# One dimensional collision of 2 unequal masses

1. Jul 13, 2010

### randombill

Imagine two particles of uneven masses colliding at relativistic velocities. What is the final velocity of each particle after the collision (no they do not stick together).

This question pertains to the relativistic treatment of a one dimensional collision. Basically I tried to derive the equations of motion for the relativistic case but I'm stuck. I attached what I've done so far.

I thought that solving for the Newtonian conservation of momentum case would be easy simply by putting in the gammas and solving for a simultaneous equation except that I end up with a simultaneous equation for two unknown masses (these are the relativistic masses) while the final rest masses are known. In the attached picture the unknowns are the final relativistic masses while the initial rest masses and the initial relativistic masses are known (please see the last picture at the last line).

Thanks.

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2. Jul 13, 2010

### starthaus

See https://www.physicsforums.com/blog.php?b=1857 [Broken], the attachment entitled "Collision"

Last edited by a moderator: May 4, 2017
3. Jul 14, 2010

### randombill

I'm looking at the final equation (3.8) where:

γ (U '1 )m1 + γ (U '2 )m2 = γ (u '1 )m1 + γ (u '2 )m2 (3.8)

which I understand to be the velocity in the "other frame" but the equation just spits back the original problem. How do I find (u '1 ) and (u '2 )?

I think these proofs only show that the conservation of momentum is frame invariant but nothing else.

Say if you had a mass of 10 kg moving along the positive x direction at .8c and a mass of 5kg moving along the negative x direction at .7c. how would I solve that using these equations in the collision.doc?

thanks.

4. Jul 14, 2010

### starthaus

This is the conservation of relativistic energy . It reduces to the conservation of relativistic mass.

Add to (3.8) the equation of conservation of relativistic momentum:

γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2= γ (u '1 )m1 u'1+ γ (u '2 )m2u'2

You now have a non-linear system of two equations with two unknowns : u'1 and u'2.

Last edited: Jul 14, 2010
5. Jul 14, 2010

### Staff: Mentor

Note that in non-relativistic mechanics you also have to start with both conservation of momentum and conservation of energy, in order to solve a one-dimensional collision. Relativistic mechanics is no different in this respect.

The main difference in the relativistic case is that momentum and energy depend on velocity in a more complicated way, so the algebra is messier.

Last edited: Jul 14, 2010
6. Jul 14, 2010

### randombill

But what about the γ's for each velocity? Can those be factored out so:

(U '1 )m1 U'1+ (U '2 )m2 U'2= (u '1 )m1 u'1+ (u '2 )m2u'2

I'm assuming that I cannot because the gammas are velocity dependent for each momentum and thats the part that makes it hard to solve. Basically that was the reason why I converted the relativistic momentum equation into the equation where:

(pc)^2 = - (m_0c^2)^2 + (mc^2)^2

(got it from here http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1)

Basically how do I solve a simultaneous equation for this mess?

NOTE: I just noticed that I used lamba's instead of gamma's in my first post photos. Sorry about that.

Right, how do I solve for it is the problem. The algebra is the part thats messing me up because each γ's have an independent velocity under a square root and Im having trouble factoring it out so I can solve for each final velocity.

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7. Jul 14, 2010

### starthaus

No, they cannot. Different speeds produce different $$\gamma$$'s.
Yet:

γ (U '1 )m1 + γ (U '2 )m2 = γ (u '1 )m1 + γ (u '2 )m2 (3.8)

gives you γ (u '2 ) (and, consequently, u '2) as a function of γ (u '1 ). Sibstitute in the equation of momentum conservation and you will get the formula for u '1.

Last edited: Jul 14, 2010
8. Jul 14, 2010

### randombill

I'm not entirely sure what you mean, could you show me that part by doing it... please, thanks.

9. Jul 14, 2010

### starthaus

OK,

$$m_1v_1\gamma_1+m_2v_2\gamma_2=A$$
$$m_1\gamma_1+m_2\gamma_2=B$$

Multiplying the second equation by $$-v_1$$ and adding with the first one we get:

$$v_1=\frac{m_2v_2\gamma_2-A}{m_2\gamma_2-B}$$

From the above, you can calculate :

$$\gamma_1=1/\sqrt{1-(v_1/c)^2}$$ as a function $$f(v_2)$$

Substitute in $$B=m_1\gamma_1+m_2\gamma_2=m_1f(v_2)+m_2\gamma_2$$ and you got yourself an algebraic equation in $$v_2$$

10. Jul 15, 2010

### randombill

Is B the final momentum equation. I'm lost as to what the initial and final velocities are.

Not sure how???

Is v2 the velocity in the final momentum equation.

Sorry but I've never solved a simultaneous equation in such a way, plus my math isn't the greatest.

