In this lecture you ll learn 簡介 How to find the center of mass of a system of particles 多粒子系統的質心問題 The principle of conservation of momentum, and how

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1 Chapter 9 Lecture Essential University Physics Richard Wolfson 2 nd Edition Systems of Particles 多粒子系統 2012 Pearson Education, Inc. Slide 9-1

In this lecture you ll learn 簡介 How to find the center of mass of a system of particles 多粒子系統的質心問題 The principle of conservation of momentum, and how it applies to

2 In this lecture you ll learn 簡介 How to find the center of mass of a system of particles 多粒子系統的質心問題 The principle of conservation of momentum, and how it applies to systems of particles 動量守恆與其應用 How to analyze collisions 分析兩個粒子 ( 物體 ) 碰撞的問題 - Inelastic collisions 彈性碰撞 - Elastic collisions 非彈性碰撞 2012 Pearson Education, Inc. Slide 9-2

3 Center of Mass 質心 The center of mass of a composite object or system of particles is the point where, from the standpoint of Newton s second law, the mass acts as though it were concentrated. 以牛頓第二定律來看, 多粒子系統的質心是各粒子質量的集中點 The position of the center of mass is a weighted average of the positions of the individual particles: 定義質心的位置就是加權之後的各粒子的平均位置 For a system of discrete particles, 對分散一個一個的多粒子系統 (? 後面有解釋 ) For a continuous distribution of matter, 物體的質量是連續分佈的時候 In both cases, M is the system s total mass. 在以上兩種情形, M 都是總 ( 的 ) 質量 m ri r cm = i M r cm = r dm 2012 Pearson Education, Inc. Slide 9-3 M

4 介紹一個方便的記號法, 例如有 3 個粒子的時候 : r = (m r1 + m r2 + m r3 ) / M cm 可以用以下的符號記法 : m 1 r1 + m 2 r2 + m 3 r3 = m ri,i =1,2,3. i 因此就有 : r cm = m i ri 3 個粒子的話, 當然 i=1,2,3. 其他的情況就依此類推 N 個粒子的時候就用 i=1,n. 來表示 ( 重要!) M,i =1,2, Pearson Education, Inc. Slide 9-4

5 當粒子數目太多或 N 太大時, 例如密密麻麻的這樣 : r cm = m i ri M r cm = r dm M 2012 Pearson Education, Inc. Slide 9-5

6 Finding the Center of Mass 我要找到質心 Example: A system of three particles in an equilateral triangle 例一 : 3 個粒子在等腰三角形的 3 個頂點 x cm = mx 1 + mx 3 4m y cm = my 1 + my 3 4m = m(x 1 x 1 ) 4m = 0 = 2my 1 4m = 1 2 y 1 = 3 4 L = 0.43L 2012 Pearson Education, Inc. Slide 9-6

Finding the Center of Mass 找到質心 Example: A system of continuous matter 質量是連續分佈, 而且是對稱的 Express the mass element dm in terms of the geometrical variable x: 將 dm 用 x 表示 dm (w/l)x dx 2x dx = = 1 M 2 wl

7 Finding the Center of Mass 找到質心 Example: A system of continuous matter 質量是連續分佈, 而且是對稱的 Express the mass element dm in terms of the geometrical variable x: 將 dm 用 x 表示 dm (w/l)x dx 2x dx = = 1 M 2 wl L 2 Evaluate the integral: x cm = 1 M x dm = 1 M 0 L " 2Mx % x dx # $ & ' = 2 L L 2 0 L 2 x2 dx 考慮對稱性,y cm =0, 就是很明顯的了 so x cm = 2 L L x 2 dx = 2 x L L = 2L3 3L = L 2012 Pearson Education, Inc. Slide 9-7

質心可以分開來計算 An object s center of mass need not lie within the object!

