Discussion: Energy principle
Write out in words:
Dp = Fnet Dt
Change of ?? = force times displacement (“work”) DE = Fnet · Dr
Change of momentum = force times time (“impulse”)
The ?? is energy. This is the second great principle, the energy principle.
If you know times, impulse and momentum change may be more useful.
If you know displacements, work and energy change may be more useful.
Discussion: Where does energy come from?
Let’s see where we energy comes from. Consider forces and displacements in the x direction (+ or ). Then dE = Fxdx = (dpx/dt)dx, which gives dE/dx = dpx/dt. Look for some quantity whose
derivative with respect to x is the same as the derivative of the x component of momentum with
respect to time. The “particle energy” turns out to be
1
E = g mc 2 =
2
1-
v
c2
mc 2
Next consider the splitting of E into rest and kinetic terms.
If v = 0, E = mc2 , one of the most famous of all equations, but for a particle this is a special case.
This is called the “rest energy.”
If speed is not zero, there is additional particle energy, “kinetic” (motional) energy K:
E = mc2 + K
So... K = E - mc 2 =
1
1- v / c
2
mc 2 - mc 2
2
Tangible: Fixing book example
Fixing book example.pdf (on pg 222-223, Figure 6.1), the drawing isn’t very well scaled.
A proton in a particle accelerator has a speed of 2.91e8 m/s.
a) What is the energy of the proton?
E = g mc 2 =
1
8 2
(2.91´10 )
1(3 ´108 )2
Etotal = 4.11Erest
(
)(
mc 2 = 4.11mc 2 = 4.11 1.7 ´10 -27 3 ´108
)
2
= 6.29 ´10 -10 J
(diagram in book is not to scale)
b) What would the energy of the proton be if it were sitting still?
(
)(
Erest = mc2 = 1.7 ´ 10 -27 3 ´ 10 8
)
2
= 1.53 ´ 10 -10 J
c) What is the kinetic energy of the moving proton?
2
K = Etotal - Erest = 4.11mc2 - mc2 = 3.11(1.7 ´ 10 -27 ) ( 3 ´ 10 8 ) = 4.76 ´ 10 -10 J
Ponderable: Activity - Rest energy vs. kinetic energy
WID 1130188 zoomer
Which is bigger, Erest or K? It depends on speed. When are they equal?
Chapter 6
1
K = E - Erest = g mc 2 - mc 2 if K = Erest = mc 2
mc 2 = g mc 2 - mc 2
1
1 = g - 1 so 2=g =
2
1-
2
1
4=
v
c2
so 4 - 4
2
2
v
v
= 1 so 3 = 4 2
2
c
c
v
c2
v = 0.75c = 0.86c = 2.6 ´ 10 8
1-
m
s
below this speed, Erest is bigger than K, above this speed K is bigger
Tangible: Activity - Where's the nameblock hiding all that energy?
WID 1130159 bigE
Calculate the rest energy of your nameblock (m = .033 kg, mc2 = 2.97x1015 J). How much is this?
Consider that 1 watt is 1 J/s, and most compact fluorescents use about 20 watts of power. How long
would you be able to light a thousand such lightbulbs using the energy stored in one nameblock?
(about 4700 years!). The solution to our energy crisis?
Now, how fast would a nameblock need to go to have that much kinetic energy (Ponderable:
Activity - Rest energy vs. kinetic energy says 2.6e8 m/s)
Discussion: Non-relativistic approximation to the energy
Binomial Expansion (pg 223):
n
1
(1 + e )n = 1 + e +
æ é v ù2 ö
ç1 - ê ú ÷
è ëcû ø
-
1
2
= 1+
+
n ( n - 1) 2 n ( n - 1) ( n - 2 ) 3
e +
e +
2 ´1
3´ 2 ´1
( - 12 ) æ - é v ù2 ö + ( - 12 ) ( - 12 - 1) æ - é v ù2 ö
ç
÷
1 è êë c úû ø
2
ç ê ú ÷
è ëcû ø
2 ´1
( - 12 ) ( - 12 - 1) ( - 12 - 2 ) æ - é v ù2 ö
3
ç ê ú ÷ +
è ëcû ø
3´ 2 ´1
2
4
6
»1+
1év ù
3é v ù
5 évù
+ ê ú + ê ú + ...
2 êë c úû
8 ë c û 16 ë c û
»1+
1év ù
2 êë c úû
2
So if v << c,
æ é v ù2 ö
E=
= ç1 - ê ú ÷
2
è ëcû ø
æ vö
1- ç ÷
è cø
mc 2
-
1
2
2
æ
1év ù ö
1
2
mc 2 » ç 1 + ê ú ÷ mc 2 = mc 2 + m v
2ë c û ø
2
è
Have students calculate the factors
and
1
1 - v / c2
2
Chapter 6
2
1+
1 v
2 c2
for (v/c) = 0.01, which is a very high speed
2
yet the approximation works very well (1.000050004 and 1.00005).
