Answer
The target flies off with $7.4\%$ of the projectile's initial kinetic energy.
Work Step by Step
The projectile, whose mass $m=0.2kg$, is fired at initial velocity $v$ and embeds into the target, whose mass is $M=2.5kg$ and who has initial velocity $V=0$. After that, the system has mass $M+m=2.7kg$ and velocity $V_f$.
According to the principle of conservation of total linear momentum, $$MV+mv=(M+m)V_f$$ $$0+mv=(M+m)V_f$$ $$V_f=\frac{m}{M+m}v=0.074v$$
We have $$\frac{KE_f}{KE_{projectile}}=\frac{\frac{1}{2}(M+m)V_f^2}{\frac{1}{2}mv^2}=\frac{(M+m)V_f^2}{mv^2}=\frac{2.7}{0.2}\times\frac{5.48\times10^{-3}v^2}{v^2}=7.4\times10^{-2}$$
So the percentage of the system's final kinetic energy compared with the projectile's initial kinetic energy is $7.4\%$
