Prinsip fizik asas pelancaran roket dan pergerakannya ditentukan

Prinsip fizik asas pelancaran roket dan pergerakannya ditentukan oleh keabadian momentum linear. 1

Ø Momentum linear dan keabadianya Ø Impulse dan mementum Ø Perlanggaran dalam 1 -D Ø Perlanggaran dalam 2 -D Ø Pusat Jisim Ø Gerakan sistem zarah-zarah Ø “Rocket Pulsulsion” 2

Mengapa momentum linear? Jasad-jasad berinteraksi di antara mereka melalui berbagai jenis daya Ø Kesan-kesan interaksi tersebut mungkin amat komplikated, melibatkan tabii daya, tempoh interkasi, kedudukan jasad-jasad yang terlibat (contoh: bayangkan perlanggaran pin-bowling), geometri sistem dsb. Ø Ø Analisa and perihaln interaksi dalam sistem yang komplikated melalui HN adalah kadang-kadang mustahil Ø Jadi kita cari jalan lain yang lebih mudah dan ekonomik Ø Guna konsep momentum linear dan keabadiannya 3

Momentum linear Ø Momentum linear suatu zarah atau objek yang extended dalam dimensi yang boleh dimodelkan sebagai suatu zarah dengan jisim m bergerak dengan halaju v ditakrifkan sebagai hasildarab jisimnya dengan halajunya: l p=mv • Kita akan guna istilah “momentum” sebagai kata ganti kepada momentum linear m v 4

Momentum linear, samb Ø Ø ML ialah suatu kuantiti vektor, arahnya sama dengan arah v Dimensi = ML/T Unit SI = kg · m / s Ia boleh dileraikan kepada komponennya: l l px = m vx py = m vy pz = m v z y p p y = py p x = px x p z = pz z 5

Quick Quiz 9. 1 Two objects have equal kinetic energies. How do the magnitudes of their momenta compare? (a) p 1 < p 2 (b) p 1 = p 2 (c) p 1 > p 2 (d) not enough information to tell 6

Quick Quiz 9. 1 Answer: (d). Two identical objects (m 1 = m 2) traveling at the same speed (v 1 = v 2) have the same kinetic energies and the same magnitudes of momentum. It also is possible, however, for particular combinations of masses and velocities to satisfy K 1 = K 2 but not p 1 = p 2. For example, a 1 -kg object moving at 2 m/s has the same kinetic energy as a 4 -kg object moving at 1 m/s, but the two clearly do not have the same momenta. Because we have no information about masses and speeds, we cannot choose among (a), (b), or (c). 7

Quick Quiz 9. 2 Your physical education teacher throws a baseball to you at a certain speed, and you catch it. Now the teacher is going to throw you a medicine ball whose mass is ten times the mass of the baseball. You are given the following choices: You can have the medicine ball thrown with (a) the same speed as the baseball, (b) the same momentum, (c) the same kinetic energy. Rank these choices from easiest to hardest to catch. (a) a, b, c (b) a, c, b (c) b, c, a (d) b, a, c (e) c, a, b 8

Quick Quiz 9. 2 Answer: (c): b, c, a. The slower the ball, the easier it is to catch. If the momentum of the medicine ball is the same as the momentum of the baseball, the speed of the medicine ball must be 1/10 the speed of the baseball because the medicine ball has 10 times the mass. If the kinetic energies are the same, the speed of the medicine ball must be the speed of the baseball because of the squared speed term in the equation for K. The medicine ball is hardest to catch when it has the same speed as the baseball. 9

Newton dan momentum Ø HN 2 dapat digunakan untuk mengkaitkan momentum suatu zarah dengan daya bersih yang bertindak padanya Ø di mana jisim zarah adalah malar Ø Interpretasi: daya bersih F pada zarah menyebabkan momentum zarah berubah dengan kadar dp/dt m Fx px berubah pada kadar Fx= dpx/dt 10

l l Sebenarnya F = dp/dt adalah bentuk asal Newton membentangkan hukum keduanya Bentuk HN 2 ini adalah lebih umum daripada F = ma Bentuk F = dp/dt juga membenarkan kira perihalkan dimamik zarah jika melibatkan jisim yang berubah Ia terutamanya adalah amat powerful untuk memerihalkan sistem zarah-zarah 11

