Transcript 7-2 Conservation of Momentum

Chapter 7
Linear Momentum
Objectives: The student will be able
to:
• Perform several investigations in order to make
conclusions about the total momentum of a
system.
• Analyze a problem and choose a system to
determine if the forces are internal or external to
that system.
• State the law of conservation of momentum and
use it to solve one-dimensional explosion and
collision problems using an equation.
Inquiry for Total Momentum Before
and After a Collision/Explosion
• Design an experiment to demonstrate the effect of a collision/explosion
on total momentum of the objects before and after using the same mass
for each cart and then for a second experiment change one of the cart’s
mass by adding mass to it.
• Make a prediction on the effect of some condition on the
total momentum before and after a collision.
• Materials
–
–
–
–
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Dynamics cart with spring bumper or plunger
Meter stick
Stop watch
Masses
Large white boards
Inquiry for Total Momentum Before and
After a Collision/Explosion
• Your group will present using the whiteboards your
design and findings.
• Did anything unusual happen? Were there any special insights you gained
and want to make a note of? Were there any changes made to your
procedure? What new questions arose?
• Were any of the results NOT what you expected?
• Which of your pre-lab ideas have you decided are now incorrect? Why?
• Did the data support your original hypothesis?
• If not, what hypothesis does the data support?
Evaluation
•
Individually, you will make a claim about each investigation with supporting
evidence, and then explain how the conservation of momentum can be applied to
these investigations.
• What conclusion(s) did you reach due to the results of this experiment?
• What evidence supports your conclusion(s)?
• Are your results reliable? How did you compensate for sources of error in the
experiment?
• Can you test the predictions? If so, do results agree with your conclusion(s)?
• What new problems/questions does the experiment bring up?
RELATE
• What are some possible applications of your conclusions to the real world
situations?
• Do the results of your experiment fit any laws/theories of physics?
•
This will be due
.
7-2 Conservation of Momentum
During a collision, measurements show that the
total momentum does not change:
(7-3)
7-2 Conservation of Momentum
More formally, the law of conservation of
momentum states:
The total momentum of an isolated system of
objects remains constant.
7-2 Conservation of Momentum
Example 7-3: Railroad cars collide:
momentum conserved.
A 10,000-kg railroad car, A, traveling at a
speed of 24.0 m/s strikes an identical car, B,
at rest. If the cars lock together as a result of
the collision, what is their common speed
immediately after the collision?
7-2 Conservation of Momentum
Momentum conservation works for a rocket as
long as we consider the rocket and its fuel to
be one system, and account for the mass loss
of the rocket.
7-2 Conservation of Momentum
Example 7-4: Rifle recoil.
Calculate the recoil velocity of a 5.0-kg
rifle that shoots a 0.020-kg bullet at a
speed of 620 m/s.
Conceptual Example 7-5: Falling on or off a
sled.
(a) An empty sled is sliding on frictionless ice
when Susan drops vertically from a tree
above onto the sled. When she lands, does
the sled speed up, slow down, or keep the
same speed?
(b) Later: Susan falls sideways off the sled.
When she drops off, does the sled speed up,
slow down, or keep the same speed?
Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each
other without colliding, such as gravity) momentum of the system
(both objects together) is conserved. This mean the total momentum
of the objects is the same before and after the collision.
(Choosing right as the +
before: p = m1 v1 - m2 v2
v2
v1
m1
direction, m2 has - momentum.)
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
after: p = - m1 va + m2 vb
va
m1
m2
vb
Directions after a collision
On the last slide the boxes were drawn going in the opposite direction
after colliding. This isn’t always the case. For example, when a bat hits
a ball, the ball changes direction, but the bat doesn’t. It doesn’t really
matter, though, which way we draw the velocity vectors in “after”
picture. If we solved the conservation of momentum equation (red box)
for vb and got a negative answer, it would mean that m2 was still moving
to the left after the collision. As long as we interpret our answers
correctly, it matters not how the velocity vectors are drawn.
v2
v1
m1
m2
m1 v1 - m2 v2 = - m1 va + m2 vb
va
m1
m2
vb
Proof of Conservation of Momentum
The proof is based on Newton’s 3rd Law. Whenever two objects collide
(or exert forces on each other from a distance), the forces involved are an
action-reaction pair, equal in strength, opposite in direction. This means
the net force on the system (the two objects together) is zero, since these
forces cancel out.
F
F
M m
force on M due to m
force on m due to M
For each object, F = (mass) (a) = (mass) (v / t) = (mass v)/ t = p / t.
Since the force applied and the contact time is the same for each mass,
they each undergo the same change in momentum, but in opposite
directions. The result is that even though the momenta of the individual
objects changes, p for the system is zero. The momentum that one
mass gains, the other loses. Hence, the momentum of the system before
equals the momentum of the system after.
Conservation of Momentum applies only
in the absence of external forces!
In the first two sample problems, we dealt with a frictionless surface.
We couldn’t simply conserve momentum if friction had been present
because, as the proof on the last slide shows, there would be another
force (friction) in addition to the contact forces. Friction wouldn’t
cancel out, and it would be a net force on the system.
The only way to conserve momentum with an external force like
friction is to make it internal by including the tabletop, floor, or the
entire Earth as part of the system. For example, if a rubber ball hits a
brick wall, p for the ball is not conserved, neither is p for the ballwall system, since the wall is connected to the ground and subject to
force by it. However, p for the ball-Earth system is conserved!
Example
An apple is originally at rest and then dropped. After falling a short
time, it’s moving pretty fast, say at a speed V. Obviously, momentum
is not conserved for the apple, since it didn’t have any at first. How can
this be?
answer: Gravity is an external force on the
apple
m
V
F
v
Earth
M
apple, so momentum for it alone is not
conserved. To make gravity “internal,” we
must define a system that includes the
other object responsible for the
gravitational force--Earth. The net force
on the apple-Earth system is zero, and
momentum is conserved for it. During the
fall the Earth attains a very small speed v.
So, by conservation of momentum:
F
mV = M v
Internal and External Forces

