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United States24670 Posts
For those who didn't read the introductory entry, head over to http://www.teamliquid.net/blogs/viewblog.php?topic_id=79275
This is entry 3.
Last week the question of the week was:
Question 2
Newton's Laws of Motion
If a can of compressed air is punctured and the escaping air blows to the right, the can will move to the left in a rocket-like fashion. Now consider a vacuum can that is punctured. The air blows in the left as it enters the can. After the vacuum is filled the can will:
a) be moving to the left
b) be moving to the right
c) not be moving
The correct answer is c. First of all, for those who were confused about which way the can was punctured, I agree the wording was a bit tricky... but it didn't matter!
It's very easy to get confused by this problem because there is going to be a momentarily slight oscillation about the center of mass. Just as the can is punctured, the pressure that was being exerted by the atmosphere on the now-absent piece of metal doesn't exist anymore... and the force acting on that surface is weaker than that acting on the opposite side of the can. However, the air rushing into the can will eventually hit the opposite wall, which will provide that force (plus a bit extra until everything equals out). These two drawings I made should clear it up:
I assign this problem a difficulty level of 3 out of 5 because it's easy to confuse the physics principles.
Congratulations to Xeofreestyler for being the first to provide the correct answer and provide a reasonable justification. Dr.Dragoon named the problem.
There were a lot of explanations that were either wrong (even among people who chose C) or not backed up. For example: saying 'C because conservation of momentum' isn't necessarily wrong, but is not sufficient in my opinion. There were a few explanations that I am not sure if they are correct or not... although nobody quite used the one that I provided. Feel free to provide further analysis.
Question 3
Momentum and Energy
An open railroad car is rolling without friction in a vertically-falling downpour and an appreciable amount of rain falls into the car and accumulates there. Consider the effect of accumulating rain on the speed, momentum, and kinetic energy of the car.
1) The speed of the car will
(a) increase
(b) decrease
(c) not change
2) The momentum of the car will
(a) increase
(b) decrease
(c) not change
3) And the kinetic energy of the car will
(a) increase
(b) decrease
(c) not change
The rain has now stopped. A drain plug is opened in the bottom of the rolling car allowing the accumulated water to run out. Consider the effects of the draining water on the speed, momentum, and kinetic energy of the rolling car.
4) The speed of the car will
(a) increase
(b) decrease
(c) remain the same
5) The momentum of the car will
(a) increase
(b) decrease
(c) remain the same
6) And the kinetic energy will
(a) increase
(b) decrease
(c) remain the same
Remember to spoiler your answers and explanations, at least on the first page!
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of the car, or of the car + water?
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+ Show Spoiler +1) The speed of the car will
(b) decrease
2) The momentum of the car will
(a) increase
3) And the kinetic energy of the car will
(c) not change
The rain has now stopped. A drain plug is opened in the bottom of the rolling car allowing the accumulated water to run out. Consider the effects of the draining water on the speed, momentum, and kinetic energy of the rolling car.
4) The speed of the car will
(a) increase
5) The momentum of the car will
(b) decrease
6) And the kinetic energy will
(c) remain the same
Assuming the only affect the water has on the train is to change its mass, this is conservation of energy, and you can just use KE = 1/2(m)(v^2). So energy stays the same regardless, velocity is inversely related to mass, and because energy depends on v^2, momentum, L = mv won't be conserved .
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United States24670 Posts
Spoiler it you fag. If I ever meet you on bnet I'm going to make fun of your mother.
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well, lets see if I get this right...
+ Show Spoiler +1b)it needs to move accelerate the vertical water, which it does effectively by colliding with it.
2c) the overall momentum of the car + contained water will not change, because the water has no horizontal momentum, and it must be conserved.
3b)for the car+contained water, mv=momentum, unchanging, v decreases. E=mv*v/2 so E decreases.
4a) nothing acts on the car +contained water, so it must stay the same velocity.
5b) for the car + contained water: it loses water, so therefore the momentum stored in that water.
6b) energy in the escaping water is not counted in the car+water, so for the car+water, it decreases.
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On September 28 2008 12:43 micronesia wrote:
There were a lot of explanations that were either wrong (even among people who chose C) or not backed up. For example: saying 'C because conservation of momentum' isn't necessarily wrong, but is not sufficient in my opinion. There were a few explanations that I am not sure if they are correct or not... although nobody quite used the one that I provided. Feel free to provide further analysis.
Question 3
Momentum and Energy
An open railroad car is rolling without friction in a vertically-falling downpour and an appreciable amount of rain falls into the car and accumulates there. Consider the effect of accumulating rain on the speed, momentum, and kinetic energy of the car.
1) The speed of the car will
(a) increase
(b) decrease
(c) not change
+ Show Spoiler +b) decrease.....if it stayed the same or sped up, it would have more energy than it initially had
2) The momentum of the car will
(a) increase
(b) decrease
(c) not change
+ Show Spoiler +c) not change...the momentum of the entire system is constant.
3) And the kinetic energy of the car will
(a) increase
(b) decrease
(c) not change
+ Show Spoiler +b) decrease.....inelastic collision
The rain has now stopped. A drain plug is opened in the bottom of the rolling car allowing the accumulated water to run out. Consider the effects of the draining water on the speed, momentum, and kinetic energy of the rolling car.
