Transcript Momentum

Momentum
p= mv
Impulse Momentum Change Theorem
• The sports announcer says "Going into the
all-star break, the Chicago White Sox have
the momentum." The headlines declare
"Chicago Bulls Gaining Momentum." The
coach pumps up his team at half-time, saying
"You have the momentum; the critical need is
that you use that momentum and bury them
in this third quarter."
Sports vs. Physics
• Momentum is a commonly used term in
sports. A team that has the momentum is on
the move and is going to take some effort to
stop. A team that has a lot of momentum is
really on the move and is going to be hard to
stop. Momentum is a physics term; it refers
to the quantity of motion that an object has. A
sports team which is "on the move" has the
momentum. If an object is in motion ("on the
move") then it has momentum.
Defining Momentum
• Momentum can be defined as "mass in motion." All
objects have mass; so if an object is moving, then it
has momentum - it has its mass in motion. The
amount of momentum which an object has is
dependent upon two variables: how much stuff is
moving and how fast the stuff is moving.
Momentum depends upon the variables mass and
velocity. In terms of an equation, the momentum of
an object is equal to the mass of the object times
the velocity of the object.
The Equations
• Momentum = mass * velocity In physics,
the symbol for the quantity momentum is the
small case "p"; thus, the above equation can
be rewritten as
• p=m*v
• where m = mass and v=velocity. The
equation illustrates that momentum is directly
proportional to an object's mass and directly
proportional to the object's velocity
Defining the Units of Measure
• The units for momentum would be mass units times
velocity units. The standard metric unit of momentum
is the kg*m/s. While the kg*m/s is the standard
metric unit of momentum, there are a variety of other
units which are acceptable (though not conventional)
units of momentum; examples include kg*mi/hr,
kg*km/hr, and g*cm/s. In each of these examples, a
mass unit is multiplied by a velocity unit to provide a
momentum unit. This is consistent with the equation
for momentum.
Vector or Scalar?
• Momentum is a vector quantity. As discussed in an
earlier unit, a vector quantity is a quantity which is
fully described by both magnitude and direction. To
fully describe the momentum of a 5-kg bowling ball
moving westward at 2 m/s, you must include
information about both the magnitude and the
direction of the bowling ball. It is not enough to say
that the ball has 10 kg*m/s of momentum; the
momentum of the ball is not fully described until
information about its direction is given.
Defining Direction
• The direction of the momentum vector is the
same as the direction of the velocity of the
ball. In a previous unit, it was said that the
direction of the velocity vector is the same as
the direction which an object is moving. If the
bowling ball is moving westward, then its
momentum can be fully described by saying
that it is 10 kg*m/s, westward. As a vector
quantity, the momentum of an object is fully
described by both magnitude and direction.
Which variable (mass or
velocity) defines momentum?
• From the definition of momentum, it becomes
obvious that an object has a large momentum if
either its mass or its velocity is large. Both variables
are of equal importance in determining the
momentum of an object. Consider a Mack truck and
a roller skate moving down the street at the same
speed. The considerably greater mass of the Mack
truck gives it a considerably greater momentum. Yet
if the Mack truck were at rest, then the momentum
of the least massive roller skate would be the
greatest; for the momentum of any object which is
at rest is 0. Objects at rest do not have momentum they do not have any "mass in motion." Both
variables - mass and velocity - are important in
comparing the momentum of two objects.
Conceptualizing Momentum
• The momentum equation can help us to think about
how a change in one of the two variables might
effect the momentum of an object. Consider a 0.5-kg
physics cart loaded with one 0.5-kg brick and moving
with a speed of 2.0 m/s. The total mass of loaded
cart is 1.0 kg and its momentum is 2.0 kg*m/s. If the
cart was instead loaded with three 0.5-kg bricks,
then the total mass of the loaded cart would be 2.0
kg and its momentum would be 4.0 kg*m/s. A
doubling of the mass results in a doubling of the
momentum.
