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GR9677 #38
Problem
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Special Relativity$\Rightarrow$}Energy

The problem gives $\gamma mc^2 = 10$ and $\gamma mv = 5$. Divide the two to get $v/c^2 = 0.5 \Rightarrow v=c/2$, as in choice (D). The hardest part of the problem is remembering the definition of relativistic energy in terms of just the rest mass and $\gamma$.

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darthvishous
2013-10-04 21:42:00
Put $c=1$ , and then use $E=\gamma m$ and $p=\gamma mv$, thereby getting $v=1/2$.
mianghazanfar786
2010-07-20 05:39:19
An easy way to solve the problem:

we know the relation b/w Relativistic Momentum and Energy is

P = Ec

where P is Momemtum and E is Energy,

Just putting the values of P and E, we can find c = v
Only the choice (D)
 gravity2010-11-04 03:29:49 This is incorrect. This relationship is only valid for photons. p = $\gamma$ m v and E = $\gamma$ m c$^2$ These are two equations you essentially need to have memorized. Divide the two and heed the notes as to the units on p.
Prologue
2009-10-20 20:34:01
For those wondering what is wrong in the solution you need to include the units when dividing equations. So when dividing

$\gamma mv=5\frac{Mev}{c}$

$\gamma mc^{2}=10Mev$

$\frac{v}{c^{2}}=\frac{1}{2c}$

as said by boundforthefloor.
tinytoon
2008-10-04 17:18:24
a much easier solution would be to use the equation Ev=pc^2
 ali002014-06-04 07:11:06 it's true for everywhere.
boundforthefloor
2006-11-23 22:33:13
After the division of the equations it should be $1/(2c) = v/c^2 \Rightarrow v = c/2$

For those wondering what is wrong in the solution you need to include the units when dividing equations. So when dividing $\gamma mv=5\frac{Mev}{c}$ $\gamma mc^{2}=10Mev$ $\frac{v}{c^{2}}=\frac{1}{2c}$ as said by boundforthefloor.
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}\$ $\frac{a}{b}$