Transcript momentum-120305090246-phpapp02 (1)x

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Momentum and Collisions
Chapter 8
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Momentum and Impulse
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What do you think?
 Imagine
an automobile collision in which
an older model car from the 1960s collides
with a car at rest while traveling at 15 mph.
Now imagine the same collision with a 2007
model car. In both cases, the car and
passengers are stopped abruptly.
 List
the features in the newer car that are
designed to protect the passenger and the
features designed to minimize damage to the
car.
 How are these features similar?
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What do you think?
 What
are some common uses of the term
momentum?
 Write
a sentence or two using the term
momentum.
 Do
any of the examples provided
reference the velocity of an object?
 Do
any of the examples reference the
mass of an object?
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Momentum
 Momentum
(p) is proportional to both mass and
velocity.
A
vector quantity, so direction matters!!
 Things moving right are positive
 Tings moving left are negative
 SI
Units: kg • m/s
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Example
A
2250kg pick-up truck has a velocity of
25m/s to the east. What is the momentum
of the truck?
 m=
2250kg
 P=
mv
 P=
(2250)(25)
 P=
56,250 kg • m/s
v=25m/s east (+)
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Momentum and Newton’s 2nd Law
 Prove
that the two equations shown below are
equivalent.
F
= ma
 Newton
 Force
and
F = p/t
actually wrote his 2nd Law as F = p/t.
depends on how rapidly the momentum
changes.
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Impulse and Momentum
The
 SI
quantity Ft is called impulse.
units: N•m or kg•m/s
Impulse
 Another
equals change in momentum.
version of Newton’s 2nd Law
 Changes in momentum depend on both the
force and the amount of time over which the
force is applied.
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Changing momentum
 Greater
changes in
momentum (p) require
more force (F) or more
time (t) .
A
loaded truck requires
more time to stop.
 Greater
p for truck with
more mass
 Same stopping force
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Movie
 file:///Volumes/Physics_mac/inquiry_p
pts/files/ch06/70598.html
 Put
Advanced CD in to work
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Example
A
1400kg car moving westward with a
velocity of 15m/s collides with a utility pole
and is brought to rest in 0.3sec. Find the force
exerted on the car during collision.
 Given:
m=1400kg
t= 0.30sec
vi= 15m/s west (-)
vf= 0m/s
F=??
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Example
 Ft=
mvf- mvi
F
mv f  mvi
t
(1400)(0)  (1400)(15)
F
0.30
 F=70,000N
to the east
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Example
A
2240kg car traveling west slows down
uniformly from 20m/s to 5m/s. how long
does it take the car to slow down if the
force on the car is 8410N to the east?
How far does the car travel during the
time it slows down?
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Example
t
 Given:
 m=
2240kg
 vi=
20m/s west (-)
 vf=
5m/s west (-)
 F=
8410N east (+)
 t=
??
d=??
mv f  mvi
F
(2240)(5)  (2240)(20)
t
8410
 t=
4 sec
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Example
 d=
½ (vi + vf)t
 d=
½ (-20-5) 4
 d=
-50 m/s or 50 m/s west
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Stopping Time
 Ft
= p = mv
 When
stopping, p is the same for rapid or
gradual stops.
 Increasing
the time (t) decreases the force (F).
 What
examples demonstrate this relationship?
 Air bags, padded dashboards, trampolines, etc
 Decreasing
 What
the time (t) increases the force (F).
examples demonstrate this relationship?
 Hammers and baseball bats are made of hard
material to reduce the time of impact.
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Conservation of Momentum
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What do you think?
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Two skaters have equal mass and are at rest.
They are pushing away from each other as
shown.
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Compare the forces on the two girls.
Compare their velocities after the push.
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How would your answers change if the girl
on the right had a greater mass than her
friend?
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How would your answers change if the girl
on the right was moving toward her friend
before they started pushing apart?
