Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 1.10 x 104 m/s. What is the distance of closest approach?
Initially the protons have only kinetic energy.
`E_k = 2*(m_p*V^2)/2 =m_p*V^2`
All this kinetic energy is transformed into potential energy at the closest point of their approach. Since the charge of the proton is `+e` and the potential generated by a charge `+e` at a distance `R` is
`U = +k*e/R` , where `k=1/(4*pi*epsilon_0)=9*10^9 (V*m)/C`
then the final potential energy is
`E_p = +e*U = k*e^2/R`
Now from the equality
`E_k =E_p` we have
`R = (k*e^2)/(m*V^2) = (9*10^9*(1.6*10^-19)^2)/((1.67*10^-27)*(1.1*10^4)^2) =1.14*10^-9 m =`