A proton and an alpha particle ( q= + 2e, m=4u ) are fired directly toward each other from far away, each with an initial speed of 0.01 c.What is their distance of closest approach, as measured between their centers?

Answer 1

This is not very different from your earlier problem with the three electrons and the proton in the middle if the triangle.
All they have in the beginning is kinetic energy, 1/2 m v^2
All they have when they stop, glaring at each other, is potential energy.
Again your potential energy is that integral of the force in from infinity k q1 q2/r

Answer 2

but i think when hey are closest their velocities are not zero since their masses are different. i tried using conservation of momentum also.

Answer 3

You are right. They do not stop. Momentum is conserved
(m1+m2) Vfinal = m1 v1 initial + m2 v2 initial

one v initial will be negative of course.
Then
final KE of system = (1/2) m Vfinal^2
That KE is smaller than the original sum of KEs
the difference is PE = k q1 q2/r

Answer 4

Holy bals this is hard