A particle moves in the X-Y plane under the i | vedclass

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(D) The force $\vec{F}(t)$ is the rate of change of linear momentum $\vec{p}(t)$.Given $\vec{p}(t) = A \cos(kt) \hat{i} - A \sin(kt) \hat{j}$.$\vec{F}

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$90$

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(D) The force $\vec{F}(t)$ is the rate of change of linear momentum $\vec{p}(t)$.Given $\vec{p}(t) = A \cos(kt) \hat{i} - A \sin(kt) \hat{j}$.$\vec{F}(t) = \frac{d\vec{p}}{dt} = \frac{d}{dt} [A \cos(kt) \hat{i} - A \sin(kt) \hat{j}] = -Ak \sin(kt) \hat{i} - Ak \cos(kt) \hat{j}$.To find the angle $	heta$ between $\vec{F}(t)$ and $\vec{p}(t)$,we calculate their dot product:$\vec{F}(t) \cdot \vec{p}(t) = (-Ak \sin(kt))(A \cos(kt)) + (-Ak \cos(kt))(-A \sin(kt))$$= -A^2 k \sin(kt) \cos(kt) + A^2 k \cos(kt) \sin(kt) = 0$.Since the dot product $\vec{F}(t) \cdot \vec{p}(t) = |\vec{F}(t)| |\vec{p}(t)| \cos 	heta = 0$ and the magnitudes are non-zero,we must have $\cos 	heta = 0$.Therefore,$	heta = 90^{\circ}$.

$A$ particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)=A[\hat{i} \cos (kt)-\hat{j} \sin (kt)]$,where $A$ and $k$ are constants. The angle between the force and the momentum is (in $^{\circ}$)

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