1. The “Big MO”

2. Momentum is the product of __________ X the _____________ of an object

3. The “Big MO” Momentum is the product of mass X the velocity of an object

4. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol:

5. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p

6. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: ?

7. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv

8. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar ???

9. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar

10. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar Unit: ?

11. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar Unit: kg m/s

12. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar Unit: kg m/s Other Forms of the momentum equation: v = ?m = ?

13. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar Unit: kg m/s Other Forms of the momentum equation: v = p / m m = ?

14. The “Big MO” Momentum is the product of mass X the velocity of an object Symbol: p Formula: p = mv Vector or Scalar Unit: kg m/s Other Forms of the momentum equation: v = p / m m = ǀ p ǀ / ǀ v ǀ

15. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity?

16. a)Given: m = 2.4 kg v = 4.0 m/s [W] p =?

17. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = ?

18. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv

19. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = ?

20. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = ?

21. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W]

22. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W] b)Given: m = 2.0 X 10 3 kg p = 8.0 X 10 4 kg m/s [S] v = ?

23. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W] b)Given: m = 2.0 X 10 3 kg p = 8.0 X 10 4 kg m/s [S] v = ? Formula: v = ?

24. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W] b)Given: m = 2.0 X 10 3 kg p = 8.0 X 10 4 kg m/s [S] v = ? Formula: v = p / m

25. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W] b)Given: m = 2.0 X 10 3 kg p = 8.0 X 10 4 kg m/s [S] v = ? Formula: v = p / m Sub: v = ?

26. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W] b)Given: m = 2.0 X 10 3 kg p = 8.0 X 10 4 kg m/s [S] v = ? Formula: v = p / m Sub: v = 8.0 X 10 4 kg m/s [S] / 2.0 X 10 3 kg = ?

27. Example: a) A 2.4 kg baseball is moving at 4.0 m/s [W]. What is its momentum? b) A 2.0 X 10 3 kg car is moving with a momentum of 8.0 X 10 4 kg m/s [S]. What is its velocity? a)Given: m = 2.4 kg v = 4.0 m/s [W] p =? Formula: p = mv Sub: p = 2.4 kg X 4.0 m/s [W] = 9.6 kg m/s [W] b)Given: m = 2.0 X 10 3 kg p = 8.0 X 10 4 kg m/s [S] v = ? Formula: v = p / m Sub: v = 8.0 X 10 4 kg m/s [S] / 2.0 X 10 3 kg = 40.0 m/s [S]

28. Alternate Form of Newton’s Second Law

29. What is Newton’s second law equation?

30. Alternate Form of Newton’s Second Law F net = m a

31. Alternate Form of Newton’s Second Law F net = m a In terms of delta notation, what is the defining equation for acceleration?

32. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t

33. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t So what is F net = ?

34. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t

35. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t What is the equation for ∆v in terms of final velocity and initial velocity?

36. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1

37. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 So what is F net = ?

38. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t

39. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ?

40. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t

41. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t But what is m v 2 = ?

42. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2

43. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 But what is m v 1 = ?

44. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1

45. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 What is F net in terms of final and initial momentum?

46. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t

47. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t What is ( p 2 – p 1 ) in terms of delta notation?

48. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t ( p 2 – p 1 ) = ∆ p

49. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t ( p 2 – p 1 ) = ∆ p So what is F net = ?

50. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t ( p 2 – p 1 ) = ∆ p therefore F net = ∆ p / ∆t

51. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t ( p 2 – p 1 ) = ∆ p therefore F net = ∆ p / ∆t Using the term “rate”, define net force in words.

52. Alternate Form of Newton’s Second Law F net = m a a = ∆v / ∆t therefore F net = m ∆v / ∆t ∆v = v 2 – v 1 therefore F net = m ( v 2 – v 1 ) / ∆t Expand the numerator F net = ( m v 2 – mv 1 ) / ∆t m v 2 = final momentum or p 2 m v 1 = initial momentum or p 1 F net = ( p 2 – p 1 ) / ∆t ( p 2 – p 1 ) = ∆ p therefore F net = ∆ p / ∆t The net force is the rate of change of momentum.

53. Alternate Form of Newton’s Second Law F net = ∆ p / ∆t The net force is the rate of change of momentum. This is another and better form or equation for Newton’s second law than F net = m a. This equation for Newton’s second law takes into consideration the case when mass changes as well as velocity, such as rocket motion.

54. Impulse

55. Let’s start with the alternate equation for Newton’s second law:

56. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t

57. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t If we multiply both sides of this equation by ∆t, what do we get?

58. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p

59. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”.

60. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. Why is “J” a vector ?

61. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p.

62. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = ?

63. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = Newton X second = Ns

64. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = Newton X second = Ns We use impulse for short time interval collisions, like cricket bat hitting ball, baseball bat hitting baseball etc.

65. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = Newton X second = Ns We use impulse for short time interval collisions, like cricket bat hitting ball, baseball bat hitting baseball etc. Equation for J = ?

66. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = Newton X second = Ns We use impulse for short time interval collisions, like cricket bat hitting ball, baseball bat hitting baseball etc. Equation J = ∆ p or F net ∆t

67. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = Newton X second = Ns We use impulse for short time interval collisions, like cricket bat hitting ball, baseball bat hitting baseball etc. Equation J = ∆ p or F net ∆t We can also calculate the impulse as the area under a “net force” vs “time of collision” curve: F net Time of contact during collision Area

68. Impulse Let’s start with the alternate equation for Newton’s second law: F net = ∆ p / ∆t F net ∆t = ∆ p “F net ∆t” is called “Impulse” and is given the symbol “J”. “J” is a vector b/c it is defined in terms of another vector ∆ p. Units of J = Newton X second = Ns We use impulse for short time interval collisions, like cricket bat hitting ball, baseball bat hitting baseball etc. Equation J = ∆ p or F net ∆t We can also calculate it as the area under a “net force” vs “time of collision” curve: The area can be approximated as two triangles as shown. F net Time of contact during collision Area

69. Example : A 50.00 g ping-pong ball moving horizontally at 6.00 m/s [right] strikes a wall. The ball makes contact with the wall for 0.0600 seconds. The ping-pong ball rebounds off the wall with a horizontal velocity of 4.00 m/s [left]. Ignoring any vertical forces, find: a) change in momentum of the ball b) impulse of the wall on the ball c) the average net horizontal force on the ball

70. a)∆ p = p 2 – p 1 = m v 2 – mv 1 = 0.05(-4.00) - 0.05(+6.00) ∆ p = - 0.200 – 0.300 = - 0.500 kg m/s or 0.500 kg m/s [L] b)J = ∆ p = 0.500 Ns [left] c)F net = ∆ p / ∆t = 0.500 kg m/s [r] /0.0600 s = 8.33 N [left]

71. Homework Try the Impulse and momentum problems handout and check answers New textbook: Read p222 – p225 Try q1,2 p223 q1,q2 p226 Q1-Q12 p227 check answers p717 Old Textbook: Read p232 – pp236 Try 6-10 p237 check same page