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Topic 5
Conservation of Linear Momentum
I.
Introduction.
a. One of the most important principles in physics is the law of conservation of
momentum, which says that the total momentum of a system and its
surroundings does not change. Whenever the momentum of a system
changes, we can account for the change by the appearance or disappearance
of momentum somewhere else.
b.
II.
In this chapter, we introduce the ideas of impulse and linear momentum, and
show how integrating Newton’s second law produces an important theorem
known as the impulse–momentum theorem. We will also determine if the
momentum of a system remains constant, and how to exploit constant
momentum to solve problems involving collisions between objects. In addition,
we examine a new reference frame, known as the center-of-mass reference
frame, and explore situations in which a system has a continuously changing
mass.
Conservation of Linear Momentum.
a. When Newton devised his second law, he considered the product of mass and
velocity as a measure of an object’s “quantity of motion.” Today, we call the
product of a particle’s mass and velocity linear momentum
b.
Linear momentum is a vector quantity, it is the product of a vector (velocity)
and a scalar (mass). Its magnitude is
and it has the same direction as
The units of momentum are units of mass times speed, so the SI units of
momentum are kg•m/s.
c.
.
Momentum may be thought of as a measurement of the effort needed to bring
a particle to rest.
i. Thus, the net force acting on a particle equals the time rate of change
of the particle’s momentum.
ii.
d.
The total momentum
of a system of particles is the vector sum of the
momenta of the individual particles.
i.
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Conservation of Linear Momentum
ii. But according to Newton’s second law for a system of particles,
equals the net external force acting on the system.
iii.
e.
When the sum of the external forces acting on a system of particles remains
zero, the rate of change of the total momentum remains zero and the total
momentum of the system remains constant.
i.
ii. This result is known as the law of conservation of momentum.
iii.
iv. This law is one of the most important in physics. It is more widely
applicable than the law of conservation of mechanical energy because
internal forces exerted by one particle in a system on another are
often not conservative. The nonconservative internal forces can
change the total mechanical energy of the system, though they effect
no change of the system’s total momentum. If the total momentum of
a system remains constant, then the velocity of the center of mass of
the system remains constant. The law of conservation of momentum is
a vector relation, so it is valid component by component.
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Conservation of Linear Momentum
f.
Finding Velocities Using Momentum Conservation (Equation 8-5)
i.
PICTURE Determine that the net external force
(or
) on
the system is negligible for some interval of time. If the net force is
determined not to be negligible, do not proceed.
ii. SOLVE
1. Draw a sketch showing the system both before and after the
time interval. Include coordinate axes and label the initial and
final velocity vectors.
2. Equate the initial momentum to the final momentum. That is,
write out the equation
.
3. Substitute the given information into the step-2 equation(s)
and solve for the quantity of interest.
iii. CHECK Make sure you include any minus signs that accompany
velocity components because they influence your final answer.
g.
Problems.
i. Example 8 – 1: A Space Repair.
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 propelled in the opposite direction. 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 the panel a shove. After the shove it is moving at
0.30 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.)
ii. Practice Problem 8 – 1.
Find the final kinetic energy of the astronaut–panel system.
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Conservation of Linear Momentum
iii. Example 8 – 2: A Runaway Railroad Car.
A runaway 14,000-kg railroad car is
rolling horizontally at 4.00 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 grain falls straight down
and that slowing due to rolling
friction or air drag is negligible.
iv. Practice Problem 8 – 2.
Suppose that there is a small vertical chute in the bottom of the car so
that the grain leaks out at 10 kg/s. Now how long does it take the car to
cover the 500 m?
v.
Example 8 – 3: A Skateboard Workout.
A 40.0-kg skateboarder on a 3.00kg board is training with two 5.00kg weights. Beginning from rest,
she throws the weights
horizontally, one at a time, from
her board. The speed of each
weight is 7.00 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?
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Conservation of Linear Momentum
vi. Practice Problem 8 – 3.
