This preview shows pages 1–17. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EXAM #2
PHYS 125
(g) 31 OCTOBER 2006 @ Name: iétgi‘rw E‘J‘st (please print legibly) Student ID #: ‘m .JM?&3 Part Score Short answer ,i
1 Problem #1 Z z m
r0 em #2 .. 6 K ' 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 my honor, I have neither given nor received any unauthorized
aid on this exam.” .f‘ 5,1,0 {‘T/U “a el 0: awe,
(signature) (date) PHYS 125 Exam #2 31 October 2006 Short Answer (40 points) Each of the ten questions below is worth 4 points. 3 1. A car makes a righthand turn as shown below at a constant speed. Draw
the average acceleration vector. If there is no average acceleration, indicate it with the zero vector 0. 4, ﬂ (0‘99 2. A piano mover raises a IOUkg piano at a constant rate using the friction
less pulley system shown here. With how much force is be pulling on the
rope? Ignore friction and assume g = 10 m/s2. (a) 2,000 N
(b) 1,500 N 413;"
(c) 1,000 N
(d) 750 N
/ (e) 500 N
(f) 200 N (V? 100 _
50 N p (h) gmpossible to determine \k/ U PHYS 125 Exam #2 31 October 2006
—————~_—__~______._.___________ & 3. In positron emission tomography, a positron and an electron annihilate at
rest producing two photons. One is emitted with the momentum vector
shown. Draw the momentum vector for the other photon. Photon 1 4. A jet plane ﬂies at constant speed in a vertical circular loop. Circle the
point(s) at which the force exerted by the seat on the pilot is greatest. compact car and a large truck collide head on and stick together. Which undergoes the larger momentum change? (a) car
(b) truck Cant tell without knowing the ﬁnal velocity of combined mass. Which vehicle undergoes the larger acceleration during the collision? @ car (b) truck
(0) Both experience the same acceleration. (d) Cant tell without knowing the ﬁnal velocity of combined mass. PHYS 125
31 October 2006 atigrnﬁon a circular path. Which is greater when it starts? tangential acceleration is greater. , c) The tangential and radial accelerations are equal. Its radial acceleration is greater. Which is greater after a long time? a Its tangential acceleration will be greater. !@ Its radial acceleration will be greater. c) The tangential and radial accelerations will be equal. 7. You have a superball (a very bouncy projectile) and a spherical lump of
clay that have the exact same mass. If you can throw each at the same
speed, which would be more effective in knocking over a bottle? (a) he superball is more effective because it will bounce off the bottle. \ ) The lump of clay is more effective because it will stick to the bottle. Both are equally effective because they carry the same momentum
before they hit the bottle. ppose there are three astronauts outside a spaceship far away from any
‘ planets or large masses, and they decide to play catch. All the astronauts
weigh the same on earth and are equally strong. The ﬁrst astronaut throws
the second one toward the third one and the game begins. Including the
ﬁrst, how many “throws” (not catches) will take place before the game must end? (d) 4
Where is no physical reason the game should ever have to end. \r “074/ PHYS 125 Exam #2 31 October 2006 Use the ﬁgure below to answer the next 2 questions. The ﬁgure shows the
potential energy function U of a particle with total mechanical energy E as a
function of position x. At what point is the magnitude of the force greatest? 5 “if”; '
At this point, does the force point in the positive or negative direction? if 155%? t W“; oi { 10. At What p0int(s) is the particle moving the fastest? «ﬂying? 4” ,X J\
x.) \ kv‘lvuc 95v .JZU‘ 1:33, ..Q,i:’ﬁ“%‘ v4 PHYS 125 Exam #2 31 October 2006 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) The Headless Horseman is sitting atop his horse just outside
the village of Sleepy Hollow. The Headless Horseman uses a 3.00~kg jack as?
