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Lecture Outline
Chapter 9
Physics, 4th Edition
James S. Walker
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Chapter 9
Linear Momentum and
Collisions
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Units of Chapter 9
• Linear Momentum
• Momentum and Newton’s Second Law
• Impulse
• Conservation of Linear Momentum
• Inelastic Collisions
• Elastic Collisions
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Units of Chapter 9
• Center of Mass
• Systems with Changing Mass: Rocket
Propulsion
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9-1 Linear Momentum
Linear momentum is defined as the product of the mass m and
velocity V of an object.
Momentum is a vector; its direction is the
same as the direction of the velocity.
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9-1 Linear Momentum
Change in momentum:
(a) mv
(b) 2mv
Change in momentum
A beanbag bear and a rubber ball with the same mass m and the same downward speed v, hit
the floor. (a) The beanbag bear comes to rest on hitting the floor. Its change in momentum is mv
upward. (b). The rubber ball bounces upward with a speed . Its change in momentum is 2mv
upward.
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9-1 Linear Momentum
At a city park, a person throws some bread into a duck pond. Two 4.0-kg ducks
and a 9.0-kg goose paddle rapidly toward the bread, as shown in our sketch. If the
ducks swim at 1.10 m/s, and the goose swims with a speed of 1.30 m/s., find the
magnitude and direction of the total momentum of the three birds.
Use x and y unit vectors to express the momentum of each bird in
vector form:
Pd1 = mavdx = (4.0kg)(1.10 m/s)x = 4.40 kg.m/s)x
Pd2 = - mavdy = - (4.0kg)(1.10 m/s)y= 4.40 kg.m/s)y
Pg = mgvg y = (9.0kg)(1.30 m/s)y = 11.7 kg.m/s)y
Ptotal = Pd1 + Pd2 + Pg
Ptotal = (4.40 kg.m/s)x + [-4.40 + 11.7 kg.m/s]y
= (4.40 kg.m/s)x + (7.30 kg.m/s) y
Calculate the magnitude of the total momentum:
Ptotal = (P2total,x + P2total,y)1/2 = ((4.40 kg.m/s)2 + (7.30 kg.m/s)2)1/2
Ptotal = 8.52 kg.m/s
Calculate the direction of the total momentum:
Θ = tan -1 ((7.30 kg.m/s) / (4.40 kg.m.s)) = 58.9O
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9-2 Momentum and Newton’s Second Law
Newton’s second law, as we wrote it before:
is only valid for objects that have constant
mass. Here is a more general form, also
useful when the mass is changing:
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9-3 Impulse
The pitcher delivers a fastball, the batter takes a swing, and with a crack of the
bat the ball that was approaching home plate at 95.0 mi/h is now heading
toward the pitcher at 115 mi/h. We say the bat has delivered an impulse, I to
the ball.
Impulse is a vector, in the same direction
as the average force.
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9-3 Impulse
We can rewrite
as
So we see that
The impulse is equal to the change in
momentum.
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9-3 Impulse
Therefore, the same
change in momentum
may be produced by a
large force acting for a
short time, or by a
smaller force acting for a
longer time.
Hitting a baseball
A batter hits a ball, sending it back toward the pitcher’s mound.
The impulse delivered to the ball by the bat changes the ball’s
momentum from –pi x to pf x
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9-4 Conservation of Linear Momentum
The net force acting on an object is the rate
of change of its momentum:
If the net force is zero, the momentum does not
change:
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9-4 Conservation of Linear Momentum
Internal Versus External Forces:
Internal forces act between objects within the
system.
As with all forces, they occur in action-reaction
pairs. As all pairs act between objects in the
system, the internal forces always sum to zero:
Therefore, the net force acting on a system is
the sum of the external forces acting on it.
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9-4 Conservation of Linear Momentum
Furthermore, internal forces cannot change the
momentum of a system.
However, the momenta of components of the
system may change.
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9-4 Conservation of Linear Momentum
An example of internal forces moving
components of a system:
Two groups of canoeists meet in the middle of a lake.
After a brief visit, a person in canoe 2 with a force of
46 N to separate the canoes. If the mass of canoe 1
and its occupants is 130-kg, and the mass of canoe 2
and its occupants is 250-kg, find the momentum of
each canoe after 1.20 s of pushing.
Choose the positive direction to point from
canoe 1 to canoe 2. With this choice, the force
exerted on canoes 2 is F2 = (46 N) x and the
force exerted on canoe 1 is F1 = (-46 N) x.