Ø Ø Ø Ø Mula-mulanya ia menghujan Kemudian hujan turun menjadi ‘hail’ yang turun pula Katakan 1) kadar titik hujan mengena payung adalah sama dengan kadar biji hail mengenai payung 2) jisim titik hujan = jisim biji hail, 3) kelajuan mereka kena permukaan payung juda sama Tanya: adalah daya yang diperlukan untuk memegang payung semasa hail sama, lebih besar atau kurang berbanding dengan kes hujan? Quick quiz 12

Ø Daya yang lebih besar dalam kes hail turun Jawapan Kerana kadar perubahan momentum biji hail adalah lebih kurang dua kali lebih besar daripda kes untuk titui hujan Ø Mengapa: kerana air hujan tidak ‘melantun’ tapi ‘splatter and run off” manakala biji hail akan melantun ke atas yang bertentangan Ø 13

Keabadian momentum Ø Bila-bila sahaja dua atau lebih zarahj dalam sistem terpencil berinteraksi, jumlah momentum sistem akan tetap malar (terabadikan) l l Jumlah momentum sistem terabadi tidak bermakna momentum individu mesti tak berubah Ini juga mengimplikasikan bahawa jumlah momentum sisterm terpencil bersamaan dengan nilai awal jumlah momentumnya 14

Keabadian momentum, samb. Boleh dibuktikan dengan HN 3 dan HN 2 bahawa jumlah momentum suatu sistem terpencil adalah malar Ø Keabadian mementum dapat dinyatakan secara matematik sebagai Ø l l l ptotal = p 1 + p 2 = pemalar, atau p 1 i + p 2 i = p 1 f + p 2 f , atau pi = pf Jumlah momentum komponen-komponen sistem zarah juga mesti masing-masing terabadi secara merdeka, iaitu Ø (p i )x = (p f )x (p i )y = (p f )y (p i ) z = (p f )z Ø Ø Keabadian momentum boleh diaplikasikan ke atas sistem yang mengandungi sebarang numbor zarah 15

Terbitan kebabadian momentum Pertimbangkan sistem terpencil dua zarah yang berinteraksi melalui Fij: Ø Mengikut HN 3, dua zarah yang berinteraksi untuk suatu sela masa dt melalui daya Fij mestilah mematuhi F 12 = - F 21 Ø Ø Ø Mengikut HN 2 pula, Pasangan daya tindakan-tindakbalas ini menghasilkan pecutan pada zarah-zarah yang ditindak oleh mereka masing-masing: m 2 a 2 = - m 1 a 1 m 2 dv 2/dt = - m 1 dv 1/dt m 2 dv 2/dt + m 1 dv 1/dt = 0 d/dt (m 2 v 2 + m 1 v 1) = 0 Tapi, mengikut definasi, (m 2 v 2 + m 1 v 1 ) = p 2 + p 1 Jadi kita sampai kepada d/dt (p 2 + p 1) = 0, atau p 2 + p 1 = pemalar =sama sebelum dan selepas interaksi 16

Quick Quiz 9. 3 A ball is released and falls toward the ground with no air resistance. The isolated system for which momentum is conserved is (a) the ball (b) the Earth (c) the ball and the Earth (d) impossible to determine 17

Quick Quiz 9. 3 Answer: (c). The ball and the Earth exert forces on each other, so neither is an isolated system. We must include both in the system so that the interaction force is internal to the system. 18

Quick Quiz 9. 4 A car and a large truck traveling at the same speed make a head-on collision and stick together. Which vehicle experiences the larger change in the magnitude of momentum? (a) the car (b) the truck (c) The change in the magnitude of momentum is the same for both. (d) impossible to determine 19

