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Prinsip fizik asas pelancaran roket dan pergerakannya
ditentukan oleh keabadian momentum linear.
1
 Momentum
linear dan keabadianya
 Impulse dan mementum
 Perlanggaran dalam 1-D
 Perlanggaran dalam 2-D
 Pusat Jisim
 Gerakan sistem zarah-zarah
 “Rocket Pulsulsion”
2
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
3
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=mv
• Kita akan guna istilah “momentum” sebagai kata
ganti kepada momentum linear
m
v

m
v
4
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 komponenkomponennya:


px = m v x py = m v y
pz = m v z
y
p
py= py y
px= px x
x
pz = pz z
z
5
Quick Quiz 9.1
Two objects have equal kinetic energies. How do the
magnitudes of their momenta compare?
(a) p1 < p2
(b) p1 = p2
(c) p1 > p2
(d) not enough information to tell
6
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).
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 1
10
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

HN 2 dapat digunakan untuk mengkaitkan
momentum suatu zarah dengan daya bersih
yang bertindak padanya
dv d  mv  dp
F  ma  m


dt
dt
dt

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
10




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
11







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

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
 Mengapa: kerana air hujan
tidak ‘melantun’ tapi ‘splatter
and run off” manakala biji
hail akan melantun ke atas
yang bertentangan
13
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
14
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 zarahzarah 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
15
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
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 p1 + p2 = constant, then it
follows that Δp1 + Δp2 = 0 and Δp1 = -Δp2. 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
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
21
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?
x
22
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
x
x
23

m1v1f + m2v2f = 0
 Jadi, v1f = -m2v2f / m1 = - (0.5/60) 50 m/s
= - 0.42 m/s x
x

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”
24

Jika pemanah memanah pada
arah yang buat suatu sudut q
dengan ufukan, bagaimanakah
halaju pemanah berubah?
Apa kata jika…
Jawapan
 Recoil velocity pemanah akan
berkurangan magnitudnya kerana
momentum komponen x anak
panah berkurangan dalam kes ini,
x
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
25
daya luar (N dan mg)
Contoh keabadian momentum
dalam Kaon



Kaon reput kepada zarahzarah 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
26
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:
tf
 dp  p f  pi  ti Fdt  I
 Kamiran ini dikenali sebagai impuls, I, daya
pada objek tersebut untuk suatu sela Dt
27
Teorem momentum-impuls
 Persamaan
ini adalah teorem impulsmomentum: Impuls daya F yang
bertindak pada suatu zarah bersamaan
dengan perubahan dalam momentum
zarah tersebut

Ini juga bentuk alternatif untuk HN2
28
Nota tamhanan tentang Impuls

Impuls kuantiti vektor
 Magnitud impuls
sama dengan luas di
bawah graf F lawan t
 Dimensi impuls ialah
ML/T
 Impuls bukan sifat
zarah tapi suatu
ukuran perubahan
momentum zarah itu
29
Impuls, terakhir

Impulse juga boleh diwakili
dengan daya min dalam sela
masa Dt


I = Dt
di mana
luas bawah lengkung
F
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
30
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 zarahzarah semasa daya impulse bertindak adalah
amat kecil dan boleh diabaikan
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) p1 < p2
(b) p1 = p2
(c) p1 > p2
(d) K1 < K2
(e) K1 = K2
(f) K1 > K2
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, K1 = K2.
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) p1 < p2
(b) p1 = p2
(c) p1 > p2
(d) K1 < K2
(e) K1 = K2
(f) K1 > K2
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 K2 > K1.
36
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.
37
Quick Quiz 9.7a
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”

Dalam suatu
“Kecelakaan jalanraya”,
halaju awal dan akhir
sebuah kereta ialah
vi  15.0x m/s dan v f  2.6x m/s
 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

Anggapkan penghampiran
impulse adalah benar

Momentum awal,
pi  mvi  1500 kg  (15.0 x m/s) =  2250 x kgm/s
p f  mv f  ...  =3900 x kgm/s

Maka impuls pada kerata
I  D p  p f  p i  ...  2640 x kgm/s

Daya min: F  D p  2640 x kgm/s  176000 N x
Dt
0.15s
40
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
41
Perlanggaran secara kontak terus


Perlanggaran boleh jadi
hasil daripada kontek
secara terus
Daya impuls mengkin
berubah-masa dengan
cara yang amat
komplikated


Deformed ball hit
Daya impuls adalah daya
dalam sistem
misalnya perlanggaran
billard ball-billard ball
42
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)
43
Jenis-jenis perlanggaran

