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Astronaut in space, calculating velocity etc

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey!
    I´m stuck on a physics problem. I had to translate this problem into English, so please excuse any grammar mistakes!
    An astronaut is on a mission in space. He is currently outside his spaceship which is standing still. The astronauts lifeline breaks, and he starts floating away from the spaceship with a velocity of 0,10 m/s. In an attempt to save himself and return back to his spaceship, the astronaut gets rid of a tool weighing 12 kg. The astronaut weighs 90 kg.
    a) At what speed and in which direction should the tool be dispatched so that the astronaut can return to the spaceship with a velocity of 0,05 m/s?
    b) What average force does the astronaut need to use if if takes 0,5 s to dispatch the tool, and the astronaut then returns to the spaceship with a velocity of 0,05 m/s?


    2. Relevant equations
    Newton II: F=m*a
    Newton III: F1=F2



    3. The attempt at a solution

    a) I know that the answer is supposed to be 1,225 m/s, but I did not get this answer when attempting to solve the question.
    I tried to solve the question in the following way:
    Astronauts velocity before getting rid of the tool =0,1 m/s
    Astronauts velocity after getting rid of the tool = 0,05 m/s
    Astronauts weight before getting rid of the tool = 90 + 12 kg = 102 kg = m1
    Astronauts weight after getting rid of the tool = 90 kg=m2
    Which means that Δv1 = (0,05-0,1) m/s = -0,05 m/s
    From Newton II we know that m1a1=m2a2 and hence m1Δv1=m2Δv2
    I plugged the above numbers into the equation m1Δv1 = m2Δv2 and solved for Δv2 and got the following answer: -0,056 m/s
    Where did I go wrong??

    b) Since I already know that the answer to question a) was supposed to be 1,225 m/s, I tried solving for the tools acceleration in order to then use Newton II´s formula F=m*a and Newton III´s law that F1=F2.

    ΔVtool/Δtime = acceleration tool
    acceleration tool = 1,225 m/s/ 0,5 s = 2,45 (m/s^2)
    F= m*a = 12 kg * 2,45 (m/s^2) = 29,4 N
    Which is also wrong! The answer is supposed to be 27 N

    I urgently need your help! Thanks!
    Astrid
     
  2. jcsd
  3. Feb 13, 2014 #2
    So to solve this problem you'll need to think about the directions of the astronaut and the tool both before and afterwards. In your answer you assume that the astronaut is traveling in the same direction both before and afterward he throws the tool. How would you correct this?
     
  4. Feb 13, 2014 #3
    Well, before he throws the tool, he´s moving in a positive direction, but I still have to take into account that the change in velocity is given by velocity after throwing away the tool - velocity before throwing away the tool? Can I correct for this by simply getting rid of the negative sign infront of Δv1, such that Δv1 = 0,05 m/s?
    But the problem is that I still get the wrong answer!
     
  5. Feb 13, 2014 #4
    [itex]\Delta v = v_f - v_i [/itex] right? You should recalculate that again.

    Also, I don't think that 1,225 m/s is the correct answer.
     
  6. Feb 13, 2014 #5
    If I recalculate ΔV = final velocity - initial velocity I´ll still get the same answer of -0,05 m/s, since the final velocity is supposed to be 0,05 and the initial velocity 0,1.
    Sorry! I´m just not very logical.
     
  7. Feb 13, 2014 #6

    BvU

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    If you want to change from 0.1 m/s away from the life-saving ship ot 0.05 m/s towards the ship, what do you think Δv1 should be? If in doubt draw a picture with arrows.

    In general, you should formulate a bit more carefully: From Newton II you know that ##\Sigma \vec F = 0 \ \ \ ## so ## m_1 \vec a_1 + m_2 \vec a_2 = 0 \enspace \Rightarrow \enspace m_1 \vec a_1 = - m_2\vec a_2##
     
  8. Feb 13, 2014 #7
    The final velocity is in an opposite direction (towards the spaceship) compared to the initial velocity (away from the spaceship). If you take the difference of the two you need to consider the signs (direction) of the velocities.

    The way you're currently taking the difference you are assuming that the velocities are in the same direction: [itex]\Delta v = (+0.05) - (+0.1) [/itex]
     
  9. Feb 13, 2014 #8
    Ok, You mean that the initial velocity away from the ship should be - 0,1 m/s and the velocity towards the ship should be + 0,05 m/s, which thus gives us a change in velocity of 0,05 m/s - - 0,1 m/s = 0,15 m/s?
    But plugging this into the equation m1a1 = - m2a2 and thus m1*v1 = m2*v2 still results in the wrong answer of -0,17 m/s?
     
  10. Feb 13, 2014 #9

    BvU

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    0.15 m/s is OK. the "and thus" escapes me. Could you explain ?
     
  11. Feb 13, 2014 #10
    Yeah, sorry! Made a mistake there. The equation is supposed to be m1Δv1 = - m2Δv2
    since acceleration (when constant) is given by a= Δv/Δt, I simply plugged in Δv1/Δt = a1 and Δv2/Δt for a2 in the equation m1a1 = -m2a2, and solved for Δv2?
     
  12. Feb 13, 2014 #11

    BvU

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    We are in deep space, no external forces. The "and thus" is wrong:
    ##\Sigma \vec F = 0 \ \ \ ## so (by integrating over time from just before to just after throwing
    -- the acceleration doesn't even have to be constant for this; in part 2 we calculate an average acceleration --) :$$ m_1 \vec a_1 + m_2 \vec a_2 = 0 \enspace \Rightarrow \enspace m_1 \Delta\vec v_1 = - m_2\Delta\vec v_2$$As you found out correctly. It comes out just right, but that isn't always a good argument. It comes out right because it's correct.

    The other way to write is as momentum conservation:
    ##\Sigma \vec F = 0 \ \ \ ## and ## \vec F = {d\vec p\over dt} \enspace \Rightarrow \enspace \Sigma \vec p = {\rm constant} \enspace \Rightarrow \enspace m_1 v_{1,{before}} = m_2 v_{1,{after}} + (m_1-m_2) v_{tool} ##
    with the proper signs !
     
  13. Feb 13, 2014 #12
    Thanks! I get the right answer (1,225 m/s) if I use the "momentum conservation" equation, but the wrong answer if I use the equation m1Δv1 = m2Δv2
    If
    m1=102 kg
    Δv1= 0,15 m/s
    and
    m2=90 kg
    solving for Δv2 results in 0,17 m/s.
    I´m wondering what went wrong...
     
  14. Feb 13, 2014 #13

    BvU

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    there is no Δv1 for m1. one part of m1 has a Δv of -0.15, the other part has a Δv of (1.225 - 0.1) m/s
     
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