Transcript Momentum and Collisions

Momentum and
Collisions
Linear Momentum

The linear momentum of a particle or an
object that can be modeled as a particle of
mass m moving with a velocity v is defined to
be the product of the mass and velocity:

p=mv
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
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We will usually refer to this as “momentum”, omitting the
“linear”.
Momentum is a VECTOR.
It has components. Don’t forget that.
Linear Momentum
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The dimensions of momentum are ML/T
The SI units of momentum are kg · m / s
Momentum can be expressed in component
form:

px = m v x
py = m v y
pz = m v z
A 3.00-kg particle has a velocity of . (a) Find its x and y
components of momentum. (b) Find the magnitude and
direction of its momentum.
Newton and Momentum

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Newton called the product mv the quantity of
motion of the particle
Newton’s Second Law can be used to relate
the momentum of a particle to the resultant
force acting on it
dv d  mv  dp
F  ma  m


dt
dt
dt
with constant mass.
Newton’s Second Law

The time rate of change of the linear momentum of
a particle is equal to the net force acting on the
particle
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This is the form in which Newton presented the Second
Law
It is a more general form than the one we used previously
This form also allows for mass changes
Applications to systems of particles are particularly
powerful
Two Particles (1 dimension for now)
dptotal dp1 dp2


dt
dt
dt
IF there are NO external forces then
the only forces acting on this system are
those that are one particle producing a
force on the other.
Continuing
dptotal dp1 dp2


dt
dt
dt
dptotal
Fexternal  0 
  Fint ernal  0
dt
NO EXTERNAL FORCES: ptotal is
constant

Force on P1 is from particle 2 and is F21
Force on P2 is from particle 1 and is F12

F21=-F12

Summary
The momentum of a SYSTEM of particles that are
isolated from external forces remains a constant of
the motion.
Conservation of Linear Momentum
Textbook Statement

Whenever two or more particles in an
isolated system interact, the total
momentum of the system remains
constant

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The momentum of the system is conserved,
not necessarily the momentum of an individual
particle
This also tells us that the total momentum of
an isolated system equals its initial momentum
Two blocks of masses M and
3M are placed on a horizontal,
frictionless surface. A light
spring is attached to one of
them, and the blocks are
pushed together with the
spring between them). A cord
initially holding the blocks
together is burned; after this,
the block of mass 3M moves
to the right with a speed of
2.00 m/s. (a) What is the speed
of the block of mass M? (b)
Find the original elastic
potential energy in the spring
if M = 0.350 kg.
An Easy One…
A particle of mass m moves with
momentum p. Show that the kinetic
energy of the particle is K = p2/2m. (b)
Express the magnitude of the particle’s
momentum in terms of its kinetic energy
and mass.
Consider a particle roaming around that is
suddenly subjected to some kind of FORCE
that looks something like the last slide’s graph.
NEWTON :
p  mv
dp
dv
m
 ma  F
dt
dt
p  constant if F  0
If F  0 then p will change
Let’s drop the vector notation and stick to
one dimension.
dp
F
dt
dp  Fdt
f
tf
i
ti
 dp   Fdt
tf
p f  pi   Fdt
ti
Change of Momentum
Impulse
NEW LAW

