When we ignore friction and mass of pulley what would be the accelerations of the two blocks `m_(1)` and `m_(2)`
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Answer

a_(1) =(2m_(2) g)/(4 m_(1) + m_(2)) ;a_(2) = (m_(2) g)/(4 m_(1) + m_(2))` .

Explanation

It is clear that when block `m_(1)`
moves to the rigth through a distance `x` block `m_(2)` moves down through distance `x//2` Therefore
`a_(2) = a_(1)//2`
Applying Newton s 2nd law, `T_(1) = m_(1) a_(1)`
`:. T_(2) = 2 T_(1)` Equation of motion of `m_(2)` is
`m_(2) g - T_(2) = m_(2) a_(2) or m_(2) g - T_(1) = m_(2) (a_(1)/(2))`
or `m_(2) g - 2 m_(1) a_(1) = m_(2) a_(1)/(2)` or `a_(1) = (2 m_(2)g)/(4m _(1) + m_(2))`
And `a_(2) = a_(1)/(2) = (m_(2) g)/(4 m_(1) + m_(2))` .