Relativistic Momentum of a particle

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Homework Help Overview

The discussion revolves around a particle decay problem in the context of relativistic momentum. The original poster presents a scenario where a particle of unknown mass decays into two particles with known masses and momenta, seeking to determine the unknown mass and its speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of conservation laws, particularly momentum and energy, in the context of particle decay. Questions are raised about the principles governing mass relationships and the implications of varying mass values.

Discussion Status

Some participants have suggested using conservation of energy and momentum as approaches to find the unknown mass and speed. There is an acknowledgment of the need to consider the decay reaction in terms of collision principles, indicating a productive direction in the discussion.

Contextual Notes

Participants are navigating the complexities of relativistic effects and the implications of mass and momentum in particle physics, with an emphasis on understanding the foundational principles rather than arriving at a definitive solution.

~Sam~
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Homework Statement


A particle of unknown mass M decays into two particles of known masses m1 = 0.5 GeV/c2 and m2 = 1.0 GeV/c2, whose momenta are measured to be p1 = 2.0 GeV/c along the positive y-axis and p2 = 1.5 GeV/c along the positive x-axis. Find the unknown mass M and its speed.


Homework Equations



p=1/sqrt(1-v^2/c^2)(mu)

The Attempt at a Solution



I tried summing the momentas, via pythagoras theorem. after that I'm not sure.
 
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What's the general principle which governs how you can find the answer? I.e. for a given mass 'M,' why wouldn't m1 = 500000 GeV and m2 = 100000 Gev?

Hint: what if you think about the decay reaction as a collision?
 
Ahh I got it...using conservation of energy.
 
use conservation of momentum and energy to get the velocity in ground frame
 

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