Chap. 8: Collisions and Momentum Conservation

Transcription

1 Chap. 8: Collisions and Momentum Conservation 1. System in Collision and Explosion C.M. 2. Analysis of Motion of System (C.M.) Kinematics and Dynamics Conservation between Before and After a) b) Energy Conservation Don t ignore impulse! 1

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4 Simple Pendulum: v B =? C A v B =? B What do we know about the motion of this pendulum? 5

5 Pendulum: Find v B What do we know about the motion of this pendulum? Phys. Topics Straight-line motion Circular motion 1D&2D kinematics Force Newton s Laws Work KE W-E theorem C Conservative force Non-cons. force PE M.E. conservation E conservation Power v B =? B F T A Math Topics Trig. geometry F g (mg ) ĵ 6

6 Pendulum: Find v B 1. F.B.D. for a ball F T and F g 2. W T(AB) =? C 3. Gravitation force is a conservative force: W g(ab) =? F T A v B =? B F g (mg ) ĵ 7

7 Newton s Cradle [Q] Where is Physics? 12

8 E Conservation: Anything Else? Impulsive Force How can we treat such an impulsive force? 13

9 Very large magnitude Impulsive Force and Impulse [Example] an impulsive force on a baseball that is struck with a bat has: <F> ~ 5000 N & Dt ~ 0.01 s The impulse concept is most useful for impulsive forces. Impulsive Force Very short time 14

10 Work vs. Impulse Work Impulse Distance, x K = (1/2) m v 2 Work-Energy Theorem Energy Conservation p = m v Impulse-Momentum Theorem 15

11 Impulse-Momentum Theorem f p f p i p t F J p t F t p t m t m m a F p p t t i f i d (1) d d d d d d ) d( d d v v i f i f i f i f ) ( Δ Δ p p t t F t t p p t p F J x F (t) F x 16

12 What is the Impulse? v f = 28 m/s What is the impulse given to the wall? Note: m = kg. v i = 28 m/s 17

13 Impulse in Colliding Balls Analysis on x(y)-component of the force F x 2 ( t) y F 2 F 1 x F x 1 ( t) 19

14 Example 1: Ballistic Pendulum Textbook Example 8.8 Express V and V in terms of m, M, g, and h. V V h Inelastic Collision in 1D and Energy Conservation after the Collision 21

15 (A) Example 1: Solution (I) Express V and V in terms of m, M, g, and h. (A) (B) 1 2 (B) Energy Conservation 22

16 (A) Same Concept with Ballistic Pendulum , (B) Energy Conservation 25

17 Same Concept with Ballistic Pendulum Problem 3 ( points) Rescuing Jane Tarzan (mass M T = 100 kg) runs at V = 8.0 m/s, picks up (collides inelastically with) Jane (mass M J = 50 kg), who is at the end of a rope (length L = 10.0 m), and swings out over a lake. He releases the rope when his velocity is zero. Assume the rope is very light. (a) (b) (c) (10 pts) What is the angle q max when he releases the rope? (10 pts) What is the tension in the rope just before he releases it? (10 pts) What is the maximum tension in the rope? (d) (10 pts-bonus) Express the tension in terms of M T, M J, V, g, q, and/or L at position A. q max L = 10.0 m q A Tarz an Jane 27

18 Consider Energy Conservation before and after Elastic Collision K A,1 K K ( ) B,1 system,1 Q 0 Elastic Collision K ( A,2 K B,2 K sysyem,2 ) K1, i K 2,i K1,f K 2,f 28

19 Consider Energy Conservation before and after Inelastic Collision K A,1 K B,1 K system, ( ) 1 Q 0 Inelastic Collision Loss of energy as thermal And other forms of energy K ( A,2 K B,2 K sysyem,2 ) If A and B stick together after collision, this is a special case, called completely inelastic collision. K 1, i K 2,i K 1,f K 2, f Q 29

