Relativistic Time Dilation and Length Contraction in Overtaking Rockets

[little neutrino]

Homework Statement

Two rockets are each 1000m long in their rest frames. Rocket A, traveling at 0.800c relative to the earth, is overtaking Rocket B, which is moving in the same direction at 0.600c.

(i) According to the crew on B, how long does A take to completely pass? I.e. how long is it from the instant the nose of A is at the tail of B until the tail of A is at the nose of B?
(ii) Would the crew of A measure the same time interval according to their clocks? If so, explain why. If not, what is the time interval measured by the crew of A?

Homework Equations

l = l₀/γ --- Eqn 1
v = (v' + u)/(1 + uv'/c²) --- Eqn 2

The Attempt at a Solution

(i) Using Eqn 2, A is moving 0.384c relative to B.
Using Eqn 1, length contracted A from B's perspective = 0.923l₀
From B's perspective, B is stationary, and A is moving past at 0.384c.
Time taken = (1+0.923)l₀/0.384c = 1.67 * 10^-5 s
Is this correct?

(ii) I am not sure, but I think the crew would measure the same time interval because of symmetry...??

Thanks!

 

[TSny]

Your answers to both parts appear to me to be correct (including the symmetry argument). Since you are not confident of (ii), try working it out with the same type of reasoning as (i).

 

[little neutrino]

Your answers to both parts appear to me to be correct (including the symmetry argument). Since you are not confident of (ii), try working it out with the same type of reasoning as (i).

Hmm I thought it would be symmetrical because according to the crew of A, A is stationary and B is moving at -0.384c relative to A. Thus the length of B is contracted to 0.923l₀. So time taken for B to pass A is (1+0.923)l₀/0.384c = 1.67 * 10^-5, which is the same as (i). Is this correct?

 

[TSny]

Yes. Exactly.