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Homework Help: Signifigant digits?

  1. Sep 28, 2005 #1
    Hi Guys,
    I think I've done the question below correctly but my result doesn't jive with a couple others I have found online for the same problem (answers of 160kg and 170kg). I want to insure I get full marks (of course) while simultaneously making sure I am understanding the concepts correctly. I'm wondering if I'm doing this right, I think the variance in answers must be rounding errors, maybe on my part or maybe on the part of the other individuals...

    The question;
    "A 7600-kg space probe is travelling through space at 120 m/s. Mission control determines that a change in course of 30.0° is necessary and, by electronic communication, instructs the probe to fire rockets perpendicular to its present direction of motion. If the escaping gas leaves the crafts rockets at an average speed of 3200 m/s, what mass of gas should be expelled?"

    My Answer and logic,
    (sorry for the crude diagram)

    ................./ |
    .............. /60|
    ........../........| A
    ....... /..........|
    ..... /............|
    ... /..............|
    ../30...... 90...|

    So I make my vector triangle, and attain B.
    B = 7600kg * 120m/s = 912,000kg *m/s = 9.12 * 10^5 kg*m/s

    now I use law of sines to get side A.

    A = (sin 30.0) * (9.12 * 10^5 kg*m/s)/(sin 60.0) = 526543.45 = 5.26 * 10^5kg*m/s

    So now I have A = 5.26 * 10kg*m/s and I devide by 3200m/s to get 164.38 = 1.64 * 10^2kg or 164kg.
    So my final answer is 164kg which apparently doesn't agree with some similar question/answers I've seen for honors physics on the net..
    Any ideas?
    Last edited: Sep 28, 2005
  2. jcsd
  3. Sep 28, 2005 #2


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    Why would you need law of sines? That's a right triangle. Assuming this change of course does not change speed, then B is just 7600*120*sin(30).
  4. Sep 28, 2005 #3
    That doesn't seem right, wouldn't it be A = C * sin 30 ..I want A not B right? I have B = 912000.
    In any case, that yields 142.5 which I'm sure is incorrect, can you explain why using law of sines wouldn't work? It's all a derivative of the the same trig functions anyway?
    I'm using law of sines becuase it is useful for computing the lengths of the unknown sides in a triangle if two angles (or more) and one side are known.
    Last edited: Sep 28, 2005
  5. Sep 28, 2005 #4


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    Okay, your picture confused me for a moment- Your 30 degree angle looks larger than the 60 degree angle!

    I said in my post that I was assuming that the speed did not change so I was working with a hypotenuse of 7600*120. If the rocket blast is momentary, or if by "perpendicular" they mean "perpendicular to the rocket ship throughout the blast" rather than perpendicular to the original path, that will be true. You are taking one leg of the triangle to be 7600*120 and solving for the other. In your case the sine law will work, it's just unnecessary. The sine law gives sin 30/sin 60= sin 30/cos 30= tan 30. That you can get directly from the right triangle.
    Last edited by a moderator: Sep 28, 2005
  6. Sep 28, 2005 #5
    Alright ..do you agree that the answer is correct? The purpose of my post was to find out if the significant digits are done correctly. As far as I know I don't use significant digits (3 of them) with sin x or cos x because they are absolute values.
    My issue is that my answer doesn't agree with one posted by a high school honours physics teacher here (question 18, answer of 160kg towards the bottom), I always get leery when my work disagrees with a teachers!
  7. Sep 28, 2005 #6
    You've got to look at the significant digits in the values you're given. 7600 kg, 120 m/s, and 3200 m/s all have 2 significant digits. Since you're multiplying and dividing these numbers, it's the number of digits that matter, so your answer should have 2, i.e., 160 kg.
  8. Sep 29, 2005 #7

    You're exactly right ..I was thinking that 7600 was 4 signifigant digits and 120 was 3 etc, forgetting that 0's to the right of the whole number are ambigious.
    Thanks so much Grogs!
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