Version 031 – midterm 03a – turner – (58425)
1
This print-out should have 19 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Ephoton = |∆E1,3 |
= |E1 − E3 |
= |−9 eV + 2.8 eV|
= 6.2 eV
001 10.0 points
The energy levels of a particular quantum
object are −9 eV, −4.2 eV, and −2.8 eV. If a
collection of these objects is bombarded by an
electron beam so that there are some objects
in each excited state, what are the energies of
the photons that will be emitted?
Ephoton = |∆E2,3 |
= |E2 − E3 |
= |−4.2 eV + 2.8 eV|
= 1.4 eV
1. 4.8 eV, 7 eV, 13.2 eV
2. 4.8 eV, 13.2 eV, 1.4 eV
002 10.0 points
A beam of high-energy π − (negative pions) is
shot at a flask of liquid hydrogen, and sometimes a pion interacts through the strong interaction with a proton in the hydrogen, in
the reaction
3. 11.8 eV, −4.8 eV, 6.2 eV
4. −4.8 eV, −6.2 eV, −1.4 eV
π − + p+ → π − + X + ,
where X + is a positively charged particle of
unknown mass.
5. −9 eV, −4.2 eV, −2.8 eV
Before
6. 13.2 eV, 11.8 eV, 7 eV
After
π−
7. 9 eV, −4.2 eV, −2.8 eV
8. 4.8 eV, 6.2 eV, 1.4 eV correct
9. 13.2 eV, 1.4 eV, 11.8 eV
35◦
π−
p+
X+
θ
10. 6.2 eV, −4.2 eV, 2.8 eV
Explanation:
4.8 eV, 6.2 eV, 1.4 eV is the correct answer. The 3 emitted photons correspond to 3
transitions between the states. Label them as
E1 = −9 eV
E2 = −4.2 eV
E3 = −2.8 eV
Then the possible transitions are given by
Ephoton = |∆E1,2 |
= |E1 − E2 |
= |−9 eV + 4.2 eV|
= 4.8 eV
The incoming pion momentum is
2900 MeV/c. The pion is scattered through
35◦ , and its momentum is measured to be
1480 MeV/c. The X + particle is scattered
through an unknown angle θ with an unknown momentum. A pion has a rest energy
of 140 MeV, and a proton has a rest energy of
938 MeV.
Find the scattering angle θ of the X + particle.
1. 28.6141
2. 28.9283
3. 29.9567
4. 28.0464
5. 29.3117
6. 30.2768
7. 29.6276
Version 031 – midterm 03a – turner – (58425)
8. 27.5308
9. 25.5217
10. 26.7025
Correct answer: 4.116 kg · m2 .
Correct answer: 26.7025 Degrees.
Explanation:
The moment of inertia of a solid about a
diameter is
Explanation:
We conserve momentum in the x and y
directions:
pπ,i =pπ,f cos 35◦ + pX cos θ
0 =pπ,f sin 35◦ − pX sin θ
Icm =
pπ,i = pπ,f cos 35◦ + pX cos θ
sin 35◦
◦
pπ,i = pπ,f cos 35 + pπ,f
tan θ
◦
pπ,f sin 35
tan θ =
pπ,i − pπ,f cos 35◦
1480 MeV/c sin 35◦
=
2900 MeV/c − 1480 MeV/c cos 35◦
⇒ θ = 26.7025◦ .
003 10.0 points
Find the moment of inertia of a solid sphere of
mass M = 6 kg and radius R = 0.7 m about
an axis that is tangent to the sphere.
m
ω
r
2
M R2 .
5
Using the parallel-axis theorem, the moment
of inertia about an axis that is tangent to the
sphere is
We rearrange the second expression to solve
for pX , and then plug it into the first expression and solve for θ.
1. 0.896
2. 4.116
3. 1.225
4. 0.875
5. 1.008
6. 2.401
7. 2.52
8. 2.016
9. 2.772
10. 1.925
2
I = Icm + M R2
2
= M R2 + M R2
5
7
= M R2
5
7
= (6 kg)(0.7 m)2
5
= 4.116 kg · m2
004 10.0 points
A skater pushes straight away from a wall.
She pushes on the wall with a force whose
magnitude is F , so the wall pushes on her
with a force F (in the direction of her motion).
