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Q. 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 :

NEETNEET 2022System of Particles and Rotational Motion

Solution:

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By conservation of momentum :
$m(0)=\frac{2 m}{5}(-v \hat{i})+\frac{2 m}{5}(-v \hat{j})+\frac{m}{5} \vec{v}^{\prime} $
$\Rightarrow \vec{v}^{\prime}=2 v \hat{i}+2 v \hat{j} $
$\Rightarrow v^{\prime}=\sqrt{(2 v)^{2}+(2 v)^{2}} $
$=2 \sqrt{2} v$