2
Note however that
for (v/c) = 0.999 is 1.5, so the approximation doesn't work at all (
1v
1+
2 c2
we found to be 22.4).
1
1- v / c
2
2
2
So at low speeds K » 1 m v 2 » p (prove this by substitution;
2
2m
v =
p
m
).
Ponderable: Review + graphical thinking
Odd tables write down definition of momentum and a graph showing magnitude of momentum vs
speed. Even tables do the same thing showing total energy vs speed. Give them guidance like,
“What is the range of speed values for the horizontal axis?”
E = g mc 2 =
1
2
v
1- 2
c
1
p = g mv =
2
1-
v
c2
mc 2
mv
Discussion: Relationship between energy and momentum
Are energy and momentum related? Yes. This true for any speed:
(
E 2 - ( pc ) = mc 2
2
)
2
E
pc
mc2
Note that right side is constant, regardless of reference frame. So the left side is also. It is called an
invariant.
Clickers: Particle Energy Q6.2a to Q6.2e
Discussion: Work
Lift a block some distance, or push it some distance; lift it twice as far, or push it twice as far, do
twice as much “work.” Or apply twice the force through the same distance, do twice as much
“work.”
Speed something up by pushing it; do positive work.
Slow something down by pushing back on it; do negative work.
Chapter 6
3
Consider three simple cases:
Suppose I apply a force in the x direction, displacement in x direction, work is Fx Dx .
Suppose instead that you apply a force in the y direction, displacement in y direction, it is clear that
work is
Fy Dy .
Now suppose that while I do work Fx Dx you’re doing work
total? Evidently
Fy Dy . How much work was done in
Fx Dx + Fy Dy .
And if while you and I are doing this some third person applies a force in the z direction, through a
displacement in the z direction, what is the total work? Evidently we have the following, WHICH
SHOULD BE MEMORIZED:
W = Fx Dx + Fy Dy + Fz Dz
For obvious reasons, this is called the “scalar product” of two vectors.
Ponderable: Activity – Scalar product
WID 1110044
scalar
Dr = rf + ri = 2 - 3,10 - 7,12 - 4 m = -1,3,8 m
W = Fx Dx + Fy Dy + Fz Dz
(
(
)) (
) (
W = 10 N ´ -1 m + -20 N ´ 3 m + 30 N ´ 8 m
)
W = -10 Nm - 60 Nm + 240 Nm = 170 Nm = 170 J
It is also called the “dot product” for calculating work, with a compact abbreviation F · Dr :
Work = F · Dr = < Fx , Fy , Fz > · < Dx, Dy, Dz >= Fx Dx + Fy Dy + Fz Dz
It is EXTREMELY important not to write F · Dr = < Fx Dx, Fy Dy, Fz Dz > , because work is a
SCALAR, not a vector.
Equivalently, F · Dr = F Dr cosq , as can be seen for example in pulling a sled, where the sled
moves in the x direction but you apply a force at an angle. This is a quick derivation of the result.
Because we so often use the triple notation for vectors, the form W = Fx Dx + Fy Dy + Fz Dz will often
be the most useful, but if you know the magnitudes and angle between two vectors, it is more
convenient to use F · Dr = F Dr cosq
Is this a scalar or a vector? And therefore is work a scalar or a vector? How nice: a principle that is
easier to calculate with, because the energy principle is a scalar principle (but we will of course
continue to use the momentum principle as well).
Chapter 6
4
Tangible: Activity - Ain’t it a drag?
WID 1130012 dragon
Stick a piece of tape to their nameblock and drag it across the table. How much work do they do?
They have to estimate amount of force and establish an angle and distance. The force they can
estimate by comparing to the force required to hold the block vertically.
W = FiDr = Fx Dx + Fy Dy + Fz Dz = Fx Dx
(
)
Dt = ( F
(
)
W = 0.10 N cos30 0.20 m = 0.017 J = 17 mJ
Dp = Fnet
Tape
)
+ FEarth Dt
(
)(
)
0,0,0 = 0, FTape - FEarth ,0 Dt = 0, FTape - 0.036 kg 9.8 kgN ,0 Dt
)(
(
)
FTape = 0.036 kg 9.8 kgN = 0.35 N
Let’s say they decide to go with a 0.10 N force at a 30˚ angle, dragging the block 20 cm. Include
friction force. Carefully guide them through the FBD showing that horizontal (first) and vertical
forces must cancel. Then just focus on the force the tape applies.
30˚
Ponderable: No work
Discuss where we see cases of a force doing no work, including changing the momentum without
changing the energy.
a) How much work done holding a 1kg book for 1 minute …
b) Circular pendulum, how much work done by string in 1 revolution …
c) How much work done by Sun on the Earth during 1 orbit
Clickers: Work Q6.3a to Q6.3i
---end of test 2--Discussion: It’s the principle of the thing
Odd tables write down momentum principle. Even tables write down energy principle.
