Two particles hv equal mass 5g each & opp. charges 4 X 10^-5 .They r released frm rest with separation of 1 m b\w them.Find speeds of the particle when the seperation is reduce to 50cm

Applying conservation of momentam bw initial and final stage

Pi = 0

Pf = mv1 +mv2

Pf = Pi

v1 = -v2..........1

Now applying energy conservation bw two points

KEi + (P.E)i = KEf  + (P.E)f

(mv1^2 -mv2^2)/2 =kq^2(2/r -1/r )

mv1^2=154/10

v1=53.66m/s=-v2

Regards

Arun (askIITians forum expert)