- #1

walking

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What is wrong with saying that since the speed of fuel goes from 0 to -313m/s (rel to ground) in 1s, then its acceleration is -313m/s^2. This leads to wrong answer according to author's solution but I don't see why.

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- Thread starter walking
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- #1

walking

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What is wrong with saying that since the speed of fuel goes from 0 to -313m/s (rel to ground) in 1s, then its acceleration is -313m/s^2. This leads to wrong answer according to author's solution but I don't see why.

- #2

add314

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- #3

BvU

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It's not just the fuel that is ejected !speed of fuel

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- #5

jbriggs444

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The length of time that it takes to accelerate a particular lump of fuel from 0 [big hint, the fuel does not start at a ground-relative speed of 0] to -313 m/sec is utterly irrelevant here. The required acceleration for that lump is also utterly irrelevant here. Irrelevant, that is, unless you can get a handle on the amount of fuel that is being accelerated at a particular point in time.View attachment 279536

What is wrong with saying that since the speed of fuel goes from 0 to -313m/s (rel to ground) in 1s, then its acceleration is -313m/s^2. This leads to wrong answer according to author's solution but I don't see why.

What is possibly relevant is the momentum at the start of a 1 second interval, the momentum at the end of said interval and the mass/momentum flows that take place during that 1 second interval. [Which is what @kuruman is trying to lead you toward as well]

- #6

add314

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I was inattentive , yes so any gases it ejects, but still gases does not get an acceleration of 313 m/s^2It's not just the fuel that is ejected !

- #7

jbriggs444

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The approach is, perhaps, not complete nonsense, although the execution was borked. If one considers the fuel that is ejected in one second, the center of mass of said fuel begins at +184 m/s (ground relative) and ends at 313 m/s (ground-relative). The center of mass of said diffuse blob of fuel would have an acceleration given by ##\frac{\Delta v}{\Delta t}##. The ##\Delta v## would be 184+313 and the ##\Delta t## would be 1 sec.I was inattentive , yes so any gases it ejects, but still gases does not get an acceleration of 313 m/s^2

Of course, by itself, this acceleration is meaningless. If one wanted to figure out a rate of momentum flow, one would need to know how much stuff is experiencing this acceleration.

Unfortunately, @walking has not seen fit to provide us with the rest of his working. Nor has he seen fit to explain his approach. So we are left to guess at what else might be wrong.

@walking: Do not assume that the problem is with the bits that you show us. The bits that you do not show us may be equally important.

- #8

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In case @jbriggs444 's comment is not quite clear, your calculation of the velocity change of the air is fine but not for the velocity change of the fuel burnt.

And you cannot derive its acceleration by dividing by 1s. You do not know how long it takes to accelerate any given molecule of exhaust. However, the correct analysis is likely to lead to an expression in which this Δv is in the numerator and the given 1s is in the denominator, so depending on how you use this "acceleration" you might still end up with a valid answer.

In short, please show all your working. (You could have done that in less time than the total taken by responders so far! )

Edit: corrected citation of earlier post.

And you cannot derive its acceleration by dividing by 1s. You do not know how long it takes to accelerate any given molecule of exhaust. However, the correct analysis is likely to lead to an expression in which this Δv is in the numerator and the given 1s is in the denominator, so depending on how you use this "acceleration" you might still end up with a valid answer.

In short, please show all your working. (You could have done that in less time than the total taken by responders so far! )

Edit: corrected citation of earlier post.

Last edited:

- #9

rsk

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I have a feeling the acceleration approach will give you the same answer IF you consider the acceleration of the fuel AND the acceleration of the air. The maths will end up being the same as the momentum-impulse approach.

- #10

walking

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Sorry about the lack of working and I appreciate everyone's detailed responses. I take down my questions as I read as I have limited internet time. Then I post down when I have written much later on, meaning I don't have time when I post them to actually go through the working for every single problem I have written a question for. I will definitely try to include more detail in my questions as I write them down next time so this problem doesn't occur.In case @jbriggs444 's comment is not quite clear, your calculation of the velocity change of the air is fine but not for the velocity change of the fuel burnt.

