Question

A proton of mass ${}^{\prime }{m}^{\prime }$ moving with a speed ${}^{\prime }{v}_0{}^{\prime }$ approaches a stationary proton that is free to move. Assume impact parameter to be zero, i.e. head-on collision. How close will the incident proton go to other proton?

• A. $\frac{{ϵ}^3}{\pi {ϵ}_0{m}^2{v}_0}$
• B. $\frac{{ϵ}^3}{\pi {ϵ}_{0}m{v}_0}$
• C. $\frac{{ϵ}^2}{\pi {ϵ}_{0}m{v}_0^2}$
• D. None of the above

Answer 1

Answer: (C)

$\frac{{ϵ}^2}{\pi {ϵ}_{0}m{v}_0^2}$

Since there is a collision, the KE keeps getting converted to PE and they approach each other.
Until, KE = PE after which they start to push each other away. I.e when $PE=\frac{1}{2}K{E}_{initial}$ is when we have closest approach.
Initial KE is: $\frac{1}{2}m{v}^2$
Half of which is $\frac{1}{4}m{v}^2$
PE is $\frac{1}{4\pi {ϵ}_0}\frac{e^2}{d}$
Equating the two, we get distance of closest approach as: $d=\frac{1}{\pi {ϵ}_0}\frac{e^2}{m{v}^2}$