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Impulse-momentum homework

  1. May 10, 2005 #1
    I'm having trouble with the following question. Can anyone please give me a push in the right direction?

    A body consists of equal masses [itex]M[/itex] of inflammable and non-inflammable material. The body descends freely under gravity from rest. The combustible part burns at a constant rate of [itex]kM[/itex] per second, where [itex]k[/itex] is a constant. The burning material is ejected vertically upwards with constant speed [itex]u[/itex] relative to the body, and air resistnace may be neglected. Show, using momentum considerations, that

    [tex]\frac{d}{dt}[(2-kt)v] = k(u-v) + g(2-kt)[/tex]

    where [itex]v[/itex] is the speed of the body at time [itex]t[/itex]. Hence show that the body descends a distance

    [tex]\frac{g}{2k^2} + \frac{u}{k} (1 - \ln 2)[/tex]

    before all the inflammable material is burnt.

    I managed to do the first part, but I have no idea how to approach the second.
  2. jcsd
  3. May 10, 2005 #2


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    I'd like to see how you got the first part. I keep getting another term in that derivative.
  4. May 10, 2005 #3
    (For simplicity, ~d = [itex]\delta[/itex].)

    (m + ~dm)(v + ~dv) - ~dm(v - u) - mv = (m + ~dm)g . ~dt

    Multiplying everything out and letting ~dt, ~dm and ~dv -> 0:
    m(dv/dt) + u(dm/dt) = mg

    dm/dt = -kM
    m = -kMt + C

    When t=0, m=2M, so C =2M. Hence:
    m = 2M - kMt = M(2-kt)

    Substituting this in the ODE:
    M(2-kt)(dv/dt) + u(-Mk) = M(2-kt)g
    (2-kt)(dv/dt) - uk = (2-kt)g

    Substract -vk from both sides:
    (2-kt)(dv/dt) - vk = uk - vk + (2-kt)g
    d[(2-kt)v]/dt = k(u-v) + g(2-kt)

    I hope that's readable. :tongue2:
    Last edited: May 10, 2005
  5. May 11, 2005 #4


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    OK. That looks good. I had totally messed up the split of the mass into 2M with the rate of conversion being a CONSTANT kM.

    I've almost got the result, but I'm off by a sign. I'm confident I'm on the right path, so see what you can do with this. Rewind your derivation to

    (2-kt)(dv/dt) - uk = (2-kt)g

    add the uk to both sides and divide by the (2-kt) factor. Then you can separate to get

    dv = f(t)dt

    Integrate and impose v(t=0)=0 to evaluate the constant. After you get the velocity, integrate from t = 0 to the time of burnout, which from your mass equation is t = 1/k, to find the distance. I get the first term fine. There is an integral of lnxdx that I think I am messing up at the moment because I have a minus in front of the 1 in the [1-ln2]term. I'll see if I can fix it, but you can probably get it.
    Last edited: May 11, 2005
  6. May 11, 2005 #5


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    Possibly, your sign flaw comes from the fact that:
    1) u is given as the fuel's SPEED relative to the body, rather than the relative velocity
    2) Thus, with positive direction parallell with gravity, the relative velocity of the fuel is -u

    At any rate, I did that mixup at first.. :blushing:
    Last edited: May 11, 2005
  7. May 11, 2005 #6


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    Thanks for the suggestion, but that's not it. That would have been a more pervasive problem. It was something very basic though. As one of my British professors used to say "The trouble with you guys (American students) is you don't know how to integrate." This was simply a matter of not correctly changing limits when I changed the integration variable. When I do it right, I get the answer given. At least I can plead "time of day" factors; that time stamp on my post is my local time :rolleyes:
  8. May 12, 2005 #7
    I appreciate the help. Thank you!

    Apparently my problem was finding the time of burnout. I was substituting m=0 instead of m=M, so I always got t=2/k which made things very messy.

    I think the reason you got a minus was because of the integral of ln(2-kt), as you said. Maybe you forgot the minus next to kt?

    Anyway... Thanks again. :smile:
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