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

Answer

C

Explanation

A small block…………..

Applying conservation of momentum.
Let velocity of the system be `v`, when the mass reaches the tap.
`mv = (eta m+m)v` , `implies v = (v)/(eta+1)`
Now, applying conservation energy.
`(1)/(2)mv^(@) = (1)/(2)(eta+1)m((v)/(eta+1))+mgh`
`implies (1)/(2)mv^(2)=(1)/(2)(m^(2))/((eta+1))+mgh` , `implies v = sqrt(((2gh)(eta+1))/(eta))`
Hence option : `(C )`