wedge block system

'le the block move towards left with acceleration 'a'

now we can see that the block will slide in reference frame of the wedge with acceleration 'a-acosα'

let the Tension be T

mgsinα -T+maCosα = ma(1- cosα)
T = m(gsinα-a-2acosα)

now considering the wdge and block system only horizontal force is T .

thus T= (M+m) a
mgsinα - ma(1+2cosa) = Ma+am

=>a= mgsinαM +2m(1-cosα)

GOTCHA!!! NICEE WORK...thnks

i meant find the relative acceleration of the block w.r.t the wedge along the slope.

a_R+ a cosα= a .

a_R = a(1-Cos α)