A particle is moving in a circular path. The | vedclass

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(D) The momentum vector $\vec{p} = m\vec{v}$ is in the direction of the velocity vector $\vec{v}$.To determine the nature of circular motion,we need t

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we cannot say anything about $a$ and $p$ only.

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(D) The momentum vector $\vec{p} = m\vec{v}$ is in the direction of the velocity vector $\vec{v}$.To determine the nature of circular motion,we need to check the angle between the velocity vector $\vec{v}$ (or $\vec{p}$) and the acceleration vector $\vec{a}$.The dot product $\vec{a} \cdot \vec{p} = (2 \hat{i} + 3 \hat{j}) \cdot (6 \hat{i} + 4 \hat{j}) = (2 	imes 6) + (3 	imes 4) = 12 + 12 = 24 \ kg \cdot m^2/s^3$.Since the dot product is positive,the angle between $\vec{a}$ and $\vec{v}$ is acute,which implies the particle is speeding up (tangential acceleration).However,the nature of motion (uniform or non-uniform) depends on whether the tangential acceleration is zero or non-zero over an interval of time. Given only instantaneous values,we cannot conclude the long-term nature of the motion.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

$A$ particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are $\vec{a} = 2 \hat{i} + 3 \hat{j} \ m/s^2$ and $\vec{p} = 6 \hat{i} + 4 \hat{j} \ kg \cdot m/s$. Then the motion of the particle is

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