`n` elastic balls are placed at rest on a smooth horizontal plane which is circular at the ends with radius `r` as shown in the figure. The masses of the balls are `m`, `m/2`, `m/2^2`, …………….`(m)/(2^(n-1))` respectively. What is the minimum velocity which should be imparted to the first ball of mass `m` such that `n^th` ball completes the vertical circle

Explanation

Let speed to be imparted to the first ball be `v_(0)`.
Consider the impact between the first two balls and `v_(1)` and `v_(2)` be the velocities of balls `1` and `2` after the impact respectively.
According to law of conservation of linear
momentum `mv_(0)=mv_(1) +(m)/(2) v_(2) rarr` (1)
According to law of conservation of kinetic energy
`(1)/(2)mv_(0)^(2) =(1)/(2) mv_(1)^(2) +(1)/(2)((m)/(2))v_(2)^(2) rarr` (2)
Solving equations (1) and (2), we get `v_(2) = (4)/(3) v_(0)`
Similarly, for `n^(th)` ball `v_(n) = ((4)/(3))^(n-1) v_(0) rarr` (3)
For the `n^(th)` ball the complete the vertical circular
motion `v_(n) = sqrt(5gr) rarr` (4)
From equations (3) and (4), we have
`((4)/(3))^(n-1) v_(0) = sqrt(5gr) , v_(0) = ((3)/(4))^(n-1) sqrt(5gr)`.