PROBLEM 1 &amp; 2 March 30 2010

# PROBLEM 1 &amp;amp; 2 March 30 2010 - v T s. SOLVING...

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SOLVING PROBLEM 1 March 30, 2010 A perfectly elastic ball hits floor with velocity = 4 m/s. What is its period? 0 v What do you know? Conservation of energy: const U K Kinetic energy 2 2 1 mv K . Potential energy . mgz U max K is when , i.e. when the ball hits the floor, 0 U 2 0 2 1 max mv K max U is when , i.e. when the ball turns around at the top of its trajectory, U . 0 K mgh max Since , max max U K 2 0 2 1 mv mgh . Cancel the m ’s to get g v h / 2 0 2 1 . Time to fall from height h is found from 2 2 1 at s which means 2 2 1 gt h . Solve for t by substituting for h : 2 2 1 2 0 2 1 / gt g v to get . 2 2 2 0 / t g v The period is 2 t (down and back up) so 8 . 0 / 2 0 g
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Unformatted text preview: v T s. SOLVING PROBLEM 2 Something (it doesnt matter that it is a tooth brush bristle!) is oscillating at 500 Hz with peak-to-peak amplitude 5 mm. What is the maximum speed? Trick 1 is the amplitude is half the peak-to-peak amplitude, or mm. Trick 2, I gave you A in mm. 5 . 2 A The velocity is ) cos( ) sin( / t A t A dt d dt dx v . The maximum speed is thus A v max . But f 2 . Thus . s m s m Af v / 85 . 7 500 0025 . 2 2 1 max...
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## This note was uploaded on 09/23/2010 for the course PH 02c taught by Professor Mile during the Spring '04 term at Riverside Community College.

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