# Momentum in centre of mass reference frame

1. Dec 10, 2015

### REVIANNA

my book says " the total momentum is zero in the centre of mass reference frame.This should not surprise you"
but ITS NOT INTUITIVE FOR ME.
I am considering a completely elastic collision.

1. I know the v_cm is const because there are no ext forces on the system of the two masses undergoing 1D elastic collision

2. I know the momentum and energy is conserved.

3. I know how to find out the v_cm and relative velocities.

Help me get a feel for it

2. Dec 10, 2015

### Staff: Mentor

Hi REVIANNA. Please remember to use the formatting template in the homework areas.

You might be able to convince yourself of the statement if you work through a change of reference frame for two moving particles comprising a system. Then you can extrapolate to any number of particles and 3D motion if like.

Suppose that you have two particles of masses m1 and m2 and velocities v1 and v2 in some initial reference frame. You should be able to write expressions for the momentum and the velocity of the center of mass in that frame of reference.

What happens to the momentum if you subtract the center of mass velocity from both the particles (i.e. change to the center of mass frame of reference)?

3. Dec 10, 2015

### REVIANNA

what's a formatting template?

yes I will be able to find the $P_t=M*v_c$
$P_t=m_1*v_1+m_2*v_2$
$v_c=(P_t)/(m_1+m_2)$

if I find the velocity of 1 wrt to the COM and velocity of 2 wrt to the COM , multiply them with there respective masses-- the momentum is indeed equal and opposite i.e. $P_t=0$
I have worked out one such numerical.I understand the mathematics what I don't understand is why it turns out the way it does?

4. Dec 10, 2015

### Staff: Mentor

The formatting template is provided for your use in the edit window when you start a thread in the homework areas. You had to have deleted it when you started this thread.
I'm not sure how to provide a better argument than mathematical proof

How about considering it from another point of view? Suppose you are an observer at rest watching a bomb. The bomb is at rest along with you so has zero momentum from your point of view. When it explodes and fragments into multiple pieces going off in various directions, you'd expect that momentum would be conserved and that the total momentum of the system would remain zero, right? It's a perfectly inelastic collision done in reverse.

Now, another observer that happens to be riding along on one of the fragments will note that for him there's a net momentum for all the remaining pieces (due to him not counting the chunk he's riding on since its momentum is zero in his frame of reference). In fact, for every other frame of reference that is not the original center of mass / center of momentum frame there will be a net non-zero momentum. That net non-zero momentum can be transformed away by accounting for every chunk and moving the observer to the center of mass / center of momentum frame of reference.

5. Dec 10, 2015

### PeroK

To try to add to the good explanations above. First, think of the system free from any collisions. It's just a set of particles moving in all directions. And, perhaps start with two particles.

You could first define a "centre of momentum" frame. I.e. find a frame where the total momentum is 0. How could you do this? You could start by choosing one particle $m_1$ and looking at the motion of the other particle $m_2$ from this frame. The situation is simple: $m_2$ is moving with a velocity $v$ is some direction. The system's momentum from $m_1$'s frame is simply $m_2v$.

Now imagine a frame moving along this direction at speed $u$. The momentum in this frame is:

$m_2(v-u) - m_1u$

As $u$ increases, the momentum of $m_2$ will decrease and the momenum of $m_1$ will increase. So, we just need to pick the right value for $u$ to get momentum = 0.

It's not hard to calculate that $u= \frac{m_2v}{m_1+m_2}$ does the trick.

So, we've found a frame in which total momentum is 0.

Now, let's go back and look at the motion of the centre of mass in $m_1$'s frame. It was simply $\frac{m_2v}{m_1+m_2}$. That's exactly what we needed for $u$ to get momentum = 0.

Bingo! Momentum of the two-particle system is 0 in precisely the frame of reference that moves with the centre of mass.

Now imagine there is a third particle $m_3$. Let's choose the centre of mass frame of $m_1, m_2$ we found above. In this frame, $m_1$ and $m_2$ are equivalnet to a single mass $m_1 + m_2$ at rest. $m_3$ must be moving at some velocity $v_3$ in this frame. We simply do the same trick again to find the centre of momentum frame in which the momentum of $m_1, m_2$ and $m_3$ is zero. And, it's precisely the centre of mass frame of all three particles.

This argument now extends to any number of particles.

Now we know that for any system of particles we can find a frame in which the overall momentum is 0, and this is precisely the centre of mass frame.

If, in this frame, there is a collision of two particles, then by Newton's 3rd law, momentum is conserved in that collision and overall momentum remains 0. The particles are free to collide with each other as much as they want, but overall momentum will remain 0.

6. Dec 10, 2015

### REVIANNA

I understand that I am looking at $m_2$ from $m_1$ and $m_2$ 's velocity is $v$ with respect to $m_1$ that's why the momentum is $m_2*v$.(from $m_1$ 's frame.

which direction is "this" direction?

7. Dec 10, 2015

### Ray Vickson

The DEFINITION of the center-of-mass frame is that total momentum = 0 in it. Intuition is not involved---just a definition.

8. Dec 10, 2015

### PeroK

The direction that $m_2$ is moving.