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GR9277 #85
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{85}
A free electron (rest mass $m_e-0.5MeV/c^2$ has a total energy of 1.5MeV. Its momentum p in units of MeV/c is about

1. 0.86
2. 1.0
3. 1.4
4. 1.5
5. 2.0

Special Relativity$\Rightarrow$}Momentum

Given a total energy of $\gamma mc^2 = 1.5MeV$ and the rest mass of the electron to be $m_e = .5MeV/c^2$, one can figure out $\gamma =3$.

The momentum is given by $p=\gamma mv = 3mv=3v/2 (MeV/c^2)$.

Solve for the velocity from $\gamma= \frac{1}{\sqrt{1-\beta^2}} \Rightarrow \gamma^2 = \frac{1}{1-\beta^2} \Rightarrow 1 = \gamma^2(1-\beta^2) \Rightarrow \beta^2 = 1-\gamma^{-2} \Rightarrow \beta^2=8/9$. Thus, the velocity is $v=2c/3$.

Plugging this into the equation for momentum, one gets $p=3/2\times \sqrt{8/9}=\sqrt{2}$, and thus its momentum is about 1.4, as in choice (C).

Alternate Solutions
unoriginal5279
2007-04-11 09:24:43
When short on time (as I always am) I try to avoid calculations whenever possible. If you remember that E$\c^2$= m$\c^2$ c$\c^4$ + (p c)$\c^2$ then you already know that the momentum must be less than the energy, but greater than the difference between the energy and m c$\c^2$ . This eliminates all but the correct answer.
Andresito
2006-03-29 15:15:31
You could always use the general equation,

E^2 = (p*c)^2 + (m*c^2)^2

Solve for p having E = 3/2, and m*c^2 = 1/2

to obtain 1.4 as in (C).
 Andresito2006-03-29 15:21:28 The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
 Andresito2006-03-29 15:24:03 The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
EPdropout
2012-11-08 17:12:29
I used T=p$\^2$/2m=($\gamma$-1)mc$\^2$, after finding $\gamma$=3
=>p$\^2$=2m*(2/3)E
=>p$\^2$=2.0 (MeV^2/c^2)
=>p =1.4 MeV/c
rrfan
2011-11-06 17:41:09
A useful shortcut on these types of relativity problems:

The solution to $\gamma=\frac{1}{\sqrt{1-\beta^2}$ is the following: $\beta=\sqrt{1-\frac{1}{\gamma^2}$.

Even if you don't memorize this, it is straightforward to derive and is often useful on multiple problems, saving some time on the algebra.
dinoco
2010-11-07 07:52:37
You say that "v=2c/3." But since $\beta^2$=8/9 then
v should be 2$\sqrt{2}$c/3
unoriginal5279
2007-04-11 09:24:43
When short on time (as I always am) I try to avoid calculations whenever possible. If you remember that E$\c^2$= m$\c^2$ c$\c^4$ + (p c)$\c^2$ then you already know that the momentum must be less than the energy, but greater than the difference between the energy and m c$\c^2$ . This eliminates all but the correct answer.
 apr20102010-04-08 12:25:33 Remarkable
 thinkexist2012-10-12 09:29:49 This is actually so simple and brilliant I cannot believe I have not thought of this before.
 justin_l2013-10-15 23:50:31 if something is remarkably brilliant, would it not be more surprising that you *do* think of it?
Andresito
2006-03-29 15:15:31
You could always use the general equation,

E^2 = (p*c)^2 + (m*c^2)^2

Solve for p having E = 3/2, and m*c^2 = 1/2

to obtain 1.4 as in (C).
 Andresito2006-03-29 15:21:28 The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
 Andresito2006-03-29 15:24:03 The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
 ramparts2009-08-06 22:54:12 Well, it's better to use in problem 70, but I think it's pretty clear that for this question, the E^2 equation takes a lot less calculation than the "official" answer.
 GREview2009-08-30 19:22:53 It may simplify things to not even think about the $c$'s: $E = \sqrt{p^2+m^2}$ Plugging $E=3/2$ and $m=1/2$, we can solve for $p$.
 Albert2009-11-05 00:32:16 Hey, I see what you did there, you used the c's in the denominator of the units right along side the values, smart work. Best solution!

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