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.

Two balls A and B each of mass m are placed on a smooth ground as show

1.	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» Solution :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.