A body of mass (4m) is lying in the x-y plane | vedclass

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(B) Let $\vec{v}'$ be the velocity of the third piece of mass $2m$.Initial momentum,$\vec{P}_i = 0$ (as the body is at rest).Final momentum,$\vec{P}_f

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$1.5$

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(B) Let $\vec{v}'$ be the velocity of the third piece of mass $2m$.Initial momentum,$\vec{P}_i = 0$ (as the body is at rest).Final momentum,$\vec{P}_f = m v \hat{i} + m v \hat{j} + 2m \vec{v}'$.According to the law of conservation of momentum,$\vec{P}_i = \vec{P}_f$.$0 = m v \hat{i} + m v \hat{j} + 2m \vec{v}'$.$\vec{v}' = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j}$.The magnitude of $v'$ is $v' = \sqrt{(-\frac{v}{2})^2 + (-\frac{v}{2})^2} = \frac{v}{\sqrt{2}}$.Total kinetic energy generated due to the explosion is the sum of the kinetic energies of the three pieces:$K.E. = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} (2m) (v')^2$.$K.E. = m v^2 + m (\frac{v}{\sqrt{2}})^2$.$K.E. = m v^2 + m (\frac{v^2}{2}) = \frac{3}{2} m v^2 = 1.5 m v^2$.

$A$ body of mass $(4m)$ is lying in the $x-y$ plane at rest. It suddenly explodes into three pieces. Two pieces,each of mass $(m)$,move perpendicular to each other with equal speeds $(v)$. The total kinetic energy generated due to the explosion is ................. $mv^2$.

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