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CH 7 Lect 2 Recoil and Elastic Collisions

CH 7 Lect 2 Recoil and Elastic Collisions. Recoil Two skaters push off from each other F T = 0 (f is small; W/N cancel) D p T = 0 p i = p f = 0 p 2 = - p 1 or p 2 + p 1 = 0 m 2 v 2 = -m 1 v 1 If m 1 = 2m 2 , m 2 v 2 = -2m 2 v 1 v 2 = -2v 1

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CH 7 Lect 2 Recoil and Elastic Collisions

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  1. CH 7 Lect 2 Recoil and Elastic Collisions • Recoil • Two skaters push off from each other • FT = 0 (f is small; W/N cancel) • DpT = 0 • pi = pf = 0 • p2 = - p1 or p2 + p1 = 0 • m2v2 = -m1v1 • If m1 = 2m2, m2v2 = -2m2v1 v2 = -2v1 • Recoil = brief interaction of 2 objects causing them to move in opposite directions • Shotgun shooting • Gun pushes back as it pushes shot forward • M(shot) small, m(gun) large: mSvS = -mGvG • Velocity of shot fast, recoil velocity is slower • Hold gun firmly, increase body/gun mass weaken recoil

  2. Rockets • -p(gas) = p(rocket) • Problems in doing rocket calculations • Blast occurs over a long time • Mass of the rocket gets smaller as it burns up its fuel • We can treat a brief rocket blast as a recoil • How does a rocket work in space? nothing to push against (propeller, jet) • Rocket pushes against its own exhaust gases to move forward • Elastic and Inelastic Collisions • Perfectly Inelastic Collision • 2 objects collide and stick together: treat these using cons. of momentum • m1 = 20,000kg v1 = 15 m/s m2 = 80,000 kg v2 = 0 p1 = m1v1 = (20,000kg)(15m/s) = 300,000kgm/s p2 = m2v2 = (80,000kg)(0m/s) = 0 vT? pT = 300,000kgm/s

  3. Conservation of Energy in Collisions • KE = ½ mv2 • We have lost KE!!! • Perfectly Inelastic Collisions always lose KE to heat, sound, deformation • Elastic Collisions • Perfectly Elastic Collision has no loss of KE • Partially Elastic Collision loses some KE, but objects still bounce Perfectly Elastic Perfectly Inelastic

  4. Pool Balls Colliding in 1 Dimension • Perfectly Elastic Collision • Cue ball stops, other ball gains same momentum • KE is conserved because m and v identical • Collisions at an Angle 1) Football Tackle: inelastic 2D collision pT

  5. 2) Pool Balls in 2D, an Elastic 2D Collision • Cue hits other ball off center • pI = cue only = pT = pf = pC + pO • Force is in the direction of m1m2 line • Since this is the only force acting on 2nd ball, it moves off in the same direction as F • PT maintains same direction as pI • pC + pO = pT (Vector addition) • How do we find the angle between 2 balls? • Conservation of Energy (Elastic Collision) • KE cue = KE cue + KE other c a b

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