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Collision of Particles

  1. Oct 10, 2005 #1
    Could someone point me in the right direction, im totally lost

    particle A moves along the line y=30m with a constant velocity v of magnitude 3.0m/s and parallel to the x axis. At the instant particle A leaves the y axis, particle B leaves the origin with zero initial speed and constant acceleration a of magnitude 0.40m/s^2. What angle between a and the positive direction of the y axis would result in a collision?
     
  2. jcsd
  3. Oct 10, 2005 #2

    mezarashi

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    Homework Helper

    A math exercise once you can visualize what's going on. You have a particle moving along y=30m. Now you are at the origin with zero velocity. You want to crash into the dude moving at y=30. The only thing is that you don't have a steering wheel. Meaning you have to calculate the angle you want to start off with knowing that you can accelerate at 0.40 m/s/s.

    Kinematics equations. You have two equations, one in the x and the other in the y.
    What are the conditions for collision?

    1. The displacement in y is equal to 30m.
    2. The displacement in x is equal to the displacement of particle A at collision time (which we don't yet know).

    With those two conditions, we write out the variables we know and are looking for in the x and y (I gave you some hints, fill in the rest):

    x-axis
    a: asin(angle)
    vinitial:
    vfinal: ?
    displacement:
    time: ?

    y-axis:
    a: acos(angle)
    vinitial:
    vfinal: ?
    displacement:
    time: ?
     
  4. Oct 10, 2005 #3
    Wouldn't x - axis be a: acos(angle) and y -axis be a: asin(angle)
     
  5. Oct 10, 2005 #4
    Ok this is what i've tried so far
    Particle B
    Y-axis
    displacement= 30m

    30 = V0(t) + 1/2a(t^2)
    t= 12.25s

    v = v0 + at
    v = (.40)12.25
    v= 4.9m/s

    In 12.25s particle A travels 3.0m/s * 12.25s = 36.75m

    X-axis
    displacement = 36.75m
    36.75 = vo + 1/2a(t^2)
    36.75 =(.2)(t^2)
    t= 13.56s

    v= vo + at
    = (.4)(13.56)
    = 5.42m/s

    I don't know if im on the right track or not but this is as far as i can get
     
  6. Oct 10, 2005 #5
    still pretty much stuck, could anyone help
     
  7. Oct 11, 2005 #6
    from this i took the position r=(36.75m)i + (30m)j
    then i found the angle relative to the x axis

    Tan(theta)= 30/36.75
    theta = arctan .82
    = 39.4

    90 - 39.4 = 50.6 which would be the angle from the y-axis


    can someone tell me if im doing this right i would appreciate it.:smile:
     
  8. Oct 11, 2005 #7

    mezarashi

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    Normally yes, except the question asks

    so you've got to be more flexible and truly understand the geometry problem instead of thinking that cosine is always an x-axis "thing".

    How did you get the value for t? The value for a is acos(angle). I thought we were just discussing how we don't know the angle and that's what we're finding? Same goes for your x-axis analysis. The conditions we DO know are:

    [tex]d_b_y = \frac{1}{2}a_1t^2 = 30m[/tex]
    [tex]d_b_x = \frac{1}{2}a_2t^2 = d_a_x = vt[/tex]

    where a1 and a2 are acos(angle) and asin(angle). How many unknowns, how many variables do we have?
     
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