A particle moves in x-y plane under the influence of a force F such that its linear momentum is p(t) = icos (kt) - jsin (kt). If k is constant, the angle between F and p will be: (a) π/2 (b) π/3 (c) π/4 (d) π/6

Answer: (a). F = dp/dt = -ksin (kt)i - kcos (kt)j F · p = -ksin ktcos kt + ksin ktcos kt = 0 So F ⊥ p. The answer is (a) π/2.

A particle moves in x-y plane under the influence of a force F such… | JEE Main 2024 (5 Apr, Shift 2)

Q12JEE Main 2024 · 5 APR · SHIFT 2Medium

Inertia, First & Second Laws

A particle moves in $x\text{-}y$ 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:

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Answer: (a)

$\vec{F} = \frac{d\vec{p}}{dt} = -k\sin(kt)\hat{i} - k\cos(kt)\hat{j}$
$\vec{F} \cdot \vec{p} = -k\sin kt\cos kt + k\sin kt\cos kt = 0$
So $\vec{F} \perp \vec{p}$ . The answer is (a) $\frac{\pi}{2}$ .

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Inertia, First & Second Laws question from JEE Main 2024 (5 Apr, Shift 2)

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