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2D Elastic collision of varying mass, velocities, and angles

  1. Nov 22, 2014 #1
    Problem statement:
    Two pucks of radius 0.05 meters collide as shown in attached picture.

    The mass of puck 1 is 0.1 kg and that of puck 2 is 0.15 kg. The initial velocities are v1i = 1 m/s and v2i= 0.6 m/s. (1) Assuming no friction between the pucks, only normal forces during collision, in what direction is the acceleration for each puck during collision?

    Relative equations:
    m1v1i_x + m2v2i_x = m1v1f_x + m2v2f_x
    m1v1i_y + m2v2i_y = m1v1f_y + m2v2f_y
    ½ m1v1i + ½ m2v2i = ½ m1v1f + ½ m2v2f

    Work done so far:
    1) In what direction is the acceleration for each puck during collision

    Both are accelerating toward the center of the pucks during the collision phase

    2) Choose your coordinate axes to be in the direction of the acceleration and perpendicular to it. Determine the components of the original velocities along these axes.

    We will choose a coordinate system that places the x axis along the pucks’ centers of mass and the y axis perpendicular to the x axis and through the center of collision.

    V1i_x = V1i cos (θ) v1i_y = v1i sin (θ)
    v2i_x = -v2i v2i_y = 0

    (3) What happens to the velocity components perpendicular to the line between centers?

    The components perpendicular to the line between the centers (i.e. the y axis) remain the same while the x components change signs.

    (4) For the components in the direction of the acceleration, use momentum and kinetic energy conservation to determine the final values for these components. {You could use the center of mass frame here.}

    I will use a constant of q = 3/2 to change the equations of momentum conservation and KE conservation:

    m1v1i_x + m2v2i_x = m1v1f_x + m2v2f_x →
    (m1)(v1i_x) + q(m1)(v2i_x) = (m1)(v1f_x) + q(m1)(v2f_x) →
    (v1i_x) + q(v2i_x) = (v1f_x) + q(v2f_x) →
    (v1i_x) + q(v2i_x) = (v1f_x) + q(v2f_x) →
    V1i cos (θ) + q(-v2i) = (v1f_x) + q(v2f_x)

    ½ (m1)(v1i^2) + ½ q(m1)(v2i^2) = ½ (m1)(v1f^2) + ½ q(m1)(v2f^2) →
    (v1i^2) + q(v2i^2) = (v1f^2) + q(v2f^2)

    Not sure at this point on how to solve this system.

    (5) Combine the results from (3) and (4) to find the final velocities for both pucks.
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2014 #2

    haruspex

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    Not sure what that means... do you mean towards the point where they touch?
    Consider the blue puck. What is the direction of the impulse applied by the red puck?
     
  4. Nov 23, 2014 #3
    I mean if we draw a line through the centers then the forces during collision would be along that line. At least this is what I think my teacher means because he says the acceleration is along the x axis on that line through the centers. Im not sure on how that works though.
     
  5. Nov 23, 2014 #4

    haruspex

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    Yes.
    Don't do that last step - keep it in terms of the x components.
    You need an equation for the y components so that you can reduce the energy equation to only mention the x components.
     
  6. Nov 23, 2014 #5
    I didn't think you could isolate components in the conservation of KE equation. But would it look something like this:

    ##(v1i_x) + q(v2i_x) = (v1f_x) + q(v2f_x)##
    ##(v1i_y) + q(v2i_y) = (v1f_y) + q(v2f_y) ##

    ##½ (m1)(v1i^2) + ½ q(m1)(v2i^2) = ½ (m1)(v1f^2) + ½ q(m1)(v2f^2) →##
    ##(v1i^2) + q(v2i^2) = (v1f^2) + q(v2f^2) →##
    ##\sqrt{(v1i_x^2 + v1i_y^2)} + q\sqrt{ (v2i_x^2 + v2i_y^2)} = \sqrt{ (v1f_x^2 + v1f_y^2)} + q\sqrt{ (v2f_x^2 + v2f_y^2)} ##
     
  7. Nov 23, 2014 #6

    haruspex

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    As a matter of principle, you can't, and I was not suggesting you could. There may be external forces which convert KE in one direction to KE in another. I meant you to apply conservation of momentum in the y direction. However, in the absence of external forces it (usually?) turns out that KE is preserved independently in the two directions.
    Now collect the q terms on one side, the other terms on the other side, and factorise both sides.
    Use the conservation of momentum equation to eliminate a factor.
    You should end up with a standard equation that only relates the before and after velocities in the x direction. The relative masses disappear. (In its more general form, i.e. with some loss of KE, it is known as Newton's Experimental Law.)
     
