Question

A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are $\overrightarrow{a}=2\overrightarrow{i}+3\overrightarrow{j}m/s^2$ and $\overrightarrow{P}=6\overrightarrow{i}-4\overrightarrow{j}$ kgm/s. Then the motion of the particle is:

• A. uniform circular motion
• B. circular motion with tangential acceleration
• C. circular motion with tangential retardation
• D. we cannot say anything from and only

Answer 1

Answer: (A)

uniform circular motion

We consider the dot-product of $\overrightarrow{a}$ and $\overrightarrow{P}$

$\overrightarrow{a}.\overrightarrow{P}=(2\hat{i}+3\hat{j})(6\hat{i}-4\hat{j})=2\times 6-3\times 4=12-12=0$

$\therefore$ $\overrightarrow{a}$ is perpendicular to $\overrightarrow{P}$

$\therefore$ motion is uniform circular.