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Unformatted text preview: EXAM #3
PHYS 125
30 NOVEMBER 2006 Part Score Short answer 1'0 y ﬁgure #3 Total Z 2 Honor Code Exams must be completed within the allotted time and with
no outside assistance. Instructors will periodically enter the classroom, and
you are permitted to ask questions of them. However, remember that your
instructor is limited in the amount of information he can give. There should be
no communication with any other persons (verbal, electronic, etc.) other than
the instructor. The use of calculators on the exam is permitted, but is limited
to numerical calculation, Any other uses (unit conversion, formulas, graphing,
symbolic manipulation, etc.) are not allowed. Please sign the honor code below as conﬁrmation that you have complied
with this policy: “On In honor I have neither 'ven DOT received an unauthorized
i y 0n exam.” PHYS 125
Exam #3 30 November 2006 Short Answer (40 points) Each of the ten questions below is worth 4 points. he graph below represents the displacement as a f notion of ' for a
simple harmoniggsgﬂlator. At which point(s) is theWe and the aeeelel‘raﬁon zero? / f 
V . '\ m...) 2. If the graph above is represented by the equation 3305) = Acos (wt + gho),
what is the phase constant ¢0?
: at");
_7T' \
7r / 2
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\(‘diw/Q
M W 3. A kilowatt—hour is a unit of (a) force. (b) energy. c power.
\
time. 4. Jupiter is 300 times more massive than the Earth, but an object on
Jupiter’s surface would “weigh” only three times more than its weight
on earth. What is the radius of Jupiter in terms of the radius of the Earth (3.) 123/100 3 ’\ 7 r2.
(b) RE/ 10 / 66; {Parr :' "if; Vﬁ'a‘f‘f‘l,
IORE r 3 \ r
(d) wan #3 x 6 ﬂ é W :J,m/ PHYS 125
30 November 2006 measuring stick. What is the mass of the measuring stick if it is balanced by a support
force at the 0.25—m mark? (a) 0.25 kg (d) 2 kg (b) 0.5 kg e 4 kg
(0) 1 kg possible to determine 1?, {veg ; wo paths lead from a parking lot up to an observation station on the top
of a mountain. One path is a very direct, but steep route. You decide,
however, to take the scenic route which is less steep but takes you longer
to reach the top. Which of the statements below are true? . f \‘Ix The average force exerted by you along the scenic route is less than
it would have been if you had taken the steep route. \IL The work done by you along the scenic route is less than it would
have been if you had taken the steep route. {{I. The power exerted by you along the scenic route is less than it would
V have been if you had taken the steep route. (a) All are true. (6) I only.
(b) I and II. (f) II only. ,_ //
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((1) II and III, one are true. (/H‘” 7. A mass—spring system begins to oscillate with a frequency that is twice _ g what it originally was. What could explain this phenomena? (a) The spring constant was doubled. (b) The mass was halved. MThe amplitude was halved. [Any of the above could explain it. "‘7 pplying either (a) or (b) could explain it. I
g? "v,
pplying both (a) and (b) together could explain it. 5’: é ‘ N (g) No combination of (a), (b), and (c) could explain it. PHYS 125 Exam #3 30 November 2006 8. Mercury’s orbit about the sun has a signiﬁcant amount of eccentricity. At
its furthest point from the sun, it is 1.5 times further than when it is at its closest point. If it travels at 40 km / s West point, what is the
speed at the closest point? / 7 7« i x L; a UR My / WW 5 i a. l’ / Below is a graph of the potential energy of a chemical system as a function
of the spacing between reactants. 10. A tightrope walker is using a long beam to stabilize herself in a balancing
act. One beam is twice as massive but half as long as another beam.
Which will be most effective at keeping her from tipping over? a) The shorter and heavier beam. The longer and lighter beam.
(0) They are equally as effective. WV j'T’ ‘ij
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Y “'eOl PHYS 125 Exam #3 30 November 2006
ME Problems (60 points) You must work problem #1, but you may choose to work either problem #2 or
problem #3. Please circle the number of the problem you want graded
on the front of the exam. If you do not, you will be penalized 2 points
on the problem. 1. (30 points) This problem continues on the next page. The two
blocks in the ﬁgure are connected by a massless rope that passes over a
pulley. The pulley is 0.50 m in diameter and has a mass of 3.0 kg. The
moment of inertia of a disk about its center is Idisk = émRz. . 2.0 kg (3.) (12 points) Draw all relevagt free—body and torque diagrams. L390 \CQ‘ £351.an E‘ngjxxknk
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' i 7 gala.’ ox PHYS 125 Exam #3 30 November 2006 (b) (12 points) Apply Newton’s second law to each diagram. (c) (6 points) What is the acceleration of each mass and the angular
acceleration of the pulley? ‘ A («a I), U z kt “(L 15(th aft 4 “$9373 “0‘55 a ,
x“? v
z r.