4) The speed of the car will
(a) increase
(b) decrease
(c) remain the same
+ Show Spoiler +c) remain the same....the rain is in the train's frame of reference and just free falling
5) The momentum of the car will
(a) increase
(b) decrease
(c) remain the same
+ Show Spoiler +b)decrease..less mass same speed
6) And the kinetic energy will
(a) increase
(b) decrease
(c) remain the same
+ Show Spoiler +decrease...less mass same speed
Remember to spoiler your answers and explanations, at least on the first page!
+ Show Spoiler +
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+ Show Spoiler +fuck, i know I got it wrong. Forgot the friction between the car and the rain, which exerts the horizontal force that slows the car down.
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fuck I've done it again. right reasoning, right answer, wrong letter.
+ Show Spoiler +
4c) nothing acts on the car +contained water, so it must stay the same velocity.
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+ Show Spoiler +
1)b Since the rain is falling vertically, it will have to be accelerated up to the speed of the railroad car
2)b Assuming that we are talking about the momentum of the car, not the car with the rain it's carrying, it will decrease because the speed will decrease. If we're talking about the car with the rain in it, then the momentum will remain the same, due to conservation of momentum
3)b Will decrease for the same reason, assuming that it's only the car under consideratoin, but it will still decrease even if it's the car and the rain, because there's a v^2 in kinetic energy
4)c I'm not sure how to explain this
5)c Again, assuming that we're talking about the car, specifically, and not the car with water in it, since the speed doesn't change, the momentum won't either. If you count the water as part of the car, then the momentum will decrease, because of the mass decrease
6)c if it's just the car under consideration, b if it's the car+water in it.
Am I right?
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+ Show Spoiler +1. C The speed will not change because velocity is not affected by mass, though force and acceleration are. The only reason that things slow down when big stuff are put on them is because of the normal force and friction, but this is a frictionless railroad car.
2. A The momentum of the car will increase as the mass increases.
3. A The kinetic energy of the car will increase as the mass increases.
For both of 2 and 3, though, where does the new momentum and KE come from...
4. C Once again, the speed will not change because the velocity is not affected by the mass.
5. B Less mass = less momentum...
6. B Less mass = less kinetic energy... goddamnit I know this doesn't even make any sense...
Geez hmm... i may edit my answer.
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United States24670 Posts
On September 28 2008 13:19 betaben wrote:fuck I've done it again. right reasoning, right answer, wrong letter. + Show Spoiler +
4c) nothing acts on the car +contained water, so it must stay the same velocity.
Go ahead and edit your earlier post if you haven't already so that it's easier for me to see what your answer was when I go through it.
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+ Show Spoiler +
I'll go with my gut and likely get tricked again
1) C - Not change. The car is rolling without friction so nothing is slowing it down, and the rain is falling vertically so I don't think it'll cause a deceleration
2) C - I'm not sure if the water counts, so I'm gonna say no it doesn't, and the momentum of the car itself should be the same
3) C - as above
4) C - Same as 1
5) C - Same as 2
6) C - Same as 3
I like my answers, nice and uniform
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btw PM me the right answer? I'm curious, and I might miss your next blog! Seeing as I didn't see any of the ones previous to this apparently.
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Assuming "car" refers to the car itself and not its contents...
+ Show Spoiler +1. Decrease. Could see raindrops as moving backwards inside the car until they hit the back end, at which point they exert a backwards force on the car until they have accelerated to the car's velocity (while decelerating the car a tiny amount - energy ain't free).
2. Some of the momentum has transferred to the raindrops (the mass of the car itself is unchanged, but its speed is lower), so it decreased.
3. The car lost kinetic energy to the raindrops while accelerating them.
4. Remain the same - the car won't speed back up; the draining water is moving at the same rate as the car and won't "give back" the speed. This is neglecting the possibility that the drain is not centered on the center of mass of the water, which actually would change the car's velocity but in a way that is unknowable without knowing the location of the drain.
5. Well, the mass of the car isn't changing, nor is its speed. The momentum is the same.
6. Nothing is accelerating or decelerating the car. The kinetic energy is the same.
Now, with the interpretation that "car" refers to both the car and what is currently inside of it:
+ Show Spoiler +
1. Decrease. The newly acquired raindrops are falling straight down, so they are accelerated by the car to the car's velocity. This decreases the speed.
2. Increase. P = mv, and m is increasing while v is decreasing. However, m increases at a linear rate and v decreases at an inverse proportion to m (e = 1/2 * mv^2).
3. Energy ain't going anywhere...
4. Assuming the drain plug is centered on the car's water holding volume's center of mass in the XY plane, the draining water won't accelerate the car in any direction. Or more precisely, the acceleration of the remaining water inside the car towards the drain will be a net 0. Thus, the force exerted by said water on the car is 0, and the car's velocity does not change.
