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I'm curious as to whether or not the velocity of the center of mass of a system comprised of two spheres can change after the two spheres collide. Looking at the equation for the velocity of the center of mass for a system of particles:

$$V_\text{CM} = \frac{m_1v_1 + m_2v_2 + \cdots + m_nv_n}{m_1 + m_2 + \cdots + m_n}$$

It looks like, if after the collision, one of the particles changes direction, and the negative terms outweigh the positive terms in the numerator, the velocity of the center of mass would change direction (and possibly magnitude). However, I can't think of any examples where this would happen in elastic or inelastic collisions. I'm not even sure if its speed can change. It would make sense that it would if kinetic energy (velocity) was lost in an inelastic collision, however, I can't come up with any conditions to make this happen.

I could really use some insight.

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    $\begingroup$ Conservation of momentum ($mv$) is more fundamental in physics than conservation of energy. The only way the velocity of the center of mass could change is if there is an external force acting on the system. If all forces are internal to the system, then the momentum cannot change. $\endgroup$
    – Greg
    Commented Nov 6, 2013 at 0:24
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    $\begingroup$ @Greg Why not write an answer? $\endgroup$
    – joshphysics
    Commented Nov 6, 2013 at 0:43
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    $\begingroup$ @Greg please, give an answer the rest of us can comment on $\endgroup$
    – Larry Harson
    Commented Nov 6, 2013 at 1:08

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Why do we even use centre of mass? In other words, why do we define it the way it is defined, and what use is it? Well, the centre of mass of a system is a point that behaves as if though all the mass of the system and all of momentum of the object were concentrated at that point. For the two momenta to be equal, we require

$p_{com} = p_{system}$

or

$(\sum m_i) \dot x_{com} = \sum_i m_i \dot x_i$

which leads directly to the familiar definition of the c.o.m (up to a constant). Note this is just a simple rearrangement of your equation. From this you can see that if momentum is conserved on the R.H.S., it must be conserved on the L.H.S. (because they are identical, by construction).

As @Greg mentioned, momentum is always conserved, although I think he meant to say that it is more fundamental than conservation of kinetic energy, rather than just energy.

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  • $\begingroup$ This was very helpful; I finally understand what's going on. Thank you! $\endgroup$
    – user32297
    Commented Nov 6, 2013 at 4:23

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