GREPhysics.NET: GR0177 #79

GR0177 #79

Problem

GREPhysics.NET Official Solution

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Special Relativity $\Rightarrow$ }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).

Alternate Solutions

Deafro 2012-01-24 15:45:45	E^2=sqrt(p^2c^2+(mc^2)^2) E=10 GeV p=8 GeV/c mc^2= rest energy Rearrange the formula and solve for mc^2. Reply to this comment
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. Reply to this comment
Ning Bao 2008-02-01 07:43:40	Set c=1 -> pythagorean triple. Reply to this comment

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.

Reply to this comment

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.

Reply to this comment

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.

Reply to this comment

Ning Bao
2008-02-01 07:43:40

Set c=1 -> pythagorean triple.

Reply to this comment

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

Reply to this comment

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.

Reply to this comment

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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.

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