Question

Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface and push them together until they touch. Each ball has a mass of 4 kg. Think carefully about what to use for the distance between the two balls. And to keep things simpler, let's assume they slide without rolling.Now consider the case where both balls are free to move once you release them. Neglecting friction, calculate the final speed of one of the balls.?

Answer 1

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

                   R = 2r

                      = $2 \times 0.1 m$

                      = 0.2 m

where,    R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

                U = $\frac{k \times Q \times Q}{R}$

                    = $\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}$

                    = 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, $\frac{U}{2}$.

Hence,   K.E = $\frac{U}{2}$

                     = $\frac{0.081}{2}$

                     = 0.0405 J

Now, we will calculate the speed of balls as follows.

                 V = $\sqrt{\sqrt{\frac{2 \times K.E}{m}}$

                     = $\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}$

                     = 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.