A ball of mass `m` is released from A inside a smooth wedge of mass `m` as shown in figure. What is the speed of the wedge when the ball reaches point B?

Options

A. `(gR/3sqrt2)^(1/2)`
B. `sqrt 2gR`
C. `(5gR/2sqrt3)^(1/2)`
D. `sqrt3/2gR`

Answer

`(1)`

Explanation

Loss of PE = Gain in KE
` mgR cos theta = 1/2 mv^(2) + 1/2 m(v_(1)cos theta - v)^(2)`
`1/2 m(v_(1) sin theta )^(2) ----(1)`
From conservation of linear momentum
`m(v_(1) cos 45^(@) - v) = mv ---(2) `
Here `v_(1)` is the velocity of ball w.r.t wedge solve to get speed of wedge (v)