Question

A particle moves in the $xy$ plane under action of a force $F$ such that the value of its linear momentum $\left(P\right)$ at any time $t$ is $P_x=2\cos t,P_y=2\sin t$. The angle $\Theta$ between $P$ and $F$ at that time $t$ will be :

• A. $0^o$
• B. ${30}^0$
• C. ${90}^o$
• D. ${180}^o$

Answer 1

The correct option is C ${90}^o$

$|\overrightarrow{P}|=\sqrt{P_x^2+P_y^2}=2$

Since $P_x$ and $P_y$ change with time and $|\overrightarrow{P}|$ is constant,

it is a case of uniform circular motion.

$v=\frac{P}{m}$

$v_y=\frac{2}{m}\sin t$

$v_x=\frac{2}{m}\cos t$

$a=\frac{dv}{dt}$

$a_x=-\frac{2}{m}\sin t$

$a_y=\frac{2}{m}\cos t$

$\overrightarrow{v}.\overrightarrow{a}=-\cos t\sin t+\sin t\cos t=0$

Hence angle between $\overrightarrow{a}and\overrightarrow{p}$ is $={90}^0$.