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Need help with question and checking of answers

  • #1
rock.freak667
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Homework Statement


http://img234.imageshack.us/img234/9536/scan0006hj0.jpg [Broken]
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.


Homework Equations



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

The Attempt at a Solution



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

http://img233.imageshack.us/img233/2581/scan0006cb8.jpg [Broken]

http://img396.imageshack.us/img396/4219/scan0007wu2.jpg [Broken]


Strangely enough,I had a hard time looking for ways to show that tan[itex]\alpha[/itex]=2
 
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Answers and Replies

  • #2
Doc Al
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Strangely enough,I had a hard time looking for ways to show that tan[itex]\alpha[/itex]=2
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.
 
  • #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?
 
  • #4
Doc Al
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But where would I get the tan from?
Use both the horizontal and vertical components.
 
  • #5
rock.freak667
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Use both the horizontal and vertical components.
[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?
 
  • #6
Doc Al
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Yes. Now relate the tangents by dividing the vertical and horizontal components.
 
  • #7
rock.freak667
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ah thanks, got it there.
 

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