Version 076 – Final Exam – swinney – (58535)
This print-out should have 30 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
3.
1
Fhand
Ff riction
001
Fgravity
10.0 points
A book is at rest on an incline as shown
below. A hand, in contact with the top of
the book, produces a constant force Fhand
vertically downward.
Fnormal
4.
Fnormal
Fhand
Ff riction
Fhand
B
oo
k
Fgravity
5.
Fnormal
Ff riction
Fhand
The following figures show several attempts
at drawing free-body diagrams for the book.
Which figure has the correct directions for
each force? The magnitudes of the forces are
not necessarily drawn to scale.
1.
Fhand
Fnormal
Ff riction
Fgravity
Fhand
Fhand
Fnormal
7.
Ff riction
Fgravity
Ff riction
Fnormal
Fhand
8.
Fnormal
Ff riction
6.
Fgravity
2.
Fgravity
correct
Fgravity
Ff riction
Fhand
Fnormal
Fgravity
Explanation:
The normal force points perpendicular to
the surface of the inclined plane. The weight
force points down. The Fhand also points
down. The friction force keeps the book from
Version 076 – Final Exam – swinney – (58535)
sliding and consequently points up the incline.
002 10.0 points
A uniform rod of mass 3.2 kg is 6 m long. The
rod is pivoted about a horizontal, frictionless
pin at the end of a thin extension (of negligible
mass) a distance 6 m from the center of mass
of the rod. The rod is released from rest at
an initial angle of 49 ◦ with respect to the
horizontal, as shown.
Fy
O
Fx
K f = Ui
O
Correct answer: 43.6942 N.
Explanation:
Let : ℓ = 6 m ,
d = ℓ = 6 m,
θ = 49◦ , and
m = 3.2 kg .
13
1
m ℓ2 + m ℓ2 =
m ℓ2 .
12
12
Since the rod is uniform, its center of mass
is located a distance ℓ from the pivot. The
vertical height of the center of mass above
horizontal is ℓ sin θ . Using conservation of energy
3.2 kg
6m
6m
What is the magnitude of the horizontal
force of the rod on the pivot at the instant
the rod is in a horizontal position? The acceleration due to gravity is 9.8 m/s2 and the
moment of inertia of the rod about its center
1
of mass is
m ℓ2 .
12
1. 43.6942
2. 23.8791
3. 60.6865
4. 30.8964
5. 14.7283
6. 59.2416
7. 15.0199
8. 62.807
9. 46.0032
10. 18.8021
β
mg
I = Icm + m d2 =
49◦
2
1
I ω 2 = m g ℓ sin θ
2
13
m ℓ2 ω 2 = m g ℓ sin θ
24
24 g sin θ
ω2 =
13
ℓ
2
ar = r ω = ℓ
Fx = m ar =
24 g sin θ
13 ℓ
=
24
g sin θ .
13
24
m g sin θ
13
24
(3.2 kg) (9.8 m/s2 ) sin 49◦
13
= 43.6942 N .
=
003 10.0 points
Suppose a box of mass m slides off of a table of height h with an initial velocity v0 in
the positive x-direction. Let the acceleration
due to gravity be equal to g in the negative
y-direction. What is the correct algebraic expression for ∆x, the distance the box travels
in the x-direction before hitting the ground
if there is a wind blowing in the negative
x-direction that applies a constant force of
magnitude Fw ?
Version 076 – Final Exam – swinney – (58535)
Fw
h
∆x
1. ∆x = v0
s
h
hFw
+
g
mg
s
2h
2. ∆x = v0
g
2h
hFw
3. ∆x = v0
−
g
mg
s
4 h 2 hFw
4. ∆x = v0
−
g
mg
s
2 h 2 hFw
5. ∆x = v0
−
g
mg
s
h
hFw
+
6. ∆x = v0
g
2mg
s
h
hFw
7. ∆x = v0
−
g
mg
s
2 h 2 hFw
+
8. ∆x = v0
g
mg
s
h
hFw
9. ∆x = v0
−
g
2mg
s
2h
hFw
10. ∆x = v0
−
correct
g
mg
Explanation:
h=
∆t =
1
g (∆t)2
2
s
2h
g
which tells us how long the box is in flight.
The acceleration of the box in the x-direction
is due entirely to the force of the wind, which
is
ax = −Fw /M
3
The equation for motion in the x-direction
is thus given by
∆x = x0 + v0 ∆t − 1/2(Fw /M )(∆t)2
s
2h
hFw
= v0
−
g
Mg
where x0 = 0 has been taken to be the edge of
the table.
