Two identical small elastic conducting balls of mass m and radius r have charges of oppsite signs Q and -q. Inditially they are held in free space with their centres at separation `lgt 2r` and then released Their speeds v long after their collision is
`v=sqrt((1)/(axxpiepsilon_(0)m){((Q+q)^(2))/(bxxr)-(Qq)/(l)})`
(Ignore non-uniformity in charge distributuion when ball come close to each other, magnetic field produced by motion of charges and the gravitational interaction between them). Find `((6xxaxxbxx)/(48)`? (a and b are lowest possible integers)

Answer

4

Explanation

`v=sqrt((1)/(4piepsilon_(0)m){((Q+q)^(2))/(8r)-(Qq)/(1)}`
kinetic energies of the ball can only change due to work done by interaction forces. The loss of energy due to redistribution of charge between the balls at time of contact will dissipate as heat and nothing to do with change in kinetic energy. Therefore, the balls gain kinetic energy at the expense of interaction energy before the collision, during collision they exchange their velcities and after the collision, they further gain kinetic energy at the expense of their interaction energy.