Question

A body of mass $5 \,m$ initially at rest explodes into $3$ fragments with mass ratio $3 : 1 : 1$. Two of fragments each of mass $'m'$ are found to move with a speed of $60 \,m/s$ is mutually perpendicular directions. The velocity of third fragment is

• A. $10\sqrt{2}$
• B. $20\sqrt{2}$
• C. $20\sqrt{3}$
• D. $60\sqrt{2}$

Answer 1

Answer: (B)

$20\sqrt{2}$

Answer 2

Using principle of conservation of linear momentum
$3m\times v= \sqrt{\left(m\times60\right)^{2} +\left(m\times60\right)^{2}}$
$= m\times60\sqrt{2}$
$\Rightarrow v = 20\sqrt{2}$ m/s

Therefore, the correct option is (B): $\Rightarrow v = 20\sqrt{2}$