Go Back   Physics Help Forum > College/University Physics Help > Special and General Relativity

Special and General Relativity Special and General Relativity Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Apr 21st 2010, 09:11 AM   #1
Junior Member
 
Join Date: Mar 2010
Posts: 10
I seriously cannot do any Special Relativity Q's, humph =(

Hi everyone.

So my problem lies in that I understand the material covered, but when I attempt any questions it results in an epic fail. Could someone please lend me a hand? For example, could I post a question and have someone walk me through it please?

Here goes:

Q. Two spaceships A and B travel along the x axis in opposite directions towards one another. They have velocities (Beta)A= 0.8 and (Beta)B= -0.6, as measured in the Earth frame, respectively. At time t=0, ship A is at position xA=0 and ship B at xB= L = 1 light-second.

(a) At what time tc and at which position xc, do the ships cross as seen in the earth frame?
(b)Use Lorentz transformations to express at what time (tc)' and at which position (xc)' the two ships cross as seen in the frame of ship A.


Thank you very much
maple_tree is offline   Reply With Quote
Old Apr 27th 2010, 02:41 PM   #2
Senior Member
 
Join Date: Mar 2010
Location: Lithuania
Posts: 105
Can you find a) ?
zzzoak is offline   Reply With Quote
Old Oct 12th 2010, 08:37 AM   #3
Senior Member
 
Join Date: Aug 2010
Posts: 434
You say that distances are measured in light seconds but don't give any units for the speeds. Are we to assume they are measured in "light seconds per second"? Assuming that:

As seen from the earth, the two ships are closing with a net speed of 1.4 ls/s. They will cover the distance between them, 1 ls, in tc= 1/1.4 seconds. In that time, ship A will have moved a distance .8/1.4= 4/7 ls so xc= 4/7 ls.

As seen from A, ship B is closing on it at (.6c+ .8c){1+ (.6)(.8)/c^2}= 1.4c/1.48= 7/74 (of course, c is 1 ls per second). Also, as seen from A, the initial distance beween A and B is sqrt(- (.8c)^2/c^2)= sqrt(.36)= .6= 3/5 ls so they will pass, according to A, in (tc)'= (3/5)/(7/74)= 224/35 seconds. In that time, A will have moved .8(224/35)= 896/175 and (xc)' = 896/175 ls.

Last edited by HallsofIvy; Mar 27th 2011 at 08:57 AM.
HallsofIvy is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > Special and General Relativity

Tags
humph, relativity, special



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
General & Special Relativity mars shaw Special and General Relativity 1 Sep 5th 2009 08:10 AM
Special Relativity brentwoodbc Special and General Relativity 0 Jan 19th 2009 05:02 PM
Special Relativity ah-bee Special and General Relativity 0 Nov 10th 2008 02:43 AM
special relativity evabern Special and General Relativity 2 Oct 6th 2008 04:26 AM
Special Relativity Q's ah-bee Special and General Relativity 3 Aug 31st 2008 04:49 PM


Facebook Twitter Google+ RSS Feed