Quick Quiz 9. 4 Answer: (c). From Equation 9. 4, if p 1 + p 2 = constant, then it follows that Δp 1 + Δp 2 = 0 and Δp 1 = -Δp 2. While the change in momentum is the same, the change in the velocity is a lot larger for the car! 20

Contoh keabadian momentum boleh diaplikasikan Ø Pemanah berdiri di atas permukaan tanpa geseran (ais) Ø Pendekatan: Ø l l l Tak boleh guna HN 2 kerana tiada maklumat F atau a Pendekatan tenaga? Tak, kerana tak ada maklumat kerja atau tenaga Tapi boleh guna meometum 21

Contoh kiraan Ø Pemanah berjisim 60 kg, berdiri di atas permukaan tanpa geseran (ais) , memanah anak panah berjisim 0. 5 kg secara mengufuk pada kelajuan 50 m/s. Apalah halaju si pemanah bergerak selepas anak panah dilepaskan? 22

Ø Ø Ø Penyelesaian Takrifkan sistem dulu: Pemanah dengan bow (zarah 1) dan anak panah (zarah 2) Tiada daya luar dalam arah x, jadi ia sistem terpencil untuk momentum dalam arah x Jumlah momentum sebelum lepaskan anak panah = 0 Jumlah momentum selepas anak panah dilepaskan ialah p 1 f + p 2 f dan mesti sama dengan Jumlah momentum sebelum lepaskan anak panah = 0: p 1 f + p 2 f = 0 23

m 1 v 1 f + m 2 v 2 f = 0 Ø Jadi, v 1 f = -m 2 v 2 f / m 1 = - (0. 5/60) 50 m/s = - 0. 42 m/s Ø Iaitu pemanah bergerak dengan arah yang bertentangan dengan arah anak pana Ø Halaju akhir pemanah juga lebih kecil kerana jisimnya besar Ø Halaju pemanah ini dipanggil “recoil velocity” Ø 24

Ø Jika pemanah memanah pada arah yang buat suatu sudut q dengan ufukan, bagaimanakah halaju pemanah berubah? Apa kata jika… Jawapan Ø Recoil velocity pemanah akan berkurangan magnitudnya kerana momentum komponen x anak panah berkurangan dalam kes ini, menjadi p 2 i = (p 2 i )x = p 2 f cosq (p 1 f )x = - (p 2 f )x = - p 2 f cosq m 1 (v 1 f )x = - m 2 (v 2 f )x = - m 2 v 2 f cosq (v 1 f )x = = - m 2 v 2 f cosq / m 1 [banding dengan (v 1 f) x = -m 2 v 2 f / m 1] p 2 f Ø (P 1 f)x (P 2 f)x Keabadian momentum sistem dalam kes ini hanya berlaku dalam arah-x saja; arah-y ada daya 25 luar (N dan mg)

Contoh keabadian momentum dalam Kaon reput kepada zarah p positif dan p negatif Ø Momentum sebelum dan selepas adalah sifar Ø Jadi, selepas reputan jumlah momentum akhir mesti sama dengan sifar Ø l Ø p+ + p- = 0 atau p+ = -p- Walaupun sistem ini amat berbeza dengan sistem pemanah tadi, mereka mematuhi hukum fizik yang sama: keabadian momentum 26

Impuls dan momentum Ø Daya bersih menyebabkan perubahan momentum suatu zarah Ø Aplikasikan HN 2 keatas suatu zarah, F = dp/dt Ø Atau, dp = Fdt Ø Kamirkannya untuk mendapatkan perubahan dalam momentum untuk suatu sela masa: Ø Kamiran ini dikenali sebagai impuls, I, daya pada objek tersebut untuk suatu sela Dt 27

Teorem momentum-impuls Ø Persamaan ini adalah teorem impuls- momentum: Impuls daya F yang bertindak pada suatu zarah bersamaan dengan perubahan dalam momentum zarah tersebut l Ini juga bentuk alternatif untuk HN 2 28

