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A high school momentun problem

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A superball with a mass of 83.6 g is dropped
    from a height of 3.1 m. It rebounds to a height
    of 2.5 m.
    The acceleration of gravity is 9.8 m/s2.
    What is the change in its linear momentum
    during the collision with the floor? Answer in unit
    of kg m/s.


    2. Relevant equations

    P=mV P=momentun m=mass V=velocity
    changing P = F X changing time.

    3. The attempt at a solution

    the question asks the momentun during the collision with the floow, so P=mV
    we know the m is 83.6 g . and do not know the velocity.
    How do i get the velocity with above known vaule?
     
    Last edited: Jan 22, 2008
  2. jcsd
  3. Jan 22, 2008 #2
    i dont get this question, can someone helps me?
     
  4. Jan 22, 2008 #3
    Thanks
     
  5. Jan 22, 2008 #4
    Use kinematics (or energy) to determine velocity when it hits the floor.

    Think about what it means to rebound with some height. What is the velocity when it rebounds to 2.5m? How can you determine the velocity it rebounded with?
     
  6. Jan 22, 2008 #5
    how can i determine the velocity when it rebounded?
     
  7. Jan 22, 2008 #6
    You have all the information you need to use one of the kinematics equations.

    Initially the ball is at x = 0 with velocity v (which you want to solve for). It is later at x = 2.5 with 0 velocity.
     
  8. Jan 22, 2008 #7
    so whats the exact solution? i am confuse....

    i understand what you are talking about, but how do i calculate the velocity going down?
     
  9. Jan 22, 2008 #8
    You're calculating the upwards velocity when it bounces. Hint: [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]
     
  10. Jan 22, 2008 #9
    oh so.. the initial velocity is 9.8, and the final velocity is 0 right?

    and then the changing velocity is final - initial?
     
  11. Jan 22, 2008 #10
    Nono, your unknown IS the initial velocity.
     
  12. Jan 22, 2008 #11
    can i please have your msn or some kind of instant tool to communicate?
     
  13. Jan 22, 2008 #12
    We're really close to finishing this... I've even given you the formula to use.

    You want to solve for [tex]v_0[/tex]. v is the final velocity, a is the acceleration, and delta x is how far it's travelled.
     
  14. Jan 22, 2008 #13
    i got this already, i asked my teacher on MSN to find out..

    the solution is actually finding the momentum downwards + upwards.

    S=1/2at^2 and use V=at. and i got V.
     
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