MHB How Can Rocket Fuel Expulsion Maximize a Car's Kinetic Energy?

[Carla1985]

Really struggling with this one. In class we used velocity not speed for starters and had $m$ and $m+\delta t$. I can't 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

\[
\frac{1}{2} M_1c^2 (ln(\frac{M_1}{M_2}))^2
\] 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]

 

[I like Serena]

Really struggling with this one. In class we used velocity not speed for starters and had $m$ and $m+\delta t$. I can't 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

\[
\frac{1}{2} M_1c^2 (ln(\frac{M_1}{M_2}))^2
\] 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]

Hey Carla! :)

Suppose at some time t we have mass m and speed v.
Then we expel an infinitesimal amount of mass dm.
As a result our speed increases by dv, while at the same time the expelled mass goes backward with a relative speed c.

Can you set up an equation for conservation of momentum before and after expelling the infinitesimal mass dm?

 

[Carla1985]

re: Rockets & fuel expulsion

I have:

$m(t)*\frac{dv}{dt}+\frac{dm(t)}{dt}* \textbf{c}=\textbf{F}$

I think $\textbf{F}$ is 0 as we aren't modelling any forces on it atm.
So its there fore equivalent to

$\frac{d}{dt}(v(t)+\textbf{c}*ln(m(t))=0$

I think that's right :/

 

[I like Serena]

re: Rockets & fuel expulsion

I have:

$m(t)*\frac{dv}{dt}+\frac{dm(t)}{dt}* \textbf{c}=\textbf{F}$

I think $\textbf{F}$ is 0 as we aren't modelling any forces on it atm.
So its there fore equivalent to

$\frac{d}{dt}(v(t)+\textbf{c}*ln(m(t))=0$

I think that's right :/

That looks about right. :)
But we're not really interested in t, which does complicate the equation.

So let's write it as:
$$m\ dv + c\ dm = 0$$
$$dv = -\frac c m dm$$
Can you integrate that?

 

[Carla1985]

re: Rockets & fuel expulsion

oh ok. yes i think:

$v=-c\ ln(m)$

 

[Carla1985]

Re: Rockets & fule expulsion

I know I need to use the kinetic energy equation here but I'm not sure exactly how

 

[I like Serena]

Re: Rockets & fule expulsion

oh ok. yes i think:

$v=-c\ ln(m)$

That should be:
$$v_{final}=\left. -c\ln(m) \phantom{\frac{}{}} \right|_{m_{initial}}^{m_{final}}$$

I know I need to use the kinetic energy equation here but I'm not sure exactly how

Final kinetic energy is:
$$\frac 1 2 m_{final} {v_{final}}^2$$
Substitute?

 

[Carla1985]

Re: Rockets & fule expulsion

That should be:
$$v_{final}=\left. -c\ln(m) \phantom{\frac{}{}} \right|_{m_{initial}}^{m_{final}}$$
Final kinetic energy is:
$$\frac 1 2 m_{final} {v_{final}}^2$$
Substitute?

Ah yes so $v_1=-cln(M_1)+cln(M_0)=cln(M_0)-cln(M_1)=cln(\frac{M_0}{M_1})$

and substitute this in for $v_1$ in the energy eqn

 

[I like Serena]

Just out of curiosity, did you realize that conservation of momentum yielded:
$$mv=(m-dm)(v+dv) + dm(v-c)$$

 

[Carla1985]

Hi, yes in some of our workings out we had that the change in momentum is

\[
\int_{t}^{t+\delta m}\textbf{F}\ ds=(m+\delta m)(v+\delta v)-\delta m(v-c)-mv
\]

which obviously if we arent modelling any forces is equal to 0 so gives your equation :) Thanks for the help btw, it really is appreciated.

 

[Carla1985]

Hi, sorry to be an absolute pain but my exam is in 2 days n there's a part of this I'm still not fully understanding. When deriving the rocket equation, this is what our notes say:

"Now we suppose that a force F is acting on the rocket during this process, then the impulse is:

$\int_{t}^{t+\delta t}F dt=F\delta t$

assuming that F is continuous in the small interval [t, t + δt]. Therefore considering the change in momentum,
(m + δm)(v + δv) − δm(v − c) − mv ≈ F δt.

If we divide by δt and let δt → 0 then we find,

$m\frac{dv}{dt}+\frac{dm}{dt}c=0$

This the rocket equation. "

I'm confused by the swapping about of d and delta. Would someone mind explaining what is happening please when they have a little time :) thanks x

 

[I like Serena]

Hey Carla1985!

The symbol $\delta t$ represents a very small time interval.
The infinitesimal $dt$ represents a time interval that is infinitely small, that is, $\delta t$ taken to the limit where it is zero.

For instance the average acceleration over interval $$\delta t$$ starting at time t is $$a_{average} = \frac {\delta v}{\delta t}$$.
The actual acceleration at time t is $$a = \frac {dv}{dt} = \lim_{\delta t \to 0} \frac {\delta v}{\delta t}$$.

Good luck! ;)

 

[Carla1985]

Aaaaah, that makes sense. Once again, thank you :)