A soccer ball of mass 250 g is moving horizontally to the left with a speed $22 \mathrm{~ms}^{-1}$. This ball is kicked towards right with a velocity $30 \mathrm{~ms}^{-1}$ at an angle $53^{\circ}$ with the horizontal in upward direction. Assuming that it took 0.01 s for the collision to take…

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A soccer ball of mass 250 g is moving horizontally to the left with a speed $22 \mathrm{~ms}^{-1}$. This ball is kicked towards right with a velocity $30 \mathrm{~ms}^{-1}$ at an angle $53^{\circ}$ with the horizontal in upward direction. Assuming that it took 0.01 s for the collision to take place, the average force acting is $\left(\cos 53^{\circ}=\frac{3}{5} ; \sin 53^{\circ}=\frac{4}{5}\right)$

Aicharya · Accepted Answer

$\Delta \mathrm{t}=0.01 \mathrm{~s}$
Change in momentum, $\begin{aligned} & \Delta \mathrm{P}_{\mathrm{x}}=\mathrm{m}\left(\mathrm{v} \cos 53^{\circ}-\mathrm{u}\right) \\ & =\frac{1}{4}\left[30 \times \frac{3}{5}-(-22)\right]=10 \mathrm{Ns} \\ & \Delta \mathrm{P}_{\mathrm{y}}=\mathrm{mv} \sin 53^{\circ}=\frac{1}{4} \times 30 \times \frac{4}{5}=6 \mathrm{Ns} \\ & \therefore \Delta \mathrm{P} \sqrt{\Delta \mathrm{P}_{\mathrm{x}}^2+\Delta \mathrm{P}_{\mathrm{y}}^2}=\sqrt{(10)^2+(6)^2}=\sqrt{136} \end{aligned}$ $\therefore \quad$ Average force, $\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\sqrt{136}}{0.01}=1166 \mathrm{~N}$

A soccer ball of mass 250 g is moving horizontally to the left with a speed $22 \mathrm{~ms}^{-1}$. This…

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