A) \[90{}^\circ \]
B) \[0{}^\circ \]
C) \[180{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: A
Solution :
Resultant momentum \[p={{p}_{x}}\hat{i}+{{p}_{y}}\hat{j}\] Given \[{{p}_{x}}=2\cos t,\] \[{{p}_{y}}=2sint\] \[\therefore \] momentum \[\vec{p}=2\cos t\,\hat{i}+2\,\sin r\,\hat{j}\] ?.(i) \[\because \] Force = Rate of change of momentum \[=\frac{d\,\vec{p}}{dt}\] \[\therefore \] \[\vec{F}=-2\sin t\hat{i}+2\cos t\hat{j}\] ??(ii) By \[\cos \,\theta =\frac{\vec{F}\,.\,\vec{p}}{|\vec{F}|.|\vec{p}|}\] \[\cos \theta =\frac{(2-sint\hat{i}+2\cos t\hat{j}).(2\cos t\hat{i}+2\sin t\hat{j})}{\sqrt{4{{\sin }^{2}}t+4{{\cos }^{2}}t}\sqrt{4{{\cos }^{2}}t+4{{\sin }^{2}}t}}\] or \[\cos \,\theta =0\] \[\Rightarrow \] \[\theta ={{90}^{o}}\]
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