IMPULSE = CHANGE IN MOMENTUM
A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at
an angle of 60.0° with the surface. It bounces off with the
same speed and angle (Fig. P9.9). If the ball is in contact
with the wall for 0.200 s, what is the average force exerted
on the ball by the wall?
LETS
TALK
ABOUT
Consider two particles:
v2
v1
m1
m2
1
1
V1
2
2
V2
What goes on during the collision? N3
Force on m2
=F(12)
Force on m1
=F(21)
The Forces
Equal and Opposite
More About Impulse: F-t The Graph
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Impulse is a vector quantity
The magnitude of the
impulse is equal to the area
under the force-time curve
Dimensions of impulse are
ML/T
Impulse is not a property of
the particle, but a measure
of the change in momentum
of the particle
Impulse
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The impulse can also
be found by using the
time averaged force
I = F Dt
This would give the
same impulse as the
time-varying force does
SKATEBOARD DEMO
Conservation of Momentum, Archer
Example
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The archer is standing on
a frictionless surface (ice)
Approaches:
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Newton’s Second Law –
no, no information about F
or a
Energy approach – no, no
information about work or
energy
Momentum – yes
Archer Example, 2
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Let the system be the archer with bow (particle 1)
and the arrow (particle 2)
There are no external forces in the x-direction, so it
is isolated in terms of momentum in the x-direction
Total momentum before releasing the arrow is 0
The total momentum after releasing the arrow is p1f
+ p2f = 0
Archer Example, final
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The archer will move in the opposite direction
of the arrow after the release
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Agrees with Newton’s Third Law
Because the archer is much more massive
than the arrow, his acceleration and velocity
will be much smaller than those of the arrow
An estimated force-time curve for a baseball struck by a bat
is shown in Figure P9.7. From this curve, determine (a) the
impulse delivered to the ball, (b) the average force exerted
on the ball, and (c) the peak force exerted on the ball.
Overview: Collisions – Characteristics
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We use the term collision to represent an event
during which two particles come close to each other
and interact by means of forces
The time interval during which the velocity changes
from its initial to final values is assumed to be short
The interaction force is assumed to be much greater
than any external forces present

This means the impulse approximation can be used
Collisions – Example 1
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Collisions may be the
result of direct contact
The impulsive forces
may vary in time in
complicated ways
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This force is internal to
the system
Momentum is
conserved
Collisions – Example 2
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The collision need not
include physical contact
between the objects
There are still forces
between the particles
This type of collision can
be analyzed in the same
way as those that include
physical contact
Types of Collisions
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In an elastic collision, momentum and kinetic
energy are conserved
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Perfectly elastic collisions occur on a microscopic level
In macroscopic collisions, only approximately elastic
collisions actually occur
In an inelastic collision, kinetic energy is not
conserved although momentum is still conserved

