chapter 9 linear momentum and collisions 1. review of newton’s third law if two objects interact,...
TRANSCRIPT
Chapter 9
Linear Momentum and Collisions
1
Review of Newton’s Third Law
If two objects interact, the force F12 exerted by object 1 on object is equal in magnitude and opposite in direction to the force F21 exerted by object 2 on object 1:
F12 = -F21
2
m1 m2
a1
a2
Linear Momentum
doesn’t change w.r.t. time t.
Call each individual mivi linear momentum.3
m1 m2
a1
a2
F12 = - F21
F12 = - F21 m1a1+m2a2=0
02
2
1
1 dt
vdm
dt
vdm
0)( 2211 vmvmdt
d
2211 vmvm
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 is defined to be the product of the mass and velocity:
The terms momentum and linear momentum will be used interchangeably in the text
v
mp v
4
Linear Momentum, cont
Linear momentum is a vector quantity Its direction is the same as the direction of the
velocity 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 vx py = m vy pz = m vz
5
Newton and Momentum
Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it
with constant mass
d md dm m
dt dt dt
vv pF a
6
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 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
7
Conservation of Linear Momentum
Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant 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
8
Conservation of Momentum, 2
Conservation of momentum can be expressed mathematically in various ways
In component form, the total momenta in each direction are independently conserved pix = pfx piy = pfy piz = pfz
Conservation of momentum can be applied to systems with any number of particles
This law is the mathematical representation of the momentum version of the isolated system model
total 1 2p = p + p = constant
1i 2i 1f 2fp + p = p + p
9
Conservation of Momentum, Archer Example
A 60-kg archer stands at rest on a frictionless surface (ice) and fire a 0.5-kg arrow horizontally at 50m/s.
With what velocity does the archer move across the ice after firing the arrow?
10
Conservation of Momentum, Archer Example
Approaches: Newton’s Second Law – no
information about F or a Energy approach – no
information about work or energy
Momentum – yes
11
Archer Example, 2
Conceptualize The arrow is fired one way and the archer recoils in the
opposite direction Categorize
Momentum 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 Analyze
Total momentum before releasing the arrow is 0
12
Archer Example, 3 Analyze, cont.
The total momentum after releasing the arrow is
Solve v1.
Finalize The final velocity of the archer is
negative Indicates he moves in a direction
opposite the arrow Archer has much higher mass than
arrow, so velocity is much lower
1 2 0f f p p
13
02211 vmvm
Impulse and Momentum
From Newton’s Second Law, Solving for gives Integrating to find the change in momentum
over some time interval
The integral is called the impulse, , of the force acting on an object over t
f
i
t
f i tdt p p p F I
d
dt
pF
dp d dtp F
I
14
Impulse-Momentum Theorem
This equation expresses the impulse-momentum theorem: The impulse of the force acting on a particle equals the change in the momentum of the particle This is equivalent to Newton’s Second Law
p I
15
More About Impulse Impulse is a vector quantity The magnitude of the
impulse is equal to the area under the force-time curve The force may vary with
time Dimensions of impulse are
M L / T Impulse is not a property of
the particle, but a measure of the change in momentum of the particle
16
Impulse, Final
The impulse can also be found by using the time averaged force
This would give the
same impulse as the time-varying force does
t I F
17
Impulse Approximation
In many cases, one force acting on a particle acts for a short time, but is much greater than any other force present
When using the Impulse Approximation, we will assume this is true Especially useful in analyzing collisions
The force will be called the impulsive force The particle is assumed to move very little during
the collision represent the momenta immediately before
and after the collisioni fandp p
18
Impulse-Momentum: Crash Test Example
A car of mass 1500kg collides with a wall.
Vi=-15 m/s and Vf=2.6 m/s. The collision lasts 0.15 seconds.
Find the impulse caused by the collision and the average force exerted on the car.
19
Impulse-Momentum: Crash Test Example
Categorize Assume force exerted by
wall is large compared with other forces
Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum
Can apply impulse approximation
20
Crash Test Example, 2
Analyze The momenta before and after the collision between the car
and the wall can be determined Find
Initial momentum Final momentum Impulse Average force
Finalize Check signs on velocities to be sure they are reasonable
21
f
i
t
f i tdt p p p F I
ii vmP ff vmP
if vmvmPI
t
PF average
Collisions – Characteristics
We use the term collision to represent an event during which two particles come close to each other and interact by means of forces May involve physical contact, but must be generalized to
include cases with interaction without physical contact The time interval during which the velocity changes
from its initial to final values is assumed to be short The interaction forces are assumed to be much
greater than any external forces present This means the impulse approximation can be used
22
Collisions – Example 1
Collisions may be the result of direct contact
The impulsive forces may vary in time in complicated ways This force is internal to
the system Observe the variations in
the active figure Momentum is
conserved
23
Collisions – Example 2
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
24
Types of Collisions
In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic
collisions actually occur Generally some energy is lost to deformation, sound, etc.
