## Essential University Physics: Volume 1 (3rd Edition)

$\frac{m_1}{m_2}=5.83$
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: $v_{1f}=\frac{2m_2}{m_1+m_2}v_{2i}$ $v_{1f}=\frac{2m_2}{m_1+m_2}\sqrt{\frac{2m_1}{m_2}}\times v_{1f}$ $1=\frac{2m_2}{m_1+m_2}\sqrt{\frac{2m_1}{m_2}}$ $m_1+m_2=2m_2\sqrt{\frac{2m_1}{m_2}}$ $\frac{m_1}{m_2}=5.83$