GREPhysics.NET: GR9677 #38

GR9677 #38

Problem

GREPhysics.NET Official Solution

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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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Comments

darthvishous
2013-10-04 21:42:00

Put $c=1$ , and then use $E=\gamma m$ and $p=\gamma mv$ , thereby getting $v=1/2$ .

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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)

gravity
2010-11-04 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.

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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.

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tinytoon
2008-10-04 17:18:24

a much easier solution would be to use the equation Ev=pc^2

ali00
2014-06-04 07:11:06

it's true for everywhere.

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boundforthefloor
2006-11-23 22:33:13

After the division of the equations it should be $1/(2c) = v/c^2 \Rightarrow v = c/2$

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