1. “The quality of Motion”

2. Momentum A vector quantity defined as the product of an objects mass and velocity.

3. In every day speech, momentum is often misused- in science, momentum has a very specific meaning Momentum = mass x velocity p = mv units are kgm/s momentum is a vector quantity that acts in the same direction as the velocity

4. p.209 Practice 6A Problems 1-3

5. Delta V V p = m v Watch the change in velocity when there is a change in direction. V = V f – V i Example; in a car crash, a car going 50mph west then 50mph east has a V of 100mph.

6. Impulse= force x time Impulse is what is needed to bring an object to a stop. Impulse is equivalent to momentum Look at Newton's 2 nd Law –F = ma F= m ( Δ V/ Δ T) –F Δ t = m Δ V –Impulse= momentum The units for impulse are Newton Seconds –Shouldn’t impulse & momentum share the same units…?

7. They do! NS = Kg m/s Kg m/s 2 sec Kg m/s = kg m/s F Δ t = Δ p This is the impulse momentum theorem. Often the force applied is NOT constant. If so we will use the average force over the time interval

8. Which is worse, 50mph car crash in a single crash with a wall, or 2 cars (of the same mass) in opposite directions going 50mph? http://www.youtube.com/watch?v=r8E 5dUnLmh4http://www.youtube.com/watch?v=r8E 5dUnLmh4 6min clip

9. Car Crash –Inertia –Momentum –Impulse –air bag vs. windshield http://www.youtube.com/watch ?v=d7iYZPp2zYY&feature=rela ted

10. Airbag Deployment Real time http://www.youtube.com/watch?v=_Av0 WGrlTGY Slow motion http://www.youtube.com/watch?v=A2fAg W_1nD0

11. Example Problem A 144g baseball is pitched horizontally at +38m/s. After it is hit by a bat it moves horizontally at – 38m/s a.what impulse did the bat deliver to the ball? b.If the bat and ball were in contact.8 milliseconds, what was the average force the bat exerted on the ball? c.Find the average acceleration of the ball during the contact with the bat

12. A Given m=.144 kg V i = +38m/s V f = -38m/s t =.0008sec Formula F Δ t = m Δ V Ft = m(V f - V i) Solution Ft =.144 (-38 – (+38) =.144(-76m/s) = -10.9NS (in direction of hit ball)

13. B Given Δ p = -10.9kgm/s Formula Δ p = F Δ t F= Δ p / Δ t Solution F = -10.9/.0008 F = -13,625N

14. C Given F= -13, 625N Formula F= ma a= F/m Solution a= -13,625/.144 a = -94,618 m/s 2

16. Bungee Jump

20. Take a look at how stopping time affects force.

21. p. 211 Practice 6B Problems 1-3 p.213 probs 1 & 2 6C

22. Momentum is a way to describe collisions In all collisions we must define our system (often this is naming the objects involved) Example: batter hits ball –we would likely define our system as the bat and ball –However, hands, batter body, shoes, dirt, & air can all affect momentum. –When we study momentum changes, we will use a closed, isolated system

23. Two Kinds of Collisions Elastic: objects bounce off each other after the collision or pass through. Inelastic: objects stick together or couple after they collide In BOTH collisions momentum is conserved Another type of interaction is an explosion. Two or more objects are together and then they move apart.

24. Law of conservation of momentum The momentum of any closed, isolated system, does NOT change P before collision = P after collision P tot = P’ tot

25. V = 2.2m/s 1000kg V = 0 Before m = 2000kg After V = 1.1m/s

26. P total = P’ total ’ means after collision P A + P B = P’ A + P’ B M A V A + M B V B = M A V’ A + M B V’ B (elastic) (M A +M B ) V’ AB (inelastic)

27. Look at previous box cars… 1000kg (2.2) + 1000kg (0) = (1000 + 1000) V’ MAMA VAVA MBMB VBVB MAMA MBMB V’ AB = 1000(2.2) + 1000 (0) 1000 + 1000 V’ AB = 2200 + 0 2000 V’ AB = V’ AB = 1.1m/s

28. Explosion Example A 3.8kg gun is fired. The 42 gram bullet leaves the muzzle at 470m/s. Find the guns recoil velocity.

