Special Relativity: Find speed of 3rd object.

[Ken Miller]

Homework Statement

Rest length of Ship B is twice that of Ship A.
Ship A travels at 0.5c relative to a "fixed" observer.
Ship B travels at such a speed that the same "fixed" observer measures Length B = Length A.
How fast is Ship B traveling?

2. Homework Equations [/B]
All given in statement

The Attempt at a Solution

I have no experience with Latex, so pardon my clumsy equations.
I know that the answer is that Ship B travels at 0.9c. But when I go through the math, I get things wrong.
1) BetaA=0.5
2) LA = gammaA * L0A (Length A = gammaA * Rest length A)
3) LB = gammaB * L0B.

But
4) L0B=2 * L0A (rest length B = twice that of A)
5) LB = LA (length B is measured to be same as that of A)

6) So LB = LA = gammaB * (2 * L0A)

Combining eqns 2) and 6), I get
gammaA = 2 * gammaB.

But is should be gammaA = 0.5 * gammaB.

I'm clearly making a bad conceptual mistake. Can you point it out to me?[/B]

 

[nrqed]

Homework Statement

Rest length of Ship B is twice that of Ship A.
Ship A travels at 0.5c relative to a "fixed" observer.
Ship B travels at such a speed that the same "fixed" observer measures Length B = Length A.
How fast is Ship B traveling?

2. Homework Equations [/B]
All given in statement

The Attempt at a Solution

I have no experience with Latex, so pardon my clumsy equations.
I know that the answer is that Ship B travels at 0.9c. But when I go through the math, I get things wrong.
1) BetaA=0.5
2) LA = gammaA * L0A (Length A = gammaA * Rest length A)
3) LB = gammaB * L0B.
[/B]

Watch out, it is the other way around (recall, length *contraction*) so LB=LOB /gammaB and LA=L0A/gammaA

 

[Ken Miller]

Oh, my goodness, of course. Blush! :)
Thank you for getting me out of my rut!