How?

[wifiemail]

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!)

 

[Deleted member 4993]

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!)
View attachment 34059

Do you know:

Taylor's series expansion ? Remember (v/c)² << 1

 

[Dr.Peterson]

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!)
View attachment 34059

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.

Please show whatever work you have done. We need to see where you need help.

Do you know:

Taylor's series expansion ? Remember (v/c)² << 1

This is not needed.

 

[wifiemail]

 

[Otis]

Hi. When Dr. Peterson said denominator, he was referring to the big ratio – the one whose numerator is 1.

So, to begin, multiply both sides by [imath]\sqrt{1 - \frac{v^2}{c^2}}[/imath].

We can't multiply by [imath]c^2[/imath] because it's part of the radicand (the expression inside the radical sign). In other words, a radical sign is a grouping symbol; its contents go together.


[imath]\;[/imath]

 

[JeffM]

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.

 

[Deleted member 4993]

This is not needed.

Yes - that's a over-kill.

Since "no -work" shown by OP and no indication of level of knowledge (of the student) was given for intermediate/advanced algebra - that was my (probably mis-guided) way of getting a response.

 

[wifiemail]

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.

Thank you for the help - appreciated. Bit of practise needed on my behalf- . Regards.