The initial acceleration of a 3-body system

Assume that we let the system of the picture free. What will the initial acceleration of m be?

In a small time interval Δt, the length of the rope on top of object M is X-x, where X is the coordinate of M and x the coordinate of m. But if we call y the vertical coordinate the 3rd object (m), then the length L should satisfy L= X-x+y. So for the accelerations, α(m) + α(Μ) = α'(m), where α(Μ),α(m) and α'(m) are the accelerations of M, horizontal m and vertical m. 

But, T=mα(m) and mg - T = mα'(m). That is a α(m) + α'(m) = g. In addition, from conservations of momentum, we get that 3mα(Μ)=mα(m) or α(m)=3α(Μ). 

So combining the relationships we get that, α'(m)= 4α(m). So α(m)=g/5   

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