- #1

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I would figure that you could use conservation of energy in the sense that the energy of the system initially is the kinetic energies of the two particles combined (Enet1 = Kp + Ka). At the point of closest approach, their speeds should be zero, and hence Enet2 = Uelec = Kq1q2/r. From here it should be straightforward:

Enet1 = Enet2

Kp + Ka = Kq1q2/r

Then solve for "r".

However, this is incorrect. Perhaps my assumption that the alpha particle (4 times the mass, 2 times the charge) stops completely is wrong. At this point, I really have no idea.