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(C) The mass of the third fragment is $M - (M/4 + M/4) = M/2$.Let the velocities of the first two fragments be $\vec{v}_1 = 3\hat{i} \ m/s$ and $\vec{

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$2.5$

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(C) The mass of the third fragment is $M - (M/4 + M/4) = M/2$.Let the velocities of the first two fragments be $\vec{v}_1 = 3\hat{i} \ m/s$ and $\vec{v}_2 = 4\hat{j} \ m/s$.The momentum of the first fragment is $\vec{p}_1 = (M/4)(3\hat{i}) = (3M/4)\hat{i}$.The momentum of the second fragment is $\vec{p}_2 = (M/4)(4\hat{j}) = M\hat{j}$.According to the law of conservation of linear momentum,the total initial momentum is zero,so $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.Therefore,$\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -(3M/4)\hat{i} - M\hat{j}$.The magnitude of the momentum of the third fragment is $p_3 = \sqrt{(3M/4)^2 + M^2} = \sqrt{9M^2/16 + M^2} = \sqrt{25M^2/16} = 5M/4$.Since $p_3 = (M/2)v_3$,we have $(M/2)v_3 = 5M/4$.Solving for $v_3$,we get $v_3 = (5M/4) 	imes (2/M) = 2.5 \ m/s$.

$A$ body of mass $M$ is at rest. It explodes into three fragments. Two of the fragments,each of mass $M/4$,fly off perpendicular to each other with velocities of $3 \ m/s$ and $4 \ m/s$ respectively. What is the velocity of the third fragment in $m/s$?

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