$A$ ball $P$ moving with a speed of $v \ ms^{-1}$ collides directly with another identical ball $Q$ moving with a speed $10 \ ms^{-1}$ in the opposite direction. $P$ comes to rest after the collision. If the coefficient of restitution is $0.6$,the value of $v$ is (in $ms^{-1}$)

An electron collides with a free molecule initially in its ground state. The collision leaves the molecule in an excited state that is metastable and does not decay to the ground state by radiation. Let $K$ be the sum of the initial kinetic energies of the electron and the molecule,and $p$ be the sum of their initial momenta. Let $K^{\prime}$ and $p^{\prime}$ represent the same physical quantities after the collision. Then,

$A$ sphere of mass $m$ moving with velocity $v$ collides head-on with another sphere of the same mass which is at rest. The ratio of the final velocity of the second sphere to the initial velocity of the first sphere is (where $e$ is the coefficient of restitution and the collision is inelastic).

$A$ sphere strikes a wall and rebounds with a coefficient of restitution $e = 1/3$. If it rebounds with a velocity of $0.1 \ m/s$ at an angle of $60^{\circ}$ to the normal to the wall,the percentage loss of kinetic energy is:

$A$ ball falls freely from rest from a height of $6.25 \ m$ onto a hard horizontal surface. If the ball reaches a height of $81 \ cm$ after the second bounce from the surface,the coefficient of restitution is

$A$ body of mass $m$ moving with velocity $v$ collides with another body of mass $nm$ moving in the same direction with velocity $kv$. If the first body comes to rest after the collision,what is the velocity of the second body?