Tangible: Get that rebound
(similar to Lab: Throw Ball Up)
Topic of Ch 10 is collisions: big forces for short times. This chapter does not
introduce any new fundamental principles. Rather we will apply the familiar
momentum and energy principles to a novel kind of situation, collisions.
Students bounce racquetball from 1 m and use energy principle to determine speeds
just before and after impact. Show just ball (even tables) and also universe (odd
tables) as system choices, as below. It hits at 4.4 m/s and typically bounces to 0.75 m
for an upward speed of 3.3 m/s.
DE = Wexternal + Q
K f - K i = mgDy
1
2
mv 2f = mgyi
DE = Wexternal + Q
DK + DU gravity = 0
1
2
mv 2f - mgDy = 0
v f = 2gDy
v f = 2gDy
Students do:
- get a racquetball and a ruler and drop the ball from 1 m
- Use energy principle: Find the speed just before and just after impact with
floor
- Now use momentum principle: Find the force of the floor
o You need to estimate the collision time—HOW? Just guess?
o No, estimate compression distance (time of approx. 1 millisecond)
o Assuming constant force, average speed is ½ initial from just before
to full stop
o Note large force, much greater than mg. Typical value is 400 N = 100
lb.
Dp = Fnet, external Dt
p yf - p yi = Fy Dt
(
)
m v yf - v yi = Fy Dt
Fy =
Chapter 10
(
m v yf - v yi
Dt
)
1
- Use momentum principle on system of ball and earth. What is the change in
momentum of the earth during the collision? What is the change in speed of
the earth during the collision?
Discussion: Collisions
Important special case of a collision between objects in an isolated system: total
momentum of the system doesn't change if there is no net force, since d ptotal dt = Fnet .
No net force, no change in total momentum.
Example of midair collision of two baseballs. Gravitational impulse during collision
is negligible compared to the huge impulses due to the contact forces. So from just
before to just after the short-time-duration collision, no significant change in the total
momentum of the system consisting of the two balls, even though over longer times
there is a change (see Figure 9.2 on p. 304).
Collisions in physics sense need not involve contact; e.g. two charged particles.
Distinguish between elastic (no change in internal energy) and inelastic collisions.
FOR THE DROPPED BALL COLLISION YOU JUST DID:
-Ki_earth about 0, so is Kf_earth. What’s Ki_ball & Kf_ball? Is K conserved during
the collision?
Tangible: Crashing the party
Have students collide two magnetic bumper carts (using extra masses) on low
friction track. Use the Colliding Carts Handout to help them organize their data. Two
carts, track, and masses needed for each table.
Even numbered tables:
Do the cases in the top row: vinbound,x > 0, vtarget,x < 0, for the five mass cases
Odd numbered tables:
Do the cases in the bottom row: vinbound,x > 0, vtarget = 0, for the five mass cases
Come to agreement at the table. Report on:
vinbound,x > 0, vtarget,x < 0
Table 1: m=m
Table 3: m_in>m_target
Table 5: m_in<m_target
Table 7: minbound >> mtarget
Table 9: minbound << mtarget
vinbound,x > 0, vtarget = 0
Table 2: m=m
Table 4: m_in>m_target
Table 6: m_in<m_target
Table 8: minbound >> mtarget
Table 10: minbound << mtarget
Discuss results: Key idea is that momentum is conserved, and in this case of repulsive
Chapter 10
2
magnets colliding, K is conserved too—try it for a few cases (eg. equal masses, target
really large). These are all (mostly) elastic collision.
Now let’s try something different: Pass out Matter Models. Place one on each cart.
Don’t go through all the cases again, just try one or two.
Question: are the collisions now elastic? NO!
Where is the kinetic energy going? Into thermal energy.
Demo/Discussion: Ping-pong ball/basketball collision
Drop a ping-pong ball onto a basketball.
Discuss mainly qualitatively the collision of a ping-pong ball and a bowling ball.
Throw the ping-pong ball at the bowling ball and see it bounce back with nearly the
same kinetic energy but BIG change in (vector) momentum. Draw diagram for before
and after momenta of ping-pong ball and invite students to draw a correct vector for
the final momentum of the bowling ball if in outer space. Students are surprised that
the bowling ball gets a lot of momentum (nearly 2p). Say that the textbook does a
quantitative treatment in the case of no change in total kinetic energy, but emphasize
that this is a very special case, not some general set of formulas for collisions in
general.
Throw tennis ball against the wall, invite students to think about change in momentum
and change in kinetic energy for the ball, then later for the Earth, and for the Earthball system. Somewhat amusing to consider both Earth and ball as we go through the
whole cycle of throwing the ball, then the ball in the air headed toward the wall, then
the collision, then coming back in the air, then being caught: end up as we were.
Ponderable: Rutherford scattering
Consider the Rutherford experiment with an alpha particle of mass m with known
initial momentum < p1, 0, 0 > interacting with a stationary gold nucleus and scattering
through an observed (and therefore known) scattering angle . (Note: px = p cos q
and p y = p sin q .) Assume low speeds (nonrelativistic) and elastic collision. Write
down the equations that can be derived starting from the momentum principle and the
energy principle. Show that there are enough equations to solve (in principle!) for the
x and y components of the momentum of the gold nucleus. Don't attempt to solve the
equations.
Chapter 10
3
p1i
m
(alpha)
p2i = 0 M (gold)
before
after
p1 f
p2 f
Only looking at the x components:
p1i + 0 = p1f + p2f and K1i + 0 = K1f + K2f.
