Two masses $m_{a}$ and $m_{b}$ moving with velocities $v_{a}$ and $v_{b}$ in opposite directions collide elastically. After the collision,$m_{a}$ and $m_{b}$ move with velocities $v_{b}$ and $v_{a}$ respectively. The ratio $m_{a} : m_{b}$ is:

An object $A$ of mass $20 \ kg$ and travelling at $20 \ m \ s^{-1}$ crashes into another object $B$ of mass $200 \ kg$ and travelling at $10 \ m \ s^{-1}$, in the same direction. After the collision, object $A$ bounces back in the opposite direction at a speed of $10 \ m \ s^{-1}$. The speed of the object $B$ after the collision is: (in $m \ s^{-1}$)

$A$ block having mass $m$ collides with another stationary block having mass $2m$. The lighter block comes to rest after the collision. If the initial velocity of the first block is $v$,then the value of the coefficient of restitution must be:

$A$ body of mass $5\, kg$ moving with a velocity $10\, m/s$ collides with another body of mass $20\, kg$ at rest and comes to rest. The velocity of the second body due to the collision is ............ $m/s$.

$A$ particle of mass $m_1$ collides with a particle of mass $m_2$ at rest. After the elastic collision,the two particles move at an angle of $90^{\circ}$ with respect to each other. The ratio $\frac{m_2}{m_1}$ is

$A$ ball $A$ of mass $m$ moving along the positive $x$-direction with kinetic energy $K$ and momentum $p$ undergoes an elastic head-on collision with a stationary ball $B$ of mass $M$. After the collision,the ball $A$ moves along the negative $x$-direction with kinetic energy $K/9$. The final momentum of $B$ is: