when the objects stick together (perfectly inelastic). In brief: for a head-on elastic collision in one dimension, the relative speed of approach equals the relative speed of separation. The coefficient of restitution e = (speed of separation) ÷ (speed of approach) is 1 for elastic, 0 for perfectly inelastic, and between 0 and 1 for real collisions.

Billiard balls collide almost elastically, while a car crash is highly inelastic. This guide defines the types of collisions, gives the formulas for one-dimensional collisions, introduces the coefficient of restitution, compares the types in a table, works through numerical examples and corrects common mistakes.

What is an elastic collision?

In an elastic collision, there is no net loss of kinetic energy: the energy is stored briefly as elastic potential energy while the objects deform, and then fully returned. Collisions between atoms and molecules in a gas, and between billiard balls, are almost elastic. For two objects with masses m₁ and m₂, a head-on collision in one dimension with m₂ initially at rest, the final velocities are

v₁ = (m₁ − m₂) ÷ (m₁ + m₂) × u₁, v₂ = 2m₁ ÷ (m₁ + m₂) × u₁.

Special cases: for equal masses, the first object stops and the second moves off with the velocity of the first (Newton’s cradle); when a light object hits a very heavy one, it bounces back with nearly the same speed; when a heavy object hits a light one, the heavy one continues almost unchanged and the light one moves at about twice its speed.

What is an inelastic collision?

In an inelastic collision, part of the kinetic energy is converted to thermal energy, sound, and deformation of the objects. Momentum is still conserved (if the system is isolated). In a perfectly inelastic collision the objects stick together and move with a common velocity v = (m₁u₁ + m₂u₂) ÷ (m₁ + m₂). This is the case with the greatest possible loss of kinetic energy in the center-of-mass frame. Examples: a bullet embedding in a block, railcars coupling, and a car crash. Most everyday collisions are inelastic.

What is the coefficient of restitution?

The coefficient of restitution e = (v₂ − v₁) ÷ (u₁ − u₂) compares the relative speed of separation with the relative speed of approach in a head-on collision. e = 1 for a perfectly elastic collision, and e = 0 when the objects stick together. For a ball dropped on a floor, e = √(h_bounce ÷ h_drop), since the speed is proportional to the square root of the height. A steel ball on steel has e of about 0.9 or higher, a tennis ball on concrete about 0.7 to 0.75, and a lump of clay about 0.

Collision types compared

Elastic, inelastic and perfectly inelastic collisions
Type	Momentum	Kinetic energy	e	Example
Elastic	Conserved	Conserved	1	Billiard balls, gas molecules
Inelastic	Conserved	Partly lost	0 < e < 1	Bouncing tennis ball
Perfectly inelastic	Conserved	Maximum loss	0	Clay balls sticking, coupling railcars

Kinetic energy of the two carts in the worked example. It is 36 J before the collision, still 36 J after an elastic collision, and only 12 J when the carts stick together, so 24 J (two-thirds) is converted to other forms. Momentum is 12 kg·m/s in every case.

Worked example: elastic and inelastic collisions

Elastic. A 2.0 kg cart at 6.0 m/s collides head-on with a stationary 4.0 kg cart. v₁ = (2.0 − 4.0) ÷ 6.0 × 6.0 = −2.0 m/s (it rebounds). v₂ = 2 × 2.0 ÷ 6.0 × 6.0 = +4.0 m/s.
- Momentum before: 2.0 × 6.0 = 12 kg·m/s. After: 2.0 × (−2.0) + 4.0 × 4.0 = −4 + 16 = 12 ✓.
- Kinetic energy before: ½ × 2.0 × 36 = 36 J. After: ½ × 2.0 × 4 + ½ × 4.0 × 16 = 4 + 32 = 36 J ✓.
- Relative speeds: approach 6.0, separation 4.0 − (−2.0) = 6.0, so e = 1.
Perfectly inelastic. The same carts stick together. v = 12 ÷ 6.0 = 2.0 m/s. Kinetic energy after = ½ × 6.0 × 4.0 = 12 J. Loss = 36 − 12 = 24 J (two-thirds).
Bouncing ball. A ball is dropped from 1.0 m and rises to 0.64 m. e = √(0.64 ÷ 1.0) = 0.80. The fraction of kinetic energy retained at each bounce is e² = 0.64.
Ballistic pendulum. A 0.010 kg bullet embeds in a 1.0 kg block hanging from strings, and the block rises 0.050 m. Just after impact, the speed of the block and bullet is v = √(2gh) = √(2 × 9.8 × 0.050) = 0.99 m/s. Momentum conservation: 0.010 × u = 1.010 × 0.99, so u = 100 m/s. The kinetic energy of the bullet is ½ × 0.010 × 100² = 50 J, while after the impact, it is ½ × 1.010 × 0.99² = 0.50 J, so about 99 percent is lost.

How were collisions studied?

In 1668 the Royal Society asked John Wallis, Christopher Wren and Christiaan Huygens to study collisions. Wren and Huygens correctly described elastic collisions, and Wallis the inelastic ones. Newton defined the coefficient of restitution through experiments with pendulum bobs of different materials in the Principia. The ballistic pendulum was invented by Benjamin Robins in 1742 to measure the speeds of bullets.

Which mistakes are common?

Mistake 1: Saying that momentum is lost in inelastic collisions

Momentum is conserved in all collisions of isolated systems. Kinetic energy is what is lost.

Mistake 2: Applying the kinetic energy equation to inelastic collisions

Only use it for elastic collisions.

Mistake 3: Believing that kinetic energy is destroyed

It is converted to other forms such as thermal energy and sound.

Mistake 4: Thinking that sticking is the only inelastic collision

Any collision with a loss of kinetic energy is inelastic, even if the objects bounce apart.