Transcript lec09a

Retrospective
Before we continue with momentum and impulse, let’s
step back and think about where we have come from.
What tools do you have in your toolbox?
 English!
 algebra and trigonometry
 kinematics (motion without worrying about forces)
 dynamics (Newton’s laws, forces)
rotational dynamics is a subset of this
 conservation of energy (work and energy)
We have just added another tool—conservation of
momentum.
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.
In 1905, mathematician Emmy Noether proved the
following theorem:
For every continuous symmetry of the laws of physics,
there must exist a conservation law.
For every conservation law, there must exist a
continuous symmetry.
The conservation law corresponding to time translational
symmetry is the Law of Conservation of Energy (we have
seen a “special case” of this law—conservation of
mechanical energy).
The conservation law corresponding to space
translational symmetry is the Law of Conservation of
Momentum.
These conservation laws “emerge from symmetry
concepts far deeper than Newton’s laws.”
Impulse Example: 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.
Important: here is your litany for momentum problems.
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
m
x
x
before
after
Step 3: choose axes, lightly draw in components of any
vector not parallel to an axis.
Step 4: OSE.
OSE:
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,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
Conservation of Momentum Example: 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
Vf
M
M
M M
before
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.
Vi1
Vi2=0
M
M
x
Vf
x
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 Fext=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.
Vi1
Vi2=0
M
M
Vf
x
x
M M
before
after
Step 5 will be applicable later.
Step 6: write out initial and final sums of momenta (not
velocities). Zero out where appropriate.
Pfx = Pix
0
p2mfx = p1ix + p2ix
Vi1
Vi2=0
M
M
Vf
x
x
M M
before
after
Step 7: substitute values based on diagram and solve.
(2m)(+Vf ) = m(+Vi1)
Vf = mVi1 / 2m
Vf = Vi1 / 2