Two balls of masses `m_(1)` and `m_(2)` are placed on top of one over the other (with a small gap between them) and then dropped on to the ground. What is the ration `m_(1)//m_(2)` for which the upper ball ultimately receives the largest possible fraction of the total energy? Take all collisions as elastic. Neglect air resistance

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1:4`

Answer

C

Explanation

Let they fall with speed `'v'`. `(m_(2)-m_(1))v = m_(1)u` and `(m_(2)+m_(1))(v^(2))/(2) = m_(1)(u^(2))/(2)`. `'u'` is the speed of the top ball after collision.