Two Particles Collide - Momentum and Energy

[TristanJones]

Homework Statement

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

Homework 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.

 

[vela]

Try working with momenta and energies instead of velocities.

 

[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

 

[ChrisVer]

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

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

 

[TristanJones]

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

 

[ChrisVer]

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

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

 

[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

 

[PeroK]

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

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

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

 

[ChrisVer]

These problems never work out if you try to solve for vvv and γγ\gamma!

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).

 

[ChrisVer]

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

 

[Ray Vickson]

Homework Statement

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

Homework 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.

If Particle 4 is the one at rest after the collision we have
[tex] \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}
[/tex]
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.

 

[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 (?)

 

[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!

 

[ChrisVer]

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!

seems legit...

 

[TristanJones]

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

 

[ChrisVer]

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

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 [itex]\frac{p}{E}= v[/itex] for massive particles, and it's also the same for massless particles too...

 

[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

 

[ChrisVer]

there are no massless particles in this equation though ..

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.

I'm just not entirely sure if you're agreeing with my end result

it seems I agree...

 

[TristanJones]

welp, cheers!