Two balls `A` and `B` each of mass `m` are placed on a smooth ground as shown in the figure. Another ball `C` of mass `M` arranged to the right of ball `B` as shown. If a velocity `v_(1)` is given to ball `A` in rightward direction, consider two cases. Case-I `M gt m` and case-II `M lt m`. Take all the collisions perfectly elastic (`e = 1`), find the number of collision in case-I and case-II.

Answer

a. 2, b. 3`

Explanation

a. First collision will be between balls `A` and `B`. Since both the balls are of same mass, after collition `A` will come to rest and `B` will move with `v_(1)`, now it will collide to `C`. If after this collision, velocities of balls `B` and `C` are `v_(B)` and `v_(C)` respectively, we have
`v_(R)=((m-M)/(m+M))v_(I),v_(C)=((2m)/(m+M))v_(1)`
Here as we have `Mltm,ltv_(C)` and both are positive thus both the balls are moving forward and will not collide again, hence there are total two collision.
b. If in second collision we have `Mgtm`, we get `v_(B)` negative, so ball `B` moves in backward direction after between collision and will strike again to ball `A` (which is at rest to the left of it) and come to rest and ball `A` will more to the left with speed `v_(B)`. Then now there are total three collision.