Problem Solving Momentum
• Similarly, if the 2.0-kg cart had a velocity of
8.0 m/s (instead of 2.0 m/s), then the cart
would have a momentum of 16.0 kg*m/s
(instead of 4.0 kg*m/s). A quadrupling in
velocity results in a quadrupling of the
momentum. These two examples illustrate
how the equation p=m*v serves as a "guide
to thinking" and not merely a "recipe for
algebraic problem-solving."
1. Determine the momentum
of a ...
• 60-kg halfback moving eastward at 9
m/s.
• 1000-kg car moving northward at 20
m/s.
• 40-kg freshman moving southward at 2
m/s.
1 Answers
A) p = m * v
(60 kg)(9m/s) =
540 kg * m /s
B) p = m * v
(1000 kg)(20 m/s)
20,000 kg* m/s
C) p = m * v
40 kg * 2 m/s
80 kg * m/s
2. A car possesses 20 000
units of momentum. What
would be the car's new
momentum if ...
• its velocity were doubled.
• its velocity were tripled.
• its mass were doubled (by adding more
passengers and a greater load)
• both its velocity were doubled and its
mass were doubled.
2 answers
A) 40, 000 units (doubling the velocity will
double the momentum)
B) 60, 000 units (tripling the velocity will triple
the momentum
C) 40, 000 units (doubling the mass will
double the momentum)
D) 80, 000 units (doubling the mass and
doubling the velocity will quadruple the
momentum)
3. A halfback (m = 60 kg), a
tight end (m = 90 kg), and a
lineman (m = 120 kg) are
running down the football
field. Consider their ticker
tape patterns below.
• Compare the velocities of these three
players. How many times greater is the
velocity of the halfback and the velocity of the
tight end than the velocity of the lineman?
Connecting Momentum and
Impulse
• As mentioned in the previous part of this lesson,
momentum is a commonly used term in sports.
When a sports announcer says that a team has the
momentum they mean that the team is really on the
move and is going to be hard to stop. An object with
momentum is going to be hard to stop. To stop such
an object, it is necessary to apply a force against its
motion for a given period of time. The more
momentum which an object has, the harder that it is
to stop. Thus, it would require a greater amount of
force or a longer amount of time (or both) to bring an
object with more momentum to a halt. As the force
acts upon the object for a given amount of time, the
object's velocity is changed; and hence, the object's
momentum is changed.
Momentum in sports
• The concepts in the above paragraph should not
seem like abstract information to you. You have
observed this a number of times if you have watched
the sport of football. In football, the defensive players
apply a force for a given amount of time to stop the
momentum of the offensive player who has the ball.
You have also experienced this a multitude of times
while driving. As you bring your car to a halt when
approaching a stop sign or stoplight, the brakes
serve to apply a force to the car for a given amount
of time to stop the car's momentum. An object with
momentum can be stopped if a force is applied
against it for a given amount of time.
Picture Illustration
Revisiting Force Balance
• A force acting for a given amount of time will change
an object's momentum. Put another way, an
unbalanced force always accelerates an object either speeding it up or slowing it down. If the force
acts opposite the object's motion, it slows the object
down. If a force acts in the same direction as the
object's motion, then the force speeds the object up.
Either way, a force will change the velocity of an
object. And if the velocity of the object is changed,
then the momentum of the object is changed.
Back to Newton’s Second
Law
• These concepts are merely an outgrowth of
Newton's second law as discussed in an
earlier unit. Newton's second law (Fnet=m*a)
stated that the acceleration of an object is
directly proportional to the net force acting
upon the object and inversely proportional to
the mass of the object. When combined with
the definition of acceleration (a=change in
velocity/time), the following equalities result.
Formulas of Newton’s 2nd Law
Multiply both sides by t
Examining the Equation
• This equation is one of two primary equations
to be used in this unit. To truly understand
the equation, it is important to understand its
meaning in words. In words, it could be said
that the force times the time equals the mass
times the change in velocity. In physics, the
quantity Force*time is known as the impulse.
