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Chapter 6
Preview
• Objectives
• Linear Momentum
Section 1 Momentum and
Impulse
Chapter 6
Section 1 Momentum and
Impulse
Objectives
• Compare the momentum of different moving objects.
• Compare the momentum of the same object moving
with different velocities.
• Identify examples of change in the momentum of an
object.
• Describe changes in momentum in terms of force
and time.
Chapter 6
Section 1 Momentum and
Impulse
Linear Momentum
• Momentum is defined as mass times velocity.
• Momentum is represented by the symbol p, and is a
vector quantity.
p = mv
momentum = mass  velocity
Chapter 6
Section 1 Momentum and
Impulse
Momentum
Click below to watch the Visual Concept.
Visual Concept
Sample Problem A P199
• A 2250 kg pickup truck has a velocity of 25 m/s to the
east. What is the momentum of the truck?
• Given:
• Asked:
• Formula: p = m v
Assignment
• Page 199: 1,2a,3
Chapter 6
Section 1 Momentum and
Impulse
Linear Momentum, continued
• Impulse
– The product of the force and the time over which
the force acts on an object is called impulse.
– The impulse-momentum theorem states that
when a net force is applied to an object over a
certain time interval, the force will cause a change
in the object’s momentum.
F∆t = ∆p = mvf – mvi
force  time interval = change in momentum
Chapter 6
Section 1 Momentum and
Impulse
Impulse
Click below to watch the Visual Concept.
Visual Concept
Sample Problem B P201
• A 1400kg car moving westward with a velocity of
15m/s collides with a utility pole and is brought to rest
in 0.30s. Find the force exerted on the car during the
collision.
• Given:
• Asked:
• Formula: FΔt = Δp = m vf - mvi
Assignment
• Page 199: 1,2a,3
• Page 201: 1,2,3
Chapter 6
Section 1 Momentum and
Impulse
Linear Momentum, continued
• Stopping times and distances depend on the
impulse-momentum theorem.
• Force is reduced when the time interval of an impact
is increased.
Chapter 6
Section 1 Momentum and
Impulse
Impulse-Momentum Theorem
Chapter 6
Section 1 Momentum and
Impulse
Impulse-Momentum Theorem
Click below to watch the Visual Concept.
Visual Concept
Practice Problem C P202
• A 2240kg car traveling to the west slows down
uniformly from 20.0 m/s to 5.00 m/s/ How long does
it take the car to decelerate if the force on the car is
8410N to the east? How far does the car travel
during the deceleration?
• Given:
• Asked:
• Formula: FΔt = Δp = m vf - mvi
• Δx= vavg Δt
Assignments Chapter 6
• Page 199: 1,2a,3
• Page 201: 1,2,3
• Page 203: 1, 2a,b,c
Chapter 6
Section 2 Conservation of
Momentum
Preview
• Objectives
• Momentum is Conserved
• Sample Problem
Chapter 6
Section 2 Conservation of
Momentum
Objectives
• Describe the interaction between two objects in
terms of the change in momentum of each object.
• Compare the total momentum of two objects before
and after they interact.
• State the law of conservation of momentum.
• Predict the final velocities of objects after collisions,
given the initial velocities, force, and time.
Chapter 6
Section 2 Conservation of
Momentum
Momentum is Conserved
The Law of Conservation of Momentum:
The total momentum of all objects interacting with
one another remains constant regardless of the
nature of the forces between the objects.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
total initial momentum = total final momentum
Chapter 6
Section 2 Conservation of
Momentum
Conservation of Momentum
Click below to watch the Visual Concept.
Visual Concept
Chapter 6
Section 2 Conservation of
Momentum
Sample Problem D P208
Conservation of Momentum
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?
Chapter 6
Section 2 Conservation of
Momentum
Sample Problem, continued
Conservation of Momentum
1. Define
Given:
m1 = 76 kg m2 = 45 kg
v1,i = 0
v2,i = 0
v1,f = 2.5 m/s to the right
Unknown:
v2,f = ?
Assignmments
•
•
•
•
Page 199: 1,2a,3
Page 201: 1,2,3
Page 203: 1, 2a,b,c
Page 209: 1,2,3a,b, 4
Chapter 6
Section 2 Conservation of
Momentum
Sample Problem, continued
Conservation of Momentum
2. Plan
Choose an equation or situation: Because the total
momentum of an isolated system remains constant,
the total initial momentum of the boater and the boat
will be equal to the total final momentum of the boater
and the boat.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Chapter 6
Section 2 Conservation of
Momentum
Sample Problem, continued
Conservation of Momentum
2. Plan, continued
Because the boater and the boat are initially at rest,
the total initial momentum of the system is equal to
zero. Therefore, the final momentum of the system
must also be equal to zero.
m1v1,f + m2v2,f = 0
Rearrange the equation to solve for the final velocity
of the boat.
m2 v 2,f  – m1v1,f
v 2,f
 m1 
 –
 v1,f
 m2 
Chapter 6
Section 2 Conservation of
Momentum
Sample Problem, continued
Conservation of Momentum
3. Calculate
Substitute the values into the equation and solve:
v 2,f
v 2,f
 76 kg 
 –
 2.5 m/s to the right 

