lecture_21_web_2007

Lecture_21_web_2007 - Glancing collisions 2 i = υ 1 i υ 1 f υ 2 f υ A cue ball m 1 =0.1kg is struck with velocity v 1 =0.3m/s in such a way

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Unformatted text preview: Glancing collisions 2 i = υ 1 i υ 1 f υ 2 f υ A cue ball m 1 =0.1kg is struck with velocity v 1 =0.3m/s in such a way that it bounces off of a stationary colored ball at an angle of 30 degrees and a speed of 0.15m/s. If the target ball has a mass m 2 =0.08kg Q1. What is the angle between the two final velocity trajectories? Q2. What kind of collision is this? θ φ Glancing collisions A cue ball m 1 =0.1kg is struck with velocity v 1 =0.3m/s in such a way that it bounces off of a stationary colored ball at an angle of 30 degrees and a speed of 0.15m/s. If the target ball has a mass m 2 =0.08kg Q1. What is the angle between the two final velocity trajectories? Q2. What kind of collision is this? 2 i = υ 1 i υ 1 f υ 2 f υ θ φ ix fx = p p 1 1 2 2 1 1 2 2 ix ix fx fx m m m m + = + υ υ υ υ ( 29 ( 29 1 1 2 1 1 2 cos cos f i f m m m υ υ υ φ θ = + iy fy = p p 1 1 2 2 1 1 2 2 iy iy fy fy m m m m + = + υ υ υ υ ( 29 ( 29 1 2 1 2 0 sin sin f f m m υ θ υ φ = + (2) (1) ( 29 ( 29 1 1 2 1 1 2 cos cos f i f m m m υ υ υ φ θ- = ( 29 ( 29 1 1 2 1 1 2 cos cos i f f m m m υ υ φ θ υ- = Sub (1) into (2) ( 29 ( 29 ( 29 ( 29 1 1 1 1 1 1 2 2 cos 0 sin sin cos i f f m m m m m υ υ θ υ φ φ θ - = + ÷ ÷ ( 29 ( 29 ( 29 ( 29 1 1 1 1 1 1 2 2...
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This note was uploaded on 04/07/2012 for the course PHYS 111 taught by Professor Dr.jackman during the Fall '07 term at Waterloo.

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Lecture_21_web_2007 - Glancing collisions 2 i = υ 1 i υ 1 f υ 2 f υ A cue ball m 1 =0.1kg is struck with velocity v 1 =0.3m/s in such a way

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