A particle of mass `m` moving with horizontal speed `6 m//s` as shown in the figure. If `m lt lt M` then for one - dimensional elastic collision , the speed of lighter particle after collision will be

Options

A. `2 m//s` in original direction
B. `2 m//s` opposite to the original direction
C. `4 m//s` opposite to the original direction
D. `4 m//s` in original direction

Answer

A

Explanation

`m u_(1) + M u^(2) = mv_(1) Mv_(2)`
`rArr mv_(1) + Mv_(2) = mu_(1) + Mu_(2)` (i)
`v_(1) + u_(1) = v_(2) + u_(2)`
`v_(1) - v_(2) = - u_(1) + u_(2)` (ii)
Multiplying Eq.(ii) by `M` , we get
`Mv_(1) - Mv_(2) = -Mu_(1) + Mu_(2)` (iii)
Adding (i) and (iii) , we get
`(m + M) v_(1) = (m - M)u_(1) + 2Mu_(2)`
`v_(1) = ((m - M)/(m + M)) u_(1) + ((2M)/(m + M)) u_(2)`
Since `m lt lt M`
`v_(1) ~= - u_(1) + 2 u_(2)`
`~= - 6 + 2 xx 4 = 2 m//s`