GREPhysics.NET: GR0177 #32

GR0177 #32

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Special Relativity $\Rightarrow$ }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).

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

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

å¯¹å•Šï¼Œèƒ½é‡ã€åŠ¨é‡ã€è´¨é‡çš„æ’ç‰å¼é‡Œé¢å°±æ˜¯åŠ¨è´¨é‡ï¼ŒE^2=p^2*c^2+m^2*c^4å¹¶ä¸æ˜¯mâ€†0

Reply to this comment

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
$E^2=p^2+m^2$

Reply to this comment

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

Reply to this comment

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.

Reply to this comment

Post A Comment!

You are replying to:

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!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone