Answer
The ratio of the pieces' masses is 2:1.
The more massive piece has the lower kinetic energy.
Work Step by Step
Initial momentum was zero.
Excluding outside forces (air resistance, friction), we expect the final momentum to be zero.
$\displaystyle \sum\vec{\mathrm{p}}=0$
$m_{1}v_{1}+m_{2}v_{2}= 0$
$m_{1}v_{1}=-m_{2}v_{2}$
$\displaystyle \frac{m_{1}}{m_{2}}=-\frac{v_{2}}{v_{1}}\qquad (*)$
(the speeds are in the opposite direction, so the ratio is positive)
Taking $m_{1}$ to be the piece with the lower kinetic energy, we are given
$K_{2}=2K_{1}$
$\displaystyle \frac{1}{2}m_{2}v_{2}^{2}=2\cdot\frac{1}{2}m_{1}v_{1}^{2}$
$m_{2}v_{2}^{2}=2m_{1}v_{1}^{2}$
$ (\displaystyle \frac{v_{2}}{v_{1}})^{2}=2\cdot(\frac{m_{1}}{m_{2}})$
Squaring (*), we have $(\displaystyle \frac{m_{1}}{m_{2}})^{2}=(\frac{v_{2}}{v_{1}})^{2}$ so,
$(\displaystyle \frac{m_{1}}{m_{2}})^{2}=2\cdot(\frac{m_{1}}{m_{2}})\qquad $... divide with $(\displaystyle \frac{m_{1}}{m_{2}})$
$(\displaystyle \frac{m_{1}}{m_{2}})=2$
$m_{1}=2m_{2}$
So,
$m_{1}$ is 2/3 of the initial mass, and $m_{2}$ is 1/3 of the original mass.
the more massive piece has the lower kinetic energy.
The ratio of the pieces is 2:1.
The more massive piece has the lower kinetic energy.
