$A$ sphere of mass $m$ moving with velocity $v$ collides head-on with another sphere of the same mass which is at rest. The ratio of the final velocity of the second sphere to the initial velocity of the first sphere is (where $e$ is the coefficient of restitution and the collision is inelastic).

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:

$A$ 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 $1.2 \ kg$ moving with a velocity of $12 \ ms^{-1}$ makes a one-dimensional collision with another stationary ball of mass $1.2 \ kg$. If the coefficient of restitution is $\frac{1}{\sqrt{2}}$,then the ratio of the total kinetic energy of the balls after the collision to the initial kinetic energy is:

$A$ body falls from a height of $10 \ m$ onto the ground and rebounds to a height of $2.5 \ m$. The percentage loss in kinetic energy 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?