A cricket ball of mass 50   g having velocity 50   cm   s - 1 is stopped in 0 . 5   s .…

Initial momentum of the ball is

$$\begin{gathered} =\left(50\times {10}^{-3} \mathrm{kg}\right)\times \left(50\times {10}^{-2} \mathrm{m} {\mathrm{s}}^{-1}\right) \\ =25\times {10}^{-3} \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

Final momentum $=0 \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}$

Magnitude of force applied

$$\begin{gathered} =\left|\frac{∆p}{t}\right| \\ =\left|\frac{0 \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}-25\times {10}^{-3} \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1} }{0.5 \mathrm{s}}\right| \\ =\left|-5\times {10}^{-2} \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}\right| \\ =0.05 \mathrm{N} \end{gathered}$$