12/20 do now

1. 12/20 do now • A ball with a momentum of +4.0 kg • m/s hits a wall and bounces straight back without losing any kinetic energy. What is the change in the ball’s momentum?

2. Class work 5.33 s; 53.3 m W a. 14 m/s N; b. 42 m N; c. 8.0 s a. 1.22 x 104N E; b. 53.3 m W • Page 213 practice 6C • Page 214 section review #1-5 a. 31 m/s ; a. 2.5 kg∙m/s down field; b. 130 N downfield • Page 232 #16, 47 0.010 s; 0.13 m 400 N

3. 6-2 conservation of momentum Objectives • Describe the interaction between two objects in terms of the change in momentum of each object. • Compare the total momentum of two objects before and after they interact. • State the law of conservation of momentum. • Predict the final velocities of objects after collisions, given the initial velocities. Homework – 6.2 essay – is due Extra credit on chapter 6 test – do homework on time 6.1 essay rewrite – get average of the two essays – samples are posted.

4. Homework assignments • Castle learning assignments: • Winter break reviews (5) • 04 conservation of momentum 1 • Chapter 6 project • Essay 6.3

5. The Law of Action-Reaction demo • In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction in accord with Newton’s 3rd Law. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. According to Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Bigger mass has smaller acceleration, smaller mass has bigger acceleration

6. 1. 2. 3. 4. m1(v1’ – v1) = -m2(v2’– v2) m1v1 + m2v2 = m1v1’ + m2v2’ p(before) = p(after)

7. Momentum Conservation Principle • The law of momentum conservation can be stated as follows: • For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

8. Isolated Systems • A system is a collection of two or more objects. • An isolated system is a system which is free from the influence of a net external force which alters the momentum of the system.

9. No friction The friction between the carpet and the floor and the applied force exerted by Hans are both external forces. No yes

10. In reality • Because of the inevitability of friction and air resistance in any real collision, one might can conclude that no system is ever perfectly isolated. • In any real collision, resistance forces such as friction and air resistance are inevitable. However, compared to the impact force, they are very small, and they can be ignored. Therefore, in most cases, momentum can be approximate to be conserved for every situations.

11. Consider a fullback plunges across the goal line and collides in midair with the linebacker in a football game. The linebacker and fullback hold each other and travel together after the collision. Before the collision, the fullback possesses a momentum of 100 kg∙m/s, East and the linebacker possesses a momentum of 120 kg∙m/s, West. The total momentum of the system before the collision is _____________________. Therefore, the total momentum of the system after the collision must also be __________________ 20 kg∙m/s, West 20 kg∙m/s, West. Vector diagram for the situation:

12. consider a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. • The momentum of the medicine ball is 80 kg∙m/s before the collision. The momentum of the clown is 0 kg∙m/s before the collision. The total momentum of the system before the collision is ______________ • Therefore, the total momentum of the system after the collision must also be ________________. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg∙m/s. Momentum is conserved in the collision. 80 kg∙m/s. 80 kg∙m/s. Vector diagram for the situation:

13. Using Equations for Algebraic Problem-Solving • Total system momentum is conserved for collisions between objects in an isolated system. pbefore = pafter m1v1 + m2v2 = m1v1’ + m2v2’

14. Example 1 • A 2.0-kilogram ball traveling north at 5.0 meters per second collides head-on with a 1.0 kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two balls after collision? m1v1 + m2v2 = m1v1’ + m2v2’ m1 = 2.0 kg; v1 = 5.0 m/s m2 = 1.0 kg; v2 = -8.0 m/s pafter = ? m1v1 + m2v2= pafter (2.0 kg)(5.0 m/s) + (1.0 kg)(-8.0 m/s) = pafter 2 kg∙m/s, north = pafter

15. Example 2 • A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision. It is a collision problem, momentum is conserved. m1v1 + m2v2 = m1v1’ + m2v2’ m1 = 3000 kg; v1 = 10 m/s m2 = 1000 kg; v2 = 0 v1’ = ? v2’ = 15 m/s (3000 kg)∙(10 m/s) + 0 = (3000 kg)∙v + (1000 kg)∙(15 m/s) 30000 kg∙m/s = (3000 kg)∙v + 15000 kg∙m/s v = 5 m/s

16. Example 3 • A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision. m1v1 + m2v2 = m1v1’ + m2v2’ m1 = 15 kg; v1 = 20 km/hr m2 = 60 kg; v2 = 0 v1’ = v2’ = v (15 kg)(20 km/hr) + (60 kg)(0)=(15 kg)·v + (60 kg)·v 300 kg∙km/hr = (75 kg)∙v v = 4 km/hr

17. Example 4 • A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. Determine the post-collision velocity of the mitt and ball. m1v1 + m2v2 = m1v1’ + m2v2’ m1 = 0.150 kg; v1 = 45.0 m/s m2 = 0.250 kg; v2 = 0 v1’ = v2’ = v (0.15 kg)(45.0 m/s) + 0 =(0.15 kg)·v + (0.25 kg)·v 6.75 kg∙m/s = (0.40 kg)∙v v = 16.9 m/s

18. Example 5 - Sample problem 6D • A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

19. Class work • Page 219 – practice 6D 1.90 m/s 1.66 m/s E a. 12.0 m/s; b. 9.6 m/s c. 11.7 m/s 38 kg • Page 232 #24-26, 46*, 52*, 53*, 57*, 59* 1.90 m/s 1.66 m/s E a. 12.0 m/s; b. 9.6 m/s c. 11.7 m/s 38 kg

20. Jonathan Lichtenfeld