A body at rest breaks up into 3 parts. If 2 p | vedclass

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(D) Initially,the body is at rest,so the initial momentum is $0$. By the law of conservation of linear momentum,the final momentum must also be $0$.Le

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$4\sqrt{2} \, m/s$ at $135^\circ$ from each body

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(D) Initially,the body is at rest,so the initial momentum is $0$. By the law of conservation of linear momentum,the final momentum must also be $0$.Let the two parts of mass $m$ move along the $x$ and $y$ axes with velocities $\vec{v}_1 = 12\hat{i} \, m/s$ and $\vec{v}_2 = 12\hat{j} \, m/s$.The resultant momentum of these two parts is $\vec{P}_{12} = m(12\hat{i}) + m(12\hat{j}) = 12m(\hat{i} + \hat{j})$.The magnitude of this resultant momentum is $P_{12} = \sqrt{(12m)^2 + (12m)^2} = 12m\sqrt{2}$.Let the third part have mass $3m$ and velocity $\vec{V}$. For the total momentum to be zero,the momentum of the third part $\vec{P}_3$ must be equal and opposite to $\vec{P}_{12}$.So,$3m 	imes V = 12m\sqrt{2}$.$V = \frac{12m\sqrt{2}}{3m} = 4\sqrt{2} \, m/s$.The direction of this velocity is opposite to the resultant of the first two parts,which makes an angle of $135^\circ$ with each of the first two parts.

$A$ body at rest breaks up into $3$ parts. If $2$ parts having equal masses $m$ fly off perpendicularly to each other,each with a velocity of $12 \, m/s$,then the velocity of the third part,which has $3$ times the mass of each part,is:

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