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(C) Initial mass of the particle $= 4 M$. Initial velocity $= 0$. Therefore,initial momentum $= 0$.After the explosion,the system consists of three pi

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$\sqrt{13} ~ms^{-1}$

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(C) Initial mass of the particle $= 4 M$. Initial velocity $= 0$. Therefore,initial momentum $= 0$.After the explosion,the system consists of three pieces with masses $M$,$M$,and $2 M$. Let the velocities of the pieces with mass $M$ be $\vec{v}_x = 4 \hat{i} ~ms^{-1}$ and $\vec{v}_y = 6 \hat{j} ~ms^{-1}$. Let the velocity of the piece with mass $2 M$ be $\vec{v}$.According to the law of conservation of linear momentum:$\vec{P}_{initial} = \vec{P}_{final}$$0 = M(4 \hat{i}) + M(6 \hat{j}) + 2 M \vec{v}$$0 = 4 M \hat{i} + 6 M \hat{j} + 2 M \vec{v}$Dividing by $2 M$:$0 = 2 \hat{i} + 3 \hat{j} + \vec{v}$$\vec{v} = -2 \hat{i} - 3 \hat{j} ~ms^{-1}$The magnitude of the velocity is:$|\vec{v}| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} ~ms^{-1}$.

$A$ particle of mass $4 M$ which is initially at rest explodes into three pieces of masses $M$,$M$ and $2 M$. The equal masses move along $X$ and $Y$-axes with velocities $4 ~ms^{-1}$ and $6 ~ms^{-1}$,respectively. The magnitude of the velocity of the heavier mass is

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