Conservation of Momentum

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Presentation transcript:

Conservation of Momentum

Cars on ice CONSERVATION OF LINEAR MOMENTUM According to the law of conservation of linear momentum, the total momentum in a system remains the same if no external forces act on the system. Cars on ice

EX #1: A 0.105-kg hockey puck moving at 48 m/s is caught by a 75-kg goalie at rest. With what velocity does the goalie slide on the ice after catching the puck? M1 = 0.105 kg M2 = 75 kg V1 = 48 m/s V2 = 0 m/s p before= p after m1V1 + m2V2 = (m1 +m2) Vf (0.105 kg)(48m/s) + (75kg)(0) = (0.105kg + 75Kg) Vf Vf = 0.067 m/s

ELASTIC AND INELASTIC COLLISIONS Elastic Collision: A collision in which objects collide and bounce apart with no energy loss. Inelastic Collision: A collision in which objects collide and some mechanical energy is transformed into heat energy.

The animation below portrays the inelastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.

The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables.

Before the collision, the momentum of the truck is 60 000 Ns and the momentum of the car is 0 Ns; the total system momentum is 60 000 Ns. After the collision, the momentum of the truck is 30 000 Ns and the momentum of the car is 30 000 Ns; the total system momentum is 60 000 Ns.

The animation below portrays the inelastic collision between a very massive diesel and a less massive flatcar. The diesel has four times the mass of the freight car. After the collision, both the diesel and the flatcar move together with the same velocity.

EX #2: A 0. 50-kg ball traveling at 6. 0 m/s collides head-on with a 1 EX #2: A 0.50-kg ball traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a velocity of -12.0 m/s. The 0.50-kg ball moves away at -14 m/s after the collision. Find the velocity of the second ball. M1 = 0.50 kg M2 = 1.00 kg V1 = 6.0 m/s V2 = -12.0 m/s Vf1 = -14 m/s p before = p after m1V1 + m2V2 = m1Vf1 + m2V2f (.5kg)(6m/s) + (1kg)(-12m/s) = (.5kg)(-14m/s) + (1kg)(V2f) V2f = - 2 m/s

EX #3: A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck. M1 = 3000 kg M2 = 1000 kg V1 = 5.0 km/hr V2 = -10 km/hr p before = p after m1V1 + m2V2 = (m1+ m2 )Vf (3000kg)(5km/hr) + (1000kg)(-10km/hr) = (3000kg + 1000kg) Vf Vf = 1.25 km/hr, right

For the remainder of class... PSE Chapter 4 pg 58 #7-14 due BOC 12/9