Version 076 – Test 3 – swinney – (58385)
This print-out should have 20 questions.
Multiple-choice questions may continue on
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before answering.
001
10.0 points
Two air blocks with masses 107 g and
107 g are equipped with identical springs
(k = 1330 N/m) . The blocks move toward
each other with identical speeds of 6 m/s on a
horizontal air track and collide, compressing
the springs.
6 m/s
6 m/s
1330 N/m
1330 N/m
107 g
107 g
Find the maximum compression of the
spring attached to the 107 g mass.
1. 5.98583
2. 0.983192
3. 0.664098
4. 3.01228
5. 3.59551
6. 5.38167
7. 8.09984
8. 4.66017
9. 4.45635
10. 5.87699
Correct answer: 5.38167 cm.
1
The total kinetic energy of both masses
is equal to the total potential of both compressed springs.
1
1
1
2
k x21 + k x22 = [m1 + m2 ] [v 2 − vcm
]
2
2
2
1
= [m1 + m2 ] v 2
2
r
q
v2
x21 + x22 = [m1 + m2 ]
k
=
[(107 g) + (107 g)]
1/2
(6 m/s)2
×
(1330 N/m) (1000 g/kg)
×(100 cm/m)
= 7.61084 cm .
Each of the two springs absorb the same
amount of potential energy (one-half the total
potential energy); i.e.,
x = x1 = x2 , we have
q
√
x21 + x22 = 2 x , so
r
x21 + x22
x=
2
(7.61084 cm)
√
=
2
= 5.38167 cm .
Explanation:
Let : k = 1330 N/m ,
m1 = 107 g ,
m2 = 107 g , and
kv~1 k = kv~2 k = v = 6 m/s .
Note: Only the kinetic energy in the centerof-mass system is equivalent to the potential
energy of the compressed springs.
The center-of-mass velocity vcm is
m1 − m2
v
v cm =
m1 + m2
(107 g) − (107 g)
(6 m/s)
=
(107 g) + (107 g)
= 0 m/s .
002 10.0 points
Imagine a system of fundamental particles
that is made up of 5 identical particles, each
with a mass of 1000 MeV/c2 .
If the binding energy of the system is 30
MeV, what is the mass of this system of particles?
1. 5030 MeV/c2
2. None of these
3. 5000 MeV/c2
4. 4970 MeV/c2 correct
Version 076 – Test 3 – swinney – (58385)
c
λ
= (6.62607 × 10−34 J · s)
2.99799 × 108 m/s
×
1.56 × 10−8 m
1 eV
×
1.60218 × 10−19 J
= 79.4789 eV ,
hf = h
5. 4850 MeV/c2
6. 5150 MeV/c2
Explanation:
Since the total energy of a bound system is
negative, the mass of the system is less than
the mass of the constituents that make it up.
In this case M c2 = 5 m c2 −B, where B = −E
is the binding energy and m is the mass of
an individual constituent particle. Thus, the
total mass energy of the system is
so the atom will be ionized and its kinetic
energy will be
K = 79.4789 eV − 13.6 eV
= 65.8789 eV .
5 (1000) MeV − 30 MeV = 4970 MeV .
003 10.0 points
Suppose that a hydrogen atom in the ground
state absorbs a photon of wavelength 15.6 nm .
If the atom is ionized, what will the kinetic
energy of the electron be when it gets far
away from the atom? The value of the speed
of light is 2.99799 × 108 m/s ; the value of
h is 6.62607 × 10−34 J · s ; and the quantized
energy states for hydrogen are given by
EN = −13.6 eV/N 2 , N = 1, 2, ...
2
keywords:
004
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.
1 eV = 1.60218 × 10−19 J .
1. 100.15
2. 55.2817
3. 84.8024
4. 79.6233
5. 109.159
6. 77.5669
7. 91.4738
8. 82.514
9. 69.058
10. 65.8789
Correct answer: 65.8789 eV.
Explanation:
Let : c = 2.99799 × 108 m/s ,
h = 6.62607 × 10−34 J · s ,
E0H = 13.6 eV , and
λ = 15.6 nm = 1.56 × 10−8 m .
M +m
v0
M
m
v
Find the string length in simple terms of m,
M , g and v0 .
