Step-by-step Derivation:

1. Relationship between Force and Momentum: According to Newton's second law of motion, the force $\overrightarrow F$ acting on a particle is the time derivative of its linear momentum $\overrightarrow p$. $\overrightarrow F = \frac{d\overrightarrow p}{dt}$

2. Given Momentum Vector: The problem provides the linear momentum vector as a function of time t: $\overrightarrow p(t) = A\left[ {\widehat i\cos (kt) - \widehat j\sin (kt)} \right]$
   Here, A and k are constants.

3. Calculate the Force Vector: We differentiate the momentum vector $\overrightarrow p(t)$ with respect to time t to find the force vector $\overrightarrow F(t)$.
   $\overrightarrow F(t) = \frac{d}{dt} \left( A\left[ {\widehat i\cos (kt) - \widehat j\sin (kt)} \right] \right)$
   Since A is a constant and the unit vectors $\widehat i$ and $\widehat j$ are also constant, we can write: $\overrightarrow F(t) = A \left[ \widehat i \frac{d}{dt}(\cos(kt)) - \widehat j \frac{d}{dt}(\sin(kt)) \right]$
   Using the chain rule for differentiation: $\frac{d}{dt}(\cos(kt)) = -k\sin(kt)$
   $\frac{d}{dt}(\sin(kt)) = k\cos(kt)$
   Substituting these results back into the expression for the force: $\overrightarrow F(t) = A \left[ \widehat i(-k\sin(kt)) - \widehat j(k\cos(kt)) \right]$
   $\overrightarrow F(t) = -Ak \left[ \widehat i\sin(kt) + \widehat j\cos(kt) \right]$

4. Find the Angle between Force and Momentum: The angle $θ$ between two vectors can be found using their dot product. The dot product of the force vector $\overrightarrow F$ and the momentum vector $\overrightarrow p$ is given by: $\overrightarrow F \cdot \overrightarrow p = |\overrightarrow F| |\overrightarrow p| \cos(\theta)$
   Let's calculate the dot product $\overrightarrow F(t) \cdot \overrightarrow p(t)$. The components are:

   • $p_x = A\cos(kt)$, $p_y = -A\sin(kt)$
   • $F_x = -Ak\sin(kt)$, $F_y = -Ak\cos(kt)$ The dot product is the sum of the products of corresponding components: $\overrightarrow F \cdot \overrightarrow p = (F_x)(p_x) + (F_y)(p_y)$
     $\overrightarrow F \cdot \overrightarrow p = (-Ak\sin(kt))(A\cos(kt)) + (-Ak\cos(kt))(-A\sin(kt))$
     $\overrightarrow F \cdot \overrightarrow p = -A^2k\sin(kt)\cos(kt) + A^2k\sin(kt)\cos(kt)$
     $\overrightarrow F \cdot \overrightarrow p = 0$

5. Conclusion: Since the dot product of the force and momentum vectors is zero, and neither vector is a zero vector (their magnitudes are $|\overrightarrow p| = A$ and $|\overrightarrow F| = Ak$, which are non-zero constants), the angle $θ$ between them must be such that $\cos(θ) = 0$. This implies that the angle $θ$ is $90^\circ$.

   Alternative Insight: The momentum vector corresponds to uniform circular motion (magnitude $A$ is constant). In uniform circular motion, the force (centripetal force) is always directed towards the center of the circle, while the momentum (and velocity) is always tangential to the circle. The angle between a radius and a tangent at any point on a circle is $90^\circ$. Therefore, the angle between the force and momentum is $90^\circ$.