Nota tamhanan tentang Impuls kuantiti vektor Ø Magnitud impuls sama dengan luas di bawah graf F lawan t Ø Dimensi impuls ialah ML/T Ø Impuls bukan sifat zarah tapi suatu ukuran perubahan momentum zarah itu Ø 29

Impuls, terakhir Ø Impulse juga boleh diwakili dengan daya min dalam sela masa Dt Ø Ø I = Dt di mana luas bawah lengkung F lawan t bersamaan dengan luas segiempat yang diberikan oleh daya min untuk sela masa yang sama Dt Ø Daya min tersebut memberikan kesan impuls yang sama kepada zarah dalam sela masa ini dengan kesan daya impul yang berubah-masa 30

Penghampiran impuls Dalam banyak kes, daya pada suatu zarah mungkin lebih besar daripada daya-daya lain yang juga bertindak pada zarah yang sama Ø Dalam penghampiran impuls kita sentiasa menganggap daya impulse sentiasa lebih besar daripada daya-daya lain yang juga bertindak pada zarah pada ketika yang sama Ø Juga kita menggap bahawa anjakan zarah semasa daya impulse bertindak adalah amat kecil dan boleh diabaikan Ø 31

Contoh real life di mana penghampiran impuls adalah benar 32

Quick Quiz 9. 5 Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2. When a constant force is applied to object 1, it accelerates through a distance d. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (a) p 1 < p 2 (b) p 1 = p 2 (c) p 1 > p 2 (d) K 1 < K 2 (e) K 1 = K 2 (f) K 1 > K 2 33

Quick Quiz 9. 5 Answer: (c) and (e). Object 2 has a greater acceleration because of its smaller mass. Therefore, it takes less time to travel the distance d. Even though the force applied to objects 1 and 2 is the same, the change in momentum is less for object 2 because Δt is smaller. The work W = Fd done on both objects is the same because both F and d are the same in the two cases. Therefore, K 1 = K 2. 34

Quick Quiz 9. 6 Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2. When a force is applied to object 1, it accelerates for a time interval Δt. The force is removed from object 1 and is applied to object 2. After object 2 has accelerated for the same time interval Δt, which statements are true? (a) p 1 < p 2 (b) p 1 = p 2 (c) p 1 > p 2 (d) K 1 < K 2 (e) K 1 = K 2 (f) K 1 > K 2 35

Quick Quiz 9. 6 Answer: (b) and (d). The same impulse is applied to both objects, so they experience the same change in momentum. Object 2 has a larger acceleration due to its smaller mass. Thus, the distance that object 2 covers in the time interval Δt is larger than that for object 1. As a result, more work is done on object 2 and K 2 > K 1. 36

Quick Quiz 9. 7 a Rank an automobile dashboard, seatbelt, and airbag in terms of the impulse they deliver to a front-seat passenger during a collision, from greatest to least. (a) dashboard, seatbelt, airbag (b) dashboard, airbag, seatbelt (c) seatbelt, airbag, dashboard (d) seatbelt, dashboard, airbag (e) airbag, dashboard, seatbelt (f) airbag, seatbelt, dashboard (g) All three are the same. 37

Quick Quiz 9. 7 a Answer: (g). All three are the same. Because the passenger is brought from the car’s initial speed to a full stop, the change in momentum (equal to the impulse) is the same regardless of what stops the passenger. 38

Contoh impuls-momentum: “Kecelakaan jalanraya” Ø Dalam suatu “Kecelakaan jalanraya”, halaju awal dan akhir sebuah kereta ialah Ø Soalan: jika perlanggaran itu berlaku untuk sela masa 0. 15 s, tentukan impulse yang disebabkan oleh perlanggaran itu kepada kereta, dan daya impuls min yang bertindak kepadanya. 39

Penyelesaian Ø Anggapkan penghampiran impulse adalah benar Ø Momentum awal, Ø Maka impuls pada kerata Ø Daya min: 40