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-
Newton’s 2nd law: a2,x = ∑F2,x / m2 = 46 N/250 kg = 0.18 m/s2
-
a1,x = ∑F1,x / m1 = - 46 N/130 kg = - 0.35 m/s2
-
V2,x = a2,x t = (0.18 m/s2) (1.20 s) = 0.22 m/s
-
V1,x = a1,x t = (- 0.35 m/s2) (1.20 s) = - 0.42 m/s
-
p1,x = m1 v1, x = (130 kg) (- 0.42 m/s) = - 55 kg.m / s
-
p2,x = m2 v2, x = (250 kg) (0.22 m/s) = 55 kg.m / s
-
Note that the sum of the momenta of the two canoes is zero.
The canoes start at rest with zero momentum, there is zero net
external force acting on the system, hence the final momentum
must also be zero. The final velocities do not add to zero; it is
momentum (mv) that is conserved, not velocity (v).
9-5 Inelastic Collisions
Collision: two objects striking one another
Time of collision is short enough that external
forces may be ignored
Inelastic collision: momentum is conserved but
kinetic energy is not
Completely inelastic collision: objects stick
together afterwards
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9-5 Inelastic Collisions
A completely inelastic collision:
Railroad cars collide and stick together
A moving train car collides with an identical car that is stationary.
After the collision, the cars stick together and move with the same speed.
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9-5 Inelastic Collisions
Solving for the final momentum in terms of the
initial momenta and masses:
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9-5 Inelastic Collisions
On a touchdown attempt, a 95.0-kg running back runs toward the end zone at 3.75 m/s.
A 111-kg linebacker moving at 4.10 m/s meets the runner in a head-on collision. If the
two players stick together, (a) what is their velocity immediately after the collision? (b)
What are the initial and final kinetic energy of the system?
Vf = (95.0 kg) (3.75 m/s) x + (111 kg)(-4.10 m/s) x
95.0 kg + 111 kg
Vf = (-0.480 m/s) x
(b) Calculate the initial kinetic energy of the two players
Ki = 1/2m1v12 + ½ m2v22
= ½(95.0 kg)(3.75 m/s)2 + ½(111-kg)(-4.10 m/s)2
Ki = 1600.9 J
Calculate the final kinetic energy of the two players noting that they both move with
the same velocity after the collision.
Kf = ½(m1 + m2)vf2 = ½(95-kg + 111-kg)(-0.48 m/s)2 = 23.7 J
Kf = 23.7 J
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9-5 Inelastic Collisions
Ballistic pendulum: the height h can be found
using conservation of mechanical energy after
the object is embedded in the block.
Our sketch shows the physical setup of a ballistic pendulum. Initially, only the object of m is moving,
and it moves in the positive x direction with a speed vo. Immediately after the collision, the bob and
object move together with a new speed, vf, which is determined by momentum conservation. Finally, the
pendulum continues to swing to the right until its speed decreases to zero and it comes to rest at the
height h.
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9-5 Inelastic Collisions
In a ballistic pendulum, an object of mass m is fired with an initial speed vo at the bob of a
pendulum. The bob has a mass M, and is suspended by a rod of negligible mass. After the
collision, the object and the bob stick together and swing through an arc, eventually gaining a
height h. Find the height h in terms of m, M, vo and g.
Set the momentum just before the bob-object collision equal
to the momentum just after the collision.
mvo = (M + m)vf
vf = (m / (M + m) ) vo
Calculate the kinetic energy just after the collision
Kf = ½ (M + m) vf2 = ½ (M + m) ((m / (M + m) 2 ) vo2
Kf = ½ m vo2 (m )
(M + m)
Kf = (M + m) gh = ½ m vo2 (m )
(M + m)
h = (m / M + m) 2 (vo2/2g)
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9-5 Inelastic Collisions in Two Dimensions
For collisions in two dimensions, where we must conserve the
momentum component by component. To do this, we set up a
coordinate system and resolve the initial momentum into x and y
components as the initial momentum. That is,
px,i = px, f
p y, i = py,f
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9-5 Inelastic Collisions in Two Dimensions
A car with a mass of 950 kg and a speed of 16 m/s approaches an intersection as
shown below. A 1300-kg minivan traveling at 21 m/s is heading for the same
intersection. The car and the minivan collide and stick together. Find the speed and
direction of the wrecked vehicles just after the collision, assuming external forces
can be ignored.
In our sketch, we align the x and y axes with the crossing streets. With
this choice, V1 (the car’s velocity) is in the positive x direction and V2
(the minivan's velocity) is in the positive y direction. The problem
indicates that m1 = 950 kg and m2 = 1300 kg. After the collision, the two
vehicles move together ( as a unit) with a speed Vf in a direction θ with
respect to the positive x axis.
