1. Chapter 10 Lecture

2. Chapter 10 Energy
Chapter Goal: To introduce the concept of energy and the basic energy model.
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3. Chapter 10 Preview
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4. Chapter 10 Preview
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5. Chapter 10 Preview
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6. Chapter 10 Preview
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7. Chapter 10 Preview
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8. Chapter 10 Reading Quiz
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9. Reading Question 10.1 Kinetic energy is Mass times velocity.
½ mass times speed-squared.
The area under the force curve in a force-versus-time graph.
Velocity per unit mass.
Answer: C
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10. Reading Question 10.1 Kinetic energy is Mass times velocity.
½ Mass times speed-squared.
The area under the force curve in a force-versus-time graph.
Velocity per unit mass.
Answer: C
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11. Reading Question 10.2
A method for keeping track of transformations between kinetic energy and gravitational potential energy, introduced in this chapter, is
Credit-debit tables.
Kinetic energy-versus-time graphs.
Energy bar charts.
Energy conservation pools.
Energy spreadsheets.
Answer: B
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12. Reading Question 10.2
A method for keeping track of transformations between kinetic energy and gravitational potential energy, introduced in this chapter, is
Credit-debit tables.
Kinetic energy-versus-time graphs.
Energy bar charts.
Energy conservation pools.
Energy spreadsheets.
Answer: B
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13. Reading Question 10.3 Mechanical energy is
The energy due to internal moving parts.
The energy of motion.
The energy of position.
The sum of kinetic energy plus potential energy.
The sum of kinetic, potential, thermal, and elastic energy.
Answer: B
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14. Reading Question 10.3 Mechanical energy is
The energy due to internal moving parts.
The energy of motion.
The energy of position.
The sum of kinetic energy plus potential energy.
The sum of kinetic, potential, thermal, and elastic energy.
Answer: B
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15. Reading Question 10.4 Hooke’s law describes the force of Gravity.
A spring.
Collisions.
Tension.
None of the above.
Answer: B
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16. Reading Question 10.4 Hooke’s law describes the force of Gravity.
A spring.
Collisions.
Tension.
None of the above.
Answer: B
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17. Reading Question 10.5 A perfectly elastic collision is a collision
Between two springs.
That conserves thermal energy.
That conserves kinetic energy.
That conserves potential energy.
That conserves mechanical energy.
Answer: E
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18. Reading Question 10.5 A perfectly elastic collision is a collision
Between two springs.
That conserves thermal energy.
That conserves kinetic energy.
That conserves potential energy.
That conserves mechanical energy.
Answer: E
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19. Chapter 10 Content, Examples, and QuickCheck Questions
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20. Kinetic Energy K Kinetic energy is the energy of motion.
All moving objects have kinetic energy.
The more massive an object or the faster it moves, the larger its kinetic energy.
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21. Potential Energy U
Potential energy is stored energy associated with an object’s position.
The roller coaster’s gravitational potential energy depends on its height above the ground.
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22. Thermal Energy Eth
Thermal energy is the sum of the microscopic kinetic and potential energies of all the atoms and bonds that make up the object.
An object has more thermal energy when hot than when cold.
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23. The Basic Energy Model
Within a system, energy can be transformed from one type to another.
The total energy of the system is not changed by these transformations.
This is the law of conservation of energy.
Energy can also be transferred from one system to another.
The mechanical transfer of energy to a system via forces is called work.
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24. Kinetic Energy and Gravitational Potential Energy
The figure shows a before-and-after representation of an object in free fall.
One of the kinematics equations from Chapter 2, with ay = g, is:
Rearranging:
Multiplying both sides by ½m:
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25. Kinetic Energy and Gravitational Potential Energy
Define kinetic energy as an energy of motion:
Define gravitational potential energy as an energy of position:
The sum K + Ug is not changed when an object is in free fall. Its initial and final values are equal:
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26. Kinetic Energy and Gravitational Potential Energy
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27. QuickCheck 10.1
A child is on a playground swing, motionless at the highest point of his arc. What energy transformation takes place as he swings back down to the lowest point of his motion?
