Center-of-mass velocity

Also called Velocity of the center of mass, v_cm

The velocity of the point that moves as though all of a system's mass were concentrated there, equal to the system's total momentum divided by its total mass. It stays constant whenever no net external force acts on the system.

Where the center of mass sits is one question. How fast that point moves is a separate one, and it is the one conservation of momentum is really about.

All four AP equation sheets print it in two equivalent forms:

v⃗cm=∑p⃗i∑mi=∑miv⃗i∑mi\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}

Read the first form backwards and you get the sentence that matters: the total momentum of a system is its total mass times v⃗cm\vec{v}_{\text{cm}}. EK 4.3.A.1 says as much in words, that a collection of objects with individual momenta can be described as one system with one center-of-mass velocity.

Constant, under one condition. EK 4.3.A.1.ii: the velocity of a system's center of mass is constant in the absence of a net external force. Internal forces cannot shift it, however violent they are. Two carts colliding, a firework bursting, a skater pushing off a partner: in each case every individual velocity changes and v⃗cm\vec{v}_{\text{cm}} does not.

So momentum conservation and constant v⃗cm\vec{v}_{\text{cm}} are the same statement. If you have already written ∑p⃗before=∑p⃗after\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}, you have said it. Checking that the center-of-mass velocity came out unchanged is a free check on the arithmetic.

It is a vector, and it can be zero while nothing is at rest. Two equal masses approaching each other at equal speeds have v⃗cm=0\vec{v}_{\text{cm}} = 0, which is why a perfectly inelastic head-on collision between them leaves the pair motionless.

Setting up and solving those collisions is the job of the conservation of momentum guide.

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