Transcript Linear Momentum

Linear Momentum
Linear momentum describes motion in
which the center of mass of an object or
system changes position.
We call motion where the c.o.m. changes
position linear or translational motion.
What is center of mass (sometimes
called center of gravity)?
• Center of mass is the point in any object around
which the mass of the object is evenly
distributed.
• It’s located somewhere along a vertical line
through the “balance point”.
• Translational motion is a change in position of
the c.o.m. of an object.
• (Rotation occurs when the object moves---but
the c.o.m. doesn’t go anywhere!)
On the other hand, an object can be moving and
not have linear momentum.
An example would be an object that is spinning in
position. We call this rotational or angular
momentum (which we will study in a later chapter).
Momentum is a vector quantity,
with momentum and velocity in the
same direction:
p = mv
The units for momentum would be kgm/s.
If there is no outside force to affect the
momentum, momentum remains constant.
That means that the net (total) momentum
before an interaction (such as a collision or
explosion) is equal to the net (total)
momentum afterward if no external force is
applied to the system:
po = pf
 Just means “add all the quantities or vectors”.
What is an interaction?
A collision of two astronomical
objects….
Physicists study the momenta of objects after such
interactions to determine what must have been happening
before the interactions.
A collision of two sub-atomic
particles….
Examination of the momenta of the particles and jets after
such proton-proton collisions at the Large Hadron Collider in
Cern, Switzerland, may lead us to information about dark
matter.
A collision of two automobiles…
Engineers can study the positions of cars after a collision
to determine their speeds before the collision---using
conservation of momentum principles, of course.
An explosion of a supernova…
Could this be the result of a perfectly inelastic collision of
hot wings with blue cheese…?
Possibly not
If the momentum before an interaction is zero,
then (assuming no interfering force, such as
friction or gravitational force) the momentum
afterward is equal to zero.
In this case, the momentum before the interaction
with the fire extinguisher is zero.
After the extinguisher is fired, the momentum of
gases backward is equal to the momentum of the
wagon forward.
http://www.youtube.com/watch?v=IlMtJW7reKg
Since momentum is a vector, it
must have a direction. We use
positive and negative signs to
denote directions in onedimensional motion.
So if momentum to the right is positive, momentum to the left is
negative…..or if momentum to the east is positive, momentum
to the west is negative…..etc.
Example 1: A cart with mass 1.5 kg moving at
4.0 m/s to the right collides head-on with a cart
with mass 3.0 kg moving at 4.0 m/s to the left.
After the collision, the 1.5 kg cart is moving to
the left at 1.0 m/s. What is the 3.0 kg cart doing
after the collision?
?
Example 1: A cart with mass 1.5 kg moving at 4.0 m/s to the
right collides head-on with a cart with mass 3.0 kg moving at 4.0
m/s to the left. After the collision, the 1.5 kg cart is moving to the
left at 1.0 m/s. What is the 3.0 kg cart doing?
Solution:
po = pf
m1ov1o + m2ov2o = m1fv1f + m2fv2
(1.5 kg)(4.0 m/s) + (3.0 kg)(-4.0 m/s) = (1.5 kg)(-1.0 m/s) + (3.0 kg)(v)
v = -1.5 m/s (Since the answer is -, this cart is also moving to the left
after the collision. We will demonstrate examples of such phenomena
on the air track in the laboratory.)
For most collisions without an
intervening force, momentum is
conserved but some of the kinetic
energy of the colliding objects is “lost”.
Where do this energy go?
The “lost” kinetic energy is transferred to thermal energy of
molecules and to the sound you hear from the collision (which is
actually also a form of thermal energy of molecules).
In a totally inelastic collision, the
objects stick together during the
collision and move as one object after
the collision.
Example 2: A cart with mass 1.5 kg
moving at 2.0 m/s to the right collides
head-on with a cart with mass 3.0 kg
moving at 3.0 m/s to the left. Their
bumpers lock. What are they doing after
the collision?
Example 2: A cart with mass 1.5 kg moving at 2.0 m/s
to the right collides head-on with a cart with mass 3.0 kg
moving at 3.0 m/s to the left. Their bumpers lock. What
are they doing after the collision?
Solution:
po = pf
m1ov1o + m2ov2o = (m1 + m2)v
(1.5 kg)(2.0 m/s) + (3.0 kg)(-3.0 m/s) = (4.5 kg) v
v = -1.3 m/s
They move as one object to the left.
If a collision is elastic---and only if it is
elastic---the kinetic energy is also
conserved. In this case, we can write
both a momentum equation and a kinetic
energy equation. (This means that no
energy was lost during the
collision…..pretty rare…..probably only
happens during particle collisions.)
Example 3: A cart with mass 1.5 kg moving at
4.0 m/s to the right collides head-on and
elastically with a cart with mass 3.0 kg moving at
4.0 m/s to the left. After the collision, what are
the carts doing?