K  Ug
Ug  K
Eth  K
Ug  Eth
K  Eth
Answer: D
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28. QuickCheck 10.1
A child is on a playground swing, motionless at the highest point of his arc. What energy transformation takes place as he swings back down to the lowest point of his motion?
K  Ug
Ug  K
Eth  K
Ug  Eth
K  Eth
Answer: D
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29. QuickCheck 10.2
A skier is gliding down a gentle slope at a constant speed. What energy transformation is taking place?
K  Ug
Ug  K
Eth  K
Ug  Eth
K  Eth
Answer: B
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30. QuickCheck 10.2
A skier is gliding down a gentle slope at a constant speed. What energy transformation is taking place?
K  Ug
Ug  K
Eth  K
Ug  Eth
K  Eth
Answer: B
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31. Example 10.1 Launching a Pebble
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32. Example 10.1 Launching a Pebble
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33. Example 10.1 Launching a Pebble
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34. Energy Bar Charts A pebble is tossed up into the air.
The simple bar charts below show how the sum of K + Ug remains constant as the pebble rises and then falls.
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35. Energy Bar Charts
The figure below shows how to make an energy bar chart suitable for problem solving.
The chart is a graphical representation of the energy equation Kf + Ugf = Ki + Ugi.
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36. QuickCheck 10.3
Ball A has half the mass and eight times the kinetic energy of ball B. What is the speed ratio vA/vB? 16 4 2
1/4
1/16
Answer: B
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37. QuickCheck 10.3
Ball A has half the mass and eight times the kinetic energy of ball B. What is the speed ratio vA/vB? 16 4 2
1/4
1/16
Answer: B
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38. QuickCheck 10.4
Rank in order, from largest to smallest, the gravitational potential energies of the balls.
1 > 2 = 4 > 3
1 > 2 > 3 > 4
3 > 2 > 4 > 1
3 > 2 = 4 > 1
Answer: D
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39. QuickCheck 10.4
Rank in order, from largest to smallest, the gravitational potential energies of the balls.
1 > 2 = 4 > 3
1 > 2 > 3 > 4
3 > 2 > 4 > 1
3 > 2 = 4 > 1
Answer: D
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40. The Zero of Potential Energy
Amber and Bill use coordinate systems with different origins to determine the potential energy of a rock.
No matter where the rock is, Amber’s value of Ug will be equal to Bill’s value plus 9.8 J.
If the rock moves, both will calculate exactly the same value for Ug.
In problems, only Ug has physical significance, not the value of Ug itself.
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41. Example 10.2 The Speed of a Falling Rock
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42. Example 10.2 The Speed of a Falling Rock
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43. Example 10.2 The Speed of a Falling Rock
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44. Example 10.2 The Speed of a Falling Rock
ASSESS The figure below shows energy bar charts for Amber and Bill. despite their disagreement over the value of Ug, Amber and Bill arrive at the same value for vf and their Kf bars are the same height. You can place the origin of your coordinate system, and thus the “zero of potential energy,” wherever you choose and be assured of getting the correct answer to a problem.
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45. Gravitational Potential Energy on a Frictionless Surface – Slide 1 of 4
Figure (a) shows an object of mass m sliding along a frictionless surface.
Figure (b) shows a magnified segment of the surface that, over some small distance, is a straight line.
Define an s-axis parallel to the direction of motion
Newton’s second law along the axis is:
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46. Gravitational Potential Energy on a Frictionless Surface – Slide 2 of 4
Using the chain rule, we can write Newton’s second law as:
It is clear from the diagram that the net force along s is:
So Newton’s second law is:
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47. Gravitational Potential Energy on a Frictionless Surface – Slide 3 of 4
Rearranging, we obtain:
Note from the diagram that sin ds = dy, so:
Integrating this from “before” to “after”:
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48. Gravitational Potential Energy on a Frictionless Surface – Slide 4 of 4
With K = ½ mv2 and Ug = mgy, we find that:
The total mechanical energy for a particle moving along any frictionless smooth surface is conserved, regardless of the shape of the surface.
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49. QuickCheck 10.5
Starting from rest, a marble first rolls down a steeper hill, then down a less steep hill of the same height. For which is it going faster at the bottom?
Faster at the bottom of the steeper hill.
Faster at the bottom of the less steep hill.
Same speed at the bottom of both hills.