If the objects stick together after the collision, it is a
perfectly inelastic collision
Collisions, cont
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In an inelastic collision, some kinetic energy
is lost, but the objects do not stick together
Elastic and perfectly inelastic collisions are
limiting cases, most actual collisions fall in
between these two types
Momentum is conserved in all collisions
Perfectly Inelastic Collisions
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Since the objects stick
together, they share the
same velocity after the
collision
m1v1i + m2v2i =
(m1 + m2) vf
A 10.0-g bullet is fired into a stationary block of wood (m =
5.00 kg). The relative motion of the bullet stops inside the
block. The speed of the bullet-plus-wood combination
immediately after the collision is 0.600 m/s. What was the
original speed of the bullet?
. Velcro couplers make the carts stick together after colliding.
Find the final velocity of the train of three carts. (b) What If ?
Does your answer require that all the carts collide and stick
together at the same time? What if they collide in a different
order?
Most of us know intuitively that in a head-on collision between a
large dump truck and a subcompact car, you are better off being
in the truck than in the car. Why is this? Many people imagine
that the collision force exerted on the car is much greater than
that experienced by the truck. To substantiate this view, they
point out that the car is crushed, whereas the truck is only
dented. This idea of unequal forces, of course, is false.
Newton’s third law tells us that both objects experience forces
of the same magnitude. The truck suffers less damage because
it is made of stronger metal. But what about the two drivers?
Do they experience the same forces? To answer this question,
suppose that each vehicle is initially moving at 8.00 m/s and that
they undergo a perfectly inelastic head-on collision. Each driver
has mass 80.0 kg. Including the drivers, the total vehicle masses
are 800 kg for the car and 4 000 kg for the truck. If the collision
time is 0.120 s, what force does the seatbelt exert on each driver?
SPRING BREAK
BREAK
March 20, 2006
Welcome Home!!
• The pace now quickens!
• 5 weeks (+ 1 session) remain in the semester
• Next Exam March 31
– Details on next 2 slides
• This week we return to momentum
• Watch for another WA soon
• Quiz on Friday - momentum
APPROXIMATE SCHEDULE
20Mar
Ch 9: Momentum
Ch 9: Momentum
27Mar
Ch 10: Rotation
Ch 10: Rotation
3-Apr Ch 10: Rotation
Ch 9: Momentum
EXAM #3
Ch 11: Angular
Mom.
Ch 11: Angular
Mom.
10Apr
Ch 11: Angular
Mom.
Ch 11: Angular
Mom.+ Chapt 12.
Ch 12: Statics
17Apr
Ch 12: Statics
Ch 15: Oscillations
Ch 15: Oscillations
24Apr
Ch 15: Oscillations
1May
FINAL EXAM
No Class
No Class
No Class
No Class
Next Exam
• Will Include
– Potential Energy Chapter
– Conservation of Momentum
– Rotation – Material covered prior to
exam
• Includes Wednesday before exam! SORRY!
Now Where Were We???
Oh yeah …. Elastic Collisions
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Both momentum and
kinetic energy are
conserved
m1v1i  m2 v 2 i 
m1v1 f  m2 v 2 f
1
1
2
m1v1i  m2 v 22i 
2
2
1
1
2
m1v1 f  m2 v 22 f
2
2
Elastic Collisions, cont
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Typically, there are two unknowns to solve for and so
you need two equations
The kinetic energy equation can be difficult to use
The energy equation, along with conservation of
momentum, can be used to solve for the two unknowns
 It can only be used with a one-dimensional, elastic
collision between two objects
The solution is shown on pages 262-3 in the
textbook. (Lots of algebra but nothing all that
difficult.
We will look at a special case.
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One Dimension
Energy is Conserved
Let’s look at the case of equal masses.
Second Particle initially at rest
Explains the demo!
Elastic Collisions, final
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Example of some special cases
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m1 = m2 – the particles exchange velocities
When a very heavy particle collides head-on with a very
light one initially at rest, the heavy particle continues in
motion unaltered and the light particle rebounds with a
speed of about twice the initial speed of the heavy particle
When a very light particle collides head-on with a very
heavy particle initially at rest, the light particle has its
velocity reversed and the heavy particle remains
approximately at rest
Collision Example – Ballistic
Pendulum
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Perfectly inelastic collision –
the bullet is embedded in
the block of wood
Momentum equation will
have two unknowns
Use conservation of energy
from the pendulum to find
the velocity just after the
collision
Then you can find the
speed of the bullet
As shown in the figure, a bullet of mass m and speed v passes
completely through a pendulum bob of mass M. The bullet
emerges with a speed of v/2. The pendulum bob is suspended
by a stiff rod of length and negligible mass. What is the
minimum value of v such that the pendulum bob will barely
swing through a complete vertical circle?
Two Dimensional Collisions
Two-Dimensional Collisions
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The momentum is conserved in all directions
Use subscripts for
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identifying the object
indicating initial or final values
the velocity components
If the collision is elastic, use conservation of kinetic
energy as a second equation
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Remember, the simpler equation can only be used for onedimensional situations
Two-Dimensional Collision, example
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Particle 1 is moving at
velocity v1i and particle
2 is at rest
In the x-direction, the
initial momentum is
m1v1i
In the y-direction, the
initial momentum is 0
Two-Dimensional Collision, example
cont
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After the collision, the
momentum in the xdirection is m1v1f cos q 
m2v2f cos f
After the collision, the
momentum in the ydirection is m1v1f sin q 
m2v2f sin f
Problem-Solving Strategies – TwoDimensional Collisions

Set up a coordinate system and define your
velocities with respect to that system


