今日課程內容 CH8: 能量守恆 機械能與機械能守恆 能量守恆定律 重力位能 功率 CH9: 動量 動量、動量與力的關係 動量守恆、衝量 碰撞(一維與二維；彈性與非彈性)

8-4 Problem Solving Using Conservation of Mechanical Energy Conceptual Example 8-5: Speeds on two water slides. Two water slides at a pool are shaped differently, but start at the same height h. Two riders, Paul and Kathleen, start from rest at the same time on different slides. (a) Which rider, Paul or Kathleen, is traveling faster at the bottom? (b) Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length. Figure 8-9. Answer: a. The speeds are the same, due to conservation of energy. b. At every point, Kathleen has less potential energy than Paul, and therefore more kinetic energy. This means her speed is faster (until the end) and she reaches the bottom first.

8-4 Problem Solving Using Conservation of Mechanical Energy Example 8-6: Pole vault. Estimate the kinetic energy and the speed required for a 70-kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90 m off the ground and reaches its maximum height at the level of the bar itself. Figure 8-10. Caption: Transformation of energy during a pole vault. Solution: Assume the vaulter’s body is at rest at 9.0 m, and equate that total energy to the total energy when the running vaulter starts placing the pole. This gives a speed of about 9 m/s.

8-4 Problem Solving Using Conservation of Mechanical Energy For an elastic force, conservation of energy tells us: Example 8-7: Toy dart gun. A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (with spring stiffness constant k = 250 N/m and ignorable mass) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its natural length (x = 0), what speed does the dart acquire? Figure 8-12. Caption: Example 8-7. (a) A dart is pushed against a spring, compressing it 6.0 cm. The dart is then released, and in (b) it leaves the spring at velocity v2. Solution: Use conservation of energy, with the initial energy as elastic potential energy; v = 3.0 m/s.

8-4 Problem Solving Using Conservation of Mechanical Energy Example 8-8: Two kinds of potential energy. A ball of mass m = 2.60 kg, starting from rest, falls a vertical distance h = 55.0 cm before striking a vertical coiled spring, which it compresses an amount Y = 15.0 cm. Determine the spring stiffness constant of the spring. Assume the spring has negligible mass, and ignore air resistance. Measure all distances from the point where the ball first touches the uncompressed spring (y = 0 at this point). Figure 8-13. Solution: Conservation of energy again; as we are given both h and Y, can just equate energies in (a) and (c). Then k = 1590 N/m.

8-4 Problem Solving Using Conservation of Mechanical Energy Example 8-9: A swinging pendulum(擺). This simple pendulum consists of a small bob of mass m suspended by a massless cord of length l. The bob is released (without a push) at t = 0, where the cord makes an angle θ = θ0 to the vertical. (a) Describe the motion of the bob in terms of kinetic energy and potential energy. Then determine the speed of the bob (b) as a function of position θ as it swings back and forth, and (c) at the lowest point of the swing. (d) Find the tension in the cord, T. Ignore friction and air resistance. Figure 8-14. Caption: Example 8–9: a simple pendulum; y is measured positive upward. Solution: a. The bob swings back and forth in a symmetric arc as shown. Its speed is zero at the extreme points, and maximum in the middle. b. For any angle θ, equating the total energy there to the total energy at θ = θ0 gives v = [2gl(cos θ – cos θ0)]1/2 c. At the lowest point, y = 0, so v = [2gl(1 – cos θ0)]1/2 d. From the force diagram, remembering that the vertical component of FT equals mg, we find FT = (3 cos θ – 2 cos θ0)mg

8-5 The Law of Conservation of Energy(能量守恆定律) Nonconservative, or dissipative, forces: Friction Heat Electrical energy Chemical energy and more do not conserve mechanical energy. However, when these forces are taken into account, the total energy is still conserved:

8-5 The Law of Conservation of Energy The law of conservation of energy is one of the most important principles in physics. The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant.

8-6 Energy Conservation with Dissipative Forces: Solving Problems Example 8-10: Friction on the roller-coaster car. The roller-coaster car shown reaches a vertical height of only 25 m on the second hill before coming to a momentary stop. It traveled a total distance of 400 m. Figure 8-17. Caption: Example 8–10. Because of friction, a roller-coaster car does not reach the original height on the second hill. (Not to scale) Solution: The difference in the initial and final energies is the thermal energy produced, 147000 J. This is equal to the average frictional force multiplied by the distance traveled, so the average force is 370 N. Determine the thermal energy produced and estimate the average friction force (assume it is roughly constant) on the car, whose mass is 1000 kg.

