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