Question

A particle is moving in a circular path. The acceleration and momentum of the particle of a certain moment are
$\overrightarrow{a}=(4\hat{i}+3\hat{j})m/s^2$ and $\overrightarrow{p}=(8\hat{i}-6\hat{j})kg-m{s}^{-1}$.
The motion of the particle is:

• A. uniform circular motion
• B. acceleration circular motion
• C. deaccelerated circular motion
• D. we cannot say anything with $\overrightarrow{a}$ and $\overrightarrow{p}$

Answer 1

The correct option is B acceleration circular motion

Given,

$\overrightarrow{a}=(4\hat{i}+3\hat{j})m/s^2$

$\overrightarrow{p}=(8\hat{i}-6\hat{j})kgm/s$

The dot product $\overrightarrow{a}$ and $\overrightarrow{p}$ is,

$\overrightarrow{a}.\overrightarrow{p}=(4\hat{i}+3\hat{j}).(8\hat{i}-6\hat{j})$

$\overrightarrow{a}.\overrightarrow{p}=4\times 8-3\times (-6)$

$\overrightarrow{a}.\overrightarrow{p}=24-18=6$

$\overrightarrow{a}$ is not perpendicular to $\overrightarrow{p}$.

The motion of the particle is not uniform circular motion.

Dot product is positive so angle between $\overrightarrow{a}$ & $\overrightarrow{p}$ is less than ${90}^0$ So motion will be acceleration circular motion.