Direct central impact and restitution

Direct central impact

Two bodies collide in direct central impact when their centres of mass lie on the line of impact, which is also the common normal at the contact point, and the velocities before and after impact are along this line. If the surfaces are smooth, the impulses act only along the line of impact.

  before:  (m1)--> u1      (m2)--> u2      u1 > u2
  after :  (m1)--> v1      (m2)--> v2      v2 >= v1

The impact has two phases: deformation (from first contact to maximum compression, when both bodies have a common velocity) and restitution (the bodies regain shape and separate).

Coefficient of restitution

$e$ is the ratio of the impulse during restitution to the impulse during deformation, which for a direct central impact equals

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

$e = 1$ for a perfectly elastic impact and $e = 0$ for a perfectly plastic impact; $0 < e < 1$ in practice.

Final velocities

Conservation of momentum (no external impulse):

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (1)$$

Definition of $e$: $v_2 - v_1 = e(u_1 - u_2)$ (2)

Solving (1) and (2):

$$v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)}{m_1 + m_2} \qquad v_2 = \frac{m_1 u_1 + m_2 u_2 + m_1 e(u_1 - u_2)}{m_1 + m_2}$$

Loss of kinetic energy

$$\Delta T = \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 - \tfrac12 m_1 v_1^2 - \tfrac12 m_2 v_2^2$$

Substituting the velocities and simplifying:

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

For $e = 1$, $\Delta T = 0$ (kinetic energy conserved); for $e = 0$ the loss is maximum and both bodies move together with velocity $(m_1u_1 + m_2u_2)/(m_1+m_2)$. The lost energy becomes heat, sound and permanent deformation.