29. Recoil – Explosion solution. M g x v g = M b x v b V g =.042kg x 470m/s 3.8kg V g = 5.2 m/s

30. Elastic Example problem Lab cart A of mass.355kg moves along with a velocity of.095m/s. Cart A collides with cart B of mass.71kg moving in the same direction with a velocity of.045m/s. After collision cart A has velocity of.035m/s. What is the velocity of cart B after they collide?

31. Given M A =.355kg V A =.095m/s V B =.045m/s M B =.71kg V’ A =.035m/s V’ B = ? elastic collision Formula P A + P B = P’ A + P’ B M A V A + M B V B = M A V’ A + M B V’ B M A V A + M B V B - M A V’ A = M B V’ B M A V A + M B V B - M A V’ A = V’ B M B Solution.355(.095) +.71(.045) -.355(.035) = V’ B.71.0337 +.032 -.0124 = V’ B.71 =.075m/s

32. Prac Prob #5 In a hockey game, a 75kg goalie catches a 105g puck in his leather glove. The puck was traveling at 48m/s when he caught it. The goalie was at rest when he caught it. Calculate his velocity after the catch.

33. 5 Given M P =.105 kg V P = 48m/s M G = 75kg V G = 0 V’ P+G = ? Inelastic Formula M P V P + M G V G = (M P + M G ) V PG M P V P + M G V G = V PG M PG Solution.105(48) + 0 75.105kg V PG =.067m/s

34. Prac. Prob #6 A bullet was fired into a block of wood. The bullet embedded into the wood and moved off with a velocity of 8.6m/s to the right. Find the bullets velocity before striking the wood. The bullets mass is 35 grams and the blocks mass is 5kg.

35. 6 Given M B =.035kg M W = 5kg V’ BW = 8.6m/s V W = 0 V B =? Inelastic Formula M B V B + M W V W = (M B + M W )V’ BW V B = (M B + M W )V BW – M W V W M B Solution V B = 5.035(8.6) – 5(0).035 V B = 1237.2m/s

36. Prac Prob #7 Another gun fires a bullet at a wooden block. This time the bullet goes through the block. Calculate the velocity of the wooden block after the bullet passes through. You know the bullets mass 35g, blocks mass 2.5kg, bullets initial speed 475m/s, and bullets final speed 275m/s.

37. 7 Given M B =.035kg V B = 475 m/s M W = 2.5kg V’ B = 275 m/s V W = 0m/s V’ W = ? elastic Formula M B V B + M W V W = M B V’ B + M W V’ W M B V B + M W V W – M B V’ B = V’ W M W Solution.035(475) + 2.5(0) -.035kg(275m/s) 2.5 16.63 + 0 – 9.63 2.5 2.8 m/s

38. Practice Problems p. 219,224, 226 p.219 1 and 2 P.224 1&2 p.226 1&2

39. Head-On Collision

40. Car “Rear Ends” Truck

41. Truck “Rear-Ends” Car

42. The Cart and the Brick

43. Overview Know –Momentum –Impulse –Collisions Law of Conservation of momentum Difference between elastic and inelastic collisions How mass, velocity, force, and time are inter- related. Remember old formulas from past chapters.

45. Momentum = Impulse

46. Force – time graph Time (sec) Force (N)

47. Dr. Weatherly jumps off the high dive (7m) and splashes into the swimming pool. It was measured that it took the water.7 seconds to bring him to a stop underwater. His mass is 90kg. A- What is his velocity as he hits the water? B- What is his momentum as he strikes the water? C- What is the average force the water puts on our Principal to bring him to rest underwater?