# Help with length contraction and relativistic momentum please!

1. May 21, 2009

### shamille

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
L=Lo √1 - (v2/c2)
p=mv/√1 - (v2/c2)

3. 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...

2. May 21, 2009

### Cyosis

You have p and m so you can solve the momentum equation for v.

3. May 21, 2009

### shamille

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

4. May 21, 2009

### shamille

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...

5. May 21, 2009

### Cyosis

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?

6. May 21, 2009

### shamille

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

7. May 21, 2009

### Cyosis

You're welcome.