$A$ neutron with velocity $v$ and kinetic energy $E$ collides head-on with a stationary nucleus of mass number $A$. What is the fraction of kinetic energy lost by the neutron?

Two balls of mass $5 \ kg$ are moving in opposite directions with the same speed of $5 \ m/s$. They undergo a head-on elastic collision. Find the final velocity of the balls in $m/s$.

$A$ neutron moving with velocity $u$ collides elastically with an atom of mass number $A$. If the collision is head-on and the initial kinetic energy of the neutron is $E$, then the final kinetic energy of the neutron after the collision is:

Statement-$1$: In an elastic collision between two bodies,the relative speed of the bodies after the collision is equal to the relative speed of the bodies before the collision.
Statement-$2$: In an elastic collision,the linear momentum of the system is conserved.

Consider an elastic collision of a particle of mass $m$ moving with a velocity $u$ with another particle of the same mass at rest. After the collision, the projectile and the struck particle move in directions making angles $\theta_1$ and $\theta_2$ respectively with the initial direction of motion. The sum of the angles, $\theta_1 + \theta_2$, is equal to: (in $^\circ$)

$A$ billiard ball of mass $M$,moving with velocity $v_1$ collides with another ball of the same mass but at rest. If the collision is elastic,the angle of divergence after the collision is (in $^{\circ}$)