$A$ wooden block of mass $m$ moves with velocity $v$ and collides with another block of mass $4m$,which is at rest. After the collision,the block of mass $m$ comes to rest. The coefficient of restitution will be:

$A$ body falls freely onto a hard horizontal surface. If the coefficient of restitution between the surface and the body is $e = 0.8$,then the ratio of the maximum height to which the body rises after the second impact to the initial height of the body is:

Particle $A$ moving with a velocity $v = 10 \ m/s$ experiences a head-on collision with a stationary particle $B$ of the same mass. As a result of the collision,the kinetic energy of the system decreases by $1 \%$. The speed of particle $A$ after the collision is: (in $m/s$)

$A$ 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 of mass $m$ moving with velocity $v$ collides head-on with a second ball of mass $m$ at rest. If the coefficient of restitution is $e$,the velocity of the first ball after collision is $v_1$,and the velocity of the second ball after collision is $v_2$,then:

$A$ 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