LON-CAPA Basketball Player × Chegg Study Guided Sciutic × cregg Study Guided Sal

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LON-CAPA Basketball Player × Chegg Study Guided Sciutic × cregg Study Guided Salute × c secure https://capa9 phy ohi.edu/res/msu/plough physlibrary/momentum ehs prob05problem?symb-uploaded%2fhiou%2t91391804150295a49oucapa2%2tdefault 13509 178. ? O Megan Michalak(Student- section: 101) Physics 2001/Spring 2018 l Messages Courses Help Logout Main Menu Contents Grades Course Contents» * Assignment 10Basketball Player Collision Timer Notes . Evaluate-Feedback-Mint Oil? Basketball Player Collision Due this Friday, Mar 30 at 11:59 pm (EDT) Here's a basketball problem: A 80.6 kg basketball player is running in the positive direction at 7.2 m/s. She is met head-on by a 106.4 kg player traveling at 5.5 m/s toward her. If the 106.4 kg player is knocked backwards at 3.6 m/s, what is the resulting velocity of the 80.6 kg player? 4.81 m/s Remember, momentum must be conserved. Start with your formula for conservation of momentum Submit Answer Incorrect. Tries 2/10 Previous Tries This discussion is closed Send Feedba

Explanation / Answer

Given, m1 = 80.6 kg , m2 = 106.4 kg, u1 = 7.2 m/s ,

u2 = -5.5m/s , v2 = 3.6 m/s

Using conservation of energy ,

m1*u1 + m2*u2 = m1*v1 + m2*v2

(80.6*7.2) + (106.4*(-5.5)) = (80.6*v1) + (106.4*3.6)

80.6*v1 = -387.92

v1 = -387.92/80.6

v1 = - 4.813 m/s ( minus sign indicates negative direction )

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