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Conservation of Momentum of astronaut leaving ship

  1. Jun 19, 2015 #1
    1. The problem statement, all variables and given/known data
    An astronaut is pushed from her ship at a velocity of 2m/s. Her weight including her tool belt is 120kg. Remembering newtons 3rd law, she takes 5 seconds to decide to detach her belt and throw it away from her. The 20kg tool belt leaves her suit as she throws it in front of her 10m/s. If she has 20 minutes of air left in her tank, does she make it?

    2. Relevant equations
    Conservation of Momentum
    mvi=mvf

    Distance
    d=vt

    3. The attempt at a solution
    I say that the astronaut survives.

    Because of distance equation= (2)(5)=10m
    Her initial momentum is 240kg m/s
    And by throwing the belt in front of her she is propelled at -.4m/s
    Since she only traveled 10m she would make it back on time.
     
  2. jcsd
  3. Jun 19, 2015 #2
    I would start by revisiting conservation of momentum and assigning each of the velocity vectors a sign depending on whether they are directed towards the ship or away from the ship. I think that will really clear things up.
     
  4. Jun 19, 2015 #3

    gneill

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    Staff: Mentor

    The question as stated seems to be a bit ambiguous to me. Is the speed that the tool belt is thrown meant to be taken relative to the astronaut or the initial reference frame? If she happened to be pushed out at a speed of 10 m/s instead of 2 m/s, would that mean she couldn't throw the tool belt away from herself at all?
     
  5. Jun 20, 2015 #4

    haruspex

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    There are three reference frames to choose from for the 10m/s: the ship, the astronaut before throwing the belt, the astronaut after throwing the belt. For the question to be at least a bit interesting, I think it has to be the second.
    Tyrone, how do you get the -.4m/s?
     
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