As she moves away from the wall, her center
of mass moves a distance d. Consider the
following statements regarding energy.
I ∆Ktrans + ∆Einternal = F d
II. ∆Ktrans + ∆Einternal = −F d
III. ∆Ktrans + ∆Einternal = 0
IV. ∆Ktrans = F d
V. ∆Ktrans = −F d
What is the correct form of the energy principle for the skater as a real system and as a
point particle (PP) system?
1. Real: I,
2. Real: IV,
PP: IV
PP: IV
3. Real: V,
PP: I
4. Real: II,
PP: V
Version 031 – midterm 03a – turner – (58425)
3
2. A,B
5. Real: I,
PP: V
3. C,D,E
6. Real: III,
PP: V
4. A,B,D,E
7. Real: IV,
PP: III
5. A,B,C,E
8. Real: III,
PP: IV correct
6. A
9. Real: II,
PP: IV
7. A,D,E
10. Real: V,
PP: IV
Explanation:
As a real system, the wall does no work
on the skater since the contact point of the
skater’s hand and the wall does not move.
Also, the wall does not transfer energy to the
skater, i.e. by cooling down. The right hand
side of the energy principle must therefore
be zero for the real system. Furthermore,
the left hand side must contain the skater’s
change in internal energy for a real system.
III is correct.
As a point particle system, the skater’s center of mass moves a distance d while a force
F is applied. Thus, the work done is F d.
Translational kinetic energy is the only type
of energy a point particle system can have. IV
is correct.
005 10.0 points
Which of the following forces are conservative
forces?
A) Gravitational force between the Earth and
the Sun
B) Spring force
C) Air resistance
D) A force that is constant in magnitude and
direction
E) Friction force
1. C,E
8. B
9. A,B,C,D,E
10. A,B,D correct
Explanation:
The work due to a conservative force only
depends on the initial and final positions, being completely independent of the path taken.
This is true of the gravitational force, the
spring force, and any constant force.
The directions of the forces of friction and
air resistance both depend on the direction
in which the object in question is moving,
and hence the particular path taken from the
initial position to the final position is important. These two are therefore not conservative
forces.
006 10.0 points
Consider the head-on collision of two masses,
m1 = 8 kg moving to the right with speed
3 m/s and m2 = 16 kg moving to the left
with speed 1.5 m/s. Given that they stick together after the collision, what is the increase
in internal energy of the system during the
collision? Assume no energy is lost to the
surroundings and neglect any external forces
acting on the two masses.
1. 93.75
2. 48.0
3. 288.0
4. 33.75
5. 60.0
6. 54.0
7. 216.0
8. 384.0
Version 031 – midterm 03a – turner – (58425)
4
Ewoman stand for the following energy terms
associated with the woman:
9. 81.0
10. 110.25
Correct answer: 54 J.
Explanation:
We know that the velocity of the center of
mass is given by
vCM =
m1 v1 + m2 v2
m1 + m2
By substituting the values of m1 , m2 , v1 and
v2 we have
(m)(v0 ) + (2m)(−
vCM =
v0
)
2
m + 2m
Ewoman = Echemical,woman
+ Kwoman
+ Ugrav,woman+Earth
+ Ethermal,woman
The change in the kinetic energy of the
barbell is
1
1
mbb v 2 − 0 = mbb v 2 .
2
2
The general statement of the energy principle
is:
vCM = 0
Thus, the velocity of the center of mass is
zero.
We then apply the energy principle to this
problem
∆E = ∆K + ∆Eint = 0
∆Eint = − ∆K = Ki − Kf
But, the final kinetic energy is zero since the
velocity of the center of mass is zero. Thus,
we have
∆Eint = Ki
2
1
1
v0
2
Ki = m v0 + (2m)
2
2
2
1
1
1+
m v02
∆Eint =
2
2
∆Eint
∆Eint =
3
= m v02
4
3
(8 kg) (3 m/s)2
4
∆Eint = 54 J
007 10.0 points
Starting from rest, a woman lifts a barbell of
mass mbb with a constant force F through a
distance h, at which point she is still lifting,
and the barbell has acquired a speed v. Let
∆Esys = Wsurr
For which of the following systems will the
left hand side of this equation have ONLY the
1
terms +mbb gh and mbb v 2 ?