8 More on Center of Mass The center of mass of a composite object can be found from the CMs of its individual parts. 質心可以分開來計算 An object s center of mass need not lie within the object! 質心的位置可以是在物體的外面! Which point is the CM? 就像這個的質心呢? 2012 Pearson Education, Inc. Slide 9-8

9 More on Center of Mass The high jumper clears the bar, but his CM doesn t. 選手過, 質心沒有過 ( 可以很有效地利用體力 ) 勝! 2012 Pearson Education, Inc. Slide 9-9

Motion of the Center of Mass 質心的運動 The center of mass obeys Newton s second law:質心的運動是遵守牛頓第二定律的 Fnet external = Macm Here most parts of the skier s body

10 Motion of the Center of Mass 質心的運動 The center of mass obeys Newton s second law:質心的運動是遵守牛頓第二定律的 Fnet external = Macm Here most parts of the skier s body undergo complex motions, but his center of mass describes the parabolic trajectory of a projectile:很複雜而高難度的動作, 但是質心只是做很簡單的拋體運動 2012 Pearson Education, Inc. Slide 9-10

Motion of the Center of Mass In the absence of any external forces on a system, the center of mass motion remains unchanged; if it s at rest, it remains in the same place no matter what internal

11 Motion of the Center of Mass In the absence of any external forces on a system, the center of mass motion remains unchanged; if it s at rest, it remains in the same place no matter what internal forces may act. 一個 ( 多粒子 ) 系統, 未受外力時, 依照第一定律運動. Example: Jumbo, a 4.8-t elephant, walks 19 m toward one end of the car, but the CM of the 15-t rail car plus elephant doesn t move. This allows us to find the car s final position: x cm = m J x Ji + m c x ci M = m J x Ji M x cm = m J x Jf + m c x cf M x cf = (19 m)m J (m J + m c ) = m J (x Ji +19 m + x cf ) + m c x cf M = (19 m)(4.8 t) (15 t t) = 4.6 m 2012 Pearson Education, Inc. Slide 9-11

12 2012 Pearson Education, Inc. Slide 9-12

13 Momentum and the Center of Mass 動量與質心 The center of mass obeys Newton s law, which can be written or, equivalently, with where is the total momentum of the system: 總動量 v cm P the velocity of the center of mass, and 質心的速度 F net external = M a cm P = F net external = d P dt m vi i = M v cm a cm = d v cm dt 一切如舊!!! 2012 Pearson Education, Inc. Slide 9-13

14 Conservation of Momentum When the net external force is zero, dp dt = 0. Therefore the total momentum of the system is unchanged: P = constant This is the conservation of linear momentum. Example: A system of three billiard balls: Initially two are at rest; all the momentum is in the left-hand ball: Now they re all moving, but the total momentum remains the same: 2012 Pearson Education, Inc. Slide 9-14

15 Collisions 碰撞 A collision is a brief, intense interaction between objects. 在短時間內, 物體間互相極其力的作用 Examples: balls on a pool table, a tennis ball and racket, baseball and bat, football and foot, an asteroid colliding with a planet. The collision time is short compared with the timescale of the objects overall motion. 作用時間很短 ( 相對於觀察的時間 ) Internal forces of the collision are so large that we can neglect any external forces acting on the system during the brief collision time. 交互的作用力很強, 相對的可以忽略外力 Therefore linear momentum is essentially conserved during collisions. 所以動量是守恆的, 質心是沿直線運動或靜止 2012 Pearson Education, Inc. Slide 9-15

Impulse 衝量 If F is the average force acting on one object during a collision that lasts for time Δt, then Newton s second law reads F = Δp / Δt,, or Δ p = FΔt The product of average force and

16 Impulse 衝量 If F is the average force acting on one object during a collision that lasts for time Δt, then Newton s second law reads F = Δp / Δt,, or Δ p = FΔt The product of average force and time that appears in this equation is called impulse. J = FΔt The force in a collision usually isn t constant and can fluctuate wildly, so Δ p = J = J F(t) dt 2012 Pearson Education, Inc. Slide 9-16

17 Elastic and Inelastic Collisions 彈性與非彈性碰撞 In an elastic collision, the internal forces of the collision are conservative. 彈性碰撞時交互作用力為保守力 Therefore an elastic collision conserves kinetic energy as well as linear momentum. 動能與線性動量都守恆 In an inelastic collision, the forces are not conservative and mechanical energy is lost. 非保守力, 動能改變 In a totally inelastic collision, the colliding objects stick together to form a single composite object. 合為一體時稱為完全非彈性碰撞 But if a collision is totally inelastic, that doesn t ( 澄清一下 ) necessarily mean that all kinetic energy is lost Pearson Education, Inc. Slide 9-17

