$A$ body of mass $4\ kg$ collides head-on elastically with another body of mass $2\ kg$ kept at rest in free space. The time of collision is $0.02\ s$ and the average impulsive force acting on each body is $100\ N$. Find the velocity of the $2\ kg$ body after the impact. (in $m/s$)

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:

$A$ particle of mass $m$,moving with a velocity $v$,makes an elastic collision in one dimension with a stationary particle of mass $m$. During the collision,they remain in contact with each other for an extremely small time $T$. Their force of contact with time is shown in the figure. Then $F_0 =$

What is a head-on collision?

$A$ body of mass $m$ moving with velocity $v$ collides with a stationary body of mass $2m$. The fraction of kinetic energy lost by the body of mass $m$ is:

$A$ sphere of mass $0.1 \ kg$ is attached to a string of length $1 \ m$. It is released from a horizontal position and collides with another sphere of equal mass $m$ at rest. Find the kinetic energy gained by the second sphere. The collision is perfectly elastic. (in $J$)