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!