$A$ $1.0 \ kg$ 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

$A$ sphere of mass $m$ moving with velocity $u$ collides with another sphere of the same mass $m$ at rest. If the coefficient of restitution is $e = \frac{1}{2}$, what is the ratio of their velocities after the collision?

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:

Ball $P$ of mass $m$ moving with velocity $v$ collides with another ball $Q$ of mass $2m$, which is at rest. If $v_P$ and $v_Q$ are the final velocities of $P$ and $Q$ respectively after the collision, then (Assume the coefficient of restitution is $e = 1/3$):

$A$ 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:

$A$ ball strikes a smooth horizontal ground at an angle of $45^o$ with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision? (Assume $e \leq 1$ ).