Two identical spheres move in opposite directions with speeds $v_1$ and $v_2$ and pass behind an opaque screen,where they may either cross without touching (Event $1$) or make an elastic head-on collision (Event $2$).

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$. Three identical spheres of same mass undergo one-dimensional motion as shown in the figure with initial velocities $v_{A} = 5 \ m/s$,$v_{B} = 2 \ m/s$,$v_{C} = 4 \ m/s$. If we wait sufficiently long for elastic collisions to happen,then $v_{A} = 4 \ m/s$,$v_{B} = 2 \ m/s$,$v_{C} = 5 \ m/s$ will be the final velocities.
Reason $(R)$: In an elastic collision between identical masses,two objects exchange their velocities. In the light of the above statements,choose the correct answer from the options given below:

Collision takes place between two solid spheres denoted as $1$ and $2$. The initial velocities of the spheres are $u_1 = 3 \ m/s$ and $u_2 = 1.5 \ m/s$,and the final velocities are $v_1 = 2.5 \ m/s$ and $v_2 = 3.5 \ m/s$. The coefficient of restitution between the materials of the spheres is nearly:

$A$ body of mass $M$ moves with velocity $v$ and collides elastically with another body of mass $m$ $(M >> m)$ at rest. Then,the velocity of the body of mass $m$ is:

$A$ body of mass $8 \ kg$ moving with a certain velocity collides elastically with another body of mass $2 \ kg$ at rest. If $E$ is the initial kinetic energy of the moving mass,what will be the remaining kinetic energy of the moving mass after the collision (in $E$)?

$A$ perfectly rigid billiard ball of kinetic energy $E_k$ collides with an identical stationary ball. After the collision,the kinetic energy of the first ball becomes $E'_k$. Then,which of the following is true?