 45 kg 
 –4.2 m/s to the right
Chapter 6
Section 2 Conservation of
Momentum
Sample Problem, continued
Conservation of Momentum
4. Evaluate
The negative sign for v2,f indicates that the boat is
moving to the left, in the direction opposite the motion
of the boater. Therefore,
v2,f = 4.2 m/s to the left
Assignments Chapter 6
•
•
•
•
Page 199: 1,2a,3
Page 201: 1,2,3
Page 203: 1, 2a,b,c
Page 209: 1,2,3a,b,4
Chapter 6
Section 2 Conservation of
Momentum
Momentum is Conserved, continued
• Newton’s third law leads
to conservation of
momentum
• During the collision, the
force exerted on each
bumper car causes a
change in momentum for
each car.
• The total momentum is
the same before and after
the collision.
Newton’s 3rd Law and Conservation of Momentum
• If two objects interact the force on one object by the
other is equal and opposite to the force of the other
on the first.
• F1 = - F2
• F1 Δt = m1 v1,f - m1 v1,i
• F2 Δt = m2 v2,f – m2 v2,i
• Δt = Δt
• F1 Δt = - F2 Δt
• m1 v1,f - m1 v1,i = -(m2 v2,f – m2 v2,i )
• m1 v1,f + m2 v2,f = m1 v1,i + m2 v2,i
Assignments
•
•
•
•
•
Page 199: 1,2a,3
Page 201: 1,2,3
Page 203: 1, 2a,b,c
Page 209: 1,2,3a,b,4
P211 Section Review 1abcd. 2abc
Chapter 6
Preview
• Objectives
• Collisions
• Sample Problem
Section 3 Elastic and Inelastic
Collisions
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Objectives
• Identify different types of collisions.
• Determine the changes in kinetic energy during
perfectly inelastic collisions.
• Compare conservation of momentum and conservation of kinetic energy in perfectly inelastic and elastic
collisions.
• Find the final velocity of an object in perfectly
inelastic and elastic collisions.
Elastic and Inelastic
• Elastic Collision- the objects are initially deformed
(think rubber balls) but return to their original shape
so any energy causing the deformation is restored
when the objects return to their original shape
• Inelastic- some of the energy of the collision is
converted to heat, sound or other form of energy and
is lost from the mechanical kinetic energy.
• Perfectly Inelastic- the objects stick together and after
the collision are one object.
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Collisions
• Perfectly inelastic collision
A collision in which two objects stick together after
colliding and move together as one mass is called
a perfectly inelastic collision.
• Conservation of momentum for a perfectly inelastic
collision:
m1v1,i + m2v2,i = (m1 + m2)vf
total initial momentum = total final momentum
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Perfectly Inelastic Collisions
Click below to watch the Visual Concept.
Visual Concept
Sample Problem E P213
• A 1850kg luxury sedan stopped at a traffic light is
struck from the rear by a compact car with a mass of
975kg. The two cars become entangled as a result
of the collision. If the compact car was moving at a
velocity of 22.0 m/s to the north before the collision,
