As initially q is at infinite distance from Q, thus \\\\(P.E_i=0\\\\)For case 1: the closest distance given is r and velocity of q is v initially.Applying conservation of energy, \\\\(P.E_i+K.E_i=P.E_f+K.E_f\\\\)\\\\(0+ \\dfrac{1}{2}mv^2= \\dfrac{KQq}{r}+ 0\\\\)\\\\(\\implies \\\\)\\\\( \\dfrac{1}{2}mv^2= \\dfrac{KQq}{r}\\\\) ................(1)For case 1: Let the closest distance be r' and velocity of q is 2v initially.Applying conservation of energy, \\\\(P.E_i+K.E_i=P.E_f+K.E_f\\\\)\\\\(0+ \\dfrac{1}{2}m(2v)^2= \\dfrac{KQq}{r'}+ 0\\\\)\\\\(\\implies \\\\)\\\\( \\dfrac{1}{2}m(2v)^2= \\dfrac{KQq}{r'}\\\\) ................(2)Solving (1) and (2), \\\\(r'=\\dfrac{r}{4}\\\\)