Transcript V p

Conservation of Energy
and Momentum
Elastic Collisions
Inelastic Collisions
Collisions in 2 or 3 Dimensions
7.4 Conservation of Energy
and Momentum in Collisions
In the previous lecture we found that external forces acting on a
system change the system momentum
Fext=P/t,
remember, bold font
means vector
and that system momentum is conserved if no external forces act
on a system
Pf = Pi if Fext=0.
The relationship between force and momentum led us to define
impulse as
J = Fnet t.
Impulse “explains” how objects or systems gain momentum.
Now we have the tools to handle collisions between objects and
systems.
A bat hitting a ball and a person jumping and striking the ground
are examples of collisions.
During the collision, squishy objects are deformed and forces are
complex. This is not an easy problem to solve.
Another example of a complex collision is two cars colliding,
deforming, and skidding.
In the case of a bat hitting a ball, the ball is deformed by the
force of the bat, and its momentum changed dramatically.
Let’s look at the ball/bat collision more closely.
Before the collision, the ball moves under the influence of gravity
and air resistance.
The force due to air resistance depends on velocity, and so is
time dependent, but over small time periods, the force due to air
resistance changes little.
Let’s graph these forces.
force
The impulse delivered to
the ball (change in
momentum) during a
time t is the area under
the F vs time curve.
Fgrav
Fair
t
time
When the ball is hit by the bat, contact time t is small, and the
force of the bat on the ball is huge compared to the forces due to
gravity and the air.
Fbat
The impulse delivered to
the ball by the bat is
typically many hundreds
force
or thousands of times
greater than the impulse
delivered by gravity and
the air.
Under these conditions,
we can ignore Jgrav and
Jair and keep only Jbat.
Fgrav
Fair
t
time
Now, back to our OSE’s:
Fext=P/t,
Pf = Pi if Fext=0,
J = Fnet t.
We can combine these into one “master equation” which
describes the change in system momentum during some process:
OSE:
Jext = Pf – Pi .
Jext (also called Jexternal, or Jnet, or Jnet, external) is the impulse
delivered to the system by all forces outside the system under
consideration.
What about internal forces, inside the system?
By Newton’s third law, every internal force is part of an actionreaction pair.
1 F12
F21
2
F12=-F21 so F12t=-F21t so J12=-J21. The net impulse
delivered by this pair of forces is J12+J21=0. Internal forces do
not change the system momentum. The particles may be
gyrating wildly around each other, but the system momentum
remains constant.
Demonstration: trust in physics, part II.
The friction force, acting over a very short time t, provided no
(or very little impulse) to the objects of the demonstration.
Back to collisions. “During the collision, squishy objects are
deformed by internal forces* and the internal forces are
complex. This is not an easy problem to solve.”
BUT if collisions happen fast enough, external forces (such as
friction) have little time to develop significant impulse. In that
case, Jext  0, and Pi = Pf (approximately).
*Here I’ve added the words in orange to the sentence on slide 2. Internal
forces are those between objects involved in the collision—not forces due to
outside objects.
We can finally write our comprehensive, all-inclusive form of the
Law of Conservation of Momentum:
OSE:
Fext = 0  Jext = 0  Pf = Pi .
This OSE says that if the net external force is zero, then the net
external impulse is zero, and system momentum is exactly
conserved.
It also says that if for some reason the net external impulse can
be approximated by zero, then system momentum is
approximately conserved (and the approximation may be
extremely accurate).
On exams or tests, I suggest you use
Jext
0
= Pf – Pi .
Ask yourself if you are justified in “zeroing out” Jext. If the
answer is yes, do it!
To be safe, somewhere in your solution you might want to make
a note why you zeroed out Jext.
I may not demand a reason for zeroing out Jext as part of your
solution. But don’t zero out Jext unless it is justified!
This section—less than a page in your text—took many slides.
We are not done with it, either!
It may be that kinetic energy is conserved in a collision. In that
case, the collision is “elastic.”
You can use Kf = Ki any time you know a collision is elastic.
Kf = Ki does not appear on your OSE sheet. It is actually the
definition of an elastic collision. Perhaps you should write that on
your next 3x5 card.
“So when can I assume a collision is elastic.”
Never! (Almost.) (The key word is “assume.”)
