Explanation
Using momentum conservation on bullet + gun as a system
$$\Rightarrow$$ initial momentum$$=$$ Final momentum
$$0=-MV+mv$$
$$\Rightarrow mv=MV$$..........(1)
Now total energy $$=E$$
$$\Rightarrow \dfrac{1}{2}MV^{2}+\dfrac{1}{2}mv^{2}=E$$
From (1)
$$V=\dfrac{mv}{M}$$
$$\Rightarrow\dfrac{1}{2}M\dfrac{m^{2}}{M^2}v^{2}+\dfrac{1}{2}mv^{2}=E$$
$$\dfrac{1}{2}mv^{2}\left ( \dfrac{m}{M}+1 \right )=E$$
$$\dfrac{1}{2}mv^{2}=\dfrac{EM}{M+m}$$
$$\downarrow $$
K.E bullet
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