Answer
The ratio of the masses is $\frac{2}{1}$
Work Step by Step
Let the masses be $m_1$ and $m_2$.
By the law of conservation of momentum:
$m_1v_1 = m_2v_2$
This can be written as:
$v_1 = \frac{m_2v_2}{m_1}$
Let's assume that $m_1$ has twice the kinetic energy. Therefore,
$\frac{1}{2}m_1v_1^2 = 2\times \frac{1}{2}m_2v_2^2$
$m_1(\frac{m_2v_2}{m_1})^2 = 2\times m_2v_2^2$
$\frac{m_2^2}{m_1} = 2m_2$
$\frac{m_2}{m_1} = 2$
The ratio of the masses is $\frac{2}{1}$, where the larger fragment with half as much kinetic energy has double the mass of the smaller fragment.
