A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio $2:2:1$. I

Question

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

Accepted Answer

Momentum of the system remains conserved. Initial momentum = 0. Final momentum should also be zero. Let masses be $2m, 2m, m$. Momentum along x-direction = $2mv\hat{i}$. Momentum along y-direction = $2mv\hat{j}$. Net momentum of these two = $\sqrt{(2mv)^2 + (2mv)^2} = 2\sqrt{2}mv$. To balance this, the third fragment of mass $m$ must have momentum $2\sqrt{2}mv$ in the opposite direction. Thus, its speed is $2\sqrt{2}v$.

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