Referring to figure calculate the downward acceleration of mass `m_1`. Assume the surfaces are frictionless and pullyes are massless.

Explanation

Let a be the acceleration of mass M and `a_(1)"and" a_(2)`, the acceleration of `M_(1)"and " M_(2)` relative to fixed pulley , P. Then from shown in figure.
`M_(1)g-T_(2)=M_(1)a_(1)` ….(1)
`M_(2)g-T_(2)=M_(2)a_(2)` ////(2)
and `T_(1) =Ma` ....(3)
Also `T_(1)=2T_(2)` ....(4)
Acceleration of `M_(1)` relative to movable pulley Q is `(a_(1)-a)`.
Acceleration of `M_(2)` relative to pulley Q`=(a_(2)-a)`
The acceleration of `M_(1) "and" M_(2)` relative to pulley Q are equal and opposite .
`:. a_(1)-a=-(a_(2)-a)`
or ` a=(a_(1)+a_(2))/2` ....(5)
substracting (2) from (1)
`(M_(1)-M_(2)) g = M_(1)a_(1)-M_(2)a_(2)` ....(6)Adding (1) and (2) , we get
`(M_(1)+M_(2))g-2T_(2)=M_(1)a_(1)+M_(2)a_(2)`
From (3) and (4) , 2T_(2)=T_(1)=Ma
Using (5), `2T_(2)=(M(a_(1)+a_(2))/2` ....(7)
substituting this value in (7), we get
`(M_(1)+M_(2))g-(M(a_(1)+a_(2)))/2=M_(1)a_(1)+M_(2)a_(2)`
or` 2(m_(1)+M_(2))g-Ma_(1)-Ma_(2) =2M_(1)a_(1)+2M_(1)a_(1)+2M_(2)a_(2)`
or `2(M_(1)+M_(2))g=(2M_(1)+M)a+(2M_(2)+M)a_(2)` ....(8)Eliminating `a_(2)` from (6) and (8) , we get `a_(1)=[(4M_(1)M_(2)+M(M_(1)-M_(2))/(4M_(1)M_(2)+M(M_(1)+M_(2))]g`