Two uniform non conducting balls A and B have identical size having radius R but made of different density material (density of A = 2 density of B). The ball A is +vely charged and ball B is -vely charged. The balls are released on the horizontal smooth surface at the separation 10R as shown in figure. Because of mutual attraction the balls start moving towards each other. They will collide at a point 

There is no external force on system (ball A and ball B). The balls moves due to mutual attraction that is internal force to the system. The centre of mass of the system will not move. The will collide at the position of centre of mass of balls A and B. It is easy to calculate the position of centre of mass at initial stage.
${\mathrm{x}}_{\mathrm{cm}} = \frac{{\mathrm{m}}_{\mathrm{A}}\times \mathrm{A}+{\mathrm{m}}_{\mathrm{B}}\times \mathrm{B}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}} = \frac{{\mathrm{m}}_{\mathrm{A}}.\mathrm{o}+{\mathrm{m}}_{\mathrm{B}}.10\mathrm{R}}{({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}})}$

$m_A = \frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_{\mathrm{A}} \mathrm{and} {\mathrm{m}}_{\mathrm{B}} = \frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_{\mathrm{B}}$

$\mathrm{As} {\mathrm{\rho }}_{\mathrm{A}} =2{\mathrm{\rho }}_{\mathrm{B}} \Rightarrow {\mathrm{m}}_{\mathrm{A}} = 2{\mathrm{m}}_{\mathrm{B}}$

Hence ${\mathrm{x}}_{\mathrm{cm}} = \frac{{\mathrm{m}}_{\mathrm{B}}\left(10\mathrm{R}\right)}{3{\mathrm{m}}_{\mathrm{B}}} = \frac{10}{3}\mathrm{R}$

When they are touching each other then also the centre of mass lie at same place.

X'_cm = $\frac{{\mathrm{m}}_{\mathrm{A}}\left(0\right)+{\mathrm{m}}_{\mathrm{B}}2\mathrm{R}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}=\frac{2\mathrm{R}}{3}$ from the first

The two are touching each other at an additional $\frac{R}{3}$ distance from CM.

Hence the collision is happening at $\frac{10R}{3}+\frac{R}{3}=\frac{11R}{3}$