Question

A bomb of mass 4 m at rest explodes into three pieces of masses in the ratio $1:1:2$ Two identical pieces fly in mutually perpendicular directions each with a velocity $V$. The magnitude of velocity of third piece is

• A. $V$
• B. $\frac{V}{2}$
• C. $\frac{V}{\sqrt{2}}$
• D. $\frac{V}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{V}{\sqrt{2}}$

Initial momentum of system is zero.Individual masses are $m,m,2m$.
Let $\overrightarrow{x}$ be the velocity of the third piece.
By using the conservation of momentum, we get,
$\therefore mv\hat{i}+mv\hat{j}+2m\overrightarrow{x}=0$

$\Rightarrow \overrightarrow{x}=-\frac{v}{2}(\hat{i}+\hat{j})\Rightarrow |\overrightarrow{x}|=\frac{v}{\sqrt{2}}$.