Answer
$47.8 \space J$
Work Step by Step
Please see the attached image first.
Here we use the principle of conservation of momentum.
The initial momentum of the system = Final momentum of the system
$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{1}$
Let's apply this equation in horizontal & vertical directions.
$\rightarrow\space 995g\times18.6\space m/s=623g\times V+372g\times 31.3 \space m/s$
$\space \space \space \space\space \space \space \space\space \space \space\space \space 6863.4g\space m/s= 623gV $
$\space \space \space \space\space \space \space \space\space \space \space \space\space \space \space \space\space \space \space \space\space \space 11\space m/s=V$
$\uparrow\space \space \space 0=623g\space u+372g\times0=\gt u=0$
Therefore the speed of the particle = 11 m/s
Total energy after burst $(E_{1})=\frac{1}{2}m_{1}V_{1}^{2}+\frac{1}{2}m_{2}V_{2}^{2}$
$E_{1}=\frac{1}{2}(372\times10^{-3}kg)(31.3\space m/s)^{2}+\frac{1}{2}(623\times10^{-3}kg)(11\space m/s)^{2}$
$E_{1}= (182.2+37.7)\space J=219.9\space J-(1)$
Total energy before burst $(E_{2})=\frac{1}{2}mV^{2}$
$E_{2}=\frac{1}{2}\times995\times10^{-3}kg\times(18.6\space m/s)^{2}$
$E_{2}= 172.1\space J-(2)$
$(1)-(2)=\gt$
Energy gain when the rocket burst $= 219.9\space J-172.1\space J=47.8\space J$
