$A$ ball falls freely from a height $h$ on a rigid horizontal plane. If the coefficient of restitution is $e$,then the total distance travelled by the ball before hitting the plane the second time is

$A$ ball of mass $m$,moving with a speed $2v_0$,collides inelastically $(e > 0)$ with an identical ball at rest. Show that
$(a)$ For head-on collision,both the balls move forward.
$(b)$ For a general collision,the angle between the two velocities of scattered balls is less than $90^o$.

$A$ body of mass $2 \,kg$ collides head-on with another body of mass $4 \,kg$. If the relative velocities of the bodies before and after collision are $10 \,ms^{-1}$ and $4 \,ms^{-1}$ respectively, the loss of kinetic energy of the system due to the collision is (in $\,J$)

$A$ 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 $\,ms^{-1}$)? $\left[g=10 \,ms^{-2}\right]$

$A$ 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

$A$ ball of mass $M$ falls from a height $h$ on a floor. If the coefficient of restitution is $e$,the height attained by the ball after two rebounds is