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Kinematics 5 -- Vectors to two particles that are going to collide...

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  1. Jul 11, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-11_22-3-15.png

    2. Relevant equations


    3. The attempt at a solution
    The two particles are supposed to move along the same line.
    And the two radius vector could be anything.

    Is this correct?
     
  2. jcsd
  3. Jul 11, 2017 #2

    TSny

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    Two cars do not need to be moving along the same line in order to collide.
    Can you write down a vector equation using r1, v1, and t that would represent the position of particle 1 at time t?
     
  4. Jul 12, 2017 #3
    Yes

    r1 (t) = r1 + v1 t
    r2 (t) = r2 + v2 t ,for uniform motion

    Now , collision of the two particles at time t means r1 (t) = r2 (t)
    r1 - r2 = t (v2 - v1)

    What to do next?
     
  5. Jul 12, 2017 #4

    TSny

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    Good.

    Not much left to do except interpret what you have.
     
  6. Jul 12, 2017 #5
    Yes, even if the trajectory of the two cars make V shape, they will collide.

    The collision will occur only if at some time t, the relative velocity becomes equal and opposite to the initial relative displacement between the two particles.

    Should I remove t in r1 - r2 = t (v2 - v1) ?
     
  7. Jul 12, 2017 #6

    TSny

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    That can't be exactly right, since relative velocity cannot equal displacement (they have different units).

    No, you can't just drop the time. However, the time of the collision could be any time. So, can you express what this equation says about the relation between the four vectors r1, r2, v1, and v2 in order for a collision to eventually occur?
     
  8. Jul 14, 2017 #7
    Yes, that is not exactly right.
    But, the time of collision can't be anytime.
    The time of collision has to be such that this time multiplied with relative velocity becomes equal and opposite to the initial relative displacement.

    I didn't understand what I have to express here.
    Please, give me some hint.
     
  9. Jul 14, 2017 #8
    Why can't I drop the time?
     
  10. Jul 14, 2017 #9
    "Dropping" time would be like setting t=1 but there is no reason to only consider collisions occuring at t=1...


    Look at this equation:
    This is an equation of vectors, with time being a scalar. Are you able to solve this equation for the scalar t, and then put that back into the equation for t? That will give a relationship between the 4 vectors which does not involve time, which is as good as you can get. (That relation also has a simple intuitive interpretation which could've skipped past any considerations of time.)
     
  11. Jul 14, 2017 #10

    haruspex

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    It depends what you mean by that. You do not care when they collide, only whether they collide.
    One difficulty is that the question does not specify the form of the answer. You could answer it in terms of an existence criterion, or you could look for purely algebraic conditions. (You will have to be careful about future versus past.)
    What does your equation tell you about the geometric relationship between the vectors ##\vec r_1-\vec r_2## and ##\vec v_1-\vec v_2##?
     
  12. Jul 18, 2017 #11
    Yes, I am.
    r1 - r2 = t (v2 - v1)


    (|r1 - r2|) = t |(v2 - v1)|

    t = (|r1 - r2|)/ |(v2 - v1)|

    r1 - r2 / (|r1 - r2|) = (v2 - v1) / |(v2 - v1)|
    Is following the intuitive interpretation?

    direction of initial relative displacement is opposite to the direction of relative velocity.
     
  13. Jul 18, 2017 #12
    @Pushoam Yes, that is all correct, and that is the right interpretation; the two unit vectors (direction of relative displacement / direction of relative velocity) must be opposite each other.
     
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