Question

Bob is watching Anna fly by in her new high-speed plane, which Anna knows to be 60 m in length. As a greeting, Anna turns on two lights simultaneously, one at the front and one at the tail. According to Bob, the lights come on a different times, 40 ns apart.a) Which comes on first?
b) How fast is the plane moving?

Answer 1

Answer:

0.196*c

Explanation:

A good thing to remember is that once you have the position and the time in one frame we can figure out corresponding position and time in another frame in other frames given their relative velocities.

Anna moves in (S') frame and we are given two events synchronized in her frame with the distance between them, Hence using Lorentz transformation to show which event Bob frame (S) will see first:

Hence,

  $dt = y (dt' + \frac{v*dx}{c^2} ) \\\\dt' = 0\\dx' = 60 m\\\\t_2 - t_1 = \sqrt{\frac{1}{(1 - v^2/c^2) } } * (\frac{v*(x_2 - x_1)}{c^2} )$

we are assuming that dx' > 0 , hence we are saying that light emitted from front end of the plane is event number two and that coming from the back end is event number one. As the time in Bob's frame turns out to be positive, this means that event number two happened after event number one in his frame, in other words, the light emitted from the back end of Anna's  plane arrives first to Bob's eyes.

Given:

dt' = 40ns

dt = 0

dx' = 60

Solution:

$dt' = \sqrt{\frac{1}{(1 - v^2/c^2) } } * (\frac{v*dx'}{c^2} )\\\\\sqrt{1 - \frac {v^2}{c^2 } } = (\frac{v*dx'}{c^2*dt} )\\\\v^2 = \frac{c^4*dt^2}{c^2*dt^2 + dx'^2}\\\\v = \frac{c^2*dt}{\sqrt{c^2*dt^2 + dx'^2} }\\\\v = \frac{(3*10^8)^2*40*10^9}{\sqrt{(3*10^8)^2*(40*10^9)^2 + 60^2} }\\\\v = \frac{3.6*10^9}{61} \\\\v = 0.196 c$