004 10.0 points
A 53 kg man sits on the back end of a 3.3 m
long boat. The front of the boat touches the
pier, but the boat isn’t tied. The man notices
his mistake, stands up and walks to the boat’s
front, but by the time he reaches the front,
it’s moved 0.735 m away from the pier.
Assuming no water resistance to the boat’s
motion, calculate the boat’s mass (not counting the man).
1. 262.476
2. 32.6756
3. 241.553
4. 28.4167
5. 184.959
6. 133.929
7. 129.05
8. 125.802
9. 25.4301
10. 20.2581
Correct answer: 184.959 kg.
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 therefore
Mman ∆Xman + Mboat ∆Xboat = 0 .
Version 076 – Final Exam – swinney – (58535)
Solving this equation for the boat’s mass, we
find
∆Xman
Mboat = Mman ×
.
−∆Xboat
Now, let’s be careful about the displacements. Taking the back-to-front direction to
be positive, we have the boat moving backward, so
∆Xboat = −0.735 m < 0.
As to the man, his displacement relative to the
boat is the boat’s full length (back to front),
so
∆Xrel = +Lboat = +3.3 m,
but relative to the pier his displacement is only
∆Xman = ∆Xrel + ∆Xboat
= +3.3 m − 0.735 m = +2.565 m .
Consequently,
∆Xman
−∆Xboat
+2.565 m
= 53 kg ×
+0.735 m
= 184.959 kg.
Mboat = Mman ×
005 10.0 points
For an iterative calculation that determines a
particle’s trajectory of motion in the presence
of a nonconstant force, which of the following physical quantities must be updated in a
program loop (i.e., repeatedly)?
~ net
A) F
B) ∆t
C) ~p
D) ~r
E) mparticle
1. B,C,E
2. B,D,E
3. A,C,D correct
4. A,D,E
5. A,B,D,E
4
6. A,B,C,D
7. A,B,C,D,E
8. A,B,C,E
9. C,D
10. A,B,D
Explanation:
An iterative calculation consists of the repeated calculation of the momentum principle
and position update. The varying quantities
~ net , ~p, and ~r .
of those equations are F
006 10.0 points
A 15 kg block rests on the earth’s surface.
How much energy is required to move the
block very far from the earth, ending up at
rest again? Use the following values:
G = 6.67 × 10−11 N · m2 /kg2
MEarth = 5.97 × 1024 kg
REarth = 6.37 × 106 m .
1. 3125580000.0
2. 2688000000.0
3. 812651000.0
4. 4938420000.0
5. 5250980000.0
6. 1125210000.0
7. 3063070000.0
8. 2625490000.0
9. 4875910000.0
10. 937674000.0
Correct answer: 9.37674 × 108 J.
Explanation:
We define the system to be the object and
the earth. Then, infinitely far away, we set
U = 0, and since it’ll be stationary, we set
K = 0. Then the difference in energy will be
the amount of energy needed:
Version 076 – Final Exam – swinney – (58535)
5
and
∆E =U (∞) − U (REarth )
−G MEarth m
=0 −
REarth
G(5.97 × 1024 kg)(15 kg)
=
6.37 × 106 m
= 9.37674 × 108 J .
007 10.0 points
Object A with mass 4 kg moves at an initial
speed of 2 m/s along a frictionless horizontal
plane. A horizontal force F is applied opposite to the direction of the motion and brings
object A to a stop in a distance ∆xA . Object
B with mass 2 kg moves at an initial speed of
4 m/s along the frictionless horizontal plane.
The same horizontal force F is applied opposite to the direction of the motion of B and
brings object B to a stop in a distance ∆xB .
What is the relation between ∆xA and ∆xB ?
