Question

Two protons approach one another from very far away with an initial speed of 500 m/s. How close do they come to one another before turning around?

Answer 1

Compare initial energy with final energy (at closest point when protons are stationary):

Apply conservation of total energy \(E\) of system, which at separation \(r\) is:

\(E = T + U = 2 \times \frac{1}{2} m_p v^2 + \dfrac{k e^2}{r}\)

Initially, at large distance \(r\) apart: \(U_i \approx 0\)

At closest point: \(v_f = 0 ~~~ \therefore T_f = 0\)

\(\therefore m_p v_i^2 = \dfrac{k e^2}{r_f}\)

\(r_f = \dfrac{k e^2}{m_p v_i^2} \approx 552 ~ nm\)