Version 018 – Exam 2 – florin – (55205)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A person of mass m stands a distance r away
from the center of a merry-go-round that rotates with fixed angular speed ω. The person
to walks to a distance 2r away from the center and then stops. What is the change in the
person’s kinetic energy when they move from
r to 2r?
1
002 10.0 points
You pull a block of mass m, which is initially at rest, some distance with a constant
force across a frictionless table. You also
pull a heavier block of mass M , which is also
initially at rest, the same distance with the
same constant force. Separating your answers
by a comma, answer first which block has the
greater kinetic energy and second which block
has the greater momentum.
1. Same kinetic energy, M correct
1. 2mr 2 ω 2
2. Same kinetic energy, m
2. mr 2 ω 2
3. M , M
3.
4.
5.
6.
7.
8.
7mr 2 ω 2
2
2
mr ω 2
3
mr 2 ω 2
4
mr 2 ω 2
2
3mr 2 ω 2
4
5mr 2 ω 2
2
4. m, M
5. Same kinetic energy, same momentum
6. m, m
7. M , same momentum
8. m, same momentum
9. M , m
9. 3mr 2 ω 2
3mr 2 ω 2
correct
2
Explanation:
The kinetic energy the person has a position
x is
mx2 ω 2
mv 2
=
2
2
hence the difference in kinetic energies is
10.
K2r − Kr =
4mr 2 ω 2 mr 2 ω 2
−
.
2
2
So the work is
3mr 2 ω 2
.
W =
2
Explanation:
You do the same work, F ∆x, on each block.
According to the Energy Principle, ∆K =
F ∆x, each block will have the same ∆K.
Since they both start from rest, Kf is the
p2f
same for each block. Using Kf =
, and
2m
noting Kf is the same for each block, we can
write
p2f ∝ m
so the larger mass block will have the larger
final momentum.
003 10.0 points
A mass initially at rest, spontaneously decays
into two masses both traveling at 3/5 the
speed of light. What fraction of the original
mass was lost in the decay?
Version 018 – Exam 2 – florin – (55205)
1. 3/4
2. 1/5 correct
3. 2/3
4. 3/5
5. 1/3
6. There is not enough information.
7. 1/2
8. 1/4
9. 4/5
10. 2/5
Explanation:
Energy is conserved during the decay. The
initial energy is Ei = mi c2 and the final energy is Ef = γ1 m1 c2 + γ2 m2 c2 . Also momentum is conserved during the decay. The initial
momentum is zero so γ1 m1 v1 = γ2 m2 v2 but
since v1 = v2 = vf and thus γ1 = γ2 = γ it
is also true that m1 = m2 = 1/2mf . Now
the final energy reduces to Ef = γmf c2 .
Equating the initial and final energy gives
γmf c2 = mi c2 or in other words mf /mi =
1/γ. Now to calculate the fraction of mass
lost: (mi − mf )/m
qi = 1 − mf /mi = 1 − 1/γ,
and with 1/γ = 1 − (3/5)2 = 4/5 the proportion of mass lost is just 1 − 4/5 = 1/5.
004 10.0 points
A mass attached to a spring is given an initial
displacement x0 and an initial velocity v0 . If
the position of the mass is given by x(t) =
A cos(ωt + α), what is the phase angle α?
v0
−1
1. α = cot
ωx
0 v0
2. α = cot−1 −
ωx
v 0
0
−1
−
3. α = sin
ω
2
ωv0
4. α = cot
x
0 v0
5. α = tan−1
ωx
0 ωv0
6. α = cot−1 −
x0
v0
−1
−
7. α = tan
correct
ωx0
−1 ωv0
8. α = tan
x
0
1
9. α = cos−1
x
0
ωv0
−1
−
10. α = tan
x0
−1
Explanation:
Here, use the initial position and initial
velocity to solve for α. The initial position
is x(t = 0) = A cos(α) = x0 and the initial
velocity is v(t = 0) = −ωA sin(α) = v0 . Now
dividing v0 by x0 eliminates A to give:
−1
α = tan
v0
−
ωx0
.