1. ℓ =
v02
2
M
g 1+
m
2. None of these
3. ℓ =
2g
v2
r 0
1+
M
m
Version 076 – Test 3 – swinney – (58385)
4. ℓ =
v02
2g 1+
5. ℓ =
v02
3
3. P (t) = m (6 a t − 2 b) (a t3 + 2 b t2)
m 2
4. Zero, because the object’s velocity is instantaneously constant
M
2
M
2g 1−
m
v02
6. ℓ =
2 correct
M
2g 1+
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
5. P (t) = m (6 a t − 2 b) (a t3 − 2 b t2)
6. P (t) = −m (6 a t − 2 b) (3 a t2 − 2 b t)
7. P (t) = m (3 a t2 − 2 b t) (3 a t2 − 2 b t)
8. P (t) = m (6 a t − 2 b) (3 a t2 − 2 b t) correct
9. P (t) = −m (3 a t2 − 2 b t) (3 a t2 − 2 b t)
10. P (t) = m (6 a t + 2 b) (a t3 − 2 b t2)
Explanation:
x = a t3 − b t2
dx
= 3 a t2 − 2 b t and
vx =
dt
d2 x
ax = 2 = (6 a t − 2 b) , so
dt
Fx = m ax = m (6 a t − 2 b) .
The block rises a height of ℓ during the
upward swing, and gravity is the only force
doing work, so the total energy E = K + Ug
is conserved:
E1 = E2
1
(m + M ) v 2 + 0 = 0 + (m + M ) g h
2
2 1
2
1
m v0
1
g h = v2 =
· m
1
2
2 M +m
m2
v02
h=
2 .
M
2g 1+
m
005 10.0 points
A particle of mass m moves along the x axis
as described by x(t) = a t3 − b t2 , due to an
applied force Fx (t) .
What power P (t) does the force deliver to
the particle?
2
1. P (t) = m (6 a t + 2 b) (3 a t − 2 b t)
2
2. P (t) = m (6 a t − 2 b) (3 a t + 2 b t)
Thus
~ · ~v
P =F
= Fx vx = m (6 a t − 2 b) (3 a t2 − 2 b t).
006 10.0 points
A mass attached to a spring oscillates back
and forth as indicated in the position vs. time
plot below.
x
P
t
At point P, the mass has
1. positive velocity and negative acceleration. correct
Version 076 – Test 3 – swinney – (58385)
2. positive velocity and positive acceleration.
3. positive velocity and zero acceleration.
4. zero velocity and zero acceleration.
5. negative velocity and positive acceleration.
8. 0.19796
9. 0.1017
10. 0.311056
Correct answer: 0.20264.
Explanation:
Let :
6. negative velocity and negative acceleration.
7. negative velocity and zero acceleration.
8. zero velocity but is accelerating (positively or negatively).
m = 2.87 kg ,
k = 866 N/m ,
v0 = 3.52 m/s .
dx
= −ω A sin(ω t + φ) ,
dt
so the velocity amplitude or the maximum
speed is vmax = ωA; i.e.,
v0 = ωA
r
v0
m
A=
= v0
ω
k
s
007 10.0 points
Consider the oscillation of a mass-spring system where
= 0.20264 m .
At the time t = 0, the mass m is at x = 0
(the equilibrium point) and it is moving with
a positive velocity v0 .
v0
k
x
m
and
v=
Explanation:
The velocity is positive because the slope of
the curve at P is positive. The acceleration
is negative because the curve is concave down
at P.
x = A cos(ω t + φ) .
4
= (3.52 m/s)
(2.87 kg)
(866 N/m)
008 10.0 points
An elevator has a mass of M and carries a
maximum load of m and a constant frictional
force of f retards its motion upward.
What power P must the motor deliver at
an instantaneous speed of v if the elevator is
designed to provide an upward acceleration of
a ? The acceleration due to gravity is g .
1. P = [(M + m) (a + g) − f ]v
x=0
Let the mass be 2.87 kg, spring constant
866 N/m and the initial velocity 3.52 m/s.
Find the amplitude A.
1. 0.453848
2. 0.20264
3. 0.179252
4. 0.0852342
5. 0.225179
6. 0.169935
7. 0.383262
2. P = [(M + m) (a + g) + f ]v correct
3. P = [(M − m) g + f ]v
4. P = −[(M + m) (a + g) + f ]v
5. P = −[(M + m) (a + g) − f ]v
6. P = [(M + m) (a − g) + f ]v
7. P = [(M + m) (a − g) − f ]v
Version 076 – Test 3 – swinney – (58385)
8. P = [(M − m) (a + g) + f ]v
1. ∆Xwedge =
9. P = [(M + m) g + f ]v
2. ∆Xwedge =
10. P = [(M + m) g − f ]v
3. ∆Xwedge =
Explanation:
~ that
The motor must supply the force T
pulls the elevator upward. Applying the Momentum Principle to the elevator gives
4. ∆Xwedge =
T − f − (M + m) g = (M + m) a
5. ∆Xwedge =
6. ∆Xwedge =
7. ∆Xwedge =
T = (M + m) (a + g) + f.