1. ∆xA =6∆xB
2. ∆xB =8∆xA
3. ∆xA =4∆xB
4. ∆xA =∆xB
5. ∆xA =8∆xB
6. ∆xB =2∆xA correct
7. ∆xA =2∆xB
8. ∆xB =6∆xA
9. ∆xB =4∆xA
Explanation:
Consider each ball and the earth as three
separate systems. Using the Energy Principle
with the rest mass not changing, we know
∆K = W . The displacement is only in the
x-direction and the only force is F , which is
opposite to the direction of motion so
WA = −F ∆xA
WB = −F ∆xB
For both blocks, Kf = 0 so the Energy
Principle says:
1
2
F ∆xA = 0 − mA vA
2
and
1
2
F ∆xB = 0 − mB vB
2
Taking the ratio of the two equations gives:
2
2
(F ∆xA )/(F ∆xB ) = (mA vA
)/(mB vB
)
2
2
(∆xA )/(∆xB ) = (mA vA )/(mB vB )
= [4kg(2m/s)2]/[2kg(4m/s)2]
= 1/2 so
∆xB = 2∆xA
008 10.0 points
It isn’t very difficult to accelerate an electron
to 99.5% of the speed of light because the
electron has a very small mass. What is the
ratio of the kinetic energy to the rest energy in
this case? Note that this particular electron
would be very relativistic.
1. 10.1915
2. 9.01252
3. 4.15443
4. 4.44646
5. 4.99717
6. 7.46637
7. 5.47442
8. 6.08881
9. 6.92155
10. 4.29434
Correct answer: 9.01252.
Explanation:
The kinetic energy is the total energy minus
the rest energy:
KE =γ m c2 − m c2
=(γ − 1) m c2
Version 076 – Final Exam – swinney – (58535)
And so the ratio would be given by:
KE (γ − 1)m c2
=
RE
m c2
=γ − 1
1
=q
−1
0.995 c 2
1−
c
6
t, the moon’s momentum is p1 = hM v, 0, 0i.
What is the momentum of the after it has
moved 1/8 of the way around the planet in its
orbit?
≈ 9.01252 .
009 10.0 points
An automobile moves at constant speed over
the crest of a hill. The driver moves in a
vertical circle of radius 52.5 m. At the top of
the hill, she notices that she barely remains in
contact with the seat.
The acceleration of gravity is 9.8 m/s2 .
Find the speed of the vehicle.
1. 19.3484
2. 20.7181
3. 16.5946
4. 17.0029
5. 16.3866
6. 17.4017
7. 18.4673
8. 18.809
9. 22.6826
10. 20.9767
Correct answer: 22.6826 m/s.
Explanation:
Barely remaining in contact with the seat
means that the normal force the seat exerts is
essentially zero. Hence gravity alone provides
the centripetal force, so from Newton’s second
law
mv 2
mg =
,
R
thus
p
v = gR
q
= (9.8 m/s2 )(52.5 m)
= 22.6826 m/s .
010 10.0 points
A moon with mass M orbits a planet in a
circular path at a constant speed v in the x−y
plane, as shown in the diagram. At some time
1. p2 = h
r
1
M v,
2
r
1
M v, 0i correct
2
r
1
2
M v,
M v, 0i
2. p2 = h
3
3
1
2
3. p2 = h M v, M v, 0i
3
3
r
r
2
1
M v,
M v, 0i
4. p2 = h
3
3
r
5. None of the above
√
1
3
6. p2 = h M v,
M v, 0i
2
2
√
3
1
7. p2 = h
M v, M v, 0i
2
2
8. p2 = hM v, 0, 0i
Explanation:
The momentum of the moon is now a 45◦
angle relative to its original position, but has
the same magnitude. Therefore, the new momentum can be found by calculating the new
x and y components:
p2 = hcos θM v, sin θM v, 0i
= hcos 45◦ M v, sin 45◦ M v, 0i
.
Version 076 – Final Exam – swinney – (58535)
7
r
r
From momentum conservation ~p in the y1
1
=h
M v,
M v, 0i
direction (̂), we may solve for v2 ,
2
2
v2 =
011 10.0 points
A 0.37 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.298 kg
puck moving initially along the x axis with
a speed of 2.68 m/s. After the collision, the
0.298 kg puck has a speed of 1.68 m/s at an
angle of 25◦ to the positive x axis.
Determine the magnitude of the velocity of
the 0.37 kg puck after the collision.
1. 0.551494
2. 0.957408
3. 0.428839
4. 1.0936
5. 1.03589
6. 0.699094
7. 0.797964
8. 1.52246
9. 1.89193
10. 0.358222
m1 v1 sin θ
= 1.0936 m/s .
m2 sin φ
012
10.0 points
Correct answer: 1.0936 m/s.
