$A$ body of mass $m_1$ moving with a velocity $3 \, ms^{-1}$ collides with another body at rest of mass $m_2$. After the collision,the velocities of the two bodies are $2 \, ms^{-1}$ and $5 \, ms^{-1}$ respectively along the direction of motion of $m_1$. The ratio $\frac{m_1}{m_2}$ is:

$A$ 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?

$A$ ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it rebounds with a velocity $v_n$ and the ball rises to a height $h_n$,then the coefficient of restitution is given by

$A$ 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 $90^o$.

$A$ small steel ball bounces on a steel plate held horizontally. On each bounce, the speed of the ball arriving at the plate is reduced by a factor $e$ (coefficient of restitution) in the rebound, so that $V_{\text{upward}} = e V_{\text{downward}}$. If the ball is initially dropped from a height of $0.4 \text{ m}$ above the plate and if $10 \text{ s}$ later the bouncing ceases, the value of $e$ 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: