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Homework Help: Total time wagon needs to fully pass the hopper

  1. May 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A train wagon of mass M moves on a rail with constant velocity V (without friction). It passes a sand hopper which pours sand in the wagon at constant rate C [kg/s]. The sand falls vertically so it does not transfer any horizontal momentum to the wagon. The length of the wagon is L [m].
    Question 1: Determine the velocity of the train wagon as a function of time V(t) while it is under the sand hopper.
    Question 2: Show that the total time T that it takes for the wagon to fully pass the hopper machine is T=M/C*(exp(CL/MV)-1)

    2. Relevant equations
    • Conservation of momentum MwagonV0=(Mwagon+msand)*V1
    • F=ma=dP/dt (Newton's second law)
    • t=msand/C
    • Straight-line motion equations
    vx=v0x+axt (Equation 1)

    x=x0+v0xt+½axt2 (Equation 2)

    vx2=v0x2+2a(x-x0) (Equation 3)

    x-x0=½(v0x+vx)t (Equation 4)

    3. The attempt at a solution

    The first question was pretty easy. Because I need the answer from question 1 to solve question 2 I want to make sure that I have the right answer for question 1.
    Question 1:
    Using conservation of momentum I got MwagonV0=(Mwagon+msand)*V1 and the mass of the sand can be written as Ct. Subbing that in you get MwagonV0=(Mwagon+Ct)*Vt. This gives an equation for Vt -> Vt=Mwagon*V0/(Ct+Mwagon)

    Question 2:
    This one is tricky and it definitely needs integrating if you look at the equation. I have a feeling there are multiple ways to approach this but every approach I did ended up in chaotic equations that did not help. First I used Newton's second law --> F=dP/dt
    Pt=Mtvt with the mass being dependent of time because of Mwagon+Ct

    dP/dt=Mtdv/dt+vtdm/dt (dm/dt is the change in mass per unit time which is C)


    When isolating dV/dt and integrating that I got



    This is definitely going in the wrong direction and I got no L in the equations.

    Then I started with equation 2 with x being L --> L=V0t+½at2.

    Acceleration is dV/dt which is CMV/((Ct+M)2) but subbing this in gives again nothing I can work with.

    Then I used equation 3 which resulted in


    Isolating dV/dt gives


    This is also a pain to integrate but I used an integration calculator and I ended up with an arctan in my equation.

    This is going nowhere and I would really appreciate some help.
  2. jcsd
  3. May 10, 2015 #2


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    Hi, freutel.

    The constant acceleration equations that you wrote as relevant equations are not applicable in this problem.

    You're making it more difficult than necessary. You found from question 1 that Vt=Mwagon*V0/(Ct+Mwagon).
    Can you see what to do if you write Vt = dx/dt?
    Last edited: May 10, 2015
  4. May 10, 2015 #3
    I always fail to see the easiest way to solve the problem..
    Ok so I did Vt=dx/dt which gives x=((MwagonV)/C)*ln(Mwagon+Ct)
    The x is of course L so I isolated everything from the natural log which gives LC/(MwagonV)=ln(Mwagon+Ct)
    This results in Mwagon+Ct=exp(CL/MwagonV).
    Now I get for t --> t=(1/C)*exp((CL/MwagonV)-M/C.

    It appears I am missing a factor M at the exponential function... I'm looking for that missing M but I cannot find it. Did I oversee something or could it be that the given equation has a factor M too many?
  5. May 10, 2015 #4
    I know you are not supposed to give full answers but I really cannot find that missing M. Can you please help me?
  6. May 10, 2015 #5


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    When you integrate you can either do indefinite integrals or definite integrals.

    If you use indefinite integrals, you will have an arbitrary constant of integration that you will need to determine.

    If you use definite integrals, then make sure you use the appropriate limits of integration for x and t and evaluate the integrals at both the upper and lower limits.
  7. May 10, 2015 #6
    Yes! I made a little mistake, I forgot to set the boundaries from 0 to T but now I got it!
    Thank you very much, TSny!
  8. May 10, 2015 #7


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    OK. Good work!
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