A ball of mass 'm is released from the top of a smooth movable wedge of mass 'm'. When the ball collides with the floor, velocity is 'v'. Then the maximum height attained by the ball after an elastic collision with the floor is : (Nelect any edge at the lower end of the wedge)

Options

A. `(2v^(2))/(g)`
B. `(v^(2))/(4g)`
C. `(4v^(2))/(g)`
D. `(v^(2))/(2g)`

Answer

A

Explanation

Let 'u' be the velocity of the ball .w.r.t. wedge when it reaches the floor. Then, the x-component of velocity of the ball .w.r.t. ground will be `(v -(u)/(sqrt2))` towards right.
By momentum conservation : `0 = m(v) + m (v - (u)/(sqrt2)) rArr u = 2 sqrt2 v`
Therefore, the y-component of velocity of the ball after the elastic collision with the floor will be `u cos 45^(@)`
`= (u)/(sqrt2) = 2v` (upward)
`:.` Maximum height `= ((2v)^(2))/(2g) = (2v^(2))/(g)`