1. Chapter 6: Momentum and Collisions

2. Section 6 – 1 Momentum and Impulse

3. Momentum (p) A vector quantity defined as the product of an object’s mass and velocity. Describes an object’s motion.

4. Why “p”? –Pulse (Date: 14th century) from Latin pulsus, literally, beating, from pellere to drive, push, beat http://www.madsci.org/posts/archives/dec99/945106537.Ph.r.html

5. Momentum = mass x velocity p = mv Units: kg-m/s

7. Conceptualizing momentum Question – Which has more momentum; a semi-truck or a Mini Cooper cruising the road at 10 mph? Answer – The semi-truck has more mass. Since the velocities are the same, the semi has more momentum.

8. Conceptualizing momentum Question – Which has more momentum; a parked semi-truck or a Mini Cooper moving at 10 mph? Answer – The velocity of the semi is 0 mph. That means its momentum is 0 and this time the Mini Cooper has more momentum.

9. Conceptualizing momentum Question – Which has more momentum, a train moving at 1 mph or a bullet moving at 2000 mph? Answer – The mass of a train is very large, while the mass of a bullet is relatively small. Despite the large speed of the bullet, the train has more momentum.

10. Ex 1: Gary is driving a 2500 kg vehicle, what is his momentum if his velocity is 24 m/s?

11. G: m =2500 kg, v =24 m/s U: p =? E: p = mv S: p = (2500 kg)(24m/s) S: p = 60,000 kg-m/s

12. Ex 2: Ryan throws a 1.5 kg football, giving it a momentum of 23.5 kg- m/s. What is the velocity of the football?

13. G: m=1.5 kg, p =23.5 kg-m/s U: v = ? E: p = mv or v = p/m S:v=(23.5 kg-m/s)/(1.5kg) S: v = 15.7 m/s

14. A change in momentum pp (  p = m  v) Takes force and time.

15. Momentums do not always stay the same. When a force is applied to a moving object, the momentum changes.

16. Impulse The product of the force and the time over which it acts on an object, for a constant external force. Impulse = F  t

17. The change in the momentum is also called the impulse.

18. Impulse – Momentum Theorem F  t =  p or F  t = mv f - mv i

19. Ex 3: What is the impulse on a football when Greg kicks it, if he imparts a force of 70 N over 0.25 seconds? Also, what is the change in momentum?

20. G: F = 70 N,  t = 0.25 s U: Impulse = ? E: Impulse = F  t S: Impulse =(70 N)(0.25 s) S: Impulse = 17.5 kg-m/s F  t =  p = 17.5 kg-m/s

21. Ex 4: How long does it take a force of 100 N acting on a 50-kg rocket to increase its speed from 100 m/s to 150 m/s?

22. G: F = 100 N, v f = 150 m/s, v i = 100m/s, m = 50 kg U:  t = ? E:  t = m (v f - v i )/F S:  t =50kg(150–100m/s)/100 s S:  t = 25 sec

23. Stopping times and distances depend upon impulse- momentum theorem.

24. Ex 5: Crystal is driving a 2250 kg car west, she slows down from 20 m/s to 5 m/s. How long does it take the car to stop if the force is 8450 N to the east? How far does the car travel during this deceleration?

25. G: m = 2250 kg F = 8450 N east = 8450 N v i = 20 m/s west = - 20 m/s v f = 5 m/s west = - 5 m/s U:  t =? E: F  t =  p = m (v f - v i )  t = m (v f - v i ) / F

26. S:  t=[2250kg(-5– -20m/s)] 8450 N

27. S   t=[2250kg(-5– -20m/s)] 8450 N S: 4.0 s

28. B) U:  x = ? E:  x = ½(v i + v f )  t S:  x=½(-5+-20)m/s(4 s) S:  x = - 50 m  x = 50 m west

29. Frictional forces will be disregarded in most of the problems unless otherwise stated.

30. Section 6-2 Conservation of Momentum

31. Law of Conservation of Momentum The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between them.

32. What this means: Any momentum lost by one object in the system is gained by one or more of the other objects in the system.

33. total initial momentum total final momentum =

34. total initial momentum total final momentum =

35. total initial momentum total final momentum =

36. For objects that collide: The momentum of the individual object(s) does not remain constant, but the total momentum does.

37. Momentum is conserved when objects push away from each other. Ex 1: Jumping, Initially there is no momentum, but after you jump, the momentum of the you and the earth are equal and opposite.

38. Ex 2: 2 skateboarders pushing away from each other. Initially neither has momentum. After pushing off one another they both have the same momentum, but in opposite direction.

39. Which skateboarder has the higher velocity? The one with the smaller mass.

41. Ex 6: John, whose mass is 76 kg, is initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock with an a velocity of 2.5 m/s to the right. What is the final velocity of the boat?