5. Mass is decreasing while v is constant, so the momentum decreases.
6. Same argument; energy decreases.
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On September 28 2008 14:11 Jonoman92 wrote:
btw PM me the right answer? I'm curious, and I might miss your next blog! Seeing as I didn't see any of the ones previous to this apparently.
same here
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United States24670 Posts
On September 28 2008 14:11 Jonoman92 wrote:
btw PM me the right answer? I'm curious, and I might miss your next blog! Seeing as I didn't see any of the ones previous to this apparently.
On September 28 2008 14:39 fight_or_flight wrote:Show nested quote +On September 28 2008 14:11 Jonoman92 wrote:
btw PM me the right answer? I'm curious, and I might miss your next blog! Seeing as I didn't see any of the ones previous to this apparently. same here
I'm glad you guys are eager, but I'm going to have to ask you to just wait until tomorrow or the day after or whenever I post next. I don't want to get into the habit of having to pm 50 people with the advanced answer :p
Also, if you are worried you will miss it, just go to one of my posts and click the 'blog' link to scroll through my most recent blog posts.
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+ Show Spoiler +Hmm wait, theres no way it could speed up after draining the water, i'm pretty sure about the rest of my answers though.
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Maybe it is my bad english but is it possible to ROLL without a friction.
As for the question
+ Show Spoiler +
Some definitions should be made. First when i talk about car i mean car itself and water in the car, under the term kinetic energy of the car i understand energy related with the motion of the center of the mass. And it looks like sentense 'Consider the effect of accumulating rain on ...' implied that all droplet hitting the car will be retained (but it is not clear).
Lets begin with momentum. As far as all droplets will get into the car there will not be any internal forces so the momentum of droplets + momentum of car will not change. And considering that droplets had no horizontal component of momentum (when they were not part of the system) we can conclude that momentum of the car will not change.
The mass of the car will increase and the momentum will not change it means that velocity will decrease.
It is easuy to show that kinetic energy wil also decrease because part of it will transform into internal energy (if collisions with droplets are inelastic) or into the wave energy (if collisions are elastic).
In the second part we assume that all energy was dissipated and there are no waves inside the car otherwise the answer would be dependant on the way of draining.
The velocity will not change because if the plug in the bottom there will not be any additional forces. Momentum and kinetic energy will decrease because part of it will be taken out with water.
1)b
2)c
3)b
4)c
5)b
6)b
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United States24670 Posts
On September 28 2008 16:34 15vs1 wrote:
Maybe it is my bad english but is it possible to ROLL without a friction.
It's fairly typical for the word friction to be used interchangeably in two ways when talking about wheels.
1) The friction between the wheel surface and the ground. As long as this is large enough, the wheel will 'roll without slipping'.
2) The friction acting on the axle etc... provides a negative torque which decreases angular velocity. Does not apply in simplified cases where friction is ignored.
The problem is just saying not to worry about 2.
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On September 28 2008 12:43 micronesia wrote:+ Show Spoiler +For those who didn't read the introductory entry, head over to http://www.teamliquid.net/blogs/viewblog.php?topic_id=79275This is entry 3. Last week the question of the week was: Question 2
Newton's Laws of Motion
If a can of compressed air is punctured and the escaping air blows to the right, the can will move to the left in a rocket-like fashion. Now consider a vacuum can that is punctured. The air blows in the left as it enters the can. After the vacuum is filled the can will:
a) be moving to the left
b) be moving to the right
c) not be moving The correct answer is c. First of all, for those who were confused about which way the can was punctured, I agree the wording was a bit tricky... but it didn't matter! It's very easy to get confused by this problem because there is going to be a momentarily slight oscillation about the center of mass. Just as the can is punctured, the pressure that was being exerted by the atmosphere on the now-absent piece of metal doesn't exist anymore... and the force acting on that surface is weaker than that acting on the opposite side of the can. However, the air rushing into the can will eventually hit the opposite wall, which will provide that force (plus a bit extra until everything equals out). These two drawings I made should clear it up: I assign this problem a difficulty level of 3 out of 5 because it's easy to confuse the physics principles. Congratulations to Xeofreestyler for being the first to provide the correct answer and provide a reasonable justification. Dr.Dragoon named the problem. There were a lot of explanations that were either wrong (even among people who chose C) or not backed up. For example: saying 'C because conservation of momentum' isn't necessarily wrong, but is not sufficient in my opinion. There were a few explanations that I am not sure if they are correct or not... although nobody quite used the one that I provided. Feel free to provide further analysis.
I am sorry but your answer is wrong.
Assuming that all of the air moves at the same speed(Really makes no difference of the answer but makes it easier to reason).
Now there are two phases you have to consider which are constantly interchanging:
Phase 1:
In this phase the air have not reached the far end of the can yet, at this time there is a net force towards the hole since the hole is a net lack of force.
Phase 2:
In this phase the air have reached the far end for the first time, there is no net force except for air resistance which should be negligible in these small time frames. During this phase the amount of air in the bottle is area of hole/area of a bottle side.
Phase 1:
Now the air which hit the far end have came back to the near end again. Since we were at an equilibrium just before this moment we now get an extra force towards the hole direction. Note however that this is smaller than the first phase 1 since we lost a bit of air going out the hole, but at the same time that slows down the filling of the bottle by the same amount and at the end the total work on the bottle is directly proportional to the volume of it and the force never stops until it is completely full.
In the end we will thus receive a net force towards the hole of the bottle as long as the bottle is not full. The bottle will move, case closed.
And just to note, the fact that the air moves faster in one direction than the other and thus creates a force on the bottle as long as it is moving is air resistance, it will not be fast enough to slow it down ever and unless we count other things such as friction it will for sure be moving.
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Klockan, what happens when the particles that you are talking about reach the other side again?
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klockan3, since the can and air starts in rest (relative each other) AND ends up in rest (the air in the can has to have the same velocity as the can itself macroscopically) the can cannot possibly move AFTER the equillibrium has been reached again (i.e no net velocity of the system)
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On September 28 2008 18:21 fight_or_flight wrote:
Klockan, what happens when the particles that you are talking about reach the other side again?
Phase 2, aka a new equilibrium of forces, until the stream hits the near end again etc.
Now in reality we would not have phases since the velocity of air is a distribution, however we can look at every velocity in isolation and then add them all up to a constant net force relative to how full the bottle is.
On September 28 2008 18:31 glassmazarin wrote:
klockan3, since the can and air starts in rest (relative each other) AND ends up in rest (the air in the can has to have the same velocity as the can itself macroscopically) the can cannot possibly move AFTER the equillibrium has been reached again (i.e no net velocity of the system)
The air/outside system do not end at rest though, I explained this in my post at the relevant topic.
On September 28 2008 10:25 Klockan3 wrote:Much wrong use of the laws of Newton in this thread. If someone finds anything wrong with this please poke it, but for now I believe that most in this thread are wrong: + Show Spoiler +Actually, I am pretty sure that we would have a net force on the can during the fillup process, and once it is full we would have an equilibrium again and there is no way air resistance would stop it in an instant.
This do not break any of newtons laws since when it is absorbing air it will reflect less molecules and thus while it will be moving towards the hole the closed system would start moving in the other direction due to the air pressure getting slightly lower on one end. (Or the average of the gas will be moving in the direction the can is not moving)
The easiest way to picture this is to imagine a square. Remove one of the walls and imagine that its interior gets filled by gas molecules, until the molecules who entered fills the whole square up the air pressure on the other side of it will create a net force and thus speed it up, and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval). The net force on the square gets exactly evened out by the net loss of reflected air from the removal of one of the walls.
If you want an open system just imagine that we have an artificial air around it, all with a speed directed towards the can and only in a volume to have exactly enough air to hit the can untill it has reached the critical point were we stop observing it. In such a system we would get a gas sphere around the can except were the hole is, there we would also get an air hole in the sphere, and that is were the momentum gained by the can is taken from.
In the end, the answer is B: It is moving to the right.
I can not believe that so many missed this...
Also yes even if the hole is extremely small, the fact that there is no air resistance unless the object is moving, and that until the last of the air have moved in there is always a net force working on the can from the higher pressure from the wall wo a hole it got to be moving at the time when equilibrium is reached.
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ok, lets say the air molecules moves with speed v_air microscopically.
then as the can starts moving to the right with speed v_can, the air molecules hitting the can on the left outside wall will hit with the relative speed v_can - v_air.
when the air that enters the can starts hitting the left inside wall, the molecules will hit with speed v_can + v_air and create a higher pressure than that on the left outside wall.
the effect is that as long as the can is moving to the right, air will create a force on the can towards the left..
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On September 28 2008 18:44 glassmazarin wrote:
ok, lets say the air molecules moves with speed v_air microscopically.
then as the can starts moving to the right with speed v_can, the air molecules hitting the can on the left outside wall will hit with the relative speed v_can - v_air.
when the air that enters the can starts hitting the left inside wall, the molecules will hit with speed v_can + v_air and create a higher pressure than that on the left outside wall.
the effect is that as long as the can is moving to the right, air will create a force on the can towards the left..
Which is the air resistance, I have already covered that.
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On September 28 2008 18:31 Klockan3 wrote:
The easiest way to picture this is to imagine a square. Remove one of the walls and imagine that its interior gets filled by gas molecules, until the molecules who entered fills the whole square up the air pressure on the other side of it will create a net force and thus speed it up, and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval). The net force on the square gets exactly evened out by the net loss of reflected air from the removal of one of the walls.
i assume this is where you covered air resistance, please correct me if im wrong 
"and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval)."
this is wrong imo. as soon as air hits the left inner wall, the air will create a force on the can in the opposite direction of the cans movement, lasting until the can has stopped.
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On September 28 2008 19:13 glassmazarin wrote:
this is wrong imo. as soon as air hits the left inner wall, the air will create a force on the can in the opposite direction of the cans movement, lasting until the can has stopped.
I covered air resistance in more than one place, anyhow:
The can gets a force trying to slow it down by air resistance as long as it moves, this force is proportional to the square of the velocity.
As long as the can is filling it will get a force trying to speed it up.
Now the force of air resistance is proportional to the movement speed of the object, as such even if it took 100 million years to fill the bottle and as such the air resistance got almost infinite time to equalize with the accelerating force, the fact that there is a force until the very end of the process to fill it means that air resistance would need to slow it down to a stop in 0 time or it will be moving at the exact moment it gets full.
And that is just the extreme case of air resistance > all, normally air resistance would not have much of a say the 1/10 of a second the can is filling up.
But you are right though that the whole system converges against everything being at a relative rest, but that were not the question.
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aha.. i was assuming that the "full" condition = everything is in equillibrium again, but just as the can gets filled with the last air molecule, it will move to the right.
seems that you have the right answer there : )
EDIT: Crap, im wasting too much time on this right now >.<
If one would define the hole as being "small enough" (i.e infinitesimal) we can argue that the molecules enter the can one by one.
when the last molecule enters (filling the can) the can will be moving to the right, but as soon as the molecule has hit the opposite wall, the can will stop.
air molecules moves with speeds ~500 m/s, i.e the time between "filling" and "stopping" will be of order ~1/10000 s for a can 5 cm wide in this case. compared to the time it takes to fill the can from scratch this can be argued to be negligable and thus the can stops "at the same time" as the can is filled.
however the question did not say anything about the size of the hole, so the correct answer would still be B, but i believe this is what the question was supposed to say
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Germany2896 Posts
Assume: the car moves horizontally.
Assume: The rain which is currently in the car counts as part of the car. If that assumption is wrong please clearify the task.
+ Show Spoiler +1b 2c 3b 4c 5b 6b Reason for 2: the impact of the rain can be regarded a inelastic impact. The (horizontal component of) the momentum of the rain is 0. Conservation of momentum applies to inelastic impact. for 1: v=p/m p2=p1, m2>m1 =>v2<v1 for 3: E=0.5*p^2/, p2=p1, m2>m1 =>E2<E1 Reason for 4: In the moved system the emmited water is evenly distributed in all(horizontal) directions. Thus no force applies to the car and its speed is constant. 5: Decrease, p=v*m v2=v1 m2<m1 =>p2<p1 The total momentum is conserved because the emmited rain carries part of the momentum 6: Decrease, E=0.5*v^2*m v2=v1 m2<m1 =>E2<E1
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United States24670 Posts
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points.
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On September 29 2008 01:22 micronesia wrote:
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points.
I am also a physicist...
Anyway:
1: In an open system it is the air which it did not reflect where the hole is. It might be easier to imagine if you see it as a closed system though, imagine that the can is inside a box. As soon as the can gets punctured it reflects less air molecules against one side than the other which means that while it accelerates itself towards one direction it would at the same time accelerate the box it is inside towards the other.
2: Yes, the leftwards pressure always comes before the rightwards pressure equalizes it out. However since a system with no acting forces does not change in velocity we can sum together all of these gaps into the total mechanical work done on the can.
About the center of mass, think like this: Remove everything else from the model except just the gas molecules which are hitting the can. You can see that if it is a regular can the reflected air will create a spherical pattern around the can. However if we open a hole with vacuum we also create a hole in this sphere and the movement of the can is only to make up for this lack of molecules moving in that direction.
And no, as I said, air resistance should not even be considered in the time frame it takes to fill an ordinary sized can with gas under ordinary temperature and pressure and even if we make ridiculous assumptions which would mean it would be a factor it is still debatable what the answer would be.
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United States24670 Posts
On September 29 2008 03:55 Klockan3 wrote:Show nested quote +On September 29 2008 01:22 micronesia wrote:
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points. I am also a physicist...
This was not obvious to me.
How do you define physicist?
Anyway:
1: In an open system it is the air which it did not reflect where the hole is. It might be easier to imagine if you see it as a closed system though, imagine that the can is inside a box. As soon as the can gets punctured it reflects less air molecules against one side than the other which means that while it accelerates itself towards one direction it would at the same time accelerate the box it is inside towards the other.
2: Yes, the leftwards pressure always comes before the rightwards pressure equalizes it out. However since a system with no acting forces does not change in velocity we can sum together all of these gaps into the total mechanical work done on the can.
About the center of mass, think like this: Remove everything else from the model except just the gas molecules which are hitting the can. You can see that if it is a regular can the reflected air will create a spherical pattern around the can. However if we open a hole with vacuum we also create a hole in this sphere and the movement of the can is only to make up for this lack of molecules moving in that direction.
And no, as I said, air resistance should not even be considered in the time frame it takes to fill an ordinary sized can with gas under ordinary temperature and pressure and even if we make ridiculous assumptions which would mean it would be a factor it is still debatable what the answer would be.
The example you gave is very similar to the one provided in a book which analyzes this problem. I'll let you see the actual drawings rather than my own creations:
http://img141.imageshack.us/my.php?image=poofandfoopgx4.jpg
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On September 29 2008 05:23 micronesia wrote:Show nested quote +On September 29 2008 03:55 Klockan3 wrote:On September 29 2008 01:22 micronesia wrote:
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points. I am also a physicist... This was not obvious to me. How do you define physicist?
Someone who have worked in the field of physics? I have done work on how to utilize the inhomogenous effects of acoustic radiation pressure to guide objects through fluids on a micrometer scale.