Can’t say without knowing the mass of the marble.
Answer: D
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50. QuickCheck 10.5
Starting from rest, a marble first rolls down a steeper hill, then down a less steep hill of the same height. For which is it going faster at the bottom?
Faster at the bottom of the steeper hill.
Faster at the bottom of the less steep hill.
Same speed at the bottom of both hills.
Can’t say without knowing the mass of the marble.
Answer: D
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51. QuickCheck 10.6
A small child slides down the four frictionless slides A–D. Rank in order, from largest to smallest, her speeds at the bottom.
vD > vA > vB > vC
vD > vA = vB > vC
vC > vA > vB > vD
vA = vB = vC = vD
Answer: D
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52. QuickCheck 10.6
A small child slides down the four frictionless slides A–D. Rank in order, from largest to smallest, her speeds at the bottom.
vD > vA > vB > vC
vD > vA = vB > vC
vC > vA > vB > vD
vA = vB = vC = vD
Answer: D
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53. Example 10.3 The Speed of a Sled
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54. Example 10.3 The Speed of a Sled
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55. Example 10.3 The Speed of a Sled
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56. Problem-Solving Strategy: Conservation of Mechanical Energy
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57. QuickCheck 10.7
Three balls are thrown from a cliff with the same speed but at different angles. Which ball has the greatest speed just before it hits the ground?
Ball A.
Ball B.
Ball C.
All balls have the same speed.
Answer: D
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58. QuickCheck 10.7
Three balls are thrown from a cliff with the same speed but at different angles. Which ball has the greatest speed just before it hits the ground?
Ball A.
Ball B.
Ball C.
All balls have the same speed.
Answer: D
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59. QuickCheck 10.8
A hockey puck sliding on smooth ice at 4 m/s comes to a 1-m-high hill. Will it make it to the top of the hill?
Yes.
No.
Can’t answer without knowing the mass of the puck.
Can’t say without knowing the angle of the hill.
Answer: A
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60. QuickCheck 10.8
A hockey puck sliding on smooth ice at 4 m/s comes to a 1-m-high hill. Will it make it to the top of the hill?
Yes.
No.
Can’t answer without knowing the mass of the puck.
Can’t say without knowing the angle of the hill.
Answer: A
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61. Restoring Forces and Hooke’s Law
The figure shows how a hanging mass stretches a spring of equilibrium length L0 to a new length L.
The mass hangs in static equilibrium, so the upward spring force balances the downward gravity force.
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62. Restoring Forces and Hooke’s Law
The figure shows measured data for the restoring force of a real spring.
s is the displacement from equilibrium.
The data fall along the straight line:
The proportionality constant k is called the spring constant.
The units of k are N/m.
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63. Hooke’s Law One end of a spring is attached to a fixed wall.
(Fsp)s is the force produced by the free end of the spring.
s = s – se is the displacement from equilibrium.
The negative sign is the mathematical indication of a restoring force.
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64. QuickCheck 10.9
The restoring force of three springs is measured as they are stretched. Which spring has the largest spring constant?
Answer: A
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65. QuickCheck 10.9
The restoring force of three springs is measured as they are stretched. Which spring has the largest spring constant?
Steepest slope.
Takes lots of force for a small displacement.
Answer: A
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66. Example 10.5 Pull Until It Slips
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67. Example 10.5 Pull Until It Slips
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68. Example 10.5 Pull Until It Slips
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69. Example 10.5 Pull Until It Slips
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70. Example 10.5 Pull Until It Slips
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71. Example 10.5 Pull Until It Slips
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72. Stick-Slip Motion Earthquakes are an example of stick-slip motion.
Tectonic plates are attempting to slide past each other, but friction causes the edges of the plates to stick together.
Large masses of rock are somewhat elastic and can be “stretched”.
Eventually the elastic force of the deformed rocks exceeds the friction force between the plates.
An earthquake occurs as the plates slip and lurch forward.
The slip can range from a few centimeters in a relatively small earthquake to several meters in a very large earthquake.
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73. Elastic Potential Energy
Springs and rubber bands store potential energy that can be transformed into kinetic energy.
The spring force is not constant as an object is pushed or pulled.
