Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
Signature:
November 16 2015
Name:
1
Idnumber:
2
THIS IS FORM 4.ENTER NAME AND FM 4 IN THE SCANTRON SHEET
Failure to enter form number or name will cost 10 points.
3
PUT AN X NEXT TO YOUR DISCUSSION SECTION:
4
[ ] Dr. Alvarez 1B, Monday 9:05 - 9:55 a. m.
[ ] Dr. Huerta 1D, Monday 11:15 a. m. - 12:05 p.m.
There are 4 problems. You must do problem [I] which consists
of 12 multiple choice questions at 5 points each.
Also do two out of the last three long problems [II], [III] and [IV]
at 20 points each. If you do more the last one will be ignored.
In the boxes 2 - 3 - 4 cross out the problem you omit.
Leave the other boxes blank
Also clearly cross out the page of the question omitted.
Your signature signifies that you will obey the HONOR CODE
CALCULATORS, AND COMPUTER WATCHES ARE NOT ALLOWED
CELL PHONES MUST BE TURNED OFF
FORMULA SHEET
2
Gravity |F | = −G mr1 m
, U (r) = −G m1rm2
2
For two points on a stream line p1 + 12 ρv12 + ρgy1 = p2 + 12 ρv22 + ρgy2 , valid for a fluid at rest.
q
k
Simple Harmonic motion: small oscillations, x(t) = xm cos(ωt + φ), ω = m
, ω independent of xm
p
Simple Pendulum: small oscillations, θ(t) = θm cos(ωt + φ), ω = Lg , ω independent of θm
Traveling wave y(x, t) = f (ξ), where ξ = x ± vt and f (ξ) is any function.
Standing wave y(x, t) = sin kx cos(ωt + φ), or y(x, t) = cos kx cos(ωt + φ) .
Preceived
Power and Intensity Preceived = IAreceiver or I = Areceiver
Intensity level β(I) = 10 log10 II0 db, where Io = 10−12 W
Interference Constructive if r2 − r= ± 0, λ, 2λ, 3λ, ... Destructive r2 − r1 = ± 12 λ, 32 λ, ...
Doppler: vW speed of the wave, fS frequency generated by the source S that moves
with speed vS , fD frequency received by detector D that moves with speed vD
fR = fS
1±
1±
vD
vW
vS
vW
sphere
hoop
cylinder
Iparticle = mr2 Icm
= 25 mR2 , Icm
= mR2 , Icm
= 21 mR2 .
Physics 210 TEST 3 Form 4 Answer Key
November 16 2015
Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
November 16 2015
[I] This problem has 15 multiple choice questions. The best answers are marked with [X].
[I.1] The three containers shown in the figure to the left contain weights W1 , W2 , and W3 of
fluid respectively. The areas of their bottoms are A1 , A2 , and A3 as shown. Let P1 , P2 ,
and P3 signify the gauge pressures at the bottom. Choose the best answer.
[A] P1 A1 = W1 , P2 A2 = W2 , and P3 A3 = W3
[B] P1 A1 = W1 , P2 A2 > W2 , and P3 A3 < W3
[C] P1 = P2 = P3 [D] both A and C are correct [X] both B and C are correct
atmosphere
h
A1
A3
A2
[I.2] A manometer in the figure to the right above consists of a U-shaped tube with water in it.
One end of the U-tube is connected to a container filled with a gas at a pressure P4 . The
other end is open to the atmosphere, P1 = Patm . The water is 100 cm higher in the part
open to the atmosphere than in the part connected to the gas. Find the gauge pressure of
the gas Pgauge ≡ Pgas − Patm . Use ρwater = 103 kg/m3 .
[A] 105 Pa
[B] −110, 000 Pa
[C] 90, 000 Pa
[X] 104 Pa
[E] −104 Pa
[I.3] In the figure to the right above there is a coefficient of static friction µs in the surface
between the two masses. The surface below mass M is frictionless. For what range of
amplitude A of simple harmonic motion of the spring-blocks system will the mass m stay
on top of M without slipping?
