Version 073 – Test 3 – swinney – (58535)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
=
10 A
5B
= 6
11
x
x
x6
x11
=
10 A
5B
2A
5
x =
B
1/5
2A
x=
.
B
U (x) =
2. x =
3A
5B
1/7
1/5
11 A
3. x =
6B
1/11
A
4. x =
B
1/5
2A
correct
5. x =
B
1/3
2A
6. x =
B
5
A
7. x =
B
1/4
2B
8. x =
A
1/6
B
9. x =
5A
Explanation:
The force between the two atoms is given
by
dU
F =−
dx
10 A 5 B
= − − 11 + 6
x
x
002 10.0 points
A particle oscillates up and down in simple
harmonic motion. Its height y as a function
of time t is shown in the diagram.
5
t (s)
y (cm)
1. x = 0
1
2
3
4
5
5
At what time(s) t in the interval shown is
the particle moving with its maximum negative velocity?
1. t = 3 s
2. t = 4 s
3. t = 2 s
4. t = 0 and t = 4 s correct
5. t = 1 and t = 3 s
6. t = 1 and t = 5 s
7. t = 0 s
8. t = 1 s
10 A 5 B
− 6
x11
x
We are asked to find the separation when
the force is zero, which occurs when
001 10.0 points
The potential energy between two atoms in a
particular molecule has the form
A
B
− 5
10
x
x
where x is the separation distance between
the atoms and A and B are constants with
appropriate units.
The two atoms, initially very far apart,
are released from rest. At what separation
distance will the force between them be equal
to zero?
1
9. t = 5 s
Version 073 – Test 3 – swinney – (58535)
10. t = 2 and t = 4 s
Explanation:
This oscillation is described by
πt
y(t) = − sin
,
2
dy
π
πt
v(t) =
= − cos
dt
2
2
2
d y
a(t) = 2
dt
π 2
πt
=
.
sin
2
2
The maximum negative
velocity will occur
π
πt
when − cos
= −1, or at t = 0 and
2
2
t = 4 s.
We can also see this result from inspection
of the graph. The velocity is negative when
the slope of the tangent to the curve is negative. This occurs between 0 < t < 1 and
again between 3 < t < 5. After t = 0, the
slope is changing from some maximum value
to zero at t = 1, i.e. its speed is decreasing
so the maximum negative velocity occurs at
t = 0. The motion at t = 0 repeats at t = 4
so it also has the maximum negative velocity
at that time.
003 10.0 points
The escape speed from an asteroid whose radius is 6 km is only 15 m/s. If you throw
a rock away from the asteroid at a speed of
30 m/s, what will be its final speed? Use
G = 6.7 × 10−11 N · m2 /kg2 .
1. 24.2487
2. 8.66025
3. 22.5167
4. 20.7846
5. 12.1244
6. 19.0526
7. 17.3205
8. 10.3923
9. 13.8564
10. 25.9808
Correct answer: 25.9808 m/s.
Explanation:
2
First use the escape speed to get the mass
of the asteroid:
r
2GM
R
2
1v R
⇒M =
2 G
= 1.00746 × 1016 kg .
vesc =
Now, if vi = 30 m/s, then vf is found from
Ei = Ef :
Ui + K i = Uf + K f
1
GM m 1
+ m vi2 = 0 + m vf2
⇒−
ri
2
2
r
2GM
⇒ vf = vi2 −
R
= 25.9808 m/s .
004 10.0 points
Two electrons with charge q and mass m are
held at a distance d apart. After they are
let go, what would the final speed of each
electron
approach as time goes to infinity?
1 k=
4πǫ0
r
k q2
1. vf =
2 d2 m
2. vf = ∞
r
2 k q2
dm
4. vf = 0
r
2 k q2
d2 m
3. vf =
5. vf =
6. vf =
7. vf =
8. vf =
r
r
r
k q2
correct
dm
k q2
d2 m
k q2
2d m
Explanation:
Version 073 – Test 3 – swinney – (58535)
We take the initial state to be when the
electrons are held at a distance d apart, and
the final state to be when they are infinitely
far apart, since there is a repulsive electric
force between them. Then, energy conservation gives:
Relative to the boat, the man moves ∆Xrel =
1
Lboat , one half of the boats length (from the
2
middle of the boat to its prow). In terms of
the man’s and boat’s displacements relative
to the pier, this means
1
∆Xman − ∆Xboat = ∆Xrel = Lboat .