How fast is the skateboarder moving if, starting from rest, she throws
both weights simultaneously? The weights move at speed 7.00 m/s
relative to her after they are thrown. Does she gain more speed by
throwing them simultaneously or sequentially?
vii. Example 8 – 4: Radioactive Decay.
A thorium-227 nucleus (mass 227 u) at rest decays into a radium-223
nucleus (mass 223 u) by emitting an alpha particle (mass 4.00 u). The
kinetic energy of the α particle is measured to be 6.00 MeV. What is the
kinetic energy of the recoiling radium nucleus?
III.
Kinetic Energy of a System.
a. If the net external force on a system of particles remains zero, then the total
momentum of a system must remain constant; however, the total mechanical
energy of the system can change.
b.
c.
Kinetic Energy of System of Particles.
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Conservation of Linear Momentum
d.
Theorem Proof.
e.
If the external force is zero, the center-of-mass velocity remains
constant and the kinetic energy associated with bulk motion of the
system (½ Mv2cm) does not change. Only the relative kinetic energy
Krel can change in an isolated system.
f. Practice Problem 8 – 4.
Air-track glider A is moving at 1.0 m/s in the +x direction along a frictionless
horizontal air track. An identical glider, glider B, is parked on the track ahead of
glider A. The mass of each glider is 1.0 kg, and the system consists of the two
gliders. (a) What is the velocity of the center of mass, and what is the velocity of
each glider, relative to the center of mass? (b) What is the kinetic energy of each
glider relative to the center of mass? (c) What is the total kinetic energy relative to
the center of mass? (d) The gliders collide and stick together. What then is the
total kinetic energy relative to the center of mass?
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Conservation of Linear Momentum
IV.
Collisions.
a. During the brief time of collision, any external forces on the two objects are
usually much, much weaker than the forces of interaction between the two
objects.
i. Thus, the colliding objects can usually be treated as an isolated system
for the duration of the collision.
ii. During the collision the only significant forces are the internal
interaction forces, which are equal and opposite.
iii. As a result, momentum is conserved. That is, the total momentum of
the system the instant before the collision is equal to the total
momentum the instant following the collision.
iv. In addition, the collision time is usually so short that the displacements
of the colliding objects during the collision can be neglected.
b.
When the total kinetic energy of the two-object system is the same after the
collision as before, the collision is called an elastic collision. Otherwise, it is
called an inelastic collision. An extreme case is the perfectly inelastic
collision, during which all of the kinetic energy relative to the center of mass
is converted to thermal or internal energy of the system, and the two objects
share a common velocity (often because they stick together) at the end of the
collision.
c.
Impulse and Average Force.
i. When two objects collide, they usually exert very large forces on each
other for a very brief time.
ii. Such forces are sometimes called impulsive forces. Such forces are
sometimes called impulsive forces.
iii. The impulse
vector.
of a force
during time interval Δt = tf − ti is a
iv. The impulse is a measure of both the strength and the duration of the
collision force. The x component of the impulse of the force is the area
under its Fx-versus-t curve, and the S.I. units of impulse are newton
seconds (N • s).
v.
The net force
acting on a particle is related to the rate of change
of momentum of the particle by Newton’s second law:
.
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Conservation of Linear Momentum
vi. Impulse-Momentum Theorem for a Particle:
vii.
Also, the net impulse on a system due to the external forces acting on
the system equals the change in the total momentum of the system.
viii. Impulse-Momentum Theorem for a Particle:
ix. Average Force/Impulse and Average Force:
d. Estimating the Average Force.
i.
PICTURE To estimate the average force
we first estimate the
impulse of the force
. The impulse of the force equals the net
impulse (assuming any other forces are negligible). The net impulse is
equal to the change in momentum, and the change in momentum
equals the product of the mass m and the change of velocity
. An estimate of the change in velocity can be gotten from estimates of
both the collision time Δt and the displacement
.
ii. SOLVE
1.
