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UW-Madison PHYSICS 207 - Physics 207, Lecture 12

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Page 1Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 1Physics 207, Lecture 12, Oct. 13Goals:Goals:Assignment: Assignment: HW5 due WednesdayHW5 due WednesdayHW6 available todayHW6 available todayFor Wednesday: Read all of chapter 10For Wednesday: Read all of chapter 10••Chapter 9: Momentum & ImpulseChapter 9: Momentum & Impulse Solve problems with 1D and 2D Collisions  Solve problems having an impulse (Force vs. time)••Chapter 10Chapter 10 Understand the relationship between motion and energy Define Potential & Kinetic Energy Develop and exploit conservation of energy principlePhysics 207: Lecture 12, Pg 2Inelastic collision in 1-D: Example A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V :What is the momentum of the bullet with speed v ?vVbefore afterxPage 2Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 3Inelastic collision in 1-D: ExampleWhat is the momentum of the bullet with speed v ? Key question: Is x-momentum conserved ? vVbefore afterxaaaavrm V)( 0 M vMmm+=+BeforeBeforeAfterAfterPhysics 207: Lecture 12, Pg 4Home ExerciseInelastic Collision in 1-D with numbersice(no friction)Before: Before: A 4000 kg bus, twice the mass of the car, moving A 4000 kg bus, twice the mass of the car, moving at 30 at 30 m/sm/simpacts the car at rest. impacts the car at rest. What is the final speed after impact if they move together?What is the final speed after impact if they move together?Page 3Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 5Home exercise Inelastic Collision in 1-Dvvff=?finallymv = 0iceM = 2mVV00(no friction)initially2Vo/3 = 20 m/s000V22V Vor V)( VmmmMmMMmM+=+=+=Physics 207: Lecture 12, Pg 6Exercise Momentum ConservationA. Box 1B. Box 2C. same Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface.  The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks. Which box ends up moving fastest ?12Page 4Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 7Exercise Momentum Conservation The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks. Which box ends up moving fastest ? Notice the implications from the graphical solution: Box 1’s momentum must be bigger because the length of the summed momentum must be the same. The longer the green vector the greater the speed12Box 2+Ball 2Box 2+Ball 2Box 1+Ball 1Box 1+Ball 1Box 2Box 2Box 1Box 1Ball 2Ball 2Ball 1Ball 1AfterBeforeAfterBeforePhysics 207: Lecture 12, Pg 8Exercise Momentum Conservation Which box ends up moving fastest ? Examine the change in the momentum of the ball.In the case of box 1 the balls momentum changes sign and so its net change is largest. Since momentum is conserved the box must have the largest velocity to compensate.(A)(A) Box 1(B)(B) Box 2(C)(C) same 12Page 5Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 9A perfectly inelastic collision in 2-D Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).vv1vv2VVbefore afterm1m2m1+ m2 If no external force momentum is conserved. Momentum is a vector so px, pyand pzθPhysics 207: Lecture 12, Pg 10A perfectly inelastic collision in 2-Dvv1vv2VVbefore afterm1m2m1+ m2 x-dir px: m1v1= (m1+ m2) V cos θ y-dir py: m2v2= (m1+ m2) V sin θ If no external force momentum is conserved. Momentum is a vector so px, pyand pzare consevedθPage 6Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 11Elastic Collisions Elastic means that the objects do not stick. There are many more possible outcomes but, if no external force, then momentum will always be conserved Start with a 1-D problem.Before AfterPhysics 207: Lecture 12, Pg 12Billiards Consider the case where one ball is initially at rest. ppaθθθθppbFFPPaφφφφbeforeafterThe final direction of the red ball will depend on where the balls hit.vvcmPage 7Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 13Billiards: All that really matters is conservation momentum (and energy Ch. 10 & 11) Conservation of Momentum x-dir Px: m vbefore= m vaftercos θ + m Vaftercos φ y-dir Py: 0 = m vaftersin θ + m Vaftersin φppafterθθθθppbFFPPafterφφφφbeforeafterPhysics 207: Lecture 12, Pg 14Force and Impulse (A variable force applied for a given time)  Gravity: At small displacements a “constant” force t Springs often provide a linear force (-k x) towards its equilibrium position (Chapter 10) Collisions often involve a varying force F(t): 0 maximum 0 We can plot force vs time for a typical collision. The impulse, JJ, of the force is a vector defined as the integral of the force during the time of the collision.Page 8Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 15Force and Impulse (A variable force applied for a given time)F∫∫∫===pttpddtdtpddtFJrrrr)/( J reflects momentum transfertti tf∆t ImpulseJJ = area under this curve !(Transfer of momentum !)Impulse has units of Newton-secondsPhysics 207: Lecture 12, Pg 16Force and Impulse Two different collisions can have the same impulse since JJ depends only on the momentum transfer, NOT the nature of the collision.∆t FtFt∆t same area∆t big, FF small∆t small, FF bigPage 9Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 17Average Force and Impulse∆t FtFt∆t ∆t big, FFavavsmall∆t small, FFavavbigFFavavFFavavPhysics 207: Lecture 12, Pg 18Exercise 2Force & ImpulseA. heavierB. lighterC. sameD. can’t tell Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force Facts on each one for exactly 1 second.Which box has the most momentum after the force acts ?F FlightheavyPage 10Physics 207 – Lecture 12Physics 207: Lecture 12, Pg 19Boxing: Use Momentum and Impuse to estimate g “force”Physics 207: Lecture 12, Pg 20Back of the envelope calculation(1) marm~ 7 kg (2) varm~7 m/s (3) Impact time ∆∆∆∆t ~ 0.01 sQuestion: Are these reasonable? Impulse J = ∆∆∆∆p ~ marmvarm~ 49 kg m/s F ~ J/∆∆∆∆t ~ 4900 N(1) mhead~ 6 kg ahead = F / mhead~ 800 m/s2~ 80 g ! Enough to cause


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UW-Madison PHYSICS 207 - Physics 207, Lecture 12

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