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Linear Momentum 5-1 Linear Momentum Linear Momentum, p – defined as mass x velocity p mv The unit is kgm/s A quantity used in collisions So a small object with a large velocity could have the same momentum as a large object with a small velocity 9-1 Linear Momentum 5.2 Momentum and Newton’s Second Law Newton’s Second Law is F ma This is only true for objects with a constant mass The original form of the equation was p F t This statement is true even if the mass varies 5.2 Momentum and Newton’s Second Law 5.3 Impulse A baseball player hits a pitch Bat delivers an impulse We actually on consider average force Impulse is define as Ft p Impulse Sim 5.3 Impulse An increase in time produces a decreases in force Airbag A decrease in time produces an increase in force 5.3 Impulse 5.4 Conservation of Linear Momentum If no net external force is applied to a system Then momentum is conserved p0 p Explode Sim 5.4 Conservation of Linear Momentum External Forces will result in a change in momentum, so no conservation 1.Force added in Shuttle Launch 2.Force removed 5.4 Conservation of Linear Momentum 5.5 Inelastic Collisions Inelastic collision – two objects collide and stick together Momentum is conserved p0 p mAvA mBvB (mA mB )v Energy is not conserved 5.5 Inelastic Collisions Example: On a touchdown attempt, a 95 kg running back runs toward the end zone at 3.75 m/s. A 111kg linebacker moving at 4.10 m/s meets the runner in a head on collision. If the two players stick together what is their velocity immediately after the collision? mAvA mBvB (mA mB )v (95)(3.75) (111)( 4.10) (95 111)v v 0.48 ms 5.5 Inelastic Collisions Example: In a ballistic pendulum, a 100g bullet is fired at a velocity of 200 m/s at the bob of a pendulum. The bob has a mass of 10 kg. After the collision, the object and the bob stick together and swing through an arc. How high does it get? This is first a conservation of momentum problem (how fast does the combination of bullet and bob go after the collision) Then it is a conservation of energy problem. 5.5 Inelastic Collisions Example: In a ballistic pendulum, a 100g bullet is fired at a velocity of 200 m/s at the bob of a pendulum. The bob has a mass of 10 kg. After the collision, the object and the bob stick together and swing through an arc. How high does it get? mAvA mBvB (mA mB )v (0.1)( 200) (10)(0) (.1 10)v v 1.9 ms 1 2 1 2 mv mgy 2 (1.9) (9.8) y 2 y 0.18m 5.5 Inelastic Collisions If the collision occurs in two dimensions We need to consider the x and y axis separately mAv Ax mB vBx (mA mB )vx mAvAy mB vBy (mA mB )v y 5.5 Inelastic Collisions The we use vector addition to calculate the magnitude and velocity. v v v 2 x 2 y vy tan vx 1 5.5 Inelastic Collisions Example: A 950kg car traveling east at 16m/s collides with a 1300 kg car traveling north at 21 m/s. If the collision is inelastic, what is the magnitude and direction of the cars after the collision? mAvAyx mB vBxy (mAA mBB )vxy 2 2 )( 16 ) ( 1300 )( 0 ) ( 950 1300 ) v v(950 6 . 76 m / s v 6 . 76 12 . 1 13.9 ms x 2 2 x v v v x y v(v950 6 . 76 m / s x )( 120.)1m(1300 / s )( 21) (950 1300)v y tan 1 12.1 60.8o y v y 12.1m / s vy tan x vCollisions 5.5 Inelastic 1 6.76 5.6 Elastic Collisions Elastic collision – two objects collide and bounce apart Momentum is conserved m Av A0 mB vB0 m Av A mB vB In a perfect elastic collision, energy is conserved too 1 2 mv mv mv mv 2 A A0 1 2 2 B B0 1 2 2 A A 5.5 Inelastic Collisions 1 2 2 B B A 10 kg car moving at 2 m/s runs into a 5 kg car that is parked. What is the velocity of each car after the collision? m Av A0 mB vB0 m Av A mB vB (10)(2)20 (510 )(0v)A10 5vvBA 5vB 1 2 1 2 mv mv mv mv 2 A A0 1 2 2 B B0 1 2 2 A A 1 2 2 B B (10 (10 )()( 2)2)40 (5 )(10 0) v (10 (5)v(5)v 5(10 v)v)v 2 2 1 2 2 2 1 A 2 2 2 2 B A A 5.5 Inelastic Collisions 1 2 2 B 2 B A 10 kg car moving at 2 m/s runs into a 5 kg car that is parked. What is the velocity of each car after the collision? 20 10vA 5vB vA 2 .5vB 40 10v 5v 2 A 2 B 40 10(2 .5vB ) 5v 2 2 B 5.5 Inelastic Collisions A 10 kg car moving at 2 m/s runs into a 5 kg car that is parked. What is the velocity of each car after the collision? vA 2 .5vB 40 10(2 .5vB ) 5v 2 2 B 40 10(.25v 2vB 4) 5v 2 B 40 2.5v 20vB 40 5v 2 B 2 B 2 B 0 7.5v 20vB 2 B 5.5 Inelastic Collisions A 10 kg car moving at 2 m/s runs into a 5 kg car that is parked. What is the velocity of each car after the collision? 0 7.5v 20vB 2 B vB 2.67 ms m v 0 . 67 2020 10 vAvA 5(52s v.67 A 10 B ) Confirm 10(0.67 5(v2.67 40) 10 5)v 40.1 2 2 A 22 B 5.5 Inelastic Collisions 5.7 Center of Mass The point where the system can be balanced in a uniform gravitational field Center of mass of Triangle Uniform objects center of mass is in the center Motion of CM 5.7 Center of Mass Center of mass is not always in the object Objects balance if supported at their center of mass 5.7 Center of Mass 5.8 Systems with Changing Mass: Rockets When a rocket is launched (or a plane takes off). Fuel is used as the rocket launches This causes the mass to decrease 5.8 Systems with Changing Mass: Rockets So P m F v t t The force due to the ejected fuel is called thrust m thrust v t 5.8 Systems with Changing Mass: Rockets In a Saturn V rocket, fuel is ejected at 13,800 kg/s and at a speed of 2440 m/s m thrust v t thrust 13,8002440 33,700,000 N Since the initial weight of the rocket is 28,500,000N (or mass of 2,850,000 kg) F 33700000 28500000 5,700,000 N F 5700000 a 2 sm2 m 2850000 5.8 Systems with Changing Mass: Rockets As the rocket travels its mass drops, so the acceleration will actually increase 5.8 Systems with Changing Mass: Rockets