# Rockets and fuel expulsion

• Apr 22nd 2013, 07:05 AM
carla1985
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]
• Apr 22nd 2013, 11:40 AM
ebaines
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 $\frac {KE}{dk} = 0$.
• Apr 22nd 2013, 12:05 PM
carla1985
Re: Rockets and fuel expulsion
Quote:

Originally Posted by ebaines
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
• Apr 22nd 2013, 12:58 PM
ebaines
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?
• Apr 22nd 2013, 01:35 PM
carla1985
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 $\frac{M_0}{e^2}$?
• Apr 23rd 2013, 05:07 AM
ebaines
Re: Rockets and fuel expulsion
Yes, that's correct. And so the proportion of M_0 that should be fuel is $\frac {(M_0-M_1)}{M_0} = 1-\frac 1 {e^2}$.