A particle moves in $$x - y$$ 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 constant, the angle between $$\vec{F}$$ and $$\vec{p}$$ will be :

We are given the momentum vector at time t as $$\vec{p}(t)=\hat{i}\cos(kt)-\hat{j}\sin(kt)$$, and we wish to find the angle between the force $$\vec{F}$$ and the momentum $$\vec{p}$$.

Since Newton’s second law states that the force is the time derivative of the momentum, differentiating $$\vec{p}(t)$$ with respect to t yields $$\vec{F}=\frac{d\vec{p}}{dt}=-k\hat{i}\sin(kt)-k\hat{j}\cos(kt)$$.

Next we compute the dot product of $$\vec{F}$$ and $$\vec{p}$$, which gives

$$\vec{F}\cdot\vec{p}=(-k\sin kt)(\cos kt)+(-k\cos kt)(-\sin kt)=-k\sin kt\cos kt+k\cos kt\sin kt=0$$.

Because the dot product is zero, the vectors are perpendicular and hence the angle between them is $$\frac{\pi}{2}$$.

Therefore, the correct answer is Option (3): $$\frac{\pi}{2}$$.