Help with length contraction and relativistic momentum please

shamille
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Homework Statement


A woman is 2.0 m tall and has a mass of 60 kg. She moves past an observer with the direction of the motion parallel to her height. The observer measures her relativistic momentum to have a magnitude of 2.30x1010 kg·m/s. What does the observer measure for her height?

Homework Equations


L=Lo √1 - (v2/c2)
p=mv/√1 - (v2/c2)

The Attempt at a Solution


I'm pretty sure that these are what the variables are
Lo= 2.0m proper length
p=2.3x1010
m=60kg
we want to solve for L

my problem is I don't know what v is, if I did i could find it.

I saw that L/Lo = mv/p
so L= Lomv/p right?

but i have no idea how to get v. i haven't had math in a while! any ideas or can you help me? I've tried an online equation solver for didn't work...
 
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You have p and m so you can solve the momentum equation for v.
 
well i got the answer as 1.24 but i had to use a function grapher and play around with the x and y mins and maxes to find what speed gives a momentum of 2.3E10! which was 0.79C

but i still want to know how to do this because there will be a test and i won't have the internet to help me

thanks in advance
 
Cyosis said:
You have p and m so you can solve the momentum equation for v.

yes but this is relativistic momentum not p=mv but
p=mv/√(1 - (v2/c2))
and i couldn't solve it for v... is it because I'm crap at math?
can you walk me through how to solve for it? maybe it's more algebra than physics but...
 
shamille said:
yes but this is relativistic momentum not p=mv but
p=mv/√(1 - (v2/c2))

There are still only three variables of which you know two.You know how to solve quadratic equations I assume?

<br /> \begin{align}<br /> p=mv \gamma=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}<br /> \\ <br /> p\sqrt{1-\frac{v^2}{c^2}}=mv<br /> \\ <br /> p^2(1-\frac{v^2}{c^2})=m^2v^2<br /> \end{align}<br />

Can you solve it from here on?
 
yes! wow as soon as i wrote that last message i figured it out. I don't know where my head was before... and then I used a - instead of +... gahhh

i'm sorry!

but thank you so much
 
You're welcome.
 
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