If the colliding objects are rigid, do not stick together, do not
expend energy deforming, and no energy is lost to heat or sound
in the collision, then “elastic” is a good approximation.
Examples of elastic collisions (or close approximations):
Billiard balls colliding.
Marbles colliding.
Carts on air tracks colliding bumper-to-bumper, when the
bumpers are made of rubber bands.
Atoms, molecules, protons, electrons, etc. colliding (as long
as they do not stick together).
If kinetic energy is not conserved in a collision, the collision is
“inelastic.”
Never use Kf = Ki if you know a collision is inelastic.
If the colliding objects deform, stick together, heat up, lose heat
or sound energy to their surroundings, etc., the collision is
inelastic.
Examples of inelastic collisions:
Your car and a truck.
You and a truck.
Chewing gum thrown at the wall.
Rain falling in a bucket.
Railroad cars being coupled together.
A basketball dropped on the floor.
7.5 Elastic Collisions in One Dimensions
Example 7-7. A proton of mass M traveling with a speed Vp
collides head-on with a helium nucleus of mass 4M at rest.
What are the velocities of the proton and helium nucleus after
the collision.
This is an example of an elastic collision. Why? That’s what
section it is in. The book tells you it is. Collisions between
neutral or like-charged nuclear-type particles are usually
elastic.
“Head-on” means the collision is not at a glancing angle, which
means all motion takes place in one dimension (say, the xaxis). Makes the problem much easier.
This is a momentum problem. Remember the litany!
Step 1: draw before and after sketch.
Vp
4M
M
before
Vpf
VHef
M
4M
after
I chose velocities “positive” (after I put in the x-axis) so I won’t
be tricked into putting in an extra – sign later.
Step 2: label point masses and draw velocity or
momentum vectors (your choice).
Already done!
x
x
Vp
Vpf
VHef
M
4M
4M
M
before
after
Step 3: choose axes, lightly draw in components of
any vector not parallel to an axis.
Step 4: OSE.
not a required step,
but I thought you
should see it
Jext
0
= Pf – Pi
0
Ef – Ei = [Wother]if
0
0
0
0
0 = Kf + Ugf + Usf – Ki – Ugi - Usi
x
x
Vp
4M
M
before
Vpf
VHef
M
4M
after
Continuing from previous slide…
Pf = Pi
Kf = Ki
Step 5: zero out external impulse if appropriate.
Already done!
x
Vp
x
VHe,initial=0
4M
M
before
Vpf
VHef
M
4M
after
Step 6: write out initial and final sums of momenta
(not velocities). Zero out where appropriate.
This is a combined momentum plus energy problem, but for
this step I will just write out the momenta. I’ll put velocities in
the next step.
0
ppfx + pHefx = pp + pHe,initial
Dang! My figure doesn’t justify zeroing out the initial He
momentum. Better fix that up right now!
x
Vp
x
VHe,initial=0
4M
M
before
Vpf
VHef
M
4M
after
Step 7: substitute values based on diagram and solve.
From conservation of momentum:
M Vpfx + (4M) VHefx = M Vp
Conservation of energy:
Kf = Ki
we want to find
Vpfx and VHefx
½M(Vpfx)2 + ½(4M)(VHefx)2 = ½M(Vp)2
M Vpfx + (4M) VHefx = M Vp
½M(Vpfx)2 + ½(4M)(VHefx)2 = ½M(Vp)2
Vpfx + 4VHefx = Vp
----- (1)
(Vpfx)2 + 4(VHefx)2 = (Vp)2
----- (2)
Two equations, two unknowns. How would you solve?
You would probably solve (1) for Vpfx (or VHefx) in terms of the
other unknown, plug the result into (2), and solve the
quadratic.
Not impossibly difficult here, but what do you do about the 
sign? And what about more complex problems, where the
helium initial velocity is nonzero?
There’s a mathematical “trick” (actually, technique) that works
well here and is the only way to get an easy solution in more
complex problems.
The technique is used on page 188 to derive equation 7-7,
which Giancoli uses a number of times later on. It’s not on
your OSE sheet because it is not “fundamental.” If I want you
to use the technique (shown in a minute) I’ll let you know
(probably in a hint).
Winter/spring 2004, skip to slide 25.
