# Help with length contraction and relativistic momentum please!

## 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...

Cyosis
Homework Helper
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

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...

Cyosis
Homework Helper
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?

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

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

Cyosis
Homework Helper
You're welcome.