005 10.0 points
A ball of mass m is acted on by a constant
force with magnitude F for a time t. A second
ball of mass 4m is acted on by the same
constant force but for a time 2t. If both balls
start from rest, what is the ratio of the final
kinetic energy of the heavier ball to that of
the lighter ball?
1. 2
2.
1
4
3. 1 correct
4. 8
5.
1
2
6. 4
Version 018 – Exam 2 – florin – (55205)
7.
8.
1
8
√
3
2. 7/12 J
3. 22 J
2
1
9. √
2
Explanation:
Here we can use the momentum principle to
find the final momentum of each ball and from
that calculate the final kinetic energy of each
ball. For an object of mass m with zero initial
momentum acted on by a constant force F for
a time t, the momentum p is given by p = F t.
The kinetic energy of an object of mass m is
K = p2 /2m. Substituting the expression for
p into the expression for K gives
K=
(F t)2
2m
.
Thus for the light ball
K1 =
4. 5 J
5. −33/12 J
6. −22 J
7. −5 J
8. 8 J
9. −8 J correct
10. −7/12 J
Explanation:
The definition for work is: W =
2m
W =
Hence the ratio is unity.
006 10.0 points
A nonconstant force
~ = 3x2 , 4y 3 N
F
acts on a particle that is displaced from the
initial position
~ri = h1, 2i m
to the final position
h2,1i
Z
2
=
2
3x dx +
1
~ · d~r
F
3x2 , 4y 3 · hdx, dyi
h1,2i
and for the heavy ball
(F t)2
(F (2t))2
=
.
K2 =
2(4m)
2m
Z
~rf
~ri
which for the present case is
(F t)2
Z
Z
1
4y 3 dy.
2
Evaluating this expression gives
(23 − 13 ) + (14 − 24 ) = −8.
007
10.0 points
The sketch shows a coin at the edge of a
turntable. The weight of the coin is shown by
the vector W. Two other forces act on the
coin: the normal force and a force of friction
that prevents it from sliding off the edge.
ω
~rf = h2, 1i m.
~ as the
What is the work done by the force F
particle is displaced?
1. 33/12 J
W
Draw the two remaining force vectors.
Version 018 – Exam 2 – florin – (55205)
1.
f
correct
N
2. f
N
3. f
N
4. f
7.
008 10.0 points
Suppose a block of mass M is sliding down
an inclined plane which makes an angle θ
with the horizontal. The coefficient of kinetic
friction between the plane and the block is µ
(assume tan θ > µ). The block is initially at
rest.
What is the work done by friction as the
block travels a distance d along the plane?
1. −µ M g cos θd correct
2. µ M g sin θd
3. −µ M g tan θd
N
4. −µ M g sin θd
5. None of these graphs is correct.
6.
4
5. µ M g cos θd
6. µ M g tan θd
f
N
f
N
Explanation:
From the free-body diagram of the block,
the normal reaction force
N = M g cos θ.
The frictional force
8.
f
Ff = µ M g cos θ.
N
9. f
Since the block moves in a direction opposite
to that of friction,
Wf = −Ff d = −µ M g cos θd.
N
Explanation:
N
f
009 10.0 points
A car rounds a curve. The radius of curvature
of the road is R, the banking angle with respect to the horizontal is θ and the coefficient
of static friction is µ.
W
The friction force must point inward in order to keep the coin moving in a circle (otherwise it will fly off the edge of the disk), and the
normal force counteracts the force of gravity.