2
1. barbell + Earth correct
2. there is no such system
3. woman + Earth
4. woman only
5. Earth only
6. woman + barbell + Earth
7. woman + barbell
8. barbell only
Explanation:
We may first note that +mbb gh on the left
hand side of the energy principle equation
represents a potential energy change, which
only multibody systems can exhibit. Thus
the single body answer choices cannot be correct. Of the multibody choices, any containing “woman” must have the term ∆Ewoman on
the left side of the energy principle. The only
remaining choice is barbell + Earth. This is
Version 031 – midterm 03a – turner – (58425)
s
correct because the change in potential energy
(17N/m)(207 mN)
=
of this system is +mbb gh due to the barbell
(5 N/m)(27 mN)
rising, and the change in kinetic energy of this
= 5.11 .
1
system is mbb v 2 due to the barbell gaining
2
speed v.
008 10.0 points
Consider the spacing of vibrational energy
levels of Pb and Al based on the quantum
harmonic oscillator model for the interatomic
bound. Pb has a stiffness of ks ∼ 5N/m and
an atomic mass of 207 mN (where mN is the
mass of a nucleon). For Al, the stiffness is
ks ∼ 17N/m, and the atomic mass is 27 mN .
Determine the ratio of the energy level spac∆EAl
. Choose one :
ings,
∆EP b
1. 1.33
5
009 10.0 points
Consider the following diagram.
E3
E2
II
IV
E1
I
III
E0
Say which arrows in the above diagram
correspond to the following processes respectively: Absorption of a photon from the
ground state to the first excited state; Emission of a photon from an excited state to
another excited state.
2. 5.11 correct
3. 0.5
4. 1
5. 3.33
1. Transition II; Transition III
6. 6
2. Transition I; Transition IV
7. 4
3. Transition IV; Transition III
8. 2
4. Transition III; Transition II
Explanation:
A quantum harmonic oscillator has energy
level spacing
5. Transition III; Transition I
6. Transition IV; Transition II
EN = h̄ ω0 N + E0 .
So the level spacing is given by
r
ks
∆E = h̄ ω0 = h̄
m
Thus the ratio is
∆EAl
h̄ ω0Al
=
∆EP b
h̄ ω0P b
q
s
ksAl
ksAl mP b
mAl
= q Pb =
ksP b mAl
ks
mP b
7. Transition II; Transition IV
8. Transition III; Transition IV
9. Transition I; Transition II correct
10. Transition I; Transition III
Explanation:
Absorption sends the atom into a higher
energy state, so for the first process the arrow
Version 031 – midterm 03a – turner – (58425)
will point upward. Given that the amount
of energy is represented by the difference between E1 and E0 for the first process, it should
be clear that Transition I is the correct choice
for the first process. On the other hand, emission sends the atom into a lower energy state,
so for the second process the arrow will point
downward. The phrase “to another excited
state” tells us that the final state should be
either E2 or E1 . Given the possibilities on
the diagram, we find that Transition II is the
correct choice for the second process.
6
and the bullet, the initial speed of the bullet
and the spring constant, it is possible to find
the maximum compression of the spring.
1. A, C
2. B, D
3. A, B, E
4. D
5. A, E
010
10.0 points
6. A, B, D
7. B, D, E correct
8. A, C, E
9. A, B
In the figure, a block sitting on a frictionless
horizontal surface is attached to a rigid wall
on the right through a spring (whose axis is
horizontal). A bullet is shot at the block from
the left and gets embedded in it, causing the
block to move to the right, thus compressing
the spring. (Assume the bullet is travelling
perfectly horizontally, along the axis of the
spring, before hitting the block).
Which of the following are true?
A. The initial kinetic energy of the bullet
is completely converted to spring potential
energy when the spring reaches its maximal
compression.
B. The initial momentum of the bullet is
equal to the momentum of the bullet+block
system just after the bullet enters the block.
C. Part of the momentum of the bullet+block system is lost during the collision
(i.e. before the spring-compression starts).
D. Part of the energy of the bullet+block
system is ”lost” (no longer present as macroscopic kinetic energy) during the collision, before the spring-compression starts.