2
Ei = Ef
K i + Ui = K f + Uf
0+
3
k q2
1
= 2 · m vf2 + 0
d
r2
k q2
vf =
dm
(2)
Combining eqs. (1) and (2) together and solving for the boat’s displacement, we find
005 10.0 points
A 122 lb man sits in the middle of a 73 lb, 13 ft
long boat. The boat’s prow touches the pier,
but the boat isn’t tied to it. The man stands
up and walks towards the pier. Assume there
is negligible resistance between the boat and
water, and that the boat’s center of mass is
in the middle of the boat. By the time he
reaches the boat’s prow, what is the distance
between the prow and the pier?
1. 4.06667
2. 4.78795
3. 2.55652
4. 2.48651
5. 6.15126
6. 4.2087
7. 3.24149
8. 4.67975
9. 3.80812
10. 3.45644
∆Xboat = −
Mman
Lboat
×
Mman + Mboat
2
so the boat end up the distance
Lboat
Mman
×
Mman + Mboat
2
122 lb
13 ft
=
×
= 4.06667 ft
122 lb + 73 lb
2
D = −∆Xboat = +
away from the pier.
006 10.0 points
Consider a rigid 3-mass system with the
masses separated by two massless rods of
length L so that the total length is 2 L.
Consider a rotation axis perpendicular to
the system and passing through the point x0 ,
1
which is L from the leftmost mass 6 M .
3
x0
6M
3M
6M
Correct answer: 4.06667 ft.
Explanation:
In the absence of external forces, the center
of mass of the man–boat system remains at
rest. So if the man moves distance ∆Xman
and the boat moves distance ∆Xboat , then we
must have
Mman Xman + Mboat Xboat
∆XCM = ∆
Mman + Mboat
Mman ∆Xman + Mboat ∆Xboat
=0
=
Mman + Mboat
and hence
Mman ∆Xman + Mboat ∆Xboat = 0.
(1)
L
L
Find the moment of inertia of the 3-mass
system about this axis.
59
M L2
4
17
M L2
2. I =
4
56
3. I =
M L2
9
1. I =
Version 073 – Test 3 – swinney – (58535)
4. I = 10 M L2
E3
155
M L2
9
52
6. I =
M L2
3
56
7. I =
M L2 correct
3
E2
5. I =
I
m2 = 3 M
m3 = 6 M
Imiddle
Iright
III
Say which transitions in the above diagram
correspond to the following processes respectively: Emission of a photon when the electron transitions from an excited state to the
ground state; Absorption of a photon by an
electron in an excited state that transitions it
to another excited state.
1
L,
3
2
at r2 = L ,
3
5
at r3 = L .
3
1. Transition II; Transition IV
at r1 =
and
The moment of inertia X
of a system of point
particles is given by I =
mi ri2 . Considering the individual moments of inertia,
Ileft
IV
E0
71
M L2
9. I =
4
163
10. I =
M L2
9
Explanation:
m1 = 6 M
II
E1
8. I = 4 M L2
Let :
4
2
1
2
= (6 M )
L = M L2 ,
3
3
2
4
2
L = M L2 , and
= (3 M )
3
3
2
50
5
L =
M L2 , so
= (6 M )
3
3
I = Ileft + Imiddle + Iright
2
4
50
= M L2 + M L2 +
M L2
3
3
3
56
=
M L2 .
3
007 10.0 points
Consider the following energy level diagram
for an electron in some atom.
2. Transition IV; Transition III
3. Transition IV; Transition II
4. Transition III; Transition I
5. Transition I; Transition IV
6. Transition I; Transition II
7. Transition III; Transitions I and IV
8. Transition I; Transition III
9. Transition II; Transition III
10. Transition III; Transition IV correct
Explanation:
The only downward or emission transition
to the ground state E0 occurs in process
III. The only upward or absorption transition
from an excited state, i.e. not the E0 state,
is in process IV where the electron transitions
from the second to the third excited state.
Therefore, choice 10 is correct.
008 (part 1 of 2) 10.0 points
Consider a horizontal spring-mass system
Version 073 – Test 3 – swinney – (58535)
with stiffness 206 N/m and mass 0.7 kg. The
system is released with an initial compression
of the spring of 0.12 m and an initial speed of
the mass of 5 m/s.
What is the maximum speed of mass during
the motion, neglecting friction?
1. 5.26412
2. 4.59075
3. 3.07511
4. 3.1087
5. 5.40719
6. 4.64414
7. 4.77144
8. 5.07152
9. 6.07487
10. 4.30189
Correct answer: 5.40719 m.
