$A$ ball $P$ collides with another identical ball $Q$ at rest. For what value of the coefficient of restitution $e$ will the velocity of ball $Q$ become two times that of ball $P$ after the collision?

$A$ 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

$A$ moving ball collides head-on with a similar ball at rest. If $\frac{1}{4}$ of the kinetic energy is lost during the impact,the value of the coefficient of restitution is:

$A$ $1.0 \ kg$ 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

$A$ smooth sphere moves on a horizontal surface with a velocity vector $2\hat{i} + 2\hat{j}$ before it strikes a vertical wall. The wall is parallel to the $\hat{j}$ vector,and the coefficient of restitution between the sphere and the wall is $e = 1/2$. What will be the velocity vector of the sphere after it strikes the wall?

$A$ $1 \ kg$ ball moving with a speed of $6 \ ms^{-1}$ collides head-on with a $0.5 \ kg$ ball moving in the opposite direction with a speed of $9 \ ms^{-1}$. If the coefficient of restitution is $\frac{1}{3}$,then the energy lost in the collision is (in $J$)