# Equation for velocity of center of mass

What is the equation to find the velocity of the center of mass?

since...
$$x_{CofM}=\frac{m_1x_1+m_2x_2}{M_{total}}$$
then...
$$v_{CofM}=\frac{d(m_1x_1+m_2x_2)}{dt(M_{total})}$$
this means...
if p=momentum
$$v_{CofM}=\frac{p}{M_{total}}$$

is this correct?

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dextercioby
Homework Helper
Though i didn't understand very well what your formulas meant (i must be having a bad day :tongue2: ),i can tell you that it's the other way around:
VCM (velocity of the center of mass) results immediately by computing the total (linear) momentum in 2 ways...

Daniel.

I'm not quite sure I undersatnd what you are saying...what 2 ways?

let me explain.

$$x_{CofM}=\frac{m_1x_1+m_2x_2}{M_{total}}$$
is the equation to calcualte the distance of center of mass. I took the derivative of it to find the velocity. Therefore, my second equation was the velocity of the center of mass. My third equation went further to say that d(m1v1+m2v2)/dt was really momentum. So... VCM is really momentum over the total mass

dextercioby
Homework Helper
I know what u did.It's not wrong at all.I've just given u an alternative approach and i think much more intuitive.

Daniel.

does the momentum of the center of mass tell the total linear momentum of the system?

Or do I have to caculate the momentum of each object and then add them together?

dextercioby
Homework Helper
Of course.The total linear momentum of the system is the linear momentum if the CM.

Daniel.

in an elastice collision, the VCM is the same before and after to collisions right?

dextercioby
Homework Helper
Yes,and that's due to total linear momentum conservation.

Daniel.

dextercioby said:
Of course.The total linear momentum of the system is the linear momentum if the CM.

Daniel.
I have the velocity of the CM to be (3.00i-0.8j)m/s
would the total linear momentum be...
$$v=\sqrt{3^2+.8^2}$$
$$v=3.1m/s$$
$$p=(m_1+m_2)v$$
$$p=(3kg+2kg)3.1m/s$$
$$15.5Ns$$

Would 15.5Ns be the total linear momentum?

dextercioby