Coefficient of Restitution

Definition

The coefficient of restitution $e$ characterises how elastic a collision is:

$e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$

where $u_1, u_2$ are initial velocities and $v_1, v_2$ are final velocities (taking a consistent positive direction).

Range of $e$

$0 \leq e \leq 1$

Why $e$ cannot exceed 1

If $e > 1$, the bodies would separate faster than they approached — kinetic energy would be created from nothing. This violates energy conservation. Hence $e \leq 1$ always.

Finding final velocities using $e$

Combine the definition of $e$ with conservation of momentum:

Momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ ... (1)

Restitution: $v_2 - v_1 = e(u_1 - u_2)$ ... (2)

Two equations, two unknowns ($v_1$, $v_2$). Solve:

$v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1-u_2)}{m_1+m_2}$

$v_2 = \frac{m_1 u_1 + m_2 u_2 + m_1 e(u_1-u_2)}{m_1+m_2}$

Kinetic energy lost in terms of $e$

$\Delta KE = \frac{1}{2} \cdot \frac{m_1 m_2}{m_1+m_2}(u_1-u_2)^2(1-e^2)$

At $e = 1$: $\Delta KE = 0$ — elastic collision, no loss. At $e = 0$: $\Delta KE = \frac{m_1 m_2(u_1-u_2)^2}{2(m_1+m_2)}$ — maximum loss.

Ball bouncing on floor

A ball dropped from height $H$ bounces to height $h$:

Speed just before impact: $u = \sqrt{2gH}$

Speed just after impact: $v = \sqrt{2gh}$

Floor has infinite mass — does not move: $u_2 = v_2 = 0$

$e = \frac{v - 0}{u - 0} = \frac{v}{u} = \frac{\sqrt{2gh}}{\sqrt{2gH}} = \sqrt{\frac{h}{H}}$

After $n$ bounces, height reached: $h_n = e^{2n} H$