what is the velocity of the entangled mass after the
collision?
Assignments Chapter 6
•
•
•
•
•
Page 199: 1,2a,3
Page 201: 1,2,3
Page 203: 1, 2a,b,c
Page 209: 1,2,3a,b,4
Page 214: 1,2,3
Schedule
• Monday 12/9 P214 Problems Perfectly Inelastic
collisions
• Tuesday 12/10 KE in Perfectly Inelastic collisions
• P216 Problems/ Elastic Collisions
• Wednesday 12/11 P219 Problems Elastic Collisions
• Thursday 12/12 Section Review P220: 2a,b,3a,b,c
• Friday 12/13 Demonstration of Collisions- Lab
• Monday 12/16 Chapter Review (11,13,22,28)
• Tuesday 12/17 Chapter Review (30,31,33,39)
• Wednesday 12/18 Chapter 6 Test
Elastic and Inelastic Collisions (Page 214)
• Inelastic Collision – some of the work done to deform an
inelastic material is converted to other forms of energy
such as heat or sound.
• Kinetic Energy is not conserved in an inelastic
collision.
• Elastic collision – the work done to deform the material
during a collision is equal to the work the material does to
return to its original shape.
• Kinetic Energy is conserved in an Elastic Collision
Inelastic Collision (stick together)
• Momentum is conserved:
m1v1,i + m2v2,i = (m1 + m2)vf
total initial momentum = total final momentum
• Kinetic Energy is not conserved
• ∆KE = KEi – KEf
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem
Kinetic Energy in Perfectly Inelastic Collisions
Two clay balls collide head-on in a perfectly inelastic
collision. The first ball has a mass of 0.500 kg and an
initial velocity of 4.00 m/s to the right. The second ball
has a mass of 0.250 kg and an initial velocity of 3.00
m/s to the left.What is the decrease in kinetic energy
during the collision?
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
1. Define
Given:
m1= 0.500 kg
m2 = 0.250 kg
v1,i = 4.00 m/s to the right, v1,i = +4.00 m/s
v2,i = 3.00 m/s to the left, v2,i = –3.00 m/s
Unknown:
∆KE = ?
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
2. Plan
Choose an equation or situation: The change in
kinetic energy is simply the initial kinetic energy
subtracted from the final kinetic energy.
∆KE = KEi – KEf
Determine both the initial and final kinetic energy.
1
1
2
Initial: KEi  KE1,i  KE2,i  m1v1,i  m2v 2,2 i
2
2
1
Final: KEf  KE1,f  KE2,f   m1  m2  v f2
2
Section 3 Elastic and Inelastic
Collisions
Chapter 6
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
2. Plan, continued
Use the equation for a perfectly inelastic collision to
calculate the final velocity.
vf 
m1v1,i  m2v 2,i
m1  m2
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
3. Calculate
Substitute the values into the equation and
solve: First, calculate the final velocity, which will be
used in the final kinetic energy equation.
(0.500 kg)(4.00 m/s)  (0.250 kg)(–3.00 m/s)
vf 
0.500 kg  0.250 kg
v f  1.67 m/s to the right
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
3. Calculate, continued
Next calculate the initial and final kinetic energy.
1
1
2
2
0.500
kg
4.00
m/s

