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Special Relativity}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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Comments
physgre
2015-10-15 08:09:27
坑爹啊,m不是动质量吗
PinkieSans
2018-10-23 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
NEC
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!
neo55378008
2012-03-23 12:09:47
You can also work in units where c=1 just to make things faster
E^2=p^2+m^2
NEC
QM320
2008-11-05 20:40:21
In the original solution -- what is \beta?
p014k
2009-02-15 17:47:42
\beta is v/c
NEC
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)
Walter
2008-08-15 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.

istezamer
2009-11-01 15:09:48
lovely!! I solved it this way!! thanks! :D
testtest
2010-11-06 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.
NEC

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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!

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