Question

A particle moves in the $xy-plane$ under the action of a force $F$ such that the components of its linear momentum $p$ at any time $t$ are $p_x=2\cos t,p_y=2\sin t$. The angle between $F$ and $p$ at time $t$ is

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

Answer 1

The correct option is A ${90}^{\circ }$

Force is defined as the rate of change of momentum.

$P_x=2\cos t$

$P_y=2\sin t$

$\overrightarrow{P}=2\cos t\hat{i}+2\sin t\hat{j}$

$F=\frac{dP}{dt}$

$F_x=-2\sin t$

$F_y=2\cos t$

From law of vector dot product,

$\overrightarrow{A}.\overrightarrow{B}=AB\cos \theta$

Where $\theta$ is angle between them.

$\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{AB}$

Here,

$\cos \theta =\frac{\overrightarrow{F}.\overrightarrow{P}}{FP}$

$\cos \theta =\frac{-2\cos t\times 2\sin t+2\cos t+2\sin t}{FP}$

$\cos \theta =0$

$\theta =90°$