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Thread: 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

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


    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:

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

    Rearrange and integrate:

    $\displaystyle -C \frac {dm } m = dv$

    $\displaystyle -C \ln (m) = v$

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

    Now that you have a value for $\displaystyle \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 $\displaystyle k = \frac {M_1}{M_0}$, and the KE equation becomes

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

    Now find the value of k that maximizes KE by setting $\displaystyle \frac {KE}{dk} = 0$.
    Last edited by ebaines; Apr 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:

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

    Rearrange and integrate:

    $\displaystyle -C \frac {dm } m = dv$

    $\displaystyle -C \ln (m) = v$

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

    Now that you have a value for $\displaystyle \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 $\displaystyle k = \frac {M_1}{M_0}$, and the KE equation becomes

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

    Now find the value of k that maximizes KE by setting $\displaystyle \frac {KE}{dk} = 0$.
    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 $\displaystyle M_1$ as a variable and $\displaystyle M_0$ as a constant, then you can proceed as follows:

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

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

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

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

    Iget $M_1$ to be $\displaystyle M_0$ or $\displaystyle \frac{M_0}{e^2}$ does that mean the energy is max when its at $\displaystyle \frac{M_0}{e^2}$?
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    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 $\displaystyle \frac {(M_0-M_1)}{M_0} = 1-\frac 1 {e^2}$.
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