[A] A <
µs mg
k
[B] A >
µs (M +m)g
k
[X] A <
µs (M +m)g
k
[D] A <
µs M g
k
[E] A < µs g
kx
[I.4] At t = 0 a wave on string has the shape y(x, t = 0) = A xsin
2 +a2 . At a later time t the wave
is traveling toward the right (toward positive x). Its mathematical form is
kx
[A] y(x, t) = A xsin
2 +a2
[B] A sinxk(x−vt)
2 +a2
sin k(x+vt)
[D] y(x, t) = A (x+vt)
2 +a2
sin k(x−vt)
[X] y(x, t) = A (x−vt)
2 +a2
sin kx
[E] y(x, t) = A (x−vt)
2 +a2
[I.5] The equation of a transverse wave on a string is y(x, t) = 10 cm sin(πx − 100πt) where x
is in meters and t is in seconds. The wavelength and speed of the wave are
[A] λ = 1 m and v = 100 m/s
[X] λ = 2 m and v = 100 m/s
[C] λ = 2 m and v = 0.01 m/s [D] λ = 2 m and v = 100π m/s
[E] λ = 0.5 m and v = 330 m/s
Physics 210 TEST 3 Form 4 Answer Key
November 16 2015
Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
November 16 2015
[I.6] The figure at the left below shows a cylindrical spray can that sprays liquid of density ρ.
The gas on top of the liquid has a gauge pressure pG . The can and the tube are full of
liquid. The diameter of the tube is very small compared to the diameter of the can. The
liquid comes out of the tube with a speed v given approximately by
q
q
q
2(pG −ρg(h+H))
2(pG −ρgH)
[B]
v
≈
[C]
v
≈
[X] v ≈ 2(pG −ρgh)
ρ
ρ
ρ
q
q
[E] v ≈ 2(pG +ρgh)
[D] v ≈ 2ρgh
ρ
ρ
atmosphere
v
gas
liquid
gas
h
H
tube
[I.7] In the figure to the right above the two horses have weights of 10 and 17 Newtons (N). Find
the weight W of the horse on the left and the tension in the string marked T . Consider
ideal massless, frictionless pulleys.
[A] W = 10 N, T = 10 N [X] W = 15 N, T = 10 N [C] W = 30 N, T = 20 N
[D] W = 15 N, T = 5 N [E] W = 30 N, T = 10 N
[I.8] A satellite of mass m is in a circular orbit of radius r around a planet of mass M and
radius R. The speed v of the satellite is given by
q
q
q
q
q
GM
Gm
Gm
GM
[A] v = 2GM
[B]
v
=
[C]
v
=
[D]
v
=
[X]
v
=
R
R
R
r
r
[I.9] A simple pendulum consists of a point mass m suspended from a massless rod of length
L. When the amplitude of the angle of motion θm is doubled, its period
[X] remains the same. [B] increases
by a factor of 2
√
[C] decreases by a factor of √ 2. [D] decreases by a factor of 2.
[E] increases by a factor of 2
[I.10] A spherically symmetric point source of sound waves emits with a power of 2π W. The
intensity of the sound at a distance of 4 meters is
1
1
[A] 81 W
[B] 32
W
[C] 18 W/m2
[D] π8 W/m2
[X] 32
W/m2
Physics 210 TEST 3 Form 4 Answer Key
November 16 2015
Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
November 16 2015
[I.11] Two submarines are traveling toward the right with speeds v1 and v2 relative to the water
as shown in the figure to the left below. The sound generator in submarine #1 is set to
emit with a frequency f0 . Say the speed of sound waves in the water is vw . Find the
frequency f1 received by submarine #1 after the waves are reflected from submarine #2.
1+v2 /vw
1−v1 /vw
1+v1 /vw
1+v2 /vw
1−v2 /vw
[B] f1 = f0 1+v1 /vw
[X] f1 = f0 1−v2 /vw
[A] f1 = f0 1+v1 /vw
1−v2 /vw
1−v1 /vw
[D] f1 = f0
1−v2 /vw
1−v1 /vw
1−v2 /vw
1−v1 /vw
[E] f1 = f0
1−v2 /vw
1−v1 /vw
1+v1 /vw
1+v2 /vw
A
B
Q
water at rest
1
2
v1
v2
2m
1m
[I.12] The two speakers shown to the right above are producing coherent waves of the same
amplitude A and same wavelength λ. What is the longest wavelength λmax that will
produce minimum amplitude (destructive interference) at point Q?
