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GR0177 #32
Problem
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Special Relativity}Momentum

The total energy of a particle of mass m is , since the problem supplies that it is equal to twice the rest mass. This means that .

The relativistic momentum is . However, .

Knowing , one can quickly solve for . One finds that . Thus, the velocity is . Plug this into the momentum equation to get, choice (D).  Alternate Solutions
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physgre
2015-10-15 08:09:27

 PinkieSans2018-10-23 01:24:01 对啊，能量、动量、质量的恒等式里面就是动质量，E^2=p^2*c^2+m^2*c^4并不是m 0 neo55378008
2012-03-23 12:05:47
You may have seen in a relativity course that you can make a right triangle out of rest mass, momentum, and total energy (this comes from the invariance of four-momentum).
Using Pythagorean's formula
The good thing about the triangle is some times they'll give you numbers for a special triangle (i.e. rest mass of 3, momentum of 4, energy must be 5). Knowing your special triangles could save you some time!
 neo553780082012-03-23 12:09:47 You can also work in units where c=1 just to make things faster QM320
2008-11-05 20:40:21
In the original solution -- what is ?
 p014k2009-02-15 17:47:42 \beta is v/c joshuacc1
2005-12-02 23:50:42
There\'s a simpler way to answer this question...

E^2 = (pc)^2 + (mc^2)^2

E = 2mc^2 = sqrt((pc)^2 + (mc^2)^2)

4(mc^2)^2 = (pc)^2 + (mc^2)^2

3(mc^2)^2 = (pc)^2

pc = sqrt(3)mc^2

choice (D)
 Walter2008-08-15 16:22:27 This is a great solution, I've just laid it out again.. The question says the total energy is twice the rest energy, which gives square that Equating the two expressions for gives make the subject and you get , answer D as required, and you never even went near a , and that's got to be a good thing.
 istezamer2009-11-01 15:09:48 lovely!! I solved it this way!! thanks! :D
 testtest2010-11-06 16:31:13 Even easier if you take c=1 : or just put the c back to get the answer.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$