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

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Vector

A particle moves in

x - y

plane under the influence of a force

F

such that its linear momentum is

P (t) = \hat{i} cos (kt) - \hat{j} sin (kt)

. If k is constant, the angle between

F

and

P

will be :

Answer: (A)

P = cos (k t) \hat{i} - sin (k t) \hat{j}

has constant magnitude 1, so the speed is constant — the net force is purely centripetal, i.e. perpendicular to the velocity (and hence to

P

, since

P ∥ v

).
Explicitly,

a = \frac{d v}{d t} = - k sin (k t) \hat{i} - k cos (k t) \hat{j}

, and

F \cdot P \propto [- sin (k t) cos (k t) + (- cos (k t)) (- sin (k t))] = 0

F ⊥ P

and the angle between them is

\frac{π}{2}

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option