Dp = Fnet Dt
DEsys = Wsurr = Fnet iDr
Rotate notebooks, distinguish definition from principle and explain differences between the two
principles. Note how different the two sides of the equal sign are for the principles.
Chapter 6
5
Ponderable: Activity - Jack and Jill at work (was in practice test)
WID 1110057 rj47
Jack and Jill are maneuvering a 3000 kg boat near a dock. Initially the boat’s position is <2, 0, 3> m
and its speed is 1.3 m/s. As the boat moves to position <4, 0, 2> m, Jack exerts a force <-400, 0,
200> N and Jill exerts a force <150, 0, 300> N.
(a) How much work does Jack do?
Dr = rf - ri = 4,0,2 m - 2,0,3 m = 2,0,-1 m
WJack = FJack · Dr = -400,0,200 N · 2,0,-1 m = -800 J - 200 J = -1000 J
(b) How much work does Jill do?
WJill = FJill · Dr = 150,0,300 N · 2,0,-1 m = 0 J
(c) Assuming that we can neglect the work done by the water on the boat, what is the final speed of
the boat?
System: the boat
Surroundings: Jack, Jill
Initial state: just before they push
Final state: after the boat moved to its final position
DEsys = Wsurr
(m
boat
c2 + K
)
final
(
- mboat c 2 + K
)
initial
= WJack + WJill
K f = K i + WJack
1
2
mboat v 2f = 12 mboat vi2 + FJack · Dr
v 2f =
vf =
vf =
vf =
2
mboat
(
1
2
mboat vi2 + FJack · Dr
)
( (3000 kg)(1.3
2
( 3000 kg ) (1.3
3000 kg (
2
3000 kg
(
1
2
m
s
1
2
m
s
)
)
) - 1000 J )
)
2
- 1000 J
2
(
2
2
2535 J - 1000 J =
1535 J
3000 kg
3000 kg
)
v f = 1.0 ms
(d) Without doing any calculations, determine the angle between the (vector) force that Jill exerts
and the (vector) velocity of the boat. Explain briefly how you know this.
Work done is zero, so perpendicular
Chapter 6
6
Ponderable: Activity - Accelerating electron
WID 1130214 rk47
Apply the energy principle to Problem 5.P.72, an electron in an accelerator.
An electron is traveling at a speed of 0.95c in an electron accelerator. An electric force of 1.6e-13 N
is applied in the direction of motion while the electron travels a distance of 2 m. What is the new
speed of the electron?
System: the electron
Surroundings: the accelerator
Initial state: just before the push
Final state: after force applied for 2 meters
(
)
W = Fx Dx = 1.6 ´ 10 -31 N ( 2m ) = 3.2 ´ 10 -13 J
Ei =
1
mc 2 =
1
( 9.1 ´ 10 ) ( 3 ´ 10 )
-31
1 - ( 0.95 )
v
1 - i2
c
W = DE = E f - Ei or E f = Ei + W
2
2
E f = 2.623 ´ 10 -13 J + 3.2 ´ 10 -13 J = 5.823 ´ 10 -13 J =
8 2
= 2.623 ´ 10 -13 J
1
1-
vf
2
mc 2
c2
v f = 0.986c = 2.96 ´ 10 8 m/s
Ponderable: Activity - Electron meets some relative resistance
WID 1130241 tyw3
An electron traveling at a speed 0.99c encounters a region where there is a constant electric force
directed opposite to its momentum. After traveling 3 m in this region, the electron’s speed was
observed to decrease to 0.93c. What was the magnitude of the electric force acting on the electron?
System: the electron
Surroundings: the region with the field
Initial state: electron at initial position, with initial speed
Final state: after electron moved 3 m in the region
DESys = WSurr
E f - Ei = Fnet · Dr = - F Dx
æ
ö æ
ö
1
1
melectron c 2 ÷ - ç
melectron c 2 ÷ = - F Dx
ç
è 1 - 0.932
ø è 1 - 0.99 2
ø
( 2.72) (8.2 ´ 10
-3.6 ´ 10
-13
-14
) (
)(
)
J - 7.09 8.2 ´ 10 -14 J = - F Dx
J = - F Dx
3.6 ´ 10-13 J 2
F=
= 1.19 ´ 10 -13 N
3m
non-relativistic (wrong):
Chapter 6
7
DESys = WSurr
(m
electron
c2 + K
)
final
(
- melectron c 2 + K
)
initial
= WSurr
K f - K i = Fnet · Dr = - F Dx
1
2
1
2
melectron v 2f - 12 melectron vi2 = - F Dx
(9.1 ´ 10
3.5 ´ 10
F=
-14
)(
-31
kg 0.93 ´ 3 ´ 108
J - 4.0 ´ 10
-14
0.5 ´ 10
3m
J
-14
)
2
-
1
2
(9.1 ´ 10
-31
)(
kg 0.99 ´ 3 ´ 108
)
2
= - F Dx
J = - F Dx
= 1.57 ´ 10 -15 N
Force that is really needed is 130 times non-relativistic value. The mass is greater at high speeds, so
takes more force to change its speed.
