Bab 8

1
Bab 8 Momentum linear dan perlanggaran
Prinsip fizik asas pelancaran roket dan
pergerakannya ditentukan oleh keabadian momentum
linear.
2
Major Concepts

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

3
Mengapa momentum linear?

• Jasad-jasad berinteraksi di antara mereka melalui
  berbagai jenis daya
• 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.
• Analisa and perihaln interaksi dalam sistem yang
  komplikated melalui HN adalah kadang-kadang
  mustahil
• Jadi kita cari jalan lain yang lebih mudah dan
  ekonomik
• Guna konsep momentum linear dan keabadiannya

4
Momentum linear

• 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
• p m v
• Kita akan guna istilah momentum sebagai kata
  ganti kepada momentum linear

m
v
v
m
?
5
Momentum linear, samb

• ML ialah suatu kuantiti vektor, arahnya sama
  dengan arah v
• Dimensi ML/T
• Unit SI kg m / s
• Ia boleh dileraikan kepada komponen-komponennya
• px m vx py m vy
• pz m vz

y
p
py py
x
pz pz
px px
z
6
Quick Quiz 9.1
Two objects have equal kinetic energies. How do
the magnitudes of their momenta compare? (a) p1 lt
p2 (b) p1 p2 (c) p1 gt p2 (d) not enough
information to tell
7
Quick Quiz 9.1
Answer (d). Two identical objects (m1 m2)
traveling at the same speed (v1 v2) 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 K1 K2 but not p1 p2. 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).
8
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
9
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.
10
Newton dan momentum

• HN 2 dapat digunakan untuk mengkaitkan momentum
  suatu zarah dengan daya bersih yang bertindak
  padanya
• di mana jisim zarah adalah malar
• Interpretasi daya bersih F pada zarah
  menyebabkan momentum zarah berubah dengan kadar
  dp/dt

m
Fx
px berubah pada kadar Fx dpx/dt
11

• Sebenarnya F dp/dt adalah bentuk asal Newton
  membentangkan hukum keduanya
• Bentuk HN2 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

12
Quick quiz

• 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?

13
Jawapan

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

14
Keabadian momentum

• Bila-bila sahaja dua atau lebih zarahj dalam
  sistem terpencil berinteraksi, jumlah momentum
  sistem akan tetap malar (terabadikan)
• 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

15
Keabadian momentum, samb.

• Boleh dibuktikan dengan HN3 dan HN2 bahawa jumlah
  momentum suatu sistem terpencil adalah malar
• Keabadian mementum dapat dinyatakan secara
  matematik sebagai
• ptotal p1 p2 pemalar, atau
• p1i p2i p1f p2f , atau
• ?pi ?pf
• Jumlah momentum komponen-komponen sistem
  zarah-zarah juga mesti masing-masing terabadi
  secara merdeka, iaitu
• ?(pi)x ?(pf)x ?(pi)y ?(pf)y ?(pi) z
  ?(pf)z
• Keabadian momentum boleh diaplikasikan ke atas
  sistem yang mengandungi sebarang numbor zarah

16
Terbitan kebabadian momentum

• Pertimbangkan sistem terpencil dua zarah yang
  berinteraksi melalui Fij
• Mengikut HN3, dua zarah yang berinteraksi untuk
  suatu sela masa dt melalui daya Fij mestilah
  mematuhi F12 - F21
• Mengikut HN2 pula,
• Pasangan daya tindakan-tindakbalas ini
  menghasilkan pecutan pada zarah-zarah yang
  ditindak oleh mereka masing-masing
• m2a2 - m1a1
• m2 dv2/dt - m1dv1/dt
• m2 dv2/dt m1 dv1/dt 0
• d/dt (m2 v2 m1 v1) 0
• Tapi, mengikut definasi,
• (m2 v2 m1 v1) p2 p1
• Jadi kita sampai kepada
• d/dt (p2 p1) 0, atau p2 p1 pemalar
• sama sebelum dan selepas interaksi

17
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
18
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.
19
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
20
Quick Quiz 9.4
Answer (c). From Equation 9.4, if p1 p2
constant, then it follows that ?p1 ?p2 0 and
?p1 -?p2. While the change in momentum is the
same, the change in the velocity is a lot larger
for the car!
21
Contoh keabadian momentum boleh diaplikasikan

• Pemanah berdiri di atas permukaan tanpa geseran
  (ais)
• Pendekatan
• Tak boleh guna HN2 kerana tiada maklumat F atau a
• Pendekatan tenaga? Tak, kerana tak ada maklumat
  kerja atau tenaga
• Tapi boleh guna meometum

22
Contoh kiraan

• 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?

