A block of mass m can slide freely along the verticle surface of a bigger block of mass M as shown in fig. There is no friction anywhere in the system. The block m is connected to one end is fixed at point P.

The string between `P_(1)` and P is horizontal and other parts of the string are vertical. system is released from rest. Find the acceleration of m and M during their subsequent motion.

Explanation

Let us divde the string in three segments as shown in fig. As we release the system, the block m will try to move downwards due to which the length of segment I will increses. Then the length of segment III should decreses because the length of segment II is fixed. This will make M to move toward right. Let the acceleration of m downward be a and the acceleration of M toward right be A. Then in horizontal direction, the acceleration of m will also be A, because m is constrained to move horizontally with M.

Constraint relation: As the length of string is constant, there should not be any change in the length of string although there can be change in the individual segments of string. Let during some time, m moves downwards a distance x and M moves towards right a distance X.
Change in the length of segment I=+x
Change in the length of segment II=0
Change in the length of segment III = -X
But net change is zero, so `(+x)+0+(-X)=0 implies x=X`
Differentiating twice w.r.t. time, we get `a=A` (i)
Now finding the accelerations.
Method1 !: Analyses from ground frame:

If we make FBDs of `M` and m separately, then
From FBD of M:
`T-N=MA`...(ii)
Form FBD of m:
`N=mA`
and `mg-T=ma` ...(iv)
From (ii) and (iii), we get
`T=(M+m)A`...(v)
Equation (v) is same is (ii). Now solve to get the answers.
Method 2: Anaylzing block, w.r.t wedge
W.r.t. M, m has only vertical acceleration which is a downward. Here we have to apply pseudo force(mA in figure) because M is non-inertial frame.

From FBD of w.r.t. M, writing the equation of motion of block:
`N=mA, mg-T=ma`
These equation are same as (vi) and (vii). solve to get answer.
Method 3: Consider the motion of m and M together in horizontal direction.

`T=(M+m)A` ...(vi)
Now consider the motion of m in vertical direction only
`mg-T=ma=mA` ...(vii)
From (iii)
`T=m(g-A)` ...(viii)
From (ii) and (iv)
`m(g-A)=(M+m)A implies A=(mg)/(M+2m)=a`
Hence, the net acceleration of m
`a_(n et)=sqrt(a^(2)+A^(2))=sqrt(2)A=sqrt(2)(mg)/((M+2m))`.