Ponderable: Activity – Piece-wise work
WID 1110083 (also Clicker Q6.7a)
push
You push a crate out of a carpeted room and along a tiled hallway.
While on the carpet you exert a force of 30 N and the crate moves 2 m.
While on the tile you exert a force of 12 N and the crate moves 8 m.
How much work do you do?
W = F1 · Dr1 + F2 · Dr2 = 30 N 2 m + 12 N 8 m = 60 N × m + 96 N × m = 156 N × m
How did you figure it out?
Calculate first work where force was constant.
Calculate 2nd work where force was a different constant.
Add them up
(
)( ) (
)( )
Discussion: Work done by non-constant force
Discuss piece-wise case such as the situation where you push along a slick floor with a small force,
then on a sticky floor with a large force:
W = F1 · Dr1 + F2 · Dr2 + F3 · Dr3 + ..... = å Fi · Dri
So how would you deal with work done by non-constant force?
Again, W=sum(F dot delta r), so just break into lots of small delta r’s where force is approximately
constant
f
Work done by non-constant force: W = F · dr
ò
i
VPython: Earth Flyby
(Work done by a non-constant force)
WID 1131084 (Starting from VPython_Flyby_shell.py) zoom
VPython_Flyby.py
Chapter 6
8
Lab: Throw ball up
WID 817134 toss
This lab is designed to highlight the complementarity of the energy and momentum
principles. Students throw a ball in the air and measure the time of flight and the
maximum height. Then, they solve for the speed with which the ball was thrown using
each principle. They should solve algebraically using both principles before putting in
any numbers.
Their whiteboard work should reflect their choice of system (especially with energy):
System: ball only - no potential energy, but external work
System: ball and Earth - potential energy, but no external work
Discussion: Mass Change
Change in identity=mass change
Ef = Ei + W
mc2 + K = mc2 +K + W
Up to now, we canceled out the masses because they didn’t change, but what if they did?
Ponderable: Activity - Neutron Decay
WID 1130270 blip
Neutron stable inside nucleus. But by itself, it will decay in about 15 minutes. Get proton + electron
+ antineutrino. Energy principle still applies.
Discuss with students: What is initial and final state?
Now, you do it on whiteboard. Find the final KE of products.
Note particle physicists often use MeV (1 eV =1.6e-19 J) for unit of energy.
Neutron: 939.565 MeV
Chapter 6
9
Proton: 938.272 MeV
Electron: 0.511 MeV
Neutrino: ~0.05 eV: Very small compared to others, just say ~0
Do on whiteboards (10 min.)
Initially Ei = mnc2
Finally Ef = mpc2 + Kp + mec2 + Ke + Kantineutrino (mantineutrino very small but not quite zero). Since
the K’s aren’t zero, that says mn > mp + me so less mass than when you started.
mpc2 + Kp + mec2 + Ke + Kantineutrino = mnc2 + W
(938.3 MeV) + Kp +(0.511 MeV) + Ke + Kantineutrino = (939.6 MeV) + 0
(938.8 MeV) + Kp + Ke + Kantineutrino = (939.6 MeV) + 0
Kp + Ke + Kantineutrino = 0.8 MeV = 1.28e-13 J
That’s what happens in fission and fusion reactions. Reaction products less massive than starting
particles. Lots of energy output. Recall energy in nameblock calculation.
At first scientists didn’t see antineutrino. They thought energy principle didn’t work. Pauli predicted,
but not seen for 30 years.
Discussion: Potential energy
If the system contains more than one particle, there is “interaction” energy or “potential” energy U
associated with each pair of interacting particles. This additional energy can be positive (for objects
that push each other apart) or negative (for objects that attract each other, such as two stars).
(
)
D Eparticles + U pairs of particles = Wby external forces
Tangible: Internal and external forces and energy
Use 2 nameblocks to represent a proton and an electron. Have one student pull the proton one way
and another student pull the electron the other.
Manager: you pull on the proton
Skeptic: you pull on the electron
Recorder: you draw FBD of proton and electron
System = both proton and electron
Whole group: discuss and label forces:
- Which ones are INTERNAL forces (within the system)?
- Which ones are EXTERNAL forces (due to surroundings)?
F1
F2
W1
W2
F3
F4
W3
W4
Which forces and work are internal to system of both balls and which are external?
2 and 3 internal, 1 and 4 are external
Chapter 6
10
As proton and electron move, forces do work.
Whole group: discuss:
- Which forces do positive work?
- Which forces do negative work?
