Q97 / 120 · VITEEE 2013 · Physics

A body of mass $5 \mathrm{~m}$ 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 $60 \mathrm{~m} / \mathrm{s}$ is mutually perpendicular directions. The velocity of third fragment is

A
$10 \sqrt{2}$
Correct answer
B
$20 \sqrt{2}$
C
$20 \sqrt{3}$
D
$60 \sqrt{2}$

Solution

Using principle of conservation of linear momentum, $\begin{array}{l} 3 \mathrm{~m} \times \mathrm{v}=\sqrt{(\mathrm{m} \times 60)^{2}+(\mathrm{m} \times 60)^{2}} \\ =\mathrm{m} \times 60 \sqrt{2} \\ \Rightarrow \mathrm{v}=20 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{array}$

Hence, the correct answer is option: 'B'.

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