8-6 Energy Conservation with Dissipative Forces: Solving Problems Example 8-11: Friction with a spring. A block of mass m sliding along a rough horizontal surface is traveling at a speed v0 when it strikes a massless spring head-on and compresses the spring a maximum distance X. If the spring has stiffness constant k, determine the coefficient of kinetic friction between block and surface. Figure 8-18. Solution: The difference between the initial energy (kinetic) and the final energy (elastic potential) is the work done by friction, which is the frictional force multiplied by the distance. Solving for the coefficient gives μk = (v02/2gX) – (kX/2mg)

8-7 Gravitational Potential Energy(重力位能)and Escape Velocity(脫離速度) Far from the surface of the Earth, the force of gravity is not constant: The work done on an object moving in the Earth’s gravitational field is given by: Figure 8-19. Caption: Arbitrary path of particle of mass m moving from point 1 to point 2.

8-7 Gravitational Potential Energy and Escape Velocity Solving the integral gives: Because the value of the integral depends only on the end points, the gravitational force is conservative and we can define gravitational potential energy: Figure 8-20. Caption: Gravitational potential energy plotted as a function of r, the distance from Earth’s center. Valid only for points r > rE , the radius of the Earth.

8-7 Gravitational Potential Energy and Escape Velocity Example 8-12: Package dropped from high-speed rocket. A box of empty film canisters is allowed to fall from a rocket traveling outward from Earth at a speed of 1800 m/s when 1600 km above the Earth’s surface. The package eventually falls to the Earth. Estimate its speed just before impact. Ignore air resistance. Solution: Use conservation of energy; v = 5320 m/s. (Air resistance will reduce this considerably, of course!)

8-7 Gravitational Potential Energy and Escape Velocity If an object’s initial kinetic energy is equal to the potential energy at the Earth’s surface, its total energy will be zero. The velocity at which this is true is called the escape velocity; for Earth:

8-7 Gravitational Potential Energy and Escape Velocity Example 8-13: Escaping the Earth or the Moon. (a) Compare the escape velocities of a rocket from the Earth and from the Moon. (b) Compare the energies required to launch the rockets. For the Moon, MM = 7.35 x 1022 kg and rM = 1.74 x 106 m, and for Earth, ME = 5.98 x 1024 kg and rE = 6.38 x 106 m. Solution: a. The ratio of the escape velocities is 4.7 (so the escape velocity from the Moon is 2400 m/s). b. The energy is proportional to the square of the velocity, so it takes 22 times more energy to escape Earth.

8-8 Power(功率) Power is the rate at which work is done. Average power: Instantaneous power: In the SI system, the units of power are watts:

8-8 Power Power can also be described as the rate at which energy is transformed: In the British system, the basic unit for power is the foot-pound per second, but more often horsepower is used: 1 hp = 550 ft·lb/s = 746 W.

8-8 Power Example 8-14: Stair-climbing power. A 60-kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m. (a) Estimate the jogger’s power output in watts and horsepower. (b) How much energy did this require? Figure 8-21. Solution: a. The work is the change in potential energy, and the power is the work divided by time: P = 660 W = 0.88 hp. b. The energy is 2600 J.

8-8 Power Power is also needed for acceleration and for moving against the force of friction. The power can be written in terms of the net force and the velocity:

8-8 Power Example 8-15: Power needs of a car. Calculate the power required of a 1400-kg car under the following circumstances: (a) the car climbs a 10° hill (a fairly steep hill) at a steady 80 km/h; and (b) the car accelerates along a level road from 90 to 110 km/h in 6.0 s to pass another car. Assume that the average retarding force on the car is FR = 700 N throughout. Figure 8-22. Caption: Example 8–15: Calculation of power needed for a car to climb a hill. Solution: a. If the speed is constant, the forces on the car must add to zero. This gives a force of 3100 N. Then the power is 6.8 x 104 W = 91 hp. b. The car must exert enough force to reach the given acceleration, as well as overcome the retarding force. This gives F = 2000 N. The maximum power is required at the maximum speed, 6.12 x 104 W or 82 hp.

8-9 Potential Energy Diagrams; Stable and Unstable Equilibrium(穩定與非穩定平衡) Figure 8-23. Caption: A potential energy diagram. With energy E1, the object oscillates between x3 and x2, called turning points. An object with energy E2 has four turning points; an object with energy E0 is in stable equilibrium. An object at x4 is in unstable equilibrium.

