For the arrangement shown in figure when the system is released, find the acceleration of wedge. Pulley and string are ideal and friction is absent.

Explanation

Considering block and wedge as a system and using Newton's law for the system along x-direction

`T=Ma+m(a-bcostheta)" . . . . . . . (i)"`
F.B.D of m along the inclined plane

`mgsintheta-T=m(b-acostheta)" . . . . . . . . .(ii)"`
using string constraint equation

`l_(1)+l_(2)=` constant
`(d^(2)l_(1))/(dt^(2))+(d^(2)l_(2))/(dt^(2))=0`
`b-a=0" . . . . . .. (iii)"`
Solving above equations (i), (ii) & (iii), we get
`a=(mgsin theta)/(M+2m(1-cos theta))`