GR8677 #61
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Alternate Solutions |
casseverhart13 2019-08-30 03:31:57 | Cool! I\'ll surely be coming back for the next problem from you. custom car Orlando\r\n | | Herminso 2009-09-21 19:28:34 | Consider the system rocket of mass plus a fraction of combustible traveling at velocity just before the mass is exhausted. From a frame that moves at velocity we see the system instantaneously at rest, and then when the mass is exhausted, we see that the mass moves front at velocity , while moves backward at that is the speed of rocket's exhaust combustible. By momentum conservation
or just the equation given since is negative for the test makers.
Thus is truly the speed of rocket's exhaust combustible | | 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.
u = -V
mdv/dt + udm/dt = 0: Here is the answer. | |
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Comments |
casseverhart13 2019-08-30 03:31:57 | Cool! I\'ll surely be coming back for the next problem from you. custom car Orlando\r\n | | 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.
padkins 2011-02-06 12:44:41 |
I only fart neutrinos
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| | 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 v/t, so the constant u cannot be an instantaneous velocity, D. Therefore it is E.
varsha 2016-02-20 08:30:26 |
Common sense...Nice
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| | 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 plus a fraction of combustible traveling at velocity just before the mass is exhausted. From a frame that moves at velocity we see the system instantaneously at rest, and then when the mass is exhausted, we see that the mass moves front at velocity , while moves backward at that is the speed of rocket's exhaust combustible. By momentum conservation
or just the equation given since is negative for the test makers.
Thus is truly the speed of rocket's exhaust combustible
Quynh_Strange_Beauty 2014-10-13 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)
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| | 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.
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
realcomfy 2008-10-31 03:25:25 |
It was stated in the solution that higher order terms such as were neglected. The solution thus becomes what is already there
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| | hefeweizen 2006-11-30 12:10:53 | is there a typo in the last line?
mdv = -Vdm
blue_down_quark 2008-08-27 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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