太空探索／海王星軌道L5點 首度發現特洛伊小行星                                      更新日期:2010/10/20 09:32 生活中心／台北報導 美國卡內基研究所（Carnegie Institution）天文學家Scott Sheppard和雙子北座天文台（(Gemini NNorth Observatory）Chad Trujillo兩人合作，首度在海王星軌道的L5拉朗朗日點（重力平衡穩定點）發現一顆特洛伊群小行星（Trojan asteroid）2008 LC18。 http://en.wikipedia.org/wiki/Lagrangian_point http://tw.news.yahoo.com/article/url/d/a/101020/17/2fap7.html

長庚大學畢業生應具備之核心能力 a.服務社會、肩負責任的使命感與主流價值的道德 觀 b.本國語文及英文的溝通能力 c.資訊應用之基本能力 d.人文、社會及自然科學之基礎知識 e.蒐集資料、分析數據、書面及口頭報告的能力 f.終身自我學習的能力 g.協調、溝通及團隊合作之能力 h.國際觀及國際競爭力

工學院大學生應具備核心能力 一、 運用數學、科學及工程知識的能力。 二、 設計與執行實驗，以及分析與解釋數據的能力 。 三、 執行工程實務所需技術、技巧及使用工具之能 力。 四、 設計工程系統、元件或製程之能力。 五、 有效溝通與團隊合作的能力。 六、 發掘、分析及處理問題的能力。 七、 認識時事議題，瞭解工程技術對環境、社會及 全球的影響，並培養持續學習的習慣與能力。 八、 理解專業倫理及社會責任。

運用資工領域、數學、科學及工程知識之能力Ability to apply Computer and Information domain knowledge, including mathematics, science and engineering skills 發掘、分析及處理資訊系統運作問題的能力Ability to discover, analyze and resolute information system problems 設計與執行資訊系統實驗，與分析實驗數據之能力Ability to design and conduct experiments upon information systems, and to analyze the experimental results 運用適當工具、儀器與現存模組執行資訊系統軟硬體開發之能力Ability to use equipments, proper tools and existing modules to develop software and hardware of information systems 規劃、建立資訊系統測試環境，及執行測試與結果分析之能力Ability to develop test environments for information systems, to run the test processes, and to analyze the test results 具備撰寫與運用資訊系統設計文件之能力Ability to read and write technical documents of information systems 具備有效溝通與合作之能力Ability to coordinate and cooperate with co-workers efficiently 理解專業倫理與社會責任Ability to take social responsibility with sufficiently understanding professionalism 瞭解資訊工程技術對環境、社會及全球的影響，並培養持續學習的習慣與能力 Ability to uphold the habits of self-study with understanding how information engineering may affect us in a local and global perspective

Chapter 9 Linear Momentum(線動量) Chapter Opener. Caption: Conservation of linear momentum is another great conservation law of physics. Collisions, such as between billiard or pool balls, illustrate this vector law very nicely: the total vector momentum just before the collision equals the total vector momentum just after the collision. In this photo, the moving cue ball strikes the 11 ball at rest. Both balls move after the collision, at angles, but the sum of their vector momenta equals the initial vector momentum of the incoming cue ball. We will consider both elastic collisions (where kinetic energy is also conserved) and inelastic collisions. We also examine the concept of center of mass, and how it helps us in the study of complex motion.

9-1 Momentum and Its Relation to Force Momentum(動量) is a vector symbolized by the symbol , and is defined as The rate of change of momentum is equal to the net force: This can be shown using Newton’s second law.

9-1 Momentum and Its Relation to Force Example 9-1: Force of a tennis serve. For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mass of 0.060 kg and is in contact with the racket for about 4 ms (4 x 10-3 s), estimate the average force on the ball. Would this force be large enough to lift a 60-kg person? Figure 9-1. Solution: The average force is the change in momentum divided by the time: 800 N. A 60-kg person weighs about 600 N, so this would be large enough.

9-1 Momentum and Its Relation to Force Example 9-2: Washing a car: momentum change and force. Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it. (That is, we ignore any splashing back.) What is the force exerted by the water on the car? Figure 9-2. Solution: Every second, 1.5 kg of water hits the car and stops. The force is the change in momentum (final momentum is zero) divided by the time, or 30 N.

9-2 Conservation of Momentum During a collision(碰撞), measurements show that the total momentum does not change: Figure 9-3. Caption: Momentum is conserved in a collision of two balls, labeled A and B.