And since the quantity m*v is the momentum,
the quantity m*"Delta "v must be the change
in momentum. The equation really says that
the Impulse = Change in momentum
Collisions
• One focus of this unit is to understand the
physics of collisions. The physics of collisions
are governed by the laws of momentum; and
the first law which we discuss in this unit is
expressed in the above equation. The
equation is known as the impulse-momentum
change equation. The law can be expressed
this way:
Expressions of collisions
• In a collision, an object experiences a
force for a specific amount of time
which results in a change in momentum
(the object's mass either speeds up or
slows down). The impulse experienced
by the object equals the change in
momentum of the object. In equation
form, F * t = m * Delta v.
Objects in collisions with
impulse
• In a collision, objects experience an impulse;
the impulse causes (and is equal to) the
change in momentum. Consider a football
halfback running down the football field and
encountering a collision with a defensive
back. The collision would change the
halfback's speed (and thus his momentum).
If the motion was represented by a ticker
tape diagram, it might appear as follows:
Putting numbers in…
• At approximately the tenth dot on the diagram, the
collision occurs and lasts for a certain amount of
time; in terms of dots, the collision lasts for
approximately nine dots. In the halfback-defensive
back collision, the halfback experiences a force
which lasts for a certain amount of time to change
his momentum. Since the collision causes the
rightward-moving halfback to slow down, the force
on the halfback must have been directed leftward. If
the halfback experienced a force of 800 N for 0.9
seconds, then we could say that the impulse was
720 N*s. This impulse would cause a momentum
change of 720 kg*m/s. In a collision, the impulse
experienced by an object is always equal to the
momentum change.
Tennis Ball Collision
• Now consider a collision of a tennis ball with a wall.
Depending on the physical properties of the wall (its
elastic nature), the speed at which the ball rebounds
from the wall upon colliding with it will vary. The
diagrams below depict the changes in velocity of the
same ball. For each representation (vector diagram,
v-t graph, and ticker tape pattern), indicate which
case (A or B) has the greatest change in velocity,
greatest acceleration, greatest momentum change,
and greatest impulse. Support each answer.
Vector Diagrams
Tabulate your Answers
Greatest velocity change?
Greatest acceleration?
Greatest momentum
change?
Greatest Impulse?
Answers
Greatest velocity change? B (changes from +30 m/s to
-28 m/s which is a change of 58 m/s and A only
changes -15 m/s)
Greatest acceleration? B, because it has the greatest
velocity change and acceleration is dependent on
velocity change.
Greatest momentum change? B, because momentum
is dependent on velocity and the change in velocity is
greatest in B.
Greatest Impulse? B, impulse is momentum change
and the momentum change is greatest in B.
Graphs
Setting up your answers
Greatest velocity
change?
Greatest acceleration?
Greatest momentum
change?
Greatest Impulse?
Answers
Greatest
velocity
change?
Greatest in A because it changes from
+5 m/s to -3 m/s which is a change of 8
whereas B only changes 4 m/s
Greatest
A, acceleration is greatest because
acceleration? velocity change is greatest in A
Greatest
momentum
change?
A, momentum is dependent on velocity
change and that is greatest in A
Greatest
Impulse?
A, Impulse equals the change in
momentum
Using Ticker Tape Diagrams
Organizing the Answers
Greatest velocity
change?
Greatest acceleration?
Greatest momentum
change?
Greatest Impulse?
Answers
Greatest
velocity
change?
B, vi is the same for both, but object
B rebounds with a greater speed so it
goes from -10 to +5 m/s whereas A
goes from -10 m/s to +2 m/s.
Greatest
acceleration?
B, acceleration is dependent on
velocity change
Greatest
momentum
change?
B, Momentum is dependent on
velocity
Greatest
Impulse?
B, impulse is defined as the change
in momentum
Recapping the Examples
• Observe that each of the collisions above
involved the rebound of a ball off a wall.