10. 0.0546053
Correct answer: 0.0546053 m/s.
Explanation:
The time period of oscillation is
which implies that the maximum velocity
vmax =
=
r
s
v2 +
k 2
x
m
(5 m/s)2 +
r
m
k
= 0.366265 s .
T = 2π
The average power lost due to dissipation
W
. Therefore, the average input power to
is
T
keep the oscillation steady must be
−W
0.02 J
=
= 0.0546053 W .
T
0.366265 s
Explanation:
Similarly, when the mass reaches its maximum speed,
1
2
Ef = m vmax
,
2
5
010
10.0 points
A block of mass M is hanging from a string
of length ℓ. A bullet of mass m traveling horizontally with speed v0 strikes the block and
embeds itself inside it. The system of block
and bullet swings until the string is precisely
horizontal, at which point the system is momentarily at rest.
206 N/m
(0.12 m)2
0.7 kg
= 5.40719 m/s .
M +m
009 (part 2 of 2) 10.0 points
Suppose there is energy dissipation of W =
−0.02 J per cycle of this spring-mass system.
What is the average power input in watts
required to maintain a steady oscillation?
1. 0.0828014
2. 0.0690079
3. 0.0484835
4. 0.0670721
5. 0.0326946
6. 0.029846
7. 0.0290576
8. 0.0369843
9. 0.0243691
v0
m
M
v
Find the initial velocity v0 in simple terms
of m, M , g and ℓ.
1. None of these
M +mp
2gℓ
M
p
3. v0 = 2 g ℓ
p
4. v0 = g ℓ
2. v0 =
Version 073 – Test 3 – swinney – (58535)
s 5. v0 = 2
m
gℓ
M +m
2(M + m) p
6. v0 =
gℓ
m
2
p
m
7. v0 =
2gℓ
M +m
s m
gℓ
8. v0 = 2 2
M +m
M +mp
9. v0 =
2 g ℓ correct
m
m p
10. v0 =
2gℓ
M +m
Explanation:
Momentum is conserved during the collision, so considering the horizontal components,
pix = pf x
m v0 + 0 = (M + m)v
m v0
.
v=
M +m
The block rises a height of ℓ so the total
energy E = K + Ug is conserved:
6
1. None of these
2.
3.
4.
5.
6.
7.
5
3
7
2
3
5
3
7
2
5
2
correct
7
Explanation:
Krot
1
=
2
v 2 1
2
2
= m v2
mr
5
r
5
and
Ktot = Ktrans + Krot
1
7
1
m v2 ,
= m v2 + m v2 =
2
5
10
so
Ei = Ef
1
(m + M ) v 2 + 0 = 0 + (m + M ) g ℓ
2
1 2
v = gℓ
2
2
1
m v0
= gℓ
2 M +m
M +mp
2gℓ.
v0 =
m
011 10.0 points
A solid sphere with moment of inertia about
2
its center of mass Icm = M r 2 rolls along
5
a horizontal, smooth surface at a constant
linear speed without slipping.
What is the ratio between the rotational
kinetic energy about the center of the sphere
and the sphere’s total kinetic energy?
1
m v2
Krot
1
2
10
5
=
=
=
7
Ktot
5
7
7
m v2
10
keywords:
012 10.0 points
A block of mass m = 5 kg hangs from a rope
that is wrapped around a disk of mass m and
radius R1 = 32 cm. This disk is glued onto
another disk of again the same mass m and
radius R2 = 75 cm. The two disks rotate on a
fixed axle (see figure) without friction.
Version 073 – Test 3 – swinney – (58535)
⇒ω=
s
4gh
3r12 + r22
=
s
4(9.8 m/s2 )(1.3 m)
3(32 cm)2 + (75 cm)2
7
≈ 7.65473 rad/s.
(Don’t forget to convert your units.)
013 10.0 points
Consider a thin uniform rod of mass M and
length L.
If the block is released at a height 1.3 m
above the ground, what is the angular speed
of the two disk system just before the block
hits the ground? Use g = 9.8 m/s2 .
1. 9.25324
2. 8.62808
3. 7.65473
4. 9.53879
5. 8.37515
6. 9.04387
7. 8.4829
8. 8.78685
9. 9.71426
10. 8.20803
L
What is the rotational inertia of the rod
about a point a distance L from the nearest
end of the rod, along the line of the rod?
1.
2.
3.
