Question

A particle moves in the $x-y$ plane under the influence of a force such that its linear momentum is $\overrightarrow{p}\left(t\right)=A\left[\cos \right(kt)\hat{i}-\sin (kt\left)\hat{j}\right]$, where $A$ and $k$ are constants. The angle between the force and the momentum is :

• A. $0^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

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

Given,
$\overrightarrow{p}\left(t\right)=A\cos \left(kt\right)\hat{i}-A\sin \left(kt\right)\hat{j}$ .....(1)
As we know, $\overrightarrow{F}\left(t\right)=\frac{d\overrightarrow{p}}{dt}$
$\overrightarrow{F}\left(t\right)=\frac{d\overrightarrow{p}}{dt}$
$=-Ak\sin \left(kt\right)\hat{i}-Ak\cos \left(kt\right)\hat{j}$ ....(2)

We know, $\overrightarrow{A}.\overrightarrow{B}=|A||B|\cos \theta$
[$\theta$ is angle between $\overrightarrow{A}$ and $\overrightarrow{B}$]
Then, $\overrightarrow{F}\left(t\right).\overrightarrow{p}\left(t\right)=|\overrightarrow{F}\left(t\right)||\overrightarrow{p}\left(t\right)|\cos \theta$
Here,
$|\overrightarrow{F}\left(t\right)|=\sqrt{(-Ak\sin (kt){)}^2+(-Ak\cos \left(kt\right){)}^2}=Ak$
$|\overrightarrow{p}\left(t\right)|=\sqrt{\left(A\cos \right(kt){)}^2+(-A\sin \left(kt\right){)}^2}=A$
$\Rightarrow \overrightarrow{F}\left(t\right).\overrightarrow{p}\left(t\right)=\left(Ak\right)\left(A\right)\cos \theta ...\left(3\right)$
But we also know,
$\overrightarrow{F}\left(t\right).\overrightarrow{p}\left(t\right)=[-Ak\sin (kt)\hat{i}-Ak\cos (kt\left)\hat{j}\right].\left[A\cos \right(kt)\hat{i}-A\sin (kt\left)\hat{j}\right]=0...\left(4\right)$

From (3) and (4),
$0=(Ak$)($A$)$\cos \theta$
$\Rightarrow \cos \theta =0$
$\therefore \theta ={90}^{\circ }$
Angle between the force and momentum is ${90}^{\circ }$.