Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre),and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident. What is the loss in kinetic energy of the system in the process?

$A$ uniform rod of mass $M$ and length $L$ is suspended from a rigid support. $A$ small bullet of mass $m$ hits the rod with velocity $v$ and gets embedded into the rod. The angular velocity of the system just after impact is

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

$A$ cubical block of side $a$ is moving with velocity $v$ on a horizontal smooth plane as shown. It hits a ridge at point $O$. The angular speed of the block after it hits $O$ is

$A$ man is standing on the edge of a circular platform. The platform is rotating with a constant angular velocity about an axis passing through its center and perpendicular to its plane. If the man walks along the radius towards the axis, the angular velocity will ........ .

$A$ uniform rod of length $8 a$ and mass $6 m$ lies on a smooth horizontal surface. Two point masses $m$ and $2 m$ moving in the same plane with speed $2 v$ and $v$ respectively strike the rod perpendicularly at distances $a$ and $2 a$ from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is: