Transcript V f

Linear Momentum
Momentum and Force
Conservation of Momentum
Collisions and Impulse
Before we get into chapter 7, let’s step back and think about
where we have come from.
What’s in your Physics toolbox?
 English!
 algebra and trig
 kinematics (motion without
worrying about forces)
 dynamics (Newton’s laws, forces)
(including rotational dynamics)
 conservation of energy (work and energy)
We are about to add another tool—conservation of momentum.
Chapter 7
Linear Momentum
7.1 Momentum and its Relation to Force
The momentum is of an object of mass m is defined as
OSE :
v
p = mv
v
p
p
Momentum is a vector quantity! You must pay attention to all
our rules on the handling of vectors.
A force is required to change an object’s momentum.
The correct statement of Newton’s second law is actually
F=dp/dt (or F=p/t if you prefer to avoid calculus), from
which F=ma can be derived.
This gives another OSE:
OSE :
p
 F = t
Example 7-1. Water leaves a hose at a rate of 1.5 kg/s with
a speed of 20 m/s and is aimed at the side of a car, which
stops it without splashing it back (kind of a fake problem, but
that’s OK). What is the force exerted by the water on the car.
After we finish section 7.3 I’ll give you a litany for momentum
problems. For now, I’ll modify our “usual” steps.
Step 1: draw before and after sketch.
You can draw a fancy
sketch, but I suggest you
save time and draw point
masses. Make sure you
have SEPARATE before and
after parts.
before
after
Step 2: label point masses and draw velocity or
momentum vectors (your choice).
Hint: draw unknown velocity
(or momentum) vectors with
components that appear to be
positive, to avoid putting extra
– signs into your work.
vi
m
vf=0
x
x
before
after
Step 3: choose axes, lightly draw in components of
any vector not parallel to an axis.
Step 4: OSE.
OSE:
m
Fx=px/t
Steps 5 and 6 are not applicable to this problem.
Step 7: solve.
In a time of t=1s, m=1.5 kg of water hits the car.
FWC means force on
water by car
vi
m
vf=0
x
before
FWC m
x
during
This drawing is not part of the “official” procedure;
it is to help you visualize the force.
Steps 5 and 6 are not applicable to this problem.
Step 7: solve.
In a time of t=1s, m=1.5 kg of water hits the car.
FWC,x=pWx/t
FWC means force on
water by car
FWC,x = (PWfx – PWix) / t
FWC,x = (mvWfx – mvWix) / t
FWC,x = m (vWfx – vWix) / t
FWC,x = 1.5 kg (0 m/s – 20 m/s) / 1 s
FWC,x = -30 N  FCW,x = 30 N
By way of example, I have evidently told you it is now OK to
not include in your diagram “things” that appear in your
equations.
You can show either p or v. The one you don’t show will
appear in your equations but not your diagram. That’s now
“legal.”
You are not explicitly required to show the forces that
appear in F=p/t.
If the problem asks for a force, it is a good idea to show it.
7.2 Conservation of Momentum
The momentum of a single particle is denoted by lowercase p.
For a system of particles, we define the total momentum P by
OSE :
P =  mn v n = MVCM
n
where M is the total mass of all particles and Vcm is the
velocity of the center of mass of the system (we will define the
center of mass later in this chapter).
A note on notation: I will use arrows for vectors whenever possible, but it
is usually not convenient in a block of text. Lowercase p is the momentum
ofaa single particle. Script P is the total momentum of a collection of
particles on the Phys. 23 OSE sheet. I will use P in my notes.
It has been observed experimentally and verified over and
over that in the absence of a net external force, the total
momentum of a system remains constant.
The above is a verbal expression of the Law of Conservation of
Momentum.
It sounds like an experimental observation, which it is…
…which implies maybe we just haven’t done careful enough
experiments, and that maybe some day we will find the “law”
is not true after all.
But the Law of Conservation of Momentum is much more
fundamental than just an experimental observation.
If you assume that the laws of physics are invariant under
coordinate transformations, then the Law of Conservation of
Momentum follows mathematically and inevitably.
Every time I let you choose your coordinate system, I have
used that assumption of invariance.
If the assumption is false, then the laws of physics will be
different for everybody, and there is no point in doing
physics.
Any violation of the Law of Conservation of Momentum
would be as revolutionary (if not more so) as Einstein’s
relativity.
Most likely, any “new” laws of physics would contain all our
“old” ones, which would still work under “normal”
circumstances.
Our definition of system momentum, combined with the
relationship between force and momentum, gives
P
OSE :
F
=
 ext t
and our OSE expressing conservation of momentum is
OSE :
Pf = Pi if
F
ext
=0
Example 7-3. A moving railroad car, mass=M, speed=Vi1,
collides with an identical car at rest. The cars lock together as
a result of the collision. What is their common speed
afterward?
Step 1: draw before and after sketch.
Vi1
Vi2=0
M
M
before
Vf?
M M
after
Step 2: label point masses and draw velocity or
momentum vectors (your choice).
If I had made a pictorial sketch (i.e., drawn railroad cars), at
this point I would probably re-draw the sketch using two point
masses. For this example, I will stick with the above sketches.
x
Vi1
Vi2=0
M
M
x
Vf?
M M
before
after
Step 3: choose axes, lightly draw in components of
any vector not parallel to an axis.
Step 4: OSE.
OSE :
Pf = Pi because
F
ext
=0
Caution: do not automatically assume the net external force is
zero. Verify before using!
I will assume the friction in the wheels is negligible, so the net
force can be “zeroed out” here.
x
Vi1
Vi2=0
M
M
x
Vf?
M M
before
after
Step 5 will be applicable after we study section 7.3.
Step 6: write out initial and final sums of momenta
(not velocities). Zero out where appropriate.
Pfx = Pix
p2mfx = p1ix + p2ix
0
x
Vi1
Vi2=0
M
M
x
Vf?
M M
before
after
Step 7: substitute values based on diagram and solve.
p2mfx = p1ix
(2m)(+Vf) = m(+Vi1)
Vf = mVi1 / 2m
Vf = Vi1 / 2
7.3 Collisions and Impulse
The net force on an object is Fnet = F=p/t.
From this, we see that a force applied for a time t produces a
change of momentum p: Fnet t = p.
Impulse is defined as Fnet t. (Note that Fnet and Fexternal and
Fnet, external mean the same thing in this class.)
OSE :
J  Fnet t
I recently revised the OSE sheet to include the subscript “net”
on F.
The above equation is valid if the force changes little during
the time t.
If the force depends on time, we have to integrate (but we
won’t do that in Physics 31).
Demonstrate impulse with ball and bat.
You’ll see a more thorough discussion of impulse in the next
section.
Important: here is your litany for momentum problems.
Here is a link to a site that has lots of introductory-level
problems on momentum and impulse. (Here is a sample
problem. Click the letter of your choice. Find out if you are
correct, or what your mistake was.)