[A] λmax = 1 m
[B] λmax = 2 m
[C] λmax = 6 m
Physics 210 TEST 3 Form 4 Answer Key
[D] λmax = 3 m
[X] λmax = 4 m
November 16 2015
Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
November 16 2015
[II] The figure shows a human foot. The the heel of the foot is slightly raised above the floor,
so that the foot effectively contacts the floor only at point P . A person of mass M stands
on that foot.
[a] Find the magnitude and direction of the force exerted on the foot at point P by the floor.
[b] Find the magnitude and direction of the force exerted on the foot at point A by the calf
muscle.
[c] Find the magnitude an direction of the force exerted on the foot at point B by the lower
leg bones (consider the two lower leg bones as exerting one force at point B).
[d] Imagine a vertical cut of the heel between points A and B. Find the sheer force that acts
there on the rear part of the heel (neglect the weight of the heel).
[a] The floor exerts an upward force FP = M g at point P.
[b] The calf muscle exerts an upward force F at point A such that there is zero torque around
point B, therefore
b
FA a = FP b, therefore FA = M g.
a
[c] The leg bones exert a downward force FB at point B so the total force on the foot is zero.
If the weight of the food is mg, we have
b
b
FB + mg = FP + FA = M g 1 +
, so FB = M g 1 +
− mg
a
a
or if we neglect mg
b
FB = M g 1 +
a
[c] neglecting the weight of the heel we see that there is a downward shear force S = FA
acting on the rear part of the heel.
Physics 210 TEST 3 Form 4 Answer Key
November 16 2015
Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
November 16 2015
[III] Two particles of mass M are fixed on the (x, y) plane at the points (0, 0) and (0, a). A
particle of mass m is fixed at point (a, 0).
[a] Draw a diagram and find the total gravitational force vector (in terms of G, M, m, a, ı̂, and
̂) that acts on the particle of mass m. Also give the force in terms of its magnitude and
direction.
[b] Find the escape speed of the mass m, that is, find the minimum speed it needs to be given
to escape to ∞ from its initial position (it reaches ∞ with zero speed).
y
M
a
θ
M
a
m
x
Answers:
[a]
Mm
M m −ı̂ + ̂
Mm
̂
1
~
√
F = −G 2 ı̂ + G 2
= G 2 − 1 + √ ı̂ + √ .
a
2a
a
2
2 2
2 2
s
2 2
q
√
1
1
Mm
M
m
~
√
1+ √
+
=G 2 5+2 2
|F | = G 2
a
2a
2 2
2 2
tan θ =
1
√
1+2 2
[b] KEf = Uf = 0, so KEi + Ui = 0
1
GM m GM m
mve2 −
− √
= 0. or v2 =
2
a
2a
Physics 210 TEST 3 Form 4 Answer Key
r
√
GM
(2 + 2).
a
November 16 2015
Dr. Huerta
Phy 201-OSBJ Test 3 Form 4 ANSWER KEY
November 16 2015
[IV] A cube of material of side L having a density ρm floats on water of density ρw .
(a) Find the height of the submerged portion of the cube.
(b) Find how much mass, M, of a second material has to be placed on top of the cube so
that it just becomes completely submerged, with its top just level with the water.
Answers:
[a] Let the submerged portion have a height h. The buoyancy force FB = L2 hρw g equals the
weight of the block L3 ρm g, therefore
L2 hρw g = L3 ρm g or h = L
ρm
ρw
[a] Now the buoyancy force Fb = L3 ρw g equalls the weight of the block plus the mass M ,
therefore
L3 ρw g = L3 ρm g + M g, so M = L3 ρw − L3 ρm = L3 (ρw − ρm )
Physics 210 TEST 3 Form 4 Answer Key
November 16 2015