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Just analyze the motion. Total n-1 collisions will happen. If we consider those n balls to be our system, then one thing to notice is that the net external force on this system is zero (after we have provided velocity to the first ball). So, centre of mass of these n balls must retain its velocity (you can calculate). After first collision, velocities get redistributed as shown but we are only concerned about the last ball (nth ball). So, for every collision we must only be concerned about the second ball. Now, after first collision we found that velocity of second ball is 4v/3. Going by the pattern of mass of ball and velocity of that ball, we can find the velocity of nth ball.
One more thing to notice is that, each ball if it is not at rest always (until the last ball starts going in circle) has speed in the right direction, whether it is after collision or before collision. Also, after each collision between a pair of balls, velocity of first ball (first one in them from left) is reduced to a value (smaller than velocity of second ball after collision) and velocity of second ball is increased from zero to a value (greater than velocity of first ball after collision). So, the collisions will only happen towards right.
1st will collide with 2nd, 2nd with 3rd, and so on until (n-1)th with nth and throughout this process if any ball has some velocity its direction is towards right.
In short, each ball having some velocity (let's say v') towards right collides with another ball at rest having half of its mass and increases its velocity to 4v'/3 and reduces its own velocity to v'/3. That's why, velocity of 2nd ball after 1st collision is very less as compared to velocity of nth ball just after (n-1)th collision.
By the way, this can easily be solved just by putting n = 2 and solving the question for just two balls. n = 1 doesn't work here because all the options become equal to sqrt(5gr).
Hope, this helps...
