question_answer

                A ball of mass 3 kg moving with a speed of 100 m/s, strikes a wall at an angle \[{{60}^{\text{o}}}\] (as shown in figure). The ball rebounds at the same speed and remains in contact with the ball for 0.2 s, the force exerted by the ball on the wall is:

A)                                                                                                                                                                         \[1500\sqrt{3}\,N\]        

B)                 1500 N 

C)                 \[300\,\sqrt{3}\,N\]       

D)                 300 N

Correct Answer: A

Solution :

                Key Idea: The force imparted by the ball is equal to rate of change of linear momentum.                 The vector \[\overrightarrow{OA}\] represents the momentum of the ball before the collision and the vector \[\overrightarrow{OB}\] that after the collision. The vector \[\overrightarrow{AB}\] represents the change in momentum of the ball \[\overrightarrow{\Delta p}\].                 As the magnitude of \[\overrightarrow{OA}\] and \[\overrightarrow{OB}\] are equal the components of \[\overrightarrow{OA}\] and \[\overrightarrow{OB}\] along the wall are equal and in the same direction while those perpendicular to the wall are equal and opposite. Thus, die change in momentum is due only to the change in direction of the perpendicular components.                 Hence \[\overrightarrow{\Delta p}=OA\sin {{60}^{o}}-(-OB\sin {{60}^{o}})\]                 \[=mv\,\sin \,{{60}^{\text{o}}}\,+mv\,\sin \,{{60}^{\text{o}}}\]                 \[=2mv\,\sin \,{{60}^{\text{o}}}\]                 \[=2\times 3\times 100\times \frac{\sqrt{3}}{2}\]                 = 300\[\sqrt{3}\] kg-m/s                 The force exerted on the wall                 \[F=\frac{\overrightarrow{\Delta p}}{\Delta t}=\frac{300\sqrt{3}}{0.2}=1500\sqrt{3}\,N\]                 Note:    \[\overrightarrow{\Delta p}\] is directed perpendicular and away from the plate.