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Coefficient of Restitution Formula

Work, energy and collision formulasDimensionlessClass 11Updated 1 Nov 2026

e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}
SI unit
no unit
Dimensions
[M⁰ L⁰ T⁰]
where
  • e coefficient of restitution
  • u₁ velocity of body 1 before impact m/s
  • u₂ velocity of body 2 before impact m/s
  • v₁ velocity of body 1 after impact m/s
  • v₂ velocity of body 2 after impact m/s

The coefficient of restitution formula is e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}: the speed at which two bodies move apart after a collision divided by the speed at which they came together. It is a pure number between 0 and 1 that measures how bouncy an impact is.

A value of 1 means a perfectly elastic collision where no kinetic energy is lost. A value of 0 means the bodies stick together. Real collisions sit in between: a steel ball on a steel plate is close to 1, a lump of wet clay is close to 0. The idea belongs to the collisions part of the work, energy and power chapter and to the momentum chapter in AP Physics.

Quick facts

  • Formula

    e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}

  • Unit

    None, it is a ratio of two speeds

  • Dimensional formula

    [M⁰ L⁰ T⁰]

  • Range

    0≤e≤10 \le e \le 1

  • Bounce form

    e=h2/h1e = \sqrt{h_2/h_1}

  • Elastic collision

    e = 1, kinetic energy conserved

  • Perfectly inelastic

    e = 0, bodies move together

Coefficient of restitution formula, term by term

Take one straight line as positive and give every velocity a sign. Body 1 is the one chasing or hitting body 2.

  • u1−u2u_1 - u_2 is the relative velocity of approach. It must be positive, or the bodies would never meet.
  • v2−v1v_2 - v_1 is the relative velocity of separation. It is positive when they move apart.
  • ee is the ratio of the two. Speeds cancel, so e has no unit.

The relation is often written as Newton's law of restitution: v2−v1=−e (u2−u1)v_2 - v_1 = -e\,(u_2 - u_1). It is the same equation rearranged, which is the form the calculator on this page uses to find any one of the five quantities.

How to use it, step by step

  1. Draw the line of impact and mark one direction as positive.
  2. Write all four velocities with signs. A body moving left has a negative velocity.
  3. Use the restitution equation for one relation between v₁ and v₂.
  4. Use conservation of momentum, ∑mu=∑mv\sum mu = \sum mv, for the second relation.
  5. Solve the two equations together, then check that e lies between 0 and 1.
Most wrong answers come from step 2. If a ball bounces back, its velocity after impact has the opposite sign to its velocity before.

Worked examples

Example 1: two carts on a track

Cart 1 moves right at 5.0 m/s and catches cart 2, which moves right at 1.0 m/s. After the collision, cart 1 moves at 2.0 m/s and cart 2 at 4.0 m/s, both to the right. Find e.

Given (right positive): u₁ = 5.0, u₂ = 1.0, v₁ = 2.0, v₂ = 4.0, all in m/s.

e=4.0−2.05.0−1.0=2.04.0=0.50e = \frac{4.0 - 2.0}{5.0 - 1.0} = \frac{2.0}{4.0} = 0.50

Answer: e = 0.50, so the collision is partly elastic.

Example 2: a dropped ball (exam style, with units to convert)

A ball is dropped from 180 cm onto a hard floor and rises to 80 cm. Find the coefficient of restitution between ball and floor.

Convert: h₁ = 1.80 m, h₂ = 0.80 m. Speed just before impact: 2(9.8)(1.80)=5.94 m/s\sqrt{2(9.8)(1.80)} = 5.94\ \text{m/s}. Speed just after: 2(9.8)(0.80)=3.96 m/s\sqrt{2(9.8)(0.80)} = 3.96\ \text{m/s}.

The floor does not move, so e=3.965.94=0.667e = \frac{3.96}{5.94} = 0.667. The quicker route is e=h2/h1=0.80/1.80=0.444=0.667e = \sqrt{h_2/h_1} = \sqrt{0.80/1.80} = \sqrt{0.444} = 0.667.

Answer: e ≈ 0.67. Note that g cancels, and the ratio of heights works in any length unit as long as both match.

Special cases and energy loss

What the value of e tells you
Value of eType of collisionKinetic energy
e = 1perfectly elasticconserved
0 < e < 1inelasticpartly lost to heat, sound and deformation
e = 0perfectly inelastic, bodies stickmaximum loss consistent with momentum

For a ball bouncing repeatedly on a fixed floor, each rebound height is e2e^2 times the previous one, so after n bounces hn=e2nh0h_n = e^{2n}h_0. For two equal masses in a head on elastic collision, the bodies simply swap velocities. In every case momentum is still conserved; only kinetic energy depends on e. For a body of mass m₁ hitting a body of mass m₂ at rest, the loss of kinetic energy is 12μu12(1−e2)\frac{1}{2}\mu u_1^2(1-e^2), where μ = m₁m₂/(m₁ + m₂) is the reduced mass. The loss is zero when e = 1.

Frequently asked questions

Which formula for coefficient of restitution should I use?

e = (v₂ − v₁)/(u₁ − u₂), the relative velocity of separation after impact divided by the relative velocity of approach before impact. For a ball and a fixed floor, it reduces to e = √(h₂/h₁).

Does the coefficient of restitution have a unit?

None at all: one velocity divided by another leaves the units cancelled, so it is dimensionless, [M⁰ L⁰ T⁰] in bracket form.

Can the coefficient of restitution be greater than 1?

Not in ordinary collisions. A value above 1 would mean kinetic energy is gained, which only happens if stored energy is released, for example in an explosion.

What is e for an elastic and an inelastic collision?

For a perfectly elastic collision e = 1. For a perfectly inelastic collision, where the bodies stick together, e = 0. Everything in between is called inelastic.

Does e depend on the masses?

e depends mainly on the materials and the surfaces in contact, and somewhat on the impact speed. It is not calculated from the masses, though the masses decide the final velocities.