Elastic Collision — Final Velocities

What an elastic collision is

In a perfectly elastic collision, both momentum and kinetic energy are conserved. No energy is lost to heat, sound, or deformation. The bodies bounce off each other.

Real collisions are never perfectly elastic, but billiard ball collisions and atomic/subatomic collisions come close.

Setting up

Two bodies of masses $m_1$ and $m_2$ moving with initial velocities $u_1$ and $u_2$ (all in the same line, taking rightward as positive).

After collision, their velocities are $v_1$ and $v_2$.

Derivation

Conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{...(1)}$

Conservation of kinetic energy:

$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \quad \text{...(2)}$

Rearrange (1): $m_1(u_1 - v_1) = m_2(v_2 - u_2)$ ... (3)

Rearrange (2): $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$

Factor: $m_1(u_1-v_1)(u_1+v_1) = m_2(v_2-u_2)(v_2+u_2)$ ... (4)

Divide (4) by (3):

$u_1 + v_1 = v_2 + u_2$

$u_1 - u_2 = v_2 - v_1 \quad \text{...(5)}$

Equation (5) is key: relative velocity of approach = relative velocity of separation.

Now solve (1) and (5) simultaneously.

From (5): $v_2 = v_1 + u_1 - u_2$

Substitute into (1):

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2(v_1 + u_1 - u_2)$

$m_1 u_1 + m_2 u_2 = (m_1+m_2)v_1 + m_2 u_1 - m_2 u_2$

$(m_1-m_2)u_1 + 2m_2 u_2 = (m_1+m_2)v_1$

$\boxed{v_1 = \frac{(m_1-m_2)u_1 + 2m_2 u_2}{m_1+m_2}}$

Similarly: $\boxed{v_2 = \frac{(m_2-m_1)u_2 + 2m_1 u_1}{m_1+m_2}}$

Special cases

Equal masses ($m_1 = m_2$):

$v_1 = u_2, \quad v_2 = u_1$

The bodies exchange velocities. If $m_2$ is at rest: $m_1$ stops, $m_2$ moves off with $m_1$'s initial velocity.

Very heavy projectile ($m_1 \gg m_2$, $m_2$ at rest):

$v_1 \approx u_1 \quad v_2 \approx 2u_1$

The heavy body barely slows down. The light body shoots off at twice the projectile's speed.

Very light projectile ($m_1 \ll m_2$, $m_2$ at rest):

$v_1 \approx -u_1, \quad v_2 \approx 0$

The light body bounces back with the same speed. The heavy body barely moves. (Like a ball bouncing off a wall.)

The relative velocity result (equation 5)

For elastic collisions: relative velocity of separation = relative velocity of approach

$v_2 - v_1 = u_1 - u_2$

This is equivalent to coefficient of restitution $e = 1$ for elastic collisions.