Internal forces: the forces that the particles of the system exert on
each other (for any system)

Forces exerted on any part of the system by some object outside it called
external forces

For the system of two astronauts, the internal forces are FBonA, exerted
by particle B on particle A, and FAonB, exerted by particle A on particle B

There are NO external forces: we have isolated system

Total momentum of two particles is the vector
sum of the momenta of individual particles
What is an isolated system?
• A system is a collection of two or more objects. An isolated system is
a system that F is free from the influence of a net external force that
alters the momentum of the system. There are two criteria for the
presence of a net external force; it must be...
• a force that originates from a source other than the two objects of the
system
• a force that is not balanced by other forces.
• A system in which the only forces that
contribute to the momentum change of an
individual object are the forces acting
between the objects themselves can be
considered an isolated system.
Read the following description of a collision and
evaluate whether or not the collision occurs in an
isolated system. If it is not an isolated system, then
identify the net external force.
• Two cars collide on a gravel roadway on which
frictional forces are large.
• No. What is the net external force?
• The friction between the cars and the road is an
external force. This force contributes to a change in
total momentum of the system.
Read the following description of a collision and
evaluate whether or not the collision occurs in an
isolated system. If it is not an isolated system, then
identify the net external force.
• Hans Full is doing the annual vacuuming. Hans is
pushing the Hoover vacuum cleaner across the
living room carpet.
• No. What is the net external force?
• The friction between the cleaner and the floor and
the applied force exerted by Hans are both external
forces. These forces contribute to a change in total
momentum of the system.
Read the following description of a collision and
evaluate whether or not the collision occurs in an
isolated system. If it is not an isolated system, then
identify the net external force.
• Two air track gliders collide on a friction-free air
track.
• Yes. It is an isolated system.
• There are no external forces; the system is isolated.
The total system momentum would be expected to
be conserved.
Sample Problem 1
35 g
7 kg
700 m/s
v=0
A rifle fires a bullet into a giant slab of butter on a frictionless surface.
The bullet penetrates the butter, but while passing through it, the bullet
pushes the butter to the left, and the butter pushes the bullet just as hard
to the right, slowing the bullet down. If the butter skids off at 4 cm/s
after the bullet passes through it, what is the final speed of the bullet?
(The mass of the rifle matters not.)
35 g
v=?
4 cm/s
7 kg
continued on next slide
Sample Problem 1
(cont.)
Let’s choose left to be the + direction & use conservation of
momentum, converting all units to meters and kilograms.
35 g
p before = 7 (0) + (0.035)(700)
7 kg
= 24.5 kg · m /s
v=0
35 g
4 cm/s
v=?
p before = p after
7 kg
700 m/s
p after = 7 (0.04) + 0.035 v
= 0.28 + 0.035 v
24.5 = 0.28 + 0.035 v
v = 692 m/s
v came out positive. This means we chose the correct
direction of the bullet in the “after” picture.
Sample Problem 2
35 g
7 kg
700 m/s
v=0
Same as the last problem except this time it’s a block of wood rather
than butter, and the bullet does not pass all the way through it. How fast
do they move together after impact? Assume frictionless surface.
v
Sample Problem 2
35 g
7 kg
700 m/s
v=0
Same as the last problem except this time it’s a block of wood rather
than butter, and the bullet does not pass all the way through it. How fast
do they move together after impact?
v
7. 035 kg
(0.035) (700) = 7.035 v
v = 3.48 m/s
Note: Once again we’re assuming a frictionless surface, otherwise there
would be a frictional force on the wood in addition to that of the bullet,
and the “system” would have to include the table as well.
Sample Problem 3
A crate of raspberry donut filling collides with a tub of lime Kool Aid
on a frictionless surface. Which way on how fast does the Kool Aid
rebound?
before
3 kg
10 m/s
6 m/s
15 kg
after
4.5 m/s
3 kg
15 kg
v
Sample Problem 3
A crate of raspberry donut filling collides with a tub of lime Kool Aid
on a frictionless surface. Which way on how fast does the Kool Aid
rebound? answer: Let’s draw v to the right in the after picture.
3 (10) - 6 (15) = -3 (4.5) + 15 v
v = -3.1 m/s
Since v came out negative, we guessed wrong in drawing v to the
right, but that’s OK as long as we interpret our answer correctly.
After the collision the lime Kool Aid is moving 3.1 m/s to the left.
before
3 kg
10 m/s
6 m/s
15 kg
after
4.5 m/s
3 kg
15 kg
v
Homework
• Evaluation argument from inquiry labs.
• Problems 4, 6, 8, 10
Conclusions
• Based on the investigations, what do you think
is meant by conservation of momentum?
• Kahoot 7-2