# Two Particles Collide - Momentum and Energy

1. Feb 2, 2016

### TristanJones

1. The problem statement, all variables and given/known data
Q: A 3.000u object moving to the right through a laboratory at 0.8c collides with a 4.000u object to the left of the laboratory at 0.6c. Afterward, there are two objects, one of which is a 6.000u mass at rest.
A) what are the mass and speed of the other object?
B) determine the change in kinetic energy in the collision

2. Relevant equations
γ1m1v1 + γ2m2v2 = γ3m3v3+γ4m4v4
γ1m1c2+γ2m2c2 = γ3m3c2+m4c2

3.= The attempt at a solution
I've been struggling with this for a while, and haven't really made much headway. The part I'm finding difficulty with is using both conservation of momentum and conservation of energy formulas when I need to consider that the third particle technically isn't moving. I don't necessarily want the answers handed to me - just a few tips to get me headed in that direction would be much appreciated.

2. Feb 3, 2016

### vela

Staff Emeritus
Try working with momenta and energies instead of velocities.

3. Feb 3, 2016

### TristanJones

do you mean eliminating the v as a variable? i used the conservation of energy and momentum formulas, but there's just something not quite working. i end up with more unknowns than i know what to do with

4. Feb 3, 2016

### ChrisVer

Aren't the unknowns in your equations the gamma3 and the m3 (for 2 equations this is solvable)?

5. Feb 3, 2016

### TristanJones

and whatever V3 would be in the momentum equation (unless I'm not supposed to use the momentum equation)

6. Feb 3, 2016

### ChrisVer

The v3 and gamma3 are the "same" unknown... aren't they?

7. Feb 3, 2016

### TristanJones

gahhh, you're right. i think my issue was dealing with the one particle which was at rest (and has no momentum). i'll play around with it a bit

8. Feb 3, 2016

### PeroK

Try using $E^2 = p^2c^2 + m^2c^4$

These problems never work out if you try to solve for $v$ and $\gamma$!

9. Feb 3, 2016

### ChrisVer

Well they do, it's just a very rough way to do things... Either way you have to apply the conservation of energy/momentum at some point (in one way or another).

10. Feb 3, 2016

### ChrisVer

But the expression for the momentum conservation is not quiet right...(except for if vi are vectors)

11. Feb 3, 2016

### Ray Vickson

If Particle 4 is the one at rest after the collision we have
$$\begin{array}{rcrcrc} \gamma_1 m_1 v_1 &+& \gamma_2 m_2 v_2 &=& \gamma_3 m_3 v_3 &+ &0 \\ \gamma_1 m_1 &+& \gamma_2 m_2 &=& \gamma_3 m_3 &+& m_4 \end{array}$$
In the second equation we dropped the common factor $c^2$ from all terms on both sides.
From the known quantities, the equations allow us to find $\gamma_3 m_3$ and $\gamma_3 m_3 v_3$; so we can get $v_3$ by taking a ratio of known quantities. Once we know $v_3$ we are almost done.

12. Feb 3, 2016

### TristanJones

@Ray Vickson - ok, so for the momentum I got
γ3m3v3= 1uC
for energy I got: γ3m3= 4u
so I divide the first one by the second to get v3= 0.25c (?)

13. Feb 3, 2016

### TristanJones

ok, then I used that, calculated gamma, and inserted back into the momentum equation to solve for m3 and got 3.873 u. should be good to go. thanks!

14. Feb 3, 2016

### ChrisVer

seems legit...

15. Feb 3, 2016

### TristanJones

unless you got something entirely different and I am to to lunch, hehe

16. Feb 3, 2016

### ChrisVer

Yes I got something totally different because I omitted the m4, as if it was not there...
As for the questionmark for the v3, it's true that $\frac{p}{E}= v$ for massive particles, and it's also the same for massless particles too...

17. Feb 3, 2016

### TristanJones

there are no massless particles in this equation though ... sorry, I'm just not entirely sure if you're agreeing with my end result, or implying I should make changes

18. Feb 3, 2016

### ChrisVer

nah, I just made a general statement for your questionmarked phrase "so I divide the first one by the second to get v3= 0.25c (?)"
That p/E happens to be always equal to the velocity.

it seems I agree...

19. Feb 3, 2016

### TristanJones

welp, cheers!