Is there something wrong with this reasoning:(adsbygoogle = window.adsbygoogle || []).push({});

First the question:

"A rocket A leave earth at a speed v at a time where the clocks onboard and those on earth read 0. After a time T (for the earth's clock), a second rocket is lauched with a speed u > v. The second rocket catches up with the first one at a time t such that (t - T)u = tv, or equivalently, t = uT/(u-v)."

a) For the clocks on A, when does B leave earth?

MY ANSWER: The time between the events "A leaves" and "B leaves" is proper on earth. It is T on earth; therfor it is given on A by

[tex]t_A_1= T\gamma _{AT} = \frac{T}{\sqrt{1-(v/c)^2}}[/tex]

b) For the clocks on A, when does B catches up with A?

MY ANSWER: By Lorentz,

[tex]t_A_2 = \frac{1}{\sqrt{1-(v/c)^2}}(t-v(vt)/c^2) = t\sqrt{1-(v/c)^2}[/tex]

d) Deduce the speed of B as seen in frame A. Does the result agree with the law of addition of speed?

MY ANSWER: Speed of B = distance traveled in catching up with A / time took to do so

[tex]\Rightarrow v_{B \ in \ A} = \frac{vt_A_2}{t_A_2 - t_A_1}=\frac{vt\sqrt{1-(v/c)^2}}{t\sqrt{1-(v/c)^2} - T/\sqrt{1-(v/c)^2}} = \frac{vt}{t - T/(1-(v/c)^2)}[/tex]

[tex]=\frac{uv}{(u-v)(\frac{u}{u-v}-\frac{1}{1-(v/c)^2})}[/tex]

But according to Mapple (and to me), this is not equal to the predicted

[tex]v_{B \ in \ A} = \frac{u-v}{1-uv/c^2}[/tex]

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: SR question

**Physics Forums | Science Articles, Homework Help, Discussion**