Last updated: 2026-07-01

What the coefficient of restitution measures

The coefficient of restitution, written $e$, is a single number that captures how "bouncy" a collision is. It compares how fast two objects separate after impact with how fast they approached before it:

$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$

Because it is a ratio of two speeds, $e$ is dimensionless. For ordinary collisions it falls between 0 and 1:

• $e = 1$ — a perfectly elastic collision. The objects separate just as fast as they approached, and no kinetic energy is lost.
• $e = 0$ — a perfectly inelastic collision. The objects do not separate at all; they move off together, losing the most kinetic energy that conservation of momentum allows.

Real impacts sit between these limits. This calculator finds $e$ two ways: from a drop-and-bounce test, or directly from the velocities before and after a collision.

From bounce height

The easiest way to measure $e$ for a ball is to drop it onto a hard, fixed floor and see how high it rebounds. The floor barely moves, so the relative speeds are just the ball's own speeds at the moment of contact.

An object dropped from height $h_d$ arrives at the floor with speed $\sqrt{2 g h_d}$, and a rebound to height $h_b$ requires a leaving speed of $\sqrt{2 g h_b}$. Their ratio is the coefficient of restitution:

$e = \frac{\sqrt{2 g h_b}}{\sqrt{2 g h_d}} = \sqrt{\frac{h_b}{h_d}}$

The gravitational acceleration cancels, so only the height ratio matters. The bounce height can never exceed the drop height, which keeps $e$ at or below 1.

From velocities

In a general collision between two bodies, use the velocities along the line of impact. With approach velocities $u_1$ and $u_2$ and separation velocities $v_1'$ and $v_2'$:

$e = \frac{\left|v_2' - v_1'\right|}{\left|u_1 - u_2\right|}$

Velocities are signed. Pick one direction as positive and keep that convention for all four values, so two bodies meeting head-on have opposite signs. The absolute-value bars make the result depend only on the magnitudes of the closing and separating speeds.

Worked example: a bouncing ball

A ball is dropped from $h_d = 1\ \text{m}$ and rebounds to $h_b = 0.6\ \text{m}$.

$e = \sqrt{\frac{0.6}{1}} = \sqrt{0.6} \approx 0.775$

An $e$ of about 0.775 means roughly 60% of the kinetic energy survives each bounce, since energy scales with the square of speed and $e^2 = 0.6$. After the second bounce the ball would reach about $0.6 \times 0.6 = 0.36\ \text{m}$, and the rebound heights keep shrinking by the same factor of 0.6 each time.

Worked example: a head-on collision

Body 1 moves right at $u_1 = 5\ \text{m/s}$ toward a stationary body 2 ($u_2 = 0$). After impact body 1 rebounds at $v_1' = -1\ \text{m/s}$ and body 2 moves off at $v_2' = 2\ \text{m/s}$.

$e = \frac{\left|2 - (-1)\right|}{\left|5 - 0\right|} = \frac{3}{5} = 0.6$

The closing speed was 5 m/s and the separation speed is 3 m/s, so the collision returns 60% of the approach speed.

Typical values

Sports governing bodies often set an allowed rebound range for official equipment, which is just a way of fixing the coefficient of restitution within tight limits. Temperature, surface stiffness, and impact speed all shift $e$, which is why bounce tests are run under standardized conditions.

Relationship to collision calculators

The coefficient of restitution bridges the perfectly elastic and perfectly inelastic models. Combined with conservation of momentum, a known $e$ lets you predict both final velocities of a one-dimensional collision: the momentum equation fixes their sum and the restitution equation fixes their difference. That is how impact problems between the two ideal extremes are solved in practice.