A cricket ball of mass 100 g is moving with a velocity of 10 m s^–1 and is hit by a bat so that it turns back and moves with a velocity of 20 m s^–1. Find the impulse and the force if the force acts for 0.01 s.

Given, M = 100 g = 0.1 kg

∴ Initial velocity u = –10 m s⁻¹

Final velocity v = +20 m s⁻¹

Time t = 0.01 s

Impulse = change in momentum

Impulse = m (v − u)

= 0.1[20 - (- 10)]

$$\begin{gathered} =0.1[20+10] \\ =0.1\times 30 \mathrm{kg} \mathrm{m}{ \mathrm{s}}^{-1}=3 \mathrm{kg} \mathrm{m}{ \mathrm{s}}^{-1} \\ \mathrm{F}=\mathrm{m} \mathrm{a} \\ \text{ Force }=\frac{\text{ Change in momentum }}{\text{ time }} \\ =\frac{\text{ Impulse }}{\text{ time }}=\frac{3 \mathrm{kg}{ \mathrm{ms}}^{-1}}{0.01 \mathrm{s}} \\ \therefore \text{ Force }=\frac{0.1(20+10)}{0.01}=\frac{3}{0.01}=300 \mathrm{N} \end{gathered}$$