A particle of mass 4, m which is at rest expl | vedclass

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

Vedclass · Accepted Answer

$\frac{3}{2} \, mv^2$

Vedclass · Answer

(C) Initially,the particle 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 fragments of mass $m$ move in the $x$ and $y$ directions with velocity $v$. Their momenta are $\vec{p}_1 = mv \hat{i}$ and $\vec{p}_2 = mv \hat{j}$.The resultant momentum of these two fragments is $\vec{p}_{12} = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2} mv$.To conserve momentum,the third fragment of mass $M_3 = 4m - m - m = 2m$ must have a momentum $\vec{p}_3$ such that $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$,which means $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$.The magnitude of the momentum of the third fragment is $p_3 = \sqrt{2} mv$.Since $p_3 = M_3 v_3$,we have $2m v_3 = \sqrt{2} mv$,which gives $v_3 = \frac{v}{\sqrt{2}}$.The total energy released is equal to the final kinetic energy of the fragments:$KE = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} (2m) v_3^2$$KE = mv^2 + m \left(\frac{v}{\sqrt{2}}\right)^2 = mv^2 + m \left(\frac{v^2}{2}\right) = \frac{3}{2} mv^2$.

$A$ particle of mass $4\, m$ which is at rest explodes into three fragments. Two of the fragments,each of mass $m$,are found to move with a speed $v$ each in perpendicular directions. The total energy released in the process will be

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