Transcript Chapter 7

Chapter 7
Impulse and
Momentum
Momentum

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 are kg m / s
• Vector quantity, the direction of the
momentum is the same as the velocity’s
Momentum components
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p x  mv x and p y  mv y
Applies to two-dimensional motion
Impulse
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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, e.g.
•
v  vo  at
mv  mvo  mat  mvo  Ft
Ft  mv  mvo
• Gives an alternative statement of Newton’s
second law
Impulse cont.

When a single, constant force acts on
the object, there is an impulse
delivered to the object
• impulse  Ft  F (t 2  t1)
• is defined as the impulse
• Vector quantity, the direction is the
same as the direction of the force
• Unit N·s=kg·m/s
Impulse-Momentum Theorem

The theorem states that the impulse
acting on the object is equal to the
change in momentum of the object
•
Ft  p F  p I
• Impulse=change in momentum
(vector!)
• If the force is not constant, use the
average force applied
Impulse Applied to Auto
Collisions
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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
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Ways to increase the time
• Seat belts
• Air bags
Air Bags
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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
Example
0.05 kg ball moving at 2.0 m/s
rebounds with the same speed. If the
contact time with the wall is 0.01 s,
what is average force of the wall on
the ball?
Conservation of Momentum

Total momentum of a system equals to
the vector sum of the momenta
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
p   pi  p1  p 2  p3  ...  m1v1  m2v2  m3v3  ...
i
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When no resultant external force acts on a
system, the total momentum of the
system remains constant in magnitude
and direction.
Components of Momentum
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 not
external forces
Conservation of Momentum,
cont

The principle of conservation of
momentum states when no
external forces act on a system
consisting of two objects that
collide with each other, the total
momentum of the system remains
constant in time
• Specifically, the total momentum
before the collision will equal the total
momentum after the collision
Conservation of Momentum,
cont.

Mathematically:
m1 v1I  m2 v 2 I  m1 v1F  m2 v 2 F
• Momentum is conserved for the system of
objects
• The system includes all the objects interacting
with each other
• Assumes only internal forces are acting during
the collision
• Can be generalized to any number of objects
Example
Two skaters are initially at rest.
Masses are 80kg and 50kg. If they
push each other so that woman is
given a velocity of 2.5 m/s. What is
the velocity of the man?
Types of Collisions
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Momentum is conserved in any
collision
Perfect elastic collision
• both momentum and kinetic energy are
conserved
Ek ,before  Ek ,after
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Collision of billiard balls, steel balls
More Types of Collisions
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Inelastic collisions
• Kinetic energy is not conserved
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Some of the kinetic energy is converted into other
types of energy such as heat, sound, work to
permanently deform an object
• completely inelastic collisions occur when the
objects stick together
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Not all of the KE is necessarily lost
Actual collisions
• Most collisions fall between elastic and
completely inelastic collisions
More About Perfectly Inelastic
Collisions
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When two objects stick together after the
collision, they have undergone a perfectly
inelastic collision
Conservation of momentum becomes
m1v1  m2 v2  (m1  m2 )V
m1v1  m2 v2
V
m1  m2
Example
Railroad car (10,000kg) travels at
10m/s and strikes another railroad
car (15,000kg) at rest. They couple
after collision. Find the final velocity
of the two cars. What is the energy
loss in the collision?
Recoil
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System is released from rest
Momentum of the system is zero
before and after
m1v1F  m2v2 F  0
Example
4 kg rifle shoots a 50 grams bullet. If
the velocity of the bullet is 280 m/s,
what is the recoil velocity of the rifle?
Some General Notes About
Collisions
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Momentum is a vector quantity
• Direction is important
• Be sure to have the correct signs
More About Elastic Collisions
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Both momentum and kinetic energy
are conserved
Typically have two unknowns (1d)
m1v1I  m2v2 I  m1v1F  m2v2 F
1
1
1
1
2
2
2
2
2v12FI  vm

v
m1v1I v1I m
v

m
v
2 2F
2 I1 1F 2 F
2
2
2
2
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Solve the equations simultaneously
A Simple Case, v2i=0
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Head on elastic collision with object 2
at rest before collision.
One can show
m1  m2
v1F 
v1I
m1  m2
v2 F
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Special cases
2m1

v1I
m1  m2
Ballistic Pendulum
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Measure speed of bullet
Momentum conservation of the
collision
Energy conservation during the
swing of the pendulum
Summary of Types of Collisions
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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
Example
7.31. Balls A and B collide head-on in
a perfectly elastic collision. It is
known that mA=2mB and that the
initial velocities are +3 m/s for A and
–2 m/s for B. Find their velocities
after the collision.
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
Example
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Car, 1500 kg. SUV 2500 kg.
Find speed and direction after
collision.
Example
m1=0.15 kg, m2=0.26 kg, v1i=0.9 m/s
at 50° to y-axis,
v2i=0.54 m/s, v2f=0.7 m/s at 35°
below x-axis
Find v2f (magnitude and direction)