E-P invariant on relativistic mecahnics problem

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SUMMARY

The forum discussion centers on a relativistic mechanics problem involving two identical particles, each with mass m, where one is moving at speed \(\frac{3}{5}c\) and collides with a stationary particle. The resulting particle of mass M moves at speed v, which the user initially calculated incorrectly as \(\frac{\sqrt{2}}{2}c\). The correct speed is determined to be \(\frac{1}{3}c\) by applying the E-P invariant and conservation laws. The user is advised to incorporate conservation of energy and momentum to resolve the issue.

PREREQUISITES
  • Understanding of relativistic mechanics
  • Familiarity with the E-P invariant equation: \(E_1^2 - p_1c^2 = E_2^2 - p_2^2c^2\)
  • Knowledge of momentum and energy equations in relativistic contexts
  • Ability to manipulate and solve equations involving Lorentz factors (\(\gamma\))
NEXT STEPS
  • Study conservation of momentum in relativistic collisions
  • Learn how to apply conservation of energy in relativistic systems
  • Explore the implications of the E-P invariant in particle collisions
  • Review Lorentz transformations and their applications in relativistic physics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in mastering relativistic mechanics, particularly in understanding particle collisions and the application of conservation laws.

Froskoy
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Homework Statement


A particle of mass [itex]m[/itex] moving at speed [itex]\frac{3}{5}c[/itex] collides with an identical particle at rest, and forms a new particle of mass [itex]M[/itex] which moves off at speed [itex]v[/itex]. Find v.

Homework Equations


E-P invariant: [itex]E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}[/itex]
Momentum: [itex]p^2=m^2c^2\left({\gamma_v^2-1}\right)[/itex]
Energy: [itex]E=\sum_i\gamma_im_ic^2[/itex]

The Attempt at a Solution


There is a single particle after the collision, so [itex]E_2^2-p_2^2c^2=M^2c^4[/itex]
where [itex]E_2=\gamma_vMc^2[/itex]
and [itex]p_2^2=M^2c^2\left({\gamma_v^2-1}\right)[/itex]

so

[tex] \gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\<br /> \gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\<br /> \left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0[/tex]

Reject the [itex]\gamma_v^2=1[/itex] solution, since this would mean [itex]v=0[/itex].

The [itex]\gamma_v^2=2[/itex] solution gives [itex]v=\frac{\sqrt{2}}{2}c[/itex], which is incorrect - the answer is [itex]v=\frac{1}{3}c[/itex].

I expect there's something fundamentally wrong with my method, but not sure what? Thank you!
 
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Froskoy said:

Homework Statement


A particle of mass [itex]m[/itex] moving at speed [itex]\frac{3}{5}c[/itex] collides with an identical particle at rest, and forms a new particle of mass [itex]M[/itex] which moves off at speed [itex]v[/itex]. Find v.


Homework Equations


E-P invariant: [itex]E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}[/itex]
Momentum: [itex]p^2=m^2c^2\left({\gamma_v^2-1}\right)[/itex]
Energy: [itex]E=\sum_i\gamma_im_ic^2[/itex]


The Attempt at a Solution


There is a single particle after the collision, so [itex]E_2^2-p_2^2c^2=M^2c^4[/itex]
where [itex]E_2=\gamma_vMc^2[/itex]
and [itex]p_2^2=M^2c^2\left({\gamma_v^2-1}\right)[/itex]

so

[tex] \gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\<br /> \gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\<br /> \left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0[/tex]
You shouldn't have squared the factor ##(\gamma_v^2-1)## in the first line. With this approach, all you'll end up with is M=M, so it's not helpful.
Reject the [itex]\gamma_v^2=1[/itex] solution, since this would mean [itex]v=0[/itex].

The [itex]\gamma_v^2=2[/itex] solution gives [itex]v=\frac{\sqrt{2}}{2}c[/itex], which is incorrect - the answer is [itex]v=\frac{1}{3}c[/itex].

I expect there's something fundamentally wrong with my method, but not sure what? Thank you!
You haven't used any information from before the collision. Use conservation of energy and momentum.
 

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