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Conservation Theorems: Momentum
Luis Anchordoqui
Momentum of a particle
Originally introduced by Newton as the quantity of motion
the momentum is defined as p = mv
Using Newton´s second law we can relate
the momentum of a particle to the force acting on the particle
Substituing the force F
net
by ma
The net force acting on a particle equals
the time rate of change of the particle´s momentum
In his famous treatise Principia (1687)
Newton presents the second law of motion in this form
Luis Anchordoqui
Conservation of Momentum
The total momentum of a system of particles reads
According to Newton´s second law
If
constant
If the sum of the external forces on a system remains zero
the total momentum of the system is conserved
Luis Anchordoqui
During repair of the
Hubble Space Telescope,
an astronaut replaces a damaged
solar panel during a spacewalk
Pushing the detached panel away
into space, she is propoelled in
the opposite direction.
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The astronaut's mass is 60 kg and the panel's mass is 80 kg.
Both the astronaut and the panel initially are at rest relative to the telescope.
The astronaut then gives a panel a shove.
After the shove it is moving at 0.3 m/s relative to the telescope.
What is her subsequent velocity relative to the telescope?
( During this operation the astronaut is tethered to the ship,
for our calculations assume that the tether remains slack.)
Before
Panel
After
Astronaut
vAf = 0.4 m/s ^i
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A runaway railroad car
A runaway 14,000 kg railroad car is rolling horizontally
at 4 m/s toward a switchyard.
As it passes by a grain elevator
2000 kg of grain suddenly drops into the car
How long does it take the car to cover the 500 m distance from
the elevator to the switchyard?
Assume that the grains falls straight down and that slowing due to
rolling friction or air drag is negligible.
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A runaway railroad car (cont’d)
Δt = 143s
Before
After
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A Skateboard workout
A 40 kg skateboarder on a 3 kg board is trainning with two 5 kg weights.
Beginning from rest, she throws the weights horizontally, one at a time,
from her board.
The speed of each weight is 7 m/s relative to her after it is thrown.
Assume the board rolls without friction.
(a) How fast is she moving in the opposite direction after throwing the first weight?
(b) After throwing the second weight?
Vsg1x = 0.66 m/s
V sg2x= 1.39 m/s
Luis Anchordoqui
Radioactive decay
A thorium-227 nucleus (mass 227 u) at rest decays into radium-223
nucleus (mass 223 u) by emitting an alpha particle (mass 4 u).
The kinetic energy of the alpha particle is measured to be 6 MeV.
What is the kinetic energy of the recolling radium nucleus?
Thorium-227
Radium-223
K ra= 0.107 MeV
Luis Anchordoqui
Collisions and Impulse
When two objects collide
they usually exert very large forces on each other for very brief time
The impulse of a force exerted during a
time Δt = t - t is a vector defined as
f
i
Impulse is a measure of both
strength and duration of the collision force
Impulse momentum theorem for a particle
Impulse momentum theorem for a system
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Average force
Since the time of the collision is very short
we need not worry
about the exact time dependence of the force
and can use the average force
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You strike a golf ball with a driving iron. Hitting
a golf ball
What are reasonable estimates for the magnitude of the
(a) impulse (b) collision time (c) average force?
A typical golf ball has a mass m = 45 g and a radius r = 2 cm.
for a typical drive, the range is roughly 190 m.
Assume the ball leaves the ground at an angle Θ = 13 degrees above the horizontal
0
R = 190 m
to hole
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Hitting a golf ball (cont’d)
I x = 2.9 N s
-4
Δt = 6.1 x 10 s
Fav,x= 4.8 kN
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Perfectly Elastic and Perfectly Inelastic Collisions
In elastic collisions the kinetic energy of the system is conserved
From conservation of momentum
Taking the ratio of these two equations
Rearranging we obtain the relative velocities in an elastic collision
In perfectly inelastic collisions the objects have the same velocity after the collision
(often because they stuck together)
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A catch in space
An astronaut of mass 60 kg is on a space walk to repair a
communications satellite when he realizes he needs to consult the repair manual.
You happen to have it with you, so you throw it to him with a speed 4 m/s
relative to the spacecraft.
He is at rest relative to the spacecraft before catching the 3 kg book.
Find
(a) his velocity after catching the book
(b) the initial and final kinetic energies of the book-astronaut system
(c) the impulse exerted by the book on the astronaut.
