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Momentum and its Conservation

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data

    A 0.165kg softball is pitched to you at 20.0m/s. You hit the ball back along the same path, and at the same speed. If the bat was in contact with the ball for 9.90x10^-3 seconds,

    a.) what was the average force the bat exerted on the ball?

    b.) Could this force lift a 70.0kg person?

    2. Relevant equations

    I believe the equation to use is

    mass/change in time x (final velocity + initial velocity)

    3. The attempt at a solution

    a.)

    (.165kg (20 + 20 m/s)) / 9.90 x 10^-3
    = .165(40) / .0099
    = 666.7 N
    = 6.70 x 10^3 N

    b.)

    How do I factor in a 70kg person? Do I use 70(40)/.0099 ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 12, 2010 #2

    Doc Al

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    Staff: Mentor

    What force is required to lift a 70 kg person?
     
  4. Jul 12, 2010 #3
    Look at your equation again. Force is equal to the change in momentum over time. Change in momentum = final - initial.
     
  5. Jul 12, 2010 #4
    so I should have .165(20-20)/.0099 and get 16.7 N??
     
  6. Jul 12, 2010 #5
    you're hitting the ball back, so the final momentum should be negative.

    Can someone confirm this?
     
  7. Jul 12, 2010 #6

    Doc Al

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    That equation should be: mass/change in time x (final velocity - initial velocity)

    But as mizzy says, the sign of the velocity changes since you're hitting the ball in the opposite direction. So 20 - (-20) = 20 + 20.

    (All we care about is the magnitude of the change, so it doesn't matter which direction we call negative.)


    That's fine, except for the last step. (You have the wrong exponent.)
     
  8. Jul 12, 2010 #7
    Oooh, okay. Yes I noticed it should be ^2

    Thank you.

    To figure out b.)?? Can this be used to lift a 70kg person? Anyone?
     
  9. Jul 12, 2010 #8

    Doc Al

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    If the initial momentum was positive, then the final would be negative. (But all we really care about is the magnitude of the change.)
     
  10. Jul 12, 2010 #9

    Doc Al

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    Make sure you understand the point that mizzy was making.

    What force is required to lift a 70 kg person?
     
  11. Jul 12, 2010 #10
    F= 70kg x 9.90x10^-3
    = 70x.0099
    = .693 N

    .693 < 666.7 so less force is required to lift the person?
     
  12. Jul 12, 2010 #11
    you multiplied mass by time??
     
  13. Jul 12, 2010 #12
    or by speed? I'm not sure what to use.

    70.0kg x 20.0 m/s
    = 1400
    = 1.4 X 10^3
     
  14. Jul 12, 2010 #13
    I'm not too sure. But in your previous answer, mass x time doesn't give you F. Force is in Newtons (kgm/s^2)
     
  15. Jul 12, 2010 #14

    Doc Al

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    The answer to my question--What force is required to lift a 70 kg person?--has nothing to do with momentum, speed, or time.

    How much does the person weigh?
     
  16. Jul 12, 2010 #15
    the person weighs 70kg
     
  17. Jul 12, 2010 #16

    Doc Al

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    That's his mass, not his weight. How do you calculate the weight?
     
  18. Jul 12, 2010 #17
    into lbs? multiply by 2.20

    so 154 lbs
     
  19. Jul 12, 2010 #18

    Doc Al

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    Better calculate the weight in Newtons so you can compare it to the force you calculated in part a.
     
  20. Jul 12, 2010 #19
    oooh! so 154lb x 4.448N
    = 684.9N
    = 6.80x10^2
    and
    6.70x10^2 < 6.80x10^2
    therefore more force would be needed to life the 70kg person
     
  21. Jul 12, 2010 #20

    Doc Al

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    That's the idea.

    But the easy way to calculate the weight is by using W = mg, where g = 9.8 m/s^2.
     
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