Two billiard balls of mass $$0.05 \text{ kg}$$ each moving in opposite directions with $$10 \text{ m s}^{-1}$$ collide and rebound with the same speed. If the time duration of contact is $$t = 0.005 \text{ s}$$, then what is the force exerted on the ball due to each other?

Two billiard balls of mass $$m = 0.05 \text{ kg}$$ each, moving in opposite directions at $$10 \text{ m s}^{-1}$$, collide and rebound with the same speed. Contact time $$t = 0.005 \text{ s}$$.

Calculate the change in momentum of one ball.

Taking the initial direction of one ball as positive, its velocity changes from $$+10 \text{ m s}^{-1}$$ to $$-10 \text{ m s}^{-1}$$:

$$\Delta p = m(v_f - v_i) = 0.05 \times (-10 - 10) = 0.05 \times (-20) = -1 \text{ kg m s}^{-1}$$

Calculate the force using the impulse-momentum theorem.

$$F = \frac{|\Delta p|}{t} = \frac{1}{0.005} = 200 \text{ N}$$

The correct answer is Option B: $$200 \text{ N}$$.