Two identical particles of mass m carrying charge q each initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v the closest distance of approach be
Answers
You are assuming that particle #1 is fixed in position and cannot move. I think the question is implying that particle #1 will start moving away from particle #2 as the latter gets closer. (We must also assume that the particles are constrained to move in the same straight line.)
This is an elastic collision. Both energy and momentum are conserved. In the centre of momentum frame (= centre of mass frame) the 2 particles will be at rest at the point of closest approach. In the laboratory frame they will have the same momentum and velocity at this point (zero relative velocity), sharing the initial momentum of particle #2.
Momentum is conserved at all times. Working in the laboratory frame, at closest approach there is no relative velocity between the particles (otherwise they will be closer later/earlier) so they will share the initial momentum mv and each have velocity 12v.
Energy is also conserved so KE initially = PE + KE at closest approach :
12mv2=PE+2×12m(12v)2.
Rearranging we get
PE=kQ2r=14mv2
r=4kQ2mv2=Q2πϵ0mv2