$A$ 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:

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).

$A$ moving body with mass $m_1$ strikes a stationary mass $m_2$. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of the first body by $1.5$ times after the collision?

$A$ body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2m$,which is at rest. The collision is head-on and elastic in nature. After the collision,the fraction of energy lost by the colliding body $A$ is

Consider the following statements $A$ and $B$ and identify the correct answer:
$A$. In an elastic collision,if a body suffers a head-on collision with another of the same mass at rest,then the first body comes to rest while the other starts moving with the velocity of the first one.
$B$. Two bodies of equal mass suffering a head-on elastic collision merely exchange their velocities.

Five identical elastic balls are suspended in a row with strings of equal length such that the distance between the sides of the balls is very small. If the ball at the right end is released from one side,then: