$A$ body falls on a surface with a coefficient of restitution of $0.6$ from a height of $1 \, m$. The body rebounds to a height of ........... $m$.

Ball $P$ of mass $m$ moving with velocity $v$ collides with another ball $Q$ of mass $2m$, which is at rest. If $v_P$ and $v_Q$ are the final velocities of $P$ and $Q$ respectively after the collision, then (Assume the coefficient of restitution is $e = 1/3$):

$A$ ball of mass $10 \text{ g}$ is allowed to fall from a height of $10 \text{ m}$. After the collision with the ground, if $50\%$ of its energy is lost, then the height reached by the ball is: (in $\text{ m}$)

$A$ 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:

$A$ tennis ball is released from height $h$ above ground level. If the ball makes an inelastic collision with the ground,to what height will it rise after the third collision?

$A$ ball is dropped freely from a height $h$. It bounces repeatedly on the ground. Find the velocity of the ball after $n$ bounces.