Laws Of Motion Question 431
Question: A particle moves in the $ x-y $ plane under the action of a force $ \overrightarrow{F} $ such that the value of its linear momentum $ \overrightarrow{P} $ at any time $ t $ is $ P _{x}=2\cos t, $
$ P _{y}=2\sin t $ . The angle $ \theta $ between $ \overrightarrow{F} $ and $ \overrightarrow{P} $ at a given time t will be
Options:
A) $ 90{}^\circ $
B) $ 0{}^\circ $
C) $ 180{}^\circ $
D) $ 30{}^\circ $
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Answer:
Correct Answer: A
Solution:
[a] $ P=\sqrt{P _{x}^{2}+P _{y}^{2}}=\sqrt{{{(2\cos t)}^{2}}+{{(2\sin t)}^{2}}}=2 $
If m is the mass of the body, then kinetic energy $ =\frac{P^{2}}{2m}=\frac{{{(2)}^{2}}}{2m}=\frac{2}{m} $
Since kinetic energy does not change with time, both work done and power are zero. Now, power $ =Fv\cos \theta =0 $ As $ F\ne 0,v\ne 0 $
$ \therefore $ $ \cos \theta =0 $ or$ \theta =90{}^\circ $
As direction of $ \vec{p} $ is same that of $ \vec{v}, $
$ (\because \overrightarrow{P}=m\overrightarrow{v}), $ hence angle between $ \overrightarrow{F} $
and $ \overrightarrow{P} $ is equal to $ 90{}^\circ $ .
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