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Conservation of momentum Watch

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    Always get stuck on questions like this;

    "A proton makes a low-energy head-on collision with an unknown particle and rebounds straight back along its path with 4/9 of its initial kinetic energy. Assuming that the collision is elastic and that the unknown particle is originally at rest, calculate the mass of the unknown particle."

    I realise momentum and kinetic energy are conserved in this situation, and I've arranged and re-arranged the several conservation equations but can't seem to get the answer. It's obviously about re-arranging it the correct way but I simply can't...can anyone help?
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    Do you have an answer to this question? If you do i would like to see please
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    (Original post by soup)
    Do you have an answer to this question? If you do i would like to see please
    Nope, sorry :|
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    (Original post by Sasukekun)
    Nope, sorry :|
    Oh that's quite annoying but i'll tell you what i did (even though it's probably wrong)

    So we know that the final KE of the proton if 4/9 of the initial KE so we can say
    4/9(1/2mv^2) = 1/2mu^2
    Divide both sides by 1/2m to get
    4/9 = u^2 / v^2
    Rearrange to get
    u = 3/2 v
    So by conservation of momentum we know
    m1u1 = m1v1 + m2v2 but u1 = 3/2v1 so
    3/2 m1v1 = m1v1 + m2v2
    1/2 m1v1 = m2v2
    m1v1 = 2 m2v2
    If we put that into m1u1 = m1v1 + m2v2 we get
    m1u1 = 3m2v2
    Now we need to find a relationship that involves u1 and v2 so go back to our KE relationships and we find that
    5/9(1/2 m1u1^2) = 1/2 m2v2^2
    Rearrange to get
    u1 ^2 = (9m2v2^2) / (5m1)
    Now we square m1u1 = 3m2v2 to get
    m1 ^2 u1^2 = 9m2^2 v2^2
    Now put in u1 ^2 = (9m2v2^2) / (5m1)
    And a bit more rearranging we see
    9/45 m1 = m2
    m2 being mass of the unkown and m1 mass of proton

    Hope this explains it well but it probably wrong
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    (Original post by soup)
    Oh that's quite annoying but i'll tell you what i did (even though it's probably wrong)

    So we know that the final KE of the proton if 4/9 of the initial KE so we can say
    4/9(1/2mv^2) = 1/2mu^2
    Divide both sides by 1/2m to get
    4/9 = u^2 / v^2
    Rearrange to get
    u = 3/2 v
    So by conservation of momentum we know
    m1u1 = m1v1 + m2v2 but u1 = 3/2v1 so
    3/2 m1v1 = m1v1 + m2v2
    1/2 m1v1 = m2v2
    m1v1 = 2 m2v2
    If we put that into m1u1 = m1v1 + m2v2 we get
    m1u1 = 3m2v2
    Now we need to find a relationship that involves u1 and v2 so go back to our KE relationships and we find that
    5/9(1/2 m1u1^2) = 1/2 m2v2^2
    Rearrange to get
    u1 ^2 = (9m2v2^2) / (5m1)
    Now we square m1u1 = 3m2v2 to get
    m1 ^2 u1^2 = 9m2^2 v2^2
    Now put in u1 ^2 = (9m2v2^2) / (5m1)
    And a bit more rearranging we see
    9/45 m1 = m2
    m2 being mass of the unkown and m1 mass of proton

    Hope this explains it well but it probably wrong
    Hey thanks a lot this seems excellent, although how do you do the bolded bit? I can't seem to see it myself... xD Sorry.
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    Have you substituted u1^2 into m1 ^2 u1^2 = 9m2^2 v2^2 ???
    If you have the m1^2 divided by m1 gives m1 and then you just divide by m^2
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    (Original post by soup)
    Have you substituted u1^2 into m1 ^2 u1^2 = 9m2^2 v2^2 ???
    If you have the m1^2 divided by m1 gives m1 and then you just divide by m^2
    I see that now ^^, Thanks very much for your help.
 
 
 
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