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    I'm stuck on the first part. I managed to combine the equation and get k=3u/3u + v. how can i prove k < 1 from there?
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    (Original post by jamesw5555)
    I'm stuck on the first part. I managed to combine the equation and get k=3u/3u + v. how can i prove k < 1 from there?
    V must be greater than 0 traveling in the positive direction so your equation can never be greater than 1.
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    (Original post by jamesw5555)
    I'm stuck on the first part. I managed to combine the equation and get k=3u/3u + v. how can i prove k < 1 from there?
    Here is how I would do it:

    [EDIT: Made a mistake, see corrected version below]
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    (Original post by Middriver)
    You lost a K from the Z on the second step?
    Aargh, you're right. Let's try again:

    -Initial momentum of A = 2mu
    -Initial momentum of B = -3kmu

    -Final momentum of A = -mu
    -Final momentum of B = zkmu

    where z is a positive constant (it has to be positive as particle B is travelling in the opposite direction to A).

    So, using conservation of momentum:

    Initial momentum = final momentum
    => 2mu - 3kmu = zkmu - mu
    => 2 - 3k = kz - 1
    => 3 = k(z+3)
    =>3/(z+3) = k

    And since z > 0, then k < 1.
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    (Original post by Ollie231213)
    Aargh, you're right. Let's try again:

    -Initial momentum of A = 2mu
    -Initial momentum of B = -3kmu

    -Final momentum of A = -mu
    -Final momentum of B = zkmu

    where z is a positive constant (it has to be positive as particle B is travelling in the opposite direction to A).

    So, using conservation of momentum:

    Initial momentum = final momentum
    => 2mu - 3kmu = zkmu - mu
    => 2 - 3k = kz - 1
    => 3 = k(z+3)
    =>3/(z+3) = k

    And since z > 0, then k < 1.
    Ah okay now i understand it, i got the same thing as your last step but it was 3u/3u+z but you cancelled the mu to make it easier to work with. Thx for the help!
 
 
 
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