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

Answer

A

Explanation

Loss of `PE` = Gain in `KE`
`mgR cos theta = (1)/(2) m v^(2) + (1)/(2) m(v_(1) cos theta - v)^(2) + (1)/(2) m(v_(1) sintheta)^(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)`.