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Rocket Engine pushes puck

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data

    A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the xy plane. The 4.00-kg puck has a velocity of 3.00ˆi ms at one instant. Eight seconds later, its velocity is (8.00ˆi + 10.0ˆj) ms. Assuming the rocket engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude.

    2. Relevant equations

    ΣF = m*a
    a = ΔV/Δt

    3. The attempt at a solution

    Well, I took the all the info and applied it on the various formulas I know, and found the correct results, according to the book. So my problem doesn't lie there, as it's a simple problem of putting the correct quantities in the correct "positions". My problem lies more with "conceptualizing" the problem.

    I also checked the manual, but I can't understand one thing: It says that the force that the engine exerts on the puck is horizontal, so I take it that means the X part of the XY axis', correct? So, if this force is horizontal, then how can he ask me about its component of the net force? Does the statement mean that it's not horizontal per se, just that it acts on the first quadrant? Apart from that, how does the puck develop a velocity at the Y/vertical plane?

    From what I understand from the statement, it's a standard round puck on top of a, say floor, and I've put an engine on the back, that exerts a horizontal F force on it, causing it to move horizontally, but it somehow develops a velocity at both planes. Could anyone explain to me what exactly I'm missing?

    Thanks!
     
  2. jcsd
  3. Dec 11, 2016 #2

    gneill

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    Staff: Mentor

    In this case the Y-axis is not vertical. The entire XY plane is horizontal. If a vertical direction were involved it would (likely) be designated Z.
     
  4. Dec 11, 2016 #3
    Ah, so it's as if I'm dealing with 3D object that slides on the ground. And the problem is using the X & Z axis' of the usual XYZ axis, just renaming the Z one Y. So technically, it's moving horizontally, but not in a straight line, it skews a bit to the side. Correct?
     
  5. Dec 11, 2016 #4

    gneill

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    Staff: Mentor

    You've got the picture. The XY plane is generally taken to be horizontal when using a 3D XYZ coordinate system. Z is the usual designation for the vertical direction. X, Y, and Z are really just labels that we have come to use by convention.

    We often use a 2D coordinate system to analyze motions that occur in a plane, such as typical projectile motion, and then we usually assign the X to the horizontal direction and Y to the vertical direction. But this is not strictly necessary. All those projectile motion equations could be written using X and Z instead of X and Y. Or, other letters or symbols could be assigned (UVW, for example). It's really just a matter of convention.
     
  6. Dec 11, 2016 #5
    Got it. Yeah, I know the letters are interchangable, but I've not tackled 3D problems yet, so when I saw "horizontal" my mind went straight to the usual "X is horizontal, Y is vertical" concept.

    Thanks for clearing this up for me!
     
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