It is usually convenient to have the x-axis coincide
with one of the initial velocities
In your sketch of the coordinate system, draw
and label all velocity vectors and include all
the given information
Problem-Solving Strategies – TwoDimensional Collisions, 2
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Write expressions for the x- and y-components of
the momentum of each object before and after the
collision
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Remember to include the appropriate signs for the
components of the velocity vectors
Write expressions for the total momentum of the
system in the x-direction before and after the
collision and equate the two. Repeat for the total
momentum in the y-direction.
Problem-Solving Strategies – TwoDimensional Collisions, 3
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If the collision is inelastic, kinetic energy of
the system is not conserved, and additional
information is probably needed
If the collision is perfectly inelastic, the final
velocities of the two objects are equal. Solve
the momentum equations for the unknowns.
Problem-Solving Strategies – TwoDimensional Collisions, 4
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If the collision is elastic, the kinetic energy of
the system is conserved
Equate the total kinetic energy before the
collision to the total kinetic energy after the
collision to obtain more information on the
relationship between the velocities
Two-Dimensional Collision Example
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Before the collision, the
car has the total
momentum in the xdirection and the van
has the total
momentum in the ydirection
After the collision, both
have x- and ycomponents
Two automobiles of equal mass approach an intersection. One
vehicle is traveling with velocity 13.0 m/s toward the east and
the other is traveling north with speed v2i. Neither driver sees
the other. The vehicles collide in the intersection and stick
together, leaving parallel skid marks at an angle of 55.0o north
of east. The speed limit for both roads is 35 mi/h and the
driver of the northward-moving vehicle claims he was within
the speed limit when the collision occurred. Is he telling the
truth?
New Topic: Center of Mass
Momentum of a SYSTEM of particles.
The Center of Mass
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
There is a special point in a system or object,
called the center of mass, that moves as if
all of the mass of the system is concentrated
at that point
The system will move as if an external force
were applied to a single particle of mass M
located at the center of mass

M is the total mass of the system
Consider a SYSTEM of N particles

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


There are N point particles.
Each point “I” has a mass mi
The TOTAL mass of the system is given by M
where M=mi
Let the FORCE on particle “I” be given by Fi,
a VECTOR Quantity.
Let’s see what Newton says about such a
SYSTEM
The MATH …
Newton : Fsystem  Ma some
kind
of
average
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Fsystem is the TOTAL force on the system of
particles
asome… is some average acceleration that we
need to deal with.
We use a vector approach …
Consider a system of only two particles
external
1
F
 F2 on 1  m1a1
F2external  F1 on 2  m2a 2
F2 on 1  F1 on 2
F external  F1external  F2external
F
external
 m1a1  m2a 2
F external  m1a1  m2a 2
F external
d2
d 2 m1r1  m2r2
 2 (m1r1  m2r2 )  M 2 (
)
dt
dt
M
where
M  m1  m2
This would look like F  ma if we define
the center of mass vector R as
R CM
m1r1  m2r2

M
and
F  Ma cm
We therefore define the center of mass as
1 

R CM 
  m1ri 
M i

M   mi
i
Center of Mass, Coordinates

The coordinates of the center of mass are
xCM 

m x
i i
i
M
yCM 
m y
i
i
M
i
zCM 
m z
i i
i
M
where M is the total mass of the system
Center of Mass, position

The center of mass can be located by its
position vector, rCM
rCM 

m r
i i
i
M
ri is the position of the i th particle, defined by
ri  xi ˆi  yi ˆj  zi kˆ
Center of Mass, Example


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Both masses are on the
x-axis
The center of mass is
on the x-axis
The center of mass is
closer to the particle
with the larger mass
Center of Mass, Extended Object


Think of the extended object as a system
containing a large number of particles
The particle distribution is small, so the mass
can be considered a continuous mass
distribution
Center of Mass, Extended Object,
Coordinates

The coordinates of the center of mass of the
object are
xCM
1
1

x dm yCM 

M
M
1
zCM 
z dm

M
 y dm
Center of Mass, Extended Object,
Position

The position of the center of mass can also
be found by:
rCM

1

r dm

M
The center of mass of any symmetrical object
lies on an axis of symmetry and on any plane
of symmetry
Density r
mass
r
unit - volume
m  rV
dm  rdv
Usually, r is a constant but NOT ALWAYS!
Center of Mass, Example