Set the initial x component of momentum equal to the final x component
of momentum: m1v1 = (m1 + m2)vf cos θ .----- eq. 1
Set for the y component of momentum
m2v2 = (m1 + m2)vf sin θ ---------- (eq. 2)
Divide eq 2. by eq. 1 m2v2 = (m1+ m2)vf sin θ = tan θ
m1v1
(m1 + m2)vf cos θ
θ = tan-1 (m2v2 / m1v1) = tan -1 ((1300kg)(21 m/s) / (950 kg)(16 m/s))
θ = tan -1 (1.8) = 61O
The final speed v f = m1v1 / (m1 + m2) cos θ
= (950 kg)(16 m/s) / (950 kg + 1300 kg) cos 61o
Vf = 14 m /s
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9-6 Elastic Collisions
In elastic collisions, both kinetic energy and momentum are
conserved. That is, p i = pf and Ki = Kf
One-dimensional elastic collision:
An elastic collisions between two carts
In the case pictured, V1,f is to the right (positive), which means that m1 is greater than m2. In fact, we
have chosen m1 = 2m2 for this pilot, therefore, v1,f = Vo/3 and V2,f = 4Vo/3. if m1 were less than m2, cart 1
would bounce back toward the left, meaning that V1,f would be negative.
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9-6 Elastic Collisions
We have two equations (conservation of
momentum and conservation of kinetic energy)
and two unknowns (the final speeds). Solving
for the final speeds:
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9-6 Elastic Collisions in Two Dimensions
Two-dimensional collisions can only be solved if
some of the final information is known, such as
the final velocity of one object:
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9-6 Elastic Collisions in Two Dimensions
Two astronauts on opposite ends of a spaceship are comparing launches. One has an
apple, the other has an orange. They decide to trade. Astronauts-1 tosses the 0.130-kg
apple toward astronaut 2 with a speed of 1.11 m/s. The 1.160-kg orange is tossed from
astronaut 2 to astronaut 1 with a speed of 1.21-m/s. Unfortunately, the fruits collide,
sending the orange off with a speed of 1.16 m/s at an angle of 42.00 with respect to its
original direction of motion. Find the final speed and direction of the apple, assuming an
elastic collision. Give the apple’s direction relative to its original direction of motion.
1. Calculate the initial kinetic energy of the system: Ki = 1/2m1v1,i2 + 1/2m2v2,12
Ki = ½(0.130 kg)(1.11 m/s)2 + ½(0.160 kg)(1.21 m/s)2
Ki = 0.197 J.
2. Calculate the final Kinetic energy of the system in terms of V 1,f: Ki = 1/2m1v1f2 + 1/2m2v2,f2
Kf = ½(0.130 kg)(V1,f m/s)2 + ½(0.160 kg)(1.16 m/s)2
= ½(0.130 kg)(V1,f m/s)2 + 0.108 J
V1,f = (2(0.197 J - 0.108 J)) / 0.130) ½ = 1.17 m/s
3. Set the final y component of momentum equal to zero to determine the angle θ
0 = m1V1,f sin θ - m2V2,f sin 42.0O
sin θ = m2V2,f sin 42.0O / m1V1,f = (0.160 kg) (1.16 m/s) sin 42.0 O / (0.130 kg) (1.17 m/s)
θ = sin -1 (0.817) = 54.8 O
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9-7 Center of Mass
The center of mass of a system is the point where
the system can be balanced in a uniform
gravitational field.
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9-7 Center of Mass
For two objects:
The center of mass is closer to the more
massive object.
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9-7 Center of Mass
The center of mass need not be within the object:
Locating the center of mass
In an object of continuous, uniform mass distribution, the center of mass is located at the geometric center of the object. In
some cases, this means that the center of mass is not located within the object.
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9-7 Center of Mass
Motion of the center of mass:
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9-7 Center of Mass
The total mass multiplied by the acceleration of
the center of mass is equal to the net external
force:
The center of mass
accelerates just as
though it were a point
particle of mass M
acted on by
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9-8 Systems with Changing Mass:
Rocket Propulsion
If a mass of fuel Δm is ejected from a rocket
with speed v, the change in momentum of the
rocket is:
The force, or thrust, is
SI unit : newton, N
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Summary of Chapter 9
• Linear momentum:
• Momentum is a vector
• Newton’s second law:
• Impulse:
• Impulse is a vector
• The impulse is equal to the change in
momentum
• If the time is short, the force can be quite
large
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Summary of Chapter 9
• Momentum is conserved if the net external
force is zero
• Internal forces within a system always sum to
zero
• In collision, assume external forces can be
ignored
• Inelastic collision: kinetic energy is not
conserved
• Completely inelastic collision: the objects
stick together afterward
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Summary of Chapter 9
• A one-dimensional collision takes place along
a line
• In two dimensions, conservation of
momentum is applied separately to each
• Elastic collision: kinetic energy is conserved
• Center of mass:
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Summary of Chapter 9
• Center of mass:
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Summary of Chapter 9
• Motion of center of mass:
• Rocket propulsion:
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