Question

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

• A. $\frac{1}{2} m v^{2} \times \frac{m}{(m+M)}$
• B. $\frac{1}{-2} m v^{2} \times \frac{M}{(m+M)}$
• C. $\frac{1}{2} m v^{2} \times \frac{(M+m)}{M}$
• D. $\frac{1}{-2} M v^{2} \times \frac{m}{(m+M)}$

Answer 1

Answer: (A)

$\frac{1}{2} m v^{2} \times \frac{m}{(m+M)}$

Answer 2

By the law of conservation of momentum $m v=(m+M) V$ $\therefore V=\left(\frac{m}{m+M}\right) v$ $\therefore$ Kinetic energy of composite block $E_{k}=\frac{1}{2}(m+M) V^{2}$ $=\frac{1}{2} \frac{(m+M) m^{2} v^{2}}{(m+M)^{2}}$ $=\frac{1}{2} m v^{2}\left(\frac{m}{m+M}\right)$