Speed of proton with kinetic energy 1000 MeV

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SUMMARY

A proton with a kinetic energy of 1000 MeV has a speed that lies within the range of 0.85c ≤ v < 0.90c, as determined by the total energy and momentum relationship. The total energy is calculated by adding the rest energy (938.3 MeV) to the kinetic energy, resulting in 1938.3 MeV. The momentum is derived using the equation pc = √(E^2 + E0^2), leading to a calculated speed of 0.875c. The discrepancy in methods arose from an error in manipulating the gamma factor in the initial calculation.

PREREQUISITES
  • Understanding of relativistic kinetic energy equations
  • Familiarity with the concept of rest energy (E0 = mc^2)
  • Knowledge of the gamma factor (γ) in relativistic physics
  • Basic understanding of momentum in the context of energy (pc = √(E^2 + E0^2))
NEXT STEPS
  • Study the derivation and application of the gamma factor in relativistic mechanics
  • Learn about the relationship between total energy, kinetic energy, and momentum in special relativity
  • Explore examples of calculating speeds of particles with varying kinetic energies
  • Investigate the implications of relativistic speeds on mass and energy
USEFUL FOR

Students and educators in physics, particularly those focusing on special relativity and particle dynamics, will benefit from this discussion.

mrdoodle
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Homework Statement



A proton has a kinetic energy of 1000 MeV. In which of the following ranges does its speed v lie?
a) v < 0.80c
b) 0.80c ≤ v < 0.85c
c) 0.85c ≤ v < 0.90c
d) 0.90c ≤ v < 0.95c
e) v ≥ 0.95c

Homework Equations



K = (γ-1)mc^2
E0 = mc^2
E^2 = K + E0

The Attempt at a Solution



I tried to solve this problem using just the kinetic energy equation, K = (γ-1)mc^2, and solving for the velocity, v in the gamma factor. I used 1000 MeV as the given kinetic energy of the proton and 938.3 MeV/c^2 for the mass of the proton. Doing this, I ended up with a value of v = 0.346c. However, the answer is actually C, which my professor explained with the work below:

E = mc^2 + K = 938.3 + 1000 = 1938.3 MeV
pc = √(E^2 + E0^2) = 1696 MeV
v = (pc^2) / E = 0.875c.

I understand his work; everything seems to sense, but my question is this: why is the rest energy added to the kinetic energy of the proton when calculating the velocity? I'm a little confused as to why my method and his method yield a different result, and it seems like that question plays a part in it. Thank you in advance for your help!
 
Last edited:
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Hi, mrdoodle. Your method will work. You must have just made an error in the manipulations.

Your professor's method is based on the relationship between total energy and momentum. Total energy, E, is defined as rest energy plus kinetic energy.
 
Ah, sorry I should've checked my work more thoroughly. Turns out I forgot to subtract 1 from the gamma factor even though I had the correct equation. Thank you for the explanation!
 

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