E. If we are given the masses of the block
Explanation:
Since the bullet gets embedded in the block,
the collision is inelastic. Hence, some the the
Kinetic energy of the bullet is converted into
internal energy, heat, etc during the collision
and does not get stored up in the spring.
Therefore, A is false and D is true. However,
momentum is conserved during the collision,
since forces acting on the bullet-block system
during the short time period are negligible.
Therefore, B is true and C is false. If we are
given initial velocity and masses, we can use
momentum conservation to find velocity (and
hence kinetic energy) of bullet-block system
just after collision. We can then use energy
principle to get maximum compression of the
spring. Hence, E is true.
011 10.0 points
In a certain time interval, natural gas with energy content 11000 J was piped into a house
during a winter day. In the same time interval sunshine coming through the windows
delivered 2000 J of energy into the house.
The temperature of the house didn’t change.
What was ∆Ethermal of the house?
Ia. 11000 J
Version 031 – midterm 03a – turner – (58425)
Ib. 0 J
Ic. 13000 J
Id. 2000 J
1.
For the system of the house, what was Q,
the energy transfer between the house and the
air?
2.
IIa. −2000 J
IIb. 13000 J
IIc. 2000 J
IId. −13000 J
4.
3.
5.
6.
1. Id, IIb
7.
2. Ia, IIa
3. Ib, IIb
4. Id, IIa
5. Ic, IId
6. Ia, IIc
7. Ib, IId correct
8. Ic, IIc
Explanation:
Since the temperature doesn’t change, the
amount of thermal energy cannot change, so
∆Ethermal = 0.
Since we have no change in the thermal
energy, the energy that is being added by the
natural gas and the sunshine must transfer
out to the surrounding outside air. Hence,
Q = −13000 J, where the sign is negative
since the energy flows from the system (the
house) to the surroundings (the outside air).
012 10.0 points
A projectile of mass m1 moving with a speed
v1 in the +x direction strikes a stationary target of mass m2 head-on in an elastic collision.
Find the final velocity of the projectile m1 .
Hint: You can use the energy and momentum
principles, OR you can solve the problem in
the center of mass reference frame.
8.
7
2m1
v1
m1 + m2
m1 − m2
v1 correct
m1 + m2
m2
m1 + m2
m2
v1
2m1
m2
v1
m1
m1
v1
m2
m1 + m2
v1
m1 − m2
m1
v1
m1 + m2
Explanation:
From momentum conservation,
m1 v1 = m1 v1,f + m2 v2,f
which implies
v1,f = v1 − m2 v2,f /m1 .
From energy conservation,
2
2
m1 v12 = m1 v1,f
+ m2 v2,f
.
Plugging in the expression for v1,f in terms of
v2,f in the above expression, one can solve for
v2,f and obtain,
2m1
v1
m1 + m2
m1 − m2
v1 .
=
m1 + m2
v2,f =
⇒ v1,f
Alternate Solution: The problem can also
be solved by changing to the center of mass
reference frame.
vcm =
m1 v1
m1 + m2
′
v1i
= v1 − vcm =
m2 v1
m1 + m2
Version 031 – midterm 03a – turner – (58425)
′
v1f
m2 v1
′
= −v1i
=−
m1 + m2
′
v1f = v1f
+ vcm =
(m1 − m2 )
v1
m1 + m2
013 10.0 points
A test car of mass 730 kg is moving at a
speed of 6.6 m/s when it crashes into a wall
to test its bumper. If the car comes to rest in
0.34 s, how much average power is expended
in the process?
1. 41108.1
2. 49743.0
3. 48006.8
4. 55109.8
5. 46762.9
6. 56414.1
7. 58725.0
8. 64034.0
9. 44990.3
10. 51489.7
Correct answer: 46762.9 W.
Explanation:
The power expended is the energy lost per
unit time. The energy that is lost in the
process of stopping the car is
−∆Ecar = −∆Kcar
1
= mcar (vcar )2 .
2
Thus, the power expended will be
−∆Ecar
P =
t
1
mcar (vcar )2
= 2
t
1
(730 kg)(6.6 m/s)2
= 2
0.34 s
= 46762.9 W .
