A block A (mass 4M) is placed on the top of a wedge block B of base length l (mass=20M) as shown in figure. When the system is released from rest. Find the distance moved by the block B till the block A reaches ground Assume all surfaces are frictionless.

Explanation

Initial position of centre of mass
`=(X_(B)M_(B)+X_(A)M_(A))/(M_(B)+M_(B))=(X_(B).20M+l.4M)/(24M)=5X_(B)+(l)/(6)`

Final position of centre of mass
`=((X_(B)+x)20M+4Mx)/(24M)=(5(X_(B)+x)+x)/(6)`
since there is no horizontal force on system centre of mass initially = centre of mass finally.
`5X_(B)+l=5X_(B)=5x+x`
`l=6x`
`x=(l)/(6)`