GREPhysics.NET: GR9277 #85

GR9277 #85

Problem

GREPhysics.NET Official Solution

\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

0.86
1.0
1.4
1.5
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.

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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).

Andresito
2006-03-29 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.

Andresito
2006-03-29 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.

Reply to this comment

Comments

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

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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.

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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

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unoriginal5279
2007-04-11 09:24:43

apr2010
2010-04-08 12:25:33

Remarkable

thinkexist
2012-10-12 09:29:49

This is actually so simple and brilliant I cannot believe I have not thought of this before.

justin_l
2013-10-15 23:50:31

if something is remarkably brilliant, would it not be more surprising that you *do* think of it?

Reply to this comment

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).

Andresito
2006-03-29 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.

Andresito
2006-03-29 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.

ramparts
2009-08-06 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.

GREview
2009-08-30 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 $p$ .

Albert
2009-11-05 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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You say that "v=2c/3." But since $\beta^2$ =8/9 then v should be 2 $\sqrt{2}$ c/3

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