A body of mass M with a small block of mass placed on it reats on a smooth horizontal surface with velocity. The block is given a velocity v as shown To what height (relative to the initial velocity level) will the block rise after breaking off the body ? Neglect friction.

A body of mass M with a small block of mass placed on it reats on a sm

1.	A body of mass M with a small block of mass placed on it reats on a smooth horizontal surface with velocity. The block is given a velocity v as shown To what height (relative to the initial velocity level) will the block rise after breaking off the body ? Neglect friction.
Answer» Solution :Let it rise to a height H. Applying the law of CONSERVATION of energy, we get `(1)/(2) mv^(2) = mgh + (1)/(2) (m + M) v_(1)^(2)""...(1)` where `v_(1)` is the velocity of the combined system. Again, applying the law of conservation of momentum, we get `mv = (m + M) v_(1)` `v_(1) = (mv)/(2(m+M))""...(2)` Substituting the VALUE of `v_(1)` from eq (2) in eq (1) and solving, we get `therefore h = (Mv^(2))/(2(m+M)G)`