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GR8677 #61
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Mechanics$\Rightarrow$}Conservation of Momentum

One can easily derive the formula for rocket motion,

$\begin{eqnarray} p_{initial}&=&p_{final}\\ mv &=& (m-dm')(v+dv)+(dm')(v-V)\\ mv &=& mv + mdv - dm'v - dm'dv + dm'v - dm'V\\ mdV &\approx& dm'V\\ mdV &=& -dm V,
\end{eqnarray}$

where $m$ is the mass of the rocket and $m'$ is the mass of the fuel rejected. $v$ is the initial velocity of the rocket. $V$ is the velocity of the exhaust. The final line comes about from realizing that $dm=-dm'$.

The derivation starts with the assumption that the final mass of the rocket is its original mass minus $dm'$, the eject mass, and that its final velocity is a tiny bit faster than it was before $dv$. The exhaust mass' velocity relative to the rocket is $V$. (Higher order terms such as $dxdy$ have been thrown out to arrive at the final differential equation.)

$V$ is equivalent to the velocity of the exhaust mass relative to the rocket, the inertial reference system. Thus $u=V$ and the answer is choice (E).

(Note that firing the exhaust backwards generates a forward thrust for the rocket---motion in space is quintessentially dependent on the phenomenon of farting. So, if one is well-equipped and wanna get going in space, just let out a bit of gas, and one's good to go.)

Alternate Solutions
 Herminso2009-09-21 19:28:34 Consider the system rocket of mass $m$ plus a fraction $\triangle m$ of combustible traveling at velocity $v$ just before the $\triangle m$ mass is exhausted. From a frame that moves at velocity $v$ we see the system instantaneously at rest, and then when the mass $\triangle m$ is exhausted, we see that the mass $m$ moves front at velocity $\triangle v$, while $\triangle m$ moves backward at $u$ that is the speed of rocket's exhaust combustible. By momentum conservation $0=m\triangle v-\triangle m u$ $\Rightarrow$ $m(\triangle v/\triangle t)-(\triangle m/\triangle t) u=0$ or $m(\partial v/\partial t)-(\partial m/\partial t) u=0$ just the equation given since $u$ is negative for the test makers. Thus $u$ is truly the speed of rocket's exhaust combustibleReply to this comment kimseonta2009-01-07 01:18:00 Sorry about the one that I posted below... it is totally incorrect...so I am now going to post the solution again... The rocket is in a free space -> No external forces the momentum must be conserved. Pi=Pf mv=(m+delta m)(v+ delta v)+(V+v)( - delta m) , where the delta m should be negative when considering a rocket propulsion. And V is the velocity [Not speed] of the rocket's exhaust from the rocket's frame. [m + (delta m)](delta v) - V(delta m) = 0 take /(delta t) and limit[delta t -> 0] Then, mdv/dt - Vdm/dt = 0 Here, of course, V is negative[vector]. Let u be the speed[scalar] of the rocket's exhaust from the rocket's frame. $\Rightarrow$ u = -V mdv/dt + udm/dt = 0: Here is the answer.Reply to this comment
GPRE
2013-09-24 07:26:28
Would it be right to look at it like this: The second term suggests a change in mass and the only way the rocket's mass changes is by spending fuel.
Almno10
2010-11-11 23:11:14
If you wanted to fart in space in order to propel yourself forward, you'd have to pull your pants down first so the fart particles wouldn't collide with the seat of your pants.
 padkins2011-02-06 12:44:41 I only fart neutrinos
subwoofer911
2009-11-03 23:43:34
v is the speed of the rocket, so A, B and C are automatically wrong. And there IS an acceleration $\partial$v/$\partial$t, so the constant u cannot be an instantaneous velocity, D. Therefore it is E.
 varsha2016-02-20 08:30:26 Common sense...Nice
abacus
2009-10-06 23:39:42
Rocket equation must remain correct under gallilean symmetry
v -> v+v0

this leaves E
Herminso
2009-09-21 19:28:34
Consider the system rocket of mass $m$ plus a fraction $\triangle m$ of combustible traveling at velocity $v$ just before the $\triangle m$ mass is exhausted. From a frame that moves at velocity $v$ we see the system instantaneously at rest, and then when the mass $\triangle m$ is exhausted, we see that the mass $m$ moves front at velocity $\triangle v$, while $\triangle m$ moves backward at $u$ that is the speed of rocket's exhaust combustible. By momentum conservation

$0=m\triangle v-\triangle m u$ $\Rightarrow$ $m(\triangle v/\triangle t)-(\triangle m/\triangle t) u=0$ or $m(\partial v/\partial t)-(\partial m/\partial t) u=0$ just the equation given since $u$ is negative for the test makers.

Thus $u$ is truly the speed of rocket's exhaust combustible
 Quynh_Strange_Beauty2014-10-13 23:26:50 We can quickly see that the meaning of equation is momentum conservation (Delta (mv)_1 - Delta (mv)_2 = 0). That's enough. no more math. So u must be the speed of the exhaust. (D or E) OK. What happen if u = 0 ? the equation says rocket gain no momentum. So u must be speed of exhaust relative to rocket. (IF I picked D, my rocket can gain momentum if I choose certain frame of reference, not true)
kimseonta
2009-01-07 01:18:00
Sorry about the one that I posted below...
it is totally incorrect...so I am now going to post the solution again...

The rocket is in a free space -> No external forces
the momentum must be conserved.

Pi=Pf

mv=(m+delta m)(v+ delta v)+(V+v)( - delta m)

, where the delta m should be negative when considering a rocket propulsion. And V is the velocity [Not speed] of the rocket's exhaust from the rocket's frame.

[m + (delta m)](delta v) - V(delta m) = 0

take /(delta t) and limit[delta t -> 0]

Then,

mdv/dt - Vdm/dt = 0

Here, of course, V is negative[vector]. Let u be the speed[scalar] of the rocket's exhaust from the rocket's frame.
$\Rightarrow$ u = -V

mdv/dt + udm/dt = 0: Here is the answer.
kimseonta
2009-01-06 20:13:36
This is my own way to make the typo (?) clear...

The same idea goes: the conservation of the momentum.

Pi = Pf

mv = (m + dm)(v+dv) + (dm)V, where V <0 of course.

= mv+mdv+vdm+dmdv+(dm)v

Now there exist two appoximations that one can make.

- V >> v : one can simply assume that the speed of the exaust from a rocket should be much faster than the one of the rocket.

-dmvm is almost zero relative to the other parts of the equation.

so the simplified version will be here...

mdv+Vdm=0

m(dv/dt) + V(dm/dt) = 0
Poop Loops
2008-09-24 10:44:52
I don't get the jump from Line 2 to Line 3. Where does mdV come from? And why is it approx. equal to dm'V?

mv = mv + mdv - dm'v - dm'dv + dm'v - dm'V

Subtract mv from both sides and -dm'v and +dm'v cancel, so you get:

0 = mdv - dm'dv - dm'V

(m - dm')dv = dm'V

So I have no idea how (m - dm')dv relates to mdV

 realcomfy2008-10-31 03:25:25 It was stated in the solution that higher order terms such as $dm'dv$ were neglected. The solution thus becomes what is already there
hefeweizen
2006-11-30 12:10:53
is there a typo in the last line?

mdv = -Vdm
 blue_down_quark2008-08-27 23:44:48 I think V is assumed to be negative in the reference frame where v is positive. It's simply the exhaust velocity without its sign.

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