A proton moves with a speed u directly toward a free proton originally at rest. Find the distance of closest approach for the two protons. Given that mass of the proton is m and charge of the proton is +e.

Explanation

As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving and so the velocity of first particle will decrease while of other will increase and at closest approach, both will move with same velocity. so, if v is the common velocity of each particle at closest approach, then by 'conservation of momentum' of the two proton system.
`m u=mv+mv` i.e., `v=1/2 u`
And by conservation of energy, `1/2 m u^(2)=1/2 mv^(2)+1/2 mv^(2)+1/(4pi epsi_(0)) e^(2)/r`
`implies 1/2 m u^(2)-m (u/2)^(2) =1/(4pi epsi_(0)) e^(2)/r" "["as "v=u/2]`
`implies 1/4 m u^(2)=e^(2)/(4pi epsi_(0)r)" "implies" "r=e^(2)/(pi m epsi_(0) u^(2))`