M i f j > " —> it: T“ t 37.297; £93”? 5*»
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(\«d’mgLORQNaJUJQ j: .Ogllga AAAAAA N N ‘330 x PHYS 125 Exam #3 30 November 2006 2. (30 points) A 3.0kg cat £3.15 from the top of a LOOm tall table to the ﬂoor. While coming to a stop, the cat is contact with the ﬂoor for 0.25 s. 0 W132“ "‘7 T . E
What is the cat’s average power output during this [glowdown perlod? i
V; .‘ \A s // w‘vi H PA / 5“.) £583 a’ i: ‘J {7/3/22 \ ;\‘ r» a. P O ,3;
vaxﬁ/ W \M‘T PHYS 125
Exam #3 30 November 2006
Mm PHYS 125
Exam #3 30 November 2006
W 3. (30 points) A 1.0 x 105kg space craft is at rest on the South Pole of the
Earth. Ignoring any mass loss due to spent fuel, how much work must be done by the thrusters to move into an orbit that is 350 km above the
surface of the Earth? Show all of your work. PHYS 125 Exam #3 30 November 2006
w 10 Equation Sheet PHYS 125 Exam #3
\‘NK Kinematics:
# d§(t) ﬂ ** a N Ag
v(t) —— dt As(t) —— 12(t) dt vavg  E
ﬂ (11705) # if a # A17
t = = = _
a( ) dt Av(t) a(t) dt am At
17' = 17— I7
Constant Acceleration:
1
A3 = vi,8At + Eas(At)2 Pf”; 2 vi"? + asAt 2a3As = vfz’s — v33
Cincular Motion:
_. 0:2 __ 2 _ _ 271'
a,—— ar—wr vt—wr w~——
7‘ T
NE
Forces: do
Fuet=27i=3§ “ma FAonB:_FBonA
i IE] 5 mm lﬂl = uklﬁl If?! = mm ID! 2 w,sz w: = mg = mm Momentum, Impulse, Work, and Energy: 17:77217 J: Fmdt J=Ap‘
t1
1 ‘72
Kz—mv2 WF= ﬂag szAK
5.1
Ug = mgy Usp = $1: (As)2 3 = Jdg— WC = —AU AE = Wext
=d—VK=? v X.P‘=ABcosezAsz+AyBy
dt
\H
Gravitation:
3
111: cc 1/7, T2 a a3 : (Tinax :Tmm) Fg‘lséz = Ug’lscz : _Gm;m2 Equation Sheet PHYS 125 Exam #3
ENE. Rotation:
_ d0(t) _ _ A9
dw(t) Aw
at) = ——dt Awe) = / a0) dt (2an = E
Constant Acceleration:
A0 = wiAt + %a(At)2 wf = wi + aAt 2aA0 = L052 — 0.112
Dynamics:
— Fsin9 7' —— —Zr~~g€—>Ia
7.F — 7' AonB — monA Tnet “ 1: z — L=Iw I=/dmr2=Icm+Md2 Kwt=élw2
Lpoint = mvtr [point : mT2
Equilibrium:
Frfet =0 Fillet :0 7'net =
Simple Harmonic Motion:
a = —w2:c z(t) = Acos (wt + (150) v(t) = —Aw sin (wt + Q50)
27r k 9
w = 2W]! = ‘T 1 1 Wsp :1 a 1 wpend = E
E = K + U = Emu2 + 5km? = image“ = 5142/12 em“ 2 Au)
Good Things toKnow:
ME Mass of Earth 2 6 x 1024 kg
MS Mass of Sun 2 2 X 1030 kg
Me Mass of electron 2 9 x 10"31 kg
Mn Mass of proton/ neutron 2 (5/3) x 10‘27 kg
C'e Circumference of Earth 2 4 x 107 m
T's/e Sun—Earth separation distance 2 1.5 x 1011 m
ge Grav. acceleration at surface of Earth 2 10 m/s2
G Gravitational constant 2 g x 10‘10 N mZ/kgz
yr 1 Year 2 77 x 107 3
pair Density of air at 20°C 2 1.2 kg/m3
c Speed of light in a vacuum 2 3 X 108 m/s
223 Speed of sound in 20° C air 2 340 m/s
e Fundamental unit of charge 2 1.6 X 10‘19 C
1/47reo Electrostatic constant 2 9 X 109 N m2/C2
no Permeability constant 2 47r x 10‘7 Tm / A
BE Magnetic ﬁeld near Earth’s surface 2 5 x 10'5 T
do Approximate nucleon diameter 2 2.4fm Binomial expansion (1+x)”21+m:, fora:<<1 1 in = 2.54 cm Circumference of a circle 27r R
9 lbs 2 40 N Area of a circle 7rR2
4
9 mph 2 4 m/s Volume of a sphere 33 R3
1 MeV = 1.6 X 10‘13 J Surface area of a sphere 47r R2
1 rad = 0.010 J/kg 1 Ci E 3.7 x 1010 Bq ;—= 3.7 X 1010 decays/s Exam #3 Name: VKK. l5: Exercise 1 10 Steps '03 > 0 and a;n = 0
