Wedges `B` and `C` are smooth and they are placed in contact as shown. Block `A` is placed on wedge `B` at a height `h` above ground. Block and the two wedges are all of same mass `m`. Neglect friction every where.

The velocity of `A` when it has slide down to ground from wedge `C` is

Options

A. `0`
B. `sqrt((gh)/(2))`
C. `sqrt((gh)/(4))`
D. `sqrt(gh)/(3)`

Answer

A

Explanation

`mgh = ((1)/(2)mv^(2))2`
`v = sqrt(gh)`
From conservation of momentum
`msqrt(gh) = 2mv_("common")`
`v_("common") = sqrt(gh)/(2)`
From conservation of energy
`(1)/(2)mgh = mgh + (1)/(2)xx2mxx((gh)/(4))`
`rArr h = (h)/(4)`