A particle of mass $m$ projected with a velocity ' $u$ ' making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $\mathrm{h}$ is :
$\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}$
$\frac{\sqrt{3}}{2} \frac{m u^2}{g}$
$\frac{m u^3}{\sqrt{2} g}$
zero
Two particles, each of mass $m$ and speed $v$, travel in opposite directions along parallel lines separated by a distance $d$. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
A particle of mass $2\, kg$ is moving such that at time $t$, its position, in meter, is given by $\overrightarrow r \left( t \right) = 5\hat i - 2{t^2}\hat j$ . The angular momentum of the particle at $t\, = 2\, s$ about the origin in $kg\, m^{-2}\, s^{-1}$ is
$A$ ball of mass $m$ moving with velocity $v$, collide with the wall elastically as shown in the figure.After impact the change in angular momentum about $P$ is:
A disc of mass $M$ and radius $R$ moves in the $x-y$ plane as shown in the figure. The angular momentum of the disc at the instant shown is
The position of a particle is given by : $\overrightarrow {r\,} = (\hat i + 2\hat j - \hat k)$ and momentum $\overrightarrow P = (3\hat i + 4\hat j - 2\hat k)$. The angular momentum is perpendicular to