Explanation:
y
m1 ~v1
θ
m1 ~v
b
x
φ
m2 ~v2
Let the puck initially at rest be m2 . Using
momentum conservation, we have
x : m1 v = m1 v1 cos θ + m2 v2 cos φ
y : 0 = m1 v1 sin θ − m2 v2 sin φ
From the above two relations, we can get,
tan φ =
m2 v2 sin φ
m1 v1 sin θ
=
m2 v2 cos φ
m1 v − m1 v1 cos θ
φ = arctan
v1 sin θ
v − v1 cos θ
.
For a satellite and Earth system, which
set of curves shown above correspond to the
the case when the satellite is in a circular
orbit? Note that some sets of graphs may be
Version 076 – Final Exam – swinney – (58535)
physically impossible.
1. Diagram B correct
7. x =
2A
B
1/3
1/4
2B
8. x =
A
5
A
9. x =
B
Explanation:
The force between the two atoms is given
by
2. Diagram F
3. Diagram C
4. Diagram A
5. Diagram E
6. Diagram G
F =−
7. Diagram D
013 10.0 points
The potential energy between two atoms in a
particular molecule has the form
1/5
11 A
1. x =
6B
1/7
3A
2. x =
5B
1/6
B
3. x =
5A
1/5
2A
4. x =
correct
B
10 A
5B
= 6
11
x
x
x6
x11
=
10 A
5B
2
A
x5 =
B
1/5
2A
x=
.
B
014 10.0 points
A simple harmonic oscillator is described by
the function
x(t) = (6) cos(π t) .
where the amplitude is measured in cm and
the frequency is measured in s−1 .
What is the acceleration of this oscillator
2
at t = ?
π
1. 2 cm/s2
2. 24.6 cm/s2 correct
5. x = 0
6. x =
A
B
1/11
We are asked to find the separation when
the force is zero, which occurs when
B
A
− 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?
dU
dx
10 A 5 B
= − − 11 + 6
x
x
10 A 5 B
= 11 − 6
x
x
Explanation:
For a circular orbit, K, U and E are expected to be constant, and at the same time
E < 0. Diagram B fits this criteria.
U (x) =
8
3. Zero
4. −18.8 cm/s2
Version 076 – Final Exam – swinney – (58535)
9
Explanation:
2
5. −59.2 cm/s
6. 18.8 cm/s2
Let :
7. −24.6 cm/s2
8. −6 cm/s2
L = 32 m ,
ρ = 1030 kg/m3 ,
T1 = 4◦ C = 277 K ,
V1 = 1.4 cm3 , and
T2 = 16◦ C = 289 K .
9. 59.2 cm/s2
P V = nRT ∝ T .
10. 6 cm/s2
Explanation:
The displacement is x(t) = A cos(ω t) , so
dx
vx =
= −A ω sin(ω t) and
dt
ax =
At t =
d vx
= −A ω 2 cos(ω t) .
dt
2
,
π
2
ax ( ) = −(6 cm)(π s−1 )2 cos(2)
π
= 24.6 cm/s2 .
015 10.0 points
At 32 m below the surface of the sea (density
of 1030 kg/m3 ), where the temperature is 4◦ C,
a diver exhales an air bubble having a volume
of 1.4 cm3 .
If the surface temperature of the sea is 16◦ C,
what is the volume of the bubble immediately
before it breaks the surface? The acceleration
of gravity is 9.8 m/s2 and the atmospheric
pressure is 1.02 × 105 Pa.
1. 1.74113
2. 1.20058
3. 1.36719
4. 1.51555
5. 4.25781
6. 2.26469
7. 2.62069
8. 11.746
9. 1.87106
10. 6.08616
Correct answer: 6.08616 cm3 .
At depth L ,
P = Pa + ρ g L and
P V1 = n R T1 ;
at the surface Pa V2 = n R T2 ,
so
Pa V2
T2
=
P V1
T1
V1 T2 (Pa + ρ g L)
P V1 T2
=
V2 =
Pa T1
Pa T1
V1 T2 V1 T2 ρ g L
=
+
T1
P T
a 1
3
1.4 cm (289 K)
=
277 K 1.4 cm3 (289 K)
+
(1.02 × 105 Pa)(277 K)
× 1030 kg/m3 9.8 m/s2
× (32 m)
= 6.08616 cm3 .
016 10.0 points
A tennis ball of mass 0.1 kg falls straight
downward from very tall building at a constant velocity due to air resistance.
If the ball falls 100 m at a constant velocity
of −̂ (50 m/s), how much work is done on the
ball by air friction over that distance? The
acceleration of gravity is 10 m/s2 .