  8. Nov 23, 2014 #7
    The forces during collision along the orientation of the line through the centers, the impulse during collision is olong this orientation, so the puck 2 should move along the opposite orientation of ##\boldsymbol{v_2}##.
    Plus this, you may completely solve this problem.
     
  9. Nov 23, 2014 #8
    If you guys are seeing something I am not please speak a little more plainer. I don't see how to solve the system of equations:

    (v1i_x) + q(v2i_x) = (v1f_x) + q(v2f_x)
    (v1i_y) + q(v2i_y) = (v1f_y) + q(v2f_y)

    (v1i^2) + q(v2i^2) = (v1f^2) + q(v2f^2) →
    (v1i_x + v1i_y) + q(v2i_x + v2i_y) = (v1f_x + v1f_y) + q(v2f_x + v2f_y)

    I could eliminate q by factoring but that doesn't help.
     
  10. Nov 23, 2014 #9
    I mean add this equation
    ##|\frac{v2f_y}{v2f_x}|=|\frac{v2i_y}{v2i_x}|##.
     
  11. Nov 23, 2014 #10

    haruspex

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    You didn't do what I said. Collect all the terms involving q on one side, all others on the other side, then factorise each side. Do you know how to factorise a2-b2?
     
  12. Nov 23, 2014 #11
    Follow you, at last you got a equation by two unknown variables, so you get a relation but cann't solve it.
    Then you should find another equation for resolving it.
     
    Last edited: Nov 23, 2014
  13. Nov 24, 2014 #12
    Yes I know how to factor that: (a-b)(a+b). Do you mean to use the conservation of momentum equation in its non-component form? That is:

    q(v2i - v2f) = v1f - v1i

    I'm sorry I still don't see a solution much less a way to bring it back to component form.
     
  14. Nov 24, 2014 #13

    haruspex

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    Here's what I've asked you to do:
    (v1i2) + q(v2i2) = (v1f2) + q(v2f2)
    (v1i2) - (v1f2) = q{(v2f2) - (v2i2)}
    Can you see that both sides have expressions of the form a2-b2?
    Having factored them, can you see that you have one of the factors each side in your momentum equation?
     
  15. Nov 24, 2014 #14
    @Samuelriesterer @haruspex

    Could I ask one question about this problem?

    As fas as I understand, the unknowns are:

    ##v_1'##, ##\theta_1'##, ##v_2'## and ##\theta_2'## (The prime symbol ##(')## stands for final values).

    By assuming conserved quantities, we can have 3 equations (2 from conservation of momentum (x and y axis) and one from conservation of KE)

    Where is the other equation coming from? Or what am I not taking into account?

    Thanks !!
     
    Last edited: Nov 24, 2014
  16. Nov 24, 2014 #15
    @Jazz - All the initials are known (see attached picture). The finals are unknown.

    @haruspex - I worked the KE down after factoring at got:

    (V1i-v2i) = (v2f-v1f)

    But my text book says not to use this equation or 2D collisions.
     
  17. Nov 24, 2014 #16
    I'm indeed referring to finals values (I meant that by the prime symbol).
     
  18. Nov 24, 2014 #17
    Yes sorry I see that now.
     
  19. Nov 24, 2014 #18

    haruspex

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    @haruspex - I worked the KE down after factoring at got:

    (V1i-v2i) = (v2f-v1f)

    But my text book says not to use this equation or 2D collisions.[/QUOTE]
    You can use it applied to the velocity components along the line of collision (i.e. normal to the surfaces in contact), but not to the velocities as wholes.
     
  20. Nov 24, 2014 #19
    You guys can not see my replies or something?
    @Jazz
    No friction, and the ##\boldsymbol{F}## through the centers, so ##\theta##s are all constant.
    The unknown four variables are ##\boldsymbol{v1_f}## and ##\boldsymbol{v2_f}##, due to 2D problem.
    Now we find the other equation
    ##\boldsymbol{F} // \vec{OO'}##
    ##\Delta\boldsymbol{p}_1, \Delta\boldsymbol{p}_2 // \vec{OO'}## or ##\Delta\boldsymbol{v}_1, \Delta\boldsymbol{v}_2 // \vec{OO'}##
    For this situation, I think ##\boldsymbol{v}_{2i} // \vec{OO'}##, so it's simpler to use ## \Delta\boldsymbol{v}_2 // \vec{OO'}## than ##\Delta\boldsymbol{v}_1 // \vec{OO'}##.
     
    Last edited: Nov 24, 2014
  21. Nov 24, 2014 #20

    haruspex

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    I see them, but can't make much out of them.
    Depends what they mean. If θ1 denotes the direction of travel of the incoming puck, that changes. The other doesn't have a defined prior direction.
    Your post #7 merely seemed to confirm what had been said in post #3.
    This equation looks most odd, seems to be dividing by vectors:
    .
    Where do you get that from?
     
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