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A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio $2:2:1$. If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$, the speed of the third (lighter) fragment is:
A
$3\sqrt{2}\,v$
B
$v$
C
$\sqrt{2}\,v$
D
$2\sqrt{2}\,v$
Detailed Solution
Masses of fragments: $\frac{2m}{5}, \frac{2m}{5}, \frac{m}{5}$
By conservation of momentum (initial momentum = 0):
$0 = \frac{2m}{5}(-v\hat{i}) + \frac{2m}{5}(-v\hat{j}) + \frac{m}{5}\vec{v}'$
$\Rightarrow \vec{v}' = 2v\hat{i} + 2v\hat{j}$
$v' = \sqrt{(2v)^2 + (2v)^2} = 2\sqrt{2}\,v$