PhysicsHard68×since 2002Q7604A particle moves in xxx-yyy plane under the influence of a force F⃗\vec{F}F such that its linear momentum is p→(t)=i^cos(kt)−j^sin(kt)\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})p(t)=i^cos(kt)−j^sin(kt). If k\mathrm{k}k is constant, the angle between F→\overrightarrow{\mathrm{F}}F and p→\overrightarrow{\mathrm{p}}p will be :Aπ2\frac{\pi}{2}2πBπ3\frac{\pi}{3}3πCπ4\frac{\pi}{4}4πDπ6\frac{\pi}{6}6πCheck answerSkip