Two blocks connected by a damper and spring in a series

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:

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


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

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

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


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

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?

  • $\begingroup$ While this seems perilously close to a homework-type problem, I don't think it really qualifies. So, I'll instead ask what is the initial momentum of the system, i.e. at t=0? At any time t>0, what is the momentum of the system? What then is the momentum of the center of mass? $\endgroup$
    – Jon Custer
    Dec 21, 2015 at 14:44
  • $\begingroup$ @JonCuster It would be the system mass times the velocity of the system. Although, thanks to the reminder by Floris, the velocity of the center of mass is provided by simply calculating the time derivative of the center of the mass location given by $v=\frac{1}{2m} (m \dot x_1(0) + m \dot x_2(0)). $\endgroup$
    – Meclassic
    Dec 21, 2015 at 14:46
  • $\begingroup$ And the mass of the system is? Well, @Floris has shown you... Contemplate what will happen if the two masses are not the same value. $\endgroup$
    – Jon Custer
    Dec 21, 2015 at 14:47

1 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.


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