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 $\left(\overrightarrow{P}\right)$ 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

• A. $\theta =0^{\circ }$
• B. $\theta ={30}^{\circ }$
• C. $\theta ={90}^{\circ }$
• D. $\theta ={180}^{\circ }$

Answer 1

The correct option is C $\theta ={90}^{\circ }$

Given that,

$P_x=2\cos t$

$P_y=2\sin t$

Now,

$P=P_{x}i+P_{y}j$

$P=2\mathrm{cost}i+2\mathrm{sint}j$

Now, according to the newton’s second law

$F_x=-2\sin t$

$F_y=2\cos t$

Now,

$F=F_{x}i+F_{y}j$

$F=-2\sin ti+2\cos tj$

Now, the dot product of F and P

$FP=\mathrm{0...}\left(I\right)$

But we know that,

$F.P=FP\cos \theta .....\left(II\right)$

Now, comparing both equations

$\theta ={90}^0$

Hence, the angle is ${90}^0$