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I found a derivation for relativistic momentum on this page:

http://en.wikibooks.org/wiki/Special_Relativity/Dynamics

I understand it all the way through, except the point where they have come to this equation.

m[itex]_{red} = [/itex][itex]\frac{m_{blue}}{\sqrt{1-(\frac{u(x,r)}{c})^{2}}}[/itex]

Where(as I understand correctly) m_{blue}is the mass according to the frame moving at the same x-velocity as the blue ball and where m_{red}is the mass of the red ball according to the frame moving with the same x-velocity as the blue ball.

Then they go to:

M =[itex]\frac{m_{0}}{\sqrt{1-(\frac{u}{c})^{2}}}[/itex]

Where u is now the relative velocity of the red ball for the moving frame.

But that is what I don't understand. How can they suddenly ignore the relativistic effects of the y-velocities? The first equation gives the mass of red ball in terms of the relative x-velocity of the red ball. Then how do they get to relativistic mass in terms of the relative velocity u? Because u_{x,red}is only the x component velocity and u_{red}has both a x-component and y-component.

This is my hypothesis:

They let both u_{y,red}and u_{y, blue}approach zero. So that the mass of the blue ball will correspond to the rest mass and that the relativistic effects of the y-velocities will cancel.

Thank you.

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# Questions regarding relativistic momentum

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