$A$ body falling from a height of $10\,m$ rebounds from a hard floor. If it loses $20\%$ of its energy in the impact,then the coefficient of restitution is:

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,

$A$ ball moving with a velocity $v$ collides head-on with a stationary second ball of the same mass. After the collision, the velocity of the first ball is reduced to $0.15 v$. The kinetic energy of the system is decreased nearly by (in $\%$)

$A$ large ball of mass $M$,moving with velocity $u$,strikes a small ball of mass $m$,which is at rest. After the collision,the small ball attains velocity $u$ and the large ball attains velocity $v$. What is the value of $v$?

$A$ ball of mass $m$,moving with a speed $2v_0$,collides inelastically $(e > 0)$ with an identical ball at rest. Show that
$(a)$ For head-on collision,both the balls move forward.
$(b)$ For a general collision,the angle between the two velocities of scattered balls is less than $90^o$.

$A$ body of mass $2 \,kg$ collides head-on with another body of mass $4 \,kg$. If the relative velocities of the bodies before and after collision are $10 \,ms^{-1}$ and $4 \,ms^{-1}$ respectively, the loss of kinetic energy of the system due to the collision is (in $\,J$)