23
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 p1f p2f dan mesti sama dengan Jumlah
  momentum sebelum lepaskan anak panah 0 p1f
  p2f 0

24

• m1v1f m2v2f 0
• Jadi, v1f -m2v2f / m1 - (0.5/60) 50 m/s
• - 0.42 m/s
• Iaitu pemanah bergerak dengan arah yang
  bertentangan dengan arah anak pana
• Halaju akhir pemanah juga lebih kecil kerana
  jisimnya besar
• Halaju pemanah ini dipanggil recoil velocity

25
Apa kata jika

• Jika pemanah memanah pada arah yang buat suatu
  sudut q dengan ufukan, bagaimanakah halaju
  pemanah berubah?
• Jawapan
• Recoil velocity pemanah akan berkurangan
  magnitudnya kerana momentum komponen x anak panah
  berkurangan dalam kes ini, menjadi
• p2i ? (p2i )x p2f cosq
• (p1f )x - (p2f )x - p2f cosq
• m1 (v1f )x - m2 (v2f )x
• - m2 v2f cosq
• (v1f )x - m2 v2f cosq / m1
• banding dengan (v1f) x -m2v2f / m1

p2f
(P1f)x
(P2f)x
Keabadian momentum sistem dalam kes ini hanya
berlaku dalam arah-x saja arah-y ada daya luar
(N dan mg)
26
Contoh keabadian momentum dalam Kaon

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

27
Impuls dan momentum

• Daya bersih menyebabkan perubahan momentum suatu
  zarah
• Aplikasikan HN2 keatas suatu zarah, F dp/dt
• Atau, dp Fdt
• Kamirkannya untuk mendapatkan perubahan dalam
  momentum untuk suatu sela masa
• Kamiran ini dikenali sebagai impuls, I, daya pada
  objek tersebut untuk suatu sela Dt

28
Teorem momentum-impuls

• Persamaan ini adalah teorem impuls-momentum
  Impuls daya F yang bertindak pada suatu zarah
  bersamaan dengan perubahan dalam momentum zarah
  tersebut
• Ini juga bentuk alternatif untuk HN2

29
Nota tamhanan tentang Impuls

• Impuls kuantiti vektor
• Magnitud impuls sama dengan luas di bawah graf F
  lawan t
• Dimensi impuls ialah
• M L / T
• Impuls bukan sifat zarah tapi suatu ukuran
  perubahan momentum zarah itu

30
Impuls, terakhir

• Impulse juga boleh diwakili dengan daya min dalam
  sela masa Dt
• 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
• Daya min tersebut memberikan kesan impuls yang
  sama kepada zarah dalam sela masa ini dengan
  kesan daya impul yang berubah-masa

31
Penghampiran impuls

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

32
Contoh real life di mana penghampiran impuls
adalah benar
33
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) p1 lt p2 (b) p1 p2 (c) p1 gt p2 (d)
K1 lt K2 (e) K1 K2 (f) K1 gt K2
34
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 ?t 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, K1 K2.
35
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 ?t. The force is
removed from object 1 and is applied to object 2.
After object 2 has accelerated for the same time
interval ?t, which statements are true? (a) p1 lt
p2 (b) p1 p2 (c) p1 gt p2 (d) K1 lt K2 (e) K1
K2 (f) K1 gt K2
36
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 ?t is larger than that for object 1. As
a result, more work is done on object 2 and K2 gt
K1.
37
Quick Quiz 9.7a
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.
38
Quick Quiz 9.7a
Answer (g). All three are the same. Because the
passenger is brought from the cars initial speed
to a full stop, the change in momentum (equal to
the impulse) is the same regardless of what stops
the passenger.
39
Contoh impuls-momentum Kecelakaan jalanraya

• Dalam suatu Kecelakaan jalanraya, halaju awal
  dan akhir sebuah kereta ialah
• 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.

