Consider the following two statements:
$1.$ Linear momentum of a system of particles is zero.
$2.$ Kinetic energy of a system of particles is zero.
Then:

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicularly and comes to rest. The velocity of the centre of mass of the rod after the collision is:

Separation of motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
$(a)$ Show $p = \sum p_{i}^{\prime} + M V$,where $p$ is the total momentum of the system,$p_{i}^{\prime} = m_{i} v_{i}^{\prime}$,and $v_{i}^{\prime}$ is the velocity of the $i^{th}$ particle relative to the centre of mass. Also,prove using the definition of the centre of mass that $\sum p_{i}^{\prime} = 0$.
$(b)$ Show $K = K^{\prime} + \frac{1}{2} M V^{2}$,where $K$ is the total kinetic energy of the system,$K^{\prime}$ is the total kinetic energy of the system relative to the centre of mass,and $\frac{1}{2} M V^{2}$ is the kinetic energy of the translation of the system as a whole.
$(c)$ Show $L = L^{\prime} + R \times M V$,where $L^{\prime} = \sum r_{i}^{\prime} \times p_{i}^{\prime}$ is the angular momentum of the system about the centre of mass. Note $r_{i}^{\prime} = r_{i} - R$.
$(d)$ Show $\frac{d L^{\prime}}{d t} = \sum r_{i}^{\prime} \times \frac{d p_{i}^{\prime}}{d t}$. Further,show that $\frac{d L^{\prime}}{d t} = \tau_{ext}^{\prime}$,where $\tau_{ext}^{\prime}$ is the sum of all external torques acting on the system about the centre of mass.

Consider the following two statements for a system of particles:
$A$: The linear momentum of the system is zero.
$B$: The kinetic energy of the system is zero.

What is the total momentum of a system of particles?

$A$ uniform rod of length $L$ and mass $M$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicularly and comes to rest. The velocity of the centre of mass of the rod after the collision is: