1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Water Fall Momentum Question

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Water falls at the rate of 250 g/s from a height of 60 m into a 780 g bucket on a scale (without splashing). If the bucket is originally empty, what does the scale read after 2 s?

    2. Relevant equations

    p=mv
    F[tex]\Delta[/tex]t=[tex]\Delta[/tex]p

    3. The attempt at a solution

    So I assumed that the water was already beginning to fill the bucket at t=0, since it cant reach the bucket in 2s.

    First I found the velocity of the water right before it fills the bucket...
    vf2=vi2+2ax
    vf2= 2(9.81 m/s2)(60m)
    vf= 34.3 m/s

    Then I used the impulse momentum theorem...
    F[tex]\Delta[/tex]t=[tex]\Delta[/tex]p
    F( 2 s) = .500 kg( 0 - 34.3 m/s)
    F = 8.58 N

    Weight of the bucket is mg, which is 7.65 N...
    so 8.58 + 7.65 is 16.2 N

    im not sure about this though... can someone double check my work, I only have one try left
     
    Last edited: Nov 4, 2008
  2. jcsd
  3. Nov 5, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi asura! :smile:
    Yes, your v is correct. :smile:

    But your method after that is completely wrong.

    The impulse momentum theorem is the correct principle, but you should use it to find the instantaneous force (because the scale only measures instantaneous force, not total force).

    Then add the weight of the water already in the bucket

    (and don't forget to convert from N back into g :wink:)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Water Fall Momentum Question
  1. Water fall question (Replies: 2)

Loading...