18 Totally Inelastic Collisions 完全非彈性碰撞 Totally inelastic collisions are governed entirely by conservation of momentum 只有動量守恆的條件成立 Since the colliding objects join to form a single composite object, there s only one final velocity: 因為兩者合為一, 所以只有一個末速度 Before collision After collision Therefore conservation of momentum reads m v1 1 + m v2 2 = ( m 1 + m ) 2 v f 2012 Pearson Education, Inc. Slide 9-18

19 suppose v 2i = 0, then m v1 1 = ( m 1 + m ) 2 v f the initial kinetic energy K i is 1 m v i the final kinetic energy K f = 1 2 (m 1 + m 2 )v f2 is equal to 1 (m + m )m 2 / (m m 2 ) 2 v 2 1i = 1 m 2 2 / (m m 2 )v 1i = m 1 / (m 1 + m 2 )K i = εk i < K i 非彈性系數, 代表動能減少 2012 Pearson Education, Inc. Slide 9-19

20 Elastic Collisions 彈性碰撞 Elastic collisions conserve both momentum and kinetic energy: 動量守恆, ( 動 ) 能量守恆 Before collision After collision Therefore the conservation laws read: 要寫下兩個守恆條件 m 1 v1i + m 2 v2i = m 1 v1f + m 2 v2f 1 m v m v 2 = 1 m v m v i 2 2 2i 2 1 1f 2 2 2f 2012 Pearson Education, Inc. Slide 9-20

21 Elastic Collisions in 2 or 3 Dimensions 二維三維碰撞 In general, the conservation laws don t determine the outcome of an elastic collision. 一般, 只有守恆的條件還不能決定結果是什麼 Other information is needed, such as the direction of one of the outgoing particles. 還要知道其他的條件 ; 例如碰撞之後的方向. 這可以用兩球形體對撞時的中心點的距離 (impact parameter b) 來解釋 : 2012 Pearson Education, Inc. Slide 9-21

22 但是質量相等的物體碰撞時, 如果一物體的初速為零, 則兩物體碰撞後的速度方向成一直角 suppose that m 1 = m 2 = m, v 2i = 0 m v 1i = m v 1f + m v 2f v 1i = v 1f + v 2f ---(a) 1 mv 2 = 1 mv mv 2 v 2 2 1i 2 1f 2 2f 1i = v 2 1f + v 2 2f (b) (a), v 1i = v 1f + v 2f,square both sides v 2 1i = v 2 1f + 2v 1f v 2 2f + v 2f use (b) v 1f v 2f = 0 v 1f v 2f 2012 Pearson Education, Inc. Slide 9-22

Elastic Collisions in One Dimension 一維彈性碰撞 But for one-dimensional collisions, when particles collide head-on, then the initial velocities determine the outcome: 但是一維碰撞或 (2 or 3 D) 當頭對撞的時候,

23 Elastic Collisions in One Dimension 一維彈性碰撞 But for one-dimensional collisions, when particles collide head-on, then the initial velocities determine the outcome: 但是一維碰撞或 (2 or 3 D) 當頭對撞的時候, 初速度決定結果 Solving both conservation laws in this case gives 帶入守恆的條件, 可解得 : m1 m2 2m2 v1f = v1i + v2i m + m m + m m m m v = v + v f 1i 2i m1+ m2 m1+ m Pearson Education, Inc. Slide 9-23

24 Special Cases: 1-D Elastic Collisions; m 2 Initially at Rest 特例 : 其中 m 2 開始為靜止的情形 1) m 1 << m 2 : Incident object rebounds with essentially its incident velocity 2) m 1 = m 2 : Incident object stops; struck object moves away with initial speed of incident object 3) m 1 >> m 2 : Incident object continues with essentially its initial velocity; struck object moves away with twice that velocity 2012 Pearson Education, Inc. Slide 9-24

25 v 1f = m 1 m 2 m 1 + m 2 v 1i v 2f = 2m 1 m 1 + m 2 v 1i 2012 Pearson Education, Inc. Slide 9-25

