Answer
$\frac{K_t}{K_i}=\frac{m_1(\frac{2m_2}{m_1+m_2})^2}{m_2}$
Work Step by Step
$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$
This simplifies to:
$v_{1f}=\frac{2m_2}{m_1+m_2}v_{2i}$
We know that the initial kinetic energy is:
$K_i = \frac{1}{2}m_2v_{2i}^2 $
We know that the kinetic energy transferred is:
$K_t=\frac{1}{2}m_1(\frac{2m_2}{m_1+m_2}v_{2i})^2$
The ratio is:
$\frac{K_t}{K_i}=\frac{m_1(\frac{2m_2}{m_1+m_2})^2}{m_2}$
