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Kinematics of Particles

  1. Jan 13, 2007 #1
    1. The problem statement, all variables and given/known data

    http://img402.imageshack.us/img402/5733/problemvm7.th.jpg [Broken]

    I had to draw in that line. It was suppose to show a dotted line of the projectile, but the page did not scan well.

    2. Relevant equations

    3. The attempt at a solution

    First atempt:
    http://img222.imageshack.us/img222/9059/untitled1ym0.th.png [Broken]

    My first solution is almost correct except it seems like my answer is the inverse of the answer given in the book. I started with breaking the given velocity into x-y components and then subtituted them into one of the projectile motion equations.

    2nd atempt starting with equations of projectile motion:

    It seemed that I was getting no where so i stopped

    http://img178.imageshack.us/img178/552/untitled2db5.th.jpg [Broken]

    The image might be somewhat hard to see, so try this http://www.savefile.com/files/409621 [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 13, 2007 #2
    Okay, a good way to start off is not to jump into it first. See what you're given and what you can do with it.

    But the two things that ties these things together is that time is related to both and there's no acceleration in the x direction =).

    Try to redo your problem but write down what's happening in the x, and y direction separately. Explicitly as well, such as you showed initially the velocities in the x and y directions.

    Get into the habit of doing this type of thing here:

    X Axis
    X - Final X
    Xo - Initial X
    Vx - Final Velocity
    Vox - Initial Velocity
    a - Acceleration
    t - Time

    Same stuff but with the y axis =).

    Also, there maybe hidden tricks inside from trigonometry =). Such as cos^2 can be represented as something else and what-not. Just stuff to think about mostly.
    Last edited: Jan 13, 2007
  4. Jan 13, 2007 #3
    thanks for the response

    All my known data is written down somewhere on another paper. What i posted was me scrippling all over the place.

    Can someone confirm that my velocities are correct and that my relationship for y is correct.
    B = theta

    Vx = u sin B
    Vy = u cos B

    I put my origin at the top.
    sin B = Y/R

    Is it correct to say that R = x for the x-component of the x = (x0) + (Vx)ot ?

    Correction: Actually the 2nd image is my first atempt.
    Last edited: Jan 13, 2007
  5. Jan 13, 2007 #4
    Here's a good rule of thumb; sin -> "you *see* the side", cos -> "you're *cozy* with the side".

    Such as <)__| --> | <- that is sin, since you "see" it from the angle.
    Such as <___| --> __ <- that is cos, since it's "cozy."

    Also, it wouldn't be a bad idea to remember:
    Sin = opposite / hypotenuse
    Cos = adjacent / hypotenuse
    Tan = opposite / adjacent

    Now confirm this with your drawing and axis [where is your axis as well =). Which ways are you calling positive and which ways are you calling negative?] if you are correct or not =).

    These are little picky things that will help you along, not only this problem but future problems. Making a habit of such heh.
    Last edited: Jan 13, 2007
  6. Jan 13, 2007 #5
    DIdn;t think my question would lead to a trig review. I was getting tired and was starting to think my source of error was the trig, which it wasn't.:frown:

    After a little rest and some food, the problem is solved.
    Turns out I can't use the equation in the 2nd image. The work in the 3rd image was almost right, but it seems a small error that X= R caused things not simplify so nicely. Also a negative sign i missed somewhere.....:cry:
  7. Jan 13, 2007 #6
    You method is fine, however,
    you can get a fairly quick derivation just by using a different frame of reference:

    let x be in the direction of the incline, and y be perpendicular to the incline.

    acceleration in y:
    [tex]a_y=g \cos\theta[/tex]
    velocity in y:

    acceleration in x:
    [tex]a_x=g \sin\theta[/tex]

    the time it takes for the thing to go up then back down:

    the x distance travelled:
    [tex]x=\frac{1}{2}g\sin\theta t^2[/tex]
    [tex]x=\frac{2v^2\sin\theta} {g\cos^{2}\theta}[/tex]
  8. Jan 13, 2007 #7
    this reminds me of a very interesting physics problem I saw some time ago, the solution is very elegant using the same method (viewing things in a different frame)

    in the same situation you described, a ball is fired perpendicular to the plane, with theta unknown. after N bounces, the ball's velocity becomes horizontal. express theta in terms of N. (assume no loss of energy, and all bounces are elastic)
  9. Jan 13, 2007 #8
    Thanks for showing the alternate method. Funny that rotating the axis was my first very first atempt before I stopped and went with a regular x-y coordinate system. When i changed my reference frame I had left R = x, when it was suppose to be X = R cos (B) in the other. I should have stayed with the orignal plan of attack, because it would have saved a ton of time. One mistake, that's all it takes.:rofl:
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