A body of mass M at rest explodes into three | vedclass

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(C) Initial momentum of the body is $0$ because it is at rest.Let the masses of the three pieces be $m_1 = M/4$,$m_2 = M/4$,and $m_3 = M - (M/4 + M/4)

Vedclass · Accepted Answer

$2.5$

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(C) Initial momentum of the body is $0$ because it is at rest.Let the masses of the three pieces be $m_1 = M/4$,$m_2 = M/4$,and $m_3 = M - (M/4 + M/4) = M/2$.Let the velocities of the first two pieces be $\vec{v}_1 = 3 \hat{i} \ m/s$ and $\vec{v}_2 = 4 \hat{j} \ m/s$.The momentum of the first piece is $\vec{p}_1 = (M/4) 	imes 3 \hat{i} = (3M/4) \hat{i}$.The momentum of the second piece is $\vec{p}_2 = (M/4) 	imes 4 \hat{j} = M \hat{j}$.By the law of conservation of linear momentum,$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.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 piece is $|\vec{p}_3| = \sqrt{(3M/4)^2 + M^2} = \sqrt{9M^2/16 + 16M^2/16} = \sqrt{25M^2/16} = 5M/4$.Since $|\vec{p}_3| = m_3 	imes v_3$,we have $5M/4 = (M/2) 	imes v_3$.Solving for $v_3$,we get $v_3 = (5M/4) 	imes (2/M) = 2.5 \ m/s$.

$A$ body of mass $M$ at rest explodes into three pieces,two of which of mass $M/4$ each are thrown off in perpendicular directions with velocities of $3 \ m/s$ and $4 \ m/s$ respectively. The third piece will be thrown off with a velocity of (in $m/s$)

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