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(B) Initially,the shell is at rest,so the initial momentum is $0$.Let the masses of the three fragments be $m_1 = \frac{2m}{5}$,$m_2 = \frac{2m}{5}$,a

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$2 \sqrt{2} v$

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(B) Initially,the shell is at rest,so the initial momentum is $0$.Let the masses of the three fragments be $m_1 = \frac{2m}{5}$,$m_2 = \frac{2m}{5}$,and $m_3 = \frac{m}{5}$.The two fragments of mass $\frac{2m}{5}$ move with speed $v$ in mutually perpendicular directions. Let their velocity vectors be $\vec{v}_1 = -v \hat{i}$ and $\vec{v}_2 = -v \hat{j}$.By the law of conservation of linear momentum:$\vec{P}_{initial} = \vec{P}_{final}$$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3$$0 = \frac{2m}{5}(-v \hat{i}) + \frac{2m}{5}(-v \hat{j}) + \frac{m}{5} \vec{v}_3$$\frac{m}{5} \vec{v}_3 = \frac{2m}{5} v \hat{i} + \frac{2m}{5} v \hat{j}$$\vec{v}_3 = 2v \hat{i} + 2v \hat{j}$The speed of the third fragment is the magnitude of $\vec{v}_3$:$v_3 = |\vec{v}_3| = \sqrt{(2v)^2 + (2v)^2} = \sqrt{4v^2 + 4v^2} = \sqrt{8v^2} = 2\sqrt{2} v$.

$A$ 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:

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