2. Chapter 6 Momentum and Collisions

3. 6-1 Momentum and Impulse Momentum(p) describes the tendency of an object to continue moving (or not moving) at a constant speed. The heavier or faster an object moves, the greater its momentum. Momentum = Mass x Velocity p = mv

4. Example A 2250 kg truck has a velocity of 25m/s. What is its momentum? p = mv = (2250kg)(25m/s) p = 56,000 kg  m/s (Note the Units) On a separate sheet of paper, do problems 1-3 on page 209.

5. 6-1 (cont) A change in momentum takes a force applied over a time interval (Impulse) ∆p = F∆t A change in momentum can also be defined as (p f - p i ) or (mv f - mv i ) ∆p = mv f - mv i F∆t = mv f - mv i F = (mv f - mv i )/∆t

6. Example A 1400kg car with a velocity of 15m/s impacts a telephone pole stopping in 0.3s. How much force is exerted on the car (on Board) On a separate sheet of paper, do practice problems 1-3 on page 211.

7. Stopping Distance Remember from Chapter 4 that… –∆x = 1/2 (v i +v f )∆t Sample Problem on Board On separate sheet of paper, do practice problems 1- 3 on page 213.

8. Stopping Time and Force A change in motion over a longer time, requires less force. For example a car dashboard stops a person over a period of 0.026sec applies a significantly greater force than an airbag that stops a person over a period of.75 seconds. How much more force?

9. 6-2 Conservation of Momentum Momentum is a conserved quantity Imagine a soccer ball traveling at some velocity hits a stationary soccer ball. What would happen? It is likely that soccer ball one will slow down and soccer ball two will accelerate.

10. Conservation of Momentum Ball 1 –Before Collision m =.47kg v =.84m/s p =.40kgm/s –After Collision m =.47kg v =.04m/s p =.02kgm/s Ball 2 –Before Collision m =.47kg v = 0.0m/s p = 0.0kgm/s –After Collision m =.47kg v =.80m/s p =.38kgm/s

11. Continued… Notice that before the collision –p 1 + p 2 = (.40+0.0) =.40kgm/s After the collision –p 1 + p 2 = (.02+.38) =.40kgm/s This illustrates the Law of Conservation of Momentum. p 1i + p 2i = p 1f + p 2f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f

12. Example Problem A 76kg boater is at rest in a 45kg canoe next to a dock. The boater steps out of the canoe at a velocity of 2.5m/s to the right. What velocity will the boat move to the left at? Do Problems 1-3 on page 219 Separate sheet of paper.

13. 6-3 Elastic and Inelastic Collisions There are two basic types of collisions. The difference is whether or not the two objects separate (elastic) or stick together (inelastic).

14. Inelastic Collisions Perfectly inelastic collisions occur when two objects collide and stick together. After the collision we can treat the objects as though they were one. Because v f is the same for both objects you can factor it out –m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f becomes –m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f

15. Inelastic Collisions (cont) During inelastic collisions, some of the Kinetic energy gets converted to sound energy, internal energy, as well as elastic potential energy, as the objects deform from the force of the collision.

16. Example 1850kg Car is rear-ended by a 975kg car, the two cars entangle. If the 975kg car was moving at 22 m/s what is the final velocity of the entangled cars after the collisions? Practice problems 1-4 on own, problem 5 with class on board.

17. Example (cont) m 1 = 1850 m 2 = 975 v 1i = 0 m/s v 2i = 22.0 m/s v f = ?? (m 1 v 1i + m 2 v 2i ) / (m 1 + m 2 ) = v f (1850x0)+(975x22) / (1850+975) = v f 7.59m/s = v f

18. Elastic Collisions Elastic collisions result in the two objects moving separately after a collision. With elastic collisions both momentum and kinetic energy are conserved. To solve these problems you need to know all but one of the variables.

19. Example Sample problem 6G on board. Practice problems 1a&b and 2-4a