Transcript Chapter 13 Vibrations and Waves

Chapter 6 Momentum and
Collisions
Ying Yi PhD
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Outline
 Momentum and Impulse
 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 6.1 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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Group Problem: How good are the Bumpers?
In a crash test, a car of mass 1.5×103kg collides with a
wall and rebounds as in Figure 6.4a. The initial and
final velocities of the car are vi=-15.0 m/s and vf=2.60
m/s, respectively. If the collision lasts for 0.150 s, find
(a) the impulse delivered to the car due to the collision
and (b) the size and direction of the average force
exerted on the car.
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Conservation of Momentum
F 21t  m1V 1 f  m1V 1i
F 12 t  m2V 2 f  m2V 2i
Newton’s Third Law F 21  F12
m1V 1i  m2V 2i  m1V 1 f  m2V 2 f
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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 6.3 The Archer
An archer stands at rest on frictionless ice; his total
mass including his bow and quiver of arrows is 60.00
kg. (a) If the archer fires a 0.0300 kg arrow horizontally
at 50.0 m/s in the positive x-direction, what is his
subsequent velocity across the ice? (b) He then fires a
second identical arrow at the same speed relative to the
ground but at an angle of 30.0° above the horizontal.
Find his new speed. (c) Estimate the average normal
force acting on the archer as the second arrow is
accelerated by the bowstring. Assume a draw length of
0.800 m.
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Forces in a Collision
 The force with which
object 1 acts on object
2 is equal and opposite
to the force with which
object 2 acts on object
1
 Impulses are also
equal and opposite
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Types of collisions (Momentum is conserved)
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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 v1i  m2 v 2i  (m1  m2 )v f
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Elastic Collisions
 Both momentum and kinetic energy are conserved
 Typically have two unknowns
m1 v1i  m2 v 2i  m1 v1 f  m2 v 2 f
1
1
1
1
2
2
2
m1v1i  m2 v2i  m1v1 f  m2v22 f
2
2
2
2
 Solve the equations simultaneously
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Elastic Collisions, cont.
 A simpler equation can
be used in place of the
KE equation
v1i  v2i  ( v1 f  v2 f )
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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 6.5: The Ballistic Pendulum
The ballistic pendulum is a device used to measure the speed
of a fast-moving projectile such as a bullet. The bullet is
fired into a large block of wood suspended from some light
wires. The bullet is topped by the block, and the entire
system swings up to a height h. It is possible of obtain the
initial speed of the bullet by measuring h and the two
masses. As an example of the technique, assuming that the
mass of the bullet, m1, is 5.00 g, the mass of the pendulum,
m2, is 1.000 kg, and h is 5.00 cm. (a) Find the velocity of the
system after the bullet embeds in the block. (b) Calculate the
initial speed of the bullet.
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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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Sketches for Collision Problems
 Draw “before” and
“after” sketches
 Label each object
 Include the direction of
velocity
 Keep track of subscripts
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Sketches for Perfectly Inelastic
Collisions
 The objects stick
together
 Include all the velocity
directions
 The “after” collision
combines the masses
 Both move with the
same velocity
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Glancing Collisions
 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

m1v 1ix  m2 v 2ix  m1v 1fx  m2 v 2 fx and
m1v 1iy  m2 v 2iy  m1v 1fy  m2 v 2 fy
 Use subscripts for identifying the object, initial and final
velocities, and components
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Glancing Collisions
 The “after” velocities have x and y components
 Momentum is conserved in the x direction and in the y
direction
 Apply conservation of momentum separately to each
direction
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Problem Solving for Two-Dimensional
Collisions
 Coordinates: Set up coordinate axes and define your
velocities with respect to these axes
 It is convenient to choose the x- or y- axis to coincide
with one of the initial velocities
 Diagram: In your sketch, draw and label all the
velocities and masses
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Problem Solving for Two-Dimensional
Collisions, 2
 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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Problem Solving for Two-Dimensional
Collisions, 3
 Conservation of Energy: If the collision is elastic,
write an expression for the total energy before and
after the collision
 Equate the two expressions
 Fill in the known values
 Solve the quadratic equations
 Can’t be simplified
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Problem Solving for Two-Dimensional
Collisions, 4
 Solve for the unknown quantities
 Solve the equations simultaneously
 There will be two equations for inelastic collisions
 There will be three equations for elastic collisions
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Example 6.8 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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Rocket Propulsion
The rocket is accelerated as a result of the thrust of
the exhaust gases
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Thank you
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