As shown below, bob $A$ of a pendulum having a massless string of length $R$ is released from $60^{\circ}$ to the vertical. It hits another bob $B$ of half the mass that is at rest on a frictionless table in the center. Assuming an elastic collision, the magnitude of the velocity of bob $A$ after the collision will be (take $g$ as acceleration due to gravity):

$A$ small block of mass $200 \ g$ is placed on a horizontal slab at a height of $2 \ m$ above the floor. The block is pressed against a horizontal spring fixed at one end to compress the spring by $10.0 \ cm$. Upon releasing, the block moves horizontally until it leaves the spring. Calculate the horizontal distance covered by the block after leaving the slab and just before hitting the ground. The spring constant is $50 \ N \ m^{-1}$. (Assume $g = 10 \ m \ s^{-2}$) (in $m$)

$A$ bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$,the initial velocity of the bullet will be

$A$ spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 \ N \ m^{-1}$ and the mass of the block is $2.0 \ kg$. Ignore the mass of the spring. Initially,the spring is in an unstretched condition. Another block of mass $1.0 \ kg$ moving with a speed of $2.0 \ m \ s^{-1}$ collides elastically with the first block. The collision is such that the $2.0 \ kg$ block does not hit the wall. The distance,in metres,between the two blocks when the spring returns to its unstretched position for the first time after the collision is . . . . .

Define elastic collision and inelastic collision.

Three particles $A, B$,and $C$ of equal mass move with speed $V$ as shown to strike at the centroid of an equilateral triangle. After the collision,$A$ comes to rest and $B$ retraces its path with speed $V$. The speed of $C$ after the collision is: