Transcript Lecture-11-10

Chapter 9
Linear Momentum
1
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
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 is a vector, in the same direction
as the average force.
Impulse
We can rewrite
as
So we see that
The impulse is equal to the change in momentum.
Why we don’t dive into concrete
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.
Going Bowling II
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?
a) the bowling ball
b) same time for both
c) the Ping-Pong ball
d) impossible to say
p
p
Going Bowling II
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?
We know:
p
Fav =
t
a) the bowling ball
b) same time for both
c) the Ping-Pong ball
d) impossible to say
so p = Fav t
Here, F and p are the same for both balls!
It will take the same amount of time to
stop them.
p
p
Going Bowling III
A bowling ball and a Ping-Pong ball
a) the bowling ball
are rolling toward you with the
b) same distance for both
same momentum. If you exert the
c) the Ping-Pong ball
same force to stop each one, for
d) impossible to say
which is the stopping distance
greater?
p
p
Going Bowling III
A bowling ball and a Ping-Pong ball
a) the bowling ball
are rolling toward you with the
b) same distance for both
same momentum. If you exert the
c) the Ping-Pong ball
same force to stop each one, for
d) impossible to say
which is the stopping distance
greater?
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.
p
p
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!
With no net force:
•A vector equation
•Works for each coordinate separately
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 system
Internal forces cannot change the
momentum of a system.
However, the momenta of pieces of the
system may change.
With no net external force:
An example of internal forces moving
components of a system:
Kinetic Energy of a System
Another 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
Birth of the neutrino
Beta decay fails momentum
conservation?
First detection 1956
Pauli “fixes” it with a new ghost-like, undetectable
particle
Bohr scoffs
16
Lecture 11
Momentum, Energy, and
Collisions
Linear Momentum
Impulse
With no net external force:
Nuclear Fission I
A uranium nucleus (at rest)
a) the heavy one
undergoes fission and splits
into two fragments, one
heavy and the other light.
Which fragment has the
greater momentum?
b) the light one
c) both have the same momentum
d) impossible to say
1
2
Nuclear Fission I
A uranium nucleus (at rest)
a) the heavy one
undergoes fission and splits
into two fragments, one
heavy and the other light.
Which fragment has the
greater momentum?
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.
1
2
Nuclear Fission II
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) the heavy one
b) the light one
c) both have the same speed
d) impossible to say
1
2
Nuclear Fission II
A uranium nucleus (at rest)
undergoes fission and splits
into two fragments, one
heavy and the other light.
Which fragment has the
a) the heavy one
b) the light one
c) both have the same speed
d) impossible to say
greater speed?
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.
1
2
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.
looking from y
We know that px=0, py = 0 in initial state
v
2
above:
x
and no external forces act in the horizontal
v1
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 Mass
The 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
RECALL:
momentum of a system.
However, the momenta of pieces of the
system may change.
With no net external force:
An example of internal forces moving
components of a system:
Motion about the Center of Mass
The 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 cannon sits on a stationary
a) 0 m/s
railroad flatcar with a total mass
b) 0.5 m/s to the right
of 1000 kg. When a 10-kg
c) 1 m/s to the right
cannonball is fired to the left at
d) 20 m/s to the right
a speed of 50 m/s, what is the
e) 50 m/s to the right
recoil speed of the flatcar?
Recoil Speed
A cannon sits on a stationary
a) 0 m/s
railroad flatcar with a total mass
b) 0.5 m/s to the right
of 1000 kg. When a 10-kg
c) 1 m/s to the right
cannonball is fired to the left at
d) 20 m/s to the right
a speed of 50 m/s, what is the
e) 50 m/s to the right
recoil speed of the flatcar?
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)
Recoil Speed
A cannon sits on a
stationary railroad
a) 0 m/s
b) 0.5 m/s to the right
c) 1 m/s to the right
flatcar with a total
d) 20 m/s to the right
mass of 1000 kg.
e) 50 m/s to the right
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 cannon sits on a stationary
a) 0 m/s
railroad flatcar with a total mass
b) 0.5 m/s to the right
of 1000 kg. When a 10-kg
c) 1 m/s to the right
cannonball is fired to the left at
d) 20 m/s to the right
a speed of 50 m/s, what is the
e) 50 m/s to the right
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
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.)
