Conservation of linear momentum and velocity of a system (damper and spring in a series)

Question

This example is from a book on dynamics. Let us consider the system above formed by two blocks (each of mass $m$) connected by a linear damper and spring in a series. They slide without friction on a horizontal plane.

Let us assume the following initial conditions on position:

$x_1(0)=0,\quad x_2(0)=0,\quad x_3(0)=0$

and

$\dot x_1(0)=0,\quad \dot x_2(0)=v_0$

By using Newton's law of motion, two equations of motion are derived on $x_1$ and $x_2$:

$m x_1 = c(\dot x_3 - \dot x_1)$

and

$m x_2 = -k(x_2 - x_3)$

where $c$ is a damping coefficient of the damper and $k$ is the stiffness of the spring. Again, by Newton's law of motion, the overall motion of the system imposes that

$m \ddot x_1 + m\ddot x_2 = 0$

or equivalently

$c(\dot x_3 - \dot x_1) = k(x_2 - x_3)$.

This last equation allows to determine the initial condition $\dot x_3(0)$ which is therefore also equal to zero.

Now, the book states in a very simple manner that the system "center of mass moves at a constant velocity $\frac{1}{2}v_0$ due to conservation of linear momentum".

Judging from the way this has been stated, I suppose it is pretty obvious then. Only, I fail to see why. I do not see how come we can determine the velocity of this system only from the information that has been provided so far. Can someone explain this matter more clearly?

Answer 1

We know the initial conditions: $\dot x_1=0$ and $\dot x_2=v_0$. The mass of each object is $m$. The center of mass is therefore moving at $\frac{mv_0}{m+m} = \frac{v_0}{2}$.

There is no external force on the system (no friction between the two masses and the surface below); so the center of mass continues to move at the same velocity.