26 Summary A composite system behaves as though its mass is concentrated at the center of mass: r cm = m i ri M (discrete particles) r cm = r dm The center of mass obeys Newton s laws, so F net external = Ma cm or, equivalently, F net external = d P dt In the absence of a net external force, a system s linear momentum is conserved, regardless of what happens internally to the system. Collisions are brief, intense interactions that conserve momentum. Elastic collisions also conserve kinetic energy. Totally inelastic collisions occur when colliding objects join to make a single composite object Pearson Education, Inc. Slide 9-26 M (continuous matter)

27 Homework 13,21,25,47,59, Pearson Education, Inc. Slide 9-27

28 Chapter 9 Lecture Essential University Physics Richard Wolfson 2 nd Edition Clicker questions 2012 Pearson Education, Inc. Slide 9-28

29 Clicker Question 3 Two skaters toss a basketball back and forth on frictionless ice. Which one of the following does not change? A. The momentum of an individual skater. B. The momentum of the system consisting of one skater and the basketball. C. The momentum of the basketball. D. The momentum of the system consisting of both skaters and the basketball Pearson Education, Inc. Slide 9-29

30 Clicker Question 4 Which one of the following qualifies as an inelastic collision? A. Two magnets approach, their north poles facing; they repel and reverse direction without touching. B. A truck strikes a parked car and the two slide off together, crumpled metal hopelessly entwined. C. A basketball flies through the air on a parabolic trajectory. D. A basketball rebounds off the backboard Pearson Education, Inc. Slide 9-30

31 Clicker Question 5 Ball A is at rest on a level floor. Ball B collides elastically with Ball A, and the two move off separately, but in the same direction. What can you conclude about the masses of the two balls? A. Ball A and Ball B have the same mass. B. Ball B has a greater mass than Ball A. C. Ball A has a greater mass than Ball B. D. You cannot conclude anything without more information Pearson Education, Inc. Slide 9-31

32 Clicker Question 6 Two balls, A and B, of the same mass, are dropped from the same height, and rebound off the floor. If Ball A rebounds higher than Ball B, which ball is subjected to a greater impulse during its collision with the floor? A. Ball A B. Ball B C. The impulse is the same for both balls. D. It depends on the duration of the collision Pearson Education, Inc. Slide 9-32

33 Clicker Question 7 In a totally inelastic collision, the final total momentum A. is greater than the initial momentum. B. is smaller than the initial momentum. C. is the same as the initial momentum. D. depends on the duration of collision Pearson Education, Inc. Slide 9-33

34 Clicker Question 8 A car of mass m travels with a speed v and collides with a parked truck, whose mass is 3m. If the bumpers lock together in this head-on inelastic collision, what fraction of the initial kinetic energy is lost in this collision process? A. 1/4 B. 1/3 C. 1/2 D. 3/4 E. 2/ Pearson Education, Inc. Slide 9-34

35 Clicker Question 9 When a cannonball is fired from a cannon, why does the cannon recoil backward? A. The total energy of the cannonball and cannon is conserved. B. The energy of the cannon is greater than the energy of the cannonball. C. The momentum of the cannon is conserved. D. The momentum of the cannon is greater than the momentum of the cannonball. E. The total momentum of the cannonball and cannon is conserved Pearson Education, Inc. Slide 9-35

3 (s05q6) The diagram shows the velocity-time graph for a lift moving between floors in a building. The graph consists of straight line segments. In t

3 (s05q6) The diagram shows the velocity-time graph for a lift moving between floors in a building. The graph consists of straight line segments. In t Mechnics (9709) M1 Topic 1 : s-t and v-t graph(9) Name: Score: Time: 1 (s03q3) The diagram shows the velocity-time graphs for the motion of two cyclists P and Q, whotravel in the same direction along a

In this lecture you ll learn 簡介 How to find the center of mass of a system of particles 多粒子系統的 質心問題 The principle of conservation of momentum, and how

In this lecture you ll learn 簡介 How to find the center of mass of a system of particles 多粒子系統的質心問題 The principle of conservation of momentum, and how