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

Options

A. `sqrt(2gh)`
B. `eta sqrt(2gh)`
C. `sqrt(2gh (1 + 1//eta))`
D. `sqrt(2gh (1 - 1//eta))`

Answer

C

Explanation

Let V: common horizontal velocity when the block reaches at the top of the wedge where velocity of the block with respect to wedge just becomes zero.
`implies mu = (m + M) V " and " 1/2 mu^2 - 1/2 (m + M)V^2 = mgh and M = eta m`
`implies V = u/(1 + eta) & u^2 - (1 + eta) (u^2)/((1 + eta)^2) = 2gh " "implies (eta u^2)/(1 + eta) = 2gh implies u = sqrt(2gh(1 + 1/eta))`.