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Force related to mass, acceleration and time

by Ohoneo
Tags: acceleration, force, mass, time
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Ohoneo
#1
Feb16-11, 09:08 AM
P: 22
1. The problem statement, all variables and given/known data
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3700-kg space tug and a 6100-kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 460 m apart. How much time does it take for the ship and the asteroid to meet?


2. Relevant equations
F = ma
x = x0+ v0t + 1/2 a t2
v = v a t
v2-v02 = 2a (x - x0)


3. The attempt at a solution
I tried to solve this a couple of ways and got it wrong both times.
First, I found the acceleration of the asteroid by doing F = ma, and dividing 490 by 6100 kg. I got .0803 (approximately, I didn't round in my work, but I'll round for ease of typing) m/s2.
I then tried two equations, which both gave me the same answer:
First, I plugged into v2-v02 = 2a (x-x0)
and got v = 8.5966 for the final velocity of the asteroid.
I then plugged into v = v a t to find how long it would take to get to that velocity. I got t = 107 seconds.

When I got the answer wrong, I tried again by plugging into x = x0+ v0t + 1/2 a t2
I plugged in: 460 = 0 + 0 + 1/2 (.803)t2
and I got t = 107 seconds again.

Any help would be much appreciated :)
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AC130Nav
#2
Feb16-11, 04:39 PM
P: 153
[QUOTE=Ohoneo;3140687]1. The problem statement, all variables and given/known data
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3700-kg space tug and a 6100-kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 460 m apart. How much time does it take for the ship and the asteroid to meet?


It appears to me that you are in space and free to establish any reference you want. Make that the tug. How long would it take an object of 6100 kg to accelerate under 490 N to cover 460 m?

Maybe someone better qualified will answer.
gneill
#3
Feb16-11, 04:43 PM
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P: 11,677
Both the tug and the asteroid are free to move in space, so both will accelerate. Since all the forces in the system are internal ones (no outside forces not associated with the system), what do you think would be a good frame of reference to choose in which to solve the problem?

Ohoneo
#4
Feb17-11, 03:37 PM
P: 22
Force related to mass, acceleration and time

Quote Quote by gneill View Post
Both the tug and the asteroid are free to move in space, so both will accelerate. Since all the forces in the system are internal ones (no outside forces not associated with the system), what do you think would be a good frame of reference to choose in which to solve the problem?
I missed the class in which this was discussed and these "reference frames" aren't really mentioned in my textbook... Can I ask you to elaborate on that?
I found the acceleration for both the asteroid and the tug, but I wasn't sure what to do when I had both of them.
gneill
#5
Feb17-11, 03:59 PM
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P: 11,677
Quote Quote by Ohoneo View Post
I missed the class in which this was discussed and these "reference frames" aren't really mentioned in my textbook... Can I ask you to elaborate on that?
I found the acceleration for both the asteroid and the tug, but I wasn't sure what to do when I had both of them.
Think of a frame of reference as a coordinate system that a hypothetical observer might erect at his location in order to measure the things going on around him. Frames that are moving inertially (that is, without acceleration) are generally easier to work with, as all the laws of physics take their usual form in such frames without having to introduce pseudo-forces due to accelerations.

At some point in your physics career you must have come across the concept of the center of mass of an isolated system continuing to move inertially (in a straight line and un-accelerated) despite interactions amongst its components. Well, this applies here.

The center of mass of the tug and the asteroid will move inertially as the two of them accelerate toward it. If we declare the center of mass to be at rest (i.e., we adopt the center of mass to anchor our frame of reference), then they should both meet at the center of mass at the same time.

At this point you have a choice: Pick one of the objects (tug or asteroid), work out its distance from the center of mass, and using its acceleration determine the time to cover the distance.

Alternatively, you could realize that while the two objects individually have separate accelerations, together they have a closing acceleration (and closing velocity) that is the sum of the individual accelerations.
fandam94
#6
Feb17-11, 04:18 PM
P: 1
what is motion
Ohoneo
#7
Feb17-11, 04:22 PM
P: 22
Quote Quote by gneill View Post
At some point in your physics career you must have come across the concept of the center of mass of an isolated system continuing to move inertially (in a straight line and un-accelerated) despite interactions amongst its components. Well, this applies here.
I've never come across this =/ I don't know if it was in the lesson and I missed it or if it wasn't covered. I never took physics in high school, either.

Quote Quote by gneill View Post
The center of mass of the tug and the asteroid will move inertially as the two of them accelerate toward it. If we declare the center of mass to be at rest (i.e., we adopt the center of mass to anchor our frame of reference), then they should both meet at the center of mass at the same time.