That book is wrong though, if you consider the watercart picture 2 and instead of having a fixed inner containers you would put wheels on them too, then the outer cart and the inner cart would both accelerate in different directions which is obvious. At leasts with the small cart being filled, the other would not as much since fluids are much slower got and a much higher resistance to movement than gases and thus most of the work would just become heat.
And I already explained what he puts in the last picture, that the effects of the pressure during the time it takes for the air to bridge of the volume is certainly not negligible. As I explained in my last post the bouncing nature of the air means that we get the same total work on the objects as if we had just totally removed one of the walls, and if we just removed one of the walls and assume that the can have a volume of 1 cubic decimeter we would get 1000 newtons of force over 1/5000 seconds and a can weighting 10g would then get accelerated to 20 m/s during that fill up process.(Notice that the cans speed is still negligible compared to the speed of the particles of air)
If we just had 1 square centimeter of a hole the process would take ~200-300 times longer but the net work would be exactly the same and the air resistance can not slow such a speed at these time intervals.
Edit: And that would hardly be the first time a science book is wrong.
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The can problem turned out to be tricky but klockan's arguments looks stronger so far.
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klockan, I'm still not sure of your position.
if:
*there is no air resistance (the can moves very slow), but
*air pressure is not negligible (the air moves fast and can still exert pressure)
*open system
(no constraints on hole size and outer pressure, because these are not independent when considering the outcome.)
does the can move once a steady state has been reached? (all acceleration has stopped)
if yes, what was the total momentum before and after the puncture?
if no, what slows the can down?
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On September 30 2008 04:17 betaben wrote:
does the can move once a steady state has been reached? (all acceleration has stopped)
If we totally neglect air resistance, yes. (Even though air resistance would slow down anything to a relative near stop no matter how slow it moves if we let time go on)
On September 30 2008 04:17 betaben wrote:
if yes, what was the total momentum before and after the puncture?
Total momentum of what? The can would logically gain momentum exactly equal to the momentum of the air required to fill the can as long as the can were in relative rest compared to the air from the start and that air resistance is totally negligible, because thats how much momentum the can robbed from the outside system by refusing to reflect that much air in a specific direction.
There could be some waste due to shock waves or something else like that but they are also negligible under normal circumstances.
Also I just realized in what way the book author were thinking wrong. He were assuming that you would think that the instream of gas would create a force on the wall and as such it would accelerate due to that for some reason, which of course is a ridiculous thought. However when he made the problem he probably missed that the momental lack of force in the other direction would also make he himself wrong.
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ok, so ..
Total momentum of what?
of the entire system
The can would logically gain momentum exactly equal to the momentum of the air required to fill the can.
so, your end state is:
1). the can moving in one direction with a body of air (of the same momentum) outside of the can, moving in the opposite direction?
or,
2).with the body of air inside the can?
your wording, "required to fill the can" would imply the latter case, the air inside the can ("to fill the can") will collide with the can (as it must be moving int he opposite direction to the can to balance), nullifying the momentum of both, and the can will stop, giving the over all end state as no can movement.
so this cannot be what you mean (or is it? - do you consider the steady state to be before the collision? if so, then you are just having a disagreement with when 'end state' is called.)
So I'd guess your end state is the first of the options, which has a body of air on the other side of the can, moving away from it.
you say:
... that's how much momentum the can robbed from the outside system by refusing to reflect that much air in a specific direction.
which would support the above, with the air outside moving away from the can.
If so, how do you say again that air is propelled in that direction?
I think you're saying that in the process where the gas on the unpunctured side pushes against the can, there will be an equal amount of gas moving in the opposite direction within the gas (as it's overall momentum is beforehand zero). this does not transfer it's momentum to anything, (unlike the gas moving against the can) so continues travelling away from the can.
Is this true?
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It's basically the opposite of an explosion.
You would expect the net momentum of the system (the system being the can + the air) to be 0 before and after the puncture (conservation of momentum).
Once the can is punctured, the air particles gains net momentum. That means the can needs to move in the opposite direction to conserve the momentum.
The can won't stop moving until the net momentum of the air becomes 0 again, which occurs at equilibrium.
Arguing that the can doesn't move at all is saying air got into the can without net motion.
Right?
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On September 30 2008 08:44 Juicyfruit wrote:
It's basically the opposite of an explosion.
You would expect the net momentum of the system (the system being the can + the air) to be 0 before and after the puncture (conservation of momentum).
Once the can is punctured, the air particles gains net momentum. That means the can needs to move in the opposite direction to conserve the momentum.
The can won't stop moving until the net momentum of the air becomes 0 again, which occurs at equilibrium.
Arguing that the can doesn't move at all is saying air got into the can without net motion.
Right?
I'm interested in Klockan3's answer....
I don't think anyone is arguing that there is no motion at all.
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On September 30 2008 06:48 betaben wrote:
If so, how do you say again that air is propelled in that direction?
I think you're saying that in the process where the gas on the unpunctured side pushes against the can, there will be an equal amount of gas moving in the opposite direction within the gas (as it's overall momentum is beforehand zero). this does not transfer it's momentum to anything, (unlike the gas moving against the can) so continues travelling away from the can.
Is this true?
If we isolate the two surfaces, one were we puncture it and the other is the one exactly on the opposite side.
