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(D) The angular momentum $\vec{L}$ of a particle with respect to the origin $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{

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(D) The angular momentum $\vec{L}$ of a particle with respect to the origin $O$ is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.As shown in the figure,the particle moves parallel to the $X$-axis at a constant perpendicular distance $a$ from the origin $O$.The magnitude of the angular momentum is $|L| = m v a \sin(	heta)$,where $	heta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.Since $a = r \sin(	heta)$,we have $|L| = m v a$.Since $m$,$v$,and $a$ are all constants,the magnitude of the angular momentum remains constant.Furthermore,the direction of $\vec{L}$ (given by the right-hand rule) is perpendicular to the $X-Y$ plane (along the $Z$-axis) and does not change as the particle moves.Therefore,the angular momentum is constant.

$A$ particle of mass $m$ moves with a constant velocity $\vec{v}$ in the $X-Y$ plane. Its angular momentum with respect to the origin is:

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