GREPhysics.NET: GR9277 #70

GR9277 #70

Problem

GREPhysics.NET Official Solution

\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

mc
10mc
70mc
100mc
$10^4$ 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

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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

Andresito
2006-03-29 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

Reply to this comment

Comments

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.

Reply to this comment

evanb
2008-06-25 12:47:47

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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.

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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.

Reply to this comment

craklyn
2007-10-25 00:41:33

Thanks for the correct explanation Andresito

Reply to this comment

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

Andresito
2006-03-29 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

Reply to this comment

Post A Comment!

You are replying to:

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.

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