Each reflects molecules proportional to its surface all of the time. When we puncture one of the sides, that side will now reflect less molecules until the can is filled, as such the excess molecules reflected by the far side will then not be weighted up by the predicted reflection of molecules in the other side.
So basically the can takes its momentum from the air it reflects on the far side and it can do this due to a temporary lack of reflection on the near side. The reason the amount of momentum transfered to it should be exactly proportional to the amount of air required to fill the can is because that is exactly how much air it fails to reflect in total no matter hole size/can size/air speed etc. (Of course if we neglect air resistance and other wasted energy/transferal of momentum to the outside air)
If you want a physical system with similar forces but less abstract ones how about this:
Imagine a box. It is getting hit in even intervals by one ball from north and one ball from the south at exactly the same time, due to this the forces from the impulses always takes out each other.
Now we alter the box, we remove one of the walls, the south one. The box is made in such a way that the time it takes for one ball to traverse to the far end of the box is equal for the time it takes for a new ball to arrive at the near end. The box is much heavier than the balls and as such any movement gained by it should be negligible compared to the movement of the balls.
What would happen to the box in this system? Were did it gain its momentum from? How large is that momentum?
Edit: If you can answer that, just multiply the number of balls, make them minuscule and put them at random hit intervals and you got the old example. Just note that the smaller the hole the longer it takes for the box to fill and the less the removed force area will be so it still applies there.
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On September 30 2008 16:27 Klockan3 wrote:Show nested quote +On September 30 2008 06:48 betaben wrote:
If so, how do you say again that air is propelled in that direction?
I think you're saying that in the process where the gas on the unpunctured side pushes against the can, there will be an equal amount of gas moving in the opposite direction within the gas (as it's overall momentum is beforehand zero). this does not transfer it's momentum to anything, (unlike the gas moving against the can) so continues travelling away from the can.
Is this true?
I'm guessing by the fact that you've quoted this, you're confirming the momentum of the can is equalised by a body of air moving away from the can on the side opposite to the hole.
On September 30 2008 16:27 Klockan3 wrote:
If we isolate the two surfaces, one were we puncture it and the other is the one exactly on the opposite side.
Each reflects molecules proportional to its surface all of the time. When we puncture one of the sides, that side will now reflect less molecules until the can is filled, as such the excess molecules reflected by the far side will then not be weighted up by the predicted reflection of molecules in the other side.
So basically the can takes its momentum from the air it reflects on the far side and it can do this due to a temporary lack of reflection on the near side.
so far, this is describing a differences in pressure on either side which moves the can. this process is agreed upon - it is not the query; the query is:
the air that moves in order that can momentum is equalised in the end state - how does it get moving?
On September 30 2008 16:27 Klockan3 wrote:
The reason the amount of momentum transfered to it should be exactly proportional to the amount of air required to fill the can is because that is exactly how much air it fails to reflect in total no matter hole size/can size/air speed etc. (Of course if we neglect air resistance and other wasted energy/transferal of momentum to the outside air)
This is agreed (I'm guessing by airspeed, you are not referring to the speed of the body of air filling the can, nor the body of air on the opposite side of the can, as a small amount of air at high speed has the same momentum as a large amount of air at low speed.)
If you want a physical system with similar forces but less abstract ones how about this:
Imagine a box. It is getting hit in even intervals by one ball from north and one ball from the south at exactly the same time, due to this the forces from the impulses always takes out each other.
Now we alter the box, we remove one of the walls, the south one.
so far again, this is agreed upon.
The box is made in such a way that the time it takes for one ball to traverse to the far end of the box is equal for the time it takes for a new ball to arrive at the near end. The box is much heavier than the balls and as such any movement gained by it should be negligible compared to the movement of the balls.
What would happen to the box in this system? Were did it gain its momentum from? How large is that momentum?
It gains it's momentum from the air on the far side having more total momentum transfer than those on the hole-side (due to the hole). This is where I'm trying to interpret your view:
the air on the the far side has air moving in all directions; as one fraction of air transfers it's momentum to the can, there is a residual amount of air which is moving away from the can, (which, due to the dispersive nature of the air, would probably travel away as a shock wave from the can in the form of sound.)
This is the only way I can think of for your air to be moving away from the can on the far side.
is this your interpretation?
However, a similar process occurs on the hole side, as the air which enters the can has a component which moves away from the can, in an equal and opposite shock wave. the over all momentum of the outside air would be 0.
unless you have a different notion.
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On September 30 2008 17:21 betaben wrote:
the air on the the far side has air moving in all directions; as one fraction of air transfers it's momentum to the can, there is a residual amount of air which is moving away from the can, (which, due to the dispersive nature of the air, would probably travel away as a shock wave from the can in the form of sound.)
Wrong, the air on the far side bounces of the surface just as if it were a normal surface. The shock waves only comes from the temporary lack of reflection on the side with the hole, it creates a large negative shock wave. The right side looks just as normal all the time, the left side (With the hole) creates a negative shock wave and thus gets a negative momentum compared to normal. Now since the right side has normal momentum and the left side have less momentum than normal, the momentum balancing up the cans momentum is the fact that the right air have more momentum than the left air. This momentum difference is as I described before the same as the momentum of the air which entered the can.
