This question is only meant for the professionals

[tarekkkkk]

Modern physics special relativity

Homework Statement

Two spaceships,A and B move toward each other on a head-on collision course,According to an observer at rest,both spaceships have a velocity of 0.6c along the x-axis.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?

Homework Equations

my thought ok don't laugh... T=(2L/c)/((1-v/c)/(1+v/c))^(1/2) for the spaceship A
and T=(2L/c)/(1-v^2/c^2)^(!/2) for the observer at rest and L= 50 km when the collision happen thanks in advance

 

[tarekkkkk]

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

 

[eys_physics]

Use the formula for relative velocities (within Special Relativity)

 

[Doc Al]

I don't quite understand how you arrived at those answers. Rather than just give your final conclusion, explain your thinking step by step.

 

[Doc Al]

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

You're the one who has to solve it, not me! (I can solve it just fine. )

 

[tarekkkkk]

relative velocities what do you mean by that there's F ,T ,Energies ,... but V i can't find it i have this book concepts of modern physics 6 th edition i can't find it thanks in advance please help!

 

[tarekkkkk]

ok never mind my answer is wrong or right if yes i will explain it don't 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...

 

[Doc Al]

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.)

 

[tarekkkkk]

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 what's the formula i have to use it d=vt no thers another one what it is thanks in advance

 

[Doc Al]

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

That's correct. That's all you need to solve for the time.

From ship A's point of view, how far away is ship B and how fast is it coming at him?
100km,0.6c

Incorrect: For the distance, consider length contraction; for the speed, consider the relativistic addition of velocity to find the relative speed.

 

[tarekkkkk]

Thank youuuuuu ok L=L0(1-v^2/c^2)^(1/2) for length 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

 

[tarekkkkk]

i have to use relativistic momentum ?

 

[Doc Al]

Read this: http://hyperphysics.phy-astr.gsu.edu/Hbase/relativ/einvel.html#c1"

 

[tarekkkkk]

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

 

[tarekkkkk]

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

 

[tarekkkkk]

ok vx=0 !{*_*}!

 

[Doc Al]

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

That's the one.

 

[tarekkkkk]

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

 

[tarekkkkk]

loooooooollllllllll i thought vx=0 think again

 

[tarekkkkk]

my mistake the answer is ? for observer at rest T=50km/0.6c
and for the observer at A is T= (l=50km((1-0.6^2)^1/2))/(vx=(0.6+0.6)/(1+0.6^2)) if yes I am going to give a big thanksssssss

 

[tarekkkkk]

just tell me if yes or no because its 10:00 pm over here and tomorow i have to give it to my doctor no offense but thank you just tell me yes and i will study more...

 

[tarekkkkk]

lol i will take that for a silent yes thank youuuuuuuuuuu

 

[Doc Al]

correction

for observer at rest T=50km/0.6c

Right.

and for the observer at A is T= (l=50km((1-0.6^2)^1/2))/(vx=(0.6+0.6)/(1+0.6^2))

I was going to say almost right, but I'm afraid we neglected a key factor of this problem. (Almost right because from A's viewpoint, B must cover the entire distance not half of it.) While the relative velocity is correct, and the idea of length contraction is also correct, what is not correct is the idea that A agrees with the rest observer that B is at position x=100 km when A passes x=0. According to A, by the time A passes x=0, B has already passed the point x=100km and thus is even closer.

The efficient way to solve for the time that the collision occurs according to A is to make use of the Lorentz transformation:
$$\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$$

(I realize that this is a bit subtle if you are just learning special relativity. But that's the way it is.)

 

[tarekkkkk]

ok deltaT=T of observer at rest
v=Vx(0.6+0.6)/(1+0.6^2))=0.88c
and for the gamma the v inside it =0.88c ?
deltaX= (l=50km((1-0.6^2)^1/2))/
thanks in advance

 

[tarekkkkk]

man what to do hm.m...

 

[Doc Al]

ok deltaT=T of observer at rest
v=Vx(0.6+0.6)/(1+0.6^2))=0.88c
and for the gamma the v inside it =0.88c ?
deltaX= (l=50km((1-0.6^2)^1/2))/
thanks in advance

What you are transforming are $\Delta x$ and $\Delta t$ according to the rest observer to the time ($\Delta t'$) according to the A observer. So we need their relative velocity, which is 0.6c.

Here's a simpler way to see what's going on: The collision takes place at x = 50 km, so all we need to figure out is the time it takes for A to reach that point (according to A). This is just Distance/speed as seen by A. The speed is 0.6c and the distance is length contracted. (This gives the same answer as using the Lorentz transformation above, of course.)

 

[tarekkkkk]

ok thank you doctor womorow i will give it to my doctor and i will tell what happened with me thank you ...