Transcript Conservation of Momentum

Chapter 7 Impulse and Momentum
Ying Yi PhD
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Outline
 Impulse and Momentum Theorem
 Conservation of Momentum
 Collision
 Inelastic
 Elastic
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Momentum
 Definition: The linear momentum
p of an object of
mass m moving with a velocity is defined as the
product of the mass and the velocity v
p  mv
 SI Units: kg•m / s
 Note that: Vector quantity, the direction of the
momentum is the same as the velocity’s
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Components of Momentum
 X components:
 px = m vx
 Y components:
 p y = m vy
 Momentum is related to kinetic energy
p2
KE 
2m
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Momentum and Kinetic Energy
Definition:
Units:
Relation:
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p  mv
KE 
1
mv 2
2
J
kg•m / s
p2
KE 
2m
Change of Momentum and Force
 In order to change the momentum of an object, a
force must be applied
 The time rate of change of momentum of an
object is equal to the net force acting on it

p m(v f  v i )

 Fnet
t
t
 Gives an alternative statement of Newton’s second law
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Impulse
 When a single, constant force acts on the object, there
is an impulse delivered to the object

I  Ft
 I is defined as the impulse
 Vector quantity, the direction is the same as the
direction of the force
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Impulse-Momentum Theorem
 The theorem states that the impulse acting on the
object is equal to the change in momentum of the
object

