Ball $A$ moving at $12 \ m/s$ collides elastically with ball $B$ at rest as shown in the figure. If both balls have the same mass,what is the final velocity of ball $A$ after the collision (in $m/s$)? (Assume the collision is two-dimensional and the angle of incidence is $60^\circ$ with the vertical axis).

$A$ body of mass $m$ having an initial velocity $v$,makes a head-on collision with a stationary body of mass $M$. After the collision,the body of mass $m$ comes to rest and only the body having mass $M$ moves. This will happen only when:

$A$ ball of mass $m$ moving with velocity $V$ makes a head-on elastic collision with a ball of the same mass moving with velocity $2V$ towards it. Taking the direction of $V$ as positive,the velocities of the two balls after the collision are:

$A$ ball of mass $m$ moving with speed $v$ collides elastically with an identical stationary ball which is initially at rest. After collision,the first ball moves at an angle $\theta$ to its initial direction and has speed $(v/3)$. The second ball moves in a straight line after the collision. Then,the speed of the second ball after the collision is:

$A$ neutron travelling with a velocity $v$ and kinetic energy $E$ collides perfectly elastically head-on with the nucleus of an atom of mass number $A$ at rest. The fraction of total energy retained by the neutron is

$A$ 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: