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Math Help - Rockets and fuel expulsion

  1. #1
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    Rockets and fuel expulsion

    Really struggling with this one. In class we used velocity not speed for starters and had $m$ and $m+\delta t$. I cant figure out how to relate it to this question at all. If someone could point me in the right direction that would be great.

    A car is propelled by a rocket engine along a smooth straight horizontalroad. Initially, the car is at rest and has mass M0. The spent fuel is expelledwith a constant speed c relative to the car. Show that the kinetic energy ofthe car when all the fuel has been used up is

    \frac{1}{2}M_1c^2(ln(\frac{M_0}{M_1}))^2


    What proportion of the initial mass M0 should the initial mass of fuel be inorder to maximise the kinetic energy of the car when all the full has been usedup?


    [Hint for last part: note that the kinetic energy will be zero if either M1 = 0 orM1 = M0, so to find the value of M1 for which the kinetic energy is maximisedas M1 varies between 0 and M0, you need to differentiate the kinetic energywith respect to M1]
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    MHF Contributor ebaines's Avatar
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    Re: Rockets and fuel expulsion

    You can use impulse and momentum principles:

     m \frac {dv}{dt} = -C \frac {dm}{dt}

    Rearrange and integrate:

     -C \frac {dm } m = dv

     -C \ln (m) = v

     \Delta V = C \ln (\frac {M_0}{M_1} ) where  M_0 = initial total mass of ship plus fuel, and  M_1 = final mass of ship after all fuel is spent.

    Now that you have a value for  \Delta V , you can calculate its final kinetic energy.

    To answer the second part it would be best to introduce a variable for the ratio of the ship's mass to the ship + fuel mass. Let  k = \frac {M_1}{M_0}, and the KE equation becomes

     KE = \frac 1 2 k M_0 C^2 (\ln (\frac 1 k))^2

    Now find the value of k that maximizes KE by setting \frac {KE}{dk} = 0.
    Last edited by ebaines; April 22nd 2013 at 12:20 PM.
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    Re: Rockets and fuel expulsion

    Quote Originally Posted by ebaines View Post
    You can use impulse and momentum principles:

     m \frac {dv}{dt} = -C \frac {dm}{dt}

    Rearrange and integrate:

     -C \frac {dm } m = dv

     -C \ln (m) = v

     \Delta V  = C \ln (\frac {M_0}{M_1} where [tex] M_0 {/tex] = initial total mass of ship plus fuel, and  M_1 = final mass of ship after all fuel is spent.

    Now that you have a value for  \Delta V , you can calculate its final kinetic energy.

    To answer the second part it would be best to introduce a variable for the ratio of the ship's mass to the ship + fuel mass. Let  k  = \frac {M_1}{M_0}, and the KE equation becomes

     KE = \frac 1 2 k M_0 C^2 (\ln (\frac 1 k))^2

    Now find the value of k that maximizes KE by setting \frac {KE}{dk} = 0.
    Hi, thank you very much for the reply. I have completed the first part and understand the secong (although I didnt think of a substitution) but I'm struggling to differentiate the equation with respect to m1
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    MHF Contributor ebaines's Avatar
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    Re: Rockets and fuel expulsion

    If you treat  M_1 as a variable and  M_0 as a constant, then you can proceed as follows:

    KE = \frac {C^2} 2 M_1 (\ln(\frac {M_0}{M_1}))^2

    Now recall that the derivarive of  f(x)g(x) = f(x)g'(x) + f'(x)g(x), and also that the dierivative of  \ln(\frac A x) = \frac {-1} x . So this becomes:

     \frac {d(KE)}{dM_1} = \frac {C^2} 2 M_1 2 (\ln(\frac {M_0}{M_1})) (\frac {-1}{M_1}) + \frac {C^2} 2 (\ln(\frac {M_0}{M_1}))^2

    Set this equal to zero and solve for  M_1. What do you get?
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    Re: Rockets and fuel expulsion

    Iget $M_1$ to be M_0 or \frac{M_0}{e^2} does that mean the energy is max when its at \frac{M_0}{e^2}?
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: Rockets and fuel expulsion

    Yes, that's correct. And so the proportion of M_0 that should be fuel is \frac {(M_0-M_1)}{M_0} = 1-\frac 1 {e^2}.
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