$A$ sphere of mass $m$ travelling at constant speed $v$ strikes another sphere of the same mass. If the coefficient of restitution is $e$,then the ratio of the velocities of both spheres just after the collision is:

$300 \, J$ of work is done in sliding up a $2 \, kg$ block on an inclined plane to a height of $10 \, m$. Taking the value of acceleration due to gravity $g$ to be $10 \, m/s^2$,the work done against friction is ........ $J$.

$A$ ball is falling freely from a height. When it reaches $10 ~m$ height from the ground, its velocity is $v_0$. It collides with the ground, loses $50 \%$ of its energy, and rises back to a height of $10 ~m$. Then the velocity $v_0$ is (in $~m / s$)

Consider two blocks $A$ and $B$ of masses $m_1 = 10 \ kg$ and $m_2 = 5 \ kg$ that are placed on a frictionless table. The block $A$ moves with a constant speed $v = 3 \ m/s$ towards the block $B$ kept at rest. $A$ spring with spring constant $k = 3000 \ N/m$ is attached to the block $B$ as shown in the figure. After the collision,suppose that the blocks $A$ and $B$,along with the spring in a constant compression state,move together. Then the compression in the spring is (neglect the mass of the spring). (in $m$)

Work done by the frictional force is

$A$ chain of length $L = 2 \ m$ is placed on a smooth table such that half of its length hangs over the edge. If the chain is released,find the velocity of the chain when it just leaves the table (in $m/s$). (Take $g = 10 \ m/s^2$)