Solution:
Let the total mass be \(2m\), so \(K = mv_0^2\). Under momentum conservation, \(2mv_0 = m(-v_0) + mv_B ⇒ v_B = 3v_0\). The kinetic energy of part B is \(K_B = \frac{1}{2}m(3v_0)^2 = \frac{9}{2}mv_0^2 = \frac{9}{2}K\).
A body with kinetic energy K moving in +X direction splits up into two parts A and B of equal mass on its own. Part 'A' moves back in -X direction with a velocity equal in magnitude to the initial velocity of the body. The kinetic energy of part B will be:
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