$A$ particle of mass $m = 5$ units is moving in the $XOY$ plane along the line $Y = X + 4$ with a uniform speed $v = 3\sqrt{2}$ units. The magnitude of its angular momentum about the origin is ...... units.

$A$ thin rod of mass $M$ and length $L$ is struck at one end by a ball of clay of mass $m$,moving with speed $v$ as shown in the figure. The ball sticks to the rod. After the collision,the angular momentum of the clay-rod system about $A$,the midpoint of the rod,is:

The position vector of a particle of mass $m$ moving with a constant velocity $\vec{v} = v \hat{i}$ is given by $\vec{r} = x(t) \hat{i} + b \hat{j}$, where $b$ is a constant. At an instant, $\vec{r}$ makes an angle $\theta$ with the $x$-axis. The variation of the magnitude of the angular momentum $|\vec{L}|$ of the particle about the origin with $\theta$ will be:

The position of a particle is given by $\vec r = (\hat i + 2\hat j - \hat k)$ and its momentum is $\vec p = (3\hat i + 4\hat j - 2\hat k)$. The angular momentum is perpendicular to:

$A$ 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:

$A$ particle of mass $m$ moves in the $XY$ plane with a velocity $v$ along the straight line $AB.$ If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B,$ then