chapter 9 linear momentum
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Chapter 9
Linear Momentum
Announcements
• Assignments due Saturday
• Midterm Exam II: October 17 (chapters 6-9)
• Formula sheet will be posted
• Practice problems
• Practice exam next week
Linear Momentum
Momentum is a vector; its direction is the same as the direction of the velocity.
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):
Impulse
Impulse is a vector, in the same direction as the average force.
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.
Impulse quantifies the overall change in momentum
Impulse
We can rewrite
as
So we see that
The impulse is equal to the change in momentum.
Why we don’t dive into concreteThe 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.
Going Bowling II
p
p
a) the bowling ball
b) same time for both
c) the Ping-Pong ball
d) impossible to say
A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, which takes a longer time to bring to rest?
Going Bowling II
We know:
Here, F and p are the same for both balls!
It will take the same amount of time to stop them. p
p so p = Fav t
a) the bowling ball
b) same time for both
c) the Ping-Pong ball
d) impossible to say
A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, which takes a longer time to bring to rest?
av tp
F
Going Bowling III
p
p
A bowling ball and a Ping-Pong
ball are rolling toward you with
the same momentum. If you
exert the same force to stop each
one, for which is the stopping
distance greater?
a) the bowling ball
b) same distance for both
c) the Ping-Pong ball
d) impossible to say
Going Bowling III
p
p
Use the work-energy theorem: W = KE. The ball with less mass has the greater speed, and thus the greater KE. In order to remove that KE, work must be done, where W = Fd. Because the force is the same in both cases, the distance needed to stop the less massive ball must be bigger.
A bowling ball and a Ping-Pong
ball are rolling toward you with
the same momentum. If you
exert the same force to stop each
one, for which is the stopping
distance greater?
a) the bowling ball
b) same distance for both
c) the Ping-Pong ball
d) impossible to say
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!
•A vector equation•Works for each coordinate separately
With no net force:
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.
Momentum of components of a systemInternal forces cannot change the momentum of a system.
However, the momenta of pieces of the system may change.
An example of internal forces moving components of a system:
With no net external force:
Kinetic Energy of a SystemAnother example of internal forces moving components of a system:
The initial momentum equals the final (total) momentum.
But the final Kinetic Energy is very large
16
Birth of the neutrino
Beta decay fails momentum conservation?
Pauli “fixes” it with a new ghost-like, undetectable particle
Bohr scoffs
First detection 1956
Lecture 11Momentum, Energy, and
Collisions
Linear Momentum
Impulse
With no net external force:
Nuclear Fission I
A uranium nucleus (at rest) A uranium nucleus (at rest)
undergoes fission and splits undergoes fission and splits
into two fragments, one into two fragments, one
heavy and the other light. heavy and the other light.
Which fragment has the Which fragment has the
greater momentum? greater momentum?
a) the heavy one a) the heavy one
b) the light oneb) the light one
c) both have the same c) both have the same
momentummomentum
d) impossible to sayd) impossible to say
1 2
1 2
Nuclear Fission I
A uranium nucleus (at rest)
undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum?
a) the heavy one
b) the light one
c) both have the same
momentum
d) impossible to say
The initial momentum of the uranium
was zero, so the final total momentum
of the two fragments must also be
zero. Thus the individual momenta
are equal in magnitude and opposite
in direction.
Nuclear Fission II
a) the heavy one
b) the light one
c) both have the same speed
d) impossible to say
1 2
A uranium nucleus (at rest)
undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed?
Nuclear Fission II
We have already seen that the
individual momenta are equal and
opposite. In order to keep the
magnitude of momentum mv the
same, the heavy fragment has the
lower speed and the light fragment
has the greater speed.
a) the heavy one
b) the light one
c) both have the same speed
d) impossible to say
1 2
A uranium nucleus (at rest)
undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the
greater speed?
A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece.
We know that px=0, py = 0 in initial stateand no external forces act in the horizontal
looking from above: x
y
v1
v2
v3
An 85-kg lumberjack stands at one end of a 380-kg floating log, as shown in the figure. Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack’s speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack’s speed relative to the shore be greater than, less than, or the same as in part (a)? Explain.
Center of Mass
Center of MassThe center of mass of a system is the point where the
system can be balanced in a uniform gravitational field.
For two objects:
The center of mass is closer to the more massive object.
Think of it as the “average location of the mass”
Center of Mass
The center of mass need not be within the object
Momentum of components of a system
Internal forces cannot change the momentum of a system.
However, the momenta of pieces of the system may change.
An example of internal forces moving components of a system:
With no net external force:
RECALL:
Motion about the Center of MassThe center of mass of a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object
Motion of the center of mass
Action/Reaction pairs inside the system cancel out
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
Momentum of a composite object
Recoil Speed
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total
mass of 1000 kg. When a 10-
kg cannonball is fired to the left
at a speed of 50 m/s, what is
the recoil speed of the flatcar?
Recoil Speed
Because the initial momentum of the
system was zero, the final total
momentum must also be zero. Thus, the
final momenta of the cannonball and the
flatcar must be equal and opposite.
pcannonball = (10 kg)(50 m/s) = 500 kg-
m/s
pflatcar = 500 kg-m/s = (1000 kg)(0.5
m/s)
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total
mass of 1000 kg. When a 10-
kg cannonball is fired to the left
at a speed of 50 m/s, what is
the recoil speed of the flatcar?
