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General Force and Motion Question

  1. Sep 6, 2007 #1
    A spaceship (3500kg) and an asteroid (6200kg) both initially at rest are attached by a rope. The spaceship then pulls on the asteroid with a 490 N force. If they are initially 450 meters away, how long does it take the asteroid to hit the spaceship?

    First off I’m confused because if the spaceship was pulling the asteroid why would the asteroid ever catch up to it unless the spaceship stopped or slowed down??? The problem is unclear about this. The answer is 67 seconds.

    I tried multiple different routes to solve this problem using…
    F=ma
    Delta_x = (v_o)*t + ½*a*(t^2)

    For the asteroid to catch the spaceship it must be accelerating faster but how can that be if it weighs more???
     
  2. jcsd
  3. Sep 6, 2007 #2

    Doc Al

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    In which direction does the spaceship accelerate?
     
  4. Sep 6, 2007 #3
    It's in outer space so it doesn't matter.
     
  5. Sep 6, 2007 #4

    Doc Al

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    Of course it matters! What's the acceleration of the asteroid? What's the acceleration of the spaceship?

    Hint: Newton's 3rd law
     
  6. Sep 6, 2007 #5
    Well since the spaceship and the rock start from rest and the spaceship starts pulling the rock then wouldn't they're accerlations be in the same direction?
     
  7. Sep 6, 2007 #6

    learningphysics

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    No. The ship is not trying to move... meaning it doesn't have any thrust activated... it's just pulling on the asteroid with a rope.
     
  8. Sep 6, 2007 #7

    Doc Al

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    Well... what direction is the force on each? (A simple diagram might help you.)
     
  9. Sep 6, 2007 #8
    ohhh it is just pulling the rock towards it... I see. I did some math and still did not get the right answer though. Here it is:

    For the rock:
    a=F/m=490/6200=0.8
    Delta_x = (v_o)*t + ½*a*(t^2) => 450=0.5*0.8*t^2
    t=106 sec
    v=x/t=450/106=4.2

    For the ship:
    a=F/m=490/3500=0.14
    Delta_x = (v_o)*t + ½*a*(t^2) => 450=0.5*0.14*t^2
    t=80.2
    v=5.6

    So they're relative velocities heading toward each other is 4.2+5.6=9.82
    If that was right then using t=x/v = 450/9.8 I get around 46 secs. But the answer is 67 :(
     
  10. Sep 6, 2007 #9

    Doc Al

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    Since both asteroid and ship move, they don't both travel the full 450 (measured with respect to the initial frame). Their distances add to 450 m. Note that they both travel the same amount of time until they collide.

    There's an easier way: Consider the relative acceleration.
     
  11. Sep 6, 2007 #10

    HallsofIvy

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    You appear to have calculated the time it would take the asteroid to go 450 m, the time it would take the spaceship to go 450 m and then added them! Do you understand that if the spaceship is moving toward the asteroid, the asteroid doesn't have to cover the entire 450 m to hit it? What you want to find is the time for the asteroid and spaceship to cover a total of 450 m.
     
  12. Sep 6, 2007 #11
    Alright, this is what I just tried (didn't work)

    delta_x_r + delta_x_s = 450
    delta_x_r=0.5*0.08*t^2
    delta_x_s=0.5*0.14*t^2

    My TI-89 can solve mutiple eqns so I did not do the math wrong. Why does this not work?
     
  13. Sep 6, 2007 #12

    learningphysics

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    What answer did you get?
     
  14. Sep 6, 2007 #13

    t=64 secs
     
  15. Sep 6, 2007 #14

    Doc Al

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    Seems right to me.
     
  16. Sep 6, 2007 #15

    learningphysics

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    That's what I get also... you sure the original numbers in the question are right? Did you mix up the 490 and 450?
     
  17. Sep 6, 2007 #16
    I don't think I mixed up any numbers. Thanks for all the help guys.
     
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