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
Rolling in the Rain
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.
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
Two objects collide... and stick
mass m
mass m
No external forces... so momentum of system is conserved
initial
px = mv0
final
px = (2m)vf
mv0 = (2m)vf
vf = v0 / 2
A completely inelastic collision: no “bounce back”
Inelastic collision: What about energy?
mass m
mass m
vf = v0 / 2
initial
final
Kinetic energy is lost!
KEfinal = 1/2 KEinitial
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
Completely inelastic collision:
colliding objects stick together,
maximal loss of kinetic energy
Inelastic collision:
momentum is conserved but
kinetic energy is not
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)
Momentum Conservation:
KEfinal < KEinitial
Crash Cars I
a) I
If all three collisions below are
b) II
totally inelastic, which one(s)
c) I and II
will bring the car on the left to a
d) II and III
complete halt?
e) all three
Crash Cars I
a) I
If all three collisions below are
b) II
totally inelastic, which one(s)
c) I and II
will bring the car on the left to a
d) II and III
complete halt?
e) all three
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.
Crash Cars II
If all three collisions
a) I
b) II
below are totally
c) III
inelastic, which
e) all three
one(s) will cause the
most damage (in
terms of lost
energy)?
d) II and III
Crash Cars II
If all three collisions below are
a) I
totally inelastic, which one(s) will
b) II
cause the most damage (in terms
c) III
of lost energy)?
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.
Ballistic pendulum: the height h can be found
using conservation of mechanical energy after the
object is embedded in the block.
momentum conservation
in inelastic collision
vf = m v0 / (m+M)
energy conservation
afterwards
KE = 1/2 (mv0)2 / (m+M)
PE = (m+M) g h
hmax = (mv0)2 / [2 g (m+M)2]
Velocity of the ballistic pendulum
approximation
Pellet Mass (m): 2 g
Pendulum Mass (M): 3.81 kg
Wire length (L): 4.00 m
Inelastic Collisions in 2 Dimensions
For collisions in two dimensions, conservation of
momentum is applied separately along each axis:
Momentum is a vector equation: there is 1 conservation of momentum equation per dimension
Energy is not a vector equation: there is only 1 conservation of energy equation
Elastic Collisions
In elastic collisions, both kinetic energy
and momentum are conserved.
One-dimensional elastic collision:
Elastic Collisions in 1-dimension
For special case of v2i = 0
We have two equations:
conservation of momentum
conservation of energy
and two unknowns (the final speeds).
solving for the final speeds:
Note: relative speed is conserved for head-on (1-D) elastic collision
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
Carefully place a small rubber ball (mass m) on
a) zero
top of a much bigger basketball (mass M) and
b) v
drop these from the same height h so they
c) 2v
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?
m
v
v
M
d) 3v
e) 4v
Elastic Collisions II
Carefully place a small rubber ball (mass m) on
a) zero
top of a much bigger basketball (mass M) and
b) v
drop these from the same height h so they
c) 2v
arrive at the ground with the speed v. What is
d) 3v
the velocity of the smaller ball after the
e) 4v
basketball hits the ground, reverses direction,
and then collides with the small rubber ball?
• Remember that relative
velocity has to be equal
v
before and after collision!
Before the collision, the
basketball bounces up with v
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!!
m
M
(a)
3v
v
v
v
(b)
(c)
Elastic Collisions I
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.
a) situation 1
b) situation 2
c) both the same
In which case does the golf ball have the
greater speed after the collision?
v
at rest
at rest
1
v
2
Elastic Collisions I
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.
a) situation 1
b) situation 2
c) both the same
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!
v
1
In case 1 the bowling ball will almost
remain at rest, and the golf ball will
bounce back with speed close to v.
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.
v
2v
2