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Momentum
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Was this accurate?
What really Happens?
Does this make a difference:
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We know that it is harder to get a more
massive object moving from rest than a less
massive object.
◦ This is the concept of inertia previously
introduced.
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Building on the concept of inertia, we
ask the question, “How hard is it to
stop an object?”
◦ We call this property of an object that is related to
both its mass and its velocity, the momentum of
the object.
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The linear momentum of an object is defined as
the product of its mass and its velocity, and is
measured in “kilogram-meters per second.”
The symbol for momentum is p, as in:
p = mv
The bold-faced symbols above, momentum and
velocity, are vectors with identical direction.
Look at the equation. Can you guess why the units are
defined this way?
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So an object may have a
large momentum due to a
large mass, a large velocity,
or both. A slow battleship
and a rocket-propelled race
car (or kayak) have large
momenta.
Conceptual Question: Which
has the greater momentum,
an 18-wheeler parked at
the curb, or a Volkswagen
rolling down a hill?
A.
B.
C.
D.
19 wheeler
VW
Both the same
Can’t tell.
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The word momentum is often
used in our everyday language
in a much looser sense, but it
is still roughly consistent with
its meaning in the physics
world view; that is, something
with a lot of momentum is
hard to stop. You have
probably heard someone say,
“We don’t want to lose our
momentum!” Coaches are
particularly fond of this word.
The catcher is protected from
the baseball’s momentum by
body armor.
(v f  v0 )
v
F  ma  m
m
t
t f  t0
F
(mv f  mv0 )
t f  t0
Ft  Impulse  (mv)
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Because impulse is a product of two things, there are many
ways to produce a particular change in momentum.
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For example, two ways of changing an object’s momentum by
10 kilogram-meters per second are:
◦ exert a net force of 5 newtons on the object for 2 seconds, or
◦ exert 100 newtons for 0.1 second.
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They each produce an impulse of 10 newton-seconds (N·s)
and therefore a momentum change of 10 kilogram-meters
per second.
The units of impulse (newton-seconds) are equivalent to
those of momentum (kilogram-meters per second).
Question: Which of the following will cause the
larger change in the momentum of an object:
A. a force of 2 newtons acting for 10
seconds, or
B. a force of 3 newtons acting for 6
seconds?
C. Both the same
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Because our bones break when
forces are large, the particular
combination of force and time
interval is important.
◦ Sometimes, as in a gymnasium, a
wood floor may be enough of a
cushion; in a car, the dashboards
are made from foam rubber.
Bumpers and air bags further
increase the vehicle’s (and the
passengers’) Δt.
A pole-vaulter lands on thick pads to increase the collision time
and thus reduce the force. Otherwise, this could be a nasty fall.
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Sometimes accidents can fill in for the lack
of a parachute:
◦ During World War II, a Royal Air Force airman
jumped in just his uniform from a flaming
bomber flying at 18,000 ft. His terminal speed
(Ch. 3, p. 46) was 120 mph, but tree branches
and then 1½ feet of snow cushioned his fall.
The result: minor scratches and bruises.
◦ The record for surviving a fall without a
parachute is held by Vesna Vulovic. She was
serving as a hostess on a Yugoslavian DC-9 that
blew up at 33,330 ft in 1972.
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Play catch with a raw egg.
After each successful toss,
take one step away from your
partner.
How do you catch the egg to
keep from breaking it?
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If you are falling, (or just moving along), you
have a certain MOMENTUM, just before you
come in contact with what is going to stop
you.
Assume that your FINAL velocity (and
momentum) will be zero.
Your momentum (final momentum=0) will be
the impulse necessary to stop your fall.
Ft  (mv) just before the collision  constant
assume 10 ft floors & 50 floors  500 ft
Falls for 4 seconds
time
speed
1
32 ft/s
2
64 ft/s
3
96 ft/s
4
128 ft/s
catch
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Speed=128 ft/sec
Her weight = 100 lbs
W=mg so her mass = 100/32 ~3 English
mass units (slugs)
Momentum = mv = 3x128=384 (appropriate
units)
F x t=384
F=384/t
Time to stop
Force in pounds
.1 seconds
3840 (ouch!)
