# Lect_20 - Example Elastic collision What is the velocity of...

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Lecture 20 Center of Mass 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 1 1 2 2 1 1 2 2 mv mv mv mv + = + ( ) 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 ( ) 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 ′ = − 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’ 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 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. Between “ after ” and “ after-after ”, momentum is not conserved ( F ext 0). But only conservative forces are doing work, so E is conserved . ( ) ( ) 2 1 0 0 2 M m v M m gh + + = + + 2 2 2 2 1 1 1 Note that ( ) 2 2 2 1 (some energy is lost in the collision) v m M m gh mv mv M M m m + = = < + +
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