And you cannot derive its acceleration by dividing by 1s. You do not know how long it takes to accelerate any given molecule of exhaust. However, the correct analysis is likely to lead to an expression in which this Δv is in the numerator and the given 1s is in the denominator, so depending on how you use this "acceleration" you might still end up with a valid answer.

In short, please show all your working. (You could have done that in less time than the total taken by responders so far! )

Edit: corrected citation of earlier post.

I will try my best to include my working out for this problem as soon as possible.

- #11

Lnewqban

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As you have limited internet time, I would recommend to consider rate of change of momentum separetely for the mass of air and for the mass of fuel.Sorry about the lack of working and I appreciate everyone's detailed responses. I take down my questions as I read as I have limited internet time. Then I post down when I have written much later on, meaning I don't have time when I post them to actually go through the working for every single problem I have written a question for. I will definitely try to include more detail in my questions as I write them down next time so this problem doesn't occur.

I will try my best to include my working out for this problem as soon as possible.

Regarding the change of velocity of the air

##V_{out}-V_{in}=497-184=313~m/s##

Regarding the change of velocity of the fuel

##V_{out}-V_{in}=497-0=497~m/s##

- #12

walking

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I think this is what the author does but I still don't understand it. I think I am just misunderstanding the problem statement but here is my working out:As you have limited internet time, I would recommend to consider rate of change of momentum separetely for the mass of air and for the mass of fuel.

Regarding the change of velocity of the airrespect to the engine:

##V_{out}-V_{in}=497-184=313~m/s##

Regarding the change of velocity of the fuelrespect to the engine:

##V_{out}-V_{in}=497-0=497~m/s##

This is probably where I am going wrong as I don't completely understand the PS but I am visualizing it as follows: some air goes into the plane with mass 70.2kg. This is used to burn 2.92kg of fuel which is then ejected at -497 relative to the plane. So the change in momentum I got was after 2.92(-497+184) and before 70.2(184). So 2.92(-497+184)-70.2(184), which is wrong.

- #13

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What is the change in momentum of the fuel?2.92(-497+184)

What is the net change in momentum of the air?70.2(184)

- #14

walking

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I honestly am still confused. I think it is a problem I am having with understanding the problem statement. I feel that this is one of those problems where if someone could explain to me the PS in detail I would find the problem trivial!What is the change in momentum of the fuel?

What is the net change in momentum of the air?

- #15

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PS?I honestly am still confused. I think it is a problem I am having with understanding the problem statement. I feel that this is one of those problems where if someone could explain to me the PS in detail I would find the problem trivial!

My questions were not hard:

What mass of fuel is burnt in each second? What was its velocity wrt ground before it was burnt? What was its velocity wrt ground after it was burnt? What was its change in velocity? What was its change in momentum?

Similarly for the air, but starting with its velocity before the plane draws it in.

- #16

walking

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Right I get it now. I wasn't taking into account that not only the fuel is ejected (I only realized after re-reading BvU's comment - thanks BvU!) Also for some reason I only just realized that the fuel is actually in the plane rather than somehow converted from the air into fuel!PS?

My questions were not hard:

What mass of fuel is burnt in each second? What was its velocity wrt ground before it was burnt? What was its velocity wrt ground after it was burnt? What was its change in velocity? What was its change in momentum?

Similarly for the air, but starting with its velocity before the plane draws it in.

Anyway, as predicted the problem is very easy after understanding it properly. So we want the rate of change of momentum of the plane but the plane doesn't change so we only need consider the air and fuel. (Wrt ground:) Change of momentum of fuel in 1s is 2.92((-497+184)-184) and that of air is 70.2((-497+184)-0). Summing gives the right answer.

Of course since we only need change of momentum we can also just consider everything wrt plane as Lnewqban (and author) do. This leads to an easier calculation: 2.92(-497-0)+70.2(-497-(-184)).

It's not just the fuel that is ejected !

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