M
µ
θ
Version 018 – Exam 2 – florin – (55205)
What is the minimum speed that the car
can have without slipping assuming that if
the car were at rest, it would slide down the
plane?
s
g R (sin θ − µ cos θ)
1. vmin =
correct
cos θ + µ sin θ
s
g R (cos θ − µ sin θ)
2. vmin =
sin θ + µ cos θ
s
g R (sin θ + µ cos θ)
3. vmin =
cos θ − µ sin θ
s
g R (sin θ − µ cos θ)
4. vmin =
cos θ − µ sin θ
s
g R (sin θ + µ cos θ)
5. vmin =
cos θ + µ sin θ
s
g R (cos θ + µ sin θ)
6. vmin =
sin θ − µ cos θ
s
g R (sin θ − µ sin θ)
7. vmin =
cos θ + µ cos θ
s
g R (cos θ − µ sin θ)
8. vmin =
sin θ − µ cos θ
s
g R (cos θ + µ sin θ)
9. vmin =
sin θ + µ cos θ
s
g R (cos θ + µ cos θ)
10. vmin =
sin θ − µ sin θ
Explanation:
Since we want to calculate the minimum
speed in order for the car to not slip, we need
to consider the case where the frictional force
is Ff = µ N and is directed up the incline.
There is no acceleration in the vertical direction and thus
N cos θ+Ff sin θ = N cos θ+µ N sin θ = m g .
Solving for the normal force, we get
mg
N =
.
cos θ + µ sin θ
In the horizontal direction we have
N sin θ − Ff cos θ = N (sin θ − µ cos θ)
2
vmin
,
=m
R
5
and then using the expression we obtained for
N,
g R(sin θ − µ cos θ)
2
vmin
=
cos θ + µ sin θ
⇒ vmin =
s
g R (sin θ − µ cos θ)
.
cos θ + µ sin θ
010 10.0 points
A mass m is attached to a spring of constant
k and hung from the ceiling. If the mass is
held at rest while the spring is relaxed, i.e.,
neither stretched nor compressed, and then
dropped, what is the
p maximum speed of the
mass, where ω = k/m is the frequency of
the oscillation?
g
correct
ω
3g
=
4ω
g
=3
ω
g
=4
ω
1g
=
3ω
g
=2
ω
4g
=
3ω
3g
=
2ω
2g
=
3ω
1g
=
2ω
1. vmax =
2. vmax
3. vmax
4. vmax
5. vmax
6. vmax
7. vmax
8. vmax
9. vmax
10. vmax
Explanation:
There are three explanations.
1. Energy is conserved so E = Uspring +
Ugravity + K = constant. Let y = 0 be the
initial position, where we are given that v = 0
and the spring is neither stretched nor compressed. The total energy at any position
is
E = Uspring + Ugravity + K
Version 018 – Exam 2 – florin – (55205)
1
1
= ky 2 + mgy + mv 2 .
2
2
Initially E = 0 + 0 + 0 so E = 0 at all times.
The kinetic energy is maximum when the
potential energy is a minimum. Since
6
the negative of the change in the spring potential. Hence if the total distance the mass
falls is h then mgh = 1/2kh2 which gives
h/2 = (m/k)g which is again the amplitude
of the motion and the above analysis in (2.)
still applies.
Utotal = Uspring + Ugravity
1
= ky 2 + mgy,
2
we can find the value of y where dU/dy =
0, which corresponds to the maximum K.
Taking the derivative we have
011 10.0 points
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball
is then set in motion in a horizontal circle so
that the thread’s trajectory describes a cone.
The acceleration of gravity is 9.8 m/s2 .
dU
= ky + mg
dy
= 0,
θ
g
which yields
mg
,
k
corresponding to the minimum in U . Substituting y = −mg/k into Utotal with K =
2
1/2mvmax
, we obtain the answer,
vmax =
where
r
g
ω
k
.
m
2. The mass is held at rest while the
spring is relaxed. But if the mass were allowed to hang at rest with only the spring
and gravity acting on it, there would be a
stretch in the spring ∆y = y0′ − y0 where
y0′ is the new equilibrium point of the system. ∆y is given by the relation k∆y = mg
(since the forces balance) so ∆y = (m/k)g.
