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UT Arlington PHYS 1441 - Lecture Notes

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PHYS 1441 – Section 501 Lecture #11AnnouncementsElastic and Inelastic CollisionsExample for CollisionsTwo dimensional CollisionsExample of Two Dimensional CollisionsCenter of MassCenter of Mass of a Rigid ObjectExample 7-11Example for Center of Mass in 2-DMotion of a Diver and the Center of MassWednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu1PHYS 1441 – Section 501Lecture #11Wednesday, July 7, 2004Dr. Jaehoon Yu•Collisions •Center of Mass •CM of a group of particles•Fundamentals on Rotation•Rotational Kinematics•Relationships between linear and angular quantitiesRemember the second term exam, Monday, July 19!!Today’s homework is HW#5, due 6pm next Wednesday!!Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu2Announcements•Quiz results:–Class average: 57.2–Want to know how you did compared to Quiz #1?•Average Quiz #1: 36.2–Top score: 90•I am impressed of your marked improvement•Keep this trend up, you will all get 100% soon…Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu3Elastic and Inelastic Collisions Collisions are classified as elastic or inelastic by the conservation of kinetic energy before and after the collisions.A collision in which the total kinetic energy and momentum are the same before and after the collision. Momentum is conserved in any collisions as long as external forces negligible.Elastic CollisionTwo types of inelastic collisions:Perfectly inelastic and inelastic Perfectly Inelastic: Two objects stick together after the collision moving at a certain velocity together.Inelastic: Colliding objects do not stick together after the collision but some kinetic energy is lost.Inelastic CollisionA collision in which the total kinetic energy is not the same before and after the collision, but momentum is.Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu4Example for CollisionsA car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the two become entangled. If the lighter car was moving at 20.0m/s before the collision what is the velocity of the entangled cars after the collision?ipThe momenta before and after the collision areWhat can we learn from these equations on the direction and magnitude of the velocity before and after the collision?m120.0m/sm2vfm1m2Since momentum of the system must be conservedfipp The cars are moving in the same direction as the lighter car’s original direction to conserve momentum. The magnitude is inversely proportional to its own mass.Before collisionAfter collisioniivmvm2211ivm220 fpffvmvm2211 fvmm21 fvmm21ivm22fv 2122mmvmismii/ 67.618009000.20900Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu5Two dimensional Collisions In two dimension, one can use components of momentum to apply momentum conservation to solve physical problems.fiivmvmvmvmf2212112121121ivmm2m1v1im1v1fm2v2fConsider a system of two particle collisions and scatters in two dimension as shown in the picture. (This is the case at fixed target accelerator experiments.) The momentum conservation tells us:iivmvm2121And for the elastic conservation, the kinetic energy is conserved:What do you think we can learn from these relationships?fxfxvmvm2211coscos2211 ffvmvm iyvm110fyfyvmvm2211sinsin2211 ffvmvm ivm11ixvm112222112121ffvmvm fxfxixixvmvmvmvm22112211fyfyiyiyvmvmvmvm22112211x-comp.y-comp.Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu6Example of Two Dimensional CollisionsProton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle  to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, .Since both the particles are protons m1=m2=mp.Using momentum conservation, one obtainsm2m1v1im1v1fm2v2fipvm1Canceling mp and put in all known quantities, one obtainssmvf/1080.251From kinetic energy conservation: (3) 1050.3222125ffvv Solving Eqs. 1-3 equations, one gets(1) 1050.3cos37cos521ffvvDo this at homecoscos21 fpfpvmvm sinsin21 fpfpvmvm 0x-comp.y-comp.(2) sin37sin21ffvv smvf/1011.2520.53Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu7Center of MassWe’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situation objects have shapes with masses distributed throughout the body. Center of mass of a system is the average position of the system’s mass and represents the motion of the system as if all the mass is on the point. Consider a massless rod with two balls attached at either end.CMx �The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if all the mass of the system is concentrated on the center of mass.MFa /What does above statement tell you concerning forces being exerted on the system?m1m2x1x2The position of the center of mass of this system is the mass averaged position of the systemxCMCM is closer to the heavier object1 1 2 21 2m x m xm m++Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu8Center of Mass of a Rigid ObjectThe formula for CM can be expanded to Rigid Object or a system of many particles A rigid body – an object with shape and size with mass spread throughout the body, ordinary objects – can be considered as a group of particles with mass mi densely spread throughout the given shape of the objectiiiiinnnCMmxmmmmxmxmxmx212211iiiiiCMmymyThe position vector of the center of mass of a many particle system is CMrMxmxiiiCMCMx dmrMrCM1mirirCMiiiiiCMmzmzkzjyixCMCMCMiiiiiiiiiiimkzmjymixm MrmriiiCMMxmiiimi 0lim xdmM1Wednesday, July 7, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu9Example 7-11Thee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and x3=6.0m. Find the position of


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UT Arlington PHYS 1441 - Lecture Notes

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