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

Answer

C

Explanation

By conservation of linear momentum,
mu + 0 = `(m + nm)V_c`
`V_c = u/((1+n))`
By conservation of mechanical energy
`1/2"mu"^2+0=1/2m(1+n)u^2/((1+n)^2)+mgh`
`1/2"mu"^2[1-1/(1+n)]=mgh`
`u= sqrt(2gh(1+1/n))`