Trying to help my eldest but cant work out how you get from the following to the solution below. (If I could have any help with the steps I can more forward with him!)
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First multiply both sides by the denominator of the first equation, then square both sides. Isolate v^2, then take the square root and rearrange to the form they show.Trying to help my eldest but cant work out how you get from the following to the solution below. (If I could have any help with the steps I can more forward with him!)
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This is not needed.Do you know:
Taylor's series expansion ? Remember (v/c)2 << 1
Yes - that's a over-kill.This is not needed.
Thank you for the help - appreciated. Bit of practise needed on my behalf- . Regards.Dr. Peterson literally gave step-by-step instructions
[math] p = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} * mv \implies \\ p *\sqrt{1 -\dfrac{v^2}{c^2}} = mv \implies \\ p^2 * \left ( 1 - \dfrac{v^2}{c^2} \right ) = m^2v^2 \implies \\ p^2 * \dfrac{c^2 - v^2}{c^2} = m^2v^2 \implies \\ p^2(c^2 - v^2) = c^2m^2v^2 \implies \\ c^2p^2 - p^2v^2 = c^2m^2v^2 \implies \\ c^2p^2 = c^2m^2v^2 + p^2v^2 = v^2(c^2m^2 + p^2). [/math]
That is where the plus sign comes in. Now solve for v.