Now for the situation at hand ∆y is the amplitude of the oscillation hence the energy
2
is always E = 1/2k∆y 2 = 1/2mvmax
Thus,
2
vmax
= (k/m)∆y 2 = (m/k)g 2 = (g/ω)2. And
vmax = g/ω.
3. Alternatively, if the gravitational potential is set to zero at the initial position while
the kinetic energy is zero and the kinetic energy is zero, then after the mass is released
the kinetic energy will again be zero when the
change in gravitational potential is equal to
ω=
ℓ
y=−
r
v
m
What is the speed of the ball when it is in
circular motion? Answer in terms of g, ℓ and
θ.
1. v = g ℓ tan θ
p
2. v = g ℓ sin θ
p
3. v = g ℓ tan θ
√
4. v = ℓ tan θ
p
5. v = g ℓ cos θ
√
gℓ
6. v =
sin θ
p
7. v = g ℓ sin θ
p
8. v = g ℓ cos θ
g
tan θ
ℓ
p
10. v = g ℓ tan θ sin θ correct
9. v =
Explanation:
Version 018 – Exam 2 – florin – (55205)
7
Use the free body diagram below.
θ
2.
T
mg
The tension on the string can be decomposed into a vertical component which balances the weight of the ball and a horizontal
component which causes the centripetal acceleration, acentrip that keeps the ball on its
horizontal circular path at radius r = ℓ sin θ.
If T is the magnitude of the tension in the
string, then
Tvertical = T cos θ = m g
3.
4.
b
b
b
correct
(1)
5.
and
b
Thoriz = m acentrip
or
2
m vball
T sin θ =
.
ℓ sin θ
Solving (1) for T yields
mg
T =
cos θ
and substituting (3) into (2) gives
(2)
6. None of these
(3)
7.
b
2
m vball
m g tan θ =
.
ℓ sin θ
Solving for v yields
p
v = g ℓ tan θ sin θ .
012
10.0 points
A bug is crawling along the rim of a pizza
pan with increasing speed in the direction
indicated by the curved arrow in each figure
below.
Which sketch shows qualitatively the direction of the net force acting on the bug as it
crawls?
Explanation:
The bug has a tangential aceleration at in
the direction of its velocity, since its speed is
increasing; since it is moving in a circle it has
a radial or centripetal acceleration ar toward
the center of the circle, so its acceleration
a = ar + at is qualitatively forward and at an
angle toward the center of the circular motion.
013
1.
b
10.0 points
A dancer moves around a path like the one
shown in the figure below with a constant
speed.
Version 018 – Exam 2 – florin – (55205)
S
a
R
8.
t
tP
Q
P
8
tQ
tR
tS
tP
a
9.
t
Which of the following graphs, that have
the time the dancer passes each point labeled,
represents the magnitude of the dancer’s acceleration as a function of time t during one
trip around the path, beginning at point P ?
Assume the straight sections are perfectly
straight and each of the curved sections are
perfect semicircles both with the same radius
of curvature.
a
t
tP
tQ
tR
tS
tP
a
3.
t
tP
tQ
tR
tS
tP
tQ
tR
tS
tP
tP
t
tP
tQ
tR
tS
tP
tQ
tR
tS
tP
014 10.0 points
A spherical mass rests upon two wedges, as
seen in the figure below. The sphere and the
wedges are at rest and stay at rest. There is no
friction between the sphere and the wedges.
a
5.
tS
t
t
tP
tR
a
a
4.
tQ
Explanation:
During RS and P Q, the acceleration is zero
because the velocity is a constant both in magnitude and in direction. During QR and SP ,
even though the magnitude of the velocity is
a constant, it changes its direction, which rev2
sults in an acceleration of , where v is the
r
velocity and r is the radius of the arc. So the
acceleration is fixed at a non-zero constant
during SP and QR.
1. None of these graphs is correct.
2.
tP
tP
a
6.
