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    Why we can't apply Impulse-Momentum principle to a fixed wall ?
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    (Original post by FZka)
    Why we can't apply Impulse-Momentum principle to a fixed wall ?
    Momentum = mass times velocity.

    Question: What is the mass of a FIXED wall?
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    (Original post by FZka)
    Why we can't apply Impulse-Momentum principle to a fixed wall ?
    are you doing m1 tomorrow morning? i just got that question. If you did apply momentum principle, the speeds of the particle are exactly the same before and after because the wall's velocity is always 0.
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    Take m1 as a ball and m2 as the wall.

    If you use m1v1 +m1u1 = m2v2 + m2u2 you'd get that the m1v1= -m1u1 as the wall can not have a velocity.

    Therefore you can't use the impulse momentum as if you use

    Impulse= m1v1-m1u1 you'd get 0.
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    (Original post by SirMasterKey)
    Take m1 as a ball and m2 as the wall.

    If you use m1v1 +m1u1 = m2v2 + m2u2 you'd get that the m1v1= -m1u1 as the wall can not have a velocity.

    Therefore you can't use the impulse momentum as if you use

    Impulse= m1v1-m1u1 you'd get 0
    From what I understood i think you meant I = m2v2-m2u2 = 0 ?

    Plus. why would I use m1v1 +m1u1 = m2v2 + m2u2 ?
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    Momentum= m1(v1-u1)

    since it's fixed, v1-u1 = 0

    m1*0 = 0

    Therefore Momentum = 0

    Does that answer that bit?

    Then impulse is rate of change of momentum (if I remember correctly). So it's 0 divided by the duration of the contact which in turn equals 0

    Does that make sense?
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    (Original post by FZka)
    From what I understood i think you meant I = m2v2-m2u2 = 0 ?

    Plus. why would I use m1v1 +m1u1 = m2v2 + m2u2 ?
    Yes as long as m2 is the mass of the wall.

    Well I was simply assuming something was thrown at the wall otherwise why would there be need to find out the momentum of a wall.

    It is the conversation of momentum and is using the same principle as the impulse= m1v1-m1u1.
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    (Original post by SirMasterKey)
    Yes as long as m2 is the mass of the wall.

    Well I was simply assuming something was thrown at the wall otherwise why would there be need to find out the momentum of a wall.

    It is the conversation of momentum and is using the same principle as the impulse= m1v1-m1u1.
    It's clear now. Thanks!
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    (Original post by Snoodle)
    Momentum= m1(v1-u1)

    since it's fixed, v1-u1 = 0

    m1*0 = 0

    Therefore Momentum = 0

    Does that answer that bit?

    Then impulse is rate of change of momentum (if I remember correctly). So it's 0 divided by the duration of the contact which in turn equals 0

    Does that make sense?
    Impulse is just the change in momentum.
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    (Original post by FZka)
    Impulse is just the change in momentum.
    so your OK now?
 
 
 
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