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Inelastic collision baseball

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data
    It's the bottom of the ninth inning at a baseball game. Score is tied with a runner on 2nd when the batter gets a hit. The 85 kg runner rounds 3rd and is heading home with a speed of 8.0 m/s. Just before he reaches home plate he crashes into the catcher and the two players slide together along the base path toward home plate. Catcher's mass is 95kg. Coefficient of friction is 0.70. How far does the runner and catcher slide?



    2. Relevant equations
    m1iv1i + m2iv2i = m1fv1f + m2fv2f


    3. The attempt at a solution
    I found the velocity at which they slide together to be:

    m1iv1i +m2 = (m1 +m2)vf
    (85 kg)(8.0 m/s) + (95 kg) = (180 kg)vf
    680 kg*m/s + 95 kg = 180 kg*vf
    vf = 4.3 m/s

    So now I have mass = 85+95 = 180 kg; vi = 4.3 m/s; vf = 0; u = 0.70

    I don't know what to do next to find the displacement. The two equations I know to find x are:
    x=1/2(vf+vi)delta-t
    x=(vi)(delta-t)+1/2a(delta-t)^2

    But I don't know t. I don't know a since I don't know t, unless I'm suppose to assume its constant. even with a=0 i still don't know t. please help!
     
  2. jcsd
  3. Jul 12, 2008 #2
    you may want to go back first to where you substituted values into your original equation. what is the velocity of the catcher when the runner collides with him? if it's zero, it will cancel out the mass of the catcher entirely from the first part of the equation.
    also, maybe there's an equation you can use to substitute known values such as Vf and Vi for t...
     
  4. Jul 12, 2008 #3

    Doc Al

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    Staff: Mentor

    Careful: What's the initial speed of the catcher? (Correct this and recalculate vf.)
    You can use another kinematic formula that directly relates distance with speed, without needing the time. (Or you can just combine these two equations to eliminate t.)
    You don't need the time to calculate the acceleration. Analyze the forces. (That's why they give you the coefficient of friction.)
     
  5. Jul 12, 2008 #4
    I see my error. The vf = 308 m/s

    I'm still kinda lost on the equation bit. I was thinking F=ma

    Fy = mg-N = 0
    Fx = ma -uN ??
     
  6. Jul 12, 2008 #5

    Doc Al

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    Way off: Redo it more carefully.
    Good.
    What's the only horizontal force acting on the players and thus the net horizontal force? Set that equal to ma and solve for a.
     
  7. Jul 12, 2008 #6
    Honestly, I don't know why I typed 308 m/s.... I meant to type 3.8 m/s.

    The only force acting horizontally is u = 0.70

    So it's:
    0.70 = ma - N ??

    If so then how do I find N? mgcos(0) = (85)(9.8)cos(0) = 833?
     
  8. Jul 12, 2008 #7

    Doc Al

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    OK. (But better not to round off until the last step.)

    The only force acting horizontally is friction, which equals [itex]\mu N[/itex].

    ??

    Use the entire mass of both players. You find N by analyzing vertical forces, like you did in post #4: N-mg = 0, so N = mg.
     
  9. Jul 12, 2008 #8
    Ahhhhhh, okay.


    uN = ma
    (0.70)[(180 kg)(9.8 m/s^2)] = (180 kg)a
    a = 6.86 m/s^2

    knowing a, I can figure out t to be
    t = a/v
    t = (6.86 m/s^2) / (3.7 m/s)
    t = 1.85 s

    x = vt
    x = (3.7 m/s)(1.85 s)
    x = 6.84 m
    x = 7m


    Better?
     
  10. Jul 12, 2008 #9

    Doc Al

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    Good.

    [Edit] Not good: Correct that formula; it should be t = v/a. And be careful about rounding off.

    Careful! The speed is not constant, so you'd better use the average speed to find the distance. And you'll need to correct the time.
     
    Last edited: Jul 13, 2008
  11. Jul 12, 2008 #10
    bleh, this problem has turned into the bane of my existence...

    So average speed is v = (delta-x) / (delta-t)
    To find x I use delta-v?

    x = (delta-v)(delta-t)
    x = (vf - vi)(tf - ti)
    x = (0 -3.8 m/s)(1.8 s - 0) (t = (6.86 m/s^2)(3.8 m/s))
    x = -6.84 m

    It's the same answer, just negative. How can distance be negative?
     
  12. Jul 12, 2008 #11
    what do you guys think of this?
    I am going to solve the problem using energy
    u=Ff/Fn
    0.7*180kg*9.81m/s^2=Ff
    Ff=1236.06N

    1/2 m*v^2 = Ff *d
    1/2 180kg*(3.8m/s)^2= 1236.06N *d
    d = 1.04m

    is that correct Doc Al?
     
  13. Jul 13, 2008 #12

    Doc Al

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    I must have been asleep before. You have the formula switched around. It should be t = v/a.
     
  14. Jul 13, 2008 #13

    Doc Al

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    The equation needed is x = (average speed) x (time), not (change in speed) x (time).

    That's the one you want.

    And don't forget to redo your calculation of the time, as pointed out in posts #9 and 12.
     
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