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. (a) `((gR)/(3sqrt2))^(1//2)`
B. (b) `sqrt(2gR)`
C. (c) `((5gR)/(2sqrt3))^(1//2)`
D. (d) `sqrt(3/2gR)`

Answer

A

Explanation

Let the velocity of wedge be v.
Loss in PE=Gain in KE
`mgR cos 45^@=1/2mv^2+1/2m(v_1cos45^@-v)^2+1/2m(v_1sin45^@)^2`
From conservation of linear momentum
`m(v_1cos45^@-v)=mv`
Here `v_1` is the velocity of ball w.r.t. wedge, Solve to get
`v=((gR)/(3sqrt2))^(1//2)`