Transcript lec10a

Let’s look at a bat-ball 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.
The impulse delivered
to the ball (change in
momentum) during a
time t is the area
under the F vs time
curve.
force
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
force
hundreds 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
What about internal forces, inside the system?
By Newton’s third law, every internal force is part of an
action-reaction 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.
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.
“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 or
expend energy deforming, and no energy is lost to heat
or sound in the collision, then “elastic” may be 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.
Examples (to be worked if time permits)
Example: 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 x-axis). Makes the problem much easier.
This is a momentum problem. Back to the litany!
Step 1: draw before and after sketch.
Vp
M
before
4M
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
4M
M
Vpf
VHef
M
4M
before
after
Step 3: choose axes, lightly draw in components of any
vector not parallel to an axis.
Step 4: OSE.
not really a
necessary step,
but I thought you
should see it
0
Jext = Pf – Pi
0
Ef – Ei = [Wother]if
0 = Kf + Ugf
0
0
0
0
+ Usf – Ki – Ugi - Usi
x
x
Vp
M
4M
before
Vpf
VHef
M
4M
after
Continuing from previous slide…
Pf = Pi
Kf = K i
Step 5: zero out external impulse if appropriate.
Already done!
Vp
x
VHe,initial=0
M
before
4M
x
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!
Vp
x
VHe,initial=0
x
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 = K i
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 in the physics 31 text to derive an
equation which is used a number of times later on. It’s
not on your OSE sheet because it is not “fundamental.”
Besides, I want you to see this handy technique.
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 powerful if initial velocity of M
is nonzero).
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
----- (1)
(Vpfx+Vp) = VHefx
----- (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.
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)
Back on slide 7, I said M=4m. In the textbook problem,
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.
Sample Inelastic Collision Problems
Probably out of time today!
Example: 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.
Example: 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.
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.