Question

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

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

Answer 1

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

$\text{Step1: Calculating force}$

Linear momentum is given as:

$\overrightarrow{P}=A\cos \left(kt\right)\hat{i}-A\sin \left(kt\right)\hat{j}$

As per Newton's second law force is given by : $\overrightarrow{F}=\frac{d\overrightarrow{P}}{dt}$

$\Rightarrow \overrightarrow{F}=-Ak\left(\sin kt\right)\hat{i}-Ak\left(\cos kt\right)\hat{j}$

$\text{Step2: Calculating Angle between}\overrightarrow{P}\&\overrightarrow{F}$

Angle is gives by : $\cos \theta =\frac{\overrightarrow{P}.\overrightarrow{F}}{|\overrightarrow{P}||\overrightarrow{F}|}$

$\Rightarrow \cos \theta =\frac{\left(A\right(\cos tk)\hat{i}-A(\sin kt\left)\hat{j}\right)(-Ak\sin kt\overrightarrow{i}-Ak\cos kt)\hat{j}}{|\overrightarrow{P}||\overrightarrow{F}|}$

$\Rightarrow \cos \theta =\frac{(-A^{2}k\cos kt\sin kt)+\left(A^{2}k\sin kt\cos kt\right)}{|\overrightarrow{P}||\overrightarrow{F}|}$ $=0$

$\Rightarrow \theta ={90}^0$

Hence option $D$ is correct.