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GR0177 #32
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Special Relativity$\Rightarrow$}Momentum

The total energy of a particle of mass m is $E=2mc^2=\gamma mc^2$, since the problem supplies that it is equal to twice the rest mass. This means that $\gamma =2$.

The relativistic momentum is $p=\gamma mv=2mv$. However, $v\neq c$.

Knowing $\gamma$, one can quickly solve for $\gamma=2 = 1/\sqrt{1-\beta^2}$. One finds that $\beta^2 = 3/4$. Thus, the velocity is $v=\sqrt{3}/2 c$. Plug this into the momentum equation to get, choice (D).

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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 $E^2=p^2c^2+m^2c^4$
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 $E^2=p^2+m^2$
QM320
2008-11-05 20:40:21
In the original solution -- what is $\beta$?
 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.. $E^{2} = c^{2}p^{2} +( m_{0}c^{2})^{2}$ The question says the total energy is twice the rest energy, which gives $E = 2 m_{0}c^{2}$ square that $E ^{2}= 4 ( m_{0}c^{2})^2$ Equating the two expressions for $E^{2}$ gives $4 ( m_{0}c^{2})^2 = c^{2}p^{2} +( m_{0}c^{2})^{2}$ make $p$ the subject and you get $p=\sqrt{3}mc$, answer D as required, and you never even went near a $\gamma$, 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 : $E = 2m$ $p^2 = E^2 - m^2 = 4 m^2 - m^2 = 3 m^2$ or $p = \sqrt{3}m$ just put the c back to get the answer.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$