
#1
Feb1611, 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 3700kg space tug and a 6100kg 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 = x_{0}+ v_{0}t + 1/2 a t^{2} v = v a t v^{2}v_{0}^{2} = 2a (x  x_{0}) 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/s^{2}. I then tried two equations, which both gave me the same answer: First, I plugged into v^{2}v_{0}^{2} = 2a (xx_{0}) 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 = x_{0}+ v_{0}t + 1/2 a t^{2} I plugged in: 460 = 0 + 0 + 1/2 (.803)t^{2} and I got t = 107 seconds again. Any help would be much appreciated :) 



#2
Feb1611, 04:39 PM

P: 154

[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 3700kg space tug and a 6100kg 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. 



#3
Feb1611, 04:43 PM

Mentor
P: 11,445

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?




#4
Feb1711, 03:37 PM

P: 22

Force related to mass, acceleration and timeI found the acceleration for both the asteroid and the tug, but I wasn't sure what to do when I had both of them. 



#5
Feb1711, 03:59 PM

Mentor
P: 11,445

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 unaccelerated) 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. 



#6
Feb1711, 04:18 PM

P: 1

what is motion




#7
Feb1711, 04:22 PM

P: 22





#8
Feb1711, 04:25 PM

P: 22

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. 



#9
Feb1711, 04:35 PM

Mentor
P: 11,445

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. 



#10
Feb1711, 04:45 PM

P: 22

Which = 286.3 And then D_{a}= 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 D_{t} or D_{a}, with the associated acceleration to find time, right? 



#11
Feb1711, 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)? 



#12
Feb1711, 04:51 PM

Mentor
P: 11,445





#13
Feb1711, 05:02 PM

P: 22




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