A body of mass m_1 collides elastically and head-on with a…

A body of mass $m_1$ collides elastically and head-on with a body of mass $m_2$ at rest. How much energy is passed to the second body? Express the ratio of the first body's initial energy to the transferred energy through $k = m_1/m_2$, and discuss $k = 1$.

Explanation of Method

Energy passed on in an elastic collision.

Step 1: Energy of the second body

\[ v_2' = \frac{2m_1}{m_1 + m_2}\,v_1, \]

\[ E_2' = \tfrac12 m_2v_2'^2 = \frac{4m_1m_2}{(m_1 + m_2)^2}\,E_1. \]

\[\frac{E_2'}{E_1} = \frac{4k}{(1 + k)^2}, \qquad \frac{E_1}{E_2'} = \frac{(1 + k)^2}{4k}\]

Step 2: Equal masses

For $k = 1$ the ratio is 1: all the energy passes to the second body and the first one stops. For any other mass ratio only part of it is transferred.

$\) \(k = m_1/m_2$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $0.5$ $1$ $\frac{E_2'}{E_1} = \frac{4k}{(1 + k)^2}$

Fraction of the energy passed on in an elastic head-on collision with a body at rest. It reaches 1 only for equal masses ($k = 1$).

Result

\[\boxed{\,\frac{E_2'}{E_1} = \frac{4k}{(1 + k)^2}, \qquad \frac{E_1}{E_2'} = \frac{(1 + k)^2}{4k}\,}\]

Remark

Check. The original result is correct (the problem asks for the inverse ratio, $(1 + k)^2/4k$, which is added above).