

All Solutions of Type: Special Relativity  0  Click here to jump to the problem!  GR8677 #20

Special Relativity}Rest Energy
If one remembers the formulae from special relativity, arithmetic would be the hardest part of the problem.
The problem is to solve , where and .
The restmass of both the kaon and the proton are given in the problem. Thus, the equation reduces to .
Now, because the range of velocities vary significantly between and , one can't directly approximate that as 2. Boo. So, longdivision by hand yields approximately .
The author of this site prefers to use fractions instead of decimals. Thus . Express in terms of to get .
is about 64, so the velocity has to be greater than 0.8c. The only choice left is (E).
If one has the time, one might want to memorize the following:
, or perhaps a more elaborate list of correlations.
If one knew that beforehand, then one would immediately arrive at choice (E).

18  Click here to jump to the problem!  GR8677 #99
 Special Relativity}Conservation of Energy
The tricky part of this problem is to note that the momentum of the photon is shared equally between all three final particles. (This insight was supplied by Felipe Birk). Thus, , where p is the momentum of the photon and is the momentum of the final electron or photon.
The initial energy before the photon strikes the electron is , which is just the energy of the photon plus the rest energy of the electron.
The final energy after the collision is , which is the sum total energy of all three final particles, i.e., the positron and two electrons. (A positron is the electron's antiparticle, and thus they have the same mass.) Note that the momentum split relation mentioned above is used to equate the final particle momentum with the initial photon momentum.
Conjure up the good conservation of energy idea. Equating , one gets . Canceling the terms on both side, then solving, one arrives at , which is choice (D). 




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