# Velocity in Elastic Collision involving Center of Mass

1. Oct 28, 2013

### icesalmon

1. The problem statement, all variables and given/known data
part #1: A wooden block with m1 = 2.9kg is sliding across a frictionless surface in the −x direction at 4.5 m/s. A smaller wooden block with m2 = 1.1 kg block is traveling in the +x direction at 2.9 m/s.

part #2: Now they make a head-on elastic collision. (This means that in the COM frame, the velocity of each is reversed.)

Question #1: Transform back into the original frame by adding vcm to the velocity of each block.

What is the velocity of the m1 block in the lab frame after the collision?

2. Relevant equations

3. The attempt at a solution
V1,i* = V1,F*
where V* = Vobj,lab + Vlab,CM and Vlab,CM = -1(VCM,lab)

I would assume since the collision is Elastic that the velocities of the objects, since there masses don't change and Kinetic Energy and Momentum are Conserved to be the same as their respective initial velocities in the opposite direction they approached each other but this doesn't seem to be the case since Vm1,F != -Vm1,i so I run through the equation I have been given through pre-lectures and the answer still comes out wrong. My conceptual understanding of these relative velocities is drastically skewed and I am desperately trying to understand this. Thanks in advance for any help and I apologize if I haven't included information that is required.

Here's my work

V,1,i* = V,1,F*

V1,*= V,obj,lab - V,CM,lab

V,CM,Lab = 1/(4kg)*[2.9*4.5-1.1*2.9]
V,CM,lab = 2.465 m/s

V,m1,Lab = 4.5 - 2.465
V,m1,Lab = 2.035

wrong

I tried another method where I use the whole equation and all the velocities in the situation

V,1,i* = V,1,F*

V,Obj,Lab,i - V,CM,Lab,i = V,obj,lab,f - V,CM,F
-4.5m/s -(-2.46 m/s) + 2.46m/s = V,obj,lab,f
.42m/s = V,obj,lab,f

wrong

since they say add on the the V,CM in the question I try that and I get that V,m1,i = -V,m1,F

V,m1,F = 4.5 m/s

wrong

Okay so V,m1,F + V,CM,F is wrong
and V,m1,I + V,CM,F is wrong
And V,m1,I + V,CM,I is wrong
V,M1,F + V,CM,I is wrong

I don't think there are any other ways I could approach that question

Should I try using Conservation of Momentum?

ΔPsystem = 0

ΔPsystem,i - Δsystem,f = 0

or how about the conservation of kinetic energy?

ΔKEsytem = 0
ΔKEsystem,i - ΔKEsystem,F = 0

why would this approach be any different to the way in which I'm approaching the problem, both ideas demonstrate that the velocity of the system doesn't change in an elastic collision.

Maybe i'm reading the question wrong, or my interpretation of the question is wrong? Should I be adding on the Velocity of the Center of mass After the collision to the final velocity of the object, wrt the lab, in question? Maybe what I did above isn't the same as this in some way.

Last edited: Oct 28, 2013
2. Oct 28, 2013

### icesalmon

if there is anything else you need to know, I have answered other questions in this problem before this one that I know with 100% certainty are correct that may help, let me know.

3. Oct 28, 2013

### Simon Bridge

... so the conclusion does not always follow.

Why not do as the question suggests and solve it for the center of mass frame first?
What are their velocities in that frame?

4. Oct 28, 2013

### icesalmon

how is that possible, that the conclusion does not always follow? I'm not fighting you here, I want to understand some things before I move on with what you're asking, which I did mis-interpret. In elastic collisions, is it true to say that momentum and kinetic energy are conserved, ALWAYS? If that is true, and the system does not lose mass how is it possible, without the velocities of the objects switching that, kinetic energy and momentum are not conserved?

Also, the question says to add on the VCM, but doesn't clarify which one to use, can there be more than two in this situation? I would also assume to use the VCM of the system after the collision takes place, which I believe i've also found correctly.

actually I believe there are 3 in this case
#1 the velocity of CM relative to m1
#2 the velocity of CM relative to m2
#3 the velocity of CM relative to the observer, or the lab frame in this case, which is stationary.

And when you say "solve it for the center of mass frame first?"
are you saying solve for the velocity of m1 relative to the center of mass?

VCM,m1 = VCM,lab - VCM,CM
VCM,m1 = VCM,Lab - 0
VCM,m1 = -Vm1,CM (I'm not sure if this is true, but I really use it as a crutch when I can't picture velocities in certain reference frames but I can in the opposite reference frame)

Vm1,lab = VCM,lab + Vm1,CM

Vm1,Lab - VCM,lab = Vm1,CM

VCM,lab = (1/4kg)[m1*v1f + m2*v2f]
VCM,lab = 2.465 m/s

and I believe that Vm1,lab after the collision is 4.5 m/s

okay so Vm1,CM = 4.5 m/s - 2.465 m/s
Vm1,CM = 2.035 m/s

the sign of the answer makes sense, both the center of mass and mass 1 are moving in the positive x direction so I would assume their relative velocities be positive. As to the actual number I believe that relies on my assumption that m1 is moving away in the opposite direction at the same speed.

Last edited: Oct 28, 2013
5. Oct 28, 2013

### Simon Bridge

Isn't that what you've just found out?

If the conclusion always follows, then you would have got the right answer wouldn't you?
Or have I misread you?

There is only ever one CM - thus only one $v_{CM}$.

Yes.
Find the velocity of the center of mass in the lab frame.
Find the initial velocities of the two masses in the center-of-mass frame.
From there you can find the final velocities in the center-of-mass frame.
Then transform back to the lab frame.

I'm afraid your notation is tricky for me:
I'm reading $v_{x,y}$ to mean the velocity of x in the y frame ... so $v_{m_1,lab}$ is the velocity of mass-1 in the lab frame ... but what am I to make of $v_{CM,m_1}$? The velocity of the center of mass i the $m_1$ frame? Then yes, $v_{x,y}=-v_{y,x}$ ... you should be able to prove this.

Once you have the velocities wrt the com, and you have the velocity of the com in the lab, then you can find the velocities of m1 and m2 in the lab.

6. Oct 28, 2013

### icesalmon

VCM,lab,final = 2.465 m/s
Vm1,CM,final = -2.04 m/s

vm1,CM,F + vCM,lab,F -2.04 m/s + 2.465 m/s

i'm really not sure where i'm going wrong here.

7. Oct 28, 2013

### icesalmon

I had it right the first time it's -1*(Vm1,CM,F + VCM,lab,F) and I changed it to what I had above, my course uses a website for homework and there was glitch in rounding at the end, all is well. Thanks

8. Oct 28, 2013

Well done.