GR9277 #70



Alternate Solutions 
evanb 20080625 12:47:47  A way to do Andresito's method without math.
With a of 100, the thing is definitely ultrarelativistic, meaning that it could effectively be treated as an almostzeromass particle.
Photons have p = E/c, so this particle should have approximately p = E / c = 100 m c   Andresito 20060329 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 20060329 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

 

Comments 
createdestroy 20130918 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 20100916 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
therefore, the answer is D)   engageengage 20090116 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 dudeness 20100904 19:24:04 
In a similar vein, we can just use the two equations and . From the first equation, we get . This crazy value of tells us that v is essentially equal to c, so the second equation becomes .

his dudeness 20100904 19:26:17 
I meant to write there at the end... This site is awesome in every way, but I wish there was a way to edit posts...

  astrodoo 20081009 07:48:15  I always appreciate your endeavor, Yosun.
In this page, you got a some typo that . It should be .
Anyway, thank you.   evanb 20080625 12:47:47  A way to do Andresito's method without math.
With a of 100, the thing is definitely ultrarelativistic, meaning that it could effectively be treated as an almostzeromass particle.
Photons have p = E/c, so this particle should have approximately p = E / c = 100 m c   Ning Bao 20080129 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 20071101 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 20071025 00:41:33  Thanks for the correct explanation Andresito   Andresito 20060329 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 20060329 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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