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Momentum and horizontal speed

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    When tossed upward and hit horizontally by a batter, a .24kg softball recieves an impulse of 2.7 N*s. With what horizontal speed does the ball move away from the bat?

    2. Relevant equations

    Here are all the equations that were in our notes for this section.
    p (momentum) There was triangle P and triange t in my notes, so I wrote "change in"

    f*"change in"t="change in"p
    f*"change in"t=mvf-mvi
    "change in" P=mvf-mvi
    3. The attempt at a solution

    I really was not sure what equation I should use because I didnt really know what the impulse was. I chose f*"change in"t=mvf-mvi.
    2.7 N*s= (.24kg)(0m/s)-(.24kg)(v)
    I got v=-11.25m/s. I know this was wrong, but Im not sure which equation I was supposed to use or if I entered the information correctly in to the equation. Any help would greatly be appreciated.
  2. jcsd
  3. Nov 9, 2008 #2
    You can treat this as a collision.
  4. Nov 9, 2008 #3
    A collision? What type of equation is that?
  5. Nov 9, 2008 #4
  6. Nov 9, 2008 #5
    So, what would the 2.7 N*s be considered?
  7. Nov 9, 2008 #6
    What are the units of Ns?
  8. Nov 9, 2008 #7
    Im not really sure. I think it is Neutons*second. Im pretty sure it is the impulse.
  9. Nov 9, 2008 #8
    1 N = 1 kg*m/s/s.

    So, 2.7 N*s = 2.7 kg*m/s.

    So, the impulse has a "mass" and a "velocity."

    The ball isn't moving in the x-direction so its initial velocity is 0. Solve, the impulse goes away.
  10. Nov 9, 2008 #9
    ok so in that equation you gave me:


    Its all going to work out to be 0? I am really confused. I know you are trying so hard to help me, but Im lost... =]
  11. Nov 9, 2008 #10
    0.24kg ( 0 m/s ) + Impulse = 0.24kg ( V m/s ).
  12. Nov 9, 2008 #11
    Oh wow so I was really close with my first answer...It was just negative and I needed it to be positive! Thank you. =]
  13. Nov 9, 2008 #12
    You're welcome.
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