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Stuck on conservation of momentum

  1. Apr 13, 2009 #1
    1. The problem statement, all variables and given/known data

    a robotic space probe of mass 7600 KG is traveling through space at 120 m/s. Mission control determines that a change in course of 30.0 degrees is necessary and instructs the probe to fire rockets perpendicular to its direction of motion. If the escaping gas leaves the craft's rockets at an average speed of 3200 m/s, what mass of gas should be expelled.

    also, the correct answer is 160 kg

    2. Relevant equations

    P_{1}+P_{2}=P_{1}'+P_{2}'


    3. The attempt at a solution

    ok, So far on this one I have a diagram of a before and after of the scenario, and I used the conservation of momentum equation (P_{1}+P_{2}=P_{1}'+P_{2}'), used vector addition to find that the speed of the probe after the rockets give it a boost would be 7600.95 m/s. However I think what I have so far is wrong.


    If anyone could set me on the right track for this, It would be very much appreciated.
     
  2. jcsd
  3. Apr 13, 2009 #2
    Hi,
    well, I think I should start with a "I don't know how to solve it" :D
    But this is interesting and I'd like to try.
    First I think You should use thrust here, Did You use something about thrust?
    There is something that I can't understand(If I will, Maybe I'll solve it) , it shoots rockets perpendicular to it's direction? how the hack does that make him a 30% turn?!
    maybe I didn't understand the question right.
    well good luck.
     
  4. Apr 13, 2009 #3

    Doc Al

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    Call the original direction the x direction. What y-component of speed must the probe end up with in order to change course as needed? Then consider conservation of momentum in the y-direction.
     
  5. Apr 13, 2009 #4

    dx

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    First, did you figure out how much perpendicular velocity the probe needs to change its direction of motion by 30 degrees?

    Once you figure that out, you can figure out how much momentum it needs in that direction by multiplying that velocity by the mass. Now all you need to do is apply conservation of momentum to figure out how much mass of gas needs to be expelled.
     
  6. Apr 13, 2009 #5
    Hi
    Sorry for interrupting, Could you show how to do that ? I've got 120/tan 30.
    by making a triangle of velocity vectors.
    Is that right?
    thank You.
     
  7. Apr 13, 2009 #6

    dx

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    No, it should be 120 * tan 30.

    tan is "opposite / base".
     
  8. Apr 13, 2009 #7
    Ye ,my bad lol.
    "tan is "opposite / base"" I'm not that stupid :D
    Just wrote the 30 as the wrong angle :D stupid me.
    thanks.
    btw i completely miss understood the question, thought it shoots gas in a rate of 3200 m\s perpendicular to the direction. XD
     
  9. Apr 13, 2009 #8
    I was thinking that while the rockets are producing a thrust that the probe is still moving the original 120 m/s at 0.0 degrees and with those x and y components it cause the probe to move in the 30 degree direction. But I was just assuming, it doesn't state it in the problem. About thrust, the only thing similar that I have learned in the class is using force, but I do not know how to apply it here.



    is the y-component is the 3200 m/s produced by the rockets? That's how I set it up with a Pythagorean theorem. For the x-component, I set that as V_{x}=120 m/s and then the y-component was set to V_{y}= 3200 m/s and after using the P. Theorem, I got 3202.25 m/s but that didn't really get me anywhere.

    Could you elaborate a little on your explanation, because I'm not sure how to go about solving for that besides what I explained above.


    To figure out the perpendicular velocity, would you use the Pythagorean theorem or some equation. I'm not sure how to solve for that.


    Thank you all for your interest in helping :)
     
  10. Apr 13, 2009 #9

    Doc Al

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    No. -3200 m/s is the speed of the exhaust, not the probe.

    In order for the probe to move at 30 degrees to the original direction, what must its speed be in the y-direction? You already have its speed in the x-direction.
     
  11. Apr 13, 2009 #10

    dx

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    Draw the velocity vector of the ship at 30 degrees to the horizontal. You know that the horizontal projection of this is 120 m/s. Use this information and the fact that it is 30 degrees to figure out the vertical projection.

    HINT: You need to use trigonometry.
     
  12. Apr 13, 2009 #11

    I'm guessing that the gas is being shot at 3200 m/s, 270 degrees pushing the probe up. is that correct? Also, would the gas be pushing the probe up at 3200 m/s, 90 degrees? I set up another velocity vector triangle and I got 69.3 m/s, 90 degrees for the y-component. Is that correct?
     
  13. Apr 13, 2009 #12
    OH!! Ok, I see what I need to do. Thank you so much!! I really appreciate it.
     
  14. Apr 13, 2009 #13
    ok, another question (i'm sorry to bother you) but when I set up the conservation of momentum equation, will it look like this:

    m_{probe}v_{initial of probe}= m_{probe}(V_{x}+V_{y})+ m_{gas}V_{gas}

    (7600kg)(120 m/s)= (7600)([tex]\sqrt{120^2+69.3^2}[/tex])+m(gass)(-3200)
     
  15. Apr 13, 2009 #14

    dx

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    No you don't need to consider momentum in the horizontal direction. Momentum is conserved component-wise. Remember that momentum is a vector.
     
  16. Apr 13, 2009 #15
    alright. I got the answer, but I think i did it in an unorthodox way.

    here is my work:

    0= (7600)(69.3)+ m_{gas}(-3200)
    0=526680+(-3200)m
    -526680= -3200m
    m=164.6 (with significant digits)=160 kg

    however, i set the first half to zero which I think was the wrong way to solve it.
     
  17. Apr 13, 2009 #16

    dx

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    Why do you think it's the wrong way? I don't see anything wrong with it. The first half is zero because the initial momentum in the vertical direction is zero.
     
  18. Apr 13, 2009 #17
    Ok, that helps. The reason I thought it was wrong was because I set it equal to zero and didn't know why. Thank you so much for helping me through this. I really appreciate it.
     
  19. Apr 13, 2009 #18
    Here's my take on this, although in the time that it took me to type it out I see that others have beat me ;)

    Initial Momentum (probe and gas) = Final Momentum (probe) + Final Momentum (gas)
    p = p1' + p2'

    p2'
    --------
    |......../
    |....../
    |...../ p1'
    |30/
    |./

    Excuse my drawing, I'm sure there's a better way to show this. Initial momentum, p, is the vertical line. Final momentum of the probe is p1' , and final momentum of the gas is p2' .

    p = mv = (7600)(120) = 912000

    So p2' is (tan30)912000 = 526543.4455

    p2' = mv
    m= p2'/v = 526543.4455 / 3200 = 164.5 kg
     
  20. Apr 13, 2009 #19

    Doc Al

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    Although it won't make much difference in your answer, you might want to consider the mass of the gas as being contained in the initial mass of the probe. (Thus 7600 kg would be the total mass.)
     
  21. Apr 13, 2009 #20
    I like Your solution!
    Very nice indeed :D
    very simple!
     
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