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    i dont know - i havnt been taught that example (but then we didnt get taught much at all )

    PANTS you're right!! i had it stuck in my head it was the morning! oh thats so annoying!! wanted to get it over with! <sigh> another day to wait...
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    (Original post by Fluffstar)
    i dont know - i havnt been taught that example (but then we didnt get taught much at all )

    PANTS you're right!! i had it stuck in my head it was the morning! oh thats so annoying!! wanted to get it over with! <sigh> another day to wait...
    I'm glad it is in the afternoon! I need the extra time to study! I don't know anything! That Cosmology book is so confusing! I have done past papers, and looked in the book for the answer, but have been unable to find the answer there!

    Can somebody explain to me when you actually need to use the equation:

    Square root (1-v^2/c^2)

    It is in the book, and I'm sure in past papers I've done it's needed, but I don't know how to apply it. (if you know what I mean)

    here is an example: (from june 02 paper)

    A Pion has a mean lifetime of 26ns when at rest. They have a speed of 0.80c relative to the laboratory:

    What is the mean lifetime of the pion as measured by stationary observers in the lab?

    Any help would be so great!
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    your finding a time, so just multiply the time given by 1/root(1-v^2/c^2) where v is the time your given (must be a fraction of c, which it is)
    I think...

    (some clever uni person please come help us!!)

    tee hee cool i live just outside portsmouth!! (looked at your location ThuderCat8)
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    Why is it 1 over root... ?

    I hate maths, I hate physics, I hate cosmology. Grrr!

    Where abouts do you live then?
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    um, good question. I remember it that if you want to work out length contraction you multiply the original length by root(1-v^2/c^2), but if its time dilation or mass increase you divide it. I dont have a clue why, thats just what i was taught...and it better be right dammit i cant be learning a whole section again thisevening

    just outside pompey t'other side of portsdown.

    Is it me or is cosmology maths dressed in a physics suit?
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    (Original post by ThunderCat8)
    I'm glad it is in the afternoon! I need the extra time to study! I don't know anything! That Cosmology book is so confusing! I have done past papers, and looked in the book for the answer, but have been unable to find the answer there!

    Can somebody explain to me when you actually need to use the equation:

    Square root (1-v^2/c^2)

    It is in the book, and I'm sure in past papers I've done it's needed, but I don't know how to apply it. (if you know what I mean)

    here is an example: (from june 02 paper)

    A Pion has a mean lifetime of 26ns when at rest. They have a speed of 0.80c relative to the laboratory:

    What is the mean lifetime of the pion as measured by stationary observers in the lab?

    Any help would be so great!
    well y = 1/sqrt(1-V^2/c^2) = 1/0.6

    t = yt'
    t = (1/0.6)*26
    t = 43.3 ns
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    (Original post by Fluffstar)
    your finding a time, so just multiply the time given by 1/root(1-v^2/c^2) where v is the time your given (must be a fraction of c, which it is)
    I think...

    (some clever uni person please come help us!!)

    tee hee cool i live just outside portsmouth!! (looked at your location ThuderCat8)
    v is the relative velocity between the two frames you are looking at... so v^2/c^2 = (v/c)^2; if the two frames are at rest, then this equals zero, 1-this equals 1, and one over square root this equals 1 : t=t', there is no time dilation.

    Make sure if you can't remember which to divide and which to multiply, when the relative velocity of the frames is 0<v<c, the length of something measured in the other frame (in the direvtion of the velocity) is always smaller, and the time period between events is always longer.

    Sorry for the rushed explanation, gtg. hope this helps.
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    (Original post by mik1a)
    v is the relative velocity between the two frames you are looking at... so v^2/c^2 = (v/c)^2; if the two frames are at rest, then this equals zero, 1-this equals 1, and one over square root this equals 1 : t=t', there is no time dilation.
    Also, if v << c, obviously v/c -> 0, so you can ignore time dilation there (it does happen, but it's so small that you don't need to consider it).
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    (Original post by mik1a)
    :cool:
    Nice sunglasses.
 
 
 

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