hello Please Solve this question

Two identical particles,each having a charge of 2.0 X 10^-4 C and mass of 10 g,are kept at a seperation of 10cm and then released.What would bee the speeds of the particle When the seperation becomes large ?

Answer :600m/s

initial potential energy is (PE)i = kq² /r                  (charges are of same magnitude q)

                    k = 9 . 10⁹ units

                                        (PE)i  =3600j                 (after substituting value of charge and K)

finally when distance is maximum then (PE)=kq² /r                    

                                r=r(max)=infinity

      so final potential energy is 0...

initially kinetic energy is zero ....

finally kinetic energy is (KE)f =  mv₁² /2   +  mv_{2 ² /2}

                                         =m(v_{1² +} v_{2² )/2 ................1}

from conservation of energy ,total energy initial = total energy final

    therefore ,     m(v_{1² +v2² )/2 =3600 ..................2}

       since no external; force is acting so momentam will be conserved..

     initial momentam =0

    final momentam=mv₁ +mv₂

      mv_{1 +mv2 =0}

            v_1=-v_{2 ..................1}

    now from eq 2 putting v1=-v2

     v1=600m/s

      v2=-600m/s