# Newton's Third Law of Motion: A question for willing teachers

Discussion in 'Physics & Math' started by §lîñk€¥™, Jan 16, 2003.

1. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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My spaceship is floating freely in space (let's just say for arguments sake that it is not in any gravitational fields and can detect no apparent motion). The total mass of the spaceship is 1,000,000kg.

My intention is to "throw" a 1kg "brick" of matter out the back of the ship with a velocity of 25km/sec (we'll ignore the mechanism for this unless you tell me that we cannot).

According to Newton there should be an equal and opposite reaction. Literally, my spaceship will move in the opposite direction from the brick. ie. it will be accelerated by an amount equal to the mass and velocity of the brick.

How would one work out the acceleration of the spaceship?

Is it really as simple as this?

Force = brickmass x acceleration
25000n = 1kg x 25000m/s<sup>2</sup>

therefore

force / shipmass = acceleration
25000n/1000000kg = 0.025m/s<sup>2</sup>

The ship will be accelerated by 0.025m/s<sup>2</sup> for one second.

Or do I have to allow that my spaceship is now 1kg less massive?

therefore
25000n/999999kg = 0.025000025m/s<sup>2</sup>

I know the difference is neglibible but I'm not sure which is correct (if either).

kind regards
Paul

Last edited: Jan 16, 2003

3. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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123
I guess that leads to another question.

Same scenario. This time however, I am throwing a "brick" out every second, and will do that until I run out of bricks. I have 100,000 bricks.

I already know that each brick will give me 25000n. However, everytime I throw a brick out my spaceship gets lighter. This means that every 25000n will impart a greater acceleration on my ship as it loses mass.

How fast will I be moving once I have thrown all the bricks out?

Sheesh, I wish I'd paid attention at school..... :m:

kind regards
Paul

5. ### chrootCrackpot killerRegistered Senior Member

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2,350
This is just a simple problem dealing with the conservation of linear momentum.

The momentum is defined as mass * velocity. In the frame of the moving spaceship, the total momentum of the brick-spaceship system is zero. After the brick is expelled, the momentum gained by the brick must be equal and opposite to the momentum gained by the rocket, so that the total momentum remains zero. In terms of velocities,

v<sub>1</sub> = (m<sub>2</sub>v) / (m<sub>1</sub> + m<sub>2</sub>)

where v<sub>1</sub> is the velocity increase of the rocket, m<sub>1</sub> is the mass of the rocket, v<sub>2</sub> is the velocity increase of the brick, m<sub>2</sub> is the mass of the brick.

- Warren

7. ### CrispGone 4everRegistered Senior Member

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1,339
Hi §lîñk€¥™,

I always find it a bit tricky to use Newton's third law directly, because IMHO it does not really say much about dynamics, but on the (Galilean) relativity between different systems. Simply put: if you have two different systems and you transform from one system to the other, the third law gives you some hints on how mutual forces work (opposite direction, same strength).

Personally I would never use it to work out a problem like you did, it could work, but as chroot pointed out, it's more evident to use the conservation of momentum.

Bye!

Crisp

8. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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123
Which would make my first answer correct (albeit my reasoning was askew).

.025m/s<sup>2</sup>=(1kg x 25000m/s<sup>2</sup>) / (999999kg + 1kg)

Does this apply to my follow up question too? ie.

2500m/s<sup>2</sup>=(100000 x 1kg x 25000m/s<sup>2</sup>) / (900000kg + 100000kg)

kind regards

9. ### CrispGone 4everRegistered Senior Member

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1,339
Hi §lîñk€¥™,

Depends... If you throw out all 100000 bricks at once, it would be the same as throwing 100000 kg out at once, and then it is a matter of applying the formula.

For individual brick-throwing:
Once the spaceship is already moving, it depends on how you define the velocity of the brick. If the 25 km/s is always relative to the spaceship, there is no mathematical problem. If the 25 km/s is relative to some stationary point... which I assume it is, since if you take it relative to the spaceship, all velocities should be expressed with respect to the spaceship and you find that the spaceship always as velocity 0 with respect to itself... if the 25 km/s is relative to some stationary point, you have to take into account the increased velocity of the spaceship (increased relative to the stationary point) after throwing out every brick.

Bye!

Crisp

10. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Simply put, using my example:

(m<sub>brick</sub>v<sub>brick</sub>) - (m<sub>ship</sub>v<sub>ship</sub>) = 0

Which leads me to suspect that whether I throw 1 big 100,000kg brick out at 25km/s, or 1 kg bricks out at 25km/s each 100,000 times, I still end up with the same acceleration.

kind regards

11. ### chrootCrackpot killerRegistered Senior Member

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2,350
No. You end up with the same final velocity. The acceleration will be different, since it will take longer to throw out 100,000 bricks.

- Warren

12. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Ok, so I was wrong.

Thank you Crisp

Damn. :bugeye:

Ok, let me clarify the problem again.

I am throwing 1 brick out every second. Each brick has 1kg of mass and is thrown out at 25km/s relative to my space ship. My space ships total mass (ie. including the bricks) is 10<sup>6</sup>kg before I start throwing bricks out.

I guess this means I have to take my loss of mass into account, because my system (ship and cargo of bricks) is constantly losing mass by 1kg every second?

13. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Arrg, velocity. Thank you Warren.

14. ### chrootCrackpot killerRegistered Senior Member

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In fact you do. The device you're implying is just a rocket -- a machine that throws a certain amount of momentum out the back, and loses mass as it happens. You can integrate the change in velocity over the mass loss:

dv = -u dM/M

where dv is the differential change in velocity, u is the exhaust velocity, M is the instaneous mass of the rocket, and dM is the change in mass.

v<sub>f</sub> - v<sub>i</sub> = u ln (M<sub>i</sub>/M<sub>f</sub>)

where i indicates initial conditions, and f indicates final conditions.

- Warren

15. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Warren,

Thanks again. This is where I get envious of guys like yourself, and your ability to integrate. Can you give me some pointers on how to wise up a bit on integrating? I'm not asking you to teach me rather point me in the right direction where I can learn it! Then I can stop wasting your time further with questions of this nature.

kind regards

16. ### chrootCrackpot killerRegistered Senior Member

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Calculus books help...

and so do computer algebra systems. I am a huge fan of Maple. You can *cough cough* buy it pretty cheap if you're a student. MATLAB is good too, but is more expensive and actually a little annoying to me.

In this case, if you just study a lot of physics, you'll come across all the integrals and tools you'll ever need pretty rapidly.

For instance, natural logarithms and binomial approximations occur everywhere! Get used to those tools, and you can derive a lot of things yourself.

I also highly recommend the Feynman Lectures on Physics, because they are some of the most readable and complete books around.

- Warren

17. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Hi Warren

I had a feeling you were going to say that

I had a search on the web to see if there are any free tutorials. I found a pretty good little site (although it isn't complete yet so I will have to check some other sites too to get the full picture). I'm working through the Pre-Calculus stuff. Been quite a bit of fun actually.

I picked up a demo version from the Maple site last night but haven't had a chance to look at it. The info and screenshots made it sound and look very good. Thanks for pointing that one out to me.

That's what I figured. They seem such an integral part (pun intended) of physics which made me think they are worth wising up on in any case. I really don't mind doing the work. Finding out you can do them is where the fun is.

Ok, I'll bone up on those too. There's bound to be free tutorial sites for them also.

I had a look on Amazon for these. I found a 3 volume set. Is that the complete set?

I've got 5 of his books already. There's not a lot of math in thoe ones though, but he brings the concepts across in very interesting and graspable ways.

Thanks for the pointers. Much obliged.

kind regards

18. ### Chagur.Seeker.Registered Senior Member

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Re. Feynman's 'Lectures on Physics'

It's a three volume set and if you can find it in paperback
you can save yourself a few bucks, §lîñk€¥™?

Best

19. ### chrootCrackpot killerRegistered Senior Member

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§lîñk€¥™,

These sites are excellent references, but not so good on education:

http://scienceworld.wolfram.com/physics/
http://mathworld.wolfram.com/

My favorite physics site on the web, good for both reference and education (though it definitely does not include "thorough treatments"):

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Some elementary integral tables (you'll find a more complete table in the back of nearly any calculus text):

http://www.math2.org/math/integrals/tableof.htm
http://www.sosmath.com/tables/tables.html
http://www-ece2.engr.ucf.edu/~tomwu/course/eel4309/integral/integral.htm

The binomial expansion, which ends up being used in almost every physics derivation:

http://www.krysstal.com/binomial.html

Feynman's 3-volume Lectures:

http://search.barnesandnoble.com/te...userid=52QV23EV74&isbn=0201500647&TXT=Y&itm=1

The three-volume hardcover set is the same price as all three softcovers bought separately (they are not available in a set). Volume I contains mechanics, kinetic theory, that kinda stuff. Volume II is E&M. Volume III is quantum mechanics.

They are *full* of math AND prose in just about what I would call the perfect ratio. They are highly readable, and HIGHLY educational. Virtually every conclusion in the whole set is derived from first principles in clear, easy to understand steps. These are by far the best books I have ever come across for learning physics. If you can, go by a bookstore and just sit down with a volume or two to see how they read. If you don't buy them immediately, I will be very surprised!

Have fun. Let me know if you have q's

- Warren

20. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Hi Warren

Thanks for all the references. Much obliged.

I have an aunt who works for a book company and have already asked if she can get them for me (she gets a good discount). I'm sure they will be excellent. I can't wait to get it!

kind regards

21. ### chrootCrackpot killerRegistered Senior Member

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Awesome! You will love them. If by any chance you should reach any stumbling blocks, feel free to ask, and I'll try to help you over them.

- Warren

22. ### §lîñk€¥™Uneducated smart alecRegistered Senior Member

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Hi Warren

Excellent. Thanks. I'm sure there will be a few occasions when I might need a little hint or two.

kind regards

23. ### IggDawgRegistered Senior Member

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as far as a real world scenerio goes, I think any force you generate on the ship by throwing the brick would get bled off as heat. Think of how much mass has to be effected as a result of this. every intermolecular bond has to register this push. and every intermolecular bond will turn a little bit of this force into heat as they strech and contract ever so slightly. I think the net effect on the ship would be zero. I dunno if you are looking for this kind of answer or an ideal one. not trying to be a wang