Figure 9-47 (attatched) shows a 0.165 kg cue ball as it bounces from the rail of a pool table. The ball's initial speed is 1.60 m/s, and the angle 1 = 30.0°. The bounce reverses the y component of the ball's velocity but does not alter the x component.(adsbygoogle = window.adsbygoogle || []).push({});

(a) What is theta_2?

(b) What is the change in the ball's linear momentum in unit-vector notation?

I got the physics of part b, the trig of part a is messing me up. I went so far as to make a system:

[tex]v\cos\theta_2=-v\cos\theta_1[/tex]

[tex]v\sin\theta_2=v\sin\theta_1[/tex]

As I thought! The angle must be in the second quadrant... so that means add 90 degrees, right? (And not 180-angle?)

But I don't know how to go from the typical coordinate system used for measuring angles to the funky one shown in the diagram.

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# Homework Help: Pool ball hits table, find an angle

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