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Momentum During Collisions
 The
change in momentum for one object is equal and
opposite to the change in momentum for the other
object.
 Total
momentum is neither gained not lost during
collisions.
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Conservation of Momentum
 Total
momentum remains constant during
collisions.
 The
momentum lost by one object equals the
momentum gained by the other object.
 Conservation
solving.
of momentum simplifies problem
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Example
A
76kg boater, initially at rest, in a
stationary 45kg boat steps out of the
boat onto the dock. If the boater moves
out of the boat with a velocity of 2.5m/s
to the right, what is the final velocity of
the boat?
Given
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 m1=
76kg
 m2=
45kg
 v1i=
0m/s
 v2i=
0m/s
 v1f=
2.5 m/s
 v2f=
??
 m1v1i
+ m2v2i= m1v1f + m2v2f
 (76)(0)+(45)(0)=(76)(2.5)+(45)(v2f)
 0=
190 + 45v2f
 -190
= 45v2f
 v2f=
-4.2 m/s to the right but notice the negative
 v2f=
4.2 m/s to the left
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Now what do you think?
•
Two skaters have equal mass and are at
rest. They are pushing away from each
other as shown.
•
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Compare the forces on the two girls.
Compare their velocities after the push.
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How would your answers change if the girl
on the right had a greater mass than her
friend?
•
How would your answers change if the girl
on the right was moving toward her friend
before they started pushing apart?
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Elastic and Inelastic Collisions
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What do you think?
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Collisions are
sometimes described
as elastic or inelastic.
To the right is a list of
colliding objects. Rank
them from most elastic
to most inelastic.
What factors did you
consider when ranking
these collisions?
1.
A baseball and a bat
2.
A baseball and a glove
3.
Two football players
4.
Two billiard balls
5.
Two balls of modeling
clay
6.
Two hard rubber toy balls
7.
An automobile collision
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Collisions
 Perfectly
inelastic collisions
A
collision where 2 objects stick together
after colliding
 Perfectly
inelastic collisions are
analyzed in terms of momentum
 Due
to the fact that the objects become
one object after the collision
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Perfectly Inelastic Collisions
 Two
objects collide and stick together.
 Two
football players
 A meteorite striking the earth
 Momentum
 Masses
is conserved.
combine.
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Perfectly Inelastic Collisions
 v1i
is + (m1 is moving to the right)
 V2i
is – (m2 is moving to the left)
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Example
A
1850kg luxury sedan stopped at a
traffic light is struck from the rear by a
compact car with a mass of 975kg. Te 2
cars become entangled as a result of the
collision. If the compact car was moving
at a velocity of 22m/s to the north before
the collision, what is the velocity of the
mass after the accident?
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Example
 m1=
1850kg
 v1i=
0m/s
 m1v1i
 m2=
975kg
 v2i=
22m/s north (+)
 vf=
+ m2v2i= (m1+ m2) vf
(1850)(0)  (975)(22)
vf 
1850  975
 vf=
7.59 m/s north
??
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Classroom Practice Problem
 Gerard
is a quarterback and Tyler is a
defensive lineman. Gerard’s mass is 75.0
kg and he is at rest. Tyler has a mass of
112 kg, and he is moving at 8.25 m/s
when he tackles Gerard by holding on
while they fly through the air. With what
speed will the two players move
together after the collision?
 Answer: 4.94
m/s
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Classroom Practice Problems
 An
2.0 x 105 kg train car moving east at 21 m/s
collides with a 4.0 x 105 kg fully-loaded train
car initially at rest. The two cars stick together.
Find the velocity of the two cars after the
collision.
 Answer: 7.0
m/s to the east
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Inelastic Collisions
 Kinetic
energy is less after the collision.
 It
is converted into other forms of energy.
 Internal energy - the temperature is
increased.
 Sound energy - the air is forced to
vibrate.
 Some
kinetic energy may remain after
the collision, or it may all be lost.
