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Momentum! INELASTIC COLLISION!

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.291 kg rubber ball is dropped from a height of 3.72 m, and it rebounds to a height of 2.88 m. Assume there is no air friction.

    a) Find the magnitude of the impulse exerted by the ground on the ball.


    2. Relevant equations
    Vf^2=V0^2+2as
    MV=P
    DeltaP=I

    3. The attempt at a solution
    I used Kinematic equations to find out the the Vf down of the ball was 8.54
    and then Vavg=4.27... MVdown=1.242
    MVdown-MVup = Impulse
    MVup i did the same thing i got 1.093

    Ended up wrong
    Thank you for your help
     
  2. jcsd
  3. Oct 26, 2008 #2

    Hootenanny

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    Your value for the downward momentum is not correct. All you need to do is to multiply the downward velocity by the mass of the ball and this will give you the momentum.
     
  4. Oct 26, 2008 #3
    this is how I figured out the downward velocity:
    Since there is a netforce and an acceleration of g, it's not a constant velocity therefore I found the Vavg by finding time first x=V0t+.5gt^2 and then with t finding Vf=Vo+gt then Vavg=(Vo+Vf)/2

    Is there a flaw in my reasoning
    And I did the same for upward Vavg to find the difference in momentum to find the impulse
     
  5. Oct 26, 2008 #4

    Hootenanny

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    You calculated the velocity correctly, but miscalculate the momentum. As I said above, simply multiply the velocity by the mass to obtain the associated momentum.
     
  6. Oct 26, 2008 #5
    Thank you... but if the velocity is correct: im getting 4.27(Vavg)*.291(mass)=1.242 Kg m/s for down momentum? thats what i got earlier..
    i simply multiply my avg velocity down with the mass of the object to get Magnitude down and then i find the Magnitude up and then find the difference of both to get the impulse?
     
  7. Oct 26, 2008 #6

    Hootenanny

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    There is no need to use the average velocity here. You want to know the change of momentum over the collision with the ground, not the change in momentum of the overall fall and rebound. Therefore, you only need the velocities just before and just after the collision. In other words, you need to use the final downward momentum and the initial upward momentum.

    Does that make sense?
     
  8. Oct 26, 2008 #7
    yesss! Thank you so much...
    I forgot to realize that the Impulse is the integral of F dt from delta t right before and after the collision.
     
  9. Oct 26, 2008 #8

    Hootenanny

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    My pleasure :smile:
     
  10. Oct 26, 2008 #9
    b) If the ground exerts an average force of magnitude 296 N on the ball, find the time the ball was in contact with the ground.

    For this would i just take the impulse and divide it by the magnitude of force to get the time... since an Impulse is in N(s)
     
  11. Oct 26, 2008 #10

    Hootenanny

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    Sounds good to me :approve:
     
  12. Oct 26, 2008 #11
    im getting .2981 for the impluse and .00100722 for the time and it seems to be wrong.

    Vf(down)=8.54 Vo(up)=7.51 mass=.291 kg
     
  13. Oct 26, 2008 #12

    Hootenanny

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    These values are correct, but your calculated value for the impulse is not. You should be careful with your signs, notice that the upward velocity is in the opposite direction to the downward velocity.
     
  14. Oct 26, 2008 #13
    thank you! that was a silly mistake...
    thanks for all your help
     
  15. Oct 26, 2008 #14

    Hootenanny

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    No problem, it was a pleasure :smile:
     
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