Calculations and Questions.docx

# Calculations and Questions.docx - Calculations and...

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Calculations and Questions 1. Initial Kinetic Energy (KE i ) Case 1 Since V 2i = 0, KE i = ½ m 1 v 1i 2 M 1 = 211.4g = 0.2114kg, V 1i = 0.1/0.09768 = 1.02 m/s Therefore, K.E i = ½ * 0.2114 * 1.02 2 = 0.12 J Case 2 Since V 2i = 0, KE i = ½ m 1 v 1i 2 M 1 = 311.9 g = 0.3119 kg, V 1i = 0.1/0.1262 = 0.7924 m/s Therefore, K.E i = ½ * 0.3119 * 0.7924 2 = 0.09792 J Case 3 Since V 1f = 0, KE i = ½ m 2 v 2f 2 M 1 = 211.4g = 0.2114 kg V 1i = 0.1/0.10098 = 0.9903 m/s Therefore, K.E i = ½ * 0.2114 * 0.9903 2 = 0.1037 J Case 4 Since V 1i = V 2i = 0 K.E = 0 Case 5 Since both V 1f ≠ 0 and V 2f ≠ 0 K.E i = ½ m 1 v 1i 2 + ½ m 2 v 2i 2 M 1 = 311.9 g = 0.3119 kg, m 2 = 209.5 g = 0.2095 kg V 1i = 0.1/2.2718 = 0.044 m/s, V 2i = 0.1/0.2184 = 0.458 m/s Therefore, K.E i = ½ * 0.3119 * 0.044 2 + ½ * 0.2095 * 0.458 2 = 0.0223 J Case 6 Since V 2i = 0, KE i = ½ m 1 v i 2 M 1 = 311.9 g = 0.3119 kg

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V 1i = 0.1/0.1958 = 0.511 m/s Therefore, KE i = ½ * 0.3119 * 0.511 2 = 0.0406 J 2. Final Kinetic Energy (K.E f ) Case 1 Since V 1f = 0, KE f = ½ m 2 v 2f 2 M 2 = 209.5 g = 0.2095 kg, V 2f = 0.1/ 0.1062 = 0.9416 m/s
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• Fall '16
• kiptui

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