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GR9277 #70
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{70}
A monoenergetic beam consists of unstable particles with total energies 100 times their rest energy. If the particles have rest mass m, their momentum is most nearly

1. mc
2. 10mc
3. 70mc
4. 100mc
5. $10^4$ mc

Special Relativity$\Rightarrow$}Gamma

$E=\gamma mc^2 = 100mc^2$, where ETS supplies the total energy to be 100 times the rest energy. Thus, $p=\gamma mv = 100 mv$, but since $\gamma = 100 = \frac{1}{1-\beta^2}$, where $\beta=v/c$, one has $v\rightarrow c$, as in choice (D).

Alternate Solutions
evanb
2008-06-25 12:47:47
A way to do Andresito's method without math.

With a $\gamma$ of 100, the thing is definitely ultra-relativistic, meaning that it could effectively be treated as an almost-zero-mass particle.

Photons have p = E/c, so this particle should have approximately p = E / c = 100 m c
Andresito
2006-03-29 08:48:14
E^2 = (p*c)^2 + (m*c^2)^2

let E = 100 m*c^2

solve for p to obtain p = 100 m*c
 Andresito2006-03-29 08:51:39 E^2 = (p*c)^2 + (m*c^2)^2 let E = 100 m*c^2 since E^2 >> (mc)^2, (p*c)^2 = E^2 solve for p to obtain p = 100 m*c
createdestroy
2013-09-18 19:16:22
E_tot = 100*E_rest
E_rest = m_0 * c^2

m_0 is given as m

E_rest = mc^2
E_tot = 100*E_rest = 100mc^2
matweiss
2010-09-16 16:00:11
This question is more straight forward than it seems.

The gamma factor is what is causes the relativistic energy to be 100 times greater than the rest energy.

This same gamma factor is what influences the momentum

Thus the relativistic energy factor of 100 will apply to the relativistic momentum factor in the exact same way

engageengage
2009-01-16 18:26:19
this is a quick one once you realize that this is a 'highly relativistic' particle, since its energy by far surpasses its rest energy. Highly relativistic particles are like photons in that their energy is given by

E = pc

From this, one immediately gets the answer.
 his dudeness2010-09-04 19:24:04 In a similar vein, we can just use the two equations $E=\gamma mc^2$ and $p=\gamma mv$. From the first equation, we get $\gamma =100$. This crazy value of $\gamma$ tells us that v is essentially equal to c, so the second equation becomes $E=100mv$.
 his dudeness2010-09-04 19:26:17 I meant to write $p=100mc$ there at the end... This site is awesome in every way, but I wish there was a way to edit posts...
astrodoo
2008-10-09 07:48:15
I always appreciate your endeavor, Yosun.
In this page, you got a some typo that $\gamma = \frac{1}{1- \beta^2}$. It should be $\frac{1}{\sqrt{1- \beta^2}}$.

Anyway, thank you.
evanb
2008-06-25 12:47:47
A way to do Andresito's method without math.

With a $\gamma$ of 100, the thing is definitely ultra-relativistic, meaning that it could effectively be treated as an almost-zero-mass particle.

Photons have p = E/c, so this particle should have approximately p = E / c = 100 m c
Ning Bao
2008-01-29 12:37:57
An easier way is to use units where c=1: then E^2=p^2+m^2, and it's clear that p should be about 100m in those units, which matches the correct answer.
Jeremy
2007-11-01 15:49:13
My experience has been that solutions are more quickly obtained in problems involving relativity if you AVOID GAMMA, as it can lead to lengthy algebra. Sometimes you have no choice, but this is not one of those times. Well done, Andresito.
craklyn
2007-10-25 00:41:33
Thanks for the correct explanation Andresito
Andresito
2006-03-29 08:48:14
E^2 = (p*c)^2 + (m*c^2)^2

let E = 100 m*c^2

solve for p to obtain p = 100 m*c
 Andresito2006-03-29 08:51:39 E^2 = (p*c)^2 + (m*c^2)^2 let E = 100 m*c^2 since E^2 >> (mc)^2, (p*c)^2 = E^2 solve for p to obtain p = 100 m*c

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