Vpfx + 4VHefx = Vp
----- (1)
(Vpfx)2 + 4(VHefx)2 = (Vp)2
----- (2)
To better illustrate the technique, on the next page I will use m
in place of the proton mass, and M in place of the helium mass,
because in general the masses don’t cancel so conveniently.
mVpfx + MVHefx = mVp
----- (1)
m(Vpfx)2 + M(VHefx)2 = m(Vp)2
----- (2)
Group all terms with m on one side, all terms with M on the
other side (especially useful if initial velocity of M is nonzero,
because the quadratic would be very difficult).
mVpfx - mVp = -MVHefx
----- (1)
m(Vpfx)2 - m(Vp)2 = -M(VHefx)2
----- (2)
The left hand side of equation (2) can be factored
mVpfx - mVp = -MVHefx
m(Vpfx+Vp)(Vpfx-Vp) = -M(VHefx)2
----- (1)
----- (2)
Neither side of equation (1) is zero (why?) so divide (2) by (1)
to give…
mVpfx - mVp = -MVHefx
(Vpfx+Vp) = VHefx
----- (1)
----- (2)*
Now I can plug VHefx from (2) back into (1) to get Vpfx:
mVpfx - mVp = -M (Vpfx + Vp)
mVpfx - mVp = -MVpfx - MVp
mVpfx + MVpfx = mVp - MVp
(m+M)Vpfx = (m-M)Vp
Vpfx = (m-M) Vp / (m+M)
You could use this Vpfx in (1) or (2) to get VHefx.
*Wow! No more quadratic. Just a simple 2 equation, 2 unknown system!
mVpfx - mVp = -MVHefx
----- (1)
(Vpfx+Vp) = VHefx
----- (2)
I find it easier to solve (2) for Vpfx and get VHefx from (1):
m(VHefx-Vp) - mVp = -MVHefx
mVHefx - 2mVp = -MVHefx
mVHefx + MVHefx = 2mVp
(m+M)VHefx = 2mVp
VHefx = 2mVp / (m+M)
When I set the problem up, I said M=4m. In the text, Giancoli
uses M=4u and m=1.01u (u=1.66x10-27 kg).
I haven’t boxed my answers, because I am not quite done yet.
Vpfx = (m-M) Vp / (m+M)
VHefx = 2mVp / (m+M)
M>m so Vpfx is negative, but VHefx is positive.
That means the lightweight proton bounced back to the left
after it collided with the helium, and the helium was given a
“kick” to the right.
That makes sense, so now I put a box around my answers.
You can review the algebra that got this result. I’ll design test/quiz problems so that
you don’t have to go through such a complex algebraic process in a limited time
and under pressure.
Abbreviated Litany for Momentum Problems
Step 1: draw before and after sketch.
Step 2: label point masses and draw velocity or momentum
vectors (your choice).
Step 3: choose axes, lightly draw in components of any
vector not parallel to an axis.
Step 4: OSE.
Step 5: zero out external impulse if appropriate.
Step 6: write out initial and final sums of momenta (not
velocities). Zero out where appropriate.
Step 7: substitute values based on diagram and solve.
7.6 Inelastic Collisions
Example 7-8. A railroad car of mass M moving with a speed
V collides with a stationary railroad car of mass M. Calculate
how much kinetic energy is transformed to thermal or other
forms of energy.
In lecture 16, we used conservation of momentum to show
that the two railroad cars move together with a speed of V/2
after the collision.
x
x
Vi1=V
Vi2=0
Vf=V/2
M
M
M M
before
after
x
x
Vi1=V
Vi2=0
Vf=V/2
M
M
M M
before
after
Now we look at the change in kinetic energy…
K = Kf - Ki
K = ½(2M)(V/2)2 – ½MV2
K = ¼MV2 – ½MV2
K = -¼MV2 = -½Ki
Half the initial kinetic energy is “lost” in the collision.
Example 7-9. A bullet of mass m and speed v is fired into
(and sticks in) a hanging block of wood of mass M. The wood
is initially at rest. Calculate the height h the wood block rises
after the collision.
The collision between the bullet and block is handled using
Jext = Pf – Pi .
After the collision, the motion of the block+bullet is handled
using
Ef – Ei = [Wother]if .
I’ll work this one on the blackboard.
7.7 Collisions in Two or Three Dimensions
We’ll skip this section.
To solve a collision in more than one dimension, just solve for
each component separately.