M
t
tP
tQ
tR
tS
tP
a
7.
tP
correct
t
tQ
tR
tS
tP
The following figures show several attempts
at drawing free-body diagrams for the sphere.
Which figure has the correct directions for
each force?
Note: The magnitude of the forces are not
necessarily drawn to scale.
Version 018 – Exam 2 – florin – (55205)
1.
normal
9
weight
9.
normal
normal
weight
2.
normal
weight
Explanation:
Weight – the force of gravity – pulls the
sphere down. The normal force of the left
wedge upon the sphere acts perpendicular to
(normal to) their surfaces at the point of contact; i.e., diagonally upward and rightward.
Likewise, the normal force of the right wedge
upon the sphere acts diagonally upward and
leftward.
normal
weight
3.
weight
4.
normal
normal
correct
weight
015 10.0 points
What is the approximate angular speed, i.e.,
the order of magnitude, measured in radians
per second, of the hour hand of the clock in
the UT tower?
1. 103 s−1
2. 10−6 s−1
5.
3. 10−3 s−1
normal
weight
weight
4. 10−2 s−1
5. 10 s−1
6.
normal
normal
friction
friction
weight
6. 10−5 s−1
7. 102 s−1
8. 10−1 s−1
9. 1 s−1
7.
normal
weight
normal
weight
8. Since the sphere is not moving, no forces
act on it.
10. 10−4 s−1 correct
Explanation:
The hour hand goes around the clock once
every 12 hours or once every 12×3600 seconds.
1
1
2π
≈
≈
=
Hence ω =
12 × 3600
2 × 3600
7000
1
× 10−3 ≈ 10−4
7
Version 018 – Exam 2 – florin – (55205)
10
y
016
10.0 points
A box of mass M is pushed up a vertical
~ applied at an angle
wall by a constant force F
θ with the vertical, as shown. The acceleration of the box ~a has nonzero magnitude and
is directed up the wall. The coefficient of kinetic friction between the box and wall is µk
and g is the acceleration an object has when
only gravity is acting.
~a
M
~
F
µk
θ
If the box is pushed a distance d, what is
the total work W done on the box?
1. W = (F (cos θ − µk sin θ) − M g) d correct
2. W = (F (sin θ − µk cos θ) + M g) d
n
F sin θ
fk
F cos θ
Mg
The displacement is in the y-direction so
the total work done is W = Fnet,y d. Summing
forces in the y-direction gives
Fnet,y = F cos θ − fk − M g.
To solve for fk sum the forces in the xdirection and find
0 = F sin θ − n
so fk = µk n = µk F sin θ giving
W = (F (cos θ − µk sin θ) − M g) d.
017 10.0 points
How fast must a roller coaster car go through
a circular dip for you to feel five times as
“heavy” as usual, due to the upward force of
the seat? The center of the car moves along a
circular arc of radius R as in the figure below.
3. W = (M g − F sin θ) d
R
4. W = F d sin θ
b
b
b
b
5. W = 0
6. W = (F (sin θ − µk cos θ) − M g) d
7. W = (F (cos θ + µk sin θ) − M g) d
8. W = F d
9. W = µk M g d
10. W = (F − M g) d
Explanation:
Let x be perpendicular to the wall and y
parallel to it.
x
1. |~v| =
p
gR
3p
gR
2
1p
gR
3. |~v| =
2
p
4. |~v| = 3g R
p
5. |~v| = 2 g R correct
p
6. |~v| = 3 g R
p
7. |~v| = 2g R
2. |~v| =
~v
b
b
Version 018 – Exam 2 – florin – (55205)
8. |~v | = 2g R
9. |~v | = 3g R
10. |~v | = g R
Explanation:
At the bottom of a hill (or a “dip”), the net
force on the rider is upward. The force by the
seat on the rider is also upward, and in this
case has a magnitude of 5m g. The net force
on the rider is
Fnet = Fseat + Fgrav
m |~v |2
h0,
, 0i = h0, 5m g, 0i + h0, −m g, 0i
R
m |~v |2
= 4m g
R
p
⇒ |~v | = 2 g R .