T 556‘ o—lantern in place of his head, and it accidently rolls off his neck from a
r ,(n /\ height of 2.50 In above the ground. The jack—olantern strikes the ground 3))?
c,» / I ‘ m and bounces up to a height of 0.10 m. / X,» ‘54 .. V l’ l x ‘ (a) (20 points) Find the magnitude and direction of the impulse that the
Q 5{ ground exerted on the jacko—lantern.
(\ I R' t (“a «2
J .5 A (>4 (x434 3 \Jf 7' ’ Vz A“: K r‘ ,
Q i  46.8%?«0; “972 r/ 1
i I J . A
J’ANU 'x/fxf’jl‘iiv
.) 7 M, _n¥xr~<”'""' l V
'21 ff: V “‘1 M Z “‘1.
/ Q, "‘ H U/
f \ i/l j x \ ‘\\ / A w .. 7
J ‘” ozéfiﬂjtol) r u? ‘4.
w; 11‘ch (31%” ‘lH’Wtb ~— Iet, e/vg t
V . '1’7’x a} PM \ b r ‘ VJ : , U) or ,,,,,,,,,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,,,, a all as {P ‘Jr‘Ssrwﬁjd *“hér' avers J i ~ (b) (10 points) Beigw~is—agraphertﬁafare;damage'ty’thé~gfcsmdsn,.,/’
the jack—o—lantern as a. function of time. Note that the units on the
horizontal axis are milliseconds. What is the value of Fmax?
}\ AC, 5 Force vs. Time
._lr / ,s);‘ j «r1
/ V a M «4 (“he ‘2"! )V 7 4/ A z w“
f'x i w: a r
V ‘J \ J . :1
@L, 5 10 15 20 25 30 35 4o /, ,.
Time (ms) > AU
3 J; u
, ),. /
6 {Jo / A 1/;
€91 &
f, __ / /
x, L,, a») y
g // 0W3 PHYS 125
Exam #2 31 October 2006 M PHYS 125 Exam #2 31 October 2006
W 2. (30 points) You are helping create a haunted house for a school carnival,
You’d like to create a spooky ghost scene in which a 2.5kg object, painted
as a ghost, is suspended from the ceiling by a 2.00—m wire and acts as a
pendulum. Initially held out horizontally from the point at which the wire
is tied, the ghost then swings freely. Another volunteer tells you that they have some thin wire available that
can support tensions up to 50.0 N. However, you know from your physics
class that it if you use that wire, it will break. Find the angle 0 at which it will break. PHYS 125
Exam #2 31 October 2006 M PHYS 125 Exam #2 31 October 2006
—~————__________________ //A\. // 3. (3 points) This problem extends over two pages. Two vampires,
(\ ngelus and Spike, have a strange set of sleeping arrangements. They
\“ sleep upsidedown connected to one another by a rope attached at their
feet and slung over a pulley. Most days Angelus and Spike have the same
mass, approximately 75 kg, but today Spike has taken on an extra 10 kg
due to an all night binge on the blood of young college students. The
masses of the pulley and rope are negligible compared to our vampires’ masses, and the system is frictionless. Spike ' Angelus (a) (10 points) Draw a free body diagram for each vampire, making sure
to clearly label all axes and forces. 5" 10 PHYS 125 Exam #2 31 October 2006 (b) (10 points) With what magnitude of acceleration will the system
move? 4" i0 7;? m ow C m, a \gféflﬁ/ /7§@%2 m I? >
“2/” M r x ,0 ‘9 mo 74% #433 “<55 593‘
I I ‘», ~l I (c) (10 points) rWhat is the tension of the rope on Spike’s side TS? On
Angelus’s side Ta? //'P im¢ waivka Mr”
«i ﬂkyj/ﬁq mde {/ ‘ijﬁg 00:1 313%) EHMWﬁm A mth
Mr M]
“b i ‘\ , ‘K hr / i 3 fLS'i‘xOxLqu'yb
:.¢2;(’<?,r7:1w:3 73/791 mag/95$; A
>7 7 17%0331S/t/r? izoﬁa7fw ’2 (My: 7: '\_ “a
“mix We 9 J 11 Equation Sheet PHYS 125 Exam #2
M Kinematics:
_, dé't _, t‘ _, _, Aé’
v0?) = d2) As(t) = v(t) dt 1;an = Z?