4.
Correct answer: 7.65473 rad/s.
Explanation:
Using the energy principle,
Ktrans,f + Krot,f = Ui .
1
For a disk, I = M R2 . Adding the two mo2
ments of inertia and knowing that the speed
of the box is v = ωR1 , we solve for ω:
1 2 1
mv +
2
2
⇒
1 2 1 2
mr + mr ω 2 = mgh
2 1 2 2
r12 r12 r22
+
+
2
4
4
ω 2 = gh
5.
6.
4
M L2
3
5
M L2
3
11
M L2
3
9
M L2
4
1
M L2
2
7
M L2 correct
3
7. M (2 L)2 = 4 M L2
8.
2
M L2
3
9. Still the same, since empty space has no
mass and thus no rotational inertia.
10. Zero.
Explanation:
According to the parallel axis theorem,
Ip = Icm + M h2 , where h is the distance
from the CM to the new parallel axis pivot
Version 073 – Test 3 – swinney – (58535)
point p. In this case, the rotational inertia of
1
the rod about its center of mass is
M L2 .
12
(We don’t care about the rotational inertia
about its end; there is no theorem relating
that rotational inertia to any arbitrary axis
rotational inertia.) The additional distance h
from the center of mass to the new pivot point
L
3L
is L + =
, so the new rotational inertia
2
2
is
7
9
1
M L2 = M L2 .
+
12 4
3
014 10.0 points
What explains why so many physical systems
in nature are well-described as a simple harmonic oscillator?
1. The Energy Principle.
2. The concept of buoyancy states that
when a solid object displaces fluid in a gravitational field, the buoyant force exerted by
the fluid on the object is equal to the weight
of the displaced fluid.
3. None of the other answers are correct.
4. Many systems have a large amount of
vibrational energy.
5. Matter is really made out of many small
balls and springs.
6. When a system at rest is at a potential
energy minimum, the potential energy will increase quadratically for small displacements,
just as for a spring. correct
8
Explanation:
At a minimum in a potential energy function the derivative of the potential energy is
zero, and for small displacements away from
that minimum the potential energy increases
quadratically with the displacement; that is,
the function is parabolic about the minimum.
Objects become trapped in such ”potential
wells.” They oscillate between turning points
where all of their energy is potential energy,
but cross over the potential energy minimum
where their kinetic energy is maximized. This
exchange of energy establishes simple harmonic motion.
015 10.0 points
Suppose that a hydrogen atom in the ground
state absorbs a photon with energy K =
72.1 eV .
If the atom is ionized, what will the kinetic
energy of the electron be when it gets far away
from the atom? The quantized energy states
for hydrogen are given by
EN = −13.6 eV/N 2 , N = 1, 2, ...
1 eV = 1.60218 × 10−19 J .
1. 74.3
2. 58.5
3. 88.0
4. 52.4
5. 103.4
6. 81.0
7. 53.4
8. 84.8
9. 100.1
10. 56.4
Correct answer: 58.5 eV.
7. The principle of reciprocity for the force
of interaction between two objects.
8. Energy is quantized in microscopic systems.
Explanation:
The atom will be ionized and its kinetic
energy will be
K = 72.1 eV − 13.6 eV
= 58.5 eV .
9. Because many potential energy functions
U (x) exhibit maxima at some values of x.
10. The Momentum Principle.
keywords:
Version 073 – Test 3 – swinney – (58535)
016
9
10.0 points
K
Which of the diagrams corresponds to a system of two electrons that are fired into space
in opposite directions (i.e. they are separated
by some initial distance and have initial velocities in opposite directions)? Assume they
only interact with each other.
Note that the horizontal and vertical axes
in each plot are the separation between the
particles and energy, respectively.
K +U
4.
r
U
5. None of these graphs could represent the
energies of the system.
K +U
K
1.
K +U
K
U
6.
r
r
correct
U
K
U
K +U
U
2.
7.
K +U
r
r
K
K
3.
K +U
r
U
Explanation:
When the electrons are fired, they have
both kinetic and potential energy. They gain
kinetic energy from to their Coulomb repulsion as they move apart and their electric
potential energy decreases to zero at ∞ so
that all of their energy becomes kinetic. If
they were fired at each other and collided,
all of their energy would be potential energy
and they would turn around where the total
energy graph intersects the potential energy
graph. Thus the correct answer is 2.
017 10.0 points
A force acting on a particle has the potential
energy function U (x), shown by the graph.
