# Need help with question and checking of answers

1. May 20, 2008

### rock.freak667

1. The problem statement, all variables and given/known data
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 $2ms^{-1}$ 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 $\alpha$ and $\beta$ respectively (see diagram). The coefficient of resititution between the side and a ball is $\frac{1}{4}$. Resistances may be ignored and the balls may be considered as particles.

(i)Show that $cot\alpha + cot\beta=2.5$ and $tan\alpha=2$
(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

$e=\frac{\delta v}{\delta u}$

3. 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$\alpha$=2

Last edited by a moderator: May 3, 2017
2. May 20, 2008

### Staff: Mentor

One thing that will help with both showing that $\tan\alpha = 2$ 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. May 20, 2008

### rock.freak667

So the perpendicular comp. of the velocity before impact is $2sin\alpha$

and the perpendicular comp. of velocity after is vsin$\beta$

and

$$\frac{1}{4}=\frac{vsin\beta}{2sin\alpha}$$

But where would I get the tan from?

4. May 20, 2008

### Staff: Mentor

Use both the horizontal and vertical components.

5. May 20, 2008

### rock.freak667

$2cos\alpha$ and $vcos\alpha$, both point in the same direction.

So 2cos$\alpha$=vcos$\beta$ because the horizontal velocity should stay the same?

6. May 20, 2008

### Staff: Mentor

Yes. Now relate the tangents by dividing the vertical and horizontal components.

7. May 20, 2008

### rock.freak667

ah thanks, got it there.

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