$A$ particle of mass $m$ with an initial velocity $u\hat{i}$ collides perfectly elastically with a mass $3m$ at rest. It moves with a velocity $v\hat{j}$ after collision. Then, $v$ is given by:

$A$ smooth sphere $A$ of mass $m$ collides elastically with an identical sphere $B$ at rest. The velocity of $A$ before collision is $8 \ m/s$ in a direction making $60^{\circ}$ with the line of centers at the time of impact.
$(i)$ The sphere $A$ comes to rest after collision.
$(ii)$ The sphere $B$ will move with a speed of $8 \ m/s$ after collision.
$(iii)$ The directions of motion of $A$ and $B$ after collision are at right angles.
$(iv)$ The speed of $B$ after collision is $4 \ m/s$.
The correct option is:

The bob $A$ of a pendulum,released from a horizontal position to the vertical,hits another bob $B$ of the same mass at rest on a table as shown in the figure. If the length of the pendulum is $1\,m$,calculate:
$(a)$ The height to which bob $A$ will rise after the collision.
$(b)$ The speed with which bob $B$ starts moving.
Neglect the size of the bobs and assume the collision to be elastic.

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicularly and comes to rest. The velocity of the centre of mass of the rod after the collision is:

As shown in the figure,a pendulum $A$ is released from an angle of $30^{\circ}$ with the vertical and strikes a pendulum $B$ of equal mass. After the collision,to what height (in $m$) will pendulum $A$ rise? Neglect the size of the pendulum and assume the collision is perfectly elastic.

There are $n$ elastic balls placed on a smooth horizontal plane. The masses of the balls are $m, \frac{m}{2}, \frac{m}{2^2}, \ldots, \frac{m}{2^{n-1}}$ respectively. If the first ball hits the second ball with velocity $v_0$, then the velocity of the $n^{\text{th}}$ ball will be,