Calculate (or estimate) the impulse
and the time Δt. This
estimate assumes that during the collision, the collision force
on the object is very large compared to all other forces on the
object. This procedure works only if the displacement during
the collision can be determined.
2. Draw a sketch showing the before and after position of the
object. Add coordinate axes and label the pre– and
postcollision velocities
displacement
and
. In addition, label the
during the collision.
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Conservation of Linear Momentum
3. Calculate the change in momentum of the object during a
collision. The impulse on the object equals its change in
momentum
.
4. Use kinematics to estimate the collision time. This means
using both
and
to obtain
, and then solving for Δt.
5. Use
to calculate the average
force (Equation 8-13).
iii. CHECK Average force is a vector. Your result for average force should
have the same direction as the change in velocity vector.
e.
Problems.
i. Example 8 – 5: Karate Collisions.
With an expert karate blow, you shatter a concrete block. Consider your
hand to have a mass 0.70 kg, to be moving 5.0 m/s as it strikes the
block, and to stop 6.0 mm beyond the point of contact. (a) What impulse
does the block exert on your hand? (b) What is the approximate collision
time and the average force the block exerts on your hand?
ii. Example 8 – 6: A Crumpled Car.
A car equipped with an 80-kg crash-test dummy drives into a massive
concrete wall at 25 m/s. Estimate the displacement of the dummy
during the crash.
iii. Example 8 – 7: A Crash Test.
For the car crash described in Example 8-6, estimate the average force
that the seat belt exerts on the dummy during the crash.
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Conservation of Linear Momentum
iv. Example 8 – 8: Hitting a Golf Ball.
You strike a golf ball with a driving iron. What are reasonable estimates
for the magnitudes of the (a) impulse
, (b) collision time Δt, and (c)
average force
? A typical golf ball has a mass m = 45 g and a radius r
= 2.0 cm. For a typical drive, the range R is roughly 190m (about 210
yd). Assume the ball leaves the ground at an angle θ0 = 13° above the
horizontal.
f.
Collisions in which the colliding objects move along the same straight line, say
along the x axis, both before, during, and after the collision are called onedimensional collisions.
i. Perfectly inelastic collisions In perfectly inelastic collisions, the
objects have the same velocity after the collision, often because they
are stuck together.
1. It is sometimes useful to express the kinetic energy, K, of a
particle in terms of its momentum, p.
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Conservation of Linear Momentum
2. We can apply this to a perfectly inelastic collision where one
object is initially at rest.
3. After colliding, the objects move together as a single mass m1
+ m2 with velocity
. Momentum is conserved, so the final
momentum equals Psys.
4. Example 8 – 9: 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 speed 4.0 m/s relative to the spacecraft. He is at rest
relative to the spacecraft before catching the 3.0-kg book. Find (a)
his velocity just after he catches the book, (b) the initial and final
kinetic energies of the book–astronaut system, and (c) the impulse
exerted by the book on the astronaut.
5. Example 8 – 10: A 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’s speed. How fast was the bullet
traveling?
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Conservation of Linear Momentum
6. Example 8 – 11: A Collision with an Empty Box.
You repeat your feat of Example 8-10, this
time with an empty box as the 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?
7. Example 8 – 12: Collisions with Putty.
Mary has two small balls of equal mass, a
ball of plumber’s putty and a one of Silly
Putty. She throws the ball of plumber’s
putty at a block suspended by strings. The
ball strikes the block with a “thonk” and
falls to the floor. The block subsequently
swings to a maximum height h. If she had
thrown the ball of Silly Putty (instead of the
plumber’s putty) at the same speed, would
the block subsequently have risen to a
height greater than h? Silly Putty, unlike
plumber’s putty, is elastic and would
bounce back from the block.
ii. Elastic collisions In elastic collisions, the kinetic energy of the
system is the same before and after the collision. Elastic collisions are
an ideal that is sometimes approached but never realized in the
macroscopic world.
1. Only for elastic collisions is the kinetic energy the same after
the collision as before the collision.
