$A$ ball is dropped from a height $h$ on a floor of coefficient of restitution $e$. The total distance covered by the ball just before the second hit is

$A$ ball moving with velocity $2 \,m/s$ collides head-on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$, then their velocities (in $m/s$) after collision will be:

$A$ moving block having mass $m$ collides with another stationary block having mass $4m$. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is $v$,then the value of the coefficient of restitution $(e)$ will be:

Two particles of masses $m_1$ and $m_2$ move with initial velocities $u_1$ and $u_2$. On collision,one of the particles gets excited to a higher level after absorbing energy $\varepsilon$. If the final velocities of the particles are $v_1$ and $v_2$,then we must have:

$A$ ball is dropped from a height $h$ onto a ground level. If the coefficient of restitution is $e$,what is the height reached by the ball after the $n^{th}$ bounce?

$A$ 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).