Before
After
I by B on A = 11Ns
v f = 0.19 m/s K sys, i = 24 J K sys, f = 1.1 J
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A bullet travelling 850 m/s collides inelastically with an apple,
which disintegrates completely moments later.
Exposure time is less than a millionth of a second.
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A meteor whose mass was about 10 8 kg struck the Earth
(M
= 6 x 10
24
kg) with a speed of about 15 km/s
and came to rest in Earth
(a) What was the Earth's recoil speed?
(b) What fraction of the meteor's kinetic energy was transformed to
kinetic energy of the Earth?
(c) By how much did the Earth kinetic energy change as a result of this
collision?
-13
v = 2.5 x 10
K final Earth
K initial meteor
m/s
= 1.7 x 10
ΔK
-17
= 0.19 J
Earth
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Elastic collision of a neutron and a nucleus
A neutron of mass mn and speed v
undergoes a head-on elastic
n,i
collision with a carbon nucleus of mass m initially atrest.
C
(a) What are the final velocities of both particles?
(b)What fraction f of its initial kinetic energy does the neutron lose?
Neutron
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Conservation of Energy and Momentum in Collisions
Momentum is conserved in all collisions
Collisions in which kinetic energy
is conserved as well are called elastic collisions
and those in which it is not are called inelastic
Approach
Collision
If elastic
If inelastic
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Elastic Collisions in One Dimension
Here we have two objects colliding elastically
We know the masses and the initial speeds
Since both momentum and kinetic energy are conserved
we can write two equations
This allows us to solve for the two unknown final speeds
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Inelastic Collisions
With inelastic collisions, some of the initial kinetic energy
is lost to thermal or potential energy
It may also be gained during explosions
as there is the addition of chemical or nuclear energy
A completely inelastic collision
is one where the objects stick together afterwards
so there is only one final velocity
Luis Anchordoqui
Ballistic Pendulum
In a feat of public marksmanship, you fire a bullet into a
hanging wood block, which is a device known as a ballistic pendulum.
The block, with bullet embedded, swings upward.
Noting the height reached at the top of the swing,
you immediately inform the crowd of the bullet speed.
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Collision with an empty box
You repeat your feat, this time with an empty box as target.
The bullet strikes the box and passes through it completely.
A laser ranging device indicates that the bullet emerged with half its initial velocity.
Hearing this you correctly report how high the target must have swung.
How high did it swing?
Luis Anchordoqui
Coefficient of Restitution
A measure of inelasticity in a head-on collision of two objects is the
coefficient of restitution defined as
where
is the relative velocity of the two objects after the
collision and
is their relative velocity before it.
(a) Show that e = 1 for a perfectly elastic collision
e = 0 for a complete inelastic collision.
for a perfectly elastic collision
for a complete inelastic collision
e = 1
e = 0
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Coefficient of Restitution (cont´d)
(b) A simple method for measuring the coefficient of restitution for an
object colliding with a very hard surface like steel is to drop the
object on to a heavy steel plate, as shown in the figure.
Determine a formula fore in terms of the original height h and the
maximum height h'reached after one collision.
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Collisions in Two or Three Dimensions
Conservation of energy and momentum can also be used to analyze
collisions in two or three dimensions
but unless the situation is very simple
the math quickly becomes unwieldy
If a moving object collides with an object initially at rest
Knowing the masses and initial velocities is not enough
we need to know the angles as well in order to find the final velocities
Luis Anchordoqui
A novice pool player is faced with the corner pocket shot shown in
the figure. Relative dimensions are also given.
Should the player be worried about this being a ``scratch shot'' in
which the cue ball will also fall into a pocket?
4
Give details.
√3
1
Cue ball
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m
V’A
m
V
m
m
V’ B
In an elastic collision between two objects of equal mass with the target object
initially at rest the angle between the final velocities of the objects is 90º
Momentum conservation
V’A
Kinetic energy conservation
Θ
V’
B
Applying the law of cosines
V
Equating the two expressions for v² leads to
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1 m
θ1
3 m
√3
m
θ2
Assume that the target ball is hit correctly so that it goes in the pocket
From the geometry of the left triangle
From the geometry of the right triangle
Because the balls will separate at 90º
if the target ball goes in the
pocket this does appear to be a good possibility of a scratch shot
Luis Anchordoqui
A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees
to the horizon with a speed of 25 m/s.
