Question

A particle moves in the $x-y$ plane under the action of a force $\overrightarrow{F}$ such that the value of its linear momentum $\overrightarrow{p}$ at any instant is $\overrightarrow{p}=2(\cos t\hat{i}+\sin t\hat{j})$. The angle $\theta$ between $\overrightarrow{F}$ and $\overrightarrow{p}$ is

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=2(-\sin t\hat{i}+\cos t\hat{j})$
Dot product of $\overrightarrow{F}$ and $\overrightarrow{p}$ is
$\overrightarrow{F}.\overrightarrow{p}=4(-\sin t\hat{i}+\cos t\hat{j}).(\cos t\hat{i}+\sin t\hat{j})=0$
$\therefore \theta ={90}^{\circ }$