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A particle moves in $x-y$ plane under the influence of a force $\vec{F}$ such that its linear momentum is $\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})$. If $\mathrm{k}$ is constant, the angle between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{p}}$ will be :
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$\frac{\pi}{2}$
$\begin{aligned} & \overrightarrow{\mathrm{P}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}} ;|\overrightarrow{\mathrm{P}}|=1 \\ & \because \overrightarrow{\mathrm{P}}=\mathrm{m} \overrightarrow{\mathrm{v}} \\ & \therefore \hat{\mathrm{P}}=\hat{\mathrm{v}} \\ & \Rightarrow \hat{\mathrm{v}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}} \\ & \hat{\mathrm{a}}=\frac{-\mathrm{k} \sin (\mathrm{kt}) \hat{\mathrm{i}}-\mathrm{k} \cos (\mathrm{kt}) \hat{\mathrm{j}}}{\mathrm{k}} \\ & \Rightarrow \hat{\mathrm{a}}=-\sin k t \hat{\mathrm{i}}-\cos \mathrm{kt} \hat{\mathrm{j}} \\ & \because \hat{\mathrm{F}}=\hat{\mathrm{a}}=-\sin \mathrm{kt} \hat{\mathrm{i}}-\cos \mathrm{kt} \hat{\mathrm{j}} \\ & \cos \theta=\frac{\hat{\mathrm{F}} \cdot \hat{\mathrm{P}}}{|\hat{\mathrm{F}}||\hat{\mathrm{P}}|}=-\frac{\sin \mathrm{kt} \cos \mathrm{t}+\sin \mathrm{kt} \cos \mathrm{t}}{1 \times 1}=0 \\ & \Rightarrow \theta=\frac{\pi}{2}\end{aligned}$
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