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Rocket Problem

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A rocket on its launch pad has a mass of 20,000kg. The engine fires at t=0s and produces a constant force of 500,000 N straight up. The engines fire for 1 minute during which the entire 10,000kg of fuel on poard is consumed and expelled from the rocket at a constant rate. Ignore air resistance and assume that the force of gravity is constant. We will find the velocity of the rocket as its engines stop firing.

    a. Note the initial and final mass of the rocket. Write down an equation for m(t), the mass as a function of time. You will introduce a constant k which represents the rate at which fuell is burned. Make sure you have the correct units for k.

    b. Draw a free-body diagram, and write down the equation (or equations) that govern the motion. Note that the mass has to be m(t).

    c. From your answer to b, write down an expression for the velocity. Your answer may be left in the form of a definite integral. You do not have to evaluate the integral.​

    2. Relevant equations

    F_net = ma
    x_f = x_i + v_i*t + 1/2 at^2

    3. The attempt at a solution

    I got a linear equation for part a, m(t) = 20,000 - 166.6t, k being 166.6. For b, I plugged that in for m in the first equation above, then solved for a and plugged that into the second equation, but I'm not sure if that's right. I ended up with x(t) = (250,000t^2)/(20,000 - 166.6t). c has me really stumped though. I'd really appreciate a quick answer, as this is due tomorrow morning. Thanks!
    Last edited: Sep 21, 2007
  2. jcsd
  3. Sep 20, 2007 #2

    D H

    Staff: Mentor

    The relevant equations you cited are not relevant here. Those equations implicitly assume a constant mass. You have to use the more general form,

    [tex]\vec{F}=\frac{d}{dt}(m\vec v)[/tex]

    You will not get a quadratic. You can use the chain rule to seperate the effects of the change in mass and change in velocity. You already have an equation for the mass as a function of time. All that is left is solving for dv/dt.
  4. Sep 21, 2007 #3
    Ok, so using that equation, I get 500,000 = (20,000 - 166.6t)*d/dt(v) - 166.6v. So how do I find the derivative of v when I don't know what v is?
  5. Sep 21, 2007 #4
    Please, anyone, I'm still lost and I need this in the next 20 minutes!
  6. Sep 21, 2007 #5
    10 minutes...
  7. Sep 21, 2007 #6
    Thanks anyway guys.
  8. Sep 21, 2007 #7


    User Avatar

    Staff: Mentor

    Sorry about that, but it would be best to PM one of the HH's if one does not get a prompt response.

    this gives the mass flow rate of the propellant, which is also the rate at which the rocket looses mass.

    dm/dt = - 10,000 kg/ 1 min or 166.67 kg/s. That has been done correctly.

    The next step would be to write the rocket's equation of motion,

    m(t) dv(t)/dt = Thrust - Weight, and the weight is changing W(t) = m(t) g, if g is constant.
  9. Sep 21, 2007 #8

    D H

    Staff: Mentor

    Sean, you posted the OP at 4:49 PM CDT yesterday. The first answer was posted less than an hour later. Unless there is an ongoing discussion, that one hour lag is fairly typical. Responses tend to be a lot quicker once a discussion gets going. You should have worked on this yesterday when you had time and people around willing to help you rather than waiting until 30 minutes before the assignment was due.
  10. Sep 21, 2007 #9


    User Avatar

    Staff: Mentor

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