Lecture 12: Collisions and Explosions l Momentum Examples! è Problem Solving è Collisions (elastic & inelastic) è Explosions.

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Presentation transcript:

Lecture 12: Collisions and Explosions l Momentum Examples! è Problem Solving è Collisions (elastic & inelastic) è Explosions

Impulse and Momentum: quick review l Momentum-Impulse Theorem è F  t  I = p f - p i =  p l For single object…. è If F = 0, then momentum conserved (  p = 0) l For “system” of objects … è p total   p è Internal forces: forces between objects in system è External forces: all other forces è F ext  t =  p total è if F ext = 0, then total momentum conserved (  p total = 0) l Applications: Collisions & Explosions

Collisions “before” “after” m1m1 m2m2 m1m1 m2m2 Explosions “before” “after” M m1m1 m2m2 Draw “before”, “after” Define system so that F ext = 0 Set up axes Compute P total “before” Compute P total “after” Set them equal to each other Procedure

Some Terminology Elastic Collisions: collisions that conserve mechanical energy Inelastic Collisions: collisions that do not conserve mechanical energy * Completely Inelastic Collisions: objects stick together

Collision Example l Two cars are approaching an intersection. The first car is 1520 kg and is moving east at 18 m/s; the second car is 1380 kg and is moving north at 16 m/s. The cars have a completely inelastic collision. What is their final velocity? y x Before: After:

Collision Example l Two cars are approaching an intersection. The first car is 1520 kg and is moving east at 18 m/s; the second car is 1380 kg and is moving north at 16 m/s. The cars have a completely inelastic collision. What is their final velocity? y x x-direction: p ix = p fx (1520 kg)(18 m/s) + 0 = p fx y-direction: p iy = p fy 0 + (1380 kg)(16 m/s) = p fy

Collision Example l Two cars are approaching an intersection. The first car is 1520 kg and is moving east at 18 m/s; the second car is 1380 kg and is moving north at 16 m/s. The cars have a completely inelastic collision. What is their final velocity? y x x-direction: (1520 kg)(18 m/s) + 0 = p fx (1520 kg)(18 m/s) = (1520 kg kg) v fx y-direction: 0 + (1380 kg)(16 m/s) = p fy (1380 kg)(18 m/s) = (1520 kg kg) v fy

Collision Example l Two cars are approaching an intersection. The first car is 1520 kg and is moving east at 18 m/s; the second car is 1380 kg and is moving north at 16 m/s. The cars have a completely inelastic collision. What is their final velocity? y x x-direction: v fx = 9.43 m/s y-direction: v fy = 7.61 m/s V f = 12.1 m/s at 51º east of north

Explosion Example l A ball has a mass of 3 kg. It explodes into three equal masses, shooting a 1 kg piece in the positive y direction with speed v and another 1 kg piece in the positive x direction with speed v. What is the speed and direction of the remaining 1 kg piece? y x Before: After:

Explosion Example l A ball has a mass of 3 kg. It explodes into three equal masses, shooting a 1 kg piece in the positive y direction with speed v and another 1 kg piece in the positive x direction with speed v. What is the speed and direction of the remaining 1 kg piece? y x x-direction: p ix = p fx = 0 + (1 kg) v + (1 kg) v fx y-direction: p iy = p fy = (1 kg) v (1 kg) v fy

Explosion Example l A ball has a mass of 3 kg. It explodes into three equal masses, shooting a 1 kg piece in the positive y direction with speed v and another 1 kg piece in the positive x direction with speed v. What is the speed and direction of the remaining 1 kg piece? y x x-direction: v fx = -v y-direction: v fy = -v V f = v  2 at 45º below the –x axis

Center of Mass Center of Mass = Balance point Example 1: L mm Example 2: L m5m x CM = (0 + mL)/2m = L/2 x CM = (0 + 5mL)/6m = 5L/6 X=0 X=L

Summary Collisions and Explosions Draw “before”, “after” Define system so that F ext = 0 Set up axes Compute p total “before” Compute p total “after” Set them equal to each other Center of Mass (Balance Point)