Two particles each of mass m and charge q are separated by r1 and the…

Due to symmetry, each particle will have same speed

$E_{i}=u_{i}+v_{i}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_1^2}+0$

$E_{f}=u_{f}+v_{f}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_2^2}+\frac{1}{2} mv^2+\frac{1}{2} mu^2$

$=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_2^2}+mv^2$

At the field is conservative, 

hence applying COE,

$\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_1^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r_2^2}+mv^2$

∴ $mv^2=\frac{q^2}{4 \pi \varepsilon_0}\left(\frac{1}{r_1^2}-\frac{1}{r_2^2}\right)$

∴ $v=\frac{q^2}{r_1 r_2} \sqrt{\left(r_1^2-r_2^2\right) / 4 \pi \varepsilon_0 m}$