What is the total momentum of a system of particles?

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:

Two men $A$ and $B$ are standing on a plank of length $120 \ cm$ and mass $40 \ kg$. $B$ (mass $60 \ kg$) is at the middle of the plank and $A$ (mass $40 \ kg$) is at the left end of the plank. The system is initially at rest on a smooth horizontal surface. $A$ and $B$ start moving such that the position of $B$ remains fixed with respect to the ground until $A$ meets $B$. The point where $A$ meets $B$ is located at:

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.

Given below are two statements:
Statement $I$: For a mechanical system of many particles,the total kinetic energy is the sum of the kinetic energies of all the particles.
Statement $II$: The total kinetic energy can be expressed as the sum of the kinetic energy of the center of mass with respect to the origin and the kinetic energy of all the particles with respect to the center of mass as the reference frame.
In the light of the above statements,choose the correct answer from the options given below:

$A$ 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: