Download Momentum and Its Conservation

Momentum and Its Conservation Chapter Outline
Lesson 1: The Impulse-Momentum Change Theorem
Lesson 2: The Law of Momentum Conservation
Lesson 1: The ImpulseMomentum Change Theorem
• Momentum
• Momentum and Impulse Connection
• Real-World Applications
Momentum
• Momentum can be defined as "mass in motion." It is a
property of a moving body. The amount of momentum
that an object has is dependent upon two variables: how
much stuff is moving and how fast the stuff is moving.
• Momentum is equal to the product of the body's mass
and velocity.
Momentum = mass • velocity
p=m•v
m: mass in kg
v: velocity in m/s
p: momentum in kgm/s
Momentum is a vector quantity
• The direction of the momentum vector is
the same as the direction of the velocity
of the ball. Which is the same as the
direction that an object is moving.
Mass and velocity have equal
importance in momentum
• Momentum is directly proportional to mass and
momentum is directly proportional to velocity
• Consider a 0.5-kg physics cart loaded with one 0.5-kg
brick and moving with a speed of 2.0 m/s. 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 4.0 kg•m/s. A doubling of the mass results in a
doubling of the momentum.
• Similarly, if the 2.0-kg cart had a velocity of 4.0 m/s
(instead of 2.0 m/s), then the cart would have a
momentum of 8.0 kg•m/s (instead of 4.0 kg•m/s). A
doubling in velocity results in a doubling of the
momentum.
Example
•
Express your understanding of the concept
and mathematics of momentum by answering
the following questions Determine the
momentum of a ...
a. 60-kg halfback moving eastward at 9 m/s.
b. 1000-kg car moving northward at 20 m/s.
c. 40-kg freshman moving southward at 2 m/s.
Example
•
A car possesses 20 000 units of momentum.
What would be the car's new momentum if ...
a. its velocity was doubled.
b. its velocity was tripled.
c. its mass was doubled (by adding more
passengers and a greater load)
d. both its velocity and mass was doubled.
Example
• Four billiard balls, each of mass .5 kg, all are traveling in
the same direction on a billiard table, with speeds 2
m/s, 4 m/s, 8 m/s and 10 m/s. What is the linear
momentum of this system?
ptotal = p1 + p2 + p3 + p4
ptotal = (0.5 kg)(2 m/s + 4 m/s + 8 m/s + 10 m/s)
ptotal = 12 kg∙m/s
• If all four balls collide, what is the total momentum after
the collision?
12 kg∙m/s
A change in Momentum takes
force and time
• p = m∙v
• In order to change momentum, one must change velocity,
which means one must produce acceleration. To produce
acceleration, one must apply force.
• ∆p = m∙∆v
v
a
v  a  t
t
• ∆p = m∙a∙∆t
• ∆p = F∙∆t A change in momentum takes force and time
p  F  t
The equation states that a net external force, F, applied to an object for
a certain time interval, ∆t, will cause a change in the object’s
momentum. This change in the object’s momentum equals to the
product of F and the time interval.
Impulse-Momentum Theorem
F  t  p
F  t  mv f  mvi
The quantity F•∆t is known as impulse, represented by letter
J.
F: force in Newton
J = F∙∆t
t: time in second
J: impulse in N∙s
•Impulse is a vector quantity, its direction is the same as
the net force F.
Impulse = Change in Momentum
Impulse = Change in momentum
F  t  p
• The impulse experienced by the object is the cause of
the change in momentum.
• The impulse experience by the object equals the change
in momentum of the object.
• There are two way to determine impulse:
– F·t
– m·∆v
• There are two way to determine change in momentum:
– F·t
– m·∆v
Example
• If the halfback experienced a force of 800 N for
0.9 seconds, then we could say that the impulse
720 N•s
was _______________
• 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.
Note: the unit
N•s = kg•m/s
Example
•
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 Nforce for 0.50 seconds.
a. Which cart (#1 or #2) has the greatest acceleration?
Explain.
b. Which cart (#1 or #2) has the greatest impulse?
Explain.
c. Which cart (#1 or #2) has the greatest change in
momentum? Explain.
Example
•
Two cars of equal mass are traveling down Lake
Avenue with equal velocities. They both come to a stop
over different lengths of time. The ticker tape patterns
for each car are shown on the diagram below.
a. Which car (A or B) experiences the greatest
acceleration? Explain.
b. Which car (A or B) experiences the greatest change in
momentum? Explain.
c. Which car (A or B) experiences the greatest impulse?
Explain.
Practices
1.
2.
3.
4.
5.
In the diagram, a 60.-kilogram roller-skater exerts a 10.-newton force
on a 30.-kilogram roller-skater for 0.20 second. What is the
magnitude of the impulse applied to the 30.-kilogram rollerskater?
A 0.10-kilogram model rocket’s engine is designed to deliver an
impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75
second, what average force does it produce?
A 0.15-kilogram baseball moving at 20. meters per second is stopped
by a catcher in 0.010 second. What is the average force stopping the
ball?
A 2.0-kilogram laboratory cart is sliding across a horizontal
frictionless surface at a constant velocity of 4.0 meters per second
east. What will be the cart’s velocity after a 6.0-newton westward
force acts on it for 2.0 seconds?
A 1,000-kilogram car traveling due east at 15 meters per second is hit
from behind and receives a forward impulse of 8,000 newtonseconds. Determine the magnitude of the car’s change in momentum
due to this impulse.
6. The instant before a batter hits a 0.14-kilogram baseball, the
velocity of the ball is 45 meters per second west. The instant
after the batter hits the ball, the ball’s velocity is 35 meters
per second east. The bat and ball are in contact for 1.0 × 10−2
second. Calculate the magnitude of the average force the bat
exerts on the ball while they are in contact.
7. A 2.0-kilogram laboratory cart is sliding across a horizontal
frictionless surface at a constant velocity of 4.0 meters per
second east. What will be the cart’s velocity after a 6.0newton westward force acts on it for 2.0 seconds?
8. A 1400 kg car moving westward with a velocity of 15 m/s
collides with a utility pole and is brought to rest in 0.30 s.
Find the magnitude of the force exerted on the car during the
collision.
Real-World Applications
• The effect of collision time upon the
amount of force an object experiences
• The effect of rebounding upon the velocity
change and hence the amount of force an
object experiences.
The relationship between
Collision Time and Force of Impact
m∙∆v = F∙∆t
•The greater the time over which the collision occurs, the smaller the force
acting upon the object.
•To minimize the effect of the force on an object involved in a collision, the
time must be increased.
•To maximize the effect of the force on an object involved in a collision, the
time must be decreased.
Examples of Reducing force by increase time
• Air bags are used in automobiles because they are able
to minimize the affect of the force on an object
involved in a collision. Air bags accomplish this by
extending the time required to stop the momentum of
the driver and passenger.
• Padded dashboards also reduces force by increase
time.
• A boxer rides the punch in order to extend the time of
impact of the glove with their head.
• Nylon ropes are used in the sport of rock-climbing
because of its ability to stretch. The rock climber can
appreciate minimizing the effect of the force through
the use of a longer time of impact .
The Effect of Rebounding
• Bouncing off each other is known as rebounding.
Rebounding involves a change in the direction of an
object; rebounding situations are characterized by
– a large velocity change
F·∆t = m·∆v
– a large momentum change.
– a large impulse
• The importance of rebounding is critical to the
outcome of automobile accidents.
• Automobiles are made with crumple zones.
Crumple zones minimize the affect of the force in
an automobile collision in two ways.
– By crumpling, the car is less likely to rebound
upon impact, thus minimizing the momentum
change and the impulse.
– The crumpling of the car lengthens the time
over which the car's momentum is changed; by
increasing the time of the collision, the force of
the collision is greatly reduced.
Increase velocity by increasing time:
F·∆t = m·∆v
• In racket and bat sports, hitters are often
encouraged to follow-through when striking a
ball.
• In this situation, both the force applied (as
hard as you can) and the mass (the mass of
the ball) are constant. By following through,
the hitter increases the time, the result is
increasing the ball’s velocity.
examples
•
A constant force can act on an object for
different lengths of time. As the length of time
increases,
1. the impulse imparted to the object
a) decreases
b) increases
c) remains the same
2. The momentum of the object
a) decreases
b) increases
c) remains the same
Example
• A cannonball shot from a long-barrel cannon
travels faster than one shot from a short-barrel
cannon because the cannonball receives a
greater
a. force.
A cannonball shot from a cannon
b. impulse.
receive the same force regardless of
the length of its barrel. A long-barrel will
c. both A and B
take longer time (t) for the cannonball
d. neither A nor B to travel.
Since J = F∆t = m∆v, the longer the
time, the bigger the impulse, the faster
it will travel.
Example
•
A student drops two eggs of equal mass simultaneously
from the same height. Egg A lands on the tile floor and
breaks. Egg B lands intact, without bouncing, on a foam
pad lying on the floor. Compared to the magnitude of
the impulse on egg A as it lands, the magnitude of the
impulse on egg B as it lands is _________________
(less, greater, the same) as egg A.
Stopping times and stopping distances depend
on the impulse-momentum theorem
F·∆t = m·∆v
m
2m
•
Two identical trucks travels with the same velocity. One is
empty with mass m, and the other is loaded with mass 2m.
1. Why is loaded truck’s stopping time twice as much as the
empty truck’s when acted on by the same force?
Loaded is 2x unloaded
2. How do the stopping distances compare?
Loaded is 2x unloaded
practice
1. A 2250 car traveling to the west slows down uniformly
from 20.0 m/s to 5.00 m/s.
a. How long does it take the car to decelerate if the force on the
car is 8450 N to the east?
b. How far does the car travel during the deceleration?
2. If the maximum coefficient of kinetic friction between a
car and a road is 0.50, what is the minimum stopping
distance for car moving at 29 m/s? (about 65 mi/hr)
Lesson 2: The Law of Momentum
Conservation
1.
2.
3.
4.
The Law of Action-Reaction (Revisited)
Momentum Conservation Principle
Isolated Systems
Momentum Conservation in Collisions
– Using Equations as a "Recipe" for Algebraic
Problem-Solving
– Using Equations as a Guide to Thinking
5. Momentum Conservation in Explosions
6. Elastic and Inelastic collisions
The Law of Action-Reaction Revisited
• In a collision between two objects, both objects experience
forces that are equal in magnitude and opposite in direction in
accord with Newton’s 3rd Law.
While the forces are equal in
magnitude and opposite in
direction, the accelerations of the
objects are not necessarily equal in
magnitude.
According to Newton's second law
of motion, Bigger mass has smaller
acceleration, smaller mass has
bigger acceleration
1.
2.
3.
4.
m1(v1’ – v1) = -m2(v2’– v2)
m1v1 + m2v2 = m1v1’ + m2v2’
p(before) = p(after)
Momentum Conservation Principle
• The law of momentum conservation can be stated as
follows:
– For a collision occurring between object 1 and
object 2 in an isolated system, the total
momentum of the two objects before the collision
is equal to the total momentum of the two objects
after the collision. That is, the momentum lost by
object 1 is equal to the momentum gained by
object 2.
Isolated Systems
• A system is a collection of two or
more objects.
• An isolated system is a system
which is free from the influence
of a net external force which
alters the momentum of the
system.
• In reality, momentum can
be approximate to be
conserved for every
situations due to friction.
Collision
Description
1.
Two cars collide on a
gravel roadway on
which frictional forces
are large.
2.
Hans Full is doing the
annual vacuuming.
Hans is pushing the
Hoover vacuum
cleaner across the
living room carpet.
3.
Two air track gliders
collide on a frictionfree air track.
Isolated
System?
Yes or No
No
No
yes
If No, the
external force is
friction
The friction
between the carpet
and the floor and
the applied force
exerted by Hans
are both external
forces.
Vector diagram for conservation of momentum
Consider a fullback plunges across the goal line and collides in
midair with the linebacker in a football game. The linebacker
and fullback hold each other and travel together after the
collision. Before the collision, the fullback possesses a
momentum of 100 kg∙m/s, East and the linebacker possesses
a momentum of 120 kg∙m/s, West. The total momentum of the
system before the collision is _____________________.
20 kg∙m/s, West
Therefore, the total momentum of the system after the collision
20 kg∙m/s, West.
must also be __________________
Vector diagram for the situation:
Another vector diagram for conservation of momentum
• consider a medicine ball is thrown to a clown who is at rest
upon the ice; the clown catches the medicine ball and
glides together with the ball across the ice.
• The momentum of the medicine ball is 80 kg∙m/s before
the collision. The momentum of the clown is 0 kg∙m/s
before the collision. The total momentum of the system
before the collision is ______________
80 kg∙m/s.
• Therefore, the total momentum of the system after the
80 kg∙m/s.
collision must also be ________________.
The clown and
the medicine ball move together as a single unit after the
collision with a combined momentum of 80 kg∙m/s.
Momentum is conserved in the collision.
Vector diagram
for the situation:
Using Equations for Algebraic
Problem-Solving
• Total system momentum is conserved for collisions
between objects in an isolated system.
pbefore = pafter
m1v1 + m2v2 = m1v1’ + m2v2’
Example 1
• A 2.0-kilogram ball traveling north at 5.0 meters
per second collides head-on with a 1.0 kilogram
ball traveling south at 8.0 meters per second.
What is the magnitude of the total momentum of
the two balls after collision?
m1v1 + m2v2 = m1v1’ + m2v2’
m1 = 2.0 kg; v1 = 5.0 m/s
pafter = ?
m2 = 1.0 kg; v2 = -8.0 m/s
m1v1 + m2v2 = pafter
(2.0 kg)(5.0 m/s) + (1.0 kg)(-8.0 m/s) = pafter
2 kg∙m/s, north = pafter
Example 2
• A 3000-kg truck moving with a velocity of 10 m/s hits a
1000-kg parked car. The impact causes the 1000-kg car to
be set in motion at 15 m/s. Assuming that momentum is
conserved during the collision, determine the velocity of the
truck immediately after the collision.
It is a collision problem, momentum is conserved.
m1v1 + m2v2 = m1v1’ + m2v2’
m1 = 3000 kg; v1 = 10 m/s
v1’ = ?
m2 = 1000 kg; v2 = 0
v2’ = 15 m/s
(3000 kg)∙(10 m/s) + 0 = (3000 kg)∙v + (1000 kg)∙(15 m/s)
30000 kg∙m/s = (3000 kg)∙v + 15000 kg∙m/s
v = 5 m/s
Practices
1. A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg
person who is at rest on ice. The person catches the ball and
subsequently slides with the ball across the ice. Determine the
velocity of the person and the ball after the collision.
2. A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate
and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's
mitt immediately recoils backwards (at the same speed as the ball)
before the catcher applies an external force to stop its momentum.
Determine the post-collision velocity of the mitt and ball.
3. A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of
the boat and onto the dock. If the boater moves out of the boat with a
velocity of 2.5 m/s to the right, what is the final velocity of the boat?
Types of collisions
Inelastic collision
elastic collision
Inelastic collision
elastic collision
Elastic and inelastic collisions
• The two types of collisions are elastic and inelastic
collisions.
• To simplify matters, we will consider any collisions in
which the two colliding objects stick together and move
with the same post-collision speed to be an extreme
example of an inelastic collision.
• In an elastic collision, the two objects do not stick
together. they will bounce off each other.
Inelastic collision
• The two objects stick together and have the
same speed after collision.
m1v1 + m2v2 = m1v1’ + m2v2’
Since the two objects stick together after the collision,
v1’ = v2’ = v
m1v1 + m2v2 = (m1+ m2)v’
Example
• Cart A (50. kg) approaches cart B (100. kg initially at
rest) with an initial velocity of 30. m/s. After the collision,
cart A locks together with cart B. both travels with what
velocity?
Before collision
A
after collision
B
mA = 50 kg, vA = 30. m/s
mB = 100. kg, vB = 0
A
B
vA’ = vB’ = v’ = ?
mAvA + mBvB
= mAvA’ + mBvB’
(50.kg)(30.m/s) + (100.kg)(0) = (50.kg +100.kg)v’
v’ = 10. m/s
mass increases by 3 times (50 kg to 150 kg), speed decrease by 3
times (30 m/s to 10 m/s)
Example
• A railroad diesel engine coasting at 5.0 km/h runs into a
stationary flatcar. The diesel’s mass is 8,000. kg and the
flatcar’s mass is 2,000. kg. Assuming the cars couple
together, how fast are they moving after the collision?
Before collision
diesel
after collision
Flat car
diesel
Flat car
v’ = 4.0 km/h
mass increases by 1.25 times (8000 kg to 10000 kg), speed
decrease by 1.25 times (50 m/s to 40 m/s)
Example (pay attention to directions)
• Cart A (50. kg) moving with an initial velocity of 30. m/s
approaches cart B (100. kg moving with initial velocity of
20. m/s towards cart A. The two carts lock together and
move as one. Calculate the magnitude and the direction
of the final velocity.
Before collision
A
B
after collision
A
v’ = 3.3 m/s to the left
B
Example
• A block of mass M initially at rest on a frictionless
horizontal surface is struck by a bullet of mass m moving
with horizontal velocity v. What is the velocity of the
bullet-block system after the bullet embeds itself in the
block?
Before collision
after collision
mv + M(0)
(m + M)v’
mv + 0
=
mv / (M+m) =
(m + M)v’
v’
Example
•
1.
2.
3.
4.
A woman with horizontal velocity v1 jumps
off a dock into a stationary boat. After
landing in the boat, the woman and the boat
move with velocity v2. Compared to velocity
v1, velocity v2 has
the same magnitude and the same direction
the same magnitude and opposite direction
smaller magnitude and the same direction
larger magnitude and the same direction
Example
• A 1850 kg luxury sedan stopped at a traffic light
is struck from the rear by a compact car with a
mass of 975 kg. the two cars become entangle
as a result of the collision. If the compact car
was moving at a velocity of 220 m/s to the north
before the collision, what is the velocity of the
entangled mass after the collision?
7.59 m/s North
Explosions
Momentum before the
explosion is zero. so the
momentum after the
explosion is also zero.
m1v1 + m2v2 = m1v1’ + m2v2’
0 = m1v1’ + m2v2’
• Consider a homemade cannon.
p(after) = 0
p(before) = 0
•
•
•
•
p(cannon) +
p(ball) = 0
In the exploding cannon, total system momentum is
conserved.
The system consists of two objects - a cannon and a
tennis ball.
Before the explosion, the total momentum of the system
is zero since the cannon and the tennis ball located
inside of it are both at rest.
After the explosion, the total momentum of the system
must still be zero. If the ball acquires 50 units of forward
momentum, then the cannon acquires 50 units of
backwards momentum. The vector sum of the individual
momentum of the two objects is 0. Total system
momentum is conserved.
Example
• A 2.0-kilogram toy cannon is at rest on a frictionless
surface. A remote triggering device causes a 0.005kilogram projectile to be fired from the cannon. Which
equation describes this system after the cannon is fired?
1. mass of cannon + mass of projectile = 0
2. speed of cannon + speed of projectile = 0
3. momentum of cannon + momentum of projectile = 0
4. velocity of cannon + velocity of projectile = 0
Example
• A 2-kilogram rifle initially at rest fires a
0.002-kilogram bullet. As the bullet leaves
the rifle with a velocity of 500 meters per
second, what is the momentum of the riflebullet system?
Example
• A 56.2-gram tennis ball is loaded into a 1.27-kg homemade
cannon. The cannon is at rest when it is ignited. Immediately
after the impulse of the explosion, a photogate timer measures
the cannon to recoil backwards a distance of 6.1 cm in 0.0218
seconds. Determine the post-explosion speed of the cannon
and of the tennis ball.
Given: Cannon:
m = 1.27 kg; d = 6.1 cm; t = 0.0218 s
Ball:
m = 56.2 g = 0.0562 kg
• The strategy for solving for the speed of the cannon is to
recognize that the cannon travels 6.1 cm at a constant speed
in the 0.0218 seconds.
• vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded
• The strategy for solving for the post-explosion speed of the
tennis ball involves using momentum conservation principles.
mball • vball = - mcannon • vcannon
v = 63.3 m/s
Example
• A 60. kg man standing on a stationary 40. kg boat throws a
.20 kg baseball with a velocity of 50. m/s. With what speed
does the boat move after the man throws the ball? Assume
no friction between the water and the boat.
v’(boat) = -0.1 m/s (in the opposite direction of the baseball)
Example
• In the diagram, a 100.-kilogram clown is fired
from a 500.-kilogram cannon. If the clown's
speed is 15 meters per second after the firing,
then what is the recoil speed (v) of the cannon?
3.0 m/s
practices
1. A 1200-kilogram car moving at 12 meters per second collides with
a 2300-kilogram car that is waiting at rest at a traffic light. After the
collision, the cars lock together and slide. Eventually, the combined
cars are brought to rest by a force of kinetic friction as the rubber
tires slide across the dry, level, asphalt road surface. Calculate the
speed of the locked-together cars immediately after the collision.
2. On a snow-covered road, a car with a mass of 1.1 × 103
kilograms collides head-on with a van having a mass of
2.5 × 103 kilograms traveling at 8.0 meters per second. As
a result of the collision, the vehicles lock together and
immediately come to rest. Calculate the speed of the car
immediately before the collision. [Neglect friction.]
3. A 1,000-kilogram car traveling due east at 15 meters per
second is hit from behind and receives a forward impulse
of 7,000 newton-seconds. Determine the magnitude of the
car’s change in momentum due to this impulse.