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Projectile Problem again

  1. Sep 10, 2008 #1
    I thought I have already master the topic on projectile. Now I'm not that sure. I fail to solve the Problem 3.4 in the book An Introduction to Mechanics by Daniel Kleppner and Robert J. Kolenkow.

    The verbatim problem is as follows:
    "An instrument-carrying projectile accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is L. The projectile breaks into two pieces which fly apart horizontally. The large piece has three times the mass of the smaller piece. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station.
    How far away does the larger piece land? Neglect air resistance and effects due to the earth’s curvature."

    I have consider something like this:
    Assume that the velocity of the projectile before the explosion to be [tex]u\hat{i}[/tex] (top of the trajectory). Let the velocities of the pieces after the explosion be
    [tex]\vec{V}_1 = V_{1x}\hat{i} + V_{1y}\hat{j} [/tex] and [tex]\vec{V}_2 = V_{2x}\hat{i} + V_{2y}\hat{j} [/tex]. Also let the time for the smaller piece to return to the launching station be t1. Then

    -L = V1xt1 ------ (1)

    Now I have already introduced six variables u, V1x , V2x , V1y , V2y and t1.

    I also have to know when the larger piece landed (t2 - another variable?) in order for me to calculate where the larger piece landed. That's the seventh variable.

    From the principle of conservation of momentum, I can get another two equations. Altogether now I have only 3 equations with 7 unknowns.

    My questions:
    1. What other 4 equations can I have without introducing another unknowns.
    2. Is t1 = t2 ?
    3. Why is the momentum conserved in the vertical direction ? There is an external force, gravity acting downward.

    Really appreciate any suggestion. Thank you for your interest.
  2. jcsd
  3. Sep 11, 2008 #2


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    You don't have seven unknowns at this point in the problem; remember that the problem says that the pieces explode apart horizontally.

    But actually you don't need to know all of the values of the unknowns to answer this question.

    The fact that the pieces fly apart horizontally is important here, too. What do you think the answer to this question is? Think about what determines the values of [itex]t_1[/itex] and [itex]t_2[/itex].

    The idea is that the explosion occurs fast enough that the impulse is effectively zero, so you can ignore the effect of gravity on the explosion itself.
  4. Sep 11, 2008 #3
    I must have overlook the words explode apart horizontally. Careless me. With your hints now I can imagine writing the solution. Thanks alphysicist.

    [itex]t_1[/itex] = [itex]t_2[/itex] because immediately after explosion there are no vertical component velocities. Yes, I recognize that it is not necessary to know all the unknowns. What we are interested is only to know the value of V2xt2.

    How about the following situation? Is the vertical momentums still conserve?

    A circus acrobat of mass M leaps straight up with initial velocity u from a trampoline. As he rises up, he takes a trained monkey of mass m off a perch at a height h above the trampoline. What is the maximum height attained by the pair?
  5. Sep 11, 2008 #4


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    When you ask the question "Is the momentum conserved" you have to choose two things to focus on: what is your system and what time period are you looking at. This is because the equation that describes momentum conservation is (in terms of average forces):

    \vec F_{\rm ext} \Delta t = \Delta \vec p
    (If you don't use average forces you will need the integral form of this equation where the impulse is [itex]\int \vec F\, dt[/itex].)

    So before asking is the momentum conserved, are you asking about the momentum of (earth + man + monkey), or of (man + monkey), or just the man, or just the monkey? If your system is (earth +man+monkey), there are no external forces on the system so momentum is conserved during the entire motion. In other words, momentum is conserved because the external forces are zero.

    If you are only looking at the momentum of the (man+monkey), then it matters what time period you are looking at. For the entire motion, from when the man is launched to when he reaches the highest point, vertical momentum is not conserved, because [tex] F_{\rm ext,y} \Delta t[/tex] is not small (compared to the momentum in the problem).

    However, if you only look at the very short time period during which the man is grabbing the monkey, then the impulse from gravity that you can calculate from [tex]\vec F_{\rm ext,y} \Delta t[/tex] will be small enough (because [itex]\Delta t[/itex] is very small) that it will not affect the motion, and so vertical momentum is conserved during that very short time period. So although the vertical force is not zero, the time interval is vanishingly small and so the impulse is negligible, and it is the impulse that determines whether momentum is conserved or not.
  6. Sep 11, 2008 #5
    That answer my questions. Thank you very much alphysicist. I really enjoy and appreciate your responds.
  7. Sep 11, 2008 #6


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    Sure, glad to help!
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