Answer
$m_{1}=3m_{2}$
Work Step by Step
Please see the attached image first.
For an elastic collision, we can write the equation below.
$V_{2}=\frac{2m_{1}}{m_{1}+m_{2}}u_{1}+\frac{(m_{1}-m_{2})}{m_{1}+m_{2}}u_{2}-(1)$
In this case, $u_{2}=0$
So we get the following combination from $(1)=\gt$
$V_{2}=\frac{2m_{1}u_{1}}{(m_{1}+m_{2})}-(2)$
Kinetic energy of 685g block $(K_{1})=\frac{1}{2}m_{1}u_{1}^{2}$
Kinetic energy of 232g block $(K_{2})=\frac{1}{2}m_{2}V_{2}^{2}$
Given that, $=\frac{K_{2}}{K_{1}}=\frac{3}{4}$
$\frac{3}{4}=\frac{\frac{1}{2}m_{2}V_{2}^{2}}{\frac{1}{2}m_{1}u_{1}^{2}}-(3)$
$(2)=\gt(3)$
$\frac{3}{4} =\frac{m_{2}(\frac{2m_{1}u_{1}}{m_{1}+m_{2}})^{2}}{m_{1}u_{1}^{2}}$
$\frac{3}{4}=\frac{4m_{1}^{2}m_{2}u_{1}^{2}}{(m_{1}+m_{2})^{2}}\times\frac{1}{m_{1}u_{1}^{2}}$
$4m_{1}m_{2}=\frac{3}{4}(m_{1}+m_{2})^{2}$
$16m_{1}m_{2}=3m_{1}^{2}+6m_{1}m_{2}+3m_{2}^{2}$
$0=3m_{1}^{2}-10m_{1}m_{2}+3m_{2}^{2}$
$0=3m_{1}^{2}-9m_{1}m_{2}-m_{1}m_{2}+3m_{2}^{2}$
$0=(m_{1}-3m_{2})(3m_{1}-m_{2})$
From the above results, we get,
$m_{1}=3m_{2}$ or $3m_{1}=m_{2}$
According to the previous problem (3), we can confirm that the $m_{2}\lt m_{1}$ (more than $50\%$ of kinetic energy transferred after collision). Therefore we can choose the solution,
$$m_{1}=3m_{2}$$