1. 1000 J.
2. 10 J.
3. 125 J.
Version 076 – Final Exam – swinney – (58535)
4. −12.5 J.
10
v
5. 100 J.
t
1.
6. 12.5 J.
7. −125 J.
v
8. −1000 J.
t
2.
9. −10 J.
10. −100 J. correct
Explanation:
Let :
v
t
3.
m = 0.1 kg ,
g = 10 m/s2 ,
h = 100 m .
and
v
The work done by air friction is
Wa = Ef − Ei .
t
4.
Since K does not change, the work done by
air friction will reduce to
Wa = Ef − Ei = Ugf − Ugi
= ∆Ug = −m g h
2
v
t
5.
= −(0.1 kg) (10 m/s ) (100 m)
= −100 J .
017
v
10.0 points
A block with an initial velocity v0 slides up
and back down a frictionless incline.
v0
t
6.
correct
θ
Which graph best represents a description
of the velocity of the block versus time? The
initial position of the block is the origin;
i.e., x = 0 at t = 0 . Consider up the track
to be the positive x-direction.
v
7.
t
Version 076 – Final Exam – swinney – (58535)
v
t
8.
11
3. It would float lower, because the iron
displaces water and the overall water level
would rise.
4. It depends on the ratio of iron and wood
volumes.
Explanation:
v
t
9.
v
t
10.
Explanation:
The block’s acceleration
a = −g sin θ < 0
is down the track, hence negative in our coordinate system. It is also constant — because
the forces on the block are constant. The initial velocity of the block is up the track, hence
positive in our coordinate system, v0 > 0.
Consequently, the block’s equation of motion
is
1
g sin θ t2 ,
x(t) = v0 t −
2
whose plot is a straight line with negative
slope.
018 10.0 points
A piece of iron is fastened on top of a block of
wood floating in water.
If the block is inverted so the iron is below
the block of wood, would the wood float at
the same level, lower, or higher?
1. It would float at the same level.
2. Higher, because there is a buoyant force
acting on the iron now. correct
The iron-wood unit displaces its combined
weight and the same volume of water whether
the iron is on top or the bottom. When the
iron is on top, more wood is in the water.
When the iron is on the bottom, less wood is
in the water.
019 10.0 points
A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from
374 K to 748 K.
If v1 is the average speed of the gas
molecules before the absorption of heat and
v2 their average speed after the absorption of
v2
heat, what is the ratio ?
v1
v2
1.
=4
v1
v2
=1
2.
v1
v2 √
3.
= 2 correct
v1
v2
4.
=2
v1
v2
1
5.
=
v1
2
Explanation:
Let : T1 = 374 K and
T2 = 748 K .
For an ideal gas, the internal energy,
(and therefore average kinetic energy of the
Version 076 – Final Exam – swinney – (58535)
molecules as well), is proportional to the temperature of the gas, i.e.,
12
5.
U ∝ nRT
Kav ∝ k T .
The ratio of the kinetic energies is thus
K2
K1
0.5 m v22
0.5 m v12
v22
v12
v2
v1
k T2
=
k T1
2 T1
=
T1
6.
=2
7.
=
√
2.
020 10.0 points
A rigid circular wheel spins at constant angular velocity about a stationary axis. Choose
the picture below that correctly describes the
relative magnitudes and directions of the velocity vector of points on the wheel.
1.
correct
2.
8.
Explanation:
Any point on the wheel travels in a circle
around the axis, so the velocities of all points
are directed parallel to tangents to the wheel.
The wheel is rigid, so all points on the wheel
travel with the same angular velocity.
v = ωr
so the points farther from the axis travel faster
than the points closer to the axis.
3.
4.
keywords:
021 10.0 points
A pendulum consists of a very light but
stiff rod of length 0.8 m hanging from a nearly
frictionless axle, with a mass 0.25 kg at the
end of the rod. Suppose that you hit the
stationary mass so it has an initial speed.
What speed is needed to cause the pendulum
to oscillate between +/- 90◦ ? (Use 9.8 m/s2
for the acceleration due to gravity)
1. 3.003
Version 076 – Final Exam – swinney – (58535)
2. 6.102
3. 6.038
4. 5.422
5. 3.96
6. 4.246
7. 5.565
8. 5.048
9. 4.601
10. 5.738
What is the magnitude of the force Fl,r that
the right block exerts on the left block?
