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 $
Show Answer
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 $