Write down the energy principle for system:
(
)
D EL + ER = W1 + W2 + W3 + W4
We would like to be able to express the energy principle in terms of what happens inside the system
(on the left of the “=”) and what the surroundings do to the system (on the right of the “=”). So we
move W2 and W3 to the left side of the equation:
(
) (
)
D EL + ER - W2 + W3 = W1 + W4
Next we define the “potential energy” of the system to be something whose change is equal to the
negative of the work done by the forces internal to the system:
DU º - (W2 + W3 )
Now we can write
D ( EL + ER ) + DU = D ( EL + ER + U ) = W1 + W4
(
)
D EL + ER + U = Wby external forces
We define the energy of the system to include not only the energies of the individual particles but
also their pair-wise interactions (one U term for each pair of particles):
Esystem = EL + ER + U
And finally we have the energy principle for a multiparticle system:
DEsystem = Wexternal
Clickers: System and Surroundings Q6.5a-6.8f
Discussion: Relationship between force and potential energy
Make plausible that the force associated with an interaction has the form
dU since DU = -F × Dr .
Fr = dr
Invite students to find a function U of r such that its derivative with respect to r gives the
gravitational force. Although I repeatedly asked the students to check their guess by actually taking
the derivative of their function, almost no one had the faintest idea how to do that!
Mm which is the derivative of some function of U,
F = -G
r
r2
so … - dU
dr
= -G
Mm ,
r2
so dU = G Mm
2
dr
r
And if we integrate both sides …
Chapter 6
11
U=ò
æ r -1 ö
dU
Mm
dr = ò G 2 dr = GMm ò r -2 dr = GMm ç ÷
dr
r
è -1 ø
U = -G
Mm
r
Ponderable: Closest approach
Set up for fusion reaction. Two widely separated protons, ri. Give each some K toward each
other. What happens? They slow down until they stop and reverse direction, almost like a spring.
We can use an energy argument to figure out separation at closest approach. Write down an energy
principle for slow speeds. Why not momentum principle? No time, just distance. Define system as
both charges, so no external work.
E f = Ei + W
m1c 2 + m2 c 2 + K1 f + K 2 f + U f = m1c 2 + m2 c 2 + K1i + K 2i + U i + W
mc 2 + mc 2 + 0 + 0 + U f = mc 2 + mc 2 + K1i + K 2i + 0 + 0
æ 1 ö q1q2
= K1i + K 2i
ç 4pe ÷
è
0 ø rf
started very far apart
only unknown is rf.
What if we knew how close we wanted them to be? You could find Ki. They should be close enough
to “touch” 10e-15m so that strong force takes over.
Ponderable: Activity - It’s rocket science!
WID 2399309
grphs
Consider a rocket blasting off from a planet. (Assume the launch happens very quickly and then the
rocket coasts away from the planet.) If we know the launch speed, what is the rocket’s final speed
for some distance r from the planet’s center? We could use the momentum principle, but the force
varies—we’d have to integrate or do it on a computer. Since distance is involved, we should more
likely use the energy principle. We will pick the system to be both the planet and rocket so no
external work is done (and thus we don’t have to worry about integrating the force.)
M
m
R
Do it on whiteboards
Work it out symbolically first
Then, solve for these given values: ri = 6.8e6 m, rf = 14e6 m, vi = 1e4 m/s (10 km/s).
Assume planet is earth.
Chapter 6
12
E f = Ei + W = Ei + 0
( Mc
2
+ K M + mc 2 + K m + U Mm
) = ( Mc
f
2
+ K M + mc 2 + K m + U Mm
)
i
( K m + U Mm ) f = ( K m + U Mm )i
1
m vf
2
vf =
2
-G
Mm 1
Mm
2
= m vi - G
rf
2
ri
since v << c
æ 1 1ö
2
vi + 2GM ç - ÷
è rf ri ø
For given values, vf = ~6e3 m/s.
Slows down—does this make sense?
We are assuming the planet’s kinetic energy KM doesn’t change much. The mathematical calculation
above was straightforward, so let’s also consider how things look graphically.
Plot U
vs. r (all do this)
U
As r  , U  
GMm
, which is negative, gets closer to zero
r
R
R is planet’s
radius
Fr  
dU
Mm
 G 2 , which is negative of the slope
dr
r
Ponderable: K+U plots
3 situations, similar to example we just worked:
Spacecraft is moving at vi at distance ri from center.
Group A: When it reaches rf, v=0
Group B: When it it gets VERY far away from planet, v=0
Group C: When it it gets VERY far away from planet, v=vf
Each group: Plot U vs r, K vs r and K+U vs r, all on same graph, for your given situation.
Think about following: What does energy principle predict for K+U?
Plot K vs. r
K is decreasing due to the planet pulling on the rocket. The way we’ve drawn this graph, the
rocket never actually stops moving. As the gravitational force gets very small, the rate of change
Chapter 6
13
of the rocket’s momentum becomes negligible—the speed becomes nearly constant and so does
K.
