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

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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 2 = 2.0 m/s 1 kg 5 kg Before 11 22 mv ′′ +=+ () 12 vv −= −− 1 = 10 m/s 2 = 2.0 m/s 1 kg 5 kg Before 1 2 1 kg 5 kg After 10 10 5 10 2 − − 20 5 8 =− 8 20 5 8 2 28 4.7 m/s 6 == Answer: E 1 3.3 m/s ′=− 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 + p total, initial = mv + 0 total, final = ( + m)v’ vv Mm ′ = + = ( + Now we’ll do the same thing with the block hanging from two strings of length L. + After At rest Before L + After after L-h Between “ before ” and “ after ”, the collision is exactly the same as for the block on the table. + After At rest Before ′ = + 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 00 2 Mmv Mmgh ++ = + + 22 11 1 Note that ( ) 2 1 (some energy is lost in the collision) vm m gh   += = < +  + 2 2 h g = 2 2 2 = + + After + After after
ACT: Ballistic pendulum ACT: Ballistic pendulum The projectile is a ball with two sides. One is smooth and the other has two spikes, so the ball sticks to the wooden block. In which case will the block move higher?

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## This note was uploaded on 04/16/2008 for the course PHYSICS 221 taught by Professor Johnson during the Fall '06 term at Iowa State.

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Lect_20 - Example: Elastic collision What is the velocity...

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