Two masses $m_1$ and $m_2$ moving with velocities $v_1$ and $v_2$ in opposite directions collide elastically and after collision masses $m_1$ and $m_2$ move with velocity $v_2$ and $v_1$ respectively. The ratio $\left(\frac{m_2}{m_1}\right)$ is

$A$ 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?

$A$ body of mass $m$ moving with velocity $v$ makes a head-on elastic collision with another body of mass $2m$ which is initially at rest. What is the loss in kinetic energy of the colliding body?

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

$A$ sphere of mass $0.1\,kg$ is attached to a cord of $1\,m$ length. Starting from the height of its point of suspension,this sphere hits a block of the same mass at rest on a frictionless table. If the impact is elastic,then the kinetic energy of the block after the collision is ............. $J$.

$A$ 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