Answer includes point A
Answer only includes A ('3) —7T/2 (b) Energy (C) 10 RE (6) 1 kg (0) Less force and less power
(f) Doubling k and halving m
Speed of Mercury Speed is 60 km/s Wrong units Oscillation points Answer includes C Answer includes I Answer includes nothing else (b) Longer and lighter beam Total PHYS 125
30 November 2006 ID: SOME{825' Max Score eyewx see/xx I
H <9 nx 40 20 Exam #3 PHYS 125
30 November 2006 Problem Steps la 1b 1c Draw FBDs Either mass Weight points down A tension points up The vertical axis is labeled
No other forces present Other Mass
Same as ﬁrst mass
Tension is different 3kg Pulley A tension on the left pointing down A tension on the right pointing down cw or ccw is chosen as positive Any other forces present must act at the ro
tational axis Apply second law
Must use different labels for tensions on left and right Must use different labels for masses/weights
on left and right Accurate application of 2nd law to 4—kg mass
based on FBD, e.g. ng — TL = mLaL Accurate application of 2nd law to 2—kg mass
based on FBD, e.g. TR — mag = 7111103 Accurate application of 2’“ law to 3—kg pulley
based on FBD, erg. TLR — TRR = I a Accelerations
Accelerations of the masses must have the same magnitude Angular acceleration of pulley must be equal
to acceleration of masses divided by 0.25 m
Accurately solve for a z 2.6 m/s2 and a as
10.5 rad/$2 Correct units Total I!
K Mmc\ Score 12 I)
r,/ 1
1
1
1 [0ng MHHHD‘ \
6
1 i_
2 2 1 _L
30 " Exam #3 PHYS 125
30 November 2006 Problem Steps
2 Cat Power! Calculate work done by cat
Method #1
Apply a form of the workenergy theorem
between table and ﬂoor: W“: 2 AK or
We)“ = No change in kinetic energy: AK = 0 W9 = —AU9 = mgh a: 29 J
Wu“ = —Wg = AUg m —29 J Method #2
Use kinematics or energy conservation to ﬁnd speed at ground 1) = «2977? m 4.4 m/s
Apply a. form of the workenergy theorem from just touching ground to just at a stop:
Wnet = AK or We,“ = AE AK = —21mv2 it: —29 J Wcat r3 —29 .1 Calculate Cat Power! Apply Pavg = %
Use At = 0.25 8
Find Peat % ——120 W Correct sign
Correct units Total Max Score
30 20 wwwwwg 30 PHYS 125 Exam #3 30 November 2006
W Max Score Problem Steps Circular Motion/ Newton’s Second Law
Apply 211d law to orbital point Ewe. 2 mar
Force is that of gravity Fg = GMEm/r2
Acceleration is centripetal a, = vz/r
Distance is r = R3 + h z 6.72 x 106 In Find speed at orbit v = GMB/(RE + h)
7700 m/s or equivalently K = 20£+g
3.0 x 1012 J Z 22 22 Work/ Energy Apply a form of the workenergy theorem
between surface and orbit: Wu“ = AK or
Wext = AE Find work done by gravity, or equivalently,
the change in gravitational potential en
ergy: Wg = —AUg = GMEm/(RE + h) ~—
GMEm/RE as 3.3 x 10‘1 J Initial kinetic energy is zero Work done by thrusters is Wumm = AK —
W9 = 3.3 x 1012 J Work must be positive Correct units Total 15 wwwww Hill to N) l l 30 l l ...
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This note was uploaded on 05/02/2008 for the course PHYS 125 taught by Professor Mutchler during the Fall '08 term at Rice.
 Fall '08
 MUTCHLER

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