Question

The components of the linear momentum of a body moving in the X-Y plane are given by $P_x=Asin\left(t\right)$ and $P_y=Acos\left(t\right)$, where ${}^{\prime }{A}^{\prime }$ is a constant and ${}^{\prime }{t}^{\prime }$ is time. The angle between the linear momentum and the force acting on the body is

• A. ${45}^0$
• B. $0^0$
• C. ${180}^0$
• D. ${90}^0$

Answer 1

The correct option is D ${90}^0$

$\overrightarrow{P}={\overrightarrow{P}}_x\hat{i}+{\overrightarrow{P}}_y\hat{j}=Acos\left(t\right)\hat{i}+Asin\left(t\right)\hat{j}$
$\overrightarrow{F}=\frac{\partial \overrightarrow{P}}{\partial t}=-Asin\left(t\right)\hat{i}+Acos\left(t\right)\hat{j}$
Cosine of angle between $\overrightarrow{F}$ and $\overrightarrow{P}=\frac{\overrightarrow{F}.\overrightarrow{P}}{\left|\overrightarrow{F}\right|\left|\overrightarrow{P}\right|}$
$\overrightarrow{F}.\overrightarrow{P}=0$
$\therefore cos\theta =0$
$\therefore \theta ={90}^0$