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 (in degree) between $F$ and $p$ at time $t$ is

Answer 1

Formula Used: $\cos \theta =\frac{\overrightarrow{F}.\overrightarrow{P}}{|\overrightarrow{F}||\overrightarrow{P}|}$
Given that $\overrightarrow{p}=p_x\hat{i}+p_y\hat{j}$
$=2\cos t\hat{i}+2\sin t\hat{j}$
$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=-2\sin t\hat{i}+2\cos t\hat{j}$
Now the angle between the force and the momentum is calculated as
$\cos \theta =\frac{\overrightarrow{F}.\overrightarrow{P}}{|\overrightarrow{F}||\overrightarrow{P}|}$
$\cos \theta$
$=\frac{[2\cos t\hat{i}+2\sin t\hat{j}].[-2\sin t\hat{i}+2\cos t\hat{j}]}{|\overrightarrow{F}||\overrightarrow{P}|}$

Now, $\overrightarrow{F}.\overrightarrow{P}=0$ i.e. angle between $\overrightarrow{F}$ and $\overrightarrow{p}$ is ${90}^{\circ }$.
Final Answer: $90$