An extended object can
be considered a
distribution of small
mass elements,
Dm=rdxdydz
The center of mass is
located at position rCM
For a flat sheet, we define a mass per unit
area … s
mass
s
unit volum e
dm  sda
Linear Density m
m  mL
dm  mdx
Center of Mass, Rod



Find the center of mass
of a rod of mass M and
length L
The location is on the
x-axis (or
yCM = zCM = 0)
xCM = L / 2
A golf club consists of a shaft connected to a club
head. The golf club can be modeled as a uniform rod
of length L and mass m1 extending radially from the
surface of a sphere of radius R and mass m2. Find the
location of the club’s center of mass, measured from
the center of the club head.
R
L
m2
m1
Motion of a System of Particles
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Assume the total mass, M, of the system
remains constant
We can describe the motion of the system
in terms of the velocity and acceleration of
the center of mass of the system
We can also describe the momentum of
the system and Newton’s Second Law for
the system
Velocity and Momentum of a System
of Particles

The velocity of the center of mass of a system of
particles is
mv
vCM

dr
 CM 
dt
i
i
i
M
The momentum can be expressed as
Mv CM 
m v
i
i


i

p
i
 p tot
i
The total linear momentum of the system equals the
total mass multiplied by the velocity of the center of
mass
Acceleration of the Center of Mass

The acceleration of the center of mass can
be found by differentiating the velocity with
respect to time
aCM
dv CM
1


dt
M
m a
i i
i
Forces In a System of Particles

The acceleration can be related to a force
MaCM   Fi
i

If we sum over all the internal forces, they
cancel in pairs and the net force on the
system is caused only by the external forces
Newton’s Second Law for a System of
Particles


Since the only forces are external, the net external
force equals the total mass of the system multiplied
by the acceleration of the center of mass:
Fext = M aCM
The center of mass of a system of particles of
combined mass M moves like an equivalent particle
of mass M would move under the influence of the
net external force on the system
Momentum of a System of Particles


The total linear momentum of a system of
particles is conserved if no net external force
is acting on the system
MvCM = ptot = constant when Fext = 0
Motion of the Center of Mass,
Example



A projectile is fired into the
air and suddenly explodes
With no explosion, the
projectile would follow the
dotted line
After the explosion, the
center of mass of the
fragments still follows the
dotted line, the same
parabolic path the projectile
would have followed with
no explosion
Rocket Propulsion

The operation of a rocket depends upon the
law of conservation of linear momentum as
applied to a system of particles, where the
system is the rocket plus its ejected fuel
Rocket Propulsion, 2
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
The initial mass of the
rocket plus all its fuel is
M + Dm at time ti and
velocity v
The initial momentum
of the system is pi = (M
+ Dm) v
Rocket Propulsion, 3


At some time t + Dt, the
rocket’s mass has been
reduced to M and an
amount of fuel, Dm has
been ejected
The rocket’s speed has
increased by Dv
Rocket Propulsion, 4



Because the gases are given some momentum
when they are ejected out of the engine, the rocket
receives a compensating momentum in the opposite
direction
Therefore, the rocket is accelerated as a result of
the “push” from the exhaust gases
In free space, the center of mass of the system
(rocket plus expelled gases) moves uniformly,
independent of the propulsion process
Rocket Propulsion, 5


The basic equation for rocket propulsion is
 Mi 
v

v

v
ln
 rocket

f
i
e in
increase
M
 f 
The
speed is proportional to the
speed of the escape gases (ve)
So, the exhaust speed should be very high
The increase in rocket speed is also proportional to the natural
log of the ratio Mi/Mf
 So, the ratio should be as high as possible, meaning the mass of
the rocket should be as small as possible and it should carry as
much fuel as possible


Thrust

The thrust on the rocket is the force exerted on it by
the ejected exhaust gases
dv
dM
Thrust =
M
v
dt


e
dt
The thrust increases as the exhaust speed
increases
The thrust increases as the rate of change of mass
increases

The rate of change of the mass is called the burn rate