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?

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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

`m*V^2 =k*e^2/R`

`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 =`

`=1.14 nm`


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