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

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(C) Given the linear momentum vector $\overrightarrow{P} = P_x \hat{i} + P_y \hat{j} = (2\cos t) \hat{i} + (2\sin t) \hat{j}$.The force $\overrightarr

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$90$

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(C) Given the linear momentum vector $\overrightarrow{P} = P_x \hat{i} + P_y \hat{j} = (2\cos t) \hat{i} + (2\sin t) \hat{j}$.The force $\overrightarrow{F}$ is the rate of change of linear momentum: $\overrightarrow{F} = \frac{d\overrightarrow{P}}{dt}$.Calculating the derivative: $\overrightarrow{F} = \frac{d}{dt}(2\cos t) \hat{i} + \frac{d}{dt}(2\sin t) \hat{j} = (-2\sin t) \hat{i} + (2\cos t) \hat{j}$.To find the angle $	heta$ between $\overrightarrow{F}$ and $\overrightarrow{P}$,we use the dot product formula: $\overrightarrow{F} \cdot \overrightarrow{P} = |\overrightarrow{F}| |\overrightarrow{P}| \cos 	heta$.Calculating the dot product: $\overrightarrow{F} \cdot \overrightarrow{P} = (-2\sin t)(2\cos t) + (2\cos t)(2\sin t) = -4\sin t \cos t + 4\sin t \cos t = 0$.Since the dot product is $0$,the vectors are perpendicular,which means $\cos 	heta = 0$,so $	heta = 90^o$.

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