The motion of the mass is not constant-acceleration motion, and therefore we cannot use our old kinematics equations.
One way to analyze motion when spring force is involved is to look at energy before and after some motion.
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74. Elastic Potential Energy
The figure shows a before-and-after situation in which a spring launches a ball.
Integrating the net force from the spring, as given by Hooke’s Law, shows that:
Here K = ½ mv2 is the kinetic energy.
We define a new quantity:
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75. Elastic Potential Energy
An object moving without friction on an ideal spring obeys:
where
Because s is squared, Us is positive for a spring that is either stretched or compressed.
In the figure, Us has a positive value both before and after the motion.
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76. QuickCheck 10.10
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be
1.0 m/s.
2.0 m/s.
2.8 m/s
4.0 m/s.
16.0 m/s.
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77. QuickCheck 10.10
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be
1.0 m/s.
2.0 m/s.
2.8 m/s
4.0 m/s.
16.0 m/s.
Conservation of energy:
Double x double v
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78. QuickCheck 10.11
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be
1.0 m/s.
2.0 m/s.
2.8 m/s.
4.0 m/s.
16.0 m/s.
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79. QuickCheck 10.11
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be
1.0 m/s.
2.0 m/s.
2.8 m/s.
4.0 m/s.
16.0 m/s.
Conservation of energy:
Double k  increase v by square root of 2
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80. Example 10.6 A Spring-Launched Plastic Ball
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81. Example 10.6 A Spring-Launched Plastic Ball
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82. Example 10.6 A Spring-Launched Plastic Ball
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83. Example 10.6 A Spring-Launched Plastic Ball
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84. Example 10.6 A Spring-Launched Plastic Ball
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85. Energy Diagrams Potential energy is a function of position.
Functions of position are easy to represent as graphs.
A graph showing a system’s potential energy and total energy as a function of position is called an energy diagram.
Shown is the energy diagram of a particle in free fall.
Gravitational potential energy is a straight line with slope mg and zero y-intercept.
Total energy is a horizontal line, since mechanical energy is conserved.
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86. A Four-Frame Movie of a Particle in Free Fall
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87. Energy Diagrams
Shown is the energy diagram of a mass on a horizontal spring.
The potential energy (PE) is the parabola:
Us = ½k(x – xe)2
The PE curve is determined by the spring constant; you can’t change it.
You can set the total energy (TE) to any height you wish simply by stretching the spring to the proper length at the beginning of the motion.
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88. A Four-Frame Movie of a Mass Oscillating on a Spring
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89. Energy Diagrams Shown is a more general energy diagram.
The particle is released from rest at position x1.
Since K at x1 is zero, the total energy TE = U at that point.
The particle speeds up from x1 to x2.
Then it slows down from x2 to x3.
The particle reaches maximum speed as it passes x4.
When the particle reaches x5, it turns around and reverses the motion.
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90. Equilibrium Positions: Stable
Consider a particle with the total energy E2 shown in the figure.
The particle can be at rest at x2, but it cannot move away from x2: This is static equilibrium.
If you disturb the particle, giving it a total energy slightly larger than E2, it will oscillate very close to x2.
An equilibrium for which small disturbances cause small oscillations is called a point of stable equilibrium.
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91. Equilibrium Positions: Unstable
Consider a particle with the total energy E3 shown in the figure.
The particle can be at rest at x3, and it does not move away from x3: This is static equilibrium.
If you disturb the particle, giving it a total energy slightly larger than E3, it will speed up as it moves away from x3.
An equilibrium for which small disturbances cause the particle to move away is called a point of unstable equilibrium.
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92. Tactics: Interpreting an Energy Diagram
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93. Example 10.9 Balancing a Mass on a Spring
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94. Example 10.9 Balancing a Mass on a Spring
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95. Example 10.9 Balancing a Mass on a Spring
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96. Example 10.9 Balancing a Mass on a Spring
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97. Example 10.9 Balancing a Mass on a Spring
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98. Example 10.9 Balancing a Mass on a Spring
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99. QuickCheck 10.12
A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. In the region 1.0 m < x < 2.0 m, the particle is
Speeding up.
Slowing down.
Moving at constant speed.
I have no idea.
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100. QuickCheck 10.12
A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. In the region 1.0 m < x < 2.0 m, the particle is
Speeding up.
