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

Approved