Ball $A$ of mass $50 \ gm$ and speed $10 \ m/s$ collides with another ball $B$ of mass $10 \ gm$ and speed $15 \ m/s$ travelling in opposite directions. Determine the final speed of ball $B$,if the coefficient of restitution is $\frac{2}{5}$.

$A$ ball of mass $M$ falls from a height $h$ on a floor. If the coefficient of restitution is $e$,the height attained by the ball after two rebounds is

$A$ ball falls freely from rest from a height of $6.25 \ m$ onto a hard horizontal surface. If the ball reaches a height of $81 \ cm$ after the second bounce from the surface,the coefficient of restitution is

$A$ ball kept at $20 \,m$ height falls freely in a vertically downward direction and hits the ground. The coefficient of restitution is $0.4$. What is the velocity of the ball after the first rebound (in $\,ms^{-1}$)? $\left[g=10 \,ms^{-2}\right]$

$A$ spherical ball $A$ of mass $4 \ kg$, moving along a straight line, strikes another spherical ball $B$ of mass $1 \ kg$ at rest. After the collision, $A$ and $B$ move with velocities $v_1 \ ms^{-1}$ and $v_2 \ ms^{-1}$ respectively, making angles of $30^{\circ}$ and $60^{\circ}$ with respect to the original direction of motion of $A$. The ratio $\frac{v_1}{v_2}$ will be

$A$ 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: