$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

$A$ uniform thin rod of length $2L$ and mass $m$ lies on a horizontal table. $A$ horizontal impulse $J$ is given to the rod at one end. There is no friction. The total kinetic energy of the rod just after the impulse will be

Two equal masses each of mass $M$ are joined by a massless rod of length $L$. Now an impulse $J = MV$ is given to one of the masses,making an angle of $30^{\circ}$ with the length of the rod. The angular velocity of the rod just after imparting the impulse is:

$A$ 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: $V=-\frac{m}{M} \sqrt{2 g R}$.

$A$ rod of length $1\,m$ is standing vertically. When its upper end is released and it falls such that the lower end touches the ground without slipping,the speed of the upper end when it hits the ground is:

$A$ uniform rod of mass $M$ is hinged at its upper end. $A$ particle of mass $m$ moving horizontally strikes the rod at its mid-point elastically. If the particle comes to rest after the collision,find the value of $M/m$.