014 10.0 points
A spring of stiffness k and relaxed length
L stands vertically on a table. A person
8
takes a mass M and very slowly lets the mass
down onto the spring. On letting go, he finds
that the mass does not move and the spring
is compressed by a length d1 . The same
experiment is now repeated by letting the
mass go all of a sudden when the spring is still
unstretched. The spring is found to compress
by a length d2 when the mass momentarily
comes to rest. What is the ratio d1 /d2 ?
1.
Mg
kL
2. kL
Mg
3. 2
4. 4
5. 14
6. 12 correct
7. 1
Explanation:
When the mass is very gradually let down
onto the spring, the net work done on the
mass is 0 because at every instant while it is
being let down, the net force acting on the
mass is 0 (since the man always provides just
enough force to balance all the forces). The
final spring compression is determined by the
fact that there is no net force acting on the
mass even when the man has let it down, i.
e., the spring force and the force of gravity
cancel. Hence we can write
kd1 − M g = 0.
From this, we get
d1 =
Mg
.
k
When the mass is let go all of a sudden, it
is easy to see that when the spring has been
compressed by d1 , there will still be a downward momentum for the mass due to the fact
that it has been pulled down by the force of
gravity which was always greater in magnitude than the spring force (the work done till
Version 031 – midterm 03a – turner – (58425)
that instant would be non-zero). Hence, the
mass goes down further, until all its Kinetic
energy has been converted into potential energy. Eventually, the mass oscillates about
the equilibrium position conserving its total
energy all the time. By using conservation
of energy between the topmost point of oscillation (the point at which the man just let
the mass go) and the bottommost point of oscillation (when the mass just comes to stop
momentarily), we get
KE1+GP E1 +SP E1 = KE2 +GP E2 +SP E2 .
This translates to
1
0 + 0 + 0 = 0 − M gd2 + kd22 .
2
From this, we get
2M g
d2 =
.
k
From these two expressions, we can see that
the required ratio is 1/2.
015 10.0 points
Which energy diagram in the figure below
is appropriate for each of the following situations?
I. Vibrational states of a diatomic molecule
such as O2
II. Idealized quantized spring - mass oscillator
III. Electronic, vibrational, and rotational
states of a diatomic molecule such as O2
IV. Electronic states of a single atom such
as hydrogen
9
1. I-B, II-A, III-C, IV-D
2. I-C, II-B, III-D, IV-A
3. I-A, II-B, III-D, IV-C correct
4. I-D, II-B, III-A, IV-C
5. I-D, II-C, III-A, IV-B
6. I-B, II-D, III-C, IV-A
7. I-B, II-C, III-A, IV-D
8. I-C, II-A, III-B, IV-C
9. I-A, II-C, III-D, IV-B
10. I-A, II-D, III-A, IV-C
Explanation:
I-A, II-B, III-D, IV-C is the correct answer.
Vibrational states of diatomic molecules
have the typical potential curve, and energy
levels become closely spaced as energy goes
up to zero - (A)
Quantum oscillator has equally spaced energy levels - (B)
Electronic states have more wide separation
than vibrational states, and vibrational states
have more spacing than rotational states; so
they combine to give (D)
Electronic states of hydrogen are well
known, which is also the only remaining option - (C)
016 10.0 points
A 300 g block of aluminum at a temperature of 840 K is placed in intimate contact
with a 600 g block of iron at 300 K. The
blocks are contained within an insulated enclosure. What is the final temperature of the
two blocks? Given: The specific heat capacity of aluminum is 1 J/K/g and the specific
heat capacity of iron is 0.4 J/K/g.
Version 031 – midterm 03a – turner – (58425)
1. 483.3
2. 511.1
3. 683.3
4. 600.0
5. 538.9
6. 344.4
7. 338.9
8. 327.8
9. 400.0
10. 583.3
10
Two spheres, A and B, have the same mass
and radius. However, sphere B is made of
a dense core and a less dense shell around
it. How does the moment of inertia of sphere
A about its center of mass compare to the
moment of inertia of sphere B about its center
of mass?
Ia. IA > IB
Ib. IA < IB
Ic. IA = IB
Correct answer: 600 K.
If the two spheres are rolled down an incline
from the same height simultaneously,
Explanation:
Let : TAl
mAl
CAl
TF e
mF e
CF e
= 840 K ,
= 300 g ,
= 1 J/K/g ,
= 300 K ,
= 600 g , and
= 0.4 J/K/g .