I  Ft  p  mvf  mvi
 If the force is not constant, use the average force applied
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Average Force in Impulse
 The average force can be
thought of as the constant
force that would give the
same impulse to the object
in the time interval as the
actual time-varying force
gives in the interval
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Average Force cont.
 The impulse imparted by a force during the time
interval Δt is equal to the area under the force-time
graph from the beginning to the end of the time
interval
 Or, the impulse is equal to the average force
multiplied by the time interval,
Fav t  p
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Impulse Applied to Auto Collisions
 The most important factor is the collision time or the
time it takes the person to come to a rest
 This will reduce the chance of dying in a car crash
 Ways to increase the time
 Seat belts
 Air bags
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Typical Collision Values
 For a 75 kg person
traveling at 27 m/s
and coming to stop in
0.010 s
 F = -2.0 x 105 N
 a = 280 g
 Almost certainly fatal
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Comparison of Accelerations
About 2g
280g
About 4g
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Survival
 Increase time
 Seat belt
 Restrain people so it takes more time for them to stop
 New time is about 0.15 seconds
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Air Bags
 The air bag increases the time of the collision
 It will also absorb some of the energy from the
body
 It will spread out the area of contact
 Decreases the pressure
 Helps prevent penetration wounds
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Example 7.1 A well hit ball
A baseball (m=0.14 kg) has an initial velocity of V0=-38
m/s as it approaches a bat. We have chosen the direction
of approach as the negative direction. The bat applies
an average force F that is much larger that the weight
of the ball, and the ball departs from the bat with a final
velocity of Vf=+58 m/s. (a) Determine the impulse
applied to the ball by the bat. (b) Assuming that the
time of the contact is ∆t=1.6×10-3 s, find the average
force exerted on the ball by the bat.
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Group Problem: Teeing off
A golf ball with mass 5.0×10-2 kg is struck with a club
as in Figure 6.3. The force on the ball varies from zero
when contact is made up to some maximum value and
then back to zero when the ball leaves the club, as in the
graph of force vs. time in Figure 6.1. Assume that the
ball leaves the club face with a velocity of +44 m/s. (a)
Find the magnitude of the impulse due to the collision.
(b) Estimate the duration of the collision and the
average force acting on the ball. (Assume contacting
distance is 2cm.)
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Conservation of Momentum
（W 1  F 12）t  m1V f 1  m1V i1
（W2  F 21）t  m2V f 2  m2V i 2
Newton’s Third Law F 21  F12
m1V i1  m2V i 2  m1V f 1  m2V f 2
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How about explosion?
Momentum conserved?
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Conservation of Momentum
 Momentum in an isolated system in which a
collision occurs is conserved
 A collision may be the result of physical contact
between two objects
 “Contact” may also arise from the electrostatic
interactions of the electrons in the surface atoms of the
bodies
 An isolated system will have no external forces
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Example 7.5 Assembling a Freight Train
A freight train is being assembled in a switching yard,
and Figure 7.8 shows two boxcars. Car 1 has a mass of
m1=65×103 kg and moves at a velocity of v01=+0.80
m/s. Car 2, with a mass of m2=92×103 kg and a
velocity of v02=+1.3 m/s, overtakes car 1 and couples to
it Neglecting friction, find the common velocity vf of
the cars after they become coupled.
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Group Problem: Ice Skaters
Starting from rest, two skaters push off against each
other on smooth level ice, where friction is negligible.
One is woman (m1=54 kg), and one is man (m2=88 kg).
Part b of the drawing shows that the woman moves
away with a velocity of vf1=+2.5 m/s. Find the recoil
velocity of vf2 of the man
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Types of collisions
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Inelastic
Collision
Kinetic energy is not conserved.
Some of the kinetic energy is
converted into other types of
energy such as heat, sound, work to
permanently deform an object.
Perfectly inelastic collisions occur
when the objects stick together
Elastic
Collision
Both momentum and kinetic
energy are conserved
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Collision Examples
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Perfectly Inelastic Collisions
 When two objects stick
together after the
collision, they have
undergone a perfectly
inelastic collision
 Conservation of
momentum becomes
m1 v i1  m2 v i 2  (m1  m2 )v f
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Elastic Collisions
 Both momentum and kinetic energy are conserved
 Typically have two unknowns
m1 v i1  m2 v i 2  m1 v f 1  m2 v f 2
1
1
1
1
2
2
2
m1vi1  m2 vi 2  m1v f 1  m2v 2f 2
2
2
2
2
 Solve the equations simultaneously
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Summary of Types of Collisions
 In an elastic collision, both momentum and kinetic
energy are conserved
 In an inelastic collision, momentum is conserved
but kinetic energy is not
 In a perfectly inelastic collision, momentum is
conserved, kinetic energy is not, and the two
objects stick together after the collision, so their
final velocities are the same
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Problem Solving for One -Dimensional Collisions
 Coordinates: Set up a coordinate axis and define the
velocities with respect to this axis
 Diagram: Draw all the velocity vectors and label the
velocities and the masses
 Conservation of Momentum: Write a general
expression for the total momentum of the system before
and after the collision
 Conservation of Energy: If the collision is elastic,
write a second equation for conservation of KE, or the
alternative equation (perfectly elastic collisions)
 Solve: The resulting equations simultaneously
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Example 7.3 measuring the speed of a bullet
The ballistic pendulum shown in Figure 7.12 a consists
of a stationary 2.5 kg block of wood suspended by a
wire of negligible mass. A 0.0100 kg bullet is fired into
the block, and the block (with the bullet in it) swings to
a maximum height of 0.650 m above the initial position.
Find the speed with which the bullet is fired. (Ignore air
resistance)
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Group Problem: A truck versus a compact
A pickup truck with mass 1.80×103 kg is traveling
eastbound at +15.0 m/s, while a compact car with mass
9.00×102 kg is traveling west bound at -15.0 m/s. The
vehicles collide head-on, becoming entangled. (a) Find
the speed of the entangled vehicles after the collision.
(b) Find the change in the velocity of each vehicle. (c)
Find the change in the kinetic energy of the system
consisting of both vehicles.
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A collision in two dimensions
 For a general collision of two objects in three-
dimensional space, the conservation of
momentum principle implies that the total
momentum of the system in each direction is conserved

m1 v i1x  m2 v i 2 x  m1 v f 1x  m2 v f 2 x
m1 v i1 y  m2 v i 2 y  m1 v f 1 y  m2 v f 2 y
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Problem Solving for Two-Dimensional Collisions
 Conservation of Momentum: Write expressions
for the x and y components of the momentum of
each object before and after the collision
 Write expressions for the total momentum before
and after the collision in the x-direction and in the
y-direction
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Group Problem: collision at an intersection
A car with mass 1.50×103 kg traveling east at a speed
of 25.0 m/s collides at an intersection with a 2.50×103
kg van traveling north at a speed of 20.0 m/s. Find the
magnitude and direction of the velocity of the
wreckage after the collision, assuming that the vehicles
undergo a perfectly inelastic collision and assuming that
friction between the vehicles and the road can be
neglected.
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Homework: 4,8,12,16,17,30,38,42,
Thank you
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