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A bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be
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The correct answer is:
$\frac{1}{2} m v^2 \times \frac{m}{(m+M)}$
By conservation of momentum, $\quad m v+M \times 0=(m+M) v$ Velocity of composite block $V=\left(\frac{m}{m+M}\right) v$ K.E. of composite block $=\frac{1}{2}(M+m) V^2$ $=\frac{1}{2}(M+m)\left(\frac{m}{M+m}\right)^2 v^2=\frac{1}{2} m v^2\left(\frac{m}{m+M}\right)$
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