Perfectly Inelastic Collision

What a perfectly inelastic collision is

In a perfectly inelastic collision, the two bodies stick together and move as one combined mass after the collision. This is the maximum possible loss of kinetic energy — no collision loses more KE.

Momentum is still conserved (always). Kinetic energy is not.

Derivation

Bodies of mass $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$ collide and stick together, moving with common velocity $v$.

Conservation of momentum:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

$\boxed{v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}}$

This is simply the velocity of the centre of mass of the system — after sticking together, the combined body moves at the centre of mass velocity.

Kinetic energy lost

Initial KE: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2$

Final KE: $\frac{1}{2}(m_1+m_2)v^2$

Loss: $\Delta KE = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \frac{1}{2}(m_1+m_2)v^2$

After substitution (see KE loss entry):

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

Special case: one body at rest

$m_2$ at rest ($u_2 = 0$):

$v = \frac{m_1 u_1}{m_1 + m_2}$

The combined mass is larger, so the speed is reduced.

Fraction of KE retained:

$\frac{KE_f}{KE_i} = \frac{\frac{1}{2}(m_1+m_2)v^2}{\frac{1}{2}m_1 u_1^2} = \frac{m_1}{m_1+m_2}$

Fraction of KE lost: $\frac{m_2}{m_1+m_2}$

A bullet ($m_1$) embedding in a block ($m_2 \gg m_1$): almost all KE is lost — the block barely moves.

Ballistic pendulum

A classic application: a bullet of mass $m$ and speed $u$ embeds in a suspended block of mass $M$. The block swings up to height $h$.

Step 1 (inelastic collision):

$v = \frac{mu}{m+M}$

Step 2 (conservation of energy for the swing):

$\frac{1}{2}(m+M)v^2 = (m+M)gh$

$v = \sqrt{2gh}$

Combining: $u = \frac{(m+M)}{m}\sqrt{2gh}$ — bullet speed from measurable quantities $h$, $m$, $M$.