A proton of mass `m` moving with a speed `v_(0)` apporoches a stationary proton that is free to move. Assuming impact parameter to be zero., i.e., head-on collision. How close will be incident proton go to other proton ?

Explanation

By momentum conservation
`mv_(0) =mv + mv implies v=(v_(0))/(2)` ….(i)
By energy conservation
`(1)/(2)mv_(0)^(2) =mv^(2) +(1)/(2)mv^(2) +(1)/(4pi epsilon_(0))(e.e)/(r)` ….(ii)
`(1)/(2)mv_(0)^(2) = (1)/(2) mv^(2) +(e^(2))/(4pi in_(0)r) =m((v_(0))/(2))^(2) +(e^(2))/(4pi in_(0)r)`
`(mv_(0)^(2))/(2) - (mv_(0)^(2))/(4) =(e^(2))/(4pi in_(0)r)`
`(mv_(0)^(2))/(4) =(e^(2))/(4pi in_(0)r)`
`r=(e^(2))/(2pi in_(0)m v_(0)^(2))`
Note: At the time of minimum distance, protons move with same velocity.