How would I solve this for example:

If you had a mass of 10 kg moving along the positive x direction at .8c and a mass of 5kg moving along the negative x direction at .7c, what are the final velocities of each mass?

http://teachers.web.cern.ch/teacher...ch/mbitu/applications_of_special_relativi.htm

where the Lorentz transformation matrix is used to find the final momentum and energies but how do I find the velocities of the particles knowing the final energy for each particle?

Last edited: Jul 15, 2010
11. Jul 15, 2010

### starthaus

No, it is the initial momentum. You know it since you know the speeds before collision.

Simple, substitute the formula for $$v_1$$ that I just derived for you, in the formula for $$\gamma(v_1)$$

$$v_1$$ and $$v_2$$ are the final velocities of particle 1 and particle 2 (after the collision)

12. Jul 15, 2010

### randombill

I tried substituting A and B into

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-3.png [Broken]

and then substituted that into,

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-4.png [Broken]

and all the terms cancel out and I got back the original equation for gamma?

But then I'm not sure what you mean at this part for,

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-5.png. [Broken] I see that I can get v1 and v2 (the final velocities) by squaring gamma_1 and solving for v1.

The part I'm not getting is that B does not have any initial velocities so how would that find the final velocities?

Last edited by a moderator: May 4, 2017
13. Jul 15, 2010

### starthaus

I give up. This is basic algebra.

Last edited by a moderator: May 4, 2017
14. Jul 15, 2010

### randombill

C'mon you almost solved it. Algebra is never basic.

15. Jul 15, 2010

### starthaus

$$\gamma_1=1/\sqrt{1-1/c^2*(\frac{m_2v_2\gamma_2-A}{m_2\gamma_2-B}})^2$$

Can you take it from here?

16. Jul 15, 2010

### randombill

Yes I substituted A and B into that equation and the terms reduced under the square root to:

[PLAIN]https://www.physicsforums.com/latex_images/28/2800728-4.png [Broken]

Please let me know if there is something totally wrong with what I did, thanks.

Last edited by a moderator: May 4, 2017
17. Jul 15, 2010

### starthaus

This is impossible since

A=γ (U '1 )m1 U'1+ γ (U '2 )m2 U'2

B=γ (U '1 )m1+ γ (U '2 )m2

where U'1 and U'2 are the speeds before the collision.

Last edited by a moderator: May 4, 2017
18. Jul 16, 2010

### randombill

Oh, ok. So I retried doing that substitution for A and B and I attached two photos of the work. Let me know if the final result makes sense for v2 which is the final velocity. Do you know if there is a way to find v1 similar to Newtonian mechanics where:

$$v2f - v1f = v1i - v2i$$

These are the photos, I'm not good with tex either.

https://www.physicsforums.com/attachment.php?attachmentid=26983&stc=1&d=1279283932

https://www.physicsforums.com/attachment.php?attachmentid=26984&stc=1&d=1279283932

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19. Jul 16, 2010

### starthaus

No, it doesn't look right.

20. Jul 16, 2010

### starthaus

I showed you how to get $$\gamma_1=1/\sqrt{1-(v_1/c)^2}$$ as a function $$f(v_2)$$

Now, substitute $$\gamma_1$$ in the equation $$B=m_1\gamma_1+m_2\gamma_2$$ and you will get an algebraic equation in $$v_2$$. Solve it and you will find $$v_2$$. Substitute $$v_2$$ in the expression for $$\gamma_1$$ and you will find $$v_1$$. Try learning latex, or, at least, print your solutions.

21. Jul 16, 2010

### randombill

I tried to do this:

Take A and B and substitute into:
[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png [Broken]

and then I substituted that back into B and solved for v2. The solution was long and I had trouble with tex.

Let me know what your final solution looks like because at least then I could try to find that.

Thanks.

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22. Jul 16, 2010

### starthaus

no, it isn't right, you forgot that $$\gamma_2$$ is a function of $$v_2$$

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23. Jul 16, 2010

### randombill

It would be impossible to factor out $$\gamma_2$$ out of the equation after doing this:

[PLAIN]https://www.physicsforums.com/latex_images/28/2801452-8.png [Broken]

because $$\gamma_2$$ is mixed in with everything else. I read this book:

where he mentions toward the end of paragraph 2 on page 225 the use of numerical solutions for this problem. I guess this is why?

Basically I give up I cant solve it.

Last edited by a moderator: May 4, 2017
24. Jul 17, 2010

### starthaus

You should get an equation degree 8 in $$v_2$$. This equation does not have symbolic solutions. Nevertheless, it is a good exercise for you to finish the calculations, never give up once you start something. There is satisfaction in forming the equation. How else would you solve it numerically if you don't have the final equation?

Last edited by a moderator: May 4, 2017
25. Jul 17, 2010

### randombill

Does that mean (x)^8 = $$v_2$$ where x is the final solution?

I don't need a numerical solution, unfortunately seeing how there is no symbolic solution makes me not need this result.

Last edited: Jul 17, 2010