9-2 Conservation of Momentum Conservation of momentum can also be derived from Newton’s laws. A collision takes a short enough time that we can ignore external forces. Since the internal forces are equal and opposite, the total momentum is constant. Figure 9-4. Caption: Collision of two objects. Their momenta before collision are pA and pB, and after collision are pA’ and pB’. At any moment during the collision each exerts a force on the other of equal magnitude but opposite direction.

9-2 Conservation of Momentum For more than two objects, Or, since the internal forces (內力)cancel,

9-2 Conservation of Momentum(動量守恆) This is the law of conservation of linear momentum: when the net external force (外力)on a system of objects is zero, the total momentum of the system remains constant. Equivalently, the total momentum of an isolated system remains constant.

9-2 Conservation of Momentum Example 9-3: Railroad cars collide: momentum conserved. A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their common speed immediately after the collision? Figure 9-5. Solution: Momentum is conserved; after the collision the cars have the same momentum. Therefore their common speed is 12.0 m/s.

9-2 Conservation of Momentum Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket. Figure 9-6. Caption: (a) A rocket, containing fuel, at rest in some reference frame. (b) In the same reference frame, the rocket fires, and gases are expelled at high speed out the rear. The total vector momentum, P = pgas + procket, remains zero.

9-2 Conservation of Momentum Example 9-4: Rifle recoil. Calculate the recoil velocity of a 5.0-kg rifle that shoots a 0.020-kg bullet at a speed of 620 m/s. Figure 9-7. Solution: Use conservation of momentum. The recoil velocity of the gun is -2.5 m/s.

9-2 Conservation of Momentum Conceptual Example 9-5: Falling on or off a sled. (a) An empty sled is sliding on frictionless ice when Susan drops vertically from a tree above onto the sled. When she lands, does the sled speed up, slow down, or keep the same speed? (b) Later: Susan falls sideways off the sled. When she drops off, does the sled speed up, slow down, or keep the same speed? The sled will slow down. The sled keeps going at the same speed (as does Susan).

9-3 Collisions(碰撞) and Impulse(衝量) During a collision, objects are deformed due to the large forces involved. Since , we can write Integrating, Figure 9-8. Caption: Tennis racket striking a ball. Both the ball and the racket strings are deformed due to the large force each exerts on the other.

9-3 Collisions and Impulse This quantity is defined as the impulse, J: The impulse is equal to the change in momentum:

9-3 Collisions and Impulse Since the time of the collision is often very short, we may be able to use the average force, which would produce the same impulse over the same time interval. Figure 9-9. Caption: Force as a function of time during a typical collision: F can become very large; Δt is typically milliseconds for macroscopic collisions. Figure 9-10. Caption: The average force acting over a very brief time interval gives the same impulse (FavgΔt) as the actual force.

9-3 Collisions and Impulse Example 9-6: Karate blow. Estimate the impulse and the average force delivered by a karate blow that breaks a board a few cm thick. Assume the hand moves at roughly 10 m/s when it hits the board. Figure 9-11. Solution: Take the mass of the hand plus a reasonable portion of the arm to be 1 kg; if the speed goes from 10 m/s to zero in 1 cm the time is 2 ms. This gives a force of 5000 N.

9-4 Conservation of Energy and Momentum in Collisions Momentum is conserved in all collisions. Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic. Figure 9-12. Caption: Two equal-mass objects (a) approach each other with equal speeds, (b) collide, and then (c) bounce off with equal speeds in the opposite directions if the collision is elastic, or (d) bounce back much less or not at all if the collision is inelastic.

9-5 Elastic Collisions(彈性碰撞) in One Dimension(一維) Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds. Figure 9-13. Caption: Two small objects of masses mA and mB, (a) before the collision and (b) after the collision.

9-5 Elastic Collisions in One Dimension Example 9-7: Equal masses. Billiard ball A of mass m moving with speed vA collides head-on(正向碰撞) with ball B of equal mass. What are the speeds of the two balls after the collision, assuming it is elastic? Assume (a) both balls are moving initially (vA and vB), (b) ball B is initially at rest (vB = 0). Figure 9-14. Caption: In this multi-flash photo of a head-on collision between two balls of equal mass, the white cue ball is accelerated from rest by the cue stick and then strikes the red ball, initially at rest. The white ball stops in its tracks and the (equal mass) red ball moves off with the same speed as the white ball had before the collision. See Example 9–7. Solution: a. Use both conservation of momentum and conservation of energy; the balls exchange velocities. b. Ball A stops, and ball B moves on with ball A’s original velocity.

