# Homework Help: Quick Collision Question

1. May 9, 2010

### bon

1. The problem statement, all variables and given/known data

A particle mass m moves to the right with relativistic speed v. It collides inelastically with another particle mass 3m moving to the left also at speed v. A single body of mass 5m is produced in the collision. What is the threshold speed needed for this process to occur?

2. Relevant equations

3. The attempt at a solution

So i was just wondering what the quickest way to solve this would be.

I solved using the invariant E^2 - c^2 p^2 looking at the lab frame before the collision and CMF frame after..

i get v to be 3/root21 c..but am really unsure about it..is this right?

Also, what does 'threshold' mean in this context..? Surely the velocity is completely determined by the statement of the problem...i.e. with the given data there aren't a range of possible velocities are there? :S

Thanks

2. May 9, 2010

### Andrew Mason

Momentum is conserved in this collision. So, try analysing the collision in the lab frame and applying conservation of momentum:

$$p = \gamma m_0v$$

and:

$$p_1 + p_2 = p_f$$

AM

3. May 10, 2010

### bon

Ok so is my answer wrong?

Also could you answer my question about the "threshold" velocity in this context..thanks

4. May 10, 2010

### Andrew Mason

You will have to show your work.

What is the final speed? (hint: what is the rest mass of the combined masses? what is the ratio of relativistic mass to rest mass? => what does that tell you about the speed of the combined masses?)

From that you will be able to determine the initial velocity using conservation of momentum.

That should tell you whether there can be more than one value for v.

AM

Last edited: May 10, 2010
5. May 10, 2010

### bon

Sorry this isn't helping. Please could you explain what "threshold" means in this context..i think the sytem is determined..

6. May 10, 2010

### ehild

Threshold speed is the minimum speed so as the conditions can be fulfilled. Remember, you will have one m more than before, so the energy of the colliding particles has to be enough to produce that excess mass.

ehild

7. May 10, 2010

### bon

thanks ehild - sorry maybe i should have asked my question more carefully - you say that threshold speed in the minimum speed so as the conditions can be fulfilled. My point is this - surely if the two initial masses are m and 3m, and the final mass 5m then there is only ONE possible velocity such that the conditions are fulfilled - i.e. if the velocity was any less the final mass would be less and vica versa. yes?

8. May 10, 2010

### Andrew Mason

Can you not post your work?

The threshold speed is the minimum speed needed to form a particle of mass 5m. That is the case where no other particles are produced in the collision to carry away momentum. The resulting particle must absorb a great deal of energy (you can work it out) and this usually results in new particles flying off carrying momentum. You could have higher speed collisions producing particles of mass 5m, but not in which all the momentum is carried by the 5m mass.

AM

9. May 10, 2010

### ehild

bon,

your v=3/sqrt(21) is correct if both the momentum and the energy is conserved during the collision. It is difficult to read the mind of the person who gave this problem to you. But the collision was said inelastic. The question is what it means. It can mean that although the mechanical energy of the particles is not conserved, the total energy is, but it can mean also that some of the total energy is lost, is transformed to the energy of a photon or vibration of a crystal lattice, something which does not take away momentum, but consumes energy. In that context the threshold energy has sense.

The problem can be even simpler. From conservation of energy,

$$\frac{4mc^2}{\sqrt{1-v^2/c^2}}=\frac{5mc^2}{\sqrt{1-v'^2/c^2}}$$

(v' is the velocity of the new particle) you get a relation between (v/c)^2 and (v'/c)^2, and (v'/c)^2 >=0 involves that the original speed is at last 0.6c. Maybe, this was only the question.

ehild

10. May 10, 2010

### bon

Ahh so it is possible for the momentum not to be conserved? Since if v'=0 the momentum afterwards is lost!

11. May 10, 2010

### collinsmark

Hello bon,

Momentum will be conserved. And that's kind of important to solving this problem.

Using guidance from ehild's last post, you can get an equation (using conservation of energy) to find a relationship between v and v' (see ehild's last post).

But now you still have two unknowns, v and v'. One equation, two unknowns. Wouldn't it be nice to form a second, simultaneous equation, so you'll have two equations and two unknowns? http://www.websmileys.com/sm/love/301.gif. Hmmm, what conservation law can you use to generate the second equation?....

[Edit: btw, $v' \ne 0$.]

Last edited: May 10, 2010
12. May 10, 2010

### Andrew Mason

Momentum cannot be lost. If the momenta of the colliding masses before and after are not equal, you are missing some other particle carrying momentum. This is one way to discover new particles - eg. the neutrino.

AM

13. May 11, 2010

### ehild

The momentum is always conserved if all the participants in a collision are taken into account. I answered the question about the "threshold speed". One equation, conservation of energy, requires that the original speed of the colliding particles is at least 0.6c. The second equation is conservation of momentum, and the two conservation laws together result in a single value for the initial speed, the same you have given in your very first post:
v=3/sqrt(21) c = 0.65c.

As there is a unique solution of the problem, I still say that I can not read the mind of other people, although I have tried. Your solution was correct, you were right, but next time please write out your solutions in detail.

ehild

14. May 11, 2010

### bon

Ok I see - very clear now. Thanks for your help :)

I'll try to post full working new time.

Thanks

15. May 11, 2010

### Andrew Mason

$$p_1 + p_2 = p_f$$

(1) $$\gamma 3mv - \gamma mv = 5mv' = \gamma' 4mv'$$

Since:

$$5mv' = \gamma' 4mv'$$ then,

$$\gamma' = 5/4$$,

$$1 - v'^2/c^2 = (4/5)^2$$

$$v' = \sqrt{1 - (4/5)^2}c = \sqrt{.36} = .6c$$

Substituting into (1):

$$\gamma 3mv - \gamma mv = \gamma 2mv = 5mv' = 5m(.6c) = 3mc$$

$$\gamma v/c = 3/2$$

Squaring both sides and multiplying by [itex]1/\gamma^2 = (1 - v^2/c^2)[/tex]

$$(3/2)^2(1 - v^2/c^2) = (v/c)^2$$

$$(1 + (3/2)^2)(v/c)^2 = (3/2)^2$$

$$v^2/c^2 = (3/2)^2/(1 + (3/2)^2) = (9/4)/(1 + 9/4) = 9/(4 + 9) = 9/13$$

$$v = 3c/\sqrt{13} = .832c$$

AM

16. May 11, 2010

### bon

No - I think you are wrong because the particle that is produced has rest mass 5m but is still moving, right?

17. May 11, 2010

### Andrew Mason

The resulting particle must have a rest mass of 3m + 1m. How can the rest mass be more than the rest masses of the original particles?

AM

18. May 11, 2010

### bon

It says a single body of mass 5m is produced in the collision..it doesn't say that this is the total energy of the body..

19. May 11, 2010

### ehild

Well, the problem also does not say if the given masses are the rest masses of the particles or not. Bon assumed that all of them are rest masses.

ehild

20. May 11, 2010

### bon

so here is rest mass not conserved?