2
2
2
2
2
p
p
p
p
p
æ
ö
1
f
2
Since K = 12 mv = 12 m ç ÷ =
we have 1i + 0 =
+ f
è mø
2m
2m 2M
2m
SKIP THIS
Light projectile, heavy target (use momentum principle and energy principle, and M
at rest, and “Elastic” Eint = 0) not during collision
2
ém ± M ù
p1 f = ê
ú p1i » - p1i for m << M and
ëm + M û
(+ means alpha did not interact with gold nucleus – “passes through”)
(- means [ ] ~ -1 for m << M)
é 2M ù
p2 f = ê
ú p1i » 2 p1i for m << M
m
+
M
ë
û
(px for alpha = -2p1i)
Heavy projectile, light target
é M ± mù
p1 f = ê
ú p1i » p1i for m << M and
ëM + mû
(+ means no interactions)
(- means [ ] slightly less than +1; gold moves with slightly less
momentum)
é 2m ù
m
p2 f = ê
p1i » 2 p1i for m << M
ú
M
ëM + mû
mv2 f » 2
m
Mp1i
M
v2 f » 2v1i
Chapter 10
4
Main point as to why Rutherford was surprised: He thought he was firing bowling
balls at ping pong balls, in which case nothing should have bounced back.
VPython Demo: Scattering
alpha_on_alpha.py
Rutherford_dist.py
Rutherford_dist_pudding.py
alpha_on_electron.py
Scattering: Explain the geometry of impact parameter b and scattering angle. Show
alpha particle on a particle that is 8 proton masses (program alpha_on_alpha.py).
Start with zero impact parameter, then show b = 2e-15 m, then b = 40e-15 m.
Establish that larger impact parameter gives smaller scattering angle. Then set target
to be another alpha particle and show what happens: emphasize that the 90 degrees
between the outgoing momenta is VERY special; only for equal masses. Note that
even though the collision is elastic, during the collision the kinetic energy does
decrease (as electric potential energy increases). “Elastic” means that there is no
change in internal energy from the initial state (far away) to the final state (far away).
Set target mass to 0.1 proton masses to show again what happens when a heavy
projectile interacts with a light target.
Scattering distributions: Show program Rutherford_dist.py (could hand out red-cyan
glasses and use scene.stereo = 'redcyan'). Emphasize that the distribution tells us about
the nature of the interaction.
Show program Rutherford_dist_pudding.py in which the target is a single proton.
Footnote to the main issues: Show program alpha_on_electron.py to emphasize that
in Rutherford scattering you can forget the atomic electrons,
which are swept away due to their low mass, with little effect on
the alpha trajectory. The collision occurs deep inside the atom. (If
you let the program run long enough, you see that the two
particles orbit each other; this would not happen in the more
complicated environment of a solid.) Note that this is essentially a
Chapter 10
5
bowling ball hitting a ping-pong ball: the bowling ball’s velocity hardly changes, but
the ping-pong ball moves twice as fast as the bowling ball.
Also stress difference between nuclear scale (10–14 m) and atomic scale (10–10 m).
Show that we can ignore all but one nucleus in the scattering. At far left of
blackboard draw a dot 0.5 mm in radius to represent one gold nucleus. Have students
calculate how far away, to scale, (the next gold nucleus should be drawn. This is about
5 meters! (Factor of about 104). Draw another tiny dot far away. The essence of the
Rutherford experiment is one alpha at a time colliding (electrically) with just one gold
nucleus, and no electrons around. Even if the 79 protons were spread apart, they too
would be nearly isolated.
Clickers
Ponderable: Activity - Solar sailing
WID 1157084 (Problem 10.P.22) sail
Discuss
10.P.22 Solar sailing, with the reminder that since
E - ( pc) = (mc ) , for massless particles such as the photon E = pc, so p = E/c.
2
Prob.
2
2 2
(a) If the photon is absorbed, the sail’s momentum increases by p, the momentum of
Chapter 10
6
the absorbed photon, since the total momentum of the photon + sail system does not
change. But if the photon is reflected backwards, it is like a ping pong ball hitting a
bowling ball, and the magnitude of the photon’s final momentum is approximately p.
The change in the photon’s momentum is therefore -p-(-p ) = -2p, and the change in
the momentum of the sail is 2p, since the total momentum of the photon + sail system
does not change. Therefore we get twice the kick with a shiny sail.
(b) Each photon gives the sail an increase of 2p in the sail’s momentum. If there are N
photons hitting the sail per second, the rate of momentum change for the sail is N(2p),
and since the rate of change of the momentum is equal to the net force, we have
F=N(2p). A photon of energy E has momentum p, so we have F=N(2E/c).
E = pc incoming, so change in momentum with silver sail is 2 p
2p
E
=2
1
c
2
A comment: Rearrange to write F = ( NE ) and note that NE is the energy per
c
second hitting the sail. If the solar sail is about as far from the Sun as the Earth is, this
number is NE = 1400 J/s on an area of 1 square meter (1400 W/m2). Suppose we build
a solar sail a kilometer on a side, and the payload is 100 kg. We can calculate
acceleration of the sail:
Dt = 1 s, so from Dp = FDt, we have F =
(
)(
)(
)
2 1400 s×mJ 2 10 3 m 10 3 m
F
2
m
a= =
=
0.1
( NE ) =
m mc
s2
(100 kg ) 3 ´ 10 8 ms
(
)
That’s a small acceleration, but in a month it leads to a high speed:
æ
mö
hr ö æ
min ö æ
s ö
æ
4 m
çè 0.1 ÷ø ( 30 days ) ç 24
çè 60
÷ø çè 60
÷ø = 2.6 ´ 10
÷
è
s
day ø
hr
min
s
Chapter 10
7