Two boys each of mass `40 kg` are standing on the right side of a trolley of mass `80 kg` , initially at rest on smooth floor. The boys jumps to the right with velocity `3 m//s` with respect to the trolley before the jump. Find the final velocity of the trolley if the jumps : (a) together , (b) one after the other and (c ) one boy is on the left and the other on the right , the boy on the left jumps to the left and then the boy on the right jumps to the right. Each boy jumps with `3 m//s` with respect to the trolley before the jump.

Explanation

`v_(B//g) = 3 m//s` (Since initially the trolley is at rest)
`0 = 80 v_(1) + ( 40 + 40) xx 3`
`v_(1) = -3 m//s`
i.e. trolley will move to the left with velocity `3 m//s`.
(b) By the momentum conservation
`0 = 40 xx 3 + (80 + 40)v_(1)`
`v_(1) = -1 m//s` , i.e. trolley to the right with velocity `3 m//s` w.r.t. the trolley.
(c ) ` 0 = -40 xx 3 + (80 + 40) v_(1) rArr v_(1) = 1 m//s`
`v_(B//g) = ( 3 + 1) = 4 m//s`
`(80 + 40) xx 1 = 80 v_(1) + 40 xx ( 3 + 1)`
`v_(2) = -(1)/(2) m//s` , i.e. the trolley moves to the left.