JEE Main 2024 — Laws Of Motion Question with Solution

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:

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

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 $$t$$:

$$\frac{d\overrightarrow{\mathrm{p}}}{dt}=-k\hat{i}\sin (kt)-k\hat{j}\cos (kt)$$

So, $$\vec{F}=-k\hat{i}\sin (kt)-k\hat{j}\cos (kt)$$.

Now, to find the angle between $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$, we use the dot product formula:

$$\vec{F}\cdot \overrightarrow{\mathrm{p}}=|\vec{F}||\overrightarrow{\mathrm{p}}|\cos (\theta )$$,

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.

The dot product of $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$ is:

$$(-k\hat{i}\sin (kt)-k\hat{j}\cos (kt))\cdot (\hat{i}\cos (kt)-\hat{j}\sin (kt))=$$

$$-k\sin (kt)\cos (kt)+k\cos (kt)\sin (kt)=0$$

The result is zero, indicating that the angle between $$\vec{F}$$ and $$\overrightarrow{\mathrm{p}}$$ is indeed $$\frac{\pi }{2}$$.

Therefore, the correct option is:

Option A $$\frac{\pi }{2}$$.