Perlanggaran – sifat-sifatnya Ø Istilan “perlanggaran” merujuk kepada perkara yang mana dua zarah mendekati sata sama lain and berinteraksi melalui daya Ø Sela masa dalam mana perubahan halaju berlaku dianggap pendek berbanding dengan skala (scale) tompoh pencerapan/pengukuran eksperimen Ø Daya impuls adalah dianggap jauh lebih besar berbanding dengan daya-daya luar yang bertindak pada zarah (mislanya graviti, daya perintang etc. ) – penghampiran impuls teraplikasikan 41

Perlanggaran secara kontak terus Perlanggaran boleh jadi hasil daripada kontek secara terus Ø Daya impuls mengkin berubah-masa dengan cara yang amat komplikated Ø l Ø Deformed ball hit Daya impuls adalah daya dalam sistem misalnya perlanggaran billard ball-billard ball 42

Perlanggaran secara tidak berkontak (penyerakan) Ø Perlanggaran tidak semestinya berkontak secara fizikal di antara objek-objek Ø Kadang-kadang dikenali sebagai penyerakan (scattering) Ø Dalam kes ini daya yang mengiteraksikan mesti dapat bertindakan melalui ruang (misalnya daya elektromagnetik, daya graviti) Ø Perlanggaran sedemikan masih boleh dianalisakan sepertimana yang dilakukan ke atas sistem yang berlanggar melalui kontak fizikal (tapi mungkin lebih komplikated) 43

Jenis-jenis perlanggaran Ø Perlanggaran elastik: momentum dan tenaga kinetik terbabadikan Ø Perlanggaran elastik penuh berlaku pada skala mikroskopik (misalnya penyerakan elektron) l l Ø Dalam perlanggaran makroskopik hanya perlanggaran elastik Koefisien elastik, e = 1 utk kes ini Perlanggaran tak elastik, KE tidak terabadi, hanya momentum yang terabadi l l Jika objek terlekat bersama selepas perlanggaran, ia dikenali sebagai perlanggaran tak elastik penuh Koefisien elastik, e = 0 utk kes ini 44

Jenis-jenis perlanggaran, samb Ø Dalam perlanggaran tak elastik, terdapat KE yang terlesap (hilang), tapi objek tak lepat bersama Perlanggaran elastik penuh dan perlanggaran tak elastik penuh adalah kes-kes limit bagi perlanggaran yang lebih umum Ø Kebanyakan perlanggaran tertelak di antara dua jenis perlanggaran limit tersebut Ø 45

Mengapa mometum linear terabadi? Momentum sentiasa terabadi dalam mana-mana perlanggaran Ø sebenarnya momentum adalah terabadi dalam semua proses fizik yang pernah dicerap, tiada pengecualian) Ø Teorem Noether (teorem matematik) mengatakan setiap symmetri yang terabadi mesti bersepadanan dengan suatu kuantiti Ø Wujud dalam semester kita simetri translasi ruang (spatial translation symmetry): hukum fizik pada x = hukum fizik pada kedudukan x’. Ø Ø ini merupakan punca kepada keabadian momentum linear. 46

Perlanggaran tak elastik penuh Ø Ø Ø Oleh kerana objek terlekat bersama mereka berkongsi halaju selepas perlanggaran (v 2 f = v 1 f) m 1 v 1 i + m 2 v 2 i = (m 1 + m 2 ) v f Takrifkan koefisien perlanggaran: e = v 2 f – v 1 f / (v 1 i – v 2 i) e = 0 untuk perlanggaran tak elastik penuh Active fig. 9. 8. 47

Contoh real life Ø Kereta bertembang muka-dengan-muka adalah hampir perlanggarangan tak elastik penuh 48

Perlanggaran elastik Ø Kedua-dua momentum dan KE terabadi koefisien perlanggaran: Ø e = v 2 f – v 1 f / (v 1 i – v 2 i) = 1 untuk perlanggaran elastik penuh Ø Active fig. 9. 9. 49