The particle is moving in one dimension under
the influence of this force and has kinetic
Version 073 – Test 3 – swinney – (58535)
Potential Energy (J)
energy 1.0 Joule when it is at position x1 .
Potential Energy vs Position
10
where K is the kinetic energy of the particle.
Since kinetic energy ≥ 0 ,
1
K(x) = −U (x) ≥ 0
U (x) ≤ 0 J ,
0
so the particle oscillates between position x0
and x2 .
−1
x0 x1
x2 x3
Position
Which of the following is a correct statement about the motion of the particle?
1. It oscillates with maximum position x2
and minimum position x0 . correct
018 10.0 points
In a certain time interval, natural gas with
energy content 9000 J was piped into a house
during a winter day. In the same time interval, sunshine coming through the windows
delivered 3000 J of energy into the house.
The thermal energy of the house increased by
8000 J. For the system of the house, what was
Q, the energy transfer between the house and
the air?
2. It comes to rest at either x0 or x2 and
remains at rest.
Ia. −3000 J
Ib. 8000 J
Ic. 4000 J
Id. −4000 J
3. The particle remains at x1 because the
force is zero there.
How did the temperature of the house
change?
4. It moves to the right of x3 and does not
return.
IIa. It increased.
IIb. It decreased.
IIc. It remained the same.
5. It cannot reach either x0 or x2 .
6. It moves to the left of x0 and does not
return.
7. The particle moves to either x0 or x2 ,
where it turns around, and then moves to x1
where it stops and stays because the force is
zero there.
Explanation:
In this case, the total energy of the particle
is conserved so at any point on the axis,
K(x) + U (x) = K(x1 ) + U (x1 )
= 1.0 J + (−1.0 J) = 0
K(x) = −U (x) ,
1. Id, IIb
2. Ia, IIc
3. Id, IIc
4. Ia, IIa
5. Ic, IIb
6. Ib, IIa
7. Ib, IIc
8. Id, IIa correct
Version 073 – Test 3 – swinney – (58535)
11
10. 1.43403
9. Ic, IIa
Correct answer: 1.77515 m.
10. Ia, IIb
Explanation:
The particle reverses its direction when its
velocity and so its kinetic energy equals zero.
At that point, all of the particle’s energy is in
the form of potential energy, so
Explanation:
The energy principle states
Ethermal = Q + other
= Q + Egas + Esun
U = E = a |~x|
So,
and the magnitude of the distance is
|xmax| =
Q = Ethermal − Egas − Esun
= 8000 J − 9000 J − 3000 J
= −4000 J
=
(15 J)
(8.45 J/m)
= 1.77515 m .
The temperature increased since Ethermal
increased.
019 10.0 points
A 2.4 kg particle moves along the x-direction
under the influence of a force described by the
potential energy function U = a |~x|, where
a = 8.45 J/m and x is the position of the
particle in meters measured from the origin
as in Figure.
The total energy of the particle is E =
K + U = 15 J.
E
a
The distance itself can be both positive and
negative.
020 10.0 points
An object is placed into a tank of water as
shown in the figure.
ρw
ρ
K +U
U
U
x
−xmax
0
+xmax
Determine the magnitude of the distance
|xmax| that the particle travels from the origin
before reversing direction.
1. 2.49151
2. 1.53121
3. 1.77515
4. 1.91471
5. 2.83353
6. 1.69875
7. 1.60285
8. 1.7179
9. 1.41343
If the density of the object ρ is greater
than the density of water ρw , which one of
the following expressions is correct for the net
force on the object in the vertical direction if
upward is taken to be positive.
Use the following definitions of the magnitudes of the forces acting on the object:
FG = the gravitational force
FD = the fluid drag force
FB = the buoyant force .
1. Fnet = FB − FG − FD
Version 073 – Test 3 – swinney – (58535)
2. None of the above.
3. Fnet = FD − FG − FB
4. Fnet = FG − FB − FD
5. Fnet = FB + FG − FD
6. Fnet = FB − FG + FD correct
7. Fnet = FG − FB + FD
8. Fnet = −FB − FG − FD
9. Fnet = FB + FG + FD
10. Cannot be determined without knowing
the velocity of the object.
Explanation:
The object is more dense than the water
so it will sink to the bottom and have velocity in the negative direction. The drag force
will be opposite to this motion or vertically
upward. The buoyant force is also vertically
upward and the gravitational force is vertically downward. Thus, the correct answer
is:
Fnet = FB − FG + FD
12