40
Penyelesaian

• Anggapkan penghampiran impulse adalah benar
• Momentum awal,
• Maka impuls pada kerata
• Daya min

41
Perlanggaran sifat-sifatnya

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

42
Perlanggaran secara kontak terus

• Perlanggaran boleh jadi hasil daripada kontek
  secara terus
• Daya impuls mengkin berubah-masa dengan cara yang
  amat komplikated
• Daya impuls adalah daya dalam sistem
• misalnya perlanggaran billard ball-billard ball

Deformed ball hit
43
Perlanggaran secara tidak berkontak (penyerakan)

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

44
Jenis-jenis perlanggaran

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

45
Jenis-jenis perlanggaran, samb

• 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
• Kebanyakan perlanggaran tertelak di antara dua
  jenis perlanggaran limit tersebut

46
Mengapa mometum linear terabadi?

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

47
Perlanggaran tak elastik penuh

• Oleh kerana objek terlekat bersama mereka
  berkongsi halaju selepas perlanggaran (v2f v1f)
• m1v1i m2v2i
• (m1 m2) vf
• Takrifkan koefisien perlanggaran
• e v2f v1f / (v1i v2i)
• e 0 untuk perlanggaran tak elastik penuh

Active fig.9.8.
48
Contoh real life

• Kereta bertembang muka-dengan-muka adalah hampir
  perlanggarangan tak elastik penuh

49
Perlanggaran elastik

• Kedua-dua momentum dan KE terabadi
• koefisien perlanggaran
• e v2f v1f / (v1i v2i) 1 untuk
  perlanggaran elastik penuh

Active fig.9.9.
50
Perlanggaran elastik, samb

• Secara tipikal, ada dua unknown untuk
  diselesaikan (misalnya halaju akhir bagi dua
  objek yang terlibat)
• Kedua-dua unknow in dapat diselesaikan kerana
  kita ada dua persamaan yang merdeka daripada satu
  sama lain (K. mom dan K. KE)
• 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 v2f v1f v1i v2i
• yang diperolehi daripada e v2f v1f / (v1i
  v2i) 1
• Hanya boleh digunakan ke atas kes perlanggaran
  elastik penuh dua jasad dalam 1-D

51
algebra

• Nak tunjukkan terbitan
• v2f v1f v1i v2i

52
Elastic Collisions, final

• Example of some special cases
• m1 m2 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

53
To find fig. 8.18 of billard ball, Young
54
Contoh 9.5, Fig. 9.10, pg. 263
55
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
56
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.
57
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
58
Quick Quiz 9.9
Answer (b). Because momentum of the two-ball
system is conserved, pTi 0 pTf pB. Because
the table-tennis ball bounces back from the much
more massive bowling ball with approximately the
same speed, pTf -pTi. As a consequence, pB
2pTi. Kinetic energy can be expressed as K p2 /
2m. Because of the much larger mass of the
bowling ball, its kinetic energy is much smaller
than that of the table-tennis ball.
59
Collision Example Ballistic Pendulum

• 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

60
Ballistic Pendulum, cont

• A multi-flash photograph of a ballistic pendulum

61
Contoh 9.7, pg. 264
62
Contoh 9.8
63
Penyelesaian
64
Example 8.12, YoungGravitatinal slingshot effect
65
Two-Dimensional Collisions

• The momentum is conserved in all directions
• Use subscripts for
• 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
• Remember, the simpler equation can only be used
  for one-dimensional situations

66
Two-Dimensional Collision, example

• Particle 1 is moving at velocity v1i and particle
  2 is at rest
• In the x-direction, the initial momentum is m1v1i
• In the y-direction, the initial momentum is 0

67
Two-Dimensional Collision, example cont

• After the collision, the momentum in the
  x-direction is m1v1f cos q m2v2f cos f
• After the collision, the momentum in the
  y-direction is m1v1f sin q m2v2f sin f

Active Figure 9.13
68
Problem-Solving Strategies Two-Dimensional
Collisions

• Set up a coordinate system and define your
  velocities with respect to that system
• It is usually convenient to have the x-axis
  coincide with one of the initial velocities
• In your sketch of the coordinate system, draw and
  label all velocity vectors and include all the
  given information

69
Problem-Solving Strategies Two-Dimensional
Collisions, 2

• Write expressions for the x- and y-components of
  the momentum of each object before and after the
  collision
• 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.

70
Problem-Solving Strategies Two-Dimensional
Collisions, 3

• If the collision is inelastic, kinetic energy of
  the system is not conserved, and additional
  information is probably needed
• If the collision is perfectly inelastic, the
  final velocities of the two objects are equal.
  Solve the momentum equations for the unknowns.

