A ball of mass $m$ strikes a rigid wall with speed $u$ at an angle of $30^{\circ}$ and get reflected with the same speed and at the same angle as shown in the figure. If the ball is in contact with the wall for time t, then the force acting on the wall is

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$\frac{2 m u \cos 30^{\circ}}{t}$

$\frac{m u \cos 30^{\circ}}{t}$

$\frac{2 m u \sin 30^{\circ}}{t}$

$\frac{m u \sin 30^{\circ}}{t}$

answer is D.

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Detailed Solution

Initial momentum of the ball is ${\vec{p}}_{i} = m u \cos 30^{\circ} \hat{i} - m u \sin 30^{\circ} \hat{j}$

Final momentum of the ball is ${\vec{p}}_{f} = - mucos 30^{\circ} \hat{i} - musin 30^{0} \hat{j}$

Therefore change in momentum, $∆ \vec{p} = {\vec{p}}_{f} - {\vec{p}}_{i} = - 2 m u \cos 30^{\circ} \hat{i}$

Impulse $=$ Change in momentum $= - 2 m u \cos 30^{\circ} \hat{i}$

As impulse and force are in the same direction, therefore, force on the ball due to the wall is normal to the wall along the negative $x -$ axis. Using Newton's $3^{r d}$ law of motion the force on the wall due to the ball is normal to the wall along the positive $x -$ direction.

Therefore $F = \frac{2 m u \cos 30^{\circ}}{t}$

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