shown, both blocks are released room rest. Length of `4kg` block is `2m` and of `1kg` is `4m`. Find the time they take to cross each other Assume pulley to be light and string to be light and inelastic
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Answer

1

Explanation

From `FBD` of blocks `A` and `B` solve acceleration of each block `4g -T =4a….(1)`
`T -1g = 1xx a…(2)`
After solving eqns (1) and (2) `a = (3g)/(5)`
acceleration of A w.r.t `B a_(A//B) = (6g)/(5) =12m//s^(2)`
If A will cross `B` then distance travelled by A w.r.t `B` is `6m`
`6 = 0 + (1)/(2) xx 12 xx t^(2), t = 1sec` For
`F_(0) =4m_(2)g`
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