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Homework Help: Colliding particles, level 3 difficulty

  1. Sep 17, 2006 #1
    Halliday 7e chapt 4 #16 (with web generated values)

    In the figure (attatched), particle A moves along the line y = 33 m with a constant velocity of magnitude 3 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration of magnitude 0.47 m/s2. What angle θ between and the positive direction of the y axis would result in a collision?

    I got theta = 74 degrees, which is wrong. Here's my work:

    PARTICLE A:
    x_0=0m
    y_0=y=33m
    v_0=v=3 m/s
    a=0m/s^2

    PARTICLE B:
    x_0=0m
    y_0=0m
    v_0=0m/s
    a=0.47m/s^2
    a_x=0.57sin theta
    a_y=0.47cos theta
    (NOTE: theta is not conventional. It is measured from the y-axis)

    GENERAL:
    collision occurs when A and B have the same x and y co-ordinates.

    I did some simplifying of [itex]x=x_{0}+v_{0}t+\frac{1}{2}at^2[/itex] and arranging what I know and got the following 6 equations:

    [itex]y_{A}=y_{B}[/itex]
    [itex]x_{A}=x_{B}[/itex]
    [itex]y_{A}=y_{0A}[/itex]
    [itex]x_{A}=v_{0A}t[/itex]
    [itex]y_{B}=\frac{a\cos\theta\t^2}{2}[/itex]
    [itex]x_{B}=\frac{a\sin\theta\t^2}{2}[/itex]

    and from these I get the following two eqations:

    1) [itex]y_{0A}=\frac{a\cos\theta\t^2}{2}[/itex]
    2) [itex]v_{0A}t={a\sin\theta\t^2}{2}[/itex]

    solve equation 2 for t, and subsitute that value into eqution 1 to get:

    3) [itex]y_{0A}=\frac{a\cos\theta}{2}\(\frac{2v_{0A}}{a\sin\theta})^2[/itex]

    distribute the squared and kill all fractions in equation 3 to get:
    [itex]y_{A}a(\sin^2\theta)=2(v_{0A})^2\cos\theta[/itex]

    Replacing sin^2(theta) with 1-cos^2(theta) and moving everything to one side gives a quadratic equation in the form ax^2+bx+c=0. Plug it into the quadratic equation to get:

    [itex]\frac{-2(v_{0A})^2\pm\sqrt{4v_{0A}+(4)(y_{A}^2)(a^2)}}{2y_{A}}[/itex]

    Sove in values from the table, and voila, either 144.699 degrees or74.296 degrees. Both of which are wrong. Help?
     

    Attached Files:

    Last edited: Sep 17, 2006
  2. jcsd
  3. Sep 17, 2006 #2
    Actually, when taking the arctan of an angle, sometimes you have to subtract your answer from 180 degrees. Is this a similar situation? If so, what do you add/subtract when using arccos?

    EDIT: I would think to subtract the angle from 360 degrees, but that would give me an angle poingting below the x-axis, and in this case, that would prevent the two particles from ever colliding. So it can't be that.
     
    Last edited: Sep 17, 2006
  4. Sep 18, 2006 #3
    Anyone? Did I skip too many intermediate steps to warrant help?
     
  5. Sep 18, 2006 #4
    is the answer around 61-62 degrees?
     
  6. Sep 18, 2006 #5
    Beats me; it's a WebAssign problem. I type in the answer and the computer says yes or no. It's not 74 degrees; I think I may have punched in fifty-something degrees at one point, also.
     
  7. Sep 18, 2006 #6
    hmmm. cause i worked out the problem and got down to a tan(x)sec(x)=3.9 and had to use a graphing calculator to solve for x. it's been a while since i have done trig-ID's, so a implicit way to solve for x is escaping me.....or i might hae just done the problem wrong.
     
  8. Sep 18, 2006 #7
    shame... :sigh:. Well, the due date has passed to hand in the assignment, so I'll just get the answer in recitation... I wish I could figure this out, though!
     
  9. Sep 18, 2006 #8
    well, here is how i worked it. maybe this will help.

    (.47)sin(x)=a_y (acceleration in the y direction).
    (.47)cos(x)=a_x (acceleration in the x direction).

    (position of partical A)X_a=(3m/s)t
    (position of partical B in the y direction)33=(1/2)(a_y)t^2
    (position of partical b in the x direction)X_b=(1/2)(a_x)t^2

    X_a=X_b
    (3m/s)t=(1/2)(a_x)t^2
    t=(6/a_x)

    33=(1/2)(a_y)(6/a_x)^2
    (a_y/(a_x)^2)=66/36

    this resulted in tan(x)sec(x)=3.9

    however, i'm thinking this might be wrong since the answer is explicit as of now.
     
  10. Sep 18, 2006 #9
    But the angle we're looking for is with respect to the y-axis?

    Woof. I am too tired to understand this now(12:41 my time), I'll look it over tomorrow. Thanks, though!
     
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