When it reaches the maximum height, it is hit from below by a
15-g pellet traveling vertically upward at a speed of 200 m/s.
The pellet is embedded in the skeet.
(a) How much higher did the skeet go up?
(b) How much extra distance, Δ x, does the skeet travel because of
the collision?
ΔX
(y – y 0 ) extra= 6.5 m
Δx = 31 m
Luis Anchordoqui
Eruption of a large volcano on Jupiter´s moon
When the volcano erupts the speed of the effluence
exceeds the escape speed of Io and so a stream of
particles is projected into space
The material in the stream can collide
with and sticks to the surface of an
asteroid passing through the stream
We now consider the effect of the impact of
this material on the motion of the asteroid
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Continuously varying mass
Consider a continuous stream of matter moving at velocity u
which impacts an object of mass M that is moving with velocity v
This impacting particles stick to the object increasing
its mass by ΔM during time Δt
In addition during time Δt the velocity v changes by Δv
Applying the impulse momentum theorem to this system
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Continuously varying mass (cont’d)
Rearrenging terms
Dividing by Δt
Taking the limit Δt 0 that also means ΔM
0 and Δv
0
Rearranging terms we obtain Newton´s second law for a system
that has a continuosly changing mass
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Rocket Propulsion
The mass of the rocket changes continously
as it burns fuel and expels exhaust gas
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Rocket Propulsion (cont’d)
Momentum conservation works for a rocket as long as we consider the rocket
and its fuel to be one system, and account for the mass loss of the rocket.
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Rocket Propulsion (cont’d)
Consider a rocket moving staight up with velocity v relative to Earth and
assume the fuel is burned at constant rate α
The rocket’s mass at time t is
m = m 0 -α t
Initial mass of the rocket
The exhaust gases leave the rocket engine with
velocity uex relative to the rocket
We choose the rocket including unspent fuel as the system
Neglecting air drag the only external force on the system is that of gravity
Fnext,ext = mg
The rocket equation is then
dv
mg-αu ex = m dt
The quantity – αu is the thrust force exerted on the rocket by the exhaust gases
ex
Fth = - αu ex = -
dm
dt
u ex
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Rocket Propulsion (cont’d)
Choosing upward as the positive y direction
the direction of uex is downward so u ex,y = -u ex
Substituting gives
The acceleration is then
Integration leads to
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Saturn V: America’s Moon Rocket
Saturn V -developed at NASA’s Marshall Space Flight Centerwas the largest in a family of liquid-propellant rockets that solved the problem
of getting to the Moon
32 Saturns were launched
not one failed!!!
The Saturn V was flight-tested twice without crew, the first manned Saturn V
sent the Apollo 8 astronauts into orbit around the Moon in December 1968
After two more missions to test the lunar landing vehicle in July 16th 1969 a
Saturn V launched the crew of Apollo 11 to the first manned landing on the Moon
Apollo 11 rocket blast off
Estimate the final speed of the first stage at burnout and its vertical height
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Saturn V: America’s Moon Rocket (cont’d)
6
m 0 ≈ 2.8 x 10 kg
mean thrust ≈ 37 x 106 N
6
mass of first stage fuel ≈ 2.1 x 10 kg
u ≈ 2600 m/s
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Saturn V: America’s Moon Rocket (cont’d)
Use definition of thrust to determine the fuel burn rate
The final rocket mass
m= 2.8 x 10 6 kg – 2.1 x 10 6 kg = 0.7 x 10 6 kg
The speed of the space ship at burnout
Time to burnout
tb about two and a half minutes
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Saturn V: America’s Moon Rocket (cont’d)
Since dm/dt = -α
dt = -dm/α
Using
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Saturn V: America’s Moon Rocket (cont’d)
Evaluate C from initial conditions @ t = 0
y = 0 and m = m0
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A two stage rocket
A rocket is shot into the air as shown in the figure. At the moment
the rocket reaches it highest point, a horizontal distance d from its
starting point, a prearranged explosion separates it into two parts of
equal mass. Part I is stopped by the explosion in midair, and it falls
vertically to Earth. Where does part II land? (Assume g= constant)
d
d
After the rocket is fired the path of the CM of the system continues
to follow the parabolic trajectory of a projectile acted on by only a
constant gravitational force.
The CM will thus land at 2d from the starting point.
Because the masses I and II are equal, the CM must be midway
between them at any time
part II lands a distance 3d from the
starting point.
Luis Anchordoqui