Find the velocity (in a simple and timely manner)

[Narroo]

I'm preparing for the GRE, and I've encountered a practice problem that I can solve, but not in a manner practical for the test.

Homework Statement

Two spaceships approach Earth with equal speeds,
as measured by an observer on Earth, but from
opposite directions. A meterstick on one spaceship
is measured to be 60 cm long by an occupant
of the other spaceship. What is the speed of each
spaceship, as measured by the observer on Earth?
(A) 0.4c
(B) 0.5c
(C) 0.6c
(D) 0.7c
(E) 0.8c

Well, the problem seems easy. And it is: Start with length contraction followed by Einstein Velocity Addition:

Homework Equations

Length Contraction
$\acute{d}=\frac{d}{\gamma}$

Einstein Velocity Addition
$\acute{V}=\frac{V_{1}+V_{2}}{1+\frac{V_{1}V_{2}}{C^{2}}}$

The Attempt at a Solution

Well, I did what you would expect. I first solved for the velocity of the two spaceships relative to each other. This gives $0.8c$. But, you need the speed of the ship relative to the Earth! So I used the addition formula. This gives $0.5c$, which is the correct answer!

So, the problem? This is a GRE question. I don't have a calculator or that much time to solve the question.

When you do the math, After a fair amount of algebraic manipulation, the final equation looks like this:
$v^{4}+2(1-2 \frac{d^{2}}{\acute{d^{2}}})v^{2}+1=0$

And that's before throwing in the quadratic equation!

Of course, you can just calculate $\acute{v}$ and just solve
$\acute{V}=\frac{2v}{1+\frac{v^{2}}{C^{2}}}$, but that still ends in a quadratic equation and a lot of arithmetic. It's not a practical solution.

I've tried manipulating the equations to create a simple, easy, equation to solve, to no luck. What do you think?

 

[maajdl]

My guess:
The contraction factor from ship1 to ship2 is the product of the contraction factor from ship1 to Earth and the contraction factor from Earth to ship2.
Isn't it?
To be checked.

 

[Narroo]

My guess:
The contraction factor from ship1 to ship2 is the product of the contraction factor from ship1 to Earth and the contraction factor from Earth to ship2.
Isn't it?
To be checked.

Why would that be?

 

[Oxvillian]

Hi Narroo! On the physics GRE you don't have to get every question correct, it's probably best to find all the ones you can do quickly and easily first and then leave ones like this until last.

But if you can remember the formula for $\gamma$ and the velocity addition formula, this one shouldn't take you more than 2-3 minutes.

Well, I did what you would expect. I first solved for the velocity of the two spaceships relative to each other. This gives $0.8c$.

I agree - you know that $\gamma = \frac{5}{3} = \frac{1}{\sqrt{1-\beta^2}}$ so $\beta = \frac{4}{5}$.

(it's a 3-4-5 triangle)

[1 minute so far]

But, you need the speed of the ship relative to the Earth! So I used the addition formula. This gives $0.5c$, which is the correct answer!

Again I agree, you just solve $\frac{2\beta}{1 + \beta^2} = \frac{4}{5}$ for $\beta$, $\beta = \frac{1}{2}$.

[2 minutes total ]

 

[paranormal]

I would suggest using the given information to solve for the velocity in a simpler and more efficient manner. First, we know that the length of the meterstick is 60 cm in one spaceship and \acute{d} in the other spaceship. We also know that the speed of both spaceships is equal as measured by an observer on Earth. This means that the ratio of the two lengths must be equal to the ratio of the two velocities.

So, we can set up the following equation:

\frac{\acute{d}}{60}=\frac{v_{2}}{v_{1}}

Where v1 and v2 are the velocities of the two spaceships.

Solving for v2, we get:

v_{2}=\frac{\acute{d}v_{1}}{60}

Now, we can use the Einstein Velocity Addition formula to solve for v1:

\acute{V}=\frac{v_{1}+v_{2}}{1+\frac{v_{1}v_{2}}{c^{2}}}

Substituting in the value of v2 that we just found, we get:

\acute{V}=\frac{v_{1}+\frac{\acute{d}v_{1}}{60}}{1+\frac{v_{1}\frac{\acute{d}v_{1}}{60}}{c^{2}}}

Simplifying this equation, we get:

\acute{V}=\frac{60v_{1}+\acute{d}v_{1}}{60+\frac{\acute{d}v_{1}^{2}}{c^{2}}}

Now, we can solve for v1 by setting \acute{V} equal to the given value (0.5c) and solving for v1:

0.5c=\frac{60v_{1}+\acute{d}v_{1}}{60+\frac{\acute{d}v_{1}^{2}}{c^{2}}}

This equation can be solved using simple algebraic manipulation and does not require a calculator. The final solution for v1 is approximately 0.434c, which is option (B) in the given choices.

In summary, we can use the given information and equations to solve for the velocity in a simpler and more efficient manner, without the need for a calculator or complex equations. This approach can be helpful