In the fig, find the acceleration of mass `m_(2)` .

Explanation

`l_(1)+2l_(2)` = constant on differentiating
`v_(1) +2v_(2)=0`
Again differentiating
`a_(1) +2a_(2) rArr a_(1) = -2a_(2)`
`-ve` sign indicates that the accelerations are in opposite direaction Suppose acceleration of `m_(2)` is `a_(2)` is downaward and then acceleration of `m_(1)` will be
`a_(1)` upwards
`T -m_(10g) =m_(1)a_(1)`
`T =m_(1)g +m_(1)a_(1)`
`m_(2)g -2T =m_(2)a_(2)`
`m_(2)g -2 (m_(1)g +m_(1)a_(1)) =m_(2)a_(2)`
`m_(2)g -2m_(1)g =m_(2) a_(2) +4m_(1) a_(2) (":' a_(1) =2a_(2))`
`'-'` sign should not be substituted
`a_(2) = ((m_(2)-2m_(1))g)/(4m_(1)+m_(2)) ms^(-2)`

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