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Need help with question and checking of answers

  1. May 20, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    [​IMG]
    Two identical snooker balls are at rest at the point A and B of a snooker table, where AB+1m. The liunes AB is aparallel to a smooth side of the table and at a distance of 0.4m from that side. The ball A is projected along the table with a speed of [itex]2ms^{-1}[/itex] to hit the side of the table at C. It rebounds and subsequently hits the ball at B. The angles that AC and CB make with the table are [itex]\alpha[/itex] and [itex]\beta[/itex] respectively (see diagram). The coefficient of resititution between the side and a ball is [itex]\frac{1}{4}[/itex]. Resistances may be ignored and the balls may be considered as particles.

    (i)Show that [itex]cot\alpha + cot\beta=2.5[/itex] and [itex]tan\alpha=2[/itex]
    (ii) Show that the direction of motion of the ball turns through 90 degrees on impact at C
    (iii) find the speed of the ball after impact at C.


    2. Relevant equations

    [itex]e=\frac{\delta v}{\delta u}[/itex]

    3. The attempt at a solution

    Well here is my work.....in pictures because, typing it up would have taken a long time.

    [​IMG]

    [​IMG]


    Strangely enough,I had a hard time looking for ways to show that tan[itex]\alpha[/itex]=2
     
  2. jcsd
  3. May 20, 2008 #2

    Doc Al

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    One thing that will help with both showing that [itex]\tan\alpha = 2[/itex] and finding the speed after impact is realizing that the coefficient of restitution only applies to the component of velocity perpendicular to the surface.
     
  4. May 20, 2008 #3

    rock.freak667

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    So the perpendicular comp. of the velocity before impact is [itex]2sin\alpha[/itex]

    and the perpendicular comp. of velocity after is vsin[itex]\beta[/itex]

    and

    [tex]\frac{1}{4}=\frac{vsin\beta}{2sin\alpha}[/tex]


    But where would I get the tan from?
     
  5. May 20, 2008 #4

    Doc Al

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    Use both the horizontal and vertical components.
     
  6. May 20, 2008 #5

    rock.freak667

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    [itex]2cos\alpha[/itex] and [itex]vcos\alpha[/itex], both point in the same direction.

    So 2cos[itex]\alpha[/itex]=vcos[itex]\beta[/itex] because the horizontal velocity should stay the same?
     
  7. May 20, 2008 #6

    Doc Al

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    Yes. Now relate the tangents by dividing the vertical and horizontal components.
     
  8. May 20, 2008 #7

    rock.freak667

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    ah thanks, got it there.
     
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