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
25
Collisions, cont
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
26
Perfectly Inelastic Collisions
Since the objects stick together, they share the same velocity after the collision
Compare Ki with Kf
1 1 2 2 1 2i i fm m m m v v v
27
Elastic Collisions
Both momentum and kinetic energy are conserved
1 1 2 2
1 1 2 2
2 21 1 2 2
2 21 1 2 2
1 1
2 21 1
2 2
i i
f f
i i
f f
m m
m m
m m
m m
v v
v v
v v
v v
28
Elastic Collisions, cont Typically, there are two unknowns to solve for and so you
need two equations The kinetic energy equation can be difficult to use With some algebraic manipulation, a different equation can be
used
v1i – v2i = - (v1f - v2f) why?
This 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
29
Elastic Collisions, cont
Momentum:
kinetic energy:
30
ffii vmvmvmvm 22112211
222
211
222
211 2
1
2
1
2
1
2
1ffii vmvmvmvm
)()( 22
222
21
211 fifi vvmvvm
)()( 222111 fifi vvmvvm
fifi vvvv 2211 )( 2121 ffii vvvv ….(B)
….(A)
Elastic Collisions, cont
By (A) and (B),
31
iif vmm
mv
mm
mmv 2
21
21
21
211
2
iif vmm
mmv
mm
mv 2
21
121
21
12
2
Elastic Collisions, final
Example of some special cases 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
32
Problem-Solving Strategy: One-Dimensional Collisions
Conceptualize Image the collision occurring in your mind Draw simple diagrams of the particles before and
after the collision Include appropriate velocity vectors
Categorize Is the system of particles isolated? Is the collision elastic, inelastic or perfectly
inelastic?
33
Problem-Solving Strategy: One-Dimensional Collisions
Analyze Set up the mathematical representation of the
problem Solve for the unknown(s)
Finalize Check to see if the answers are consistent with
the mental and pictorial representations Check to be sure your results are realistic
34
Example: Stress Reliever
Conceptualize Imagine one ball coming in
from the left and two balls exiting from the right
Is this possible? Categorize
Due to shortness of time, the impulse approximation can be used
Isolated system Elastic collisions
35
Example: Stress Reliever, cont
Analyze Check to see if momentum is conserved
It is Check to see if kinetic energy is conserved
It is not Therefore, the collision couldn’t be elastic
Finalize Having two balls exit was not possible if only one
ball is released
36
Example: Stress Reliever, final
What collision is possible
Need to conserve both momentum and kinetic energy Only way to do so is with
equal numbers of balls released and exiting
37?
Example: Stress Reliever, final
What if balls 4 and 5 are glued together?
v45 = ?
Pi = Pf mv1i = mv1f + 2mv45
Ki = Kf
38
?
245
21
21 )2(
2
1
2
1
2
1vmmvmv fi
ifi vvvv 11145 3
1
3
2
Collision Example – Ballistic Pendulum
Conceptualize Observe diagram Determine the velocity of the bullet in
terms of h Categorize
Isolated system of projectile and block 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
39
Collision Example – Ballistic Pendulum
Pi = Pf m1v1A = (m1+m2)vB
KB = (1/2)(m1+m2)vB 2
= (m12 v1A
2)/2(m1+m2)
KB = UC = (m1+m2)gh
v1A =(m1+m2)2gh/m1
40
Ballistic Pendulum, cont
A multi-flash photograph of a ballistic pendulum
Analyze Solve resulting system of
equations Finalize
Note different systems involved
Some energy was transferred during the perfectly inelastic collision
41
Two-Dimensional Collisions
The momentum is conserved in all directions Use subscripts for
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 Remember, the simpler equation can only be used for one-
dimensional situations
42
Two-Dimensional Collision, example
Particle 1 is moving at velocity and particle 2 is at rest
In the x-direction, the initial momentum is m1v1i
In the y-direction, the initial momentum is 0
1iv
43
Two-Dimensional Collision, example cont
After the collision, the momentum in the x-direction is m1v1f cos m2v2f cos
After the collision, the momentum in the y-direction is m1v1f sin m2v2f sin
If the collision is elastic, apply the kinetic energy equation
This is an example of a glancing collision
44
Problem-Solving Strategies – Two-Dimensional Collisions
Conceptualize Imagine the collision Predict approximate directions the particles will
move after the collision 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
45
Problem-Solving Strategies – Two-Dimensional Collisions, 2
Categorize Is the system isolated? If so, categorize the collision as elastic, inelastic or
perfectly inelastic Analyze
Write expressions for the x- and y-components of the momentum of each object before and after the collision 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.
46
Problem-Solving Strategies – Two-Dimensional Collisions, 3
Analyze, cont 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.
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
47
Problem-Solving Strategies – Two-Dimensional Collisions, 4
Finalize Check to see if your answers are consistent with
the mental and pictorial representations Check to be sure your results are realistic
48
Two-Dimensional Collision Example
A 1500-kg car traveling east with a speed of 25 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20 m/s
Find the direction and magnitude of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.
