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

Step 1: State the given linear momentum.

The linear momentum of the particle as a function of time $t$ is given by:

$$\vec{p}(t) = A[\hat{i}\cos(kt) - \hat{j}\sin(kt)]$$

Step 2: Find the force vector.

By Newton's second law, force is the time derivative of linear momentum.

$$\vec{F} = \frac{d\vec{p}}{dt}$$

Differentiating each component with respect to $t$ yields:

$$\vec{F} = -Ak[\hat{i}\sin(kt) + \hat{j}\cos(kt)]$$

Step 3: Calculate the dot product of force and momentum.

Compute the components of the dot product $\vec{p} \cdot \vec{F}$.

$$p_x F_x = (A\cos(kt))(-Ak\sin(kt)) = -A^2 k \sin(kt)\cos(kt)$$ $$p_y F_y = (-A\sin(kt))(-Ak\cos(kt)) = A^2 k \sin(kt)\cos(kt)$$ $$\vec{p} \cdot \vec{F} = p_x F_x + p_y F_y$$ $$\vec{p} \cdot \vec{F} = -A^2 k \sin(kt)\cos(kt) + A^2 k \sin(kt)\cos(kt)$$ $$\vec{p} \cdot \vec{F} = 0$$

Step 4: Determine the angle between the force and momentum.

Use the definition of the dot product to find the angle $\theta$.

$$\cos\theta = \frac{\vec{p} \cdot \vec{F}}{|\vec{p}||\vec{F}|}$$ $$\cos\theta = 0$$ $$\theta = 90^\circ$$

Step 5: Verify using geometric interpretation.

Calculate the magnitude of the momentum vector.

$$|\vec{p}|^2 = A^2\cos^2(kt) + A^2\sin^2(kt)$$ $$|\vec{p}|^2 = A^2$$ $$|\vec{p}| = A$$

Since the magnitude of $\vec{p}$ is constant, its tip moves on a circle of radius $A$. The derivative vector $\vec{F} = \frac{d\vec{p}}{dt}$ is tangent to this circle, making it perpendicular to the radius vector $\vec{p}$.

$$\vec{F} \perp \vec{p}$$ $$\theta = 90^\circ$$

Answer: The angle between the force and the momentum is $90^\circ$.