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Inelastic Collision of a football tackle

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data
    In an American football game, a 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Explain why the successful tackle constitutes a perfectly inelastic collision. (b) Calculate the velocity of the players immediately after the tackle. (c) Determine the mechanical energy that disappears as a result of the collision. Account for the missing energy.

    2. Relevant equations
    m1v1=m1+m2*v2
    v2= m1/(m1+m2)*v1

    3. The attempt at a solution
    (a) Because they are stuck together, it is a perfectly inelastic collision.

    (b)Theta= 40degrees = tan-1(3/5)
    m1=90.0kg
    m2=95.0kg
    v1=(5.00i+3.00j)m/s= [tex]\sqrt{5.00^2+3.00^2}[/tex]m/s= 5.83m/s

    v2= m1/(m1+m2)*v1
    =(90kg/185kg)*5.83 m/s =2.84 m/s
    v2x= 2.84*sin40degrees=1.82i
    v2y= 2.84*cos40degrees=2.16j
    So, the vector for the velocity after impact is (1.82i+2.16j)m/s

    (c)w=[tex]\Delta[/tex]K=(1/2)m*[tex]\Delta[/tex]v2
    w=(1/2)(m1+m2)(v22-v12)
    w=-2.405kJ
    This is the heat generated from the collision.

    I really worked out the problem, but I don't know if it's correct. Please check it for me. Thanks.
     
  2. jcsd
  3. Mar 17, 2009 #2

    rl.bhat

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    (b)Theta= 40degrees = tan-1(3/5)
    m1=90.0kg
    m2=95.0kg
    v1=(5.00i+3.00j)m/s= LaTeX Code: \\sqrt{5.00^2+3.00^2} m/s= 5.83m/s

    This is not correct.
    Find the initial momentum of each player and find the resultant momentum of the tackled players.
     
  4. Mar 17, 2009 #3

    LowlyPion

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    Your equations don't accurately reflect the collision. You are dealing with 2 dimensions here. Perhaps if you separate out the components and deal with them independently it will become clearer?

    For instance can treat them in x as though one was stationary. And again in y as though the other is stationary, yielding your 2 velocities, and then you can use Pythagoras to determine the resultant from the x,y as you were attempting.
     
  5. Mar 17, 2009 #4
    So part a and c are correct?
     
  6. Mar 17, 2009 #5

    LowlyPion

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    A is ok. Perhaps you should correct b to determine whether c is?
     
  7. Mar 17, 2009 #6
    Right.

    So, I googled the Inelastic Collision to get the formulas.
     
    Last edited: Mar 17, 2009
  8. Mar 17, 2009 #7
    Part B:

    m1v1i+m2v2i=(m1+m2)vf
    vf=[m1v1i+m2v2i/(m1+m2)
    I did the x direction and then the y direction:
    vfx=1.56m/s
    vfy=1.86m/s
    Used pythagorean theroum.
    vf=2.43m/s
    tan-1(vfy/vfx)=50.0degrees
    So, it is 50 degrees from east.

    For part C:

    w=1/2 m1+m2 (vf2-vi2)
    w=(1/2)(185kg)(2.432-5.832)m2/s2
    w=-2.60kJ

    Is this correct? Do I need to break the work into componants too? I am so confused.
     
  9. Mar 17, 2009 #8

    LowlyPion

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    You may be confused, but it looks OK.
     
  10. Mar 17, 2009 #9
    Thanks so much.
     
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