8. ∆Xwedge =
and the required instantaneous power is
P =Tv
= [(M + m) (a + g) + f ]v .
9. ∆Xwedge =
10. ∆Xwedge =
009 10.0 points
A triangular wedge H high, L base length,
and with a M mass is placed on a frictionless
table. A small block with a m mass (and
negligible size) is placed on top of the wedge
as shown in the figure below.
m
M
m+M
m
m+M
m
m+M
m
m+M
M
m+M
M
m+M
2m
m+M
m
m+M
M
m+M
2M
m+M
5
L+H
3
L correct
L+H
2
L+H
3
L
L+H
2
L
L
2
L
2
L
Explanation:
Consider the wedge and the block as a twobody system. The external forces acting on
this system — the weight of the wedge, the
weight of the block and the normal force from
the table — are all vertical, hence the net horizontal momentum of the system is conserved,
H
Pxwedge + Pxblock = constant .
M
Furthermore, we start from rest =⇒ centerof-mass is not moving, and therefore the X
coordinate of the center-of-mass will remain
constant while the wedge slides to the right
and the block slides down and to the left,
L
M
m
H
∆Xwedge
Xcm =
m Xblock + M Xwedge
= constant .
m+M
Note: Only the X coordinate of the centerAll of-mass is a constant of motion; i.e., the Ycm
surfaces are frictionless, so the block slides
accelerates downward because the Py compodown the wedge while the wedge slides sidenent of the net momentum is not conserved.
wise on the table. By the time the block slides
Constant Xcm means ∆Xcm = 0 and thereall the way down to the bottom of the wedge,
fore
how far ∆Xwedge does the wedge slide to the
m ∆Xblock + M ∆Xwedge = 0 .
right?
L
Version 076 – Test 3 – swinney – (58385)
Note that this formula does not depend on
where the wedge has its own center-of-mass;
as long as the wedge is rigid, its overall displacement ∆Xwedge is all we need to know.
Finally, consider the geometry of the problem: By the time the block slides all the way
down, its displacement relative to the wedge
is equal to the wedge length L, or rather −L
because the block moves to the left of the
wedge. In terms of displacements relative to
the inertial frame of the table, this means
∆Xblock − ∆Xwedge = −L .
Explanation:
Two of the rods in the triangle are rotating
about their ends. By the parallel axis theorem, the moment of inertia of a rod rotating
1
about its end is Irod, end = m L2 .
3
The other rod on the opposite side of the
triangle from the point of
√ rotation is rotating
3
about a point that is
L from its center
2
of mass. Using the parallel axis theorem for
opposite side rod, we get a moment of inertia
of
Consequently,
0 = m ∆Xblock + M ∆Xwedge
= m (−L + ∆Xwedge ) + M ∆Xwedge
m
L
m+M
010 10.0 points
Consider an equilateral triangle made of thin
rods of mass m and length L that are welded
together. What is the moment of inertia of
this triangle for rotation about an axis that is
perpendicular to the plane of the triangle and
through one of vertices of the triangle?
The moment of inertia of a rod rotated
1
about its center of mass is Irod, cm =
m L2 .
12
3
1. m L2 correct
2
5
2. m L2
6
1
3. m L2
2
11
4. m L2
12
2
5. m L2
3
6. m L2
17
m L2
12
7
8. m L2
3
7.
Irod, opp
1
m L2 + m
=
12
√
3
L
2
!2
1
3
m L2 + m L2
12
4
5
= m L2
6
=
and therefore
∆Xwedge =
6
We can now add these individual moments
of inertia to get the moment of inertia for the
whole triangle
1
5
1
m L2 + m L2 + m L2
3
3
6
3
= m L2 .
2
Itriangle, end =
011 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.)
Version 076 – Test 3 – swinney – (58385)
1. ω =
2. ω =
3. ω =
4. ω
5. ω
6. ω
7. ω
8. ω
9. ω
10. ω
s
s
s
4
M R2
1
F l − M v2
2
4
M R2
1
F R − M v2
2
2
M R2
1
F R − M v2
2
s
correct
4
FR
=
3M R2
s
2
1
2
=
Fl − Mv
M R2
2
s
4
Fl
=
3M R2
s
1
=
FR
M R2
s
1
=
Fl
M R2
s
2
Fl
=
3M R2
s
2
FR
=
3M R2
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
7
012 10.0 points
One particle, of mass 3m , moves with a speed
v/2 in the x-direction, and another particle,
of mass 2 m , moves with a speed v/3 in the
y-direction.
What is the velocity of the center of mass
of these two particles?
