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Transfer of momentum?

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A hose can produce a horizontal stream of water at the rate of 600kg every 10 seconds as the water flows with constant speed of 30 m/s. Assume the water strikes the flat vertical back of a 2000kg truck. The water splashes out along the back of the truck without rebounding and then runs down to the ground. The truck is free to move without friction. What will be the acceleration of the truck in 30 seconds?


    2. Relevant equations
    thrust=|vdM/dt|
    F=ma


    3. The attempt at a solution

    okay, so we know that dM/dt=600/10 kg/sec=60kg/sec.

    we also know that as the truck speeds up, the relative speed of the stream of water will slow down, so the acceleration is not constant. this means that the thrust applied by the water is also not constant, but dependent on the v of the truck.

    so here goes:

    thrust=(relative v of water to truck)(dM/dt)=(30-v)(60)kgm/s^2

    acceleration of truck=thrust/mass of truck=(1800-60v)/2000

    a=dv/dt, so

    dv/dt=(1800-60v)/2000
    dt=2000dv/(1800-60v)

    with a bit of integrating:

    t=-(100/3)ln(90-3v)

    and rearranging:

    v=-(1/3)e^(-3t/100)-30

    but when you plug in 30 seconds for t into this equation, you get a number greater than 30, which is physically impossible because the speed of the truck has v=30 as its asymptote.

    what am i doing wrong?

    thanks for your help.
     
  2. jcsd
  3. Feb 10, 2009 #2
    and another thing: assuming that i have a conceptual error and not a mathematical error, i will also need assistance solving part b of the question:

    What will be the acceleration of the truck in 30 seconds if the water can enter the truck through a hole and remain inside?

    thanks again.
     
  4. Feb 10, 2009 #3

    Delphi51

    User Avatar
    Homework Helper

    Integrating your dt=2000dv/(1800-60v)
    I get integral from v = 0 to v of 2000/60*dv/(30-v)
    Let u = 30 - v, du = -dv to get
    integral from u = 30 to 30-v of du/u
    t = -2000/60[ln(30-v) - ln(30)]
    60t/2000 = ln[30/(30-v)]
    30/(30-v) = e^(-60t/2000)
    v = 30 - 30*e^(-60t/m)
    Note that at time zero, v = 0. In your answer you get v = -1/3 - 30 at time zero.
     
  5. Feb 10, 2009 #4
    ahhhh! there we go. so errors with integration...

    thank you for correcting me! it works now.



    now off to figure out part b.

    thanks again!
     
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