Question

A gun of mass $M$ fires a bullet of mass $m$ with maximum speed $v$. Given that $m<M$. The kinetic energy of the gun will be :

• A. $\frac{1}{2}M{v}^2$
• B. $\frac{1}{2}m{v}^2$
• C. More than $\frac{1}{2}M{v}^2$
• D. Less than $\frac{1}{2}M{v}^2$

Answer 1

Answer: (D)

Less than $\frac{1}{2}M{v}^2$

From Law of conservation of momentum:

$MV=mv$

$⟹V=\frac{m}{M}v$

So, Kinetic Energy of mass $M$ is given by:

$KE=\frac{1}{2}M{V}^2=\frac{1}{2}M(\frac{mv}{M}{)}^2=(\frac{m}{M}{)}^2\times \left(\frac{1}{2}M{v}^2\right)$

Now, as $m<M$.

So, $\frac{m}{M}<1$

So, $KE<\frac{1}{2}M{v}^2$