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Impulse/momentum problem

  1. Jun 8, 2004 #1
    I have no clue on where to start on this. I have spent time playing with that cart thing and the momemtum balls and I have ideas floating around, but nothing concrete.

    There is a 248.2kg Soviet satellite travelling due east at a velocity of 1025km/hr and a 149.9kg US satellite travelling due west at a velocity of 1181.8km/hr. The colission is perfectly elastic. What happens afterwards?
     
  2. jcsd
  3. Jun 8, 2004 #2

    arildno

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    Use conservation of momentum (no external forces) and conservation of energy (fully elastic collision)
     
  4. Jun 9, 2004 #3
    From what you told me, I used

    m1v1 + m2v2 = m1v1' + m2v2'
    (mv1^2)/2 + (mv2^2)/2 = (mv1'^2)/2 + (mv2'^2)/2

    I solved both for V1' then I plugged in the numbers.After that, it came out to

    5364792208 - 35930354.61v2' +59675.19v2'^2

    I then plugged it into the quadratic equation, getting 328.08333m/s which is v2, and 274.0153711m/s, which is not the answer my teacher gave me
     
  5. Jun 9, 2004 #4

    arildno

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    Did you take into account that the velocities are in different directions?
     
  6. Jun 9, 2004 #5

    arildno

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    Now, let's solve this problem in full:
    I will assume that you gave me the right information, in particular that the collision was fully ELASTIC.

    The equations for two objects, 1 and 2, to be solved are:
    [tex]m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}[/tex]

    This is conservation of momentum, as you had; b and a refer to initial and final velocities (before and after collision).

    Energy conservation:
    [tex]\frac{1}{2}m_{1}v_{1,a}^{2}+\frac{1}{2}m_{2}v_{2,a}^{2}=\frac{1}{2}m_{1}v_{1,b}^{2}+\frac{1}{2}m_{2}v_{2,b}^{2}[/tex]

    This is also, I believe, what you meant.
    Note that the energy equation can be simplified in the following manner:
    [tex]m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a})+m_{2}(v_{2,b}-v_{2,a})(v_{2,b}+v_{2,a})=0[/tex]

    Rewrite the momentum equation as:
    [tex]m_{2}(v_{2,b}-v_{2,a})=-m_{1}(v_{1,b}-v_{1,a})[/tex]

    Hence, the energy equation may be written as:
    [tex]m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a}-(v_{2,b}+v_{2,a}))=0[/tex]

    The root [tex]v_{1,b}=v_{1,a}[/tex] corresponds to no collision, so we have the following system to solve:
    [tex]v_{1,b}+v_{1,a}=v_{2,b}+v_{2,a}[/tex]
    [tex]m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}[/tex]

    Solving for [tex]v_{1,a},v_{2,a}[/tex] yields:
    [tex]v_{1,a}=\frac{(m_{1}-m_{2})v_{1,b}+2m_{2}v_{2,b}}{m_{1}+m_{2}}[/tex]
    [tex]v_{2,a}=\frac{(m_{2}-m_{1})v_{2,b}+2m_{1}v_{1,b}}{m_{1}+m_{2}}[/tex]

    Now, we have simplified the symbolic expressions maximally; it is time to enter in the values!
    We set:
    [tex]m_{1}=149.9,v_{1,b}=1181.8[/tex]
    [tex]m_{2}=248.2,v_{2,b}=-1025[/tex]

    The answers should be quite different from the ones you gave
     
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