$A$ ball $P$ moving with a speed of $v \ ms^{-1}$ collides directly with another identical ball $Q$ moving with a speed $10 \ ms^{-1}$ in the opposite direction. $P$ comes to rest after the collision. If the coefficient of restitution is $0.6$,the value of $v$ is (in $ms^{-1}$)

$A$ 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?

$A$ smooth sphere is moving on a horizontal surface with a velocity vector $(2 \hat{i} + 2 \hat{j}) \ m/s$ immediately before it hits a vertical wall. The wall is parallel to the vector $\hat{j}$ and the coefficient of restitution between the sphere and the wall is $e = 1/2$. The velocity of the sphere after it hits the wall is:

$A$ rubber ball is released from a height of $5\, m$ above the floor. It bounces back repeatedly,always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take $g = 10\, m/s^2$)

$A$ ball of mass $1.2 \ kg$ moving with a velocity of $12 \ ms^{-1}$ makes a one-dimensional collision with another stationary ball of mass $1.2 \ kg$. If the coefficient of restitution is $\frac{1}{\sqrt{2}}$,then the ratio of the total kinetic energy of the balls after the collision to the initial kinetic energy is:

$A$ $2 \,kg$ ball moving at $24 \,ms^{-1}$ undergoes inelastic head-on collision with a $4 \,kg$ ball moving in the opposite direction at $48 \,ms^{-1}$. If the coefficient of restitution is $2/3$, their velocities in $ms^{-1}$ after impact are