Kinematics Question 276
Question: A particle moves in the x-y plane under the action of a force $ \overrightarrow{F} $ such that the value of it’s linear momentum $ (\overrightarrow{P}) $ at anytime 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 [MNR 1991; UPSEAT 2000]
Options:
A) $ \theta =0{}^\circ $
B) $ \theta =30{}^\circ $
C) $ \theta =90{}^\circ $
D) $ \theta =180{}^\circ $
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Answer:
Correct Answer: C
Solution:
$ P _{x}=2\cos t $ , $ P _{y}=2\sin t $ $ \vec{P}=2\cos t\ \hat{i}+2\sin t\ \hat{j} $
$ \vec{F}=\frac{d\vec{P}}{dt}=-2\sin t\ \hat{i}+2\cos t\ \hat{j} $
$ \vec{F}.\vec{P}=0 $ $ \theta =90{}^\circ $
BETA
