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(A) Given linear momentum $\vec{P}(t) = \cos(kt) \hat{i} - \sin(kt) \hat{j}$.Force $\vec{F}$ is the rate of change of momentum: $\vec{F} = \frac{d\vec

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$\frac{\pi}{2}$

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(A) Given linear momentum $\vec{P}(t) = \cos(kt) \hat{i} - \sin(kt) \hat{j}$.Force $\vec{F}$ is the rate of change of momentum: $\vec{F} = \frac{d\vec{P}}{dt}$.$\vec{F} = \frac{d}{dt} [\cos(kt) \hat{i} - \sin(kt) \hat{j}] = -k \sin(kt) \hat{i} - k \cos(kt) \hat{j}$.To find the angle $	heta$ between $\vec{F}$ and $\vec{P}$,we use the dot product: $\vec{F} \cdot \vec{P} = |\vec{F}| |\vec{P}| \cos 	heta$.$\vec{F} \cdot \vec{P} = (-k \sin(kt))(\cos(kt)) + (-k \cos(kt))(-\sin(kt))$.$\vec{F} \cdot \vec{P} = -k \sin(kt) \cos(kt) + k \sin(kt) \cos(kt) = 0$.Since the dot product is $0$,$\cos 	heta = 0$,which implies $	heta = \frac{\pi}{2}$.

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

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