A block of mass m is arranged on the wedge as shown in figure . The wedge angle is ` theta `. If the masses of pulley and thread and negligible and friction is absent , find the acceleration of the wedge

Explanation

It is obvious that when block m moves downward along the incline of wedge, the wedge moves to the right . As the length of thread is constant, the distance traversed by wedge along the incline is equal to distance traversed by wedge to the right . This implies that acceleration of wedge to the right is equal to downward acceleration of wedge .
Let a be the acceleration of wedge to the right .
Then the force acting on the block m are
(i) weight mg acting vertically downward
(ii) normal reaction `R_(1)`
(iii)tension T (up the incline )
(iv)Fictitious force ma to the left .

The free body diagram of mass m is shoen in figure .
For motion of block .m. on inclined plane
`mg sin theta +ma cos theta -T = ma ` ....(1)
As mass .m. does not breaks off the inclined plane , therefore for forces on .m. normal to inclined plane
`R_(1) +ma sin theta =mg cos theta ` ....(2)
The forces acting on the wedge are
(i) weight Mg downward
(ii) normal reactionof ground on wedge =`R_(2)`
(iii) normal reaction of block on wedge =`R_(1)`
(iv) tension (T,T) in string .
The free body diagram of wedge is shown in figure .
For motion of wedge in horizontal direction
` R_(1) sin theta +T-T cos theta = mg` ....(4)
From (1),
`T=mg sin theta + ma cos theta -ma` ....(5)
From (2), `R_(1) =mg cos theta - ma sin theta ` ....(6)
substituting these values in (3) , we get
`(mg cos theta - ma sin theta ) sin theta +(mg sin theta +ma cos theta - ma ) ( 1-cos theta )=Ma`
or `{M+m sin ^(2) theta +m(1-cos theta)^(2)}a`
`=mg cos theta sin theta +mg sin theta (1-cos theta )`
`a= (mg sin theta )/(M+m sin^(2) theta + m(1-2 cos theta+cos^(2) theta))=(mg sin theta)/ (M+m(sin^(2) theta +1 -2 cos theta+cos^(2) theta))`
`:. a=(mg sin theta)/(M+2m(1-cos theta))`