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Conservation of Linear Momentum
2. Example 8 – 13: Elastic Collision of Two Blocks.
A 4.0-kg block moving to the right at 6.0 m/s undergoes an elastic
head-on collision with a 2.0-kg block moving to the right at 3.0
m/s. Find their final velocities.
3. Example 8 – 14: Elastic Collision of a Neutron and a Nucleus.
A neutron of mass mn and speed
undergoes a head-on elastic
collision with a carbon nucleus of mass mc initially at rest. (a) What
are the final velocities of both particles? (b) What fraction ƒ of its
initial kinetic energy does the neutron lose?
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Conservation of Linear Momentum
4. Practice Problem 8 – 5.
Consider an elastic head-on collision between a moving object
(object 1) and a second moving object of equal mass (object 2).
Use Equation 8-19 and Equation 8-23 to show that the two
objects exchange velocities. That is, show that the final velocity of
object 2 equals the initial velocity of object 1, and vice versa.
5. The final velocity of the incoming particle
and that of the
originally stationary particle
are related to the initial
velocity of the incoming particle.
6. The coefficient of restitution Most collisions lie somewhere
between the extreme cases of elastic, in which the relative
velocities are reversed, and perfectly inelastic, in which there
is no relative velocity after the collision. The coefficient of
restitution e is a measure of the elasticity of a collision. It is
defined as the ratio of the speed of recession to the speed of
approach.
7. Practice Problem 8 – 6.
From the picture of the golf club
hitting the golf ball, estimate the
coefficient of restitution of the golf
ball–golf club interaction.
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Conservation of Linear Momentum
g.
Collisions in Two and Three Dimensions.
i. For one-dimensional collisions, the directions of the initial and final
velocity vectors can be specified by a + or a −. For two– or threedimensional collisions, this is not the case. During such a collision,
momentum is conserved in each of the x, y, and z directions.
1. Inelastic collisions For collisions in two or three dimensions,
the total initial momentum is the sum of the initial momentum
vectors of each object involved in the collision.
2. Example 8 – 15: A Car-Truck Collision.
You are at the wheel of a 1200-kg car traveling east through an
intersection when a 3000-kg truck traveling north through the
intersection crashes into your car. Your car and the truck remain
stuck together after impact. The
driver of the truck claims you were
at fault because you were speeding.
You look for evidence to disprove
this claim. First, there are no skid
marks, indicating that neither you
nor the truck driver saw the
accident coming and braked hard;
second, the posted speed limit for
the road on which you were driving
is 80 km/h; third, the speedometer
of the truck was smashed on
impact, leaving the needle stuck at
50 km/h; and fourth, the wreck
initially skidded from the impact zone at an angle of 59° north of
east. Does this evidence support or undermine the claim that you
were speeding?
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Conservation of Linear Momentum
3. Example 8 – 16: A Glancing Collision.
An object with mass m1 and with an
initial speed of 20 m/s undergoes
an off-center collision with a second
object of mass m2. The second
object is initially at rest. After the
collision the first object is moving at
15 m/s at an angle of 25° with the
direction of the initial velocity of the
first object. In what direction is the
second object moving?
ii. Elastic collisions Elastic collisions in two and three dimensions are
more complicated than those we have already covered.
1. An off-center collision is often referred to as a glancing
collision (as opposed to a head-on collision).
2. The distance b between the centers measured perpendicular to
the direction of
is called the impact parameter.
3. After the collision, object 1 moves off with velocity
,
making an angle θ1 with the direction of its initial velocity, and
object 2 moves with velocity
, making an angle θ2 with
. The final velocities depend on the impact parameter and on
the type of force exerted by one object on the other.
4. Linear momentum is conserved.
5. We can apply the law of conservation of momentum in
component form to give us two of the needed relations among
these quantities.
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Conservation of Linear Momentum
6. Because the collision is elastic, we can use the conservation of
kinetic energy to find a third relation.
