$A$ cricket ball hits the ground with a speed of $5 \,m \,s^{-1}$ making an angle of $30^{\circ}$ with the horizontal. If the coefficient of restitution is $0.2$, the speed of the reflected ball in $m \,s^{-1}$ is

$A$ 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 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

$A$ 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 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 $e \leq 1$ ).

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