Special relativity spaceships time question

[thenewbosco]

Here is the problem and my solution,

Two spaceships each measuring 100m in its own rest frame, pass by each other traveling in opposite directions. Instruments on board spaceship A determine that the front of spaceship B takes 5x10^-6 s to traverse the full length of A.
a) what is the relative velocity of the two spaceships
b) how much time elapses on a clock on spaceship B as it traverses the full length of A?

So for this i thought that since A is doing the measurement, it takes 5 x 10^-6 s for B to pass 100 m. Since according to A, it is still 100m. then the velocity is just 100/5 x 10^-6. is this correct?

then i took this speed and put it into the time dilation equation with the 5x10^-6s to get the answer for part b.

is this correct and if not how should i go about this problem, thanks.

 

[P3X-018]

The first one seems correct.
In b if you by time dilation equation mean $T = \gamma T_0$ where T is the time obsered in b, then this will be incorrect. I had a similar problem (thread: Relativity). You must remember that the events don't occur at the same location, that is $\Delta x \neq 0$ and $\Delta x' \neq 0$. You have $\Delta x =100$ and $\Delta t =5\times 10^{-6}$ in A's reference frame, and since B is moving with a relative velocity v, you must use the Lorentztransformation to transform this time $\Delta t$ to B's referenceframe, which will be

$$\Delta t' = \gamma(v)\left(\Delta t-\Delta x\,\frac{v}{c^2}\right)$$

You can also solve this problem by considering the event from B. In B referenceframe the length of A's spaceship will be $L = L_0/\gamma(v)$, and since A is moving with a relative velocity v with respect to B, the time it will take to for B to traverse THAT length of A, will be

$$\Delta t' = \frac{L}{v} =\frac{L_0}{v\gamma(v)}$$

 

[kyle8921]

I can confirm that your approach to this problem is correct. Special relativity states that the relative velocity between two objects can never exceed the speed of light, so your calculation of 100/5x10^-6 is correct. This means that the relative velocity between the two spaceships is 2x10^10 meters per second.

For part b, you correctly used the time dilation equation to calculate the time elapsed on a clock on spaceship B as it traverses the full length of A. This equation, t' = t/sqrt(1-v^2/c^2), takes into account the relative velocity between the two objects and the speed of light. Using the velocity you calculated in part a, you should get a time dilation factor of approximately 1.0000000000000000000005. This means that the clock on spaceship B will experience a very small amount of time dilation, but it will still be noticeable.

Overall, your solution to this problem is correct and demonstrates a good understanding of special relativity. Keep up the good work!