Explanation:
Given : m = 0.25 kg ,
g = 9.8 m/s2 ,
l = 0.8 m ,
Treating the pendulum and the earth as
one system, the energy principle implies that
∆K + ∆U = 0
K f + K i = Uf + Ui
The final kinetic energy is zero and the final
potential energy is Uf = mgl. The initial
potential energy is zero and the kinetic energy
1
is mv 2 Thus, we have
2
1 2
mv = mgl
2
v =
q
2gl = 2(9.8 m/s2 )(0.8 m) = 3.96 m/s
022 10.0 points
Two blocks lay on a surface that has negligible
friction. As shown in the picture: a force F
is applied to the left block at angle θ and the
left block pushes against the right block. The
right block has mass m and the left block has
mass M = 4 m.
F
θ
3F
5
2 F sin θ
2. Fl,r =
5
F cos θ
3. Fl,r =
2
2 F cos θ
4. Fl,r =
5
3 F cos θ
5. Fl,r =
5
F cos θ
6. Fl,r =
4
3 F sin θ
7. Fl,r =
5
F sin θ
8. Fl,r =
correct
5
F sin θ
9. Fl,r =
4
F
10. Fl,r =
5
Explanation:
Let Fl,r be the magnitude of the force of
the small box pushing on the large block, but
by reciprocity that is equal in magnitude to
the force of the large block on the small block,
i.e. Fl,r = Fr,l . Let right be the positive
x-direction.
For the combined system of both blocks
1. Fl,r =
Correct answer: 3.96 m/s.
p
13
M
m
F sin θ = (M + m) a
F sin θ
a=
5m
The contact force supplies the acceleration
on the smaller box:
Fr,l = m a
F sin θ
=
5
= Fl,r
023 10.0 points
A ball of putty with mass m falls vertically
onto the outer rim of a horizontal turntable
Version 076 – Final Exam – swinney – (58535)
I 0 ωi
I0 + m R 2
ωi
=
.
m R2
1+
I0
of radius R and moment of inertia I0 that is
rotating freely with angular speed ωi about
its vertical axis.
What is the post-collision angular speed of
the turntable plus putty?
ωi
m R4
2+
I0
ωi
2. ωf =
mR
1+
I0
ωi
3. ωf =
mR
2+
I0
ωi
4. ωf =
m R2
2+
I0
ωi
5. ωf =
m R3
2m +
I0
ωi
correct
6. ωf =
m R2
1+
I0
ωi
7. ωf =
m R3
3m +
I0
ωi
8. ωf =
m R4
1+
I0
ωi
9. ωf =
m R3
m+
I0
ωi
10. ωf =
m R3
1+
I0
Explanation:
The final rotational inertia of the turntableplus-putty is
1. ωf =
If = I0 + Iblob = I0 + m R2 .
Since there is no external torque on the system
of the putty plus the turntable, we know Lf =
L i = I 0 ωi .
I f ωf = I 0 ω0
I 0 ωi
ωf =
If
14
ωf =
024 10.0 points
An ice cube melts on the warm sidewalk on
a hot summer day. Let the entropies of the
ice cube, of the pavement and of the ice cubesidewalk system be Sice , Ssw and Ssys .
What happens to these entropies?
1. Sice decreases, Ssw increases, Ssys decreases
2. Sice increases, Ssw decreases, Ssys does
not change.
3. Sice increases, Ssw increases, Ssys increases
4. Sice increases, Ssw decreases, Ssys increases correct
Explanation:
Heat is added to the ice cube, increasing its
entropy. Heat is removed from the sidewalk,
decreasing its entropy. The total entropy
change, however, is positive.
025 10.0 points
An oscillator is on the way to equilibrium (x =
0) from its point of maximum displacement
A
(x = A), and happens to be at x = .
3
At this point, how much of its total energy
K
is kinetic; i.e., what is the fraction ?
E
1
1.
2
1
2.
4
2
3.
3
4. Zero; it will be momentarily at rest at this
point.
5.
3
4
Version 076 – Final Exam – swinney – (58535)
8
correct
9
5
7.
9
1
8.
3
1
9.
9
Explanation:
6.
15
Sx − Px = 6 × 1010 − (−3 × 1010 ) = 9 × 1010
Sy − Py = −6 × 1010 − 7 × 1010 = −1.3 × 1011
Sz − Pz = 4 × 1010 − (−5 × 1010 ) = 9 × 1010
So
~ =S
~ −P
~
R
1
1
At any point, E = K + k x2 = k A2 .
