These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. Outline of Last Lecture I. Power II. The Force of Gravity III. Center of Mass IV. Momentum Outline of Today’s Lecture I. Problems II. Elastic Collisions III. Inelastic Collisions IV. Conservation of Momentum Today’s Lecture Problem 1: A hunting rifle fires a bullet of mass 0.012 kg with a velocity of 600 m/s to the right. The rifle has a mass of 4 kg. What is the recoil speed of the rifle as the bullet leaves the rifle? Pi=pf M1V1 = -m2v2 Vf = -m2V2f/ m1 Vf = -1.8 m/s Problem 2: A SUV with mass 1.80 x 10^3 kg is traveling east bound at +14.0 m/s while a compact car with mass 8.00 x 10^2 kg is traveling west bound at -14.0 m/s. The cars collide head-on, becoming entangled. (a) Find the speed of the entangled cars after the collision. Pi = pf M1v1= m2 v2 = (m1 + m2)(vf) Vp = 5.38 m/s (b) Find the change of velocity of each car. SUV = vi = 14 m/s Vf = 5.38 m/s Car= v2 = -14 m/s V2f = 5.38 m/s DV = 19.38 m/s (c) Find the change of kinetic energy of the system consisting of both cars. Ki = ½ mv^2 + ½ m2 v^2 Ki = 254,800 J Kf = ½ (m1+m2) vf^2 PHY 317K Markert, Christina 2013 Spring Week 6 Lecture 10 February 19, 2013Kf = 37,628 J DK = kf – ki = -217,172 J Elastic collision: - Energy is conserved (kinetic) - No loss of kinetic energy - Momentum conservation Inelastic collision: - momentum conserved - Perfect inelastic collision: two objects stuck together and lose energy o Momentum conserved - Kinetic energy is changed to some other form of energy Conservation of Momentum: - Summation of pf = summation of pi - Momentum is denoted by “P”, so final momentum is Pf and initial momentum is Pi - Vf = (m1-m2)/(m1+m2) vi + (2m2/(m1+m2))v2i - V2f = (2m1/ (m1+m2))v1i + (m2-m1)/(m1 + m2))
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