0.250
kg
–3.00
m/s





  5.12 J
2
2
1
2
KEf   0.500 kg  0.250 kg1.67 m/s   1.05 J
2
KEi 
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Kinetic Energy in Perfectly Inelastic Collisions
3. Calculate, continued
Finally, calculate the change in kinetic energy.
KE  KEf – KEi  1.05 J – 5.12 J
KE  –4.07 J
4. Evaluate The negative sign indicates that kinetic
energy is lost.
Assignments Chapter 6
•
•
•
•
•
•
Page 199: 1,2a,3
Page 201: 1,2,3
Page 203: 1, 2a,b,c
Page 209: 1,2,3a,b,4
Page 214: 1,2,3
Page 216: 1,2,3
Elastic and Inelastic Collisions (Page 214)
• Inelastic Collision – some of the work done to deform an
inelastic material is converted to other forms of energy
such as heat or sound.
• Kinetic Energy is not conserved in an inelastic
collision.
• Elastic collision – the work done to deform the material
during a collision is equal to the work the material does to
return to its original shape.
• Kinetic Energy is conserved in an Elastic Collision
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Elastic Collisions
• Elastic Collision
A collision in which the total momentum and the
total kinetic energy are conserved is called an
elastic collision.
• Momentum and Kinetic Energy Are Conserved in
an Elastic Collision
m1v1,i  m2v 2,i  m1v1,f  m2v 2,f
1
1
1
1
2
2
2
2
m1v1,i  m2v 2,i  m1v1,f  m2v 2,f
2
2
2
2
Two Extremes- Reality is in Between P217
• Elastic and perfectly inelastic collisions are limiting
cases, most collisions actually fall into a category
between these two extremes.
• In this third category, called inelastic collisions, the
colliding objects bounce and move separately after
the collision, but the total kinetic energy decreases.
• For problems in this book we consider all collisions in
which the objects do not stick together to be elastic.
• Therefore we will assume that the total momentum
and the total KE will each stay the same before and
after a collision that is not perfectly inelastic.
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Types of Collisions
Click below to watch the Visual Concept.
Visual Concept
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem G P218, continued
Elastic Collisions
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 0.180 m/s. After
the collision, the smaller marble moves to the left at
0.315 m/s. Assume that neither marble rotates before
or after the collision and that both marbles are
moving on a frictionless surface.What is the velocity
of the 0.030 kg marble after the collision?
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Elastic Collisions
1. Define
Given: m1 = 0.015 kg m2 = 0.030 kg
v1,i = 0.225 m/s to the right, v1,i = +0.225 m/s
v2,i = 0.180 m/s to the left, v2,i = –0.180 m/s
v1,f = 0.315 m/s to the left, v1,i = –0.315 m/s
Unknown:
v2,f = ?
Assignments Chapter 6
•
•
•
•
•
•
•
•
•
•
Page 199: 1,2a,3
Page 201: 1,2,3
Page 203: 1, 2a,b,c
Page 209: 1,2,3a,b,4
Page 214: 1,2,3
Page 216: 1,2,3
Page 219: 1a,b,2a,b, 3a,b
Chapter Review Page 223-225
11a,b, 13, 22a,b, 28, 30a,b, 31a,b, 33, 39
5 Points Extra Credit Problem 43 Page 226
Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Elastic Collisions
2. Plan
Choose an equation or situation: Use the equation for
the conservation of momentum to find the final velocity
of m2, the 0.030 kg marble.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Rearrange the equation to isolate the final velocity of m2.
m2 v2,f  m1v1i,  m2 v2,i – m1v1,f
v 2,f 
m1v1,i  m2 v2,i – m1v1,f
m2
Section 3 Elastic and Inelastic
Collisions
Chapter 6
Sample Problem, continued
Elastic Collisions
3. Calculate
Substitute the values into the equation and solve: The
rearranged conservation-of-momentum equation will
allow you to isolate and solve for the final velocity.
 0.015 kg 0.225 m/s   0.030 kg  –0.180 m/s  – 0.015 kg –0.315 m/s 
v 
2,f
v 2,f
v 2,f
0.030 kg
3.4  10


–3
 
 
kg  m/s  –5.4  10 –3 kg  m/s – –4.7  10 –3 kg m/s
2.7  10 –3 kg  m/s

3.0  10 –2 kg
v2,f  9.0  10 –2 m/s to the right
0.030 kg

Chapter 6
Section 3 Elastic and Inelastic
Collisions
Sample Problem, continued
Elastic Collisions
4. Evaluate Confirm your answer by making sure kinetic
energy is also conserved using these values.
1
1
1
1
m1v1,2i  m2v 2,2 i  m1v1,2f  m2v 2,2 f
2
2
2
2
1
1
2
2
KEi   0.015 kg  0.225 m/s    0.030 kg  –0.180 m/s 
2
2
 8.7  10 –4 kg  m2 /s2  8.7  10 –4 J
1
1
2
2
KEf   0.015 kg  0.315 m/s    0.030 kg 0.090 m/s 
2
2
 8.7  10 –4 kg  m2 /s2  8.7  10 –4 J