Answer
$\frac{m_1}{m_2}=5.83$
Work Step by Step
We know that half of the Kinetic energy is transferred, so:
$\frac{1}{4}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2$
$v_{2i}=\sqrt{\frac{2m_1}{m_2}}\times v_{1f}$
We know the following equation for elastic collisions:
$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$
We know that object one is initially at rest, so:
$v_{1f}=\frac{2m_2}{m_1+m_2}v_{2i}$
$v_{1f}=\frac{2m_2}{m_1+m_2}\sqrt{\frac{2m_1}{m_2}}\times v_{1f}$
$1=\frac{2m_2}{m_1+m_2}\sqrt{\frac{2m_1}{m_2}}$
$m_1+m_2=2m_2\sqrt{\frac{2m_1}{m_2}}$
$\frac{m_1}{m_2}=5.83$