.5 seconds
768
1 second
384
How Fast is the
Stop??
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Imagine standing on a giant
skateboard, at rest.
◦ The total momentum of you and the
skateboard must be zero, because
everything is at rest.
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Now suppose that you walk on the
skateboard. What happens to the skateboard?
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When you walk in one direction, the skateboard
moves in the other direction, as shown in the figure.
An analogous thing happens when you fire a rifle:
the bullet goes in one direction, and the rifle recoils
in the opposite direction.
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The force you exert on the skateboard is, by Newton’s
third law, equal and opposite to the force the skateboard
exerts on you.
◦ Because you and the skateboard each experience the same force
for the same time interval, you must each experience the samesize impulse and, therefore, the same-size change in
momentum.
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Because the impulses are in opposite directions
(red arrows, foot/skateboard & you), the changes in the
momenta are also in opposite directions.
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Thus, your momentum and that of the
skateboard still add to zero.
Even though you and the skateboard are moving
and, individually, have nonzero momenta,
the total momentum remains zero.
http://www.youtube.com/watch?v=KL8-PbdRYY0
Conservation of Momentum
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Because the changes in the momenta of the two objects are
equal in size and opposite in direction, the value of the total
momentum does not change. We say that the total momentum is
conserved.
We can generalize these findings. Whenever any object is acted
on by a force, there must be at least one other object involved.
◦ This other object might be in actual contact with the first,
◦ or it might be interacting at a distance of 150 million km, but it is
there.
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Consider the objects as a system. Whenever there is no net force
acting on the system from the outside (that is, the system is
isolated, or closed), the forces that are involved act only
between the objects within the system. As a consequence of
Newton’s third law, the total momentum of the system remains
constant.
This generalization is known as the law of conservation of linear
momentum.
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Consider a “system” of “interacting” particles.
At one instant of time add up all the
momenta of all of the particles in the system.
◦ Vector addition, of course, but let’s not worry much
about that.
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Wait some amount of time.
After that time, the sum of the momenta is
the same as it was when initially measured.
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MOMENTUM IS CONSERVED
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The Law of Conservation of Linear Momentum:
◦ The total linear momentum of a system does not change
if there is no net external force.
A GAS??
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You experience conservation of
momentum firsthand when you try to step
from a small boat onto a dock.
◦ As you step toward the dock, the boat moves away
from the dock, and you may fall into the water.
Although the same effect occurs when we
disembark from an ocean liner, the large
mass of the ocean liner reduces the speed
given it by our stepping off.
◦ A large mass requires a small change in velocity to
undergo the same change in momentum.
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If a rifle of mass 4.5 kg fires 10-gram bullets
at a speed of 900 m/s (2000 mph), what will
be the speed of its recoil?
◦ The magnitude of the momentum p of the bullet
is:
p = mv = (0.01 kg)(900 m/s)
= 9 kg·m/s
◦ Because the total momentum of the bullet and rifle
was initially zero, conservation of momentum
requires that the rifle recoil with an equal
momentum in the opposite direction. If the mass
of the rifle is M, the speed V of its recoil is given
by
=2 m/s
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If you do not hold the rifle snugly against
your shoulder, the rifle will hit your shoulder
at this speed (2 m/s = 4.5 mph) and hurt
you.
Question: Why does holding the rifle snugly
reduce the recoil effects?
Answer: Holding the rifle snugly increases
the recoiling mass (your mass is now added
to that of the rifle) and therefore reduces the
recoil speed.
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Interacting objects don’t need to be initially at rest for
conservation of momentum to be valid. Suppose a ball moving
to the left with a certain momentum crashes head-on with an
identical ball moving to the right with the same-size
momentum.
◦ Before the collision, the two momenta are equal
in size but opposite in direction, and because
they are vectors, they add to zero.
◦ After the collision the balls move apart with
equal momenta in opposite directions.
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Because the balls are identical and their masses are the
same, the speeds after the collision are also the same.
These speeds depend on the ball’s physical properties.
