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Rocket Momentum Problems

  1. Jun 25, 2006 #1

    My class has just started going in to learning about momentum and I’ve found that I’ve been having quite a time putting all the formulas together to figure out homework problems… in fact I haven’t been able to answer one correctly yet.

    My first problem is:

    “A 4200-kg rocket is traveling in outer space with a velocity of 120m/s toward the Sun. It needs to alter its course by 23 degrees, which can be done by shooting is rockets briefly in a direction perpendicular to it’s original motion. If the rocket gases are expelled at a speed of 2200 m/s relative to the rocket, what mass of gas must be expelled?”

    So far I’ve gone about solving the problem by assuming that the final momentum will be equal to the initial momentum and that the initial momentum will be equal to 504,000 N (4,200 kg * 120 m/s). I’m then setting up a free body diagram which has a force (with v = to 2,200 m/s) shooting perpendicular to the rocket’s movement. Then when it comes time to actually figure out the answer I get stuck.

    I’m assuming I need to find the new momentum in the X and Y (which I find to be 464,000 N and 197,000 N respectively) but I’m having trouble figuring out the mass of gas I’d need from this information. I’m guessing a derivative needs to be taken somewhere but I can’t figure out where, also I’m assuming the Sun plays no role which might be a bad asumption.

    So I get stumped on the first problem. No big deal I’ll just move on to the second… except I find myself getting stuck yet again.

    “A 200 kg projectile, fired with a speed of 100 m/s at a 60 degree angle breaks into three pieces of equal mass at the highest point of it’s arc. Two of the fragments move with the same equal speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally. Determine (a) the velocity of the third fragment immediately after the explosion and (b) the energy released in the explosion.”

    This problem was the one that made me realize I needed help. I can see the answers in the back of the book as a) (100 m/s)i + (50 m/s)j (b) 3.3 x 10^5 J. But I can’t even figure out (a).

    I go about the problem by first finding out how fast the projectile will be traveling when it reaches it’s highest point. I do this by first figuring out the velocities in the X and Y direction (50 m/s and 87 m/s) and I realize that at the highest arc point there will be no more movement in the Y direction so the velocity will be 50 m/s meaning the total momentum pre-explosion will be 10,000 N (50 m/s * 200 kg). I then set up my equation:

    10,000 N = 2m(V1) + m(V2)

    Knowing that M will be (200/3) kg I can get to this point:

    150 m/s = 2(V1) + (V2)

    Which looks somewhat similar to the answer I’m supposed to be getting but I don’t understand where the i or j come in. Vectors I’m thinking but I don’t understand why that would even matter. The more I think about it the more I believe that all the velocities should be equal since all of the objects are of equal mass but the books answer tells me otherwise.

    Starting to get confused and frustrated so any help would be greatly appreciated, I’ve still got about 20 more problems to go…

  2. jcsd
  3. Jun 25, 2006 #2

    Andrew Mason

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    What is the change in momentum of the rocket ship? (think vector subtraction).

    What is the impulse (vector quantity) received from the rocket firing?

    How is the impulse related to the change in momentum?

  4. Jun 25, 2006 #3
    Looking in my book it says

    The rate of change of momentum of an object is equal to the net force applied to it.

    My understanding was that momentum didn't change but from your question I'm now assuming momentum is a vector not a quantity... which would make sense because it is a force.

    But even with this information I don't understand what the "change" would be other than direction. Isn't it the same magnitude?

    My book tells me that the change of momentum is dP/dt but I don't have a time slice that I know of yet. That's probably what I need to figure out first... hmm.

    My book tells me the change in momentum of an object is equal to the impulse acting on it. Does this work on a single axis?

    If so then I'd guess 200,000 N in the opposite direction of thrust. Not sure on that though.

    Unfortunately I'm still not clear on how this connects to finding the mass. Thanks for the help though.
  5. Jun 25, 2006 #4


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    Momentum is not a number: it is a vector.
  6. Jun 25, 2006 #5
    I understand that now. Am I correct in assuming that the magnitude will be the same and that only change will be direction?
    Last edited: Jun 25, 2006
  7. Jun 25, 2006 #6

    Andrew Mason

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    Correct. Draw the velocity vector before and the velocity vector after the impulse. The change in velocity is the vector from the head of the first to the head of the second. What is the magnitude of that vector? That is the magnitude of the change of momentum. Work out how the magnitude is related impulse. Work out the force provided by the rocket thrust. Then determine the time needed for that force to provide the needed impulse.

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