Slowing down.
Moving at constant speed.
I have no idea.
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101. QuickCheck 10.13
A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. Where is the particle’s turning point?
1.0 m.
2.0 m.
5.0 m.
6.0 m.
It doesn’t have a turning point.
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102. QuickCheck 10.13
A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. Where is the particle’s turning point?
1.0 m.
2.0 m.
5.0 m.
6.0 m.
It doesn’t have a turning point.
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103. QuickCheck 10.14
A particle with this potential energy could be in stable equilibrium at x =
0.0 m.
1.0 m.
2.0 m.
Either A or C.
Either B or C.
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104. QuickCheck 10.14
A particle with this potential energy could be in stable equilibrium at x =
0.0 m.
1.0 m.
2.0 m.
Either A or C.
Either B or C.
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105. Molecular Bonds
Shown is the energy diagram for the diatomic molecule HCl (hydrogen chloride).
x is the distance between the hydrogen and the chlorine atoms.
The molecule has a stable equilibrium at an atomic separation of xeq = 0.13 nm.
When the total energy is E1, the molecule is oscillating, but stable.
If the molecule’s energy is raised to E2, we have broken the molecular bond.
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106. Elastic Collisions
During an inelastic collision of two objects, some of the mechanical energy is dissipated inside the objects as thermal energy.
A collision in which mechanical energy is conserved is called a perfectly elastic collision.
Collisions between two very hard objects, such as two billiard balls or two steel balls, come close to being perfectly elastic.
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107. A Perfectly Elastic Collision
Consider a head-on, perfectly elastic collision of a ball of mass m1 and initial velocity (vix)1, with a ball of mass m2 initially at rest.
The balls’ velocities after the collision are (vfx)1 and (vfx)2.
Momentum is conserved in all isolated collisions.
In a perfectly elastic collision in which potential energy is not changing, the kinetic energy must also be conserved.
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108. A Perfectly Elastic Collision
Simultaneously solving the conservation of momentum equation and the conservation of kinetic energy equations allows us to find the two unknown final velocities.
The result is:
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109. A Perfectly Elastic Collision: Special Case 1
Consider a head-on, perfectly elastic collision of a ball of mass m1 and initial velocity (vix)1, with a ball of mass m2 initially at rest.
Case 1: m1 = m2.
Equations give vf1 = 0 and vf2 = vi1.
The first ball stops and transfers all its momentum to the second ball.
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110. A Perfectly Elastic Collision: Special Case 2
Consider a head-on, perfectly elastic collision of a ball of mass m1 and initial velocity (vix)1, with a ball of mass m2 initially at rest.
Case 2: m1 >> m2.
Equations give vf1  vi1 and vf2  2vi1.
The big first ball keeps going with about the same speed, and the little second ball flies off with about twice the speed of the first ball.
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111. A Perfectly Elastic Collision: Special Case 3
Consider a head-on, perfectly elastic collision of a ball of mass m1 and initial velocity (vix)1, with a ball of mass m2 initially at rest.
Case 3: m1 << m2.
Equations give vf1 ≈ −vi1 and vf2 ≈ 0.
The little first rebounds with about the same speed, and the big second ball hardly moves at all.
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112. Perfectly Elastic Collisions: Using Reference Frames
Equations assume ball 2 is at rest.
What if you need to analyze a head-on collision when both balls are moving before the collision?
You could solve the simultaneous momentum and energy equations, but there is an easier way.
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113. Using Reference Frames: Quick Example
A 200 g ball moves to the right at 2.0 m/s. It has a head-on, perfectly elastic collision with a 100 g ball that is moving toward it at 3.0 m/s. What are the final velocities of both balls?
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114. Using Reference Frames: Quick Example
Figure (a) shows the situation just before the collision in the lab frame L.
Figure (b) shows the situation just before the collision in the frame M that is moving along with ball 2.
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115. Using Reference Frames: Quick Example
We can use Equations to find the post-collision velocities in the moving frame M:
Transforming back to the lab frame L:
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116. Chapter 10 Summary Slides

117. General Principles
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118. General Principles
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119. Important Concepts
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120. Important Concepts
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121. Important Concepts
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