Q 11-04-078JEE MainJEE Main 2024 (5 Apr, Shift 2)Medium
A particle moves in the $x$-$y$ 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
Answer: (C) $\dfrac{\pi}{2}$
$$\vec F = \frac{d\vec p}{dt} = -k\sin(kt)\,\hat i - k\cos(kt)\,\hat j$$
$$\vec F\cdot\vec p = -k\sin(kt)\cos(kt) + k\sin(kt)\cos(kt) = 0$$
So $\vec F$ is perpendicular to $\vec p$: the angle is $\dfrac{\pi}{2}$.
Solution by Sreeraj P, M.Sc Physics