Mathematics in Physics

JEE Main level, single correct, medium

42 of 121

Question 42 of 121

A particle moves in x−yx-y plane under the influence of a force F⃗\vec{F} such that its linear momentum is p→(t)=i^cos⁡(kt)−j^sin⁡(kt)\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt}). If k\mathrm{k} is constant, the angle between F→\overrightarrow{\mathrm{F}} and p→\overrightarrow{\mathrm{p}} will be :

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Answer: (C) π2\frac{\pi}{2}

P→=cos⁡(kt)i^−sin⁡(kt)j^;∣P→∣=1∵P→=mv→∴P^=v^⇒v^=cos⁡(kt)i^−sin⁡(kt)j^a^=−ksin⁡(kt)i^−kcos⁡(kt)j^k⇒a^=−sin⁡kti^−cos⁡ktj^∵F^=a^=−sin⁡kti^−cos⁡ktj^cos⁡θ=F^⋅P^∣F^∣∣P^∣=−sin⁡ktcos⁡t+sin⁡ktcos⁡t1×1=0⇒θ=π2\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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