Physics data analysis 5

# Physics data analysis 5 - Θ Θ-Θ-rad u P P P P P P P P P...

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Sample Calculations: Mass Measurements: μ M (kg) = 0.0001 kg because +/-0.1 g is the precision of the balance Speed Before collision: μ v (m/s) = +/- 0.5mm/ 1/20 seconds = 10 mm/sec = 0.01 m/s 009 . 0 516 . 0 180 (deg) 180 u = Π = Π = Θ Θ u Mass 1 before collision: P x = mvcos Θ = 0.5652(0.582)cos37 = 0.2654 kg*m/s μ p x = = + + = Θ + Θ + Θ Θ 2 2 2 2 2 2 ) 009 . 0 * 37 sin 588 . 0 * 5652 . 0 ( ) 01 . 0 * 37 cos 5652 . 0 ( ) 00005898 . 0 * 37 cos 588 . 0 ( ) sin ( ) cos ( ) cos ( υ mv m v v m 0.00119 Puck before the collision: P x = P x,mass1 + P x,mass2 = 0.26543 + 0.192073 = 0.45751 kg*m/s μ p x = 00180 . 0 00173 . 0 0018 . 0 2 2 2 , 2 1 , 2 = + = + mass Px mass Px System before the collision: 77561 . 0 ) ( 180 (deg) 01353 . 0 01557 . 0 ) 26543 . 0 200018 . 0 1 ( ) 1 ( 01557 . 0 22165 . 0 002388 . 0 07849 . 0023889 . 0 07849 . 200018 . 0 597 . 40 ) 457505 . 0 / 39209 . 0 ( tan ) / ( tan 0018 . 0 60253 . 0 265432 . 0 60253 . 0 39209 . 0 45751 . 0 1 2 / 1 2 2 2 2 2 / 1 1 2 2 2 2 2 2 2 2 2 = = = + = + = = + - - = + = = = = Θ + = + = = + = + =

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Unformatted text preview: Θ Θ--Θ--rad u P P P P P P P P P P P P s m kg P P P Px Py x y y y x Px x y Px Py x y py y px x p y x π Kinetic energy: K = 1/2Mv 2 = ½(0.56523kg)(0.588m/s) 2 = 0.0977J 5 ^ 774 . 1 5 75 . 1 6 8843 . 2 081926 . 03301 . 0489 . 6 76 . 5 01 . ) 588 . * 56523 . ( 5 897 . 5 2 588 . ) ( 2 2 2 2 2 , 2 1 , 2 1 2 2 2 2 2 2 2 2 2-=-+-= + = = + = + =-= +-= + = E E E J K K K E E Mv v mass K mass K k mass mass v m K υ...
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## This note was uploaded on 11/27/2011 for the course PHYS 105 taught by Professor Walker during the Fall '08 term at UNC.

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Physics data analysis 5 - Θ Θ-Θ-rad u P P P P P P P P P...

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