Perlanggaran elastik, samb Ø Secara tipikal, ada dua unknown untuk diselesaikan (misalnya halaju akhir bagi dua objek yang terlibat) Ø Kedua-dua unknow in dapat diselesaikan kerana kita ada dua persamaan yang merdeka daripada satu sama lain (K. mom dan K. KE) Ø Persamaan kinetik mungkin susah diselesaikan kerana melibatkan kuasadua halaju Sebagai alternatif yang lebih mudah dalam menyelesaikan persamaan serentak K. mom dan K. KE, kita boleh guna persamaan alternatif yang lebih mudah untuk menggantikan persamaan K. KE: dangan v 2 f – v 1 f = v 1 i – v 2 i Ø yang diperolehi daripada e = v 2 f – v 1 f / (v 1 i – v 2 i) = 1 l Hanya boleh digunakan ke atas kes perlanggaran elastik penuh dua jasad dalam 1 -D Ø 50

algebra ØNak tunjukkan terbitan v 2 f – v 1 f = v 1 i – v 2 i 51

Elastic Collisions, final Ø Example of some special cases l l l m 1 = m 2 – the particles exchange velocities When a very heavy particle collides head-on with a very light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains approximately at rest 52

To find fig. 8. 18 of billard ball, Young 53

Contoh 9. 5, Fig. 9. 10, pg. 263 54

Quick Quiz 9. 8 In a perfectly inelastic one-dimensional collision between two objects, what condition alone is necessary so that all of the original kinetic energy of the system is gone after the collision? (a) The objects must have momenta with the same magnitude but opposite directions. (b) The objects must have the same mass. (c) The objects must have the same velocity. (d) The objects must have the same speed, with velocity vectors in opposite directions 55

Quick Quiz 9. 8 Answer: (a). If all of the initial kinetic energy is transformed, then nothing is moving after the collision. Consequently, the final momentum of the system is necessarily zero and, therefore, the initial momentum of the system must be zero. While (b) and (d) together would satisfy the conditions, neither one alone does. 56

Quick Quiz 9. 9 A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and bounces back along the same line. After the collision, compared to the bowling ball, the table-tennis ball has (a) a larger magnitude of momentum and more kinetic energy (b) a smaller magnitude of momentum and more kinetic energy (c) a larger magnitude of momentum and less kinetic energy (d) a smaller magnitude of momentum and less kinetic energy (e) the same magnitude of momentum and the same kinetic energy 57

Quick Quiz 9. 9 Answer: (b). Because momentum of the two-ball system is conserved, p. Ti + 0 = p. Tf + p. B. Because the table-tennis ball bounces back from the much more massive bowling ball with approximately the same speed, p. Tf = -p. Ti. As a consequence, p. B = 2 p. Ti. Kinetic energy can be expressed as K = p 2 / 2 m. Because of the much larger mass of the bowling ball, its kinetic energy is much smaller than that of the tabletennis ball. 58

Collision Example – Ballistic Pendulum Ø Ø 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 59

Ballistic Pendulum, cont Ø A multi-flash photograph of a ballistic pendulum 60

Contoh 9. 7, pg. 264 61

Contoh 9. 8 62

Penyelesaian 63

Example 8. 12, Young Gravitatinal slingshot effect 64

Two-Dimensional Collisions The momentum is conserved in all directions Ø Use subscripts for Ø l l l Ø identifying the object indicating initial or final values the velocity components If the collision is elastic, use conservation of kinetic energy as a second equation l Remember, the simpler equation can only be used for one-dimensional situations 65

Two-Dimensional Collision, example Particle 1 is moving at velocity v 1 i and particle 2 is at rest Ø In the x-direction, the initial momentum is m 1 v 1 i Ø In the y-direction, the initial momentum is 0 Ø 66

Two-Dimensional Collision, example cont After the collision, the momentum in the xdirection is m 1 v 1 f cos q + m 2 v 2 f cos f Ø After the collision, the momentum in the ydirection is m 1 v 1 f sin q + m 2 v 2 f sin f Ø Active Figure 9. 13 67

