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(A) The angular momentum $L$ of a particle about a point $O$ is given by the cross product $L = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Th

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remains constant

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(A) The angular momentum $L$ of a particle about a point $O$ is given by the cross product $L = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.The magnitude of angular momentum is $L = mvr \sin(	heta)$,where $r$ is the position vector of the particle from the origin $O$,$v$ is the velocity,and $	heta$ is the angle between $\vec{r}$ and $\vec{v}$.This can also be written as $L = mv d$,where $d = r \sin(	heta)$ is the perpendicular distance from the origin $O$ to the line of motion of the particle.Since the particle is moving along a straight line,the perpendicular distance $d$ from the origin $O$ to the line of motion remains constant throughout the motion.Since the mass $m$ and the speed $v$ are also constant (uniform motion),the product $mvd$ remains constant.Therefore,the angular momentum of the particle about $O$ remains constant during the motion from $A$ to $B$.

$A$ particle is moving uniformly along a straight line as shown in the figure. During the motion of the particle from $A$ to $B$,the angular momentum of the particle about '$O$' is:

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