`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

Options

A. `sqrt5gr`
B. `(3/4)^(n-1)sqrt5gr`
C. `(3/4)^(n)sqrtgr`
D. `(4/3)6(n)sqrt5gr`

Answer

`(2)`

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) rightarrow (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) rightarrow (2)`
Solving equations (1) and (2) , we get `v_(2) = 4/3v_(0)`
Similarly , for `n^(th)` ball `_(n) = ((4)/(3))^(n-1) v_(0) rightarrow (3)`
For the `n^(th)` ball to complete the vertical circular motion `v_(n) = sqrt(5gr) rightarrow (4)`
From equations (3) and (4), we have
`((4)/(3))^(n-1) v_(0) = sqrt(5gr),v_(0) = ((3)/(4))^(n-1) sqrt(5gr)`