7. We need an additional equation to be able to solve for the
unknowns. The fourth relation depends on the impact
parameter b and on the type of interacting force exerted by
the colliding objects on each other. In practice, the fourth
relation is often found experimentally, by measuring the angle
of deflection or the angle of recoil. Such a measurement can
then give us information about the type of interacting force
between the bodies.
8. Now let us consider the interesting special case of the glancing
elastic collision of two objects of equal mass when one is
initially at rest.
9. Practice Problem 8 – 7.
In a friendly game of pool, the cue ball moving at speed
glances off the initially stationary 2 ball. The collision is elastic, the
2 ball is at rest prior to the collision, and the cue ball is deflected
30° from its pre-collision path. What is the speed of the 2 ball
following the collision? (The 2 ball and the cue ball have the same
mass.)
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Conservation of Linear Momentum
V.
Collisions in the Center of Mass Reference Frame.
a. If the net external force on a system remains zero, the velocity of the center
of mass remains constant in any inertial reference frame.
i. It is often convenient to do calculations in an alternate reference frame
that moves with the center of mass.
ii. Relative to the original reference frame, called the laboratory reference
frame, this reference frame moves with a constant velocity
relative to the laboratory frame.
iii. A reference frame that moves at the same velocity as the center of
mass is called the center-of-mass reference frame.
iv. If a particle has velocity relative to the laboratory reference frame,
then its velocity relative to the center-of-mass reference frame is
.
v.
Because the total momentum of a system equals the total mass times
the velocity of the center of mass, the total momentum is also zero in
the center-of-mass frame. Thus, the center-of-mass reference frame is
also a zero-momentum reference frame.
b.
The mathematics of collisions is greatly simplified when considered from the
center-of-mass reference frame.
c.
After a perfectly inelastic collision, the objects remain at rest. However, for an
elastic head-on collision the direction of each velocity vector is reversed
without changing its magnitude.
d.
Consider a simple two-particle system in a reference frame in which one
particle is moving with a velocity and a second particle is moving with a
different velocity.
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Conservation of Linear Momentum
e. Example 8 – 17: The Elastic Collision of Two Blocks.
Find the final velocities for the elastic head-on collision in Example 8-13 (in which
a 4.0-kg block moving right at 6.0 m/s collides elastically with a 2.0-kg block
moving right at 3.0 m/s) by transforming their velocities to the center-of-mass
reference frame.
f. Practice Problem 8 – 8.
Verify that the total momentum of the system both before the collision and after
the collision is zero in the center-of-mass reference frame.
VI.
AP Problems.
a. Find the rebound speed of a 0.5-kg ball falling straight down that hits the floor
moving at 5 m/s, if the average normal force exerted by the floor on the ball
was 205 N for 0.02 s.
b.
A 5-kg mass moving at 10 m/s in the +x direction is acted upon by a force
acting in the –x direction with magnitude given as a function of time by the
graph below. Determine the speed and the direction of the mass after the
force has stopped acting.
F (x103 N)
7
0
0
2
10
t (x10-3 s)
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Conservation of Linear Momentum
c.
Consider a one-dimensional example in which a 4-kg mass moving in the +x
direction at 10 m/s collides head-on with a 2-kg mass moving at 6 m/s in the
–x direction. What are the final velocities for the elastic collision?
d.
Consider a one-dimensional example in which a 4-kg mass moving in the +x
direction at 10 m/s collides head-on with a 2-kg mass moving at 6 m/s in the
–x direction. What are the final velocities for the inelastic collision and what
fraction of the original kinetic energy is lost?
e.
Consider 4-kg mass moving in the -y direction collides and sticks to a 2-kg
mass moving at 6 m/s in the +x direction. Find the final velocity
components.
f.
A 6-kg spherical mass is currently moving horizontally at 20 m/s at a height of
45 m. A small explosion breaks the mass into a 2-kg and 4-kg piece, with the
4-kg piece falling straight down with 0 initial speed. How far does the 2-kg
piece travel?
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