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Special Relativity}Rest Mass

The total relativistic energy is 10GeV=\gamma mc^2=\sqrt{p^2c^2+m^2c^4}.
The total relativistic momentum is p=8GeV/c = \gamma mv.

Plugging the momentum into the first equation, one has (8^2+m^2c^2)c^2=10^2 \Rightarrow m^2c^4 = 100-64 = 36 \Rightarrow m = 6GeV/c^2, as in choice (D).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
Deafro
2012-01-24 15:45:45
E^2=sqrt(p^2*c^2+(m*c^2)^2)

E=10 GeV
p=8 GeV/c
mc^2= rest energy

Rearrange the formula and solve for mc^2.
Alternate Solution - Unverified
mike
2009-11-06 18:34:27
Another way of doing it:

\gamma m v = 8 /c

and

\gamma m c^2 = 10. Multiply the first eqn by c^2 and divide the two to get:

 v = \frac{4}{5}c

Plug in to \gamma to find \gamma = \frac{5}{3}.

Plug back in to the momentum equation:

\frac{5}{3} m \frac{4}{5} c = 8 / c

thus  m = 6.

I omit units, but it should be easy enough to follow. It seems a little longer, but its much less thinking once one remembers the relativistic energy/momenta equations.
Alternate Solution - Unverified
Ning Bao
2008-02-01 07:43:40
Set c=1 -> pythagorean triple.Alternate Solution - Unverified
Comments
Deafro
2012-01-24 15:45:45
E^2=sqrt(p^2*c^2+(m*c^2)^2)

E=10 GeV
p=8 GeV/c
mc^2= rest energy

Rearrange the formula and solve for mc^2.
Alternate Solution - Unverified
mike
2009-11-06 18:34:27
Another way of doing it:

\gamma m v = 8 /c

and

\gamma m c^2 = 10. Multiply the first eqn by c^2 and divide the two to get:

 v = \frac{4}{5}c

Plug in to \gamma to find \gamma = \frac{5}{3}.

Plug back in to the momentum equation:

\frac{5}{3} m \frac{4}{5} c = 8 / c

thus  m = 6.

I omit units, but it should be easy enough to follow. It seems a little longer, but its much less thinking once one remembers the relativistic energy/momenta equations.
Alternate Solution - Unverified
astrodoo
2008-09-17 01:38:23
manasi, the value of 64 (<- the square of momentum) has a dimension of GeV^{2}/c^{2}, so you can easily catch the yosun's solution if you consider the dimension of each terms. NEC
Ning Bao
2008-02-01 07:43:40
Set c=1 -> pythagorean triple.Alternate Solution - Unverified
Gaffer
2007-10-24 08:42:58
There is a typo in the soln, which is why manasi is confused. Yosun used the \c^2  in p^2 c^2 to cancel the GeV/c in the momentum. But then she pulled it out again inthe next line.

No biggie.

It should go:

Plugging the momentum into the first equation, one has \8^2 (c/c)^2  + m^2 c^4 = 10^2
and then continue as she concludes
Typo Alert!
manasi
2007-09-28 12:05:48
hi, if i m not mistaken shudnt it be m^2 c^4 = 100-64c^2 ????

hope it appears rite m new to latex!!
hey thanks a lot yosun.. ur solutions are very helpful!! :)
astrodoo
2008-09-17 01:46:59
manasi, the value of 64 (<- the square of the momentum) has a dimension of GeV^{2}/c^{2}, so you can easily catch Yosun's solution in which c^{2} term is disappeared if you look dimensions of each term.
NEC

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