Q. A $3 \, kg$ ball strikes a heavy rigid wall with a speed of $10 \, m \, s^{- 1}$ at an angle of $60^\circ $ with the wall. It gets reflected with the same speed at $60^\circ $ with the wall. If the ball is in contact with the wall for $0.2 \, s$ , the average force exerted on the ball by the wall is
Question

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Solution:

Change in momentum along the wall
$=mvcos\left(60 ^\circ \right)-mvcos\left(60 ^\circ \right)=0$ .
Change in momentum perpendicular to the wall
$=mvsin\left(60 ^\circ \right)-\left(- m v sin \left(60 ^\circ \right)=2mvsin\left(60 ^\circ \right)$
$\therefore $ Applied force $=\frac{Change \, in \, momentum}{Time}$
$=\frac{2 m v sin \left(60 ^\circ \right)}{0.20}$
$=\frac{2 \times 3 \times 10 \times \sqrt{3}}{2 \times 0 . 20}=50\times 3\sqrt{3}$
$=150\sqrt{3}N$ .