PHY 317K 1st Edition Lecture 9Outline of Last LectureI. PowerII. The Force of GravityIII. Center of MassIV. MomentumOutline of Current LectureI. Problems II. Elastic CollisionsIII. Inelastic CollisionsIV. Conservation of MomentumCurrent LectureProblem 1:A hunting rifle fires a bullet of mass 0.012 kg with a velocity of 600 m/s to the right. The rifle hasa mass of 4 kg. What is the recoil speed of the rifle as the bullet leaves the rifle?Pi=pfM1V1 = -m2v2Vf = -m2V2f/ m1 Vf = -1.8 m/sProblem 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 = pfM1v1= m2 v2 = (m1 + m2)(vf)Vp = 5.38 m/s(b) Find the change of velocity of each car.SUV = vi = 14 m/sVf = 5.38 m/sCar= v2 = -14 m/sV2f = 5.38 m/sDV = 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^2Ki = 254,800 JKf = ½ (m1+m2) vf^2Kf = 37,628 JDK = kf – ki = -217,172 JElastic collision:- Energy is conserved (kinetic)- No loss of kinetic energy- Momentum conservationInelastic collision:- momentum conserved- Perfect inelastic collision: two objects stuck together and lose energyo Momentum conserved- Kinetic energy is changed to some other form of energyConservation 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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