# Homework Help: Relativity collision problem

1. May 17, 2012

### snellslaw

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

According to an Earth observer, rocket 1 has speed 0.800c, rocket 2 has speed 0.600 c and rockets are initially 2.52x10^12 m apart. The rockets are approaching each other.
According to each Rocket, how long is it before they collide?

Rocket 1: 60 min
Rocket 2: 80 min

2. Relevant equations

From velocity addition equations, I found that the velocity of each rocket according to the other rocket is 0.946.

3. The attempt at a solution

Now I don't understand why the rockets think that the collision occurs in different times. Isn't it 2.52x10^23 / 0.946c = 8879 s for each? (I know this is wrong since it doesn't match the answer but not why.)
I believe that each rocket thinks the other is 2.52x10^12 m away... is this wrong? (I think it is but I don't know why)

2. May 18, 2012

### vela

Staff Emeritus
You have to be more careful. The distance $2.52\times 10^{12}\text{ m}$ is measured in the Earth rest frame. The speed you calculated is measured in the rockets' rest frame.

You also have to remember that events simultaneous in the Earth frame aren't necessarily simultaneous in the rockets' frames.

Last edited: May 18, 2012
3. May 18, 2012

### snellslaw

Thanks! How would I calculate the distance in the rocket's rest frame (for each rocket)?

4. May 18, 2012

### snellslaw

By the way, I know how to solve this problem by first calculating the time of collision as seen from the Earth. I just want to know how to calculate it using distances and speeds from the point of view of the rockets as well.

5. May 18, 2012

### vela

Staff Emeritus
Try calculating the space-time coordinates of the other ship as observed in the rest frame of the first ship.

6. May 18, 2012

### snellslaw

By length contraction, ship A thinks B is a distance 2.52x10^12/√(1-0.8^2) away
thus the time is 2.52x10^12/√(1-0.8^2) / (0.946c) = 14799... still not right :(

7. May 19, 2012

### vela

Staff Emeritus
No, you need to calculate both the space and time coordinates of the other ship. It's not simple length contraction.

8. May 19, 2012

### snellslaw

9. May 19, 2012

### vela

Staff Emeritus
Use the Lorentz transformations.

10. May 21, 2012

### vela

Staff Emeritus
Let S denote the rest frame of Earth and S', the rest frame of rocket 1.

In the Earth's frame, the rockets are initially $2.52\times 10^{12}\text{ m}$ apart. So we'll set up the coordinate systems to rocket 1 is passing through x=0 when t=0. We'll call this space-time point event A. Rocket 2 is therefore at $x=2.52\times 10^{12}\text{ m}$ when t=0. This is event B.

In rocket 1's frame, we'll set up the coordinates so that event A is at x'=0 and t'=0. When we set up the coordinate systems this way (so that the origins of the two coordinate systems coincide), the coordinates are related by the Lorentz transformations:
\begin{align*}
x' &= \gamma(x-\beta ct) \\
ct' &= \gamma(ct - \beta x)
\end{align*} where $\beta = v/c$ is the velocity of S' relative to S and $\gamma = 1/\sqrt{1-\beta^2}$. Using these equations, you can calculate the space-time coordinates of event B in S'. Once you have them, you can calculate when the two rockets will collide, as seen by an observer on rocket 1.