In the previous problem another block `C` of mass `m` is placed at distance `1 m` right of block `B`. Find `(a)` the distance travelled by the combined mass if the collision is completely inelastic and `(b)` separation between the blocks when they stop after head - on elastic collision.

Explanation

Speed of block `B` after traveling a distance `1 m`.
`u_(1)^(2) = (2)^(2) - 2 as = 4 - 2 xx 1 xx 1 = 2 rArr mu_(1) = sqrt(2) m//s`
`(a = mu g = 0.1 xx 10 = 1 m//s^(2))`
(a) Now , the collision is completely elastic , the blocks move together.
By the momentum conservation
`3m sqrt(2) + m xx 0 = ( 3 m + m) V rArr V = (3 sqrt(2))/(4) m//s`
The distance travelled by the combined mass on a rough surface
`0 = V^(2) - 2ad rArr 0 = ((3 sqrt(2))/(4))^(2) - 2 xx 1 xx d`
`d = (9)/(16) m`
(b) If the collision is head - on elastic , the blocks will move with different velocities
`3m sqrt(2) + m xx 0 = 3mv_(1) + mv_(2)` (i)
`v_(1) + sqrt(2) = v_(2) + 0` (ii)
Solving , `v_(1) = (1)/(sqrt(2)) m//s , v_(2) = (3)/(sqrt(2)) m//s`
Let the distance travelled by blocks `B` and `C` are `d_(1)` and `d_(2)` before stop, in the same direction.
`0 = v_(1)^(2) - 2 ad`
`0 = (1)/(2) - 2 xx 1 xx d_(1) rArr d_(1) = (1)/(4) m`
`0 = v_(2)^(2) - 2ad_(2)`
`0 = ((3)/(sqrt(2)))^(2) - 2 xx 1 xx d_(2)`
`rArr d_(2) = (9)/(4) m`
Separation between blocks `= d_(2) - d_(1) = (9)/(4) - (1)/(4) = 2m`