$A$ moving block having mass $m$ collides with another stationary block having mass $4m$. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is $v$,then the value of the coefficient of restitution $(e)$ 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:

$A$ mass $m$ moves with a velocity $v$ and collides inelastically with another identical mass at rest. After the collision,the $1^{st}$ mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{nd}$ mass after the collision.

$A$ ball is dropped from some height and after the first collision with the ground,if it reaches $\frac{3}{4}$ of its original height,then the percentage loss of its energy is:

An electron collides with a free molecule initially in its ground state. The collision leaves the molecule in an excited state that is metastable and does not decay to the ground state by radiation. Let $K$ be the sum of the initial kinetic energies of the electron and the molecule,and $p$ be the sum of their initial momenta. Let $K^{\prime}$ and $p^{\prime}$ represent the same physical quantities after the collision. Then,

$A$ 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?