018 10.0 points
Consider a typical situation: A 1 m long wire
with cross sectional area 1 mm2 that is made
with a material whose interatomic spring constant is 10 N/m and whose interatomic bond
length is 10−10 m, is attached to the ceiling
and a 10 kg mass is hung from it. What is the
approximate elongation of the wire?
11
Explanation:
(Use g ≈ 10 m/s2 )
Let : ka = 10 N/m ,
d = 10−10 m ,
L = 1m,
m = 10 kg ,
A = 10−6 m2 ,
g = 10 m/s2 .
and
We know that the Young’s modulus is given
by
Y =
ka
10 N/m
= −10 = 1011 N/m2
d
10
m
By using the formula below, we can determine
∆L and we use the fact that F = m g ≈
102 N.
∆L
F
=Y
A
L
By making use of the fact that
L
F
∆L =
A
Y
2
10 N
1m
=
10−6 m2
1011 N/m2
= 10−3 m
= 1 mm
1. 1 mm correct
2. 100 µm
3. 1 m
4. 10 µm
5. 1 nm
6. 1 µm
7. 10 nm
8. 10 mm
9. 100 mm
10. 100 nm
019 10.0 points
The carnival ride shown in the figure below
is designed to rotate sufficiently fast so that a
rider remains stuck to the wall without sliding
up or down. The minimum angular speed the
ride with radius R can operate at and keep a
rider supported without falling is ω.
R
Version 018 – Exam 2 – florin – (55205)
What is the minimum angular speed ω ′ the
ride would have to operate at to keep a rider
supported if everything were the same except
the radius were 2R rather than R?
12
1
m
µs
′
1. ω = 4 ω
ω
2. ω ′ = √ correct
2
ω
′
3. ω =
2
√
′
4. ω = 2 ω
5. ω ′ = 2 ω
6. ω ′ =
ω
4
7. ω ′ = ω
2
2m
F
µk
Find F , the magnitude of the maximum
force that can be applied so that the blocks
accelerate together.
1. F = 3 m g (µk − µs )
2. F = 3 m g (µs + µk ) correct
Explanation:
To support a rider the force of static friction
must, by the momentum principle, equal the
gravitational force, i.e., fs = mg. The force
of static friction satisfies fs ≤ µN where µ is
the coefficient of static friction and N is the
normal force exerted by the wall on the rider.
Thus mg = fs ≤ µN which implies that the
minimum normal force N that will support
the rider is Nmin = mg/µ which is the same
for both rides. The normal force gives rise to
a centripetal acceleration mv 2 /R hence, with
v = Rω,
3. F = m g (µs + 2 µk )
Nmin = mRω 2 = m(2R)ω ′2.
10. F = m g (3 µk − µs )
Solving this expression for ω ′ gives
√
ω ′ = ω/ 2.
4. F = m g (3 µs + µk )
5. F = m g (µs + 3 µk )
6. F = m g (2 µs + µk )
7. F = m g (µs − 3 µk )
8. F = 2 m g (µk − µs )
9. F = m g (2 µk − µs )
Explanation:
a
1
a
2
m
020 10.0 points
Two blocks of the same size and shape are
one atop the other. The top block has mass
m and the bottom block has mass 2m. A constant horizontal force F is applied to the lower
block. The coefficient of kinetic friction between the lower block and the level tabletop
on which the blocks slide is µk , and the coefficient of static friction between the two blocks
is µs
F
fs
3m
fk
The only force acting horizontally on the
top block is static friction, therefore the maximum value for m a is given by m a = µs m g.
Taking the whole 3 m as a system, it must also
be true that F will satisfy F − fk = µs 3 m g.
With fk = µk 3 m g, solving for F gives
F = 3 m g (µs + µk )