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Systems with Varying Mass (Momentum)

  1. Feb 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Q77.png

    2. Relevant equations
    Momentum conservation and Energy conservation equations

    3. The attempt at a solution
    Ok First I translated the units, ##1000 \frac {kg} {min}=16.66\frac {kg} {s}##
    ##10\frac {km} {h}=2.77\frac {m} {s}##
    ##20\frac {km} {h}=5.55\frac {m} {s}##

    Now I tried to write an momentum equation for a barge which has small speed (##2.77\frac {m} {s}##,lets call it A)
    İnitial momentum is ##p_i=Mv## final momentum will be ##(M-dm)(v+dv)##.
    so ##Mv=(M-dm)(v+dv)## or I write ##Mv=(M-dm)v+Fdt##

    I wrote like this cause İnitial momentum is ##Mv## thats for sure.After time Our mass decreases by ##dm## and our speed must increase by ##dv## for momentum conservation.

    Actually I am stucked here.I couldnt write a proper equation.If Its true and I continue I get this (For both equations). The questions says last and inital velocity must be same so...thats why v are same

    ## \frac {dm} {dt}.v=F##
    which gives me ##46N## which answer is "none"

    interesting thing is the answer for fast barge is ##46N## so where I did wrong ??
     
    Last edited: Feb 5, 2017
  2. jcsd
  3. Feb 5, 2017 #2
    Does that mean that no additional force is required for the slower barge? I'm picturing a guy shoveling coal off of the slower boat. And if he is shoveling it off directly to the side (perpendicular to the velocity of the barge), then there is no change in momentum of the coal, and thus no change in momentum of the slower boat. The mass of the coal flying through the air as it leaves the shovel is the same as the boat, isn't it? But when that shovel full of coal lands in the faster barge, it requires an increase in its momentum, supplied by the faster barge, to stay in the faster barge.
     
  4. Feb 5, 2017 #3
    Ok I see,why its none for slow one,How can I calculate the needed force for faster one ?
     
  5. Feb 5, 2017 #4

    haruspex

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    Your original calculation is correct for the faster barge.
    The key point is the forward speed of the coal as it is being transferred. It leaves the slower barge at the forward speed of the slower barge, so it takes only its share of momentum with it and does not affect the speed of that barge. It arrives at the faster barge still with the forward speed of the slow barge, so requires a force to bring it up to speed.
     
  6. Feb 5, 2017 #5
    I didnt understand the idea...Could you describe it in equations ?
     
  7. Feb 5, 2017 #6
    The way I though of it was as the shovels of coal as an impulse. You can easily calculate the change in momentum of 16.67 kg of coal increasing its speed by the velocity delta of the two barges.
     
  8. Feb 5, 2017 #7
    .
    and
    I am confused...But I understand your idea...
     
  9. Feb 5, 2017 #8

    haruspex

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    Tom's "no change in momentum of the coal" referred to the process of shovelling the coal from the slow barge and getting it in transit to the other barge. His other remark was in the context of the coal arriving at the fast barge.
     
  10. Feb 5, 2017 #9
    I see just one question.If my equation is correct then whats the v in there ? If its the speed of faster burge it deosnt give the right answer from that equation ?
     
  11. Feb 5, 2017 #10
    The bottom line is that all of the mass of coal has to increase its velocity by 10 km/hr. There are unlimited ways that can be done. A few examples: (1) The guys who are shoveling it could throw it forward at a speed of 10 km/hr faster than they are traveling. If they did that, then the slower barge's momentum would experience a momentum change in the opposite direction. And once it landed in the faster barge, it would not require a momentum change by the faster barge. (2) They could throw it forward 5 km/hr faster than they are traveling. Then the increase in momentum of the coal would be shared between the slower barge and the faster barge. (3) They could throw the coal directly off the side of the boat, resulting in no change in momentum of the coal while it was in the air, and thus no change in momentum of the slower barge. Once the coal lands in the faster barge, the faster barge's momentum would be affected in the reverse direction. So the faster barge would have to do all the work. (4) They could throw it backward at 10 km/hr, thus increasing the momentum of the slower barge. But then the faster barge would have to do even more work to compensate for that.

    Does that make sense?

    It's not the speed of the faster barge that matters; it is the difference in speed of the faster and slower barge.
     
  12. Feb 5, 2017 #11
    Thanks a lot,I understand now...Sometimes I overthink
     
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