(20) Center of Mass

(20) Center of Mass - Lecture 20 Center of Mass Example...

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Lecture 20 Center of Mass
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Example: Elastic collision A. 2.0 m/s, 10 m/s B. 5.3 m/s, 2.7 m/s C. -5.3 m/s, 2.7 m/s D. -3.3 m/s, 2.7 m/s E. -3.3 m/s, 4.7 m/s What is the velocity of each block if the collision is elastic? v 1 = 10 m/s v 2 = 2.0 m/s 1 kg 5 kg Before
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1 1 2 2 1 1 2 2 mv mv mv mv + = + ( 29 1 2 1 2 v v v v - = - - v 1 = 10 m/s v 2 = 2.0 m/s 1 kg 5 kg Before v 1 v 2 1 kg 5 kg After ( 29 1 2 1 2 10 10 5 10 2 v v v v + = + - = - - 1 2 1 2 20 5 8 v v v v = - = - 1 2 8 v v = - 2 2 20 5 8 v v - = - 2 28 4.7 m/ s 6 v = = Answer: E 1 3.3 m/ s v = -
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Example: Ballistic pendulum Example: Ballistic pendulum Let’s go back to the bullet and block example…
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v m M At rest Before v’ After M +m p total, initial = mv + 0 p total, final = (M + m)v’ m v v M m = + mv = (M + m)v’
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Now we’ll do the same thing with the block hanging from two strings of length L. v’ M + m After v m M At rest Before L M + m After after L-h
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Between “ before ” and “ after ”, the collision is exactly the same as for the block on the table. v’ M + m After v m M At rest Before L m v v M m = + Linear momentum is conserved.
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Between “ after ” and “ after-after ”, momentum is not conserved (F ext 0). But only conservative forces are doing work, so E is conserved . ( 29 ( 29
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(20) Center of Mass - Lecture 20 Center of Mass Example...

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