??
Example 2: A cart with mass 1.5 kg moving at 4.0 m/s to the right collides head-on and
elastically with a cart with mass 3.0 kg moving at 4.0 m/s to the left. After the collision,
what are the carts doing?
Solution: Since we don’t know the velocity of either cart after the
collision, there are two unknowns—so we’ll need two equations.
Since the collision is elastic, we have two equations---conservation of
momentum and conservation of kinetic energy.
po = pf
m1ov1o + m2ov2o = m1fv1f + m2fv2
(1.5 kg)(4.0 m/s) + (3.0 kg)(-4.0 m/s) = (1.5 kg)(v1) + (3.0 kg)(v2)
also:
KE10 + KE2o = KE1f + KE2f
½ (1.5 kg)(4.0 m/s)2 + ½ (3.0 kg)(-4.0 m/s)2 = ½ (1.5 kg)(v1)2 + ½ (3.0
kg)(v2)2
Remember: We can only write that energy equation b/c the collision is elastic!
Now solve the two equations simultaneously for
the two unknown variables:
1. (1.5 kg)(4.0 m/s) + (3.0 kg)(-4.0 m/s) = (1.5 kg)(v1) + (3.0 kg)(v2)
2. ½ (1.5 kg)(4.0 m/s)2 + ½ (3.0 kg)(-4.0 m/s)2 = ½ (1.5 kg)(v1)2 + ½ (3.0 kg)(v2)2
1. -6 = 1.5 v1 + 3 v2
2. First cancel all the ½ ‘s in equation 2:
72 = 1.5 v12 + 3v22
1. Solve for one variable in equation 1: v1 = -4 – 2v2
2. Substitute into equation 2: 72 = 1.5 (-4 – 2v2)2 + 3v22
72 = 1.5 (16 + 16v2 + 4v22) + 3v22
72 = 24 + 24 v2 + 6v22 + 3v22
0 = 9v22 + 24v2 – 48
0 = 3v22 + 8v2 - 16
(go on)
Your Algebra II teacher would be proud of you…but don’t forget
to F-O-I-L!
Now solve this quadratic:
0 = 3v22 + 8v2 - 16
v2 = -1.95 m/s or 0.177 m/s
Then, by subsitution:
v1 = 0.1 m/s when v2 = -1.95 m/s
or
v1 = -4.35 m/s when v2 = 0.177 m/s
Here’s the
answer!
That’s not
possible! How
can car 1 move
to the right and
car 2 move to
the left after
the collision--when they
can’t pass
through each
other!
Impulse, which is also a vector
(sometimes labeled J or I ), is
defined as “change in momentum of
an object or system”.
J = p = mv
Since changing both m and v at the same time
will require calculus, all of our problems will have
a constant mass, so: J = m(vf – vo).
Example: What is the impulse of a cart of
mass 2.0 kg moving at 3.0 m/s if the cart is
moving at -2.5 m/s after hitting the block?
Example: What is the impulse of a cart of mass 2.0 kg
moving at 3.0 m/s if the cart is moving at -2.5 m/s after
hitting the block?
Answer:
J = m(vf – vo)
J = (2.0 kg)(-2.5 m/s – 3.0 m/s) = -11 kgm/s
[Notice that impulse has the same units as momentum.]
Since we already know that an external
force is required to change the momentum
of an object, impulse can be related to force.
As Isaac Newton formulated in his second
law of motion: “Force is proportional to the
rate in change of momentum.”
F = p/t
Newton wrote it in the above form, but we have
simplified it to F = ma, which is equal to the
above if you substitute.
Example: What is the force exerted on a
cart of mass 2.0 kg moving at 3.0 m/s if the
cart is moving at -2.5 m/s after hitting the
block and being in contact with it for 1.0 ms?
Example: What is the force exerted on a cart of mass 2.0
kg moving at 3.0 m/s if the cart is moving at -2.5 m/s after
hitting the block and being in contact with it for 1.0 ms?
Answer:
F = m(vf – vo)/t
F = (2.0 kg)(-2.5 m/s – 3.0 m/s)/(1.0 x 10-3 s) = -11,000 N
Q: What is the force exerted by the cart on the block?
Answer? The same---but in the opposite direction.
The cart exerts an 11,000 N force on the block.
Don’t forget Newtons’ third law of motion…action
and reaction.
A corollary to Conservation of Momentum is “Conservation
of the Velocity of Center of Mass”:
The original motion of each shell is a parabola. After the
shell explodes, the individual pieces move in such a way
that if you add their positive and negative momenta, they
still follow the original parabolic path.
The Grand Finale
During this display, watch for conservation of
momentum in the form of “conservation of the velocity
of center of mass of the system”.
http://www.youtube.com/watch?v=K3THhUGBNJE