Two smooth spherical non conducting shells each of radius $$R$$ having uniformly distributed charge $$Q$$ & $$-Q$$ on their surfaces are released on a smooth non-conducting surface when the distance between their centers is $$10R$$. The mass of A is $$m$$ and that of B is $$2m$$. The speed of A just before A and B collide is: [Neglect gravitational interaction]

The electric potential energy of the two charged shells is the same as that of two point charges placed at their centres.

Initially, the distance between their centres is $$10R$$ and finally, just before collision, it is $$2R$$.

Since the system is released from rest, conservation of energy gives

$$\frac{KQ^2}{10R}-\frac{KQ^2}{2R}=-(K_{\text{total}})$$

$$K_{\text{total}}=\frac{2KQ^2}{5R}$$

Let the speeds of A and B just before collision be $$v_A$$ and $$v_B$$ respectively.

Since there is no external horizontal force, momentum is conserved.

$$mv_A=2mv_B$$

$$v_B=\frac{v_A}{2}$$

Therefore, the total kinetic energy is

$$K_{\text{total}}=\frac{1}{2}mv_A^2+\frac{1}{2}(2m)v_B^2$$

$$K_{\text{total}}=\frac{1}{2}mv_A^2+mv_B^2$$

$$K_{\text{total}}=\frac{1}{2}mv_A^2+\frac{1}{4}mv_A^2=\frac{3}{4}mv_A^2$$

Equating the kinetic energy to the decrease in potential energy,

$$\frac{3}{4}mv_A^2=\frac{2KQ^2}{5R}$$

$$v_A^2=\frac{8KQ^2}{15mR}$$

Therefore,  $${v_A=Q\sqrt{\frac{8K}{15mR}}}$$

Hence, the correct option is C.