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Unformatted text preview: EXAM #3
PHYS 125
30 NOVEMBER 2006 Name: { um Mi
(please print legibly) Student ID #: gjomggg é§ Part Score Short answer Problem #1 Z Z 55 9/;‘7Problem #2 26~ CL Problem #3 ﬂ 4,
Total ’1' 'ﬁw f. 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 HOT received any unauthorized
) g1 OH exam.” (A r 1/ m a, .m/ MW, 30 gape
1’ (si W A PHYS 125
Exam #3 30 November 2006 Short Answer (40 points) Each of the ten questions below is worth 4 points. 1. The graph below represents the displacement as a function of time for a
simple harmonic oscillator. At which point(s) is the velocity positive and
the acceleration zero? z 2. If the graph above is represented by the equation :z:(t) 2 Aces (wt + (to),
what is the phase constant 450? (a) “W 47%
W
(c) 0
(d) 7r/2 (e) Tr 3. A kilowatt—hour is a unit of (a) force. , ener . F i a] T . ’” ff”? "5" power. f
(d) 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
BE? MY R13/100 @mRE ’ .5: ($5222ny I,
(d) lOORE LA I; y
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I!” '“ ' 5;, w PHYS 125
Exam #3 30 November 2006 5. A 1—kg rock is suspended by a massless string from one end of a l—m
measuring stick. What is the mass of the measuring stick if it is balanced by a support
force at the 0.25m mark? MM
(a) 0.25 kg (d) 2 kg M5 r f u
(b 0.5 kg (e) 4 kg A
c )1 kg (f) Impossible to determine 5 3’ r l:
x .... l 6. Two 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? I. The average force exerted by you along the scenic route is less than
it would have been if you had taken the steep route. V’ The work done by you along the scenic route is less than it would have been if you had taken the steep route. III. The power exerted by you along the scenic route is less than it would
have been if you had taken the steep route. :/' (a) All are true. (e) I only. (b) I and II. (f) II only.
@I and III. (g) III only. (d) II and III. (h) None are true. 7. A massspring system begins to oscillate with a frequency that is twice
what it originally was. What could explain this phenomena? 7%) The spring constant was doubled.
> (b) The mass was halved.
M The amplitude was halved.
(d) Any of the above could explain it.
(e) Applying either (a) or (b) could explain it.
@pplying both (a) and (b) together could explain it. (g) No combination of (a), (b), and (c) could explain it. PHYS 125
Exam #3 30 November 2006
EH 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 at the furthest point, what is the
speed at the closest point? H0*.GV % x ‘2“ f f‘JI‘
.jdﬁjc qqp/ 9. Below is a graph of the potential energy of a chemical system as a function
of the spacing between reactants. About which point(s) could the system undergo oscillatory motion? (Xi. 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? (al The shorter and heavier beam. The longer and lighter beam.
(c) They are equally as effective. A ' "‘ "I"? I, Li L i5 PHYS 125 Exam #3 30 November 2006
M 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
M 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 = %mR2. (a) (12 points) Draw all relevant freebody and torque diagrams. *‘Z
l x
“A; .5,
‘ \IJE/ é
a} , 1'
L I M?
l r 5 J " 1; __ ., _ iv  bets +,
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7 °" 25! "l i ‘5. _
b" I
J .9 "xa’J
M\
'7‘“ f7 ' l x,
‘45“ t?"”";“?, V & PHYS 125 Exam #3 30 November 2006
W (b) (12 points) Apply Newton’s second’law to each diagram.
‘T. K *T;@ ‘IgL M Ni 4 /»\ ‘5 . '7’ o{
w l.« “‘M v , x —' W )ln’g :1 4“U M J Z: “:7 V‘ﬂ‘r @ j} / =/ y . q itwujgaz’rs/wa/m
A 01.6 / ,  A
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l , If a A“ A
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‘én m! Wham: f/ “'* 'v ‘7’ ,5; 3‘ “3% (c) (6 points) What is the acceleration of each mass and the angular
acceleration of the pulley? PHYS 125 Exam #3 30 November 2006 2. (30 points) A 3.0kg cat big from the top of a. 1.00—m tall table to the
ﬂoor. While coming to a stop, the cat is contact with the ﬂoor for 0.25 3.
