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Playing catch in space

  1. Mar 5, 2008 #1
    I have no idea where to begin there is so much information, just way over my head. Can anyone help?


    Two astronauts, floating in space at rest with respect to their ship, are initially 6.0 m apart. They decide to play catch with a 0.490 kg asteroid. Paul, whose mass is 101.0 kg, heaves the asteroid at 18.4 m/s toward Heather, whose mass is 63.0 kg. She catches it and heaves it back 2.5 s later at 18.4 m/s (with respect to the ship) toward Paul. Just before Paul catches it a second time, how far apart are the astronauts?
     
  2. jcsd
  3. Mar 6, 2008 #2
    Well conservation of momentum rules should apply

    Newton's law of "equal and opposite" also applies.

    The momentum caused by the first astronaut throwing the rock needs to cancel out. the rock moves forward, while he moves backwards. Make sense? (equal and opposite)

    so the momentum (mv) of the rock has to equal the momentum (mv, again) of the astronaut.

    assume that Paul starts on (0,0) and Heather (6,0)

    .490*18.4=101*|V|
    |V|= .08926.... m/s

    but remember, he's moving from (0,0) towards (-1,0), so therefore, his v is actually negative.

    V=-.08926...m/s

    then the other astronaut catches it, but that means that momentum is transferred and she gets a velocity in the positive direction.

    m(rock)*v(rock)=m(rock+heather)*V(both)

    she holds on to it for 2.5 sec, remember? so you need to add their mass because after she catches it, they both travel at the same speed :)

    18.4*.490=(63+.490)V

    here, V is positive because both start traveling towards (7,0) from (6,0)

    V=.14200....m/s

    then, some time passes (2.5 sec, but that's irrelevant for now), and she throws it back at 18.4 m/s RELATIVE TO THE SHIP! so relative to her, the rock will be traveling at .14200...+18.4=18.542... m/s

    so 18.542*.49=63*V

    V is positive again. V=.14421.... m/s

    now let's go back to paul. we left him traveling at -.08926 m/s. but how long was he there?

    well, we know that the initial distance from where he threw the rock to where she caught the rock was 6m, and the rock was traveling at 18.4 m/s, so the time taken was .3260....s, but remember, she held on to the rock for 2.5s, so he was floating along at his merry speed for 2.8260....s.

    in that time, he traveled 2.8260....s*-.08926 m/s, which is -.25225...m.

    but remember, he's still traveling at -.08926 m/s.

    so how long does it take the rock to get to him?

    the moment she threw it, the distance between them was 6.25225... m. the rock is traveling at 18.4 m/s and he's floating along at -.08926 m/s. so relative to one of them, the other is traveling at 18.31074....m/s. and it needs to cover 6.25225... m. it takes the rock .34145...s to do that.

    so the total distance traveled by paul is -.08926 m/s*3.1674....s=|.28272...m|

    so he's now at (-.28272...,0)

    but what about heather?

    she was left with a velocity of .14421...m/s. and after she throws the rock, till the time paul catches it, there's .34145...s for her to travel. she moves .04924....m.

    so she's at (6.04924...,0)
    again, he's at (-.28272...,0)

    distance between them?

    if i did this all right, 6.3320...m

    the ellipses are for extra digits and rounding. i suggest you do the calculations again to make sure. but that's the logic.

    hope that helps!
     
  4. Mar 6, 2008 #3
    oh. i was too interested in helping you, that i forgot to say

    Welcome to PhysicsForums!

    i'm new too, for the record.
     
  5. Mar 6, 2008 #4
    well, relatively.

    no pun intended
     
  6. Mar 6, 2008 #5
    Thank you for your help.
     
  7. Mar 6, 2008 #6
    is that the right answer?
     
  8. Mar 6, 2008 #7
    But unfortunately your calculations are incorrect, fortunately this was just a bonus question our physics teacher threw at us. But none the less, two astronauts have died today=[
     
  9. Mar 6, 2008 #8
    wait, i did something wrong? can you explain so that i can learn, please?
     
  10. Mar 6, 2008 #9
    I really don't know how to do it, the way we turn our answer in is on a website so we know right away whether we are correct or incorrect. But the good news is that i have 5 attempts remaining to solve it. When i find out how to do it or the answer I will be sure to post it.
     
  11. Mar 6, 2008 #10
    can you check my calculations? maybe i made a dumb calculator mistake
     
  12. Mar 6, 2008 #11
    i don't see anything.... maybe someone else'll come along and check.
     
  13. Mar 6, 2008 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi pchalla! :smile:

    I'm a bit confused as to what you did there. :confused:

    It's a lot easier to make all calculations in the rest-frame of the ship - especially since the examiner deliberately made that easier for you by giving the asteroid's second speed relative to the ship!

    Then you'll have .49x(18.4 + .142) = 63(V - .142), or:
    V = (.49x18.542)/63 +.142
    = .2862

    A simpler method is to remember that the total momentum (including Paul's) is zero, so: 101x.08926 + 18.4x.49 = 63V,
    or .49x18.4x2 =63V, or V = .2862

    Does that help? :smile:
     
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