A) 48 N
B) 24 N
C) 12 N
D) 96 N
Correct Answer: B
Solution :
The vector \[\overrightarrow{OA}\] represents the momentum of the object before the collision, and the vector \[\overrightarrow{OB}\] that after the collision. The vector \[\overrightarrow{AB}\] represents the change in momentum of the object \[\Delta \vec{p}\]. As the magnitudes 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, the change in momentum is due only to the change in direction of the perpendicular components. Hence, \[\Delta p=OB\sin {{30}^{o}}-(-OA\sin {{30}^{o}})\] \[=mv\sin {{30}^{o}}-(-mv\sin {{30}^{o}})\] \[=2\,mv\sin {{30}^{o}}\] Its time rate will appear in the form of average force acting on the wall. \[\therefore F\times t=2mv\sin {{30}^{o}}\] \[orF=\frac{2mv\sin {{30}^{o}}}{t}\] Given, m = 0.5 kg, v = 12 m/s, t = 0.25 s \[\theta ={{30}^{o}}\] Hence, \[F=\frac{2\times 0.5\times 12\sin {{30}^{o}}}{0.25}=24\,N\]You need to login to perform this action.
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