In all cases the two momenta are the same size and in
opposite directions, and the total momentum remains
zero.
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A question on the final exam asks, “What do we mean when
we claim that the total momentum is conserved during a
collision?” The following two answers are given:
◦ Answer 1: Total momentum of the system stays the same before
and after the collision.
◦ Answer 2: Total momentum of the system is zero before and after
the collision.
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Which answer (if either) do you agree with?
Answer: While we have considered several examples in which
the total momentum of the system is zero, this is not the
most general case. The momentum of a system can have any
magnitude and any direction before the collision. If
momentum is conserved, the momentum of the system always
has the same magnitude and direction after the collision.
Therefore, answer 1 is correct. This is a very powerful
principle because of the word always.
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A boxcar traveling at 10 meters per second approaches a string
of four identical boxcars sitting stationary on the track. The
moving boxcar collides and links with the stationary cars, and
the five boxcars move off together along the track.
What is the final speed of the five cars immediately after the
collision?
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Conservation of momentum tells us that the total momentum
must be the same before and after the collision.
◦ Before the collision, one car is moving at 10 meters per second.
◦ After the collision, five identical cars are moving with a common final
speed.
◦ Because the amount of mass that is moving has increased by a factor
of five, the speed must decrease by a factor of five. The cars will have
a final speed of 2 meters per second.
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Notice that we did not have to know the mass of each boxcar,
only that they all had the same mass.
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We can use the conservation of momentum to
measure the speed of fast-moving objects. For
example, consider determining the speed of an arrow
shot from a bow.
We first choose a movable, massive target—a wooden
block suspended by strings. Before the arrow hits the
block (a), the total momentum of the system is equal
to that of the arrow (the block is at rest).
After the arrow is embedded in the block (b), the two
move with a smaller, more measurable speed.
 The final momentum of the block and arrow just after
the collision is equal to the initial momentum of the
arrow. Knowing the masses, the arrow’s initial speed
can be determined.
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Accident investigators use conservation of
momentum to reconstruct automobile
accidents.
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Newton’s laws can’t be used to analyze
the collision itself because we do not
know the detailed forces involved.
Conservation of linear momentum,
however, tells us that regardless of the
details of the crash, the total
momentum of the two cars remains
the same.
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The total momentum immediately
before the crash must be equal to that
immediately after the crash.
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As an example, consider a rear-end collision. Assume that
the front car was stopped and the two cars locked bumpers
on impact.
From an analysis of the length of the skid marks made after
the collision and the type of surface, the total momentum
of the two cars just after the collision can be calculated.
◦ (We will see how to do this in Chapter 7; for now, assume we
know their total momentum.)
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Because one car was stationary, the total momentum before
the crash must have been due to the moving car. Knowing
that the momentum is the product of mass and velocity
(mv), we can compute the speed of the car just before the
collision. We can thus determine whether the driver was
speeding.
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Let’s use conservation of momentum to analyze such a
collision.
◦ For simplicity assume that each car has a mass of 1000 kg (a
metric ton) and that the cars traveled along a straight line.
◦ Further assume that we have determined that the speed of the two
cars locked together was 10 m/s (about 22 mph) after the crash.
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The total momentum after the crash was equal to the total
mass of the two cars multiplied by their combined speed:
But because momentum is conserved, this was also the value
before the crash. Before the crash, however, only one car was
moving. So if we divide this total
momentum by the mass of the
moving car, we obtain its speed:
The car was therefore traveling at 20 m/s (~45 mph) at the
time of the accident.
How fast would you have to throw a baseball (m = 145 g)
to give it the same momentum as a 10-g bullet traveling
at 800 m/s?
a. 0.55 m/s
b. 1.8 m/s
c. 55 m/s
d. 800 m/s
What change in momentum occurs when a force of 20 N
acts for 4 s?
a. 5 kg·m/s
b. 16 kg·m/s
c. 24 kg·m/s
* d. 80 kg·m/s
A.
B.
C.
D.
a. 500 kg·m/s to the right
b. 500 kg·m/s to the left
c. 500 kg·m/s
d. zero