Because there are no external forces acting on the system, the total energy K + U is constant.
That tells us the shape of the K curve is just like that of U, except “flipped over” so that the two
curves added up will produce a horizontal line corresponding to the constant energy of the
system.
K
K
d
vf =
vi
2
æ 1 1ö
+ 2GM ç - ÷
è r f ri ø
r
r
R
Plot a separate graph with both K and U on it. It is easiest to plot one point at large
near zero, so K + U is just under the K graph.
r , where U is
K
K + U = constant
R
What would it look like with a smaller initial speed, so that K + U is negative? The rocket will
eventually slow to a stop and then fall back to the planet. This is called a bound state. The area to
the right of the red and blue lines is called the forbidden region. Again, the easy way to do this is to
pick some r such as the turnaround point. Here K = 0, so K + U is just the value of U at this point.
Chapter 6
14
K
moving away from planet
falling back toward planet
R
r
r
K can’t go negative
K + U = constant < 0
U =-
GMm
r
r
Bound State, K + U < 0
What happens if the launch speed is “just right” so that K approaches 0 at a very, very large distance
from the planet? Since we know U goes to zero when we are infinitely far from the planet, so must K
+ U. Since this is a constant value, the total energy is zero for any distance r from the planet.
K i + Ui = 0
Escape speed
1
Mm
2
m vi - G
=0
2
R
2GM
vescape =
R
K+U=0
K
R
In this borderline case, where the initial speed is just big enough to make K + U not negative (so this
is not a bound state) the initial speed is called the “escape speed.” For Earth, vescape = 11 km/s.
Chapter 6
15
Ponderable: Electrical force
Consider two protons that are initially a distance
r apart.
System is both charges
r
+
+
The electrical force is very similar in form to the gravitational force:
Felec =
1
Qq
4pe 0 r
2
(
-19
2
æ
ö 1.6 ´ 10 C
9 Nm
= ç 9 ´ 10
2
C2 ÷ø
è
r
)
2
This gives us the electrical potential energy as:
U=+
1 Qq
4pe 0 r
Since there are again no external forces, the total system energy is constant. Start by graphing the U
function, which is positive this time. It still approaches zero as the separation increases. When the
separation is very small, the potential energy is very large. As the protons move away from each
other, their kinetic energies rise, and their interaction energy (electric potential energy) falls.
Algebraically:
E f = Ei + W
m1c 2 + K1 f + m2 c 2 + K 2 f + U f = m1c 2 + K1 i + m2 c 2 + K 2 i + U i + 0
K1 f + K 2 f + U f = U i
K1 f + K 2 f = K f = U i
U
f
0
K + U = constant
K
U 
1 Qq
4 0 r
r
If time, do this problem: Two protons are fired straight at each other, each with a speed of 4e6 m/s.
How close do they get?
Plot U, total K, and K+U for this system.
Chapter 6
16
Ponderable: Activity - Typical ionization energy
WID 978510
eeVV
Typical ionization energy
System: outer electron and atom.
Initial state: electron in orbit r = 1e-10 m
Final state: electron at infinity at rest
Assume initial kinetic energy is small. Single ionization leaves ion with +e charge.
E f = Ei + W
mion c 2 + K ion f + me c 2 + K e f + U f = mion c 2 + K ion i + me c 2 + K e i + U i + W
U f = Ui + W
W=
1 qion qe
1 qion qe
= 0 - 9 ´ 109
4pe 0 rf
4pe 0 ri
(1.6 ´ 10 C) ( -1.6 ´ 10 C)
-19
N×m 2
C2
-19
1 ´ 10
-10
m
æ
1 eV ö
W = 2.3 ´ 10 -18 J ç
= 14 eV
è 1.6 ´ 10-19 J ø÷
Connect to real value of –13.4eV
Discussion: Review of formulae for potential energy and forces
Point out the similarities and differences.
Should memorize these 6 formulas.
Fs = ks s
Fg = G
U s = 12 ks2
m1 m2
r
æ 1 ö q1q2
Fel = ç
÷
è 4pe 0 ø r 2
2
U g = -G
m1 m2
r
æ 1 ö q1q2
U el = ç
÷
è 4pe 0 ø r
VPython: Space Voyage 3 (energy from the Earth to the Moon)
WID 1131056 (Starting from VPython: Space Voyage 2 and adding energy graphs)
SKIPPED in Spring 2012
VPython_Space_Voyage_3.py
Clickers: Q6.11aPonderable: Activity - Nuclear Fission
WID 1131015 (unscaffolded) or WID 1131019, a scaffolded version of the problem used in M&I
labs) boom
Applying the energy principle to a complex phenomenon (6.P.100 Nuclear Fission)
Chapter 6
17
Uranium-235 fissions when it absorbs a slow-moving neutron. The two fission fragments can be
almost any two nuclei whose charges Q1 and Q2 add up to 92e (where e is the charge on a proton, e =
1.6e-19 coulomb), and whose nucleons add up to 236 protons and neutrons (U-236; U-235 plus a
neutron). One of the possible fission modes involves nearly equal fragments, palladium nuclei (Pd)
each with electric charge Q1 = Q2 = 46e. The rest masses of the two palladium nuclei add up to less
than the rest mass of the original nucleus. (In addition to the two main fission fragments there are
typically one or more free neutrons in the final state; in your analysis make the simplifying
assumption that there are no free neutrons, just two palladium nuclei.)
