$A$ ball is dropped from a height $h$ on a floor of coefficient of restitution $e$. The total distance covered by the ball just before the second hit is

Two identical balls $P$ and $Q$ moving in the $x-y$ plane collide at the origin $(x=0, y=0)$ of the coordinate system. Their velocity components just before the moment of impact were,for ball $P$,$v_x = 6 \ m/s, v_y = 0$; for ball $Q$,$v_x = -5 \ m/s, v_y = 2 \ m/s$. As a result of the collision,ball $P$ comes to rest. The velocity components of ball $Q$ just after the collision will be:

$A$ ball of mass $m$ moving with velocity $v$ collides head-on with a second ball of mass $m$ at rest. If the coefficient of restitution is $e$,the velocity of the first ball after collision is $v_1$,and the velocity of the second ball after collision is $v_2$,then:

$A$ ball is dropped from a height $h$. If the coefficient of restitution is $e$,to what height will it rise after jumping twice from the ground?

$A$ ball of mass $m$ is moving with a speed $V$ as shown in the figure. It undergoes an inelastic collision with a ball of mass $2m$ which was initially at rest. The velocity of the ball of mass $2m$ after the collision is given by

$A$ ball kept at $20 \,m$ height falls freely in a vertically downward direction and hits the ground. The coefficient of restitution is $0.4$. What is the velocity of the ball after the first rebound (in $\,ms^{-1}$)? $\left[g=10 \,ms^{-2}\right]$