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Question: A ball of mass m strikes a rigid wall with speed u at an angle of \(30^{o}\)and get reflected with t...

A ball of mass m strikes a rigid wall with speed u at an angle of 30o30^{o}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 :

A

musin⁡30ot\frac{mu\sin 30^{o}}{t}

B

2musin⁡30ot\frac{2mu\sin 30^{o}}{t}

C

mucos⁡30ot\frac{mu\cos 30^{o}}{t}

D

2mucos⁡30ot\frac{2mu\cos 30^{o}}{t}

Answer

2mucos⁡30ot\frac{2mu\cos 30^{o}}{t}

Explanation

Solution

Initial momentum of the ball is

pi→=mucos⁡30∘i∧−musin⁡θj∧\overset{\rightarrow}{p_{i}} = mu\cos 30{^\circ}\overset{\land}{i} - mu\sin\theta\overset{\land}{j}

Final momentum of the ball is

pf→=−mucos⁡30∘i∧−musin⁡30j∧\overset{\rightarrow}{p_{f}} = - mu\cos 30{^\circ}\overset{\land}{i} - mu\sin 30\overset{\land}{j}

∴\thereforeChange in momentum

Δp→=pf→−pi→=−2mucos⁡30∘i∧\Delta\overset{\rightarrow}{p_{}} = \overset{\rightarrow}{p_{f}} - \overset{\rightarrow}{p_{i}} = - 2mu\cos 30{^\circ}\overset{\land}{i}

Impulse = Change in momentum = −2mucos⁡30∘i∧- 2mu\cos 30{^\circ}\overset{\land}{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 3rd3^{rd}law of motion the force on the wall due to the ball is normal to the wall along the positive x- direction

∴F=2mucos⁡30∘t\therefore F = \frac{2mu\cos 30{^\circ}}{t}