Perlanggaran elastik: momentum dan tenaga
kinetik terbabadikan

Perlanggaran elastik penuh berlaku pada skala
mikroskopik (misalnya penyerakan elektronelektron)



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
44
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
45
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.
46
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.
47
Contoh real life

Kereta bertembang
muka-dengan-muka
adalah hampir
perlanggarangan tak
elastik penuh
48
Perlanggaran elastik

Kedua-dua momentum
dan KE terabadi
m1v1i  m2 v 2 i 
m1v1 f  m2 v 2 f
1
1
2
m1v1i  m2 v 22i 
2
2
1
1
m1v12 f  m2 v 22 f
2
2

koefisien perlanggaran:
 e = v2f – v1f / (v1i – v2i) = 1
untuk perlanggaran
elastik penuh
Active fig.9.9.
49
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
50
algebra
Nak
tunjukkan terbitan
v2f – v1f = v1i – v2i
dari K. Mom:
m1 (v1i  v1 f )  m2 (v2 f  v2i )
Eq(1),
Dari K. KE:
m1 (v1i 2  v1 f 2 )  m2 (v22 f  v22i )
m1 (v1i  v1 f )(v1i  v1 f )  m2 (v2 f  v2i )(v2 f  v2 i ), Eq(2).
Eq(2) / Eq(1): v1i  v1 f  v2 f  v2i
51
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
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, 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 tabletennis ball.
58
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
59
Ballistic Pendulum, cont

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
 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
65
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
66
Two-Dimensional Collision,
example cont

After the collision, the
momentum in the xdirection is m1v1f cos
q  m2v2f cos f
 After the collision, the
momentum in the ydirection is m1v1f sin q
 m2v2f sin f
Active Figure 9.13
67
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
68
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.
69
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.
70
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
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
 After the collision,
both have x- and ycomponents
Young’s Applet, 6.5
car collision
72
Contoh 9.10
73
Contoh 9.11, Perlanggaran protonproton
 Find
figures of proton crash with proton
74
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
75
Ilustrasi C.M. dalam kes benar
76
Center of Mass, Coordinates
 The
coordinates of the center of mass are
xCM 

m x
i i
i
M
yCM 
m y
i
i
M
i
zCM 
m z
i i
i
M
where M is the total mass of the system
77
Center of Mass, position
 The
center of mass can be located by its
position vector, rCM
rCM 
m r
i i
i
M
 ri
is the position of the i th particle, defined
by
ri  xi ˆi  yi ˆj  zi kˆ
78
Active Figure 9.16
79
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
80
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
81
Center of Mass, Extended
Object, Coordinates
 The
coordinates of the center of mass of
the object are
xCM
1
1

x dm yCM 

M
M
1
zCM 
z dm

M
 y dm
82
Center of Mass, Extended
Object, Position
 The
position of the center of mass can
also be found by:
rCM
 The
1

r dm

M
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
 The center of mass is
located at position rCM
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
 The location is on the
x-axis (or
yCM = zCM = 0)
 xCM = L / 2
89
Contoh 9.14
90
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 Newton’s Second Law
for the system
91
Velocity and Momentum of a
System of Particles

The velocity of the center of mass of a system of
particles is
 mi vi
dr
vCM 

CM
dt

i
M
The momentum can be expressed as
Mv CM   mi vi   pi  p tot

i
i
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
 The
acceleration of the center of mass can
be found by differentiating the velocity with
respect to time
aCM
dv CM
1


dt
M
m a
i i
i
93
Forces In a System of Particles
 The
acceleration can be related to a force
MaCM   Fi
i
 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:
Fext = 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
95
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 Fext = 0
96
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, Fext = M aCM  Fg = M g
97
Pembatalan daya letupan:
daya dalam tiada sumbangan dalam Fext

Tapi, daya letupan
bukanlah daya luar
 Atau dalam kata lain dayadaya letupan membatalkan
satu sama lain dan
sumbangan bersih mereka
kepada fext ialah sifar
 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
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
103
Contoh 9.17
104
Conceptual example 9.16
105
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
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
 The initial momentum
of the system is pi =
(M + Dm) v
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
 The rocket’s speed
has increased by Dv

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
 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
110
Rocket Propulsion, 5

The basic equation for rocket propulsion is
M
v f  vi  ve ln  i
Mf


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
111
Thrust

The thrust on the rocket is the force exerted on it
by the ejected exhaust gases
Thrust =
dv
dM
M
dt
 ve
dt

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
112
Fig. 8.28 Young
113
114
Contoh 9.19
115
Contoh 8.16 Young
116
Quick quiz
 Pg.
313 Young, Test your understanding
117
118