Two blocks of masses M and m are held on a smooth wedge. The wedge is placed on smooth plane and is free to move on plane. Now both blocks are released, and it is found that block m remains stationary w.r.t. the wedge, the ratio of M/m is 6k. Find the value k.

Explanation

A : acceleration of the wedge w.r.t. the ground. (in leftward direction)
a : acceleration of M w.r.t. the wedge

Equation of mass m
`N cos 60^(@)=mA " " ...(1)`
`N sin 60^(@)=mg" " ...(2)`
From equation (2)/(1)
`A= g cot60^(@)=g//sqrt(3)" "...(3)`
Solving (1) and (2)
`N=2mg//sqrt(3)" " ...(4)`
Equation of the wedge: `N_(1)sin30^(@)-N sin30^(@)=5mA` (w.r.t. wedge)
`(N_(1))/(2)-(mg)/(sqrt(3))=(5mg)/(sqrt(3)) rArr N_(1)=(12mg)/(sqrt(3))`
Equation of mass `M:N_(1)+MA cos60^(@)=Mg cos30^(@)`, perpendicular to the wedge.
`(12mg)/(sqrt(3))+(Mg)/(2sqrt(3))=(sqrt(3)Mg)/(2) rArr M=12m`
`M:m=12:1`