The rest mass of the U-236 nucleus (formed from U-235 plus a neutron) is 235.996 u (unified atomic
mass units), and the rest mass of each of the two Pd-118 nuclei is 117.894 u, where 1 u = 1.66e-27
kg (approximately the mass of one nucleon).
Keep at least 6 significant figures, because the calculations involve subtracting large numbers from
each other, leaving a small difference.
Don’t plug in numbers until the end.
(a) Calculate the final speed v, when the palladium nuclei have moved very far apart (due to their
mutual electric repulsion).
(b) Using energy considerations, calculate the distance between centers of the palladium nuclei just
after fission, when they are momentarily at rest.
(c) A proton or neutron has a radius of roughly 1 x 10-15 m, and a nucleus is a tightly packed
collection of nucleons. Experiments show that the radius of a nucleus containing N nucleons is
approximately (1.3 x 10-15 m) x N1/3. What is the approximate radius of a palladium nucleus? Draw a
sketch of the palladium nuclei in part (b), and label the distances you calculated in parts (b) and (c).
d) As a check, and for further practice, find the distance between centers of the palladium nuclei just
after fission, when they are momentarily at rest using state 2 as your initial state and state 3 as your
final state. This should agree with your results in part (b).
GATHER:
MU-236 = 235.996 u; MPd-118= 117.894 u; 1 u = 1.66e-27 kg;
QPd-118 = 46 e; e = 1.6e-19 coulomb
It is ok to use the nonrelativistic formulas, but must check that the calculated v is indeed small
compared to c. (The large kinetic energies of these palladium nuclei are eventually dissipated into
thermal energy of the surrounding material. In a nuclear reactor this hot material boils water and
drives an electric generator.)
ORGANIZE:
Chose a system: all the particles (no external work)
Identify initial and final states (there are three choices):
The U-238 nucleus before it fissions
Just after fission when two palladium nuclei are close together and momentarily at rest.
The palladium nuclei are very far away from each other, traveling at high speeds
Chapter 6
18
Initial State: State 1
Final State: State 3
Draw appropriate diagrams
Identify appropriate Fundamental Principle: Energy Principle
ANALYZE:
The analysis can be thought of as a diamond.
Write a compact statement of the energy principle for your choice of system and initial and final
states.
Expand to include all the possible energy terms.
Rewrite with appropriate subscripts for the situation.
Contract by evaluating specific terms.
Solve for the unknown quantity of interest.
E f = Ei + Wext
( m c + K ) + ( m c + K ) + ... + U
= ( m c + K ) + ( m c + K ) + ... + U
2
1f
2f
2
1i
2f
12 f
+ ...
2i
12i
+ ... + Wext
2
1f
2
1i
2i
Rewrite with appropriate subscripts for the particular situation.
Cross out any terms that are zero; write specific potential energy terms.
Solve for unknown.
Plug in numbers.
LEARN: Do the units make sense? Did the speed of each palladium nucleus turn out to be small
2
enough that 1 mv 2 or p was an adequate approximation for the kinetic energy of one of the
2
2m
palladium nuclei? Is the final speed a high speed? (High speed goes with lots of heating of the metal,
which can ruin electric generators.)
Ponderable: CH 6 HW 6 (2216238
Completely redo CH 6 HW6 in class. Go through it a problem at a time and discuss. They need to
practice writing every step: system, states, principle, solution. Have each individual write in the
notes and then pass/reflect. This way they will have a good copy.
Problem 1:
Chapter 6
19
Problem 2:
(
1 q1q2
U=
= 9 ´ 10 9
4pe 0 r
N×m
C2
2
)
(1.6 ´ 10
-19
C
-9
3 ´ 10 m
)
2
= 7.68 ´ 10 -9 J
Like compressing a spring.
With proton and electron, sign of U changes (like lowering a rock toward the Earth)
Consider attraction, so U looks like this
with slope positive. Force = -slope, so to the left.
Consider repulsion, so U looks like this
with slope negative. Force = -slope, so to the right.
Like rolling down a hill.
Problem 3:
at
rest
attract
Chapter 6
attract
repel
No K < 0
Doesn’t
add up
repel
20
Problem 4:
G: System is two protons, far apart.
Initially Ui = 0 Ki = 2(0.19 MeV)
Finally Kf = 0 What is rf ?
Guess something small
O: Can use Energy Principle to find Uf, which then gives rf. Since no change in identity, we can
ignore the mc2 terms.
