$A$ particle of mass $m$ moves with a velocity $v$ along the line $PC$ as shown in the figure. What is the angular momentum of the particle about $O$?

$A$ particle of mass $M$ moves with a constant velocity parallel to the $X$-axis. What happens to its angular momentum with respect to the origin?

$A$ mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is $L$. If the length of the string is now halved, keeping the angular velocity the same, then the angular momentum will be:

$A$ body of mass $m$ is thrown with velocity $u$ from the origin of a coordinate system at an angle $\theta$ with the horizontal. The magnitude of the angular momentum of the particle about the origin at the time $t$ when it is at the maximum height of the trajectory is proportional to

The position vector of a $1\,kg$ object is $\overrightarrow{r} = (3\hat{i} - \hat{j})\,m$ and its velocity is $\overrightarrow{v} = (3\hat{j} + \hat{k})\,m/s$. The magnitude of its angular momentum is $\sqrt{x}\,N\cdot m\cdot s$,where $x$ is:

Find the components along the $x, y, z$ axes of the angular momentum $\vec{l}$ of a particle whose position vector is $\vec{r}$ with components $x, y, z$ and momentum is $\vec{p}$ with components $p_x, p_y, p_z$. Show that if the particle moves only in the $x-y$ plane,the angular momentum has only a $z$-component.