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Distance and time of an Angry bird

  1. Mar 18, 2015 #1

    SGE

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    1. The problem statement, all variables and given/known data
    An angry bird of mass 2m is launched at theta = 45 degrees with initial velocity = 10 m/s above the horizontal. It then drops an egg of mass m= 1 kg vertically down with velocity 5sqrt(2) m/s when it is at the topmost point in its trajectory. (Bird's mass after egg drop is m.) How much further does the bird now land? How much time from launch to land?

    2. Relevant equations
    p=mv
    { and I think the horizontal momentum: p=mvcos(theta)}
    Change in momentum = net force X change in time

    3. The attempt at a solution
    I know that horizontal momentum will be conserved so I have solved for the horizontal momentum before reaching the highest point:
    p=2m(10cos45)
    p=10sqrt(2)
    and for the horizontal momentum after
    p=m5sqrt(2)cos45
    p=5
    but I'm not sure how to translate momentum into distance travelled.
     
  2. jcsd
  3. Mar 18, 2015 #2

    haruspex

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    The question is not quite clear. Is the dispatch of the egg vertical relative to the ground or relative to the moving bird? From what you wrote in words
    it seemed like you were going to take it as the former. That may be what is intended, but it would mean the bird effectively fired the egg backwards relative to itself, which seems unlikely.
    However, you then calculated p after dropping the egg as though the horizontal speed had not changed. That is, you have taken vertical as relative to the bird's motion.
    Separately from all that, you need to look at vertical momentum conservation.
     
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