A ball of mass $$0.15 \text{ kg}$$ hits the wall with its initial speed of $$12 \text{ m s}^{-1}$$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $$100 \text{ N}$$, calculate the time duration of the contact of ball with the wall.

A ball of mass $$0.15 \text{ kg}$$ hits the wall at $$12 \text{ m s}^{-1}$$ and bounces back with the same speed under a force of $$100 \text{ N}$$, so we need to find the contact time.

Taking the initial direction of motion as positive, the initial momentum is $$p_i = m \times v = 0.15 \times 12 = 1.8 \text{ kg m s}^{-1}$$. Next, the final momentum after bouncing back is $$p_f = m \times (-v) = 0.15 \times (-12) = -1.8 \text{ kg m s}^{-1}$$. From this, the change in momentum is $$\Delta p = p_f - p_i = -1.8 - 1.8 = -3.6 \text{ kg m s}^{-1}$$.

Since $$F \cdot \Delta t = |\Delta p|$$ and the magnitude of the change in momentum is $$|\Delta p| = 3.6 \text{ kg m s}^{-1}$$, substituting gives $$\Delta t = \dfrac{|\Delta p|}{F} = \dfrac{3.6}{100} = 0.036 \text{ s}$$.

The correct answer is Option B: $$0.036 \text{ s}$$.