2
2
2
x 2
1
1
1
2
2
=E
, so
kx = kA
2
2
A
3
E=K+
8
E =K.
9
1
E
9
026 10.0 points
A planet is located at
~ = h−3 × 1010 , 7 × 1010 , −5 × 1010 i.
P
A star is located at
~ = h6 × 1010 , −6 × 1010 , 4 × 1010 i.
S
~ be the vector pointing from the planet
Let R
to the star. Find R̂z , i.e. the z-component of
the unit vector R̂.
1. 0.494685
2. 0.446663
3. 0.575435
4. 0.392541
5. 0.327561
6. 0.50128
7. 0.64957
8. 0.594635
9. 0.530669
10. 0.384111
Correct answer: 0.494685.
Explanation:
~ we subtract S
~ −P
~ by respective
To find R,
components:
= h9 × 1010 , −1.3 × 1011 , 9 × 1010 i .
~ To find R
, we use the Pythagorean theorem.
q
~ R = (9 × 1010 )2 + (−1.3 × 1011 )2 + (9 × 1010 )2
=
p
3.31 × 1022
= 1.81934 × 1011 .
~ z by the magnitude:
To get R̂z ,we divide R
~z
R
9 × 1010
R̂z = =
= 0.494685
1.81934 × 1011
~
R
027 10.0 points
Masses M and m attract each other with a
gravitational force of magnitude F . Mass m
is replaced with a mass of 3m, and is moved
four times farther away. Now, what is the
magnitude of the force?
1. 8/3 F
2. 3/16 F correct
3. 4/3 F
4. 3/8 F
5. 9/16 F
6. F
7. 3/4 F
8. 9/4 F
Version 076 – Final Exam – swinney – (58535)
16
Consider just four of the energy levels in a
certain atom, as shown in the diagram below.
9. 16/3 F
10. 4/9 F
Explanation:
From Newton’s law of gravitation we originally have
GM m
F =
r2
n=4
n=3
n=2
Replacing m by 3m and r by 4r gives
Fnew
GM 3m
=
= 3/16F
(4r)2
028 10.0 points
When you are moving up at constant speed
in an elevator, there are two forces acting on
you: the floor pushing up on you (F1 ) and
gravity pulling down (F2 ).
What is the relationship between the magnitude of F1 and F2 and the physical principle
that explains this relationship?
1. F1 = F2 from the principle of reciprocity.
2. F1 > F2 from the principle of reciprocity.
3. F1 < F2 from the principle of reciprocity.
4. F1 = F2 from the momentum principle.
correct
n=1
If this atom is emitting photons, how many
spectral lines will result from all possible transitions among these levels? Which transition
corresponds to the highest- frequency light
emitted? Which transition corresponds to
the lowest-frequency?
1. three; level 4 to level 3 transition; level 2
to level 1 transition.
2. six; level 4 to level 1 transition; level 4 to
level 3 transition. correct
3. three; level 4 to level 1 transition; level 4
to level 3 transition.
4. six; level 2 to level 1 transition; level 4 to
level 1 transition.
5. six; level 4 to level 1 transition; level 2 to
level 1 transition.
5. F1 < F2 from the momentum principle.
6. three; level 4 to level 2 transition; level 4
to level 3 transition.
6. It depends on which direction the elevator
is moving.
7. three; level 2 to level 1 transition; level 4
to level 3 transition.
7. F1 > F2 from the momentum principle.
8. six; level 2 to level 1 transition; level 4 to
level 3 transition.
Explanation:
Since the speed is constant, there is no
change in momentum. The momentum principle states that the net force must therefore
be zero, which requires F1 = F2 .
029
10.0 points
Explanation:
Six transitions are possible, as shown.
Version 076 – Final Exam – swinney – (58535)
n=4
n=3
n=2
n=1
The highest-frequency transition is from
quantum level 4 to level 1. The lowestfrequency transition is from quantum level
4 to level 3.
030 10.0 points
Choose the fundamental interaction that is
responsible for the force below:
Protons and neutrons attract each other in
a nucleus.
1. Centripetal
2. Magnetic
3. Cosmological
4. Electrostatic
5. Weak
6. Gravitational
7. Electromagnetic
8. Normal
9. Tension
10. Strong correct
Explanation:
The force that holds the nucleus together
must be strong enough to overcome the repulsion between protons due to the electromagnetic force. This is a strong interaction.
17