_, dz? t _, tr _, ,, A27
a(t) = d: ) Av(t) = a(t) dt aavg 2 Z—t
17’ = 5— I7
Constant Acceleration:
1
As : ‘UijsAt + §a3(At)2 v” = viys + asAt 2a3As = 142,8 — v33
 ‘ x, J
Czrcular M otzon: 3?), byng '03 K0 at} 2 3 ‘ 27r
ar : _ (1r = w 'I‘ ’Ut = wr w : ._._
7' T
Forces: dd
FinetzzT3Ft1l: ﬁAonB=ﬁleonA
. d w _‘ _. _‘ _. 1 _‘ —¢ a
ifs! S #slnl 1ka = min! lfrl = Hrlnl lDl E apairsz M = mg Fsp = 49433 Momentum, Impulse, and Energy: 1;: m J: F(t)dt f: A}? W Good Things tDKHQWI 7 ME Mass of Earth ” “2'6x 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
Ce Circumference of Earth 2 4 x 107 m rs/e Sun—Earth separation distance 2 1.5 x 1011 m 96 Grav. acceleration at surface of Earth 2 10 m/s2 G Gravitational constant 2 g x 10‘10 N mz/kg2
yr 1 Year 2 7r x 107 8 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 ’Us 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 mZ/C2
p0 Permeability constant 2 471’ x 10‘7 T m / A
BE Magnetic ﬁeld near Earth’s surface 2 5 X 10’5 T do Approximate nucleon diameter 2 2.4 fm Equation Sheet
xam #2 PHYS 125 E
M Binomial expansion (1 + 2:)" 2 1 + mm, for :1: << 1
1 in = 2.54 cm Circumference of a circle 27R
9 lbs 2 40 N Area of a circle 7r R2
9 mph 2 4 m/s Volume of a sphere 337: 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 E 3.7 X 1010 decays/s PHYS 125
Exam #2 31 October 2006 ——————————.___________—__— Name: gig (Ziggdi [‘3 ID: L) (ZMSE 2g Exercise Steps Max Score
1 [i should point up and to the left at 4
45°. 2 (e) 500 N 4 3 Photon 2 vector 4
points opposite to photon 1 2
approximately equal in length to photon 1 2 4 Greatest force 4
Answer includes point 8 1
and includes only 8 3 5 Truck / car collision
(c) Momentum changes are the same
(a) The car undergoes greater acceleration tow“; 6 Circular Motion
(a) Tangential is greater initially
(b) Radial is greater at later time mm“; 8 (b) Two throws 4
9 Force 4
Force magnitude is greatest at 4 2 Force points in positive direction at 4 2 10 Particle is fastest at 5 4 Total 40 i
0
1
_2_
l
+
O
__.?__
__0_
_2.__.
7 (a) Superball 4 {t
0
O
_._O___
i
A PHYS 125 Exam #2 31 October 2006
M Problem Steps 1a 1b Find magnitude and direction of im pulse
Accurately use conservation of energy or
kinematics to ﬁnd velocity of object be fore hitting the ground 111'), = — g i 2
—7.0 m/s Accurately use conservation of energy or
kinematics to ﬁnd velocity of object af
ter hitting the ground 112,}, = «277% 2:
1.4 m/s Magnitude of impulse is difference in
momentum J = lmvgyy — mum] 2
25.2 kg m/s Direction of impulse is upward Find Fm!“ , w ,, ,,
Use area. of triangle for area under F vs. t curve J: det = gFmaxAt Use time of At = 35 ms: 00% 3 Use impulse from part (a