Register to reply 
This question is only meant for the professionals 
Share this thread: 
#1
Mar2508, 01:25 PM

P: 27

1. The problem statement, all variables and given/known data
Two spaceships,A and B move toward each other on a headon collision course,According to an observer at rest,both spaceships have a velocity of 0.6c along the xaxis.At the time of observation(t=0),spaceship A has the same x value (x=0) as the observer,and spaceship B is at a distance of 100 km away from him. At what time will the collision occur for the observer at rest and for an observer on spaceship A? 2. Relevant equations my thought ok dont laugh.... T=(2L/c)/((1v/c)/(1+v/c))^(1/2) for the spaceship A and T=(2L/c)/(1v^2/c^2)^(!/2) for the observer at rest and L= 50 km when the collision happen thanks in advance 


#2
Mar2508, 02:00 PM

P: 27

well thanks no one can solve it what can i say this question is meant for the professionals lol



#3
Mar2508, 02:02 PM

P: 64

Use the formula for relative velocities (within Special Relativity)



#4
Mar2508, 02:03 PM

Mentor
P: 41,454

This question is only meant for the professionals
I don't quite understand how you arrived at those answers. Rather than just give your final conclusion, explain your thinking step by step.



#5
Mar2508, 02:04 PM

Mentor
P: 41,454




#6
Mar2508, 02:07 PM

P: 27

relative velocities what do you mean by that theres F ,T ,Energies ,... but V i cant find it i have this book concepts of modern physics 6 th edition i cant find it thanks in advance please help!!!!



#7
Mar2508, 02:10 PM

P: 27

ok never mind my answer is wrong or right if yes i will explain it dont worry my doctor will ask me the same thing uve asked and if yes i will reply how i thought about it if no let me think again about the question...



#8
Mar2508, 02:10 PM

Mentor
P: 41,454

Here's a hint: Distance = speed * time. From the observer at rest's point of view, what the distance that each ship must travel and what's the speed of each ship? From ship A's point of view, how far away is ship B and how fast is it coming at him? (That's the relative speed.)



#9
Mar2508, 02:16 PM

P: 27

From the observer at rest's point of view, what the distance that each ship must travel and what's the speed of each ship?
50 km,0.6c From ship A's point of view, how far away is ship B and how fast is it coming at him? 100km,0.6c ok but whats the formula i have to use it d=vt no thers another one what it is thanks in advance 


#10
Mar2508, 02:20 PM

Mentor
P: 41,454




#11
Mar2508, 02:27 PM

P: 27

Thank youuuuuu ok L=L0(1v^2/c^2)^(1/2) for lenght contration ok for the velocity i will try my best to find out how it will change any other hint it will be great (for the speed) thank youuuuuuuuu



#12
Mar2508, 02:30 PM

P: 27

i have to use relativistic momentum ?



#13
Mar2508, 02:30 PM

Mentor
P: 41,454

Read this: Einstein Velocity Addition



#14
Mar2508, 02:31 PM

P: 27

0.6c +0.6c but >c what can i do ?



#15
Mar2508, 02:38 PM

P: 27

i have to use this formula vx=(0.6+0.6)/(1+0.6^2) ?



#16
Mar2508, 02:43 PM

P: 27

ok vx=0 !!!!{*_*}!!!!!!



#18
Mar2508, 02:53 PM

P: 27

yoooooooooopiiiiiiii thank you then the answer for observer at rest T=50km/0.6c
and for the observer at A is T= (l=50km((10.6^2)^1/2))/0=infinty if yes im going to give a big thanksssssss 


Register to reply 
Related Discussions  
Communication 101 A must read for future professionals.  General Discussion  11  
Upcoming Conferences for Students & Young Professionals  Nuclear Engineering  0  
Question for Professionals  Academic Guidance  2  
For you professionals out there  Aerospace Engineering  5 