10v
15v
x̂ +
ŷ
3
2
3v
v
2.
x̂ + ŷ
5
5
1.
3. v x̂ + v ŷ
3v
2v
x̂ +
ŷ
15
10
3v
2v
5.
x̂ +
ŷ correct
10
15
2v
3v
x̂ +
ŷ
6.
2
3
v
v
7. x̂ +
ŷ
5
10
v
v
8. x̂ + ŷ
7
5
2v
3v
9.
x̂ +
ŷ
3
2
v
10. v x̂ + ŷ
2
4.
Explanation:
Let :
m1 = 3m ,
v
~v1 = x̂ ,
2
m2 = 2 m ,
v
~v2 = ŷ .
3
and
The velocity of the center of mass of the
two particles is
~CM = m1 ~v1 + m2 ~v2
V
m1 + m2
3m (v/2) x̂ + 2 m (v/3) ŷ
=
3m+ 2m
2v
3v
=
x̂ +
ŷ
10
15
Version 076 – Test 3 – swinney – (58385)
013 10.0 points
A simple harmonic oscillator has amplitude
0.43 m and period 3.2 sec.
What is the maximum acceleration?
8
I
II
III
IV
1. 0.0419922 m/s2
2. 1.65779 m/s2 correct
3. 0.134375 m/s2
4. 0.828893 m/s2
5. 5.30491 m/s2
1. I
6. 0.263845 m/s2
Explanation:
2. II and IV
Let : A = 0.43 m and
T = 3.2 sec .
For a simple harmonic oscillator, the displacement is
2π
t+φ ,
x = A cos
T
so the acceleration is
2
d2 x
2π
2π
a = 2 = −A
t+φ .
cos
dt
T
T
Since −1 < cos α < 1, the maximum acceleration is
Amax =
4 π2 A
T2
4 π 2 (0.43 m)
=
(3.2 sec)2
= 1.65779 m/s2 .
014 10.0 points
Given that the y axis represents energy and
the x axis represents separation, which graph
correctly shows the potential energy U for two
interacting electrons?
3. IV
4. I and II
5. II correct
6. I and IV
7. III
8. None
9. II and III
10. III and IV
Explanation:
Electrons interact repulsively, so the correct
potential energy graph will decrease to zero
at large separation. Only graph B has this
feature.
015 10.0 points
Brass is an alloy made from copper and zinc.
A 0.70 kg brass sample at 99.0◦ C is dropped
into 2.65 kg of water at 4.5◦ C.
If the equilibrium temperature is 6.8◦ C,
what is the specific heat capacity of brass?
The specific heat of water is 4186 J/kg ·◦ C .
1. 823.003
Version 076 – Test 3 – swinney – (58385)
2. 564.477
3. 462.753
4. 425.551
5. 518.364
6. 495.343
7. 624.997
8. 395.316
9. 478.259
10. 584.057
to the mass m passes horizontally over a diskshaped pulley that has radius R, and inertia
I = m R2 . The string wraps around the
3
pulley and a mass m hangs from it.
2
m
µk
Correct answer: 395.316 J/kg ·◦ C.
Explanation:
Let :
9
mb
Tb
mw
Tw
Tf
cp,w
= 0.70 kg ,
= 99.0◦ C ,
= 2.65 kg ,
= 4.5◦ C ,
= 6.8◦ C , and
= 4186 J/kg ·◦ C .
∆Tw = Tf − Tw = 6.8◦ C − 4.5◦ C
= 2.3◦ C and
∆Tb = Tb − Tf = 99◦ C − 6.8◦ C
= 92.2◦ C , so
Qgained,water = Qlost,brass
mw cp,w ∆Tw = mb cp,b ∆Tb
mw cp,w ∆Tw
mb ∆Tb
(2.65 kg) (4186 J/kg ·◦ C)
=
(0.7 kg) (92.2◦ C)
× (2.3◦ C)
cp,b =
= 395.316 J/kg ·◦ C .
016 10.0 points
A mass m is held at rest on a horizontal table.
The friction coefficient between the mass and
the table is µk . A massless string connected
3/2m
If the system is released from rest, what
is the speed v of the hanging mass after it
has fallen a distance d? Assume the string
is inextensible and doesn’t slip on the pulley.
Also assume the pulley spins freely on its axle.
r
3
1. v = 2gd ( − µk ).
2
r
2
3
gd ( − µk ).
2. v =
7
2
r
2
3
gd ( + µk ).
3. v =
7
2
r
3
1
gd ( − µk ).
4. v = 2
5
2
r
3
5. v = 2gd ( + µk ).
2
r
2
3
gd ( + µk ).