Problem-Solving Strategies – Two-Dimensional Collisions Ø Set up a coordinate system and define your velocities with respect to that system l It is usually convenient to have the x-axis coincide with one of the initial velocities Ø In your sketch of the coordinate system, draw and label all velocity vectors and include all the given information 68

Problem-Solving Strategies – Two-Dimensional Collisions, 2 Ø Write expressions for the x- and y-components of the momentum of each object before and after the collision l Ø Remember to include the appropriate signs for the components of the velocity vectors Write expressions for the total momentum of the system in the x-direction before and after the collision and equate the two. Repeat for the total momentum in the y-direction. 69

Problem-Solving Strategies – Two-Dimensional Collisions, 3 Ø If the collision is inelastic, kinetic energy of the system is not conserved, and additional information is probably needed Ø If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknowns. 70

Problem-Solving Strategies – Two-Dimensional Collisions, 4 Ø If the collision is elastic, the kinetic energy of the system is conserved Ø Equate the total kinetic energy before the collision to the total kinetic energy after the collision to obtain more information on the relationship between the velocities 71

Two-Dimensional Collision Example Before the collision, the car has the total momentum in the xdirection and the van has the total momentum in the ydirection Ø After the collision, both have x- and ycomponents Ø Young’s Applet, 6. 5 car collision 72

Contoh 9. 10 73

Contoh 9. 11, Perlanggaran proton Ø Find figures of proton crash with proton 74

The Center of Mass Ø 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 Ø The system will move as if an external force were applied to a single particle of mass M located at the center of mass l M is the total mass of the system 75

Ilustrasi C. M. dalam kes benar 76

Center of Mass, Coordinates Ø The coordinates of the center of mass l are where M is the total mass of the system 77

Center of Mass, position Ø The center of mass can be located by its position vector, r. CM Ø ri is the position of the i th particle, defined by 78

Active Figure 9. 16 79

Center of Mass, Example Both masses are on the x-axis Ø The center of mass is closer to the particle with the larger mass Ø Active Figure 9. 17 80

Center of Mass, Extended Object Ø Think of the extended object as a system containing a large number of particles Ø The particle distribution is small, so the mass can be considered a continuous mass distribution 81

Center of Mass, Extended Object, Coordinates Ø The coordinates of the center of mass of the object are 82

Center of Mass, Extended Object, Position Ø The position of the center of mass can also be found by: Ø The center of mass of any symmetrical object lies on an axis of symmetry and on any plane of symmetry 83

Center of Mass, Example An extended object can be considered a distribution of small mass elements, Dm Ø The center of mass is located at position r. CM Ø 84

Fig. 9. 19 85

Quick Quiz 9. 10 A baseball bat is cut at the location of its center of mass as shown in the figure. The piece with the smaller mass is (a) the piece on the right (b) the piece on the left (c) Both pieces have the same mass. (d) impossible to determine 86

Quick Quiz 9. 10 Answer: (b). The piece with the handle will have less mass than the piece made up of the end of the bat. To see why this is so, take the origin of coordinates as the center of mass before the bat was cut. Replace each cut piece by a small sphere located at the center of mass for each piece. The sphere representing the handle piece is farther from the origin, but the product of less mass and greater distance balances the product of greater mass and less distance for the end piece: 87

Contoh 8. 14, Young 88

Center of Mass, Rod Find the center of mass of a rod of mass M and length L Ø The location is on the x-axis (or y. CM = z. CM = 0) Ø x. CM = L / 2 Ø 89

Contoh 9. 14 90

Motion of a System of Particles Ø Assume the total mass, M, of the system remains constant Ø We can describe the motion of the system in terms of the velocity and acceleration of the center of mass of the system Ø We can also describe the momentum of the system and Newton’s Second Law for the system 91

Velocity and Momentum of a System of Particles Ø The velocity of the center of mass of a system of particles is Ø The momentum can be expressed as Ø The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass 92

Acceleration of the Center of Mass Ø The acceleration of the center of mass can be found by differentiating the velocity with respect to time 93

