# What is the velocity of the center of mass of the system?

a(n) proton (mass 1.67x10^-27 kg) of kenetic energy 1.58x10^-12 J is moving toward a(n) proton (mass 1.67x10^-27) at rest. what is the velocity of the center of mass of the system

I used the equation k=1/2mv^2 -> sqrt(2k/m)=v
sqrt(2*1.58x10^-12 /1.67x10^-27)=4.34x10^7

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What would be the instantaneous position of the center of mass provided you know the positions of the two protons?

What is the relation between position (vector) and velocity?

would the position be zero.

The velocity of the center of mass is zero only in its own frame of reference. Stick to the frame of reference where the velocity of the proton is as given in the problem.

Okay, what formula gives the position of center of mass in this frame of reference?

Xcm=(x*y........)/total mass

Xcm=(x*y........)/total mass
Yes that is correct (I assume y is the mass of an object there).

Now if you differentiate that equation with respect to time, what would you get?

Which one is suppose to be time?

I think you got me wrong,

$X_{CM}=\frac{\Sigma m_ix_i}{\Sigma m_i}$ right?

Where $x_i$ and $m_i$ are the position vector and mass of the $i^{th}$ object.

Now differentiate wrt time i.e. find $\frac{dX_{CM}}{dt}$.

P.S. This helps you derive the relation between velocity of CM and velocity of the constituent particles.

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The derivative would be m/M

Let's go back to the basics.

First of all what does dx/dt represent?

Secondly when you differentiate kx (where k is a constant, and x is a position vector) with respect to time, what do you obtain?

dx/dt is the instantaneous rate of change of x with respect to t.

kx would be k

Doc Al
Mentor
dx/dt is the instantaneous rate of change of x with respect to t.
Right.

kx would be k
d(kx)/dx = k, but what about d(kx)/dt?

would it be -1/t^2

Doc Al
Mentor
would it be -1/t^2
Much simpler than all that. Just use the chain rule:
d(y)/dt = d(y)/dx * dx/dt

If kx were to change by a small amount into k(x+dx) in the time interval dt, what would be the rate of change of kx with respect to time?

I know how to take the derivative but im confused, is the d suppose to be coefficient or number where it can be ignored?

I know how to take the derivative but im confused, is the d suppose to be coefficient or number where it can be ignored?
No. d by itself doesn't have any meaning. dx means an infinitesimal change in the quantity x. It's like sinx. Sin by itself doesn't have any meaning.

so derivative is k/t?

HallsofIvy
Homework Helper
a(n) proton (mass 1.67x10^-27 kg) of kenetic energy 1.58x10^-12 J is moving toward a(n) proton (mass 1.67x10^-27) at rest. what is the velocity of the center of mass of the system

I used the equation k=1/2mv^2 -> sqrt(2k/m)=v
sqrt(2*1.58x10^-12 /1.67x10^-27)=4.34x10^7
I can see no reason to use the derivative at all. Since the two protons have the same mass, the center of mass will be exactly half way between them and so its "velocity" will be half the velocity of the moving proton.

I used the equation k=1/2mv^2 -> sqrt(2k/m)=v sqrt(2*1.58x10^-12 /1.67x10^-27)=4.34x10^7[/QUOTE said:
I did half of what I solved for and it is correct, Thank you!

Doc Al
Mentor
You would still be wise to understand the general argument that Sunil Simha was leading you towards. Then you can solve the general problem with arbitrary masses and speeds.

In words:

Start with the definition of center of mass: The mass-weighted average position of the masses.

Take the derivative with respect to time to find the velocity of the center of mass and realize that it equals the mass-weighted average velocity of the masses.

So for the derivative method, which derivative exactly should I take?

so derivative is k/t?
No. Well, it is like this:
You see, consider the situation when you are measuring a quantity, say x, which is the position vector of a particle. When you saw it at time t, It's position was x. Then you saw it after a time t + dt(note that d is not a coefficient or a constant as I had said earlier). The position had become x+dx. So, now the rate of change of x with respect to time is

$\frac{(x+dx)-x}{(t+dt)-t}$ which is $\frac{dx}{dt}$ which is also called velocity

I'm sure you knew the above stuff but it never hurts to go back to the basics.

Now take the case of kx. To make some sense, let's assume the particle moving up an inclined plane, which starts with 0 height at the origin and becomes taller as you move to the such that its height x units away form the origin is kx.( picturize this in your mind or see my attachment for the diagram). Right, now your objective is to find the vertical component of velocity.

Now, say at time t you saw the body at the X coordinate x. So its Y coordinate would be kx according to our earlier assumption.

At time t+dt, you see that its x coordinate is x+dx ok? So then its Y coordinate would be k(x+dx) right?

Now you define the vertical component of velocity as rate of change of Y coordinate with time.
And also differentiating y w.r.t. time.
So this would be $\frac{k(x+dx)-kx}{(t+dt)-t}$

that is $k\frac{dx}{dt}$ (note that dx and dt don't cancel each other)

Followed me so far? Now we come to the real business: The position vector of center of mass of your system (of total mass M and m1, m2 ...etc. are masses of the particles in the system) is defined as

$X_{CM}=\frac{m_1x_1 +m_2x_2 +....}{M}$

this can also be written as $X_{CM}=\frac{m_1x_1}{M} + \frac{m_2x_2}{M} ....$

Now the velocity of the center of mass is defined as rate of change of XCM with time.
So using the argument that I have previously used in my height example,as each of those $\frac{m_i}{M}x_1$ (because $\frac{m_i}{M}$ is a constant like k) change with respect to time as $\frac{m_i}{M}\frac{dx_i}{dt}$ okay?

So $\frac{dX_{CM}}{dt}= \Sigma \frac{m_i}{M}\frac{dx_i}{dt}$

This means that the velocity of the center of mass is the same as the sum of the products of the particle products and velocities divided by the total mass.

Just for an exercise to help you understand better, try getting the relation between the acceleration of the center of mass and the individual accelerations of the systems particles.

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