9-5 Elastic Collisions in One Dimension Example 9-8: Unequal masses, target at rest. A very common practical situation is for a moving object (mA) to strike a second object (mB, the “target”) at rest (vB = 0). Assume the objects have unequal masses, and that the collision is elastic and occurs along a line (head-on). (a) Derive equations for vB’ and vA’ in terms of the initial velocity vA of mass mA and the masses mA and mB. (b) Determine the final velocities if the moving object is much more massive than the target (mA >> mB). (c) Determine the final velocities if the moving object is much less massive than the target (mA << mB). Solution: a. Both momentum and kinetic energy are conserved. The rest is algebra. b. In this case, vA′ doesn’t change much, and vB′ = 2 vA. c. In this case, vB’ remains zero, and mass A reverses its direction at the same speed.

9-5 Elastic Collisions in One Dimension Example 9-9: A nuclear(核子) collision. A proton (p) of mass 1.01 u (unified atomic mass units) traveling with a speed of 3.60 x 104 m/s has an elastic head-on collision with a helium (He) nucleus (mHe = 4.00 u) initially at rest. What are the velocities of the proton and helium nucleus after the collision? Assume the collision takes place in nearly empty space. Figure 9-15. Caption: Example 9–9: (a) before collision, (b) after collision. Solution: We don’t need to know what u is; all we need are the relative masses of the proton and the helium nucleus. Momentum and kinetic energy are conserved; solving the equations gives the helium nucleus’s velocity as 1.45 x 104 m/s and the proton’s as -2.15 x 104 m/s (backwards).

9-6 Inelastic Collisions(非彈性碰撞) With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one in which the objects stick together afterward, so there is only one final velocity.

9-6 Inelastic Collisions Example 9-10: Railroad cars again. A 10,000-kg railroad car, A, traveling at a speed of 24.0 m/s strikes an identical car, B, at rest. If the cars lock together as a result of the collision, how much of the initial kinetic energy is transformed to thermal or other forms of energy? Figure 9-5. Solution: Momentum is conserved; the initial kinetic energy is 2.88 x 106 J and the final kinetic energy is 1.44 x 106 J, so 1.44 x 106 J of kinetic energy have been transformed to other forms. Before collision After collision

9-6 Inelastic Collisions Example 9-11: Ballistic pendulum. The ballistic pendulum is a device used to measure the speed of a projectile, such as a bullet. The projectile, of mass m, is fired into a large block of mass M, which is suspended like a pendulum. As a result of the collision, the pendulum and projectile together swing up to a maximum height h. Determine the relationship between the initial horizontal speed of the projectile, v, and the maximum height h. Figure 9-16. Solution: This has two parts. First, there is the inelastic collision between the bullet and the block; we need to find the speed of the block. Then, the bullet+block combination rises to some maximum height; here we can use conservation of mechanical energy to find the height, which depends on the speed.

9-7 Collisions in Two or Three Dimensions Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities. Figure 9-18. Caption: Object A, the projectile, collides with object B, the target. After the collision, they move off with momenta pA’ and pB’ at angles θA’ and θB’. The objects are shown here as particles, as we would visualize them in atomic or nuclear physics. But they could also be macroscopic pool balls.

9-7 Collisions in Two or Three Dimensions Example 9-12: Billiard ball collision in 2-D. Billiard ball A moving with speed vA = 3.0 m/s in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45° to the x axis, ball A above the x axis and ball B below. That is, θA’ = 45° and θB’ = -45 °. What are the speeds of the two balls after the collision? Figure 9-19. Solution: Apply conservation of momentum; the masses are the same, as are the outgoing angles. The final speeds are equal; both are 2.1 m/s.

9-7 Collisions in Two or Three Dimensions Example 9-13: Proton-proton(質子) collision. A proton traveling with speed 8.2 x 105 m/s collides elastically with a stationary proton in a hydrogen target. One of the protons is observed to be scattered at a 60° angle. At what angle will the second proton be observed, and what will be the velocities of the two protons after the collision? Figure 9-18. Solution: We have three unknowns – the two outgoing velocities and the angle of the second proton – and three equations – momentum conservation in x and y, and conservation of kinetic energy. Solving gives the speed of the first proton to be 4.1 x 105 m/s, of the second to be 7.1 x 105 m/s, and the angle of the second to be -30°.

習題 Ch8:33, 36, 39, 47, 50, 57, 65, 68, 72,