P150_Exam2_F07sol

# P150_Exam2_F07sol - Physics 150 1 2 3 4 5 6 7 8 9 10 Answer...

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To the right. F average 480 N = F average J t := To the right. J p B := Impulse = change in momentum: p B 9.6kgmsec -1 = p B m B v Bf v Bi () := t 0.02 sec := D and E) Hence this is an elastic collision. K f 392 kgm 2 sec -2 = K f 1 2 m A v Af 2 1 2 m B v Bf 2 + := K i 392 kgm 2 sec -2 = K i 1 2 m A v Ai 2 1 2 m B v Bi 2 + := In elastic collision the kinetic energy is the same before and after: C) To the right. v Af 7.6msec -1 = v Af m A v Ai m B v Bi + m B v Bf m A := m A v Ai m B v Bi + m A v Af m B v Bf + := Conservation of momentum in one dimension: A and B) v Bf 9.6 m sec := v Bi 8 m sec := v Ai 10 m sec := m B 6kg := m A 4kg := Problem #1: Physics 150 Exam #2 Solution 11/5/07
E top K A U + := K A E bottom U := K A 1.208 10 5

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## This note was uploaded on 04/10/2008 for the course PHYSICS 150 taught by Professor Naik during the Winter '08 term at University of Michigan.

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P150_Exam2_F07sol - Physics 150 1 2 3 4 5 6 7 8 9 10 Answer...

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