Momentum Halliday Pembahasan SOla
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Transcript of Momentum Halliday Pembahasan SOla
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IMPULS & MOMENTUM
M. Djamil
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Momentum From Newton’s laws: force must be present to change an
object’s velocity (speed and/or direction) Wish to consider effects of collisions and corresponding
change in velocity
Method to describe is to use concept of linear momentum
scalar vector
Linear momentum = product of mass velocity
Golf ball initially at rest, so some of the KE of club transferred to provide motion of golf ball and its change in velocity
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GAYA IMPULS.• Suatu gaya yang bekerja dalam waktu singkat disebut gaya
impuls; misal. Bola yang menumbuk tembok, tumbukan dua benda, dan bentuk tumbukan yag lain.
• Secara matematis, dari Newton II dapat dirumuskan sebagai berikut :
F = m a F = m dv/dt F dt = m dV
IMPULS - MOMENTUM
F
t
tIImpuls Momentum
Cek dimensi & satuannya !
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Impulse
In order to change the momentum of an object (say, golf ball), a force must be applied
The time rate of change of momentum of an object is equal to the net force acting on it
Gives an alternative statement of Newton’s second law
(F Δt) is defined as the impulse Impulse is a vector quantity, the direction is the
same as the direction of the force
tFporamt
vvm
t
pF net
ifnet
:)(
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Momentum
Vector quantity, the direction of the momentum is the same as the velocity’s
Applies to two-dimensional motion as well
yyxx mvpandmvp
vmp
Size of momentum: depends upon mass depends upon velocity
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Example: Impulse Applied to Auto Collisions
The most important factor is the collision time or the time it takes the person to come to a rest This will reduce the chance of dying in a car crash
Ways to increase the time Seat belts Air bags
The air bag increases the time of the collision and absorbs some of the energy from the body
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ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?
1. It takes less time to stop the ping-pong ball.2. Both take the same time.3. It takes more time to stop the ping-pong ball.
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Answer
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare?
1. It takes less time to stop the ping-pong ball.2. Both take the same time.3. It takes more time to stop the ping-pong ball.
Note: Because force equals the time rate of change of momentum, the two balls loose momentum at the same rate. If both balls initially had the same momenta, it takes the same amount of time to stop them.
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Problem: Teeing Off
A 50-g golf ball at rest is hit by “Big Bertha” club with 500-g mass. After the collision, golf leaves with velocity of 50 m/s.
a) Find impulse imparted to ballb) Assuming club in contact with
ball for 0.5 ms, find average force acting on golf ball
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Problem: teeing off
Given:
mass: m=50 g = 0.050 kgvelocity: v=50 m/s
Find:
impulse=?Faverage=?
1. Use impulse-momentum relation:
2. Having found impulse, find the average force from the definition of impulse:
smkg
smkg
mvmvpimpulse if
50.2
050050.0
N
s
smkg
t
pFthustFp
3
3
1000.5
105.0
50.2,
Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball:
iiff
ifif
R
VMvmVMvm
orVMVMvmvm
ortFtF
,
, of club
CONSERVATION OF MOMENTUM
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Conservation of Momentum
Definition: an isolated system is the one that has no external forces acting on it
A collision may be the result of physical contact between two objects
“Contact” may also arise from the electrostatic interactions of the electrons in the surface atoms of the bodies
Momentum in an isolated system in which a collision occurs is conserved (regardless of the nature of the forces between the objects)
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Conservation of Momentum
The principle of conservation of momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision
v1
v2v’2
v’1
m1m2
F’F
m1 v’1 + m2 v’2 = m1 v1 + m2 v2
Hk Kek. Momentum
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Concept TestSuppose a person jumps on the surface of Earth. The Earth
1. will not move at all2. will recoil in the opposite direction with tiny velocity3. might recoil, but there is not enough information provided to see if that could happened
Please fill your answer as question 23 of General Purpose Answer Sheet
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Concept Test
Suppose a person jumps on the surface of Earth. The Earth
1. will not move at all2. will recoil in the opposite direction with tiny velocity3. might recoil, but there is not enough information provided to see if that could happened
Note: momentum is conserved. Let’s estimate Earth’s velocity after a jump by a 80-kg person. Suppose that initial speed of the jump is 4 m/s, then:
smkg
smkgV
smkgVMp
smkgp
Earth
EarthEarth
2324
103.5106
320
so,320:Earth
320:Person
tiny negligible velocity, in opposite direction
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Types of Collisions
Momentum is conserved in any collision
what about kinetic energy?
Inelastic collisionsKinetic energy is not conserved
○ Some of the kinetic energy is converted into other types of energy such as heat, sound, work to permanently deform an object
Perfectly inelastic collisions occur when the objects stick together○ Not all of the KE is necessarily lost
energylost fi KEKE
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Perfectly Inelastic Collisions:
When two objects stick together after the collision, they have undergone a perfectly inelastic collision
Suppose, for example, v2i=0. Conservation of momentum becomes
fii vmmvmvm )( 212211
.20105.2
105
,)2500(0)50)(1000(
:1500,1000ifE.g.,
3
4
21
smkg
smkgv
vkgsmkg
kgmkgm
f
f
fi vmmvm )(0 2111
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Perfectly Inelastic Collisions:
What amount of KE lost during collision?
