A boy of mass `m_1` , standing on a smooth horizontal surface, throws a sphere of mass `m_2` parallel to the surface. After a time t, if the separation between them becomes x, then show that the work done by the boy in throwing the sphere
`=1/2 (x/t)^2((m_1m_2)/(m_1+m_2))`

Explanation

Let v denote the velocity of the sphere, and V denote the velocity of the boy, due to reaction. From the law of conservation of momentum ,
`m_1V=m_2v` ……..(1)
Separation between them after a time t.
x=(V+v)t ……..(2)
From (1) and (2) we get,
`(m_2)/(m_1)v+v=x/t or, v=x/t(m_1)/(m_1+m_2)`
Similarly, `V=x/t*(m_2)/(m_1+m_2)`
Work done in throwing the sphere
= sum of kinetic energies of the sphere and the boy
`=1/2 m_1V^2+1/2 m_2v^2`
`=1/2 m_1(x/t)^2*(m_2^2)/((m_1+m_2)^2) +1/2 m_2(x/t)^2 (m_1^2)/((m_1+m_2)^2)`
`=1/2 (x/t)^2(m_1m_2)/((m_1+m_2)^2)(m_1+m_2) `
`=1/2 (x/t)^2*(m_1m_2)/(m_1+m_2)`