1. PHY 151: Lecture 8 8.1 Linear Momentum 8.2 Isolated System (Momentum) 8.3 Nonisolated System (Momentum) 8.4 Collisions in One Dimension 8.5 Collisions in Two Dimensions 8.6 The Center of Mass 8.7 Motion of a System of Particles 8.8 Context Connection: Rocket Proposal

2. PHY 151: Lecture 8 Momentum and Collisions 8.1 Linear Momentum

3. Two particles which interact obey Newton’s third law: –We can express this as –Replacing force: –Replacing acceleration: Linear Momentum - 1

4. Since masses are constant, bring them inside the derivative operation: Since the derivative of the sum is zero, the sum must be constant: The sum of the product of masses and velocities for a particles in an isolated system is conserved. Linear Momentum - 2

5. Linear Momentum - 3 –Linear momentum is a vector quantity –SI unit: kg·m/s –In three dimensions, components are

6. Linear Momentum - 4 We can relate linear momentum to the net force acting on the particle: The time rate of change of momentum of a particle is equal to the net force acting on the particle If the mass is constant:

7. PHY 151: Lecture 8 Momentum and Collisions 8.2 Isolated System (Momentum)

8. Isolated System (Momentum) - 1 In an isolated system, the total momentum is conserved: The momentum components of an isolated system are independently constant:

9. Example 8.1 We claimed that we can ignore the kinetic energy of the Earth when considering the energy of a system consisting of the Earth and a dropped ball. Verify this claim. Set up the ratio of the KE of the Earth to the ball: Apply the isolated system (momentum) model:

10. Example 8.1 Solve for the ratio of speeds: Substitute into Equation (1): Substitute order-of-magnitude numbers for masses:

11. Example 8.2 A 60-kg archer stands at rest on frictionless ice and fires a 0.030-kg arrow horizontally at 85 m/s. With what velocity does the archer move across the ice after firing the arrow? Use the isolated system (momentum) model: Solve for the velocity and substitute numerical values:

12. Example 8.3 One type of nuclear particle, called the neutral kaon (K 0 ), decays into a pair of other particles called pions (  + and   ), which are oppositely charged but equal in mass. Assuming that the kaon is initially at rest, prove that the two pions must have momenta that are equal in magnitude and opposite in direction.

13. Example 8.3 Write an expression for decay of the kaon: Find an expression for the final momentum of the system: Incorporate the conservation of momentum:

14. PHY 151: Lecture 8 Momentum and Collisions 8.3 Nonisolated System (Momentum)

15. Nonisolated System (Momentum)- 1 From Newton’s second law, Integrating to find the change in momentum over some time interval The impulse is defined as

16. Nonisolated System (Momentum) -2 Impulse-momentum theorem: –The impulse of the force acting on a particle equals the change in the momentum of the particle –This is equivalent to Newton’s second law

17. Nonisolated System (Momentum) - 3 An impulse given to a particle means that momentum is transferred from an external agent to the particle The equation is called the conservation of momentum equation Left side represents the change in momentum Right side is a measure of how much momentum crosses the boundary of the system due to the net force being applied

18. Nonisolated System (Momentum) - 4 This represents nonisolated system (momentum) Net force on a particle can vary in time, so define a time-averaged net force: Which gives us

19. Nonisolated System (Momentum) - 5 Impulse approximation: assume that one of the forces exerted on a particle acts for a short time but is much greater than any other force present This force is called the impulsive force

20. Example 8.4 In a particular crash test, a car of mass 1500 kg collides with a wall. The initial and final velocities of the car are If the collision lasts 0.150 s, find the impulse caused by the collision and the average net force exerted on the car.

21. Example 8.4 The collision time is short We can image the car being brought to rest very rapidly and then moving back in the opposite direction with a reduced speed Assume net force exerted on the car by wall and friction with the ground is large compared with other forces Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum

22. Example 8.4 Evaluate the initial and final momenta of the car: Find the impulse on the car: Evaluate the average net force exerted on the car:

23. PHY 151: Lecture 8 Momentum and Collisions 8.4 Collisions in One Dimension

24. Collisions in One Dimension - 1 We use the term collision to represent an event during which two particles come close to each other and interact by means of forces The time interval during which the velocity changes from its initial to final values is assumed to be short The interaction force is assumed to be much greater than any external forces present This means the impulse approximation can be used

25. Collisions in One Dimension - 2 Collisions may be the result of physical contact The impulsive forces may vary in time in complicated ways. This force is internal to the system Momentum is conserved Collisions can also occur for particles that were never in contact For example, two positively charged ions repel

26. Collisions in One Dimension - 3 Elastic collision: momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic collisions actually occur Inelastic collision: kinetic energy is not conserved, momentum is still conserved If the objects stick together after the collision, it is a perfectly inelastic collision In an inelastic collision, some kinetic energy is lost, but the objects do not stick together

27. Collisions in One Dimension - 4 Elastic, and perfectly inelastic, collisions are limiting cases, most actual collisions fall in between these two types Momentum is conserved in all collisions

28. Collisions in One Dimension - 5 For perfectly inelastic collisions, use conservation of momentum:

29. Collisions in One Dimension - 6

30. Collisions in One Dimension – 6a For perfectly elastic collisions, momentum and kinetic energy are conserved: which leads to

31. Collisions in One Dimension - 7 For elastic collision in one dimension, particle 2 initially at rest:

32. Example 8.5 We claimed that the maximum amount of kinetic energy was transformed to other forms in a perfectly inelastic collision. Prove this statement mathematically for a one-dimensional two-particle collision. Find an expression for the ratio of the final kinetic energy after the collision to the initial kinetic energy:

33. Example 8.5 Minimize the fraction f:

34. Example 8.5 Differentiate the conservation of momentum equation: Substitute this expression for the derivative into (1):

35. Example 8.6 An 1800-kg car stopped at a traffic light is struck from the rear by a 900-kg car. The two cars become entangled, moving along the same path as that of the originally moving car. If the smaller car were moving at 20.0 m/s before the collision, what is the velocity of the entangled cars after the collision?

36. Example 8.6 Set the initial momentum of the system equal to the final momentum of the system: Solve for the final velocity and substitute numerical values:

37. Example 8.7 In a nuclear reactor, neutrons are produced when uranium-235 atoms split in a process called fission. These neutrons are moving at about 10 7 m/s and must be slowed down to about 10 3 m/s before they take part in another fission event. They are slowed down by being passed through a solid or liquid material called a moderator. The slowing-down process involves elastic collisions.

38. Example 8.7 Let us show that a neutron can lose most of its kinetic energy if it collides elastically with a moderator containing light nuclei, such as deuterium (in “heavy water,” D 2 O).

39. Example 8.7 Find an expression for the initial kinetic energy of the neutron: Find an expression for the final kinetic energy of the neutron:

40. Example 8.7 Find an expression for the fraction of the total kinetic energy possessed by the neutron after the collision: Find an expression for the kinetic energy of the moderator nucleus after the collision:

41. Example 8.7 Find an expression for the fraction of the total kinetic energy transferred to the moderator nucleus:

42. Example 8.8 A block of mass m 1 = 1.60 kg initially moving to the right with a speed of 4.00 m/s on a frictionless, horizontal track collides with a light spring attached to a second block of mass m 2 = 2.10 kg initially moving to the left with a speed of 2.50 m/s. The spring constant is 600 N/m.

43. Example 8.8 (A) Find the velocities of the two blocks after the collision Apply conservation of momentum: Since collision is elastic: Multiply Equation (2) by m 1 : Add Equations (1) and (3):

44. Example 8.8 Solve for v 2f : Substitute numerical values: Solve for v 1f and substitute numerical values:

45. Example 8.8 (B) Determine the velocity of block 2 during the collision, at the instant block 1 is moving to the right with a velocity of +3.00 m/s. Apply conservation of momentum:

46. Example 8.8 Solve for v 2f : Substitute numerical values:

47. Example 8.8 (C) Determine the distance the spring is compressed at that instant. Write a conservation of mechanical energy equation for the system: Evaluate the energies:

48. Example 8.8 Substitute: Solve for x:

49. PHY 151: Lecture 8 Momentum and Collisions 8.5 Collisions in Two Dimensions

50. Collisions in Two Dimensions - 1 For a general collision of two objects in three-dimensional space: total momentum in each direction is conserved If equation is elastic:

51. Example 8.9 A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.50×10 5 m/s and makes a glancing collision with the second proton (at close separations, the protons exert a repulsive electrostatic force on each other.)

52. Example 8.9 After the collision, one proton moves off at an angle of 37.0  to the original direction of motion and the second deflects at an angle of  to the same axis. Find the final speeds of the two protons and the angle .

53. Example 8.9 Use the isolated system models for energy and momentum: Rearrange Equations (1) and (2):

54. Example 8.9 Square these equations and add them: Use the sum of the squares of sine and cosine equals 1:

55. Example 8.9 Substitute Equation (4) into Equation (3): Solve for v 1f :

56. Example 8.9 Find v 2f : Find  :

57. Example 8.10 A 1500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg truck traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

58. Example 8.10 Equate the initial and final momenta of the system in the x direction: Equate the initial and final momenta of the system in the y direction: Divide Equation (2) by Equation (1):

59. Example 8.10 Solve for  and substitute numerical values: Use Equation (2) to find the value of v f and substitute numerical values:

60. PHY 151: Lecture 8 Momentum and Collisions 8.6 The Center of Mass

61. The Center of Mass - 1 There is a special point in a system or object, called the center of mass, that moves as if all of the mass of the system is concentrated at that point The system will move as if an external force were applied to a single particle of mass M located at the center of mass

62. The Center of Mass - 2 The coordinates of the center of mass are where M is the total mass of the system

63. The Center of Mass - 3 The center of mass can also be located by its position vector: where for the ith particle: If the number of elements approach infinity (and the size and mass of each element approach zero):

64. The Center of Mass - 4 The center of mass is often confused with the center of gravity The center of gravity is the average position of the gravitational forces on all parts of the object It can be determined experimentally

65. Example 8.11 A system consists of three particles located as shown. Find the center of mass of the system. The masses of the particles are m 1 = m 2 = 1.0 kg and m 3 = 2.0 kg.

66. Example 8.11 Use the defining equations for the coordinates of the center of mass: Write the position vector of the center of mass:

67. Example 8.12 (A) Show that the center of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length Find an expression for x CM : Substitute:

68. Example 8.12 (B) Suppose a rod is nonuniform such that its mass per unit length varies linearly with x according to the expression =  x, where  is a constant. Find the x coordinate of the center of mass as a fraction of L. Find an expression for x CM :

69. Example 8.12 Find the total mass of the rod: Substitute:

70. PHY 151: Lecture 8 Momentum and Collisions 8.7 Motion of a System of Particles

71. Motion of a System of Particles - 1 The velocity of the center of mass of a system is The acceleration of the center of mass of a system is The net force on a system is due only to external forces, so Newton’s second law for a system of particles is

72. Motion of a System of Particles - 2 The center of mass of a system of particles having combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system

73. Motion of a System of Particles - 3 The impulse-momentum theorem for a system of particles is The total linear momentum of a system of particles is constant if no external forces act on the system:

74. Example 8.13 A rocket is fired vertically upward. At the instant it reaches an altitude of 1000 m and a speed of v i = 300 m/s, it explodes into three fragments having equal mass. One fragment moves upward with a speed of v 1 = 450 m/s following the explosion. The second fragment has a speed of v 2 = 240 m/s and is moving east right after the explosion. What is the velocity of the third fragment immediately after the explosion?

75. Example 8.13 Use the isolated system (momentum) model, and express momenta in terms of masses and velocities: Solve for the velocity: Substitute numerical values:

76. PHY 151: Lecture 8 Momentum and Collisions 8.8 Rocket Propulsion

77. Rocket Propulsion - 1 The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel

78. Rocket Propulsion - 2 Suppose at some time t the momentum of the rocket (mass M) plus fuel (mass m) is (M +  m) During time interval  t, rocket ejects fuel  m and speed increases to v +  v

79. Rocket Propulsion - 3 Equating initial and final momenta: Taking the limit, and since dm =  dM:

80. Rocket Propulsion - 4 Integrating, we get The thrust on the rocket is the force exerted on the rocket by the ejected exhaust gases:

81. Example 8.14 A rocket moving in space, far from all other objects, has a speed of 3.0×10 3 m/s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket’s motion at a speed of 5.0×10 3 m/s relative to the rocket.

82. Example 8.14 (A) What is the speed of the rocket relative to the Earth once the rocket’s mass is reduced to half its mass before ignition? –Solve for the final velocity and substitute:

83. Example 8.14 (B) What is the thrust on the rocket if it burns fuel at the rate of 50 kg/s? Use the equation for thrust:

84. Momentum Change Example – A1 A 0.015-kg rubber bullet traveling with a speed of 150 m/s in +x-direction hits a wall Bullet bounces straight back with speed of 120 m/s What is change in momentum of bullet? –  p = p f – p i = mv f – mv i –  p = (0.015 kg)[-120 – (+150)] = -4.05 kg(m/s)

85. Impulse-Momentum Theorem Example A2 A 0.20-kg softball is tossed upward and hit horizontally by a batter Softball receives an impulse of 3.0 Ns With what horizontal speed does the ball move away from the bat?  J = p f – p i = mv f - mv i  Initial momentum is 0  J = mv f  v f = J/m = 3/0.20 = 15 m/s

