$A$ ball falls vertically onto a floor with momentum $p$ and then bounces repeatedly. If the coefficient of restitution is $e$, then the total momentum imparted by the ball on the floor is:

$A$ particle of mass $m_1$ moving along the $X$-axis collides with a stationary particle of mass $m_2$ and deviates by an angle $30^{\circ}$ to the $X$-axis as shown in the figure. If the percentage change in kinetic energy of the combined system of these two particles reduces by $50 \%$,then the ratio of the masses $\frac{m_2}{m_1}$ is

$A$ particle of mass $m$ moving with speed $2v$ collides with a mass $2m$ moving with speed $v$ in the same direction. After the collision,the first mass is stopped completely,while the second one splits into two particles each of mass $m$,which move at an angle of $45^o$ with respect to the original direction. The speed of each of the moving particles will be:

$A$ ball of mass $2 \ g$ moving with a velocity of $2 \ ms^{-1}$ collides with another ball of mass $8 \ g$ which is at rest and comes to rest after collision. Then the coefficient of restitution is

$A$ particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution,the total distance travelled before rebounding has stopped is

An object of mass $m$ moving with velocity $u$ collides with another stationary object of mass $M$ and stops just after the collision. The coefficient of restitution is