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Chapter 5: Momentum Momentum: a measure of motion Force: a cause of change in motion What changes when a force is applied? v F ma m ; but mass does not change t mv F ; mv changes wi th time t Linear Momentum: p = mv the tendency of an object to pursue straight line motion Kinetic Energy in terms of momentum: p2 KE 2m p150c5:1 Impulse: the change in motion I p mv since v at and F ma I Ft p mv Example: The head of a golf club is in contact with a 46 g golf ball for 0.50 ms, which results in the golf ball flying off at 70 m/s. Find the impulse and the average force acting on the ball during the impact. p150c5:2 Conservation of momentum two bodies + action/reaction + no other forces FAB = - FBA => equal but opposite impulses => pA + pB = 0 When the vector sum of the external forces acting on a system of particles equals zero, the total linear momentum remains constant. p1 + p2 + p3 + ... is constant p150c5:3 An application of conservation of momentum: explosions initially: m mv = 0 after explosion: m1v1 + m2v2 + ...= 0 m m v or (for two fragments) m1v1 = -m2v2 same size p, opposite directions 1 1 2 v2 Example: An astronaut just outside of the space shuttle throws her 800 g camera away when it jams. If she and her space suit have a combined mass of 100 kg and the speed of the camera is 20 m/s, how far is she from the shuttle in 10 seconds? p150c5:4 Rocket propulsion (note: no problems) “continual explosion” continual conservation of momentum: ( mv ) m F v t t combustion chamber ejected gases: m at speed v All space propulsion systems depend upon conservation of momentum. most require some “reaction mass” p150c5:5 Collisions Elastic Collisions conserve KE (total KE is same before and after collision) Inelastic Collisions some KE is lost during collision (heat, sound, etc.) Completely Inelastic Collisions objects stick together maximum possible loss of KE In all collisions, the total momentum is conserved! p150c5:6 Example 5.4 A 5.0 kg lump of clay that is moving at 10 m/s to the left strikes a 6.0 kg lump of clay moving 12 m/s to the right. The two lumps stick together after they collide. Find the final speed of the composite lump of clay and the kinetic energy lost during the collisions. Example 5.5 (discussion only) A 60 kg man sliding east on a frictionless surface of a frozen pond at a velocity of .50 m/s is struck by a 1 kg snowball whose velocity is 20 m/s towards the north. If the snowball sticks to the man, what is his final velocity? p150c5:7 Elastic collisions conservation of KE + conservation of momentum => initial relative velocity = -(final relative velocity) v1 - v2 = -(v1' - v2 ') For an object striking head-on a second object (initially at rest) v2 0 v1 - v2 -(v1 '-v2 ' ) m1v1 m2 v2 m1v1 ' m2 v2 ' algebra m1 - m2 2m1 v1 ' v1 v2 ' v1 m1 m2 m1 m2 KE 2 ' 4(m2 m1 ) KE1 (1 m2 m1 ) 2 p150c5:8 Example: A 5.0 Kg mass moving at 10 m/s collides elastically head-on with a stationary 10 Kg mass. What are the final velocities of both masses? What is the initial KE of the 5.0 kg mass? What is the final KE of the 10 kg mass? Example: A 10 Kg mass moving at 10 m/s collides elastically head-on with a stationary 5.0 Kg mass. What are the final velocities of both masses? What is the initial KE of the 10 kg mass? What is the final KE of the 5.0 kg mass? p150c5:9

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