Conservatino of Momentum: Bowling Ball and Earth

  1. 1. The problem statement, all variables and given/known data
    A 7.00kg bowling ball is dropped from rest at an initial height of 3.00m.
    (a) What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Use 5.98 x 10^23 kg as the mass of the Earth
    (b)Use your answer to part (a) to justify ifnoring t he motion of the Earth when dealing with the motions ofs terrestrial objects.


    2. Relevant equations
    p=mv
    F=p/t
    m1v1i+m2v2i=m1v1f+m2v2f


    3. The attempt at a solution
    So here is my attempt. i thought I should find the velocity of the bowling ball when the it comes in contact with the Earth first so using Vf^2=Vi^2 + 2ad:

    Vf^2 = 0 +2(9.8)(3)
    Vf^2 = 58.8
    Vf = 7.67 m/s

    so after that, i have:
    m1=7.00kg
    v1i=2.67 m/s
    m2=5.8 x 10^24

    it looks like I'm looking for V2i for part (a) so i need to find V1f and V2f.
    Thats where I'm stuck. Anyone know this one?
     
  2. jcsd
  3. Doc Al

    Staff: Mentor

    Apply conservation of momentum. Assume that both earth and bowling ball start out at rest.
     
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