A) \[{{60}^{o}}\]
B) \[{{30}^{o}}\]
C) \[{{45}^{o}}\]
D) \[{{90}^{o}}\]
Correct Answer: D
Solution :
From Newton's second law of motion the force acting is defined as rate of change of momentum. \[F=\frac{d\vec{p}}{dt}\] Given, \[\vec{p}=3\,\cos \,4t.\,\hat{i}+3\,\sin \,4t.\hat{j}\] \[\vec{F}=\frac{d\,\vec{p}}{dt}=-12\,\sin \,4t.\,\hat{i}+12\,\cos \,4t.\,\hat{j}\] Also, \[\vec{F}=\vec{p}=|\vec{F}|\,|\vec{p}|\,\cos \theta \] \[=(-12\,\sin \,4t.\,\hat{i}+12\,\cos \,4t.\hat{j}).(3\,\cos \,4t.\,\hat{i}+3\sin \,4t.\hat{j})\]\[=36(-\sin \,4t\,\cos \,4t\,+\,\cos \,4t\,\sin \,4t)=0\] \[\Rightarrow \] \[|\vec{F}|.|\vec{P}|\,\cos \,\theta =0\] Since, \[|\vec{F}|\ne 0,\,\,\,\,|\vec{p}|\ne 0\] \[\therefore \] \[\cos \,\theta =0\] Hence, \[-\,\theta ={{90}^{o}}\]
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