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Impact and collision - having problems

  1. Apr 6, 2014 #1
    Impact and collision -- having problems

    A smooth sphere A impinges obliquely with an identical smooth sphere B which is at rest. the direction of A before and after impact makes angles 60 and X , respectively, with the line of centres at impact. the coefficient of restitution is e.
    prove that tan x=2(root)3/1-e
     
  2. jcsd
  3. Apr 6, 2014 #2

    Simon Bridge

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    Welcome to PF;
    That is a good question - how have you been attempting it?
    Have you sketched out the geometry yet?
     
  4. Apr 7, 2014 #3
    this is the problem i just dont quite understand the Q,im not sure where the 60 and X is
     
  5. Apr 7, 2014 #4

    Simon Bridge

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    Draw two circles, representng the spheres, on a bit of paper, so that they touch.
    These are the spheres in the moment of contact.
    Label one A and the other B.

    Draw a dotted line through the centers of the circles - this is the reference line for the angles.

    Draw lines through each center to represent the trajectory before the contact ... remember that the collision is oblique, as opposed to head on. A is the only one moving... it's trajectory makes an angle of 60deg to the reference line. There are 4 ways you can get a 60degree angle in there but at most two will make much sense, and it does not matter which of those you pick.

    Similarly for the trajectories after the impact.
     
  6. Apr 8, 2014 #5
    ok thx for your time
     
  7. Apr 8, 2014 #6

    Simon Bridge

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    No worries - let us know how you got on.
     
  8. Apr 13, 2014 #7

    bobie

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    Seems overpower's gone. I'm interested in the problem, if I am allowed,

    first of all :
    - is the solution [itex]\frac{2\sqrt{3}}{1-e}[/itex]; if it is an elastic collision then: [itex]\frac{2\sqrt{3}}{0}[/itex]?
    - m1= m2?
    If we consider the x-axis the direction of A and the two balls at the origin at -60°, if the balls have equal mass then the direction of X should be +30° with reference the x-axis and 90° with the line of impact, right?
    tan 30° = 1/√3
     
    Last edited: Apr 13, 2014
  9. Apr 13, 2014 #8

    Simon Bridge

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    The question is a tad ambiguous because it is not clear what the angles are with respect to.
    I suppose you could proceed by having the initial velocity of A to be at 60deg to the line joining the centers of A and B at the point of impact.

    You should show your working and reasoning along with your solution.
    i.e. Why would X be 90deg to the line of impact?
    Have a play with some billiard balls and see what happens (air-hockey pucks would be better)
     
  10. Apr 13, 2014 #9

    bobie

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    In this regard the OP is not ambiguous.
    As the balls are identical the scattering angle X must always be roughly at 90° from the line of impact,because the ball at rest B will always move a 0°
    What I do not know is if or how the coefficient e affects the angle X, does it?.
    But anyway the solution in OP seems wrong to me,
    what does 2√3 mean? and
    what 2√3/0? is it a correct representation of tan 90°?
     
  11. Apr 14, 2014 #10

    Simon Bridge

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    "with line of the centers of impact" shazbot!
    managed to misread that, then read it, then miss it!

    The coefficient of restitution affects things by determining the rebound velocity.
    Check the definition.
     
  12. Apr 14, 2014 #11

    bobie

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    I'd appreciate if you give me a lodestar saying that I did not misread also the solution in the OP and that it actually it reads 2√3/1-e.
    I interpreted "the line od centers of impact" as the line joining the 2 centers of mass of the balls. Is that wrong?
    I do not understand, also, why the coefficient e appears in the value of the tangent of angle X if it affects only velocity
     
  13. Apr 14, 2014 #12

    Simon Bridge

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    $$\tan X = \frac{2\sqrt{3}}{1-e}$$ ... is how I read it.

    No - I'd say that was correct.

    You'd have to ask OP.
    However the angle X will depend on the direction of the velocity vectors, which will be determined by their magnitudes via conservation of momentum.
     
  14. Apr 14, 2014 #13

    bobie

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    I have the impression you agreed that OP is wrong.
    The angle of rebound is always at a right angle when masses are identical.
    Tan X is 1/√3, vB is v*cos60, and vA is v*cos30* e, right?
     
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