A body of mass m_1 hits a body of mass m_2 at rest in a…

A body of mass $m_1$ hits a body of mass $m_2$ at rest in a one-dimensional, perfectly inelastic collision. What fraction of the kinetic energy is turned into internal energy, as a function of $k = m_1/m_2$?

Explanation of Method

Energy lost in a perfectly inelastic collision.

Step 1: Common speed and energy

\[ v = \frac{m_1v_1}{m_1 + m_2}, \]

\[ E_2 = \tfrac12(m_1 + m_2)v^2 = \frac{m_1}{m_1 + m_2}\,E_1. \]

Step 2: Fraction lost

\[\frac{\Delta E}{E_1} = \frac{m_2}{m_1 + m_2} = \frac{1}{1 + k}\]

$\) \(k = m_1/m_2$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $0.5$ $1$ $\frac{\Delta E}{E_1} = \frac{1}{1 + k}$

Fraction of the kinetic energy turned into internal energy in a perfectly inelastic collision with a body at rest. A light projectile hitting a heavy target (small $k$) loses almost everything.

Result

\[\boxed{\,\frac{\Delta E}{E_1} = \frac{m_2}{m_1 + m_2} = \frac{1}{1 + k}\,}\]

Remark

Check. The original result is correct. For equal masses half the energy is lost; a heavy body hitting a light one (large $k$) loses little.