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Collisions between two balls in 2d-motion

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball A is dropped from rest from a
    height of 2.0 m above the floor. Meanwhile that ball A is released, an other ball B is pushed away with the starting speed v_0 from the position shown in Figure. Which angle α is needed for B to collide with ball A?

    6B3IqsR.png

    2. Relevant equations

    Equations of motion which I present below.

    3. The attempt at a solution
    Yes, one solution is that it does depend on v_0 but a second solution is at an angle 45 degrees, which I want to find.

    I lay out the equations in the x/y direction and then determine A/B equations of motion from these.

    X: a(t)=0 Y: a(t)=-g
    v(t)=v_0cosα v(t)= -gt+v_0sinα
    x(t)=v_0cosαt+x_0 y(t)= -1/2gt^2+v_0sinα+s_0

    The equations x(t) and y(t) for A and B are now determined;

    A: x_1(t) = 1 I have decided that x_0 refers to x_0=0 for B.
    y_1(t) = 2-1/2gt^2

    B: x_2(t) = v_0cosαt
    y_2(t) = 1-1/2gt^2+v_0sinαt

    For collision we need that x_1(t)+y_(1)t = x_2(t)+y_2(t) Right?

    But that leads to v_0t(cosα+sinα)=v_0t(√2sin(α+π/4))=2 which clearly is impossible for any α...
     
  2. jcsd
  3. May 13, 2013 #2

    DrClaude

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    Staff: Mentor

    You need to rethink the condition for collision.
     
  4. May 13, 2013 #3
    Well at the point of collision, the x-coordinate and y-coordinate must be equal for A and B at the same time. Hence;

    [tex]v_0cosα=1[/tex]
    [tex]1-1/2gt^2+v_0tsinα=2-1/2gt^2 ⇔ v_0tsinα=1[/tex]

    Thus we end up with the system of equations: [tex]\left\{\begin{matrix}
    v_0\cos\alpha=1 & \\
    v_0t\sin\alpha=1 &
    \end{matrix}\right.[/tex] which clearly does not have solutions.
     
    Last edited: May 13, 2013
  5. May 13, 2013 #4

    DrClaude

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    Staff: Mentor

    You're missing a ##t## in there, from which you can get ##t_c##, the time at collision.
     
  6. May 13, 2013 #5
    Actually, 45 degrees does not seem reasonable at all. Consider the right-angled triangle BAC where C is the point 1 meter below A. It is a triangle with sides 1,1,√2 and angle ABC is 45 degrees. When ball A starts moving downward, the side AC gets smaller and hence the angle gets smaller.

    In order for the angle to be 45 degrees, the ball B must hit A immediately and thus requiring that v_0→∞. Is this analysis correct?

    Edit;

    You're right;
    [tex]\left\{\begin{matrix}
    v_0t\cos\alpha=1 & \\
    v_0t\sin\alpha=1 &
    \end{matrix}\right.[/tex] leads to [tex]\tan\alpha=1 \Rightarrow \alpha=\frac{\pi}{4}[/tex]

    In pure interest; What does actually x_1(t)+y_1(t)=x_2(t)+y_2(t) mean geometrically?(In the case that there exist a solution). Yes, totally forgot gravity. Thanks a lot DrClaude.
     
    Last edited: May 13, 2013
  7. May 13, 2013 #6

    DrClaude

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    Staff: Mentor

    You're forgetting gravity.
     
  8. May 13, 2013 #7
    No need for so much analysis! Consider the relative motion of B with respect to A(consider A to be at rest). The relative acceleration is zero(as both are under influence of gravity). This implies that the path of B as seen by a will be a straight line also it will have constant velocity. The collision is therefore only possible if the velocity vector of B passes through initial position of A giving angle of 45 degrees directly!
     
  9. May 13, 2013 #8
    Nice! I like this approach better.
     
  10. May 13, 2013 #9
    This is one of the examples where relative velocity is insanely advantageous.
     
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