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Impulse problem.

  1. Apr 10, 2005 #1
    A baseball player hits a 82 mph fastball, sending it back at 120 mph. The ball has a mass of 150 g. The contact time is 0.001 s. What is the average force on the bat?

    Okay, so I converted the miles to kilometres, and then used the equation F_avg*(t_2-t_1) = m_1*v_2 - m_2*-v_2, but I can't get the right answer. I even eliminated the negative sign and tried that, but it is still wrong. I can't see what else to add, so please help.
     
  2. jcsd
  3. Apr 10, 2005 #2

    Doc Al

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    Assuming you meant [itex]F_{ave} \Delta t = \Delta (mv) = m (v_f - v_i)[/itex], and you did your unit conversions properly, that should work. Note that if [itex]v_i = + 82\ \mbox{mph}[/itex], then [itex]v_f = - 120\ \mbox{mph}[/itex]. (Signs matter, since momentum is a vector.)
     
  4. Apr 10, 2005 #3

    dextercioby

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    [tex] \bar{\vec{\mbox{F}}}=:\frac{\vec{\mbox{p}}_{f}-\vec{\mbox{p}}_{i}}{\Delta t} [/tex]

    U know the momentum both initially and finally and u know the time of impact (in which the momentum is being transfered).

    Daniel.
     
  5. Apr 10, 2005 #4
    82*1.6=131.2
    120*1.6=192
    .15*192-.15*131.2=9.12
    9.12/.001=9120

    That's the answer I got the first time, but the computer says that it is wrong. Did I do everything right?
     
  6. Apr 10, 2005 #5

    dextercioby

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    Check the units.U need to convert everything to SI-mKgs.

    Daniel.
     
  7. Apr 10, 2005 #6
    82*1.6=131.2 km
    120*1.6=192 km
    .15kg*192km-.15kg*131.2km=9.12kg*km
    9.12kg*km/.001sec=9120
    If I convert it into kgm/s it becomes 9.12*10^6N. Hmm, still wrong... Argh!
     
  8. Apr 10, 2005 #7

    Doc Al

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    It looks like you're converting miles to km; what you should be doing is converting miles/hour to meters/second.
    Reread my comments about signs. Realize that the ball reverses direction. For example: if it comes towards the bat at 10 mph, then leaves the bat at 15 mph, the change in velocity would be: 15 - (-10) = 25 (not 15 - 10 = 5).
     
  9. Apr 10, 2005 #8

    dextercioby

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    Initial momentum (negative by a choise of axis) [tex] -0.15 \ \mbox{Kg} \cdot \frac{(82\cdot 1.6) \cdot 1000 \ \mbox{m}}{3600 \ \mbox{s}} [/tex]

    Final momentum (positive) [tex] +0.15\ \mbox{Kg} \cdot \frac{(120\cdot 1.6)\cdot 1000 \ \mbox{m}}{3600 \ \mbox{s}} [/tex]

    Compute the 2 #-s and then subtract the negative one from the positive one.The result should be divided by the time interval.

    Daniel.
     
  10. Apr 10, 2005 #9
    Oh... damnnit! Thanks guys, I will try that out now.
     
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