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PHYS_2014_Lecture_15-1 - Mid-term Exam on 19 Oct 2010 5:30...

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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 15, Slide 1 The class will be divided into three groups based on their recitation TA. Each group will take the exam in a different room as shown in the table below. Based on your recitation section, please go directly to your designated room. Note that rooms for Mr. Patra’s and Mr. Stoian’s students have been switched. Mid-term Exam on 19 Oct. 2010 5:30 - 7:00 pm Recitation Section TA Exam Room 1, 2, 3, 19, 20, 21, 22, 23, 24 Ayon Patra PS-141 4, 5, 6, 7, 8, 9, 16, 17, 18, Ben Grossman PS-103 10, 11, 12, 13, 14, 15, 25, 26, 27, 28, 29, 30 Razvan Stoian PS-110
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 15, Slide 2 Regarding Exam 2 Homework Set No. 6 will not be collected or graded, since there is no recitation on Friday, 10/15/2010. Solutions to Homework Set No. 6 will be posted on Friday. There will be a help session on Wednesday, 10/13/2010 at 5:30 in PS052. Exam 2 will cover everything we’ve covered up through Homework Set No. 6. Emphasis will be on material in Homework Sets Nos. 4, 5, and 6. Orbital motion will be included on Exam 2 ; collisions will not be included on Exam 2 . The exam will be of the same format and follow the same rules as Midterm Exam No. 1.
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 15, Slide 3 Lecture 15: Elastic and Inelastic Collisions In elastic collisions , not just total energy, but kinetic (mechanical) energy is conserved . In inelastic collisions , some kinetic energy coming into the collision is converted to another form (e.g. heat, chemical “binding” energy, sound, etc.) and is not conserved . Collisions (stuff hitting stuff) can be thought of as being of one of two types, elastic and inelastic . In both elastic and inelastic collisions , energy and linear momentum are conserved .
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 15, Slide 4 Let’s look at our 1-D billiard ball collision in more detail. 4 4 4 , , i i m v p arrowrightnosp arrowrightnosp 3 3 3 , 0, 0 i i m v p = = arrowrightnosp arrowrightnosp 4 4 4 , , f f m v p arrowrightnosp arrowrightnosp 3 3 3 , 0, 0 f f m v p arrowrightnosp arrowrightnosp 4 on 3 F arrowrightnosp at t = t i at t = t c to t c + Δ t at t = t f
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 15, Slide 5 We can see that between t i and t f , the momenta of both the 4 ball and the 3 ball changed. 4 4 4 i f p p p Δ = arrowrightnosp arrowrightnosp arrowrightnosp 3 3 i p p Δ = arrowrightnosp arrowrightnosp 3 f p arrowrightnosp The force of the 4 ball on the 3 ball is over a very short interval of time, Δ t . This is an example of an impulsive force . Newton’s 2 nd Law of Motion: ( ) c c t t i f t dp F dt dp Fdt p p p F t dt = = Δ = = arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp arrowrightnosp
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 15, Slide 6 We define a new quantity, the impulse , as: ( ) c c t t t p J F t dt Δ = = arrowrightnosp arrowrightnosp arrowrightnosp This is the impulse-momentum theorem and it states that:
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