The below figure shows a ball of mass 0.6 kg rebounding between two walls separated by a distance of 2 cm. The ball rebounds between the walls every 2 seconds with a uniform velocity. Find the magnitude of each impulse?

Correct option is A

​$$\text{Given:} \\\text{Mass of the ball (m) = 0.6 kg} \\\text{Distance between the walls (x) = 2 cm = 0.02 m} \\\text{Time to travel the ball from one wall to another (t) = 2 s} \\\text{The magnitude of impulse (J) can be calculated using the formula:} \\J = F \cdot \Delta t = \Delta P \\\text{Where } F = \dfrac{d(mv)}{dt} \, \text{(Impulse-momentum equation)}$$

​​$$\text{ the velocity of the ball:} \\v = \dfrac{dx}{dt} = \dfrac{2 \, \text{cm}}{2 \, \text{s}} = 1 \, \text{cm/s} = 0.01 \, \text{m/s} \\\text{The change in momentum is the product of mass and velocity change. Since the ball rebounds, the velocity changes sign:} \\J = m \times (v - (-v)) = 2 \times m \times v \\J = 2 \times 0.6 \times 0.01 = 0.012 \, \text{kg.m/s} \\\text{Therefore, the magnitude of each impulse is:} \\J = 1.2 \times 10^{-2} \, \text{kg.m/s}$$​​