20 What Is Completely Inelastic Collision? When two objects collide and stick together after a collision, the maximum possible fraction (fraction < 100%) of the initial kinetic energy is transformed by conserving the momentum of the system of two objects. This collision is called completely (or perfectly or totally) inelastic. This maximum possible fraction does not necessary mean K f = 0 (fraction = 100%) in a case where p i(system) is not zero. The figure in the left is such an example. Q: Can you specify a type of collision where 100% of the initial kinetic energy is transformed? 30

21 Check Example 8.6 and E 8.31 Example 8.6 E 8.31 This can be solved by using momentum (P) conservation. Now we examine a problem using both P and E conservations. 31

22 Example 2: Particle Collision Proton-proton elastic collision: m 1 = m 2 = 1.67 x kg v 1 = 8.20 x 10 5 m/s q 1 = 60.0 o v 2 =? q 2 =? 32

23 The overall motion of a mechanical system can be described in terms of a special point called center of mass of the system: 34

24 Can we use Newton s 2 nd Law? Example Center of Mass (c.m. or CM) The overall motion of a mechanical system can be described in terms of a special point (x) called center of mass of the system: Fsystem M system a where F is the vector exerted system on the system. cm sum of all [Q] How do Momentum we define Conservation the system? the forces 35

25 x y 1 F 2 F t p t v M t v M a M F d d d d d d system cm system cm system cm system system ) ( 2 1 system 2 1 system cm cm cm M M M F F F m m a m a m a m m v m v m v m m r m r m r System just before collision System right after collision System at collision Define the system [Q] How does the total force on the system change? 36

26 We have applied the momentum conservation because F system = 0!

27 How to Solve Textbook P.8-94 F system, x ( ) p ( ) v system, x cm, x 0 p system, x constant M system vcm, x constant 0 because the initial (General Conclusion) velocity is zero. (Special Conclusion for P.8-94) x cm (before) = x cm (after) 39

28 Example 3: x cm F system = 0 Wearing golf shoes 50 kg 1.0 m 1.5 m 1.5 m x 1.0 m 40 kg Frictionless surface x x 40 kg d 40

29 Example 3: Analysis 1 Wearing golf shoes 1.0 m 1.5 m Frictionless Step 1: System = Person + Plate surface Step 2: F system(external) = 0 a cm = 0 Newton s 1 st Law Step 3: x cm = unchanged. Calculate x cm x 40 kg x cm initial = [M person x x person + M plate x x plate ] / [M person + M plate ] = [ (50 kg)( 1.5 m) + (40 kg)(0 m) ] / [50 kg + 40 kg] = 0.83 m x 41

30 Example 3: Analysis 2 Wearing golf shoes x x x 1.0 m 1.5 m 1.5 m 40 x kg 40 kg 40 kg 40 kg 1.0 m Frictionless surface x x 40 kg d 42

31 Example 3: Analysis 3 Step 4: Calculate x cm x cm final = [M person x x person + M plate x x plate ] / [M person + M plate ] = [ (50 kg)( d+1.5 m) + (40 kg)( d) ] / [50 kg + 40 kg] = 0.83 m d = 1.67 m 1.5 m 1.0 m x 40 kg d 43

32 Summary: 1. System C.M. 2. Kinematics and Dynamics 3. Analysis of Motion of C.M. 4. Collision and Explosion 5. (Don t ignore impulse!) 44

33 Problem 4: Two-Stage Rocket A 1000-kg two stage rocket is traveling at a speed of 5.00x10 3 m/s away from the Earth when a pre-designed explosion separates the rocket into two sections of 100 kg and 900 kg. (We assume that a loss of mass due to the explosion is negligible.) The 900-kg section moves in a direction perpendicular to the original line of motion with a speed of 1.00x10 3 m/s. Ignore any gravitational forces from the Earth and other planets. (a) What is the speed and direction of the 100-kg section (relative to the original line of motion) after the explosion? (b) How much energy was supplied by the explosion? 45

