A block of mass `m` is pushed towards the movable wedge of mass `M` and height `h`, with a velocity `v_(0)`. All surfaces are smooth . The minimum value of `v_(0)` for which the block will reach the top of the wedge is

Options

A. `sqrt(2 gh)`
B. `sqrt((2 ghm)/(M))`
C. `sqrt((2gh (m + M))/(M))`
D. `sqrt((2ghm)/((m + M)))`

Answer

C

Explanation

When the block is moving over wedge , let their common horizontal velocity be `v`
By the momentum conservation
`mv_(0) = (m + M) v` (i)
`(1)/(2) mv_(0)^(2) = (1)/(2) (m + M) v^(2) + mgh` (ii)
`(1)/(2) mv_(0)^(2) = (1)/(2) (m + M) ((mv_(0))/(m + M))^(2) + m gh`
`v_(0)^(2) (1 - (m)/(m + M)) = 2gh`
`v_(0) = sqrt(2 gh(m + M))/(M)`