But yeah a really minor shock wave will also be created due to the can accelerating but if we ignore air resistance we should also ignore that since they are a part of the same phenomenon.
Edit: If you still do not understand ask again and I can try to describe it in a new way.
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you are continually using terms which are vague. "normal momentum" -what is normal? net momentum 0? "more momentum" - without a direction, this is unclear as net momentum is not scalar (unless you mean the air is heated up). please define a net momentum direction for the air. "negative shockwave" does this mean to the left? have you defined an axis, with negative = left?
Also I don't understand:
Wrong, the air on the far side bounces of the surface just as if it were a normal surface.
and
The right side looks just as normal all the time
if the air on the right size never changes overall momentum, it never transfers net momentum to the can, and the can won't move. It is not the same as 'bouncing off a normal surface' because it makes the surface move, transferring momentum. the can's leftwards momentum can only come from this side.
I think when you say:
Now since the right side has normal momentum and the left side have less momentum than normal, the momentum balancing up the cans momentum is the fact that the right air have more momentum
again, you must define the right hand sides airs' 'more momentum'. if it balances the cans momentum it must be momentum away from the can.
you mean precisely what I wrote when you said "wrong":
the air on the the far side has air moving in all directions; as one fraction of air transfers it's momentum to the can, there is a residual amount of air which is moving away from the can
as a minor point:
+ Show Spoiler +let's forget about shock waves, we'll just talk about air momentum. but still...
But yeah a really minor shock wave will also be created due to the can accelerating but if we ignore air resistance we should also ignore that since they are a part of the same phenomenon.
this I think is incorrect - shock waves are a pressure phenomenon, nothing to do with air resistance which involves an object colliding with the air and dissipating the objects' energy within the air as heat. shock waves should lose energy slowly, just have the energy dispersed as it spreads out. air resistance would dissipate the energy as random isotropic movement in the air - i.e heat.
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On October 01 2008 05:22 betaben wrote:
if the air on the right size never changes overall momentum, it never transfers net momentum to the can
...
Air pressure transfers a huge amount of momentum against everything all of the time, it is just that it usually gets that back exactly due to an exactly balanced other side also transferring momentum. The momentum in this case going to the left comes from the balance being broken by lessening the momentum transferring on the left side, and thus the right sides momentum transfer wont be neutralized and thus the can will receive a net momentum from the air on the right side but the only sides air which changed is the left side.
I repeat: Normal circumstances is that air transfers 10 tonne*meter momentum per second and square meter. Normal circumstances is also that this process gets exactly equaled out by other similar forces. However if we removed the amount transfered on one side, the other side would then be transferring all of this unhindered and thus the can can get a momentum from the right side without changing the air on the right side at all since having the air transferring momentum to the can is the normal situation, the left side stopping transferring momentum to the can is the altered situation and as such any change to the air momentum occurs on the left side and it takes its form as a negative shock wave which carriers negative momentum compared to normal air.
It is not really negative momentum, it is just that normally in air it have a net momentum of 0, but if you remove all molecules going one way we would thus get a lower than normal momentum in that direction and as such we have a negative momentum compared to normal air and if this shock wave hit anything it would propel it towards the direction the shock wave came from.
On October 01 2008 05:22 betaben wrote:
this I think is incorrect - shock waves are a pressure phenomenon, nothing to do with air resistance which involves an object colliding with the air and dissipating the objects' energy within the air as heat. shock waves should lose energy slowly, just have the energy dispersed as it spreads out. air resistance would dissipate the energy as random isotropic movement in the air - i.e heat.
Air resistance comes from really small shock waves being created which gets larger the faster you move compared to the air. Shock waves are just relatively compressed/thinned out air, you get that the instant you move in any form of gas.
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Dunn dunn dunn! Come back tomorrow for more monstrous PHYSICISTS slugging it out!
WHO WILL WIN?
(oh come on you know i'm just poking fun... this is what i do when i'm lost)
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+ Show Spoiler +
1c
just a feeling, once the car is moving the speed should not change with changing mass. as long as we dont look at friction at least
2a
p = mv , the velocity stays the same but the mass increases, thus the momentum increases
3a
E = 1/2 mv² - same as above
1c
2b
3b
Conservation of Energy, everything should go back to when the rain started. Maybe thats a bit easy though..
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I think I'm coming round to your view - please tell me if I'm incorrect:
you view the air as having infinite momentum - capable of transferring momentum in one direction without losing it itself. thus, you believe there is no net momentum change in the air on the right side, and the can's momentum is not balanced by air on this side. (here i was confused because I thought by quoting my option with this in it, I thought you were advocating it.) The can's momentum is equaled by a wave of low pressure traveling in the same direction as the can, which you justify as having momentum opposite to it's direction of movement by the fact that it will move objects towards itself as it hits.
please directly say which parts of this are correct/incorrect
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EPT![[image loading]](http://img81.imageshack.us/img81/3089/watercartkx1.png)
![[image loading]](http://img81.imageshack.us/img81/3171/puncturedcansa1.png)