34 Line of original motion Analysis - Visualization y Direction M v 0 q x DK = K f(system) K i(system) > 0 46

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36 Question: When should we use law of energy conservation or/and law of momentum conservation? 48

37 Step 1: Defining System and F system System just before collision y F 2 F 1 System at collision x 1 2 System right after collision 49

38 Step 2: Remember work vs. impulse Change of the status (K) of system over position Change of the status (p) of system over time Work Impulse Distance, x K = (1/2) m v 2 Work-Energy Theorem Energy Conservation p = m v Impulse-Momentum Theorem 50

39 Glossary 1. K: Energy associated with the motion of an object. 2. U: Energy stored in a system of objects Can either do work or be converted to K. 3. Q: Thermal Energy (Internal Energy) The energy of atoms and molecules that make up a body. 51

40 Step 3: Analysis of E and P Conservation B Skeet+Pellet A C 52

41 Step 3: Analysis of E and P Conservation B Skeet+Pellet A C AB : Motion under gravity (We ignore air resistance.) System = skeet Read textbook about F ext 53

42 (A) Same Concept with Ballistic Pendulum B Skeet+Pellet A C 1 2 (B) Energy Conservation 57

43 See how I modified the problem. [Textbook] m 1 =m 2, elastic collision Final velocity of m 1? Problem 2: (25 points) Two billiard balls of masses m 1 (= 4M) and m 2 (= M) move at right angles and meet at the origin of an x-y coordinate system. One is moving upward along the y axis at 2.00 m/s, and the other is moving to the right along the x axis with speed 3.70 m/s. After the collision (assumed elastic), the second ball is moving along the positive y axis. What is the final direction of the first ball? What are their two speeds? 58

44 1D Collision m M m M 61

45 2D Collision 62

46 1D/2D Explosion 1 2 (or more) 63

47 Example 2 Before collision After collision (totally inelastic collision) 64

48 Example 2: Solution Before collision After collision (totally inelastic collision) m v 1 + m v 2 = m v 1 + m v 2 v 1 = v 2 65

49 m 1 = m 2 = m 3 = m CM Position (2D) 66

50 CM Position (2D) m 3 y cm = 0.50 m X m 1 + m 2 X m 1 x cm = 1.33 m m 2 + m 3 67

51 Path of C.M. ( ) in 1 2 is stopped and falls vertically v I = 0 m/s M system m I m II m 2 m 1 M system 68

52 Analysis of C.M. ( ) in 1 2 F system M system a cm (0) iˆ ( mg ) ˆj x direction :Motion w/ y direction :Motion w/ constant velocity constant acceleration 69

53 Problem 8 A 20.0-kg projectile is fired at an angle of 60.0 o above the horizontal and with a speed of 240 m/s. At the highest point of its trajectory explodes into two fragments, one of which has a mass of 15.0 kg and falls vertically with an initial speed of 10.0 m/s. Neglect air resistance. (a) How far from the point of firing does the other fragment strikes if the terrain is level? (b) How much energy is released during the explosion? (c) Find the position (x and y) of the center of mass system of two fragments at time of 10 seconds after the explosion. 70

54 Analysis - Visualization m 2 V 1 = 10 m/s M system q V 0 = 240 m/s m 1 71

55 Problem 9 A kg projectile is fired at an angle of 30.0 o above the horizontal and with a speed of 30.0 m/s. When it reaches the maximum height, it is hit from below by a kg bullet traveling vertically upward at a speed of 200 m/s. The bullet is embedded in the projectile. Neglect air resistance. (a) Calculate the maximum height just before the collision. (b) Calculate the velocity of the projectile at the maximum height right after collision. (c) How much energy is given or lost by the collision? (d) How much higher (after the collision) did the projectile go up? (e) Calculate how far away the projectile hits the ground. 72

56 Analysis - Visualization 73