Jsmkg
vmvmKE iibefore
62
222
211
1025.1)50)(1000(2
12
1
2
1
Jsmkg
vmmKE fafter
62
221
1050.0)20)(2500(2
1
)(2
1
JKElost61075.0
lost in heat/”gluing”/sound/…
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More Types of Collisions Elastic collisions
both momentum and kinetic energy are conserved
Actual collisionsMost collisions fall between elastic and
perfectly inelastic collisions
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More About Elastic Collisions
Both momentum and kinetic energy are conserved
Typically have two unknowns
Solve the equations simultaneously
222
211
222
211
22112211
2
1
2
1
2
1
2
1ffii
ffii
vmvmvmvm
vmvmvmvm
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Elastic Collisions:• Using previous example
(but elastic collision is assumed)
smkg
smkgsmkg
vmvmP iibefore
4
2111
100.2
)20)(1500()50)(1000(
smv
smv
f
f
1.31
7.26
2
1
For perfectly elastic collision:
J
JJ
vmvmKE iibefore
6
56
222
211
1055.1
1031025.1
2
1
2
1
222
211
6
22114
2
1
2
11055.1
100.2
ff
ff
vmvmJ
vmvmsmkg
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Problem Solving for One Dimensional Collisions
Set up a coordinate axis and define the velocities with respect to this axis It is convenient to make your axis coincide
with one of the initial velocities
In your sketch, draw all the velocity vectors with labels including all the given information
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Sketches for Collision Problems
• Draw “before” and “after” sketches
• Label each object – include the direction of
velocity– keep track of subscripts
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Sketches for Perfectly Inelastic Collisions
• The objects stick together
• Include all the velocity directions
• The “after” collision combines the masses
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Problem Solving for One-Dimensional Collisions.
Write the expressions for the momentum of each object before and after the collision Remember to include the appropriate signs
Write an expression for the total momentum before and after the collision Remember the momentum of the system is what is
conserved
If the collision is inelastic, solve the momentum equation for the unknown Remember, KE is not conserved
If the collision is elastic, you can use the KE equation to solve for two unknown
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Glancing Collisions
For a general collision of two objects in three-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved
Use subscripts for identifying the object, initial and final, and components
fyfyiyiy
fxfxixix
vmvmvmvm
andvmvmvmvm
22112211
22112211
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Glancing Collisions
• The “after” velocities have x and y components
• Momentum is conserved in the x direction and in the y direction
• Apply separately to each direction
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Problem Solving for Two-Dimensional Collisions
Set up coordinate axes and define your velocities with respect to these axes It is convenient to choose the x axis to coincide with
one of the initial velocities
In your sketch, draw and label all the velocities and include all the given information
Write expressions for the x and y components of the momentum of each object before and after the collision
Write expressions for the total momentum before and after the collision in the x-direction Repeat for the y-direction
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Problem Solving for Two-Dimensional Collisions, final
Solve for the unknown quantities If the collision is inelastic, additional
information is probably required If the collision is perfectly inelastic, the final
velocities of the two objects is the same If the collision is elastic, use the KE equations
to help solve for the unknowns
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Tugas Group :
Group : Holliday: Giancoli Ket.
1 & 6 9.38 7.04
2 & 7 9.35 7.17
3 & 8 9.33 7.20
4 & 9 9.32 7.27
5 & 10 9.30 7.29
Holliday hal : 232 - 233Giancolli hal : 188 - 189
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See You Next Week
Prepare :Torque
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Rocket Propulsion The operation of a rocket depends on the
law of conservation of momentum as applied to a system, where the system is the rocket plus its ejected fuel This is different than propulsion on the earth
where two objects exert forces on each other road on car train on track
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Rocket Propulsion, cont.
The rocket is accelerated as a result of the thrust of the exhaust gases
This represents the inverse of an inelastic collision Momentum is conserved Kinetic Energy is increased (at the expense of
the stored energy of the rocket fuel)
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Rocket Propulsion
The initial mass of the rocket is M + Δm M is the mass of the rocket m is the mass of the fuel
The initial velocity of the rocket is v
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Rocket Propulsion
The rocket’s mass is M The mass of the fuel, Δm, has been ejected The rocket’s speed has increased to v + Δv
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v
- 12
'1
'2
vv
ve
Koefisien Tumbukan / Restitusi :
1 e 0 Harga e
e = 1 elastis sempurna, sedang e = 0 tak elastis !
Contoh : 1. Sebuah bola dijatuhkan dari ketinggian h1 menumbuk lantai dan mantul naik setinggi h2.
Hitunglah koef. restitusinya.Penyelesaian :Kecepatan bola saat akan menumbuk lantai dapat dihitung dariteori kinematika, diperoleh :
v1 = (2 gh1)½
Kecepatan setelah menumbuk lantai dan mengakibatkan bola naik setinggi h2 adalah
v’1 = (2 gh2) ½. v1
v2
h1
h2
v
- 12
'1
'2
vv
ve
Karena v’2 = v2 = 0 - 1
'1
v
ve
1
2
1
2
h g 2
h g 2 -
h
he
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2. Dua buah bola masing-masing bermassa m1 = 2 kg, m2 = 3 kg bergerak dalam arah berlawanan dengan kecepatan v1 = 10 m/s dan v2 = 6 m/s. Jika koef. restitusi e = 0.4. Hitunglah kecepatan masing-masing benda setelah tumbukan dan berapa energi sistem yang hilang ?.
m1
m1 m2
m2
v1 v2
v’2v’1
Hukum kekekalan momentum : m1 v1 + m2 v2 = m1 v’1 + m2 v’2
2.10 + 3.(-6) = 2 . v’1 + 3 v’2
2 [kg m/det] = 2 v’1 + 3 v’2 . . . . . . . .(1)
1,6 [ m/det] = v’1 – v’2 . . . . . . . . . . . (2)
Dari kedua pers. maka v’1 dan v’2 dapat diperoleh.Untuk menghitung energi yang hilang berarti dihitung energi kinetik sebelum dan setelah tumbukan.
v
- 12
'1
'2
vv
ve
106
v - 4,0
'1
'2
v