$A$ 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?

$A$ $1.0 \ kg$ ball drops vertically onto a floor from a height of $25 \ cm$. It rebounds to a height of $4 \ cm$. The coefficient of restitution for the collision is

$A$ spherical ball $A$ of mass $4 \ kg$, moving along a straight line, strikes another spherical ball $B$ of mass $1 \ kg$ at rest. After the collision, $A$ and $B$ move with velocities $v_1 \ ms^{-1}$ and $v_2 \ ms^{-1}$ respectively, making angles of $30^{\circ}$ and $60^{\circ}$ with respect to the original direction of motion of $A$. The ratio $\frac{v_1}{v_2}$ will be

$A$ 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:

$A$ ball falls from a height $h$ and rebounds after striking the floor. The coefficient of restitution is $e$. The total distance covered by the ball before it comes to rest is

$A$ body of mass $m$ moving with velocity $v$ collides inelastically with another body of the same mass. After the collision,the first body moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. What will be the speed of the second body after the collision?