Two balls of equal mass undergo head-on collision while each was moving with speed $6 \, m/s$ in opposite directions. If the coefficient of restitution is $e = 1/3$,the speed of each ball after impact will be ............ $m/s$.

$A$ ball $P$ collides with another identical ball $Q$ at rest. For what value of the coefficient of restitution $e$ will the velocity of ball $Q$ become two times that of ball $P$ after the collision?

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).

$A$ ball is dropped from a height of $5 \ m$ on a planet. It rebounds to a height of $1.8 \ m$. What is the fraction of velocity lost by the ball?

$A$ body falls freely onto a hard horizontal surface. If the coefficient of restitution between the surface and the body is $e = 0.8$,then the ratio of the maximum height to which the body rises after the second impact to the initial height of the body is: