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(B) Let the mass of each piece be $m$. Since the bomb is initially at rest,the initial momentum is $0$. By the law of conservation of linear momentum,

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$v \sqrt{2}$

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(B) Let the mass of each piece be $m$. Since the bomb is initially at rest,the initial momentum is $0$. By the law of conservation of linear momentum,the final momentum must also be $0$. Let the velocities of the three pieces be $\vec{v}_1, \vec{v}_2,$ and $\vec{v}_3$. Given $\vec{v}_1 = v \hat{i}$ and $\vec{v}_2 = v \hat{j}$. The conservation of momentum equation is: $m \vec{v}_1 + m \vec{v}_2 + m \vec{v}_3 = 0$. Dividing by $m$,we get $\vec{v}_1 + \vec{v}_2 + \vec{v}_3 = 0$. Substituting the known values: $v \hat{i} + v \hat{j} + \vec{v}_3 = 0$. Therefore,$\vec{v}_3 = -v \hat{i} - v \hat{j}$. The speed of the third piece is the magnitude of $\vec{v}_3$: $|v_3| = \sqrt{(-v)^2 + (-v)^2} = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v \sqrt{2}$.

$A$ bomb at rest explodes into three pieces of equal masses. If two pieces move perpendicular to each other,each with a speed $v$,then the speed of the third piece is:

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