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Example
2
clay balls collide in a perfect inelastic
collision. The 1st ball has a mass of 0.500kg
and an initial velocity of 4.0m/s to the right.
The 2nd ball has a mass of 0.250 kg. and an
initial velocity of 3.0m/s to the left. What is
the decrease in kinetic energy during the
collision?
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Example
 m1=
 v1=
4m/s right (+)
 m2=
 v2=
0.500 kg
0.250kg
3m/s left (-)
 KE=
 KE=
KEf – KEi
 KEi=
½ m1v1i + m2v2i
 KEf=
½ (m1+m2)vf2
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Example
 m1v1i
+ m2v2i= (m1+ m2) vf
m1v1i  m2 v2i (0.5)(4)  (0.25)( 3)
vf 

(m1  m2 )
(0.5  0.25)
 vf=
1.67 m/s
 KEi=
½ (0.5)(42) + ½ (0.25)(-32)= 5.125J
 KEf=
½ (0.5 + 0.25) 1.672 = 1.05J
 KE=
1.05 – 5.125 = -4.08J
 Means that KE is lost
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Elastic Collisions
 Objects
collide and return to their original shape.
 Kinetic
energy remains the same after the
collision.
 Perfectly
elastic collisions satisfy both
conservation laws shown below.
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Elastic Collisions
 Two
billiard balls collide head on, as
shown. Which of the following possible final
velocities satisfies the law of conservation
of momentum?
 vf,A
= 2.0 m/s, vf,B = 2.0 m/s
 vf,A = 0 m/s, vf,B = 4.0 m/s
 vf,A = 1.5 m/s, vf,B= 2.5 m/s
 Answer: all
m = 0.35 kg
m = 0.35 kg
three
v = 4.0 m/sv= 0 m/s
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Elastic Collisions
 Two
billiard balls collide head on, as
shown. Which of the following possible final
velocities satisfies the law of conservation
of kinetic energy?
 vf,A
= 2.0 m/s, vf,B = 2.0 m/s
 vf,A = 0 m/s, vf,B = 4.0 m/s
 vf,A = 1.5 m/s, vf,B= 2.5 m/s
 Answer: only vf,A
vf,B = 4.0 m/s
m = 0.35 kg
m = 0.35 kg
= 0 m/s,
v = 4.0 m/s
v= 0 m/s
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Example
A
0.015 kg marble moving to the right at
0.225 m/s makes an elastic head on collision
with a 0.030 kg shooter marble moving to the
left at 0.180m/s. After the collision, the
smaller marble moves to the left at 0.315m/s.
Assume neither marble rotates at all and
both marbles are on a frictionless surface.
What is the velocity of the 0.030kg marble
after the collision?
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 m1=
0.015kg
 m2=
0.030kg
 v1i=
0.225m/s right (+)
 v2i=
0.18m/s left (-)
 v1f=
0.315m/s left (-)
 vf=
 m1v1i

??
+ m2v2i= m1v1f + m2v2f
(0.015)(0.225) + (0.030)(-0.18) = (0.015)(-0.315) + (0.030)(vf)
 0.003375
+ -0.0054 = -0.004725 + 0.030 vf
 -0.002025=
 0.0027
 vf=
- 0.004725 + 0.030 vf
= 0.030vf
0.09 m/s right
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Types of Collisions
Click below to watch the Visual Concept.
 file:///Volumes/Physics_mac/inquiry_p
pts/files/ch06/70600.html
 Put
Advanced CD in to work
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Types of Collisions
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Now what do you think?
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To the right is a list of
colliding objects. Rank
them from most elastic
to most inelastic.
What factors did you
consider when ranking
these collisions?
1.
A baseball and a bat
2.
A baseball and a glove
3.
Two football players
4.
Two billiard balls
5.
Two balls of modeling
clay
6.
Two hard rubber toy balls
7.
An automobile collision