42. G: m john =76 kg, m boat = 45kg, v john, i = v boat,i = 0 m/s, v john,f = 2.5 m/s U: v boat = ? E: Momentum is conserved. P J,i + p b,i = p J,f + p b,f 0 + 0 = m j v J,f + m b v b,f

43. v b,f = (m J v J,f ) /-m b

44. S: v b,f = (76kg x 2.5m/s) - 45kg

45. v b,f = (m J v J,f ) /-m b S: v b,f = (76kg x 2.5m/s) - 45kg S: v b,f = - 4.2 m/s

46. v b,f = (m J v J,f ) /-m b S: v b,f = (76kg x 2.5m/s) - 45kg S: v b,f = - 4.2 m/s or 4.2 m/s to the left

47. Newton’s 3 rd Law leads to a conservation of momentum. Open books to page 219-220

48. Forces in real collisions are not constant. They vary throughout the collision, but are still opposite and equal.

49. Section 6 – 3 Elastic and Inelastic Collisions

50. Perfectly Inelastic Collisions A collision in which two objects stick together and move with a common velocity.

52. Ex 7: A toy engine having a mass of 5.0 kg and a speed of 3 m/s, east, collides with a 4 kg train at rest. On colliding the two engines lock and remain together.

53. (a) What is the velocity of the entangled engines after the collision?

54. G: m 1 = 5.0 kg, m 2 =4.0kg v 1 = 3 m/s, & v 2 = 0 m/s U: v 1+2 = ? E: p 1,i + p 2,i = p 1+2 m 1 v 1 + m 2 v 2 =m 1+2 v 1+2

55. v 1+2 =[m 1 v 1 +m 2 v 2 ]/m 1+2 v 1+2 = [(5 x 3)+(4 x 0)]/9 v 1+2 = 1.66 m/s, east

56. KE is not constant in inelastic collisions. Some of the KE is converted into sound energy and internal energy as the objects are deformed.

57. This  KE can be calculated from the equation in Ch 5.  KE = KE f – KE i  KE =1/2mv f 2 - 1/2mv i 2

58. Ex 8: Two clay balls collide head on in a perfectly inelastic collision. The 1 st ball with a mass of 0.5 kg and an initial velocity of 4 m/s to the right. The 2 nd ball with a mass of 0.25 kg and an initial speed of 3 m/s to the left. a)What’s the final speed of the new ball after the collision? b)What’s the decrease in KE during the collision?

59. a) G:m 1,i =0.5 kg, v 1,i =4 m/s, m 2,i =0.25 kg, v 2,i =-3 m/s U: v f = ? E: m 1 v 1 + m 2 v 2 =(m 1 +m 2 )v f v f, =[m 1 v 1 +m 2 v 2 ]/m 1 +m 2

60. v f =[(0.5x4)+(0.25x-3)] 0.75

61. v 1+2,f = 1.67 m/s, to the right

62. b) U:  KE = ? E:  KE = KE f – KE i We need to find the combined final KE and both initial KE’s. Using the KE = ½ mv 2.

63. KE i = ½ mv 1,i 2 +½ mv 2,i 2 KE i =½(0.5)(4) 2 +½(0.25)(-3) 2 KE i = 5.12 J KE f = ½(m 1 +m 2 )v f 2 KE f = ½(0.5+0.25)(1.67) 2 KE f = 1.05J

64. S:  KE=1.05 J – 5.12 J S:  KE = - 4.07 J

66. Elastic Collisions A collision in which the total momentum and the total KE remains constant. Also, the objects separate and return to their original shapes.

67. In the real world, most collisions are neither elastic nor perfectly inelastic.

68. Total momentum and total KE remain constant through an elastic collision. m 1 v 1,I + m 2 v 2,I = m 1 v 1,f + m 2 v 2,f

69. ½m 1 v 1,i 2 + ½m 2 v 2,i 2 = ½m 1 v 1,f 2 + ½m 2 v 2,f 2

70. Ex 9: An 0.015 kg marble moving to the right, at 0.225 m/s has an elastic collision with a 0.03 kg moving to the left at 0.18 m/s. After the collision the smaller marble moves to the left at 0.315 m/s. Disregard friction. A) What is the velocity of the 0.03 kg marble after the collision? B) Verify answer by confirming KE is conserved.

71. G: m 1 =0.015 kg,m 2 =0.03 kg v 1,i = 0.225 m/s v 2,i = - 0.18 m/s v 1,f = - 0.315 m/s U: v 2,f = ?

72. E: m 1 v 1,i + m 2 v 2,i =m 1 v 1,f +m 2 v 2,f m 1 v 1,i =[m 1 v 1,f +m 2 v 2,f -m 2 v 2,i ]

73. E: m 1 v 1,i + m 2 v 2,i =m 1 v 1,f +m 2 v 2,f m 2 v 2,f =[m 1 v 1,i +m 2 v 2,i -m 1 v 1,f ] v 2,f = [m 1 v 1,i + m 2 v 2,i -m 1 v 1,f ] m2m2

74. S: v 2,f = [(0.015 x 0.225) + (0.03 x – 0.18) – (0.015 x – 0.315)] / 0.03 S: v 2,f = 0.09 m/s (right)

75. KE i = ½m 1 v 1,i 2 + ½m 2 v 2,i 2 KE i = ½(0.015)(0.225) 2 + ½(0.03)(-0.18) 2 KE i = 0.000866 J KE f = ½m 1 v 1,f 2 + ½m 2 v 2,f 2

76. KE f = ½(0.015)(0.315) 2 + ½(0.03)(0.09) 2 KE f = 0.000866 J Since KE i = KE f,, KE is conserved.