E f = Ei + W
A:
m f c 2 + K f + U f = mi c 2 + K i + U i + W
U f = Ki
1 q1q2
= K i = 2 0.19 ´ 10 6 eV 1.6 ´ 10 -19
4pe 0 rf
(
( 9 ´ 10
9 N×m 2
C2
)(
1.6 ´ 10 -19 C
rf
)(
) = 2 0.19 ´ 10
(
6
J
eV
)
)(
eV 1.6 ´ 10 -19
J
eV
)
rf = 3.79 ´ 10 -15 m
L: Units work out. Pretty small value. This is a good way to get protons close together! (Handy for
fusion.)
Problem 5:
(analogous to the scaffolded version of
Ponderable: Activity - Nuclear Fission)
Chapter 6
21
System: All particles. Initial state: proton and deuterium nucleus far apart, moving toward each other. Final
state: almost touching, still moving. In this process kinetic energy decreases, electric potential energy
increases, rest energy does not change (no change of particle identity). Change in kinetic energy of particles
plus change in electric potential energy = 0.
G: System is deuterium nucleus and a proton
Initially KDi = 6.2e-14 J, Kpi = 1.23e-13 J
Finally rf = 2(0.2e-15 m), Kf = ? no identity change
Since talking about two tiny particles, expect small number
O: Use Energy Principle to find final kinetic energy
A:
E f = Ei + W
mDf c 2 + mpf c 2 + K f + U f = mDi c 2 + mpi c 2 + K i + Ui
K f + U f = Ki
K f = Ki - U f = Ki -
1 q1q2
4pe 0 r
(
) (
1.6022 ´ 10 C ) (1.6022 ´ 10
)(
2 ( 0.9 ´ 10 m )
-19
) (
(
K f = 6.2 ´ 10 -14 J + 1.23 ´ 10 -13 J - 8.9875 ´ 10 9
)
N×m 2
C2
-19
C
)
-15
K f = 1.85 ´ 10 -13 J - 1.2817 ´ 10 -13 J = 5.6830 ´ 10 -14 J
Chapter 6
22
L: Units work. Number small as expected.
Initial state: proton and deuteron just before touching, still moving. Final state: Helium 3 nucleus and photon,
heading away from each other with equal and opposite momenta (we know this from the momentum principle).
No significant external forces, so external work is zero. In this process rest energy decreases, electric potential
energy decreases, kinetic energy increases. Change in rest energy plus change in kinetic energy plus change in
electric potential energy = 0.
G: System is proton and deuterium nucleus, which then fuse.
Initially, Ki = 5.6830e-14 J Ui = 1.2817e-13 J
Finally, Helium and gamma ray fly off with some Kf. Uf = 0
O: Use the Energy Principle to solve for Kf. This time, account for identity change.
A:
E f = Ei + W
mHef c 2 + K f + U f = mDi c 2 + mpi c 2 + K i + Ui
mHef c 2 + K f = mDi c 2 + mpi c 2 + K i + Ui
(
)
K f = mDi + mpi - mHef c 2 + Ki + Ui
(
K f = ( 2.0136 u +1.0073 u - 3.0155 u ) 1.6605 ´10 -27
kg
u
) ( 2.9979 ´10 )
8 m 2
s
+5.6830 ´10 -14 J +1.2817 ´10 -13 J
(
)ùû ( 2.9979 ´10 ) +1.85 ´10
kg ) ( 2.9979 ´10 ) +1.85 ´10 J
K f = éë( 0.0054 u ) 1.6605 ´10 -27
(
K f = 8.967 ´10 -30
8 m 2
s
kg
u
8 m 2
s
-13
J
-13
K f = 8.07 ´10 -13 J +1.85 ´10 -13 J = 9.92 ´10 -13 J
L: Units right. Small amount of energy is ok since only one particle and a photon.
Subtract kinetic energy of proton and deuteron (when far apart) from kinetic energy of helium and photon
(
K f - K i = 9.92 ´ 10 -13 J - 6.2 ´ 10 -14 J + 1.23 ´ 10 -13 J
K f - K i = 9.92 ´ 10
Chapter 6
-13
J - 1.23 ´ 10
-13
J = 8.07 ´ 10
-13
)
J
23
( 8.07 ´ 10 J ) ( 6.02 ´ 10 ) = 4.84 ´ 10
-13
23
11
J
Kinetic energy gain for one mole of deuterium is gain for one atom multiplied by Avogadro's number
VPython: Spring energy
WID 1132367 (Starting from VPython: Spring and adding energy graphs)
SKIPPED in Spring 2012
VPython_Spring_energy.py
K+U should not be constant for the system of the mass and spring because the Earth does work on
the system.
Students can also model with the Earth in the system (using approximation for the gravitational
potential energy near the surface of the earth). When done correctly, this gives a K+U=constant.
Chapter 6
24