A particle moves in $x$-$y$ plane under the influence of a force $\vec{F}$ such that its linear momentum is $$\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})$$. If $\mathrm{k}$ is constant, the angle between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{p}}$ will be :

<p>To find the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$, we first need to understand the relationship between force and momentum. The force $\vec{F}$ acting on a particle is related to the rate of change of its linear momentum $\overrightarrow{\mathrm{p}}$ with respect to time, as described by Newton's second law of motion:</p> <p>$\vec{F} = \frac{d\overrightarrow{\mathrm{p}}}{dt}$</p> <p>Given the expression for the momentum $\overrightarrow{\mathrm{p}}(t) = \hat{i} \cos (kt) - \hat{j} \sin (kt)$, we can find $\vec{F}$ by differentiating $\overrightarrow{\mathrm{p}}$ with respect to $t$:</p> <p>$$\frac{d\overrightarrow{\mathrm{p}}}{dt} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$$</p> <p>So, $\vec{F} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$.</p> <p>Now, to find the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$, we use the dot product formula:</p> <p>$$\vec{F} \cdot \overrightarrow{\mathrm{p}} = |\vec{F}| |\overrightarrow{\mathrm{p}}| \cos(\theta)$$,</p> <p>where $\theta$ is the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$. However, in this case, it's more insightful to see if $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ are orthogonal (at a $\frac{\pi}{2}$ angle to each other), because the dot product of two perpendicular vectors is zero.</p> <p>The dot product of $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ is:</p> <p>$$(-k\hat{i} \sin (kt) - k\hat{j} \cos (kt)) \cdot (\hat{i} \cos (kt) - \hat{j} \sin (kt)) =$$</p> <p>$- k \sin (kt) \cos (kt) + k \cos (kt) \sin (kt) = 0$</p> <p>The result is zero, indicating that the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ is indeed $\frac{\pi}{2}$.</p> <p>Therefore, the correct option is:</p> <p>Option A $\frac{\pi}{2}$.</p>