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. $\frac{1}{2}mv^{2} \times \frac{m}{(m + M)}$
• B. $\frac{1}{2}mv^{2} \times \frac{M}{(m + M)}$
• C. $\frac{1}{2}mv^{2} \times \frac{(M + m)}{M}$
• D. $\frac{1}{2}Mv^{2} \times \frac{m}{(m + M)}$

Answer 1

Answer: (A)

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

Answer 2

By conservation of momentum,

Momentum of the bullet (mv) = momentum of the composite block (m + M)V

⇒ Velocity of composite block $V = \frac{mv}{m + M}$

∴ Kinetic energy

}{= \frac{1}{2}mv^{2}\left( \frac{m}{m + M} \right).}$$