71
Problem-Solving Strategies Two-Dimensional
Collisions, 4

• If the collision is elastic, the kinetic energy
  of the system is conserved
• 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

72
Two-Dimensional Collision Example

• Before the collision, the car has the total
  momentum in the x-direction and the van has the
  total momentum in the y-direction
• After the collision, both have x- and y-components

Youngs Applet, 6.5 car collision
73
Contoh 9.10
74
Contoh 9.11, Perlanggaran proton-proton

• Find figures of proton crash with proton

75
The Center of Mass

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

76
Ilustrasi C.M. dalam kes benar
77
Center of Mass, Coordinates

• The coordinates of the center of mass are
• where M is the total mass of the system

78
Center of Mass, position

• The center of mass can be located by its position
  vector, rCM
• ri is the position of the i th particle, defined
  by

79
Active Figure 9.16
80
Center of Mass, Example

• Both masses are on the x-axis
• The center of mass is on the x-axis
• The center of mass is closer to the particle with
  the larger mass

Active Figure 9.17
81
Center of Mass, Extended Object

• Think of the extended object as a system
  containing a large number of particles
• The particle distribution is small, so the mass
  can be considered a continuous mass distribution

82
Center of Mass, Extended Object, Coordinates

• The coordinates of the center of mass of the
  object are

83
Center of Mass, Extended Object, Position

• The position of the center of mass can also be
  found by
• The center of mass of any symmetrical object lies
  on an axis of symmetry and on any plane of
  symmetry

84
Center of Mass, Example

• An extended object can be considered a
  distribution of small mass elements, Dm
• The center of mass is located at position rCM

85
Fig. 9.19
86
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
87
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
88
Contoh 8.14, Young
89
Center of Mass, Rod

• Find the center of mass of a rod of mass M and
  length L
• The location is on the x-axis (or
• yCM zCM 0)
• xCM L / 2

90
Contoh 9.14
91
Motion of a System of Particles

• Assume the total mass, M, of the system remains
  constant
• We can describe the motion of the system in terms
  of the velocity and acceleration of the center of
  mass of the system
• We can also describe the momentum of the system
  and Newtons Second Law for the system

92
Velocity and Momentum of a System of Particles

• The velocity of the center of mass of a system of
  particles is
• The momentum can be expressed as
• The total linear momentum of the system equals
  the total mass multiplied by the velocity of the
  center of mass

93
Acceleration of the Center of Mass

• The acceleration of the center of mass can be
  found by differentiating the velocity with
  respect to time

94
Forces In a System of Particles

• The acceleration can be related to a force
• 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

95
Newtons 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 aCM
• 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

96
Momentum of a System of Particles

• The total linear momentum of a system of
  particles is conserved if no net external force
  is acting on the system
• MvCM ptot constant when SFext 0

97
C.M. roket yang letup

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

98
Pembatalan daya letupan daya dalam tiada
sumbangan dalam SFext

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

99
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
100
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.
101
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
102
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!
103
Motion of the Center of Mass, Example

• A projectile is fired into the air and suddenly
  explodes
• With no explosion, the projectile would follow
  the dotted line
• 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

104
Contoh 9.17
105
Conceptual example 9.16
106
Rocket Propulsion

• 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

107
Fig. 9.27
108
Rocket Propulsion, 2

• The initial mass of the rocket plus all its fuel
  is M Dm at time ti and velocity v
• The initial momentum of the system is pi (M
  Dm) v

109
Rocket Propulsion, 3

• At some time t Dt, the rockets mass has been
  reduced to M and an amount of fuel, Dm has been
  ejected
• The rockets speed has increased by Dv

110
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
• Therefore, the rocket is accelerated as a result
  of the push from the exhaust gases
• In free space, the center of mass of the system
  (rocket plus expelled gases) moves uniformly,
  independent of the propulsion process

111
Rocket Propulsion, 5

• The basic equation for rocket propulsion is
• The increase in rocket speed is proportional to
  the speed of the escape gases (ve)
• 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
• 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

112
Thrust

• 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
• The thrust increases as the rate of change of
  mass increases
• The rate of change of the mass is called the burn
  rate

113
Fig. 8.28 Young
114
(No Transcript)
115
Contoh 9.19
116
Contoh 8.16 Young
117
Quick quiz

• Pg. 313 Young, Test your understanding

118
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