461_Physics ProblemsTechnical Physics

# 461_Physics ProblemsTechnical Physics - Chapter 15 *P15.62...

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Chapter 15 463 *P15.62 As it passes through equilibrium, the 4-kg object has speed vA k m A max . == = = ω 100 4 21 0 0 Nm kg m m s . In the completely inelastic collision momentum of the two-object system is conserved. So the new 10-kg object starts its oscillation with speed given by 46 0 1 0 400 kg 10 m s kg kg ms bg b g b g += = v v max max . (a) The new amplitude is given by 1 2 1 2 22 mv kA max = 10 4 100 126 2 2 kg m s N m m b g = = A A . Thus the amplitude has decreased by 200 0735 .. . m m −= (b) The old period was T m k = 4 ππ kg 100 N m s . The new period is T 2 10 100 199 π s s 2 . The period has increased by 0730 . m s (c) The old energy was 1 2 1 2 41 0 2 0 0 2 2 mv max kg m s J b g The new mechanical energy is
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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