Special relativity:2 ships pass eachother

  • Thread starter Thread starter bfusco
  • Start date Start date
  • Tags Tags
    Ships
AI Thread Summary
Two spaceships, A and B, are traveling towards each other, with A measuring the time it takes for B's tip to pass as 5x10^-7 seconds. The discussion centers on using Lorentz transformations to calculate the relative speed of the ships and the time observed by someone on ship B as A passes. The initial calculations suggest that the first part can be solved without relativity, focusing on distance and time to find B's velocity in A's frame. For the second part, relativity is necessary to determine the length of A in B's frame and the corresponding time. The key takeaway is that measuring relative velocity involves plotting time against distance in the observer's frame.
bfusco
Messages
126
Reaction score
1

Homework Statement


Two space ships A (90m long) and B (200m long) travel towards each other. The person in ship A observes that it takes 5x10-7s for the tip of ship B to pass his ship. What’s the relative speed of the two ships? How long does a person sitting at the tip of ship B observe for his ship to pass ship A?

The Attempt at a Solution


at the moment in class we are on the topic of Lorentz transformations. so i am going to assume that this problem is to be done in that way.

i am having a hard time in general setting up any equations to these word problems as this new idea of reference frames is confusing me in terms of writing the equations.

so what i got so far is this (with help from my textbook, not really from any understanding of my own): i am going to start by indicating that spaceship A will be the fixed system K, and spaceship B be will be the moving system K'. since i am given Δt i figure that i would have to solve for the apparent velocity of the system u'. u'=dx'/dt', using the lorentz transformation i plug in dρ(x-vt)/dρ(t-vx/c^2) ρ=1/sq.rt.(q-(v/c)^2) and after some algebra i get u'=u-v/1-vx/c^2).

the time it takes for person sitting at the tip of ship B observe his ship to pass ship A I am going to call Δt'=t2'-t1'. using the lorentz transformation i get ρ(t2-vx2/c^2)-ρ(t1-vx1/c^2) and after some algebra i got ρ[(t2-t1) - (v/c^2)(x2-x1)]. t2-t1=Δt=5 x 10^-7s and x2-x1=90m (i think), and i would think i answered the question, but as i said before i kinda just got it from the textbook if someone could help explain some of this to me.
 
Physics news on Phys.org
To solve the first part, you don't need relativity. You know that in A's reference frame a particular point on B takes a given time to travel a given distance. From that it is trivial to find B's velocity in A's frame.

For the second part, you do need relativity. Use the velocity calculated above, find the length of A in B's frame, and find the time from that.
 
The first part boils down to the following question: If you are at rest in the K frame of reference, how do you measure the relative velocity of the K' frame of reference relative to your own frame of reference if the only measurement tools available to you are meter sticks and synchronized clocks? Well you focus on a specific point x'=constant in the K' frame of reference, and record the time t that this material point passes various locations x in your frame of reference. You then plot a graph of x versus t, and fit a straight line to the data. The relative velocity is slope of this line. That is,
v=(\frac{\Delta x}{\Delta t})_{x'=const.}
Or, in terms of the symbology of partial derivatives,v=(\frac{\partial x}{\partial t})_{x'}
In the case of your problem, the point you are focusing upon is the tip of ship B, and you are measuring the time interval between when the tip is at x = 0 and x = 90 m.

Chet
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top