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  GR9677 #38
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Special Relativity}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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2013-10-04 21:42:00
Put c=1 , and then use E=\gamma m and p=\gamma mv, thereby getting v=1/2.NEC
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)
2010-11-04 03:29:49
This is incorrect.

This relationship is only valid for photons.

p = \gamma m v


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.
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


as said by boundforthefloor.
2008-10-04 17:18:24
a much easier solution would be to use the equation Ev=pc^2
2014-06-04 07:11:06
it's true for everywhere.

2006-11-23 22:33:13
After the division of the equations it should be 1/(2c) = v/c^2 \Rightarrow v = c/2Typo Alert!

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