6. v =
5
2
r
1
3
gd ( + µk ).
7. v = 2
5
2
r
3
8. v = 2 gd ( + µk ).
2
r
1
3
9. v = 2
gd ( + µk ).
7
2
r
1
3
gd ( − µk ). correct
10. v = 2
7
2
Explanation:
Let the two masses and the pulley be a
system. Then gravity and friction do work on
the system. That work is equal to the change
Version 076 – Test 3 – swinney – (58385)
10
in the system’s kinetic energy. Also the speed
A 1.36 kg skateboard is coasting along the
of the string must be the speed of the rim of
pavement at a speed of 6.40 m/s when a 0.710
the pulley. Hence
kg cat drops from a tree vertically downward
onto the skateboard.
1
13
1
3
m v2 +
m v 2 + m R2 ω 2 = −ffric d + m g d
What is the speed of the skateboard-cat
2
22
2
2
combination?
7
3
m v 2 = m g d ( − µk )
1. 3.99304
4
r 2
2. 4.20483
1
3
v=2
mgd ( − µk ) 3. 3.74044
7
2
4. 3.1806
017 10.0 points
You drop a single coffee filter of mass 1.7 g
from a very tall building. It takes 54 second
to reach the ground. Assume the time taken
to reach the terminal speed to be very small
fraction of the total falling time. The terminal velocity for the present case is v1 = 3 m/s,
where the subscript ”1” indicates the 1-filter
case. Assume the force due to the air resistance is given by Fair = C v 2 , where v is the
terminal velocity and C is a constant. Now
consider the case of 5-filters, where 5 filters
are stacked up together.
Determine the terminal velocity for this
case, v5 .
1. 2.46
2. 3.58
3. 4.02
4. 4.25
5. 4.47
6. 4.7
7. 6.26
8. 2.91
9. 6.71
10. 5.37
Correct answer: 6.71 m/s.
Explanation:
For 1-filter case, the force of air resistance
is F1 = C v12 . This force is to balance out the
weight of the filter. For the 5-filters case,
F5 = C v52 = 5 C v12
v52 = 5 v12
√
v5 = 5 v1
√
= 5 (3 m/s) = 6.71 m/s .
018
10.0 points
5. 3.58663
6. 4.03478
7. 3.6385
8. 3.9491
9. 3.43243
10. 4.13855
Correct answer: 4.20483 m/s.
Explanation:
Basic Concept: All of the motion related
to the momentum is horizontal:
m1~v1,i = (m1 + m2 )~vf
since v2,i = 0 m/s.
Given:
m1 = 1.36 kg
vi,1 = 6.40 m/s
m2 = 0.710 kg
Solution:
m1 vi,1
m1 + m2
(1.36 kg)(6.4 m/s)
=
1.36 kg + 0.71 kg
= 4.20483 m/s
vf =
019
10.0 points
Version 076 – Test 3 – swinney – (58385)
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
If the two spheres are rolled down an incline
from the same height simultaneously,
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. Ib,IIc
2. Ic,IIc
3. Ic,IIa
4. Ic,IIb
5. Ib,IIb
6. Ia,IIa
7. Ia,IIc
8. Ib,IIa
9. Ia,IIb correct
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.
Let β be the unitless parameter in the mo-
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ment 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
Therefore, B reaches the bottom first.
020 10.0 points
A solid cylinder with a mass m and a radius
R starts from rest and rolls without slipping
down a slope inclined at an angle θ with the
ground.
What is the translational speed of the cylinder after it travels a distance l measured along
the incline? The acceleration of gravity is g.
r
2
1. vf =
g l cos θ
3
r
1
g l sin θ correct
2. vf = 2
3
r
2
gl
3. vf =
3
r
2
g l sin θ
4. vf =
3
p
5. vf = 2 g l cos θ
p
6. vf = 2 g l sin θ
r
1
g l cos θ
7. vf =
3
r
1
8. vf =
g l sin θ
3
r
1
gl
9. vf = 2
3
r
1
g l cos θ
10. vf = 2
3
Explanation:
The moment of inertia of the solid cylinder
1
is I = m r 2 , vi = 0 m/s and hf = 0 m, so
2
by conservation of energy,
Ui = K f
Version 076 – Test 3 – swinney – (58385)
Ui = Ktrans,f + Krot,f
1
1
m g l sin θ = m vf2 + I ωf2
2
2
1
1 1 2 vf 2
2
m g l sin θ = m vf +
mr
2
2 2
r
3 2
1 2 1 2
g l sin θ = vf + vf = vf
2
4
4
4
vf2 = g l sin θ
3
r
4
vf =
g l sin θ
3
keywords:
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