Answer
$75.6\%$
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, $m_{1}=685\space g,\space m_{2}=232\space g,\space u_{2}=0$
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 $(K_{2})=\frac{1}{2}m_{2}V_{2}^{2}$
Percentage of transferred kinetic energy $=\frac{K_{2}}{K_{1}}\times 100\%$
$\frac{K_{2}}{K_{1}}\times 100\%=\frac{\frac{1}{2}m_{2}V_{2}^{2}}{\frac{1}{2}m_{1}u_{1}^{2}}\times100-(3)$
$(2)=\gt(3)$
$\frac{K_{2}}{K_{1}}\times 100\% =\frac{m_{2}(\frac{2m_{1}u_{1}}{m_{1}+m_{2}})^{2}}{m_{1}u_{1}^{2}}\times100\%$
$\frac{K_{2}}{K_{1}}\times 100\%=\frac{4m_{1}^{2}m_{2}u_{1}^{2}}{(m_{1}+m_{2})^{2}}\times\frac{1}{m_{1}u_{1}^{2}}\times100$
Let's plug known values into this equation,
$\frac{K_{2}}{K_{1}}\times 100\%=\frac{(4\times232g\times685g)}{(232g+685g)^{2}}\times100\%=75.6\$$
