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).

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Comments

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

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