Laws Of Motion Question 293

Question: A particle moves in the X-Y plane under the influence of force such that its linear momentum is$ \vec{p}(t)=-A[\hat{i}\cos (kt)-\hat{j}\sin (kt)] $ , where A and k are constants. The angle between the force and the momentum is

Options:

A) $ 0{}^\circ $

B) $ 30{}^\circ $

C) $ 45{}^\circ $

D) $ 90{}^\circ $

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Answer:

Correct Answer: D

Solution:

[d] $ \vec{p}( t )=A[ \hat{i}cos( kt )-\hat{j}\sin ( kt ) ] $

$ F=\frac{d\vec{p}}{dt}=Ak[-\hat{i}cos(kt)-\hat{j}\sin (kt)] $

Here, $ \vec{F}.\vec{P}=0 $ but $ \vec{F}.\vec{P}=Fp\cos \theta $

$ \therefore \cos \theta =0\Rightarrow \theta =90{}^\circ $



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