A particle moves in the x-y plane under the a | vedclass

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(C) The linear momentum vector is given by $\vec P = P_x \hat i + P_y \hat j = (2 \cos t) \hat i + (2 \sin t) \hat j$.According to Newton's second law

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$	heta = 90^\circ$

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(C) The linear momentum vector is given by $\vec P = P_x \hat i + P_y \hat j = (2 \cos t) \hat i + (2 \sin t) \hat j$.According to Newton's second law,the force $\vec F$ is the time derivative of linear momentum: $\vec F = \frac{d\vec P}{dt}$.$\vec F = \frac{d}{dt} (2 \cos t \hat i + 2 \sin t \hat j) = -2 \sin t \hat i + 2 \cos t \hat j$.To find the angle $	heta$ between $\vec F$ and $\vec P$,we calculate their dot product: $\vec F \cdot \vec P = (-2 \sin t)(2 \cos t) + (2 \cos t)(2 \sin t) = -4 \sin t \cos t + 4 \sin t \cos t = 0$.Since the dot product $\vec F \cdot \vec P = |\vec F| |\vec P| \cos 	heta = 0$,it follows that $\cos 	heta = 0$,which means $	heta = 90^\circ$.

$A$ particle moves in the $x-y$ plane under the action of a force $\vec F$ such that the value of its linear momentum $\vec P$ at any time $t$ is $P_x = 2 \cos t$ and $P_y = 2 \sin t$. The angle between $\vec F$ and $\vec P$ at a given time $t$ will be:

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