Observe that the greater the rebound effect,
the greater the acceleration, momentum
change, and impulse. A rebound is a special
type of collision involving a direction change;
the result of the direction change is large
velocity change.
Elastic
Collisions
• On occasions in a rebound collision, an
object will maintain the same or nearly the
same speed as it had before the collision.
Collisions in which objects rebound with the
same speed (and thus, the same momentum
and kinetic energy) as they had prior to the
collision are known as elastic collisions. In
general, elastic collisions are characterized
by a large velocity change, a large
momentum change, a large impulse, and a
large force.
Momentum and collisions
• Use the impulse-momentum change principle
to fill in the blanks in the following rows of the
table. As you do, keep these three major
truths in mind:
• the impulse experienced by an object is the
force*time
• the momentum change of an object is the
mass*velocity change
• the impulse equals the momentum change
Create this Chart
Force
(N)
time
(s)
1.
0.010
2.
0.100
3.
0.010
4.
-20
000
5.
-200
Mom.
Impuls
Chg.
e
(kg*m
(N*s)
/s)
-40
Mass
(kg)
Vel.
Chg.
(m/s)
10
-4
10
-200
50
-200
1.0
-8
50
Answers
Force (N)
time
(s)
Mom.
Impuls
Chg.
Mass
e
(kg*m (kg)
(N*s)
/s)
1.
-4000 N
0.010
-40
-40
10
-4
2.
-400 N
0.100
-40
-40
10
-4
-200
-200
50
-4
4. -20 000
.010 s -200
-200
25
-8
5.
1.0
-200
50
-4
3. -20, 000 N 0.010
-200
-200
Vel.
Chg.
(m/s)
Reflections on the Table
• There are a few observations which can be made in the
above table which relate to the computational nature of the
impulse-momentum change theorem. First, observe that the
answers in the table above reveal that the third and fourth
columns are always equal; that is, the impulse is always
equal to the momentum change. Observe also that the if
any two of the first three columns are known, then the
remaining column can be computed; this is true because
the impulse=force*time. Knowing two of these three
quantities allows us to compute the third quantity. And
finally, observe that knowing any two of the last three
columns allows us to compute the remaining column; this is
true since momentum change = mass*velocity change.
Additional Thoughts
•
There are also a few observations which can be made
which relate to the qualitative nature of the impulsemomentum theorem. An examination of rows 1 and 2 show
that force and time are inversely proportional; for the same
mass and velocity change, a tenfold increase in the time of
impact corresponds to a tenfold decrease in the force of
impact. An examination of rows 1 and 3 show that mass and
force are directly proportional; for the same time and
velocity change, a fivefold increase in the mass corresponds
to a fivefold increase in the force required to stop that mass.
Finally, an examination of rows 3 and 4 illustrate that mass
and velocity change are inversely proportional; for the same
force and time, a twofold decrease in the mass corresponds
to a twofold increase in the velocity change.
Examples
Express your understanding of the
impulse-momentum change theorem by
answering the following questions.
Example 1
• 1. A 0.50-kg cart (#1) is pulled with a
1.0-N force for 1 second; another 0.50
kg cart (#2) is pulled with a 2.0 N-force
for 0.50 seconds. Which cart (#1 or #2)
has the greatest acceleration? Explain.
Answer 1
• Cart 2 has the greatest acceleration.
Recall that force is dependent on
acceleration and mass. Both carts have
the same mass but cart 2 has the
greatest force.
Example 2
• Which cart (#1 or #2) has the greatest
impulse? Explain.
Answer 2
• The impulse is the same for both carts.
Impulse is force * time which calculates
each cart to 1 N * s for each.
Example 3
• Which cart (#1 or #2) has the greatest
change in momentum? Explain.
Answer 3
• The momentum change is the same for
both carts. Momentum change equals
the impulse. If both carts have the
same impulse then they both have the
same momentum change.