What is the cat’s average power output during this slowdown period? D x 9L.“ j: "
‘1 [I l‘ x r{ if / W‘s) 0\ KT: k
g y : 5,5253?“
\ "2 t 1 ' “’ﬂé’fi'ﬂggsmvm/
an; 2 f / 0 . d f F r M I  K] WW‘WHWM d K H N C a  .m} K
gin/ﬁt} ‘,20,«»waag 7 :51; J { z ¢‘\ 2 ,u Wei],
if ' ‘
W A J/ ‘ j :2 1 t,
/\j”c{l~ K p k / ﬁ/(( fv’i 3‘ ‘w JL
'  ‘ = / ﬂ: 9‘?
s“ r}; '1: 3‘ a, y”: 14/” r ,1
h I ‘ A z "+0 ,,,,,,,,,,,,, A, ,,,,,,,,,,,,,,,, ,7
792 0 (Jhkgf (“I v")
” “’ 1  ~ 2 an 4, :/ ~. : z z; J ‘ t ' ' I Y ; v. e .
at! :_, 06’“ 8f / I PHYS 125 Exam #3 30 November 2006
W PHYS 125
Exam #3 30 November 2006 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. kl
3; A K a“ P ﬁres. y. 24mg PHYS 125 Exam #3 30 November 2006
W 10 Equation Sheet PHYS 125 Exam #3
MR— Kinematics:
e d§(t) ~ “ N e A5
v(t) — dt As(t) —— A v(t) dt '0an —— Z?
a dim) a t‘ a e A17
a(t) — dt Av(t) — a(t) dt aavg — Z2—
17' = 17 I7
Constant Acceleration:
1
A3 = Ui,3At + 5%(At)2 om = vi‘s + asAt 2asAs = 1133 — v58
Circular Motion:
_ “t2 _ 2 _ _ 2”
a, ~ — a, — w r rut — wr w —  T T
NEE
Forces: d _' Ftnet:2:_Fki=a§“Vina: FAonB=_?BonA i
a —* ~ ~ _' —; q 1 —.~ “ —»
Us, ~<— “Slnl = #kln' = .ur'nl ID, :1 ZpairAv2 = mg Fsp = “19133 Momentum, Impulse, Work, and Energy: 13'sz f: t 13mm: J=A;5‘
1 8‘2
K=§mv2 WF= g1 Fdg WnetzAK
Ug = mgy Usp = $1: (As)2 F3 = —% Wc = —AU AB = Wext
P=Ci¥= .17 X§=ABCOSQ=AIB3+AyBy
NH
Gravitation:
vt QC 1/7 T2 (x a3 ___ Fg’lm : G'ngmg Ug’wz : _Gmr1m2 Equation Sheet PHYS 1 25 Exam #3
M Rotation:
d6(t) A9
W) = ——dt A005) = / w(t) dt wavg = Z7
dw(t) __ _ Aw
(1(t) —— —di—— — Clt aavg —— Z?
Constant Acceleration:
A6 = wiAt + %a(At)2 w; = w; + aAt 2aA9 = w? — w?
Dynamics:
~Fsi9 '7’ —— 7— —ZT.—(_j£_.,ja
TF—‘T n AonB“ 7BonA net— i 1" dt
L210 I=/dmr2=Icm+Md2 mop21.1w
Lpoint = mvt'r Ipoint = mrz
Equilibrium:
Fifet = 0 Frizz : 7'net :
Simple Harmonic Motion:
a = —w2a: $(t) = A cos (wt + 450) 0(t) = —Aw sin (wt + (150)
27r k
w z z T 1 wsp :1 a 1 wpend = £
E = K + U = émvz + 5ka = 5mg“ = 5M2 um = Aw
Good Things toKnow:
ME Mass of Earth 2 6 X 1024 kg
M:3 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
rS/e Sun~Earth separation distance 2 1.5 x 1011 m
99/ Grav. acceleration at surface of Earth 2 10 rn/s2
2
G Gravitational constant 2 § x 10“10 Nmz/kg2
yr 1 Year 2 7r X 107 s
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
vs Speed of sound in 20° C air 2 340 m/s
e Fundamental unit of charge 2 1.6 X 10”19 C
1/471’60 Electrostatic constant 2 9 x 109 N mz/C2
no Permeability constant = 47r X 10“7 Tm/ A
BE Magnetic ﬁeld near Earth’s surface 2 5 X 10"5 T
do Approximate nucleon diameter 2 2.4 fm
Binomial expansion (1 + x)" 2 1 + 7251:, for 3: << 1
1 in = 2.54 cm Circumference of a circle 27r R
9 lbs 2 40 N Area of a circle 7rR2
9 mph 2 4 m/s Volume of a sphere 4—32 R3
1 MeV = 1.6 x 10“13 J Surface area of a sphere 47r R2
1 rad = 0.010 J/kg 1 Ci :— 3.7 X 1010 Bq E 3.7 X 1010 decays/s ...
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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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