Chapter 9 Linear Momentum, Systems of Particles, and Collisions.
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Transcript of Chapter 9 Linear Momentum, Systems of Particles, and Collisions.
![Page 1: Chapter 9 Linear Momentum, Systems of Particles, and Collisions.](https://reader036.fdocuments.in/reader036/viewer/2022081502/56649d005503460f949d297e/html5/thumbnails/1.jpg)
Chapter 9
Linear Momentum, Systems of Particles, and Collisions
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Linear momentum (Ch. 4)
• Linear momentum (or, simply momentum) of a point-like object (particle) is
• SI unit of linear momentum is kg*m/s
• Momentum is a vector, its direction coincides with the direction of velocity
vmp
p
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Newton’s Second Law revisited (Ch. 4)
• Originally, Newton formulated his Second Law in a more general form
• The rate of change of the momentum of an object is equal to the net force acting on the object
• For a constant mass
dt
pdFnet
dt
pdFnet
dt
vmd )(
dt
vdm
am
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Center of mass
• In a certain reference frame we consider a system of particles, each of which can be described by a mass and a position vector
• For this system we can define a center of mass:
M
rm
m
rmr i
ii
ii
iii
CM
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Center of mass of two particles
• A system consists of two particles on the x axis
• Then the center of mass is
21
2211
mm
xmxmxCM
M
rm
m
rmr i
ii
ii
iii
CM
21
21 00
mm
mmyCM
0
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Center of mass of a rigid body
• For a system of individual particles we have
• For a rigid body (continuous assembly of matter) with volume V and density ρ(V) we generalize a definition of a center of mass:
ii
iii
CM m
rmr
M
dmr
dV
dVrr
volume
volumeCM
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Chapter 9Problem 41
Find the center of mass of the uniform, solid cone of height h, base radius R, and constant density shown in the figure. (Hint: Integrate over disk-shaped mass elements of thickness dy, as shown in the figure.)
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Newton’s Second Law for a system of particles
• For a system of particles, the center of mass is
• ThenM
rm
m
rmr i
ii
ii
iii
CM
i
iiCM rmrM
2
2
2
2 )(
dt
rMd
dt
rdMaM CMCM
CM
2
2
dt
rmdi
ii
i
ii dt
rdm
2
2
iiiam
i
iF
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Newton’s Second Law for a system of particles
• From the previous slide:
• Here is a resultant force on particle i
• According to the Newton’s Third Law, the forces that particles of the system exert on each other (internal forces) should cancel:
• Here is the net force of all external forces that act on the system (assuming the mass of the system does not change)
i
iCM FaM
netCM FaM
iF
netF
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Newton’s Second Law for a system of particles
netCM FaM
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Linear momentum for a system of particles
• We define a total momentum of a system as:
• Using the definition of the center of mass
• The linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the center of mass
i
iii
i vmpP
i
iivmP
dt
rmd ii
i
dt
rdM CM
CMvM
i
ii dt
rdm
M
rm
m
rmr i
ii
ii
iii
CM
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Linear momentum for a system of particles
• Total momentum of a system:
• Taking a time derivative
• Alternative form of the Newton’s Second Law for a system of particles
CMvMP
dt
vdM
dt
Pd CM
CMaM
netF
netFdt
Pd
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Conservation of linear momentum
• From the Newton’s Second Law
• If the net force acting on a system is zero, then
• If no net external force acts on a system of particles, the total linear momentum of the system is conserved (constant)
• This rule applies independently to all components
netFdt
Pd
0dt
Pd
constP
constPF xxnet 0,
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Chapter 9Problem 17
A popcorn kernel at rest in a hot pan bursts into two pieces, with masses 91 mg and 64 mg. The more massive piece moves horizontally at 47 cm/s. Describe the motion of the second piece.
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Impulse
• During a collision, an object is acted upon by a force exerted on it by other objects participating in the collision
• We define impulse as:
• Then (momentum-impulse theorem)
dttFpd net )(
)(tFdt
pdnet
f
i
f
i
t
t net
t
tdttFpd )(
f
i
t
t net dttFJ )(
Jpp if
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Elastic and inelastic collisions
• During a collision, the total linear momentum is always conserved if the system is isolated (no external force)
• It may not necessarily apply to the total kinetic energy
• If the total kinetic energy is conserved during the collision, then such a collision is called elastic
• If the total kinetic energy is not conserved during the collision, then such a collision is called inelastic
• If the total kinetic energy loss during the collision is a maximum (the objects stick together), then such a collision is called perfectly inelastic
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Elastic collision in 1D
constK constP
2222
222
211
222
211 ffii
vmvmvmvm ffii vmvmvmvm 22112211
)()( 222111 fifi vvmvvm ))(( 11111 fifi vvvvm
))(( 22222 fifi vvvvm
iif vmm
mv
mm
mmv 2
21
21
21
211 )(
2
)(
)(
iif vmm
mmv
mm
mv 2
21
121
21
12 )(
)(
)(
2
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Elastic collision in 1D: stationary target
• Stationary target: v2i = 0
• Then
iif vmm
mv
mm
mmv 2
21
21
21
211 )(
2
)(
)(
iif vmm
mmv
mm
mv 2
21
121
21
12 )(
)(
)(
2
if vmm
mmv 1
21
211 )(
)(
if vmm
mv 1
21
12 )(
2
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Perfectly inelastic collision in 1D
constP
fii vmmvmvm )( 212211
21
2211
mm
vmvmv ii
f
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Collisions in 2D
constP constPx
constPy
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Chapter 9Problem 86
In a ballistic pendulum demonstration gone bad, a 0.52-g pellet, fired horizontally with kinetic energy 3.25 J, passes straight through a 400-g Styrofoam pendulum block. If the pendulum rises a maximum height of 0.50 mm, how much kinetic energy did the pellet have after emerging from the Styrofoam?
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Questions?
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Answers to the even-numbered problems
Chapter 9
Problem 122.5 m
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Answers to the even-numbered problems
Chapter 9
Problem 164680 km
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Answers to the even-numbered problems
Chapter 9
Problem 18– 10.6 iˆ – 2.8 jˆ m/s
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Answers to the even-numbered problems
Chapter 9
Problem 787.95 s