Example 4
• 2. In a phun physics demo, two
identical balloons (A and B) are
propelled across the room on horizontal
guide wires. The motion diagrams
(depicting the relative position of the
balloons at time intervals of 0.05
seconds) for these two balloons are
shown below.
Question 4a
• Which balloon (A or B) has the greatest
acceleration? Explain.
Answer 4a
• Balloon B has the greatest acceleration
The rate at which the velocity changes
is greatest for balloon B, this is shown
by the fact that the speed
(distance/time) changes most rapidly.
Example 4B
• Which balloon (A or B) has the greatest
final velocity? Explain.
Answer 4B
• Balloon B has the greatest final velocity.
At the end of the diagram, the distance
traveled in the latest interval is greatest
for Balloon B.
Question 4C
• Which balloon (A or B) has the greatest
momentum change? Explain.
Answer 4C
• Balloon B has the greatest momentum
change. The final velocity is greatest for
Balloon B, its velocity change is also
the greatest. Momentum change
depends on velocity change. The
balloon with the greatest velocity
change will have the greatest
momentum change.
Question 4D
• Which balloon (A or B) experiences the
greatest impulse? Explain.
Answer 4D
• The impulse is the same for each car.
The impulse equals the momentum
change. If the momentum change is the
same for each car, then so must be the
impulse.
Question 5
• The diagram to the
right depicts the
before- and aftercollision speeds of a
car which undergoes
a head-on-collision
with a wall. In Case
A, the car bounces
off the wall. In Case
B, the car "sticks" to
the wall.
Question 5A
• In which case (A or B) is the change in
velocity the greatest? Explain.
Answer 5A
• Case A has the greatest velocity
change. The velocity change is -9 m/s
in case A and only -5 m/s in case B.
Question 5B
• In which case (A or B) is the change in
momentum the greatest? Explain.
Answer 5B
• Case A has the greatest momentum
change. The momentum change is
dependent on the velocity change; the
object with the greatest velocity change
has the greatest momentum change.
Question 5C
• In which case (A or B) is the impulse
the greatest? Explain.
Answer 5C
• The impulse is greatest for Car A. The
impulse equals the momentum change.
If the momentum change is greatest for
car A then the impulse is greatest.
Question 5D
• In which case (A or B) is the force
which acts upon the car the greatest
(assume contact times are the same in
both cases)? Explain.
Answer 5D
• The impulse is greatest for car A. The
force is related to impulse (I = F * t).
The bigger impulse for car A is
attributed to the greatest force upon car
A. Recall that the rebound effect is
characterized by larger forces; car A is
the car which rebounds.
Question 6
• Rhonda, who has a mass of 60.0 kg, is
riding at 25.0 m/s in her sports car
when she must suddenly slam on the
brakes to avoid hitting a dog crossing
the road. She strikes the air bag, which
brings her body to a stop in 0.400 s.
What average force does the seat belt
exert on her?
Answer 6
• F = m x change in velocity)/time
• F = (60 kg x 25 m/s)/ 0.400 s
• F = 3750 N
Question 6B
• If Rhonda had not been wearing her
seat belt and not had an air bag, then
the windshield would have stopped her
head in 0.001 s. What average force
would the windshield have exerted on
her?
Answer 6B
• F = (m x D v ) / t
• F = (60 kg x 25 m/s) / 0.001 s
• F = 1,500,000 N
Question 7
• A hockey player applies an average
force of 80.0 N to a 0.25 kg hockey
puck for a time of 0.10 seconds.
Determine the impulse experienced by
the hockey puck.
Answer 7
• Impulse = Force x time
• I = 80 N x 0.1 s
• I=8Nxs
Ex. 8
• If a 5-kg object experiences a 10-N
force for a duration of 0.1 seconds, then
what is the momentum change of the
object?
Question 8
• Momentum change = 1.0 kg m/s
• The momentum change = mass x velocity
change
• But since velocity change is unknown,
another strategy must be used to find the
momentum change. The strategy involves
first finding the impulse (I = F x t). Since
impulse = momentum change, the answer is
1 N x s.