At this point you have a choice: Pick one of the objects (tug or asteroid), work out its distance from the center of mass, and using its acceleration determine the time to cover the distance.

Alternatively, you could realize that while the two objects individually have separate accelerations, together they have a closing acceleration (and closing velocity) that is the sum of the individual accelerations.
I don't think I quite understand the first two paragraphs here (well, how to apply them). But am I right in restating what you say as that I can add the two accelerations together and then find how long it takes something going at that acceleration to cover the distance stated in the problem?
Ohoneo
#8
Feb17-11, 04:25 PM
P: 22
Quote Quote by gneill View Post
At this point you have a choice: Pick one of the objects (tug or asteroid), work out its distance from the center of mass, and using its acceleration determine the time to cover the distance.
Wait, so would this work?
I make the tug my centre of mass, so that the asteroid has to travel 460 m (as stated in the problem). Then I find it's acceleration (the same one as in my first post) and find how long it takes, using that acceleration, to travel 460 m? Because that's what I did the first time and got it wrong.
Sorry if I'm completely missing what you're saying. Something about physics makes it take a long time to click for me.
gneill
#9
Feb17-11, 04:35 PM
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P: 11,677
Quote Quote by Ohoneo View Post
Wait, so would this work?
I make the tug my centre of mass, so that the asteroid has to travel 460 m (as stated in the problem). Then I find it's acceleration (the same one as in my first post) and find how long it takes, using that acceleration, to travel 460 m? Because that's what I did the first time and got it wrong.
Sorry if I'm completely missing what you're saying. Something about physics makes it take a long time to click for me.
No, the tug is not the center of mass. The center of mass of the system will lie somewhere along the line joining the tug and asteroid. Imagine the tug and asteroid at the opposite ends of a very long beam, and that there's a uniform gravitational field acting. The center of mass would be the point under the beam where you could place a fulcrum and have the whole lot in balance.

With just two masses it's relatively easy to find the center of mass. If Mt is the mass of the tug, and Ma the mass of the asteroid, and if d is the total distance betwixt the two, then the distance from the tug to the center of mass is:

Dt = d*Ma/(Ma + Mt)

and the distance from the asteroid to the center of mass is:

Da = d*Mt/(Ma + Mt)

Note that Dt + Da = d.
Ohoneo
#10
Feb17-11, 04:45 PM
P: 22
Quote Quote by gneill View Post
No, the tug is not the center of mass. The center of mass of the system will lie somewhere along the line joining the tug and asteroid. Imagine the tug and asteroid at the opposite ends of a very long beam, and that there's a uniform gravitational field acting. The center of mass would be the point under the beam where you could place a fulcrum and have the whole lot in balance.

With just two masses it's relatively easy to find the center of mass. If Mt is the mass of the tug, and Ma the mass of the asteroid, and if d is the total distance betwixt the two, then the distance from the tug to the center of mass is:

Dt = d*Ma/(Ma + Mt)

and the distance from the asteroid to the center of mass is:

Da = d*Mt/(Ma + Mt)

Note that Dt + Da = d.
So, Dt=460 m(6100 kg)/(6100+3700)
Which = 286.3
And then Da= 460 m(3700 kg)/(6100+3700)
Which = 173.7
Which both add up (when I don't round) to be 460.

So, now that I have those, I only really need to use either Dt or Da, with the associated acceleration to find time, right?
Ohoneo
#11
Feb17-11, 04:48 PM
P: 22
I got the right answer! Thank you so much for your help and patience :)
I just have one last question, since I never learned this center of mass stuff: can I use that in any force problem when two masses are given and both objects are accelerating (towards each other)?
gneill
#12
Feb17-11, 04:51 PM
Mentor
P: 11,677
Quote Quote by Ohoneo View Post
So, Dt=460 m(6100 kg)/(6100+3700)
Which = 286.3
And then Da= 460 m(3700 kg)/(6100+3700)
Which = 173.7
Which both add up (when I don't round) to be 460.

So, now that I have those, I only really need to use either Dt or Da, with the associated acceleration to find time, right?
Correct. And to check results, why not do the calculation for both? You should get the same time, showing that the two will meet at the center of mass. For the coup de gras, employ the closing acceleration method (sum of the individual accelerations) over the entire distance and show that it, too yields the same result.
Ohoneo
#13
Feb17-11, 05:02 PM
P: 22
Quote Quote by gneill View Post
Correct. And to check results, why not do the calculation for both? You should get the same time, showing that the two will meet at the center of mass. For the coup de gras, employ the closing acceleration method (sum of the individual accelerations) over the entire distance and show that it, too yields the same result.
This is great, thanks so much! :)


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