If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

A particle of mass $1 kg$ is subjected to a force which depends on the position as $\vec{F}=-k(x \hat{i}+y \hat{j}) kgms ^{-2}$ with $k=1 kgs ^{-2}$. At time $t=0$, the particle's position $\vec{r}=\left(\frac{1}{\sqrt{2}} \hat{i}+\sqrt{2} \hat{j}\right) m$ and its velocity $\vec{v}=\left(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+\frac{2}{\pi} \hat{k}\right) m s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5 m$, the value of $\left(x v_y-y v_x\right)$ is. . . . . $m^2 s^{-1}$

A particle of mass $m$ is moving along the side of a square of side '$a$', with a uniform speed $v$ in the $x-y$ plane as shown in the figure

Which of the following statement is false for the angular momentum $\vec L$ about the origin ?

Why $\vec v \times \vec p = 0$ for rotating particle ? 

A solid cylinder of mass $2\ kg$ and radius $0.2\,m$ is rotating about its own axis without friction with angular velocity $3\,rad/s$. A particle of mass $0.5\ kg$ and moving with a velocity $5\ m/s$ strikes the cylinder and sticks to it as shown in figure. The angular momentum of the cylinder before collision will be ........ $J-s$

$A$ uniform disc is rolling on a horizontal surface. At a certain instant $B$ is the point of contact and $A$ is at height $2R$ from ground, where $R$ is radius of disc.