Angle Between Force and Momentum for Given Time Dependent Momentum

To find the angle between the force $\vec{F}$ acting on the particle and its linear momentum $\vec{p}(t)$, we use Newton's second law of motion:
$\vec{F} = \frac{d\vec{p}}{dt}$

Given the linear momentum as a function of time:
$\vec{p}(t) = A \left[ \hat{i} \cos(kt) - \hat{j} \sin(kt) \right]$

Differentiating $\vec{p}(t)$ with respect to time $t$:
$\vec{F} = \frac{d}{dt} \left\{ A \left[ \hat{i} \cos(kt) - \hat{j} \sin(kt) \right] \right\}$
$\vec{F} = A \left[ -k \sin(kt)\hat{i} - k \cos(kt)\hat{j} \right] = -Ak \left[ \hat{i} \sin(kt) + \hat{j} \cos(kt) \right]$

Now, let us calculate the scalar (dot) product of $\vec{F}$ and $\vec{p}$:
$\vec{F} \cdot \vec{p} = \left( -Ak \sin(kt)\hat{i} - Ak \cos(kt)\hat{j} \right) \cdot \left( A \cos(kt)\hat{i} - A \sin(kt)\hat{j} \right)$
$\vec{F} \cdot \vec{p} = (-Ak \sin(kt))(A \cos(kt)) + (-Ak \cos(kt))(-A \sin(kt))$
$\vec{F} \cdot \vec{p} = -A^2 k \sin(kt) \cos(kt) + A^2 k \sin(kt) \cos(kt) = 0$

The magnitudes of $\vec{p}$ and $\vec{F}$ are:
$|\vec{p}| = A \sqrt{\cos^2(kt) + (-\sin(kt))^2} = A$
$|\vec{F}| = Ak \sqrt{(-\sin(kt))^2 + (-\cos(kt))^2} = Ak$

Since both magnitudes are non-zero, the angle $\theta$ between $\vec{F}$ and $\vec{p}$ is given by:
$\cos\theta = \frac{\vec{F} \cdot \vec{p}}{|\vec{F}| |\vec{p}|} = \frac{0}{A \cdot Ak} = 0$

Therefore,
$\theta = 90^\circ$

The correct option is (D).