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Physics 200 Class #10 Notes October 10, 2005 Reading Assignment for Wednesday Text: Chapter 6 pp. 141-160 Continuation of Chapter 5 with added notes on Energy and Blackbody Radiation Reminder: The MidTerm Exam is on October 17 We discussed force, and energy... Now what about a force acting for a given length of time, independent of distance (great examples at http://www.kungfuscience.org ) Momentum Descartes and Huygens defined momentum in order to “quantify motion”. Collision problems lead to the following definition: Linear Momentum (usually denoted by p) = mass x velocity (note, it has direction) Momentum mv units : kg m / s Conservation of Linear Momentum Principle: When two objects collide, momentum (linear momentum) may be transferred from one object to the other, but the total linear momentum of the system does not change. (we return to this, but first) What produces a change in momentum? Force But time plays a roll. It is convenient to summarize the effect as follows: Impulse = force x time interval Impulse = F t Use Newton’s second law (for a constant mass) F ma v t F t mv F m means a change in the quantity following it F t (mv) The impulse is equal to the change in momentum of the object. “Bouncing” – what effect if any? No Bounce: object moving to the right with momentum mv F t = momentum after – momentum before F t = 0 – momentum before = - mv The minus sign means that the force must act to the left Phy 200 Fall 2005 Class_10 Page 1 of 3 With Bounce: F t = momentum after – momentum before F t = - mv – mv = - 2mv For the same time interval, the force must be twice as great Collisions Class Demonstrations on Collisions on an Air Track or karate chopping boards Linear momentum of the system is conserved in collisions! Why? If we consider the colliding objects to be the system, the force between them is an internal force. The net external force (and therefore impulse) is zero. Examples Suppose we have two masses: m1 2kg m2 3kg Conservati on of momentum requires : m1v1before m2 v 2before m1v1after m2 v 2 after Case I: The masses stick together after the collision (inelastic collision) v1before 10m/s v 2before 0 (2kg)(10m / s ) (3kg)(0m / s ) (5kg)(v final ) 20kgm / s (5kg)(v final ) v final 4m / s Case II: Elastic collision with the 2kg mass moving to the left afterwards, find the speed of the 3kg mass. (2kg)(10m / s ) (3kg)(0m / s ) (2kg)( 2m / s ) (3kg)(v final ) 20kgm / s 4kgm / s (3kg) v final v final 8kgm / s Case III: There is an explosion during the collision and the 2kg mass moves to the left at 95m/s, find the speed of the 3kg mass afterwards. (2kg)(10m / s ) (3kg)(0m / s) (2kg)( 95m / s) (3kg)(v final ) 20kgm / s 190kgm / s (3kg)(v final ) v final 70m / s Now with the conservation of energy too! Total energy is conserved: E = KE + PE = constant Elastic collisions, kinetic energy is conserved on it's own - things bounce. Inelastic - energy goes out of the system into the environment, or into smashing things - but momentum still stays in the system. 1 Remember KE = mv2 2 Phy 200 Fall 2005 Class_10 Page 2 of 3 Case I: Inelastic collision of a moving car (1) with a stationary car (2). m1 = 1000 kg. (2,200 pounds) m2 = 1000 kg v1initial = 20 m/s (about 45 miles per hour) v2initial = 0 Since it's inelastic, let's assume the cars stick together so v1final = v2 final which we just call vfinal conservation of momentum: m1v2initial + m2v2initial = m1v2final + m2v2final = m1 vfinal + m2vfinal = (m1 + m2) vfinal (1000 kg)(20m/s) + (100 kg)(0m/s) = (2000 kg)(vfinal) (2000 kg m/s) = (2000 kg)(vfinal) so: vfinal = 10 m/s Let's look at what happens to the energy. It's inelastic, so some of the energy is put into smashing the cars: 1 1 1 1 m1v2 1 initial + m2v2 2 initial = m1v2 1 final + m2 v2 2 final + E put into smashing the cars 2 2 2 2 (0.5)(1000 kg)(20 m/s)2 + (0.5)(1000 kg)(0 m/s)2 = (0.5)(2000 kg)(10 m/s)2 + E smash (0.5)(1000 kg)(400 m2/s2) = (0.5)(2000 kg)(100 m2/s2) + E smash 200,000 J = 100,000 J + E smash so the energy put into deforming the metal of the cars is Esmash = 200,000 J - 100,000 J = 100,000 J That's the same amount of energy it would take you to pick up one of the cars and raise it 10 meters. That's means lifting a 2,200 lb car up to the top of a 3 story house! Case II: Work that out for an initial velocity of 40 m/s, just double the speed. Really, try it. Answer below: Answer: that doubles the total final velocity. vfinal = 20 m/s The energy equation is (0.5)(1000 kg)(40 m/s)2 + (0.5)(1000 kg)(0 m/s)2 = (0.5)(2000 kg)(20 m/s)2 + E smash (0.5)(1000 kg)(1600 m2/s2) = (0.5)(2000 kg)(400 m2/s2) + Esmash 800,000 J = 200,000 J + E smash so the energy put into deforming the metal of the cars is Esmash = 800,000 J - 200,000 J = 600,000 J That's a big difference from just doubling the speed! That's like lifting a 2,200 lb car up to the top of a 18 story house! Drive carefully. Phy 200 Fall 2005 Class_10 Page 3 of 3