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What is the weight of this bucket after time?

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Water falls without splashing at a rate of 0.230 L/s from a height of 2.40 m into a 0.630-kg bucket on a scale. If the bucket is originally empty, what does the scale read 2.90 s after water starts to accumulate in it?


    2. Relevant equations
    m1vi+m2vi=m1vf+m2vf


    3. The attempt at a solution
    This is a momentum homework problem from last week. I know the answer which is 14.29 newton, just not how it's obtained. I know that density is involved although we have not covered it in class yet. I know that we're adding the weight of the water to the weight of the bucket and after 2.9 seconds, we have 0.667 liters of water in the bucket. But how can we obtain weight from this? And what else factors into the weight of our system besides the weight of the bucket? As I'm not sure of why a height is given. Can someone please guide me through this? Thank you in advance.
     
  2. jcsd
  3. Oct 13, 2013 #2
    you'll have take two things into account to solve this problem.

    First, the amount of water already in the bucket at time t. density of water (1 kg/L) is one of the things a person is expected to know, this question is no different. This water has weight.

    Second: you know that water is falling from a given height so it has a certain momentum before reaching the bucket, so when it hits the bucket impulse acts on both bucket as well as differential mass of water. Due to this impulse, the reading of the scale would more than the weight of water actually in the bucket.

    I hope it helps!!

    Hint: convert volume rate to mass rate, for second part!!!.......don't forget about the weight of bucket!!!
     
    Last edited: Oct 13, 2013
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