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Projectile Motion- Incline vs. Horizontal Surface

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    IMG_20150510_233249_460.jpg
    Two identical metal blocks, H and A, are placed at equal heights on frictionless ramps as shown above. The blocks are released at the same time, travel down the ramp, and then slide off their respective ends of the table. Block H leaves the table from a horizontal surface. Block A leaves the table at a 30 degree incline. Both blocks leave the table from the same height.

    1) Indicate which block, if either, will be travelling faster when it reaches its highest point after leaving the table. Justify your answer qualitatively without using any equations.

    2) Justify your answer about which block, if either, will be travelling faster when it reaches its highest point quantitatively using appropriate equations.

    3) Is enough information provided to determine which block, if either, travels a greater horizontal distance before hitting the floor? If so, indicate your prediction and justify your answer qualitatively. If not, indicate what additional information is needed and explain how you would use that information to determine which block would travel a greater horizontal distance.

    4) Is enough information provided to determine which block, if either, has a greater speed at the instant it hits the floor? If so, indicate your prediction and justify your answer qualitatively. If not, indicate what additional information is needed and explain how you would use that information to determine which block would hit the ground with a greater speed.

    2. Relevant equations
    x = vxt + x0
    y = 1/2ayt + v0yt + y0

    3. The attempt at a solution
    1) Vertical velocity is 0 at highest point, so speed depends on horizontal velocity? I feel like neither block may be travelling faster, but I don't know why.

    2) VAcos30 = VAx I think, I'm not sure about VHx.

    3) There is not enough information. To find the horizontal distance of each, I would need to know their horizontal velocities and the time it takes for each to hit the floor to plug into x = vxt + x0

    4) There is not enough information. To find their final velocities, I think I would need their horizontal velocities and the time, as well as their initial vertical velocities, which would be 0 and VAsin30? Vy = ayt + V0y finds their vertical velocities, and their final speeds can be found with the Pythagorean theorem?
     
  2. jcsd
  3. May 11, 2015 #2
    When the blocks just reach the end of the wedge, they will have equal kinetic energies, since both have lost the same amount of potential energies. will the velocities of the block become different at the instant they leave contact?
     
  4. May 11, 2015 #3
    Oh, no they won't! Thank you!
     
  5. May 11, 2015 #4
    so they leave contact withe equal velocities.
    the only difference is the angle of projection.
     
  6. May 11, 2015 #5
    also, the block A loses some kinetic energy as it goes up so the magnitude of velocity has decreased since it has lost the vertical component. So which block has more velocity at highest point?
     
  7. May 11, 2015 #6
    Do the quantitative part yourself. I will give you some ideas
     
  8. May 11, 2015 #7
    So VH = VHx = VA, and VAcos30 = VHcos30 = VAx, and the velocity of H will be greater?
     
  9. May 11, 2015 #8
  10. Aug 21, 2016 #9
    how are the horizontal velocities the same if block a has a longer distance to accelerate?
     
  11. Aug 21, 2016 #10
    The Horizontal velocities aren't the same. ##v_{A_X} = v_{H_X} \cos 30##. The magnitude of their overall velocities should be the same at the point of leaving the table though, as they have both travelled through the same altitude drop - you can see this by approaching it from an Energy perspective. It's just that one block has its velocity only in the x-direction, and the other has components in x and y.
     
  12. Aug 21, 2016 #11
    yes but, how can you prove it without an energy perspective.
     
  13. Aug 21, 2016 #12
    Well, the energy argument is so much simpler, that I hadn't really considered how to prove it without using that particular tool.
    I suppose if we're discounting friction and air resistance, it is trivial to prove that the blocks are travelling at the same speed as they reach the dotted line, as they have been accelerated for the same period of time. After that, block H will continue at the same speed until it leaves the table, so we just have to prove that block A is travelling at the same speed when it leaves the table as it is when it enters the curved dip. I guess you could express the kinematic equations in terms of ##\ddot \theta## and it should turn out that the speed is unchanged at the point of leaving the table, while the direction of the velocity is altered...but it's just a thought.
     
  14. Aug 21, 2016 #13

    haruspex

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    Well, no, they leave with equal speeds. If the directions are different than necessarily the velocities differ.
     
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