Let the final temperature be Tf . Heat will
flow from the hot object to the cold object
subject to the constraint that the net thermal
energy will be conserved.
∆Ethermal = C m ∆T
∆Ethermal (Al) + ∆Ethermal (F e) = 0
CAl mAl ∆TAl + CF e mF e ∆TF e = 0
IIa. sphere A reaches the bottom first.
IIb. sphere B reaches the bottom first.
IIc. spheres A and B reach the bottom simultaneously.
Choose the correct pair of statements.
1. Ic,IIc
2. Ic,IIa
3. Ib,IIc
4. Ia,IIb correct
5. Ia,IIa
(1 J/K/g)(300 g)(Tf − 840 K)
6. Ib,IIb
+ (0.4 J/K/g)(600 g)(Tf − 300 K) = 0
7. Ib,IIa
Thus, the final temperature Tf is
Tf = 600 K
8. Ic,IIb
9. Ia,IIc
017
10.0 points
Explanation:
Since the two spheres share the same mass
and radius, a comparison of their moment of
inertias depends on how their mass is distributed relative to an axis through the center
of mass. Sphere B has a dense core, implying
that its mass is on average closer to the axis
than A’s mass. Thus, A has a greater moment
of inertia than B.
Version 031 – midterm 03a – turner – (58425)
Let β be the unitless parameter in the moment of inertia formula I = βM R2 . Sphere B
has a smaller β from the preceding argument.
As a consequence of energy conservation, it
can be shown that
s
2gh
vCM =
1+β
11
they gain kinetic energy. Thus,
Kair = Kinitial − Kf inal
= 0.0588 J − 8 × 10−5 J
= 0.05872 J .
Therefore, B reaches the bottom first.
018 10.0 points
A coffee filter of mass 1 g dropped from a
height of 6 m reaches the ground with a speed
of 0.4 m/s. How much kinetic energy Kair did
the air molecules gain from the falling coffee
filter?
1. 0.040908
2. 0.01928
3. 0.0762775
4. 0.0644105
5. 0.027328
6. 0.063466
7. 0.042768
8. 0.05872
9. 0.0320705
10. 0.0212905
Correct answer: 0.05872 J.
Explanation:
Initially, the coffee filter’s energy is
Einitial = Uinitial + Kinitial
= mgh + 0
= (1 g)(9.8 m/s2 )(6 m) ×
1kg
1000g
= 0.0588 J .
When the coffee filter reaches the ground,
it’s energy is
Ef inal = Uf inal + Kf inal
1
= 0 + m v2
2
1kg
1
= (1 g)(0.4 m/s)2 ×
2
1000g
= 8 × 10−5 J .
Since these two are not equal, energy must
be lost to the air molecules, meaning that
019 10.0 points
As seen from above in the image, a string is
wrapped around the edge of a uniform disk of
radius R and mass M which is initially resting
motionless on a frictionless table.
F
M
R
ω
The end of the string is pulled with a force
of F over a total distance l. The linear speed
of the cylinder is found to be v after pulling
this distance. Find the angular speed of the
cylinder using the energy principle. (Note
that v 6= ωR in this case.)
s
1
2
2
Fl − Mv
1. ω =
M R2
2
s
1
2. ω =
FR
M R2
s
2
Fl
3. ω =
3M R2
s
1
2
2
4. ω =
FR − Mv
M R2
2
s
1
4
2
FR − Mv
5. ω =
M R2
2
s
4
6. ω =
Fl
3M R2
s
2
FR
7. ω =
3M R2
s
4
8. ω =
FR
3M R2
Version 031 – midterm 03a – turner – (58425)
9. ω =
10. ω =
s
s
4
M R2
1
M R2
1
F l − M v2
2
correct
Fl
Explanation:
By the energy principle, ∆K = W . The
moment of inertia of the disk is I =
(1/2)M R2.
1
1
M v 2 + Iω 2 = F l
2
2
1 1
1
M v 2 + ( M R2 )ω 2 = F l
2
2 2
So,
ω=
s
4
M R2
1
2
Fl − Mv
2
Note that there is a way to eliminate v from
the final answer using the angular momentum
principle, but in that case the answer is
s
8
ω=
Fl
3M R2
12