$A$ bar of mass $M=1.00 \ kg$ and length $L=0.20 \ m$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. $A$ small mass $m=0.10 \ kg$ is moving on the same horizontal surface with $5.00 \ m \ s^{-1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L/2$ from the pivoted end and returns back on the same path with speed $v$. After this elastic collision,the bar rotates with an angular velocity $\omega$. Which of the following statements is correct?

$A$ block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially,the right edge of the block is at $x=0$,in a coordinate system fixed to the table. $A$ point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block,its position is $x$ and the velocity is $v$. At that instant,which of the following options is/are correct?
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{m R}{M+m}$.
$[B]$ The position of the point mass is: $x=-\sqrt{2} \frac{mR}{M+m}$.
$[C]$ The velocity of the point mass $m$ is: $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
$[D]$ The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.

$A$ disc and a ring of the same mass are rolling. If their kinetic energies are equal,then the ratio of their velocities will be:

$A$ rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. $A$ particle of mass $m$ travelling along the surface hits at one end of the rod with velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After the collision, the particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of $x$ will be ..... .

$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speed $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after collision. The total rotational kinetic energy about the centre of mass $c$ will be:

The angular momentum of a rotating body is $L$. When the frequency of the rotating body is tripled and its kinetic energy is made one-third,the new angular momentum becomes: