Relative velocity and conservation of momentum

[Maiia]

Homework Statement

Two particles of masses m and 3m are moving toward each other along the x-axis
with the same speed v. They undergo a head-on elastic collision and rebound along the x-
axis.

I don't seem to be getting the right answer...I would really appreciate it if someone could help me figure out what I did wrong.

I put the smaller mass as A and the bigger mass as B.
MaVia + MbVib= MaVfa + MbVfb
mv - 3mv= mVfa + 3MVfb
-2mv= mVfa + 3MVfb

then i used relative velocity formula
Vaf-Vbf= -(Vai-Vbi)
Vaf - Vbf= 2v
Vaf= 2v+Vbf

Then I substituted into the equation
-2mv= m(2v+Vbf) + 3MVfb
-2mv= 2mv + 2mVbf + 3MVbf
-4mv= 5mvbf
-4/5v= vbf

 

[Dick]

Relative velocity formula? What's that? You want to use conservation of momentum (you got that equation right). Now you want to use conservation of energy to get a second equation for Vfa and Vfb. That's what 'elastic' means.

 

[Maiia]

my teacher said we could use relative velocity for head on collisions..

 

[Dick]

my teacher said we could use relative velocity for head on collisions..

Perhaps you can, but that would be a version of conservation of momentum. You need another relation here. You really have to use conservation of kinetic energy.

 

[mukundpa]

I think you mean for elastic collision

velocity of approach = velocity of separation

 

[Dick]

I think you mean for elastic collision

velocity of approach = velocity of separation

No. I mean what I said. That's not right.

 

[mukundpa]

I think we can get this using energy conservation law. I think from here only we define coefficient of restitution.

 

[Dick]

I think we can get this using energy conservation law. I think from here only we define coefficient of restitution.

I don't know what that is. I think 'elastic'=conserve energy and momentum. Is that the 'relative velocity rule' the poster is talking about? If it is equivalent to energy conservation then it will work. But I just don't know this jargon.

 

[mukundpa]

The conservation of linear momentum and energy both leads to a relation

u1 - u2 = v2 - v1

I think this is called relative velocity of approach = relative velocity of separation and can be used with momentum conservation to avoid square terms.

Sorry, the first reply was for student not for you.

Regards

 

[Dick]

Right, and it does avoid square terms. I'm sorry, I've just never seen this before. Thanks for the enlightenment. And I've got a doctorate in physics. Shows you can always learn new things. Thanks!

 

[reallybanana]

Looking at this problem kinda helped me study for my physics final tomorrow as it is on my exam. Anyways getting back on topic. I think that because it is an elastic collision, we can assume that both objects are traveling at the same velocity after the collision in the -x direction.

If we assume that's the case, then I think it would turn out like this instead:

mVi - 3mVi = -(m + 3m)Vf
mVi - 3mVi = -4mVf
-2mVi = -4mVf
Vi = 2Vf or Vf = .5Vi

I'm not sure if this is how it should be conducted, but it seems to work and make sense.

 

[mukundpa]

then i used relative velocity formula
Vaf-Vbf= -(Vai-Vbi)
Vaf - Vbf= 2v
Vaf= 2v+Vbf

I think it should be

Vbf - Vaf = 2v