86. Impulse-Momentum Theorem Example A3 Automobile with linear momentum of 3.0x10 4 kg(m/s) in +x-direction is brought to a stop in 5.0 s What is the average braking force?  J = Ft = p f – p i  Final momentum is zero  F = – p i /t = -(+3.0 x 10 4 ) / 5 = - 6000 N

87. Impulse-Momentum Theorem Example A4 A pool player imparts an impulse of 3.2 Ns to a stationary 0.25-kg cue ball with a cue stick What is the speed of the ball just after impact?  J = F  t = p f – p i = mv f - mv i  Initial momentum is zero  J = mv f  v f = J/m = 3.2/0.25 = 12.8 m/s

88. Impulse-Momentum Theorem Example A5 A car of 1500kg is traveling in +x-direction at 20m/s Force acts on car in the +x-direction for 30 seconds Final velocity of car is 40m/sec in the +x- direction What is the force?  J = F  t = p f – p i = mv f - mv i  J = 1500(+40) – 1500(+20)  J = 60000 – 30000 = 30000  F = 30000/  t = 30000/30 = 1000 N

89. Impulse-Momentum Theorem Example A6 Machine gun fires 15 bullets (mass 0.020 kg) per second Velocity of bullets is 800 m/sec in the +x- direction What is the force on the machine gun?  Change in momentum of 1 bullet = mv f – mv i = (0.020)800 = 16  Change in momentum for all bullets in second = 16(15) = 240 N  This is average force on the bullets  Force on gun is equal and opposite, -240 N

90. Conservation of Momentum Example B1 A 60-kg astronaut floats at rest in space He throws his 0.50-kg hammer such that it moves with a speed of 10 m/s in the -x-direction What happens to the astronaut?  Astronaut is object 1 and hammer is object 2  Momentum is conserved, p f = p i  Initial momentum is zero  m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i = 0  60v 1f + 0.5(-10) = 0  v 1f = 5/60 = +0.0833 m/s

91. Conservation of Momentum Example B2 An 800-kg car travelling in the +x-direction with a velocity of 30 m/sec It hits a stationary truck of 3200 kg The car and truck stick together What is speed of combined masses after the collision?  The car is object 1 and the truck object 2  p f = p i  (m 1 + m 2 )v f = m 1 v 1i + m 2 v 2i  4000v f = 800(+30) + 3200(0) = 24000  v f = 24000 / 4000 = +6 m/s

92. Conservation of Momentum Example B3 A car of mass 1200 kg is traveling 30 m/s in the +x-direction The car collides with a truck of 3200 kg traveling in the –x-direction at 20 m/s After the collision, the truck comes to a stop What is the velocity of the car?  The car is object 1 and the truck object 2  p f = p i  m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i  (1200)v 1f = 1200(30) + (3200)(-20) = -28000  v 1f = -28000/1200 = -23.33 m/s

93. Example: Stress Reliever - 1 Imagine one ball coming in from the left and two balls exiting from the right Is this possible? Due to shortness of time, the impulse approximation can be used Categorize the system as isolated in terms of both momentum and energy Elastic collisions

94. Example: Stress Reliever - 2 For a collision to actually occur, both momentum and kinetic energy must be conserved –One way to do so is with equal numbers of balls released and exiting –Another way is to have some of the balls taped together so they move as one object

95. Collision Example–Ballistic Pendulum - 1 Observe diagram The projectile enters the pendulum, which swings up to some height where it momentarily stops Isolated system in terms of momentum for the projectile and block Perfectly inelastic collision – the bullet is embedded in the block of wood Momentum equation will have two unknowns Use conservation of energy from the pendulum to find the velocity just after the collision Then you can find the speed of the bullet

96. Ballistic Pendulum - 2 A multi-flash photograph of a ballistic pendulum Equations for the momentum and conservation of energy with kinetic and gravitational potential energies Solve resulting system of equations

97. Ballistic Pendulum - 3 When bullet collides with the block momentum is conserved The initial velocity and momentum of m 2 is 0 (m 1 +m 2 )v B = m 1 v 1A v B = m 1 v 1A /(m 1 + m 2 ) As the block rises mechanical energy is conserved Initial potential energy is 0 and final kinetic energy is 0 K B = ½(m 1 + m 2 )v B 2 K B = m 1 2 v 1A 2 /2(m 1 +m 2 ) K B = U C m 1 2 v 1A 2 /2(m 1 +m 2 ) = (m 1 + m 2 )gh v 1A = [(m 1 +m 2 )/m 1 ]sqrt(2gh)