spaceship A travelling at 0.6c is chasing another spaceship B travelling at 0.8c, which is of course trying to get away. Now spaceship A fires a missile at spaceship B, what minimum speed does the missile need (relatively to A) to have to hit B?
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spaceship A travelling at 0.6c is chasing another spaceship B travelling at 0.8c, which is of course trying to get away. Now spaceship A fires a missile at spaceship B, what minimum speed does the missile need (relatively to A) to have to hit B?
couldn't it be anything greater than 0 (considering that the missile is already moving forward due to the shiips motion)? And by the way...what is c?
yeah sorry, i didn't mention, c is the speed of light: 300 000 km/s
and for all those poll options, i meant "faster than"
0.800000000000000000000000000000000000000000000000000000000000000000000000000000000..........etc etc......0001c
:p
isn't it
0.800000000000000000000000000000000000000000000000000000000000000000000000000000000..........etc etc......00001c
chenko?
well anything above 0.8c is correct, the minumun is 0.8 then 0 for infinity and a 1 on the end.....
Hehe so who else voted for me?
relatively to A, that is the velocity at which the missile travels from A
Anything is faster than you so it was a logical answer! :p
Kedaman: So I got It right then? :)
look at this:
this is relatively to start point, the point where parksie travels from at 0.6c
[parksie]>........[chenko]>
......[parksie]>................[chenko]>
............[parksie]>..*....................[chenko]>
..................[parksie]>.....*.........................[chenko]>
........................[parksie]>........*..............................[chenko]>
this is relatively to A, in this case parksie
[parksie]>........[chenko]>
[parksie]>................[chenko]>
[parksie]>..*....................[chenko]>
[parksie]>.....*.........................[chenko]>
[parksie]>........*..............................[chenko]>
now looking at this, what speed does * need to travel to catch up chenko.
so that it will look like this:
* *. *
*...* .*
. *** .*
*...*.*
After the missile left Parksie It does not matter how fast Parksie is, so I'd say just more than the difference between the Parksie and Cenki and thats 0.2c - I'm sure there's something wrong with that so I vote for "very fast" though :)
It dosent matter how fast parksie is when it leave him.
Im still sure its just above 0.8c
try that chenko and jump out of the car when it's travelling at 90km/h
Okay, this is smelling like a trick question. Why is the answer not >.2c?
yep, but i'm waiting for a smartass to show up and tell me
Yea I might fall at near enough where I left the car, but If a missile was fired, It would be forced to accelerate, we arnt really taking speed into account now are we?Quote:
Originally posted by kedaman
try that chenko and jump out of the car when it's travelling at 90km/h
Ok I have it
Spaceship B runs out of fuel, and the missle can travel at any speed above 0 to hit it!
Spaceship B, doesn't happen to have any fuel, it still goes at 0.8c because there's nothing stopping it.
suppose that the missile have enough fuel to catch up Spaceship B
kedaman, you naughty boy. You KNOW that as velocities approach the speed of light they are no longer additive.
Chenko, close but no cigar, hehehehe
Cheers,
P.
I voted. I'm still lost, but I voted.
Answer is 0.38 C according to Principle of Relativity
As a side bar...
if you started timers on both ships and the missile at time of launch and measured time 'til impact, you would get three different readings. All of these readings would also be short of the reading taken by a "stationary" observer.
*shrug* Fun with time dilation.
This one you will have to explain.Quote:
Originally posted by Active
Answer is 0.38 C according to Principle of Relativity
There are Three Inertial Frames in the description,
We the observer is in the Third frame.
According to our Measurments
A is Moving at 0.6c
B is moving at 0.8c
The minimum speed required for a missile from A to
hit B will be the Speed of B relative to A.
