Kinetic Energy Loss in Perfectly Inelastic Collision

Derivation

After a perfectly inelastic collision, the combined body moves at:

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

Initial KE:

$KE_i = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2$

Final KE:

$KE_f = \frac{1}{2}(m_1+m_2)v^2 = \frac{1}{2}(m_1+m_2)\left(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\right)^2 = \frac{(m_1 u_1+m_2 u_2)^2}{2(m_1+m_2)}$

Loss:

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

Multiply through by $2(m_1+m_2)$:

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

Expanding the right side:

$= m_1^2 u_1^2 + m_1 m_2 u_1^2 + m_1 m_2 u_2^2 + m_2^2 u_2^2 - m_1^2 u_1^2 - 2m_1 m_2 u_1 u_2 - m_2^2 u_2^2$

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

Therefore:

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

What this tells us

The energy lost depends on:

• The reduced mass $\mu = \frac{m_1 m_2}{m_1+m_2}$ of the system
• The relative velocity $(u_1-u_2)$ at the moment of collision

When is loss maximum?

For fixed total momentum, the KE loss is maximum when the relative velocity is maximum.

For fixed total momentum $p$ and masses, the loss is maximum in the centre of mass frame — which is exactly the perfectly inelastic case (bodies come to rest in the CM frame, maximum KE is converted).

Special case: one body at rest

$u_2 = 0$:

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

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

• If $m_2 \gg m_1$ (bullet hitting massive wall): almost all KE lost
• If $m_1 \gg m_2$ (heavy body hitting light one): very little KE lost