49
Two-Dimensional Collision Example
Conceptualize See picture Choose East to be the
positive x-direction and North to be the positive y-direction
Categorize Ignore friction Model the cars as particles The collision is perfectly
inelastic The cars stick together
50
Two-Dimensional Collision Example, cont
Analyze Before the collision, the car has the total momentum in the
x-direction and the van has the total momentum in the y-direction
After the collision, both have x- and y-components Write expressions for initial and final momenta in both
directions Evaluate any expressions with no unknowns
Solve for unknowns Finalize
Check to be sure the results are reasonable
51
Two-Dimensional Collision Example
Pxi = (1500 kg)(25 m/s)
Pxf = (4000 kg)(vfcos m/s)
Pyi = (2500 kg)(20 m/s)
Pxf = (4000 kg)(vfsin m/s)
Solve vf and .
52
The Center of Mass
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
53
Center of Mass, Coordinates
The coordinates of the center of mass are
M is the total mass of the system Use the active figure to
observe effect of different masses and positions
CM
CM
CM
i ii
i ii
i ii
m xx
Mm y
yMm z
zM
54
Center of Mass, Extended Object
Similar analysis can be done for an extended object
Consider the extended object as a system containing a large number of particles
Since particle separation is very small, it can be considered to have a constant mass distribution
For x-axis, xCM (1/M)iximi
xCM = (1/M)xdm
rCM = (1/M) r dm 55
Center of Mass, position
The center of mass in three dimensions can be located by its position vector, For a system of particles,
is the position of the ith particle, defined by
For an extended object, CM
1dm
M r r
ˆ ˆ ˆi i i ix y z r i j k
CMr
1CM i i
i
mM
r r
ir
56
Center of Mass, Symmetric Object
The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry If the object has uniform density
57
Finding Center of Mass, Irregularly Shaped Object
Suspend the object from one point
The suspend from another point
The intersection of the resulting lines is the center of mass
58
Center of Gravity
Each small mass element of an extended object is acted upon by the gravitational force
The net effect of all these forces is equivalent to the effect of a single force acting through a point called the center of gravity If is constant over the mass distribution, the
center of gravity coincides with the center of mass
Mg
g
59
Center of Mass, Rod
Conceptualize 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)
Categorize Analysis problem
Analyze Use equation for xcm
xCM = L / 2 why?
60
Center of Mass, Rod
61
22
2
0
L
M
Lxdx
xdmx
L
CM
L
M
Motion of a System of Particles
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
62
Velocity and Momentum of a System of Particles
The velocity of the center of mass of a system of particles is
The momentum can be expressed as
The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass
CMCM
1i i
i
dm
dt M r
v v
CM toti i ii i
M m v v p p
63
Acceleration of the Center of Mass
The acceleration of the center of mass can be found by differentiating the velocity with respect to time
CMCM
1i i
i
dm
dt M v
a a
64
Forces In a System of Particles
The acceleration can be related to a force
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
CM ii
M a F
65
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:
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
ext CMMF a
66
Impulse and Momentum of a System of Particles
The impulse imparted to the system by external forces is
The total linear momentum of a system of particles is conserved if no net external force is acting on the system
ext CM totdt M d I F v p
0CM tot extM constant when v p F
67
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
Use the active figure to observe a variety of explosions
68
Deformable Systems
To analyze the motion of a deformable system, use Conservation of Energy and the Impulse-Momentum Theorem
If the force is constant, the integral can be easily evaluated
0system
tot ext
E T K U
dt m
I p F v
69
Deformable System (Spring) Example
Conceptualize See figure Push on left block, it moves to
right, spring compresses At any given time t, the
blocks are generally moving with different velocities
The blocks oscillate back and forth with respect to the center of mass
Find the resulting speed vCM of the center of mass of the system
70
Spring Example, cont
Categorize Non isolated system
Work is being done on it by the applied force It is a deformable system The applied force is constant, so the acceleration
of the center of mass is constant Model as a particle under constant acceleration
Analyze Apply impulse-momentum Solve for vcm
71
Spring Example, final
Analyze, cont. Find energies
Finalize Answers do not depend on spring length, spring
constant, or time interval
72
Deformable System (Spring) Example
Impulse-momentum theorem
F t = Pf – Pi = 2mvCM – 0
t = (xCMf-xCMi)/vCM
= (L+x2+x1-L)/2vCM
= (x1+x2)/2vCM
F t = F(x1+x2)/2vCM = 2mvCM
vCM = F(x1+x2)/ 2m73
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
74
Rocket Propulsion, 2
The initial mass of the rocket plus all its fuel is M + m at time ti and speed v
The initial momentum of the system is
i = (M + m) p v
75
Rocket Propulsion, 3
At some time t + t, the rocket’s mass has been reduced to M and an amount of fuel, m has been ejected
The rocket’s speed has increased by v
76
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
77
Rocket Propulsion, 5
(M+m)v = M(v+v) + m(v-ve)
Mv = vem
The increase in the exhaust mass dm corresponds to an equal decrease in the rocket mass, so dm = -dM.
Mdv = vedm = -vedM78
v-ve
f
i
f
i
M
M
e
v
v
dMM
vdv1
Rocket Propulsion, 6
The basic equation for rocket propulsion is
The increase in rocket 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
ln if i e
f
Mv v v
M
79
Thrust
The thrust on the rocket is the force exerted on it by the ejected exhaust gases
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
e
dv dMthrust M v
dt dt
80