Conservation of Momentum. CONSERVATION OF LINEAR MOMENTUM According to the law of conservation of...

Conservation of Momentum

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.

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 kgM2 = 75 kgV1 = 48 m/sV2 = 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 1.00-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 kgV1 = 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 kgV1 = 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