Question

A particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are $\overrightarrow{a}=(4\hat{i}+3\hat{j})m/s^2$ and $\overrightarrow{p}=(8\hat{i}-6\hat{j})kgm/s$. The motion of the particle is

• A. Uniform circular motion
• B. Accelerated circular motion
• C. Decelerated circular motion
• D. We cannot say anything with $\overrightarrow{a}$ and $\overrightarrow{p}$ only.

Answer 1

The correct option is B Accelerated circular motion

Let the angle between $\overrightarrow{a}$ and $\overrightarrow{p}$ be $\theta$


$\overrightarrow{a}.\overrightarrow{p}=|\overrightarrow{a}||\overrightarrow{p}|\cos \theta$
$\Rightarrow \theta ={\cos }^{-1}\frac{\overrightarrow{a}.\overrightarrow{p}}{|\overrightarrow{a}||\overrightarrow{p}|}$
$={\cos }^{-1}\left\{\frac{32-18}{\sqrt{(16+9)}\sqrt{64+36}}\right\}$
$={\cos }^{-1}\left(\frac{14}{50}\right)$
$\Rightarrow \theta ={73.73}^{\circ }$
Since $0^{\circ }<\theta <{90}^{\circ }$, the motion is an accelerated one.