If A is moving with velocity u relative to observer and B is moving with velocity w relative to observer then the velocity of B relative to A is given by,
v = (w - u)/(1 - wu/c2)
Using this formula....
v = (0.8c - 0.6C)/(1-(0.8*0.6))
v = 0.2 C / 0.52
v = 0.3846 C
--
getting v from this formula, you get 0.38c
0.6c+v
__________
1+0.6cv/c^2
i suppose you did this Active
i'm running out of time though, i have another relativistic problem including an angle, and i'm not smart enough to solve it:
A light beam travels at theta' to the x' axis of frame S' which moves at velocity v in the +x direction of frame S. If theta is the angle measured to the x axis, show that:
cos theta =
cos theta' + Beta
______________
1+Beta cos theta'
where Beta = v/c
as a hint it says: Use the Lorentz transformation and note that cos theta = dx/(cdt).
(from Harris Benson/University Physics 39-Problem 7.)
slow as usual (me) :p
i'm really sorry to get your attention like this but i thought you might help me with this problem
Okay, what about the universe causes this to be true? For this to rest well with my soul (humor intended), I have to assume that this is true for terrestrial speeds like 60mph (or 100kph). Is it? If not, why not?Quote:
Originally posted by Active
There are Three Inertial Frames in the description,
We the observer is in the Third frame.
According to our Measurments
A is Moving at 0.6c
B is moving at 0.8c
The minimum speed required for a missile from A to
hit B will be the Speed of B relative to A.
If A is moving with velocity u relative to observer and B is moving with velocity w relative to observer then the velocity of B relative to A is given by,
v = (w - u)/(1 - wu/c2)
Using this formula....
v = (0.8c - 0.6C)/(1-(0.8*0.6))
v = 0.2 C / 0.52
v = 0.3846 C
--
Well..I have to admit there is a flaw in my argument
It should be atleast more than that ...otherwise it wouldQuote:
The minimum speed required for a missile from A to hit B will be the Speed of B relative to A.
never reach B
yeah it's a bit hard to evaluate the answer, but you can get close, i don't like that kind of excercises. I guess you have no luck with the one i posted next?
Well I have to say...Gimme some more time..
BTW ...on the Question of CyberThug..
Well... 100kph is 27.778m/s while c is 3x10^8 m/sQuote:
for this to rest well with my soul (humor intended), I have to assume that this is true for terrestrial speeds like 60mph (or 100kph). Is it? If not, why not?
that means your 100kph vehicle is 9.259x10^-8 C
Now f we measure two vehicles moving at 30 m/sec
and 35m/sec . they will be like 1 x 10^-7 and 1.16 x 10^-7.
Without relativity we say B is moving at 5m/sec relative to A
Using the above formula again we get the same
because since the two values are really small their
products will be further small.
There fore... the denominator will be 1- very small value...
which is almost equal to 1
So we observe no bigger changes.
thanks for your time :) i guess youre not travelling close to c relatively to me so you might get it solved in 13,5 hours :p
No complex solution here either.
a photon travelling at C in direction theta' in a reference system S' which moves with a speed v in direction x+ in reference system S. And i need to get the angle theta in reference system S. the issue is again relativistic addition, y and z components in the systems are the same, x varies, so we need to get the Vx' component tranlated from S' to S.
Vx' is cos(theta')c and thefore
Vx=
cos(theta')c+v
__________________
1+cos(theta')c*v/c^2
now since we know that c does not vary in any reference system:
cos(theta)=Vx/c=
cos (theta') + v/c
______________
1+cos (theta')*v/c
some basic knowledge of vector calculus and relativistic addition
simple.. eh?
I failed to think like that..since the question doesn't sayQuote:
y and z components in the systems are the same
like that .... hehehe.... :D
yeah well it's a bit unclear, but "moves in x+ direction of Frame S" should be understandable :)
It wasn't specified that spaceship A was chasing spaceship B in a linear fashion. If A is travelling tangential to B on an intercept vector, then that casts a whole new light on the matter.
So nuh!
LOL, guess how much fuel is needed to turn ship A to adapt to B's course!
"As fast as jeff's 56k modem" to the poll options :rolleyes:
seeing as it took 5 mins for this thread too load up.:eek: :eek: :eek:
Depends on which end of Parksoe it's leaving:rolleyes:Quote:
Originally posted by Fox
After the missile left Parksie It does not matter how fast Parksie is, so I'd say just more than the difference between the Parksie and Cenki and thats 0.2c - I'm sure there's something wrong with that so I vote for "very fast" though :)