Recoil Speed
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total
mass of 1000 kg. When a 10-kg
cannonball is fired to the left at
a speed of 50 m/s, what is the
speed of the center of mass?
Recoil Speed
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
d) 20 m/s to the right
e) 50 m/s to the right
A cannon sits on a stationary
railroad flatcar with a total mass
of 1000 kg. When a 10-kg
cannonball is fired to the left at
a speed of 50 m/s, what is the
speed of the center of mass?
Internal forces cannot change the motion of the center of mass. The CM was originally motionless at zero, and remains so after the gun is fired.
Rolling in the Rain
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
An open cart rolls along a
frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)
Because the rain falls in vertically,
it adds no momentum to the box,
thus the box’s momentum is
conserved. However, because the
mass of the box slowly increases
with the added rain, its velocity has
to decrease.
Rolling in the Rain
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
An open cart rolls along a
frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)
Two objects collide... and stick
A completely inelastic collision: no “bounce back”
No external forces... so momentum of system is conservedinitial px = mv0
final px = (2m)vf
mv0 = (2m)vf
vf = v0 / 2
mass m mass m
Inelastic collision: What about energy?
mass m mass m vf = v0 / 2
Kinetic energy is lost! KEfinal = 1/2 KEinitial
initial
final
Collisions
This is an example of an “inelastic collision”
Collision: two objects striking one another
Elastic collision ⇔ “things bounce back”
⇔ energy is conserved
Inelastic collision: less than perfectly bouncy
⇔ Kinetic energy is lost
Time of collision is short enough that external forces may be ignored so momentum is conserved
Completely inelastic collision: objects stick together afterwards. Nothing “bounces back”. Maximal energy loss
Elastic vs. Inelastic
Inelastic collision: momentum is conserved but kinetic energy is not
Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy
Elastic collision: momentum and kinetic energy is conserved.
Completely Inelastic Collisions in One Dimension
Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses:
Completely inelastic only(objects stick together, so have same final velocity)
KEfinal < KEinitial
Momentum Conservation:
Crash Cars I
a) I
b) II
c) I and II
d) II and III
e) all three
If all three collisions below
are totally inelastic, which
one(s) will bring the car on
the left to a complete halt?
Crash Cars I
In case I, the solid wall clearly stops the car.
In cases II and III, because
ptot = 0 before the
collision, then ptot must
also be zero after the collision, which means that the car comes to a halt in all three cases.
a) I
b) II
c) I and II
d) II and III
e) all three
If all three collisions below
are totally inelastic, which
one(s) will bring the car on
the left to a complete halt?
Crash Cars II
If all three collisions below are
totally inelastic, which one(s)
will cause the most damage (in
terms of lost energy)?
a) I
b) II
c) III
d) II and III
e) all three
Crash Cars II
a) I
b) II
c) III
d) II and III
e) all three
The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.
If all three collisions below are
totally inelastic, which one(s) will
cause the most damage (in
terms of lost energy)?
Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.
vf = m v0 / (m+M)
momentum conservation in inelastic collision
KE = 1/2 (mv0)2 / (m+M)
energy conservationafterwards
PE = (m+M) g h
hmax = (mv0)2 / [2 g (m+M)2]
Velocity of the ballistic pendulum
Pellet Mass (m): 2 gPendulum Mass (M): 3.81 kgWire length (L): 4.00 m
approximation
Inelastic Collisions in 2 Dimensions
Energy is not a vector equation: there is only 1 conservation of energy equation
Momentum is a vector equation: there is 1 conservation of momentum equation per dimension
For collisions in two dimensions, conservation of momentum is applied separately along each axis:
Elastic CollisionsIn elastic collisions, both kinetic
energy and momentum are conserved.
One-dimensional elastic collision:
Elastic Collisions in 1-dimension
solving for the final speeds:
Note: relative speed is conserved for head-on (1-D) elastic collision
We have two equations: conservation of momentum conservation of energy
and two unknowns (the final speeds).
For special case of v2i = 0
Limiting cases of elastic collisions
note: relative speed conserved
Limiting cases
note: relative speed conserved
Limiting cases
note: relative speed conserved
Toy Pendulum
Could two balls recoil and conserve both momentum and energy?
Incompatible!
Elastic Collisions II
v
v
m
M
Carefully place a small rubber ball (mass m) on
top of a much bigger basketball (mass M) and
drop these from the same height h so they
arrive at the ground with the speed v. What is
the velocity of the smaller ball after the
basketball hits the ground, reverses direction,
and then collides with the small rubber ball?
a) zero
b) v
c) 2v
d) 3v
e) 4v
• Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!!
Elastic Collisions II
v
v
v
v
3v
v
(a) (b) (c)
m
M
Carefully place a small rubber ball (mass m) on
top of a much bigger basketball (mass M) and
drop these from the same height h so they
arrive at the ground with the speed v. What is
the velocity of the smaller ball after the
basketball hits the ground, reverses direction,
and then collides with the small rubber ball?
a) zero
b) v
c) 2v
d) 3v
e) 4v
Elastic Collisions I
v 2v
1at rest
at rest
a) situation 1
b) situation 2
c) both the same
Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.
In which case does the golf ball have the greater speed after the collision?
Remember that the magnitude of the relative velocity has to be equal before and after the collision!
Elastic Collisions I
v1
In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v.
v 22v
In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.
a) situation 1
b) situation 2
c) both the same
Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on.
In which case does the golf ball have the greater speed after the collision?
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