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

The correct option is A $\frac{1}{2}m{v}^2\times \frac{m}{(m+M)}$

Since There is no external force, we can apply momentum conservation

By the conservation of momentum $mv+M\times 0=(m+M)V$

Velocity of composite block $V=\left(\frac{m}{m+M}\right)v$

K.E. of composite block $=\frac{1}{2}(M+m)V^2$

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