Forces In a System of Particles Ø The acceleration can be related to a force Ø If we sum over all the internal forces, they cancel in pairs and the net force on the system is caused only by the external forces 94

Newton’s Second Law for a System of Particles Since the only forces are external, the net external force equals the total mass of the system multiplied by the acceleration of the center of mass: SFext = M a. CM Ø The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system Ø 95

Momentum of a System of Particles Ø The total linear momentum of a system of particles is conserved if no net external force is acting on the system Ø Mv. CM = ptot = constant when SFext = 0 96

C. M. roket yang letup Ø Sebelum letupan, roket mengikut lintasan projektil Ø Bagi roket yang telah letup, C. M. serpihan-serpihan tetap mengikut lintasan parabola sebagaimana dalam kes projektil untuk suatu objek yang masih bersepadu Ø Lintasan projektil kerana ada daya luar – graviti - yang bertindak dalam arah cancangan. Ø Di sini, SFext = M a. CM Fg = M g 97

Pembatalan daya letupan: daya dalam tiada sumbangan dalam SFext Ø Tapi, daya letupan bukanlah daya luar Ø Atau dalam kata lain daya letupan membatalkan satu sama lain dan sumbangan bersih mereka kepada Sfext ialah sifar Ø Jadi ia tidak akan mempengaruhi gerakan C. M. sistem serpihan semasa dan selepas letupan 98

Quick Quiz 9. 11 The vacationers on a cruise ship are eager to arrive at their next destination. They decide to try to speed up the cruise ship by gathering at the bow (the front) and running all at once toward the stern (the back) of the ship. While they are running toward the stern, the speed of the ship is (a) higher than it was before (b) unchanged (c) lower than it was before (d) impossible to determine 99

Quick Quiz 9. 11 Answer: (a). This is the same effect as the swimmer diving off the raft that we just discussed. The vessel-passengers system is isolated. If the passengers all start running one way, the speed of the vessel increases (a small amount!) the other way. 100

Quick Quiz 9. 12 The vacationers in question 11 stop running when they reach the stern of the ship. After they have all stopped running, the speed of the ship is (a) higher than it was before they started running (b) unchanged from what it was before they started running (c) lower than it was before they started running (d) impossible to determine 101

Quick Quiz 9. 12 Answer: (b). Once they stop running, the momentum of the system is the same as it was before they started running – you cannot change the momentum of an isolated system by means of internal forces. In case you are thinking that the passengers could do this over and over to take advantage of the speed increase while they are running, remember that they will slow the ship down every time they return to the bow! 102

Motion of the Center of Mass, Example A projectile is fired into the air and suddenly explodes Ø With no explosion, the projectile would follow the dotted line Ø After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion Ø 103

Contoh 9. 17 104

Conceptual example 9. 16 105

Rocket Propulsion Ø 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 106

Fig. 9. 27 107

Rocket Propulsion, 2 The initial mass of the rocket plus all its fuel is M + Dm at time ti and velocity v Ø The initial momentum of the system is pi = (M + D m) v Ø 108

Rocket Propulsion, 3 At some time t + Dt, the rocket’s mass has been reduced to M and an amount of fuel, Dm has been ejected Ø The rocket’s speed has increased by Dv Ø 109

Rocket Propulsion, 4 Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction Ø Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases Ø In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of the propulsion process Ø 110

Rocket Propulsion, 5 Ø The basic equation for rocket propulsion is Ø The increase in rocket speed is proportional to the speed of the escape gases (ve) l Ø So, the exhaust speed should be very high The increase in rocket speed is also proportional to the natural log of the ratio Mi/Mf l So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible 111

Thrust Ø The thrust on the rocket is the force exerted on it by the ejected exhaust gases Thrust = The thrust increases as the exhaust speed increases Ø The thrust increases as the rate of change of mass increases Ø l The rate of change of the mass is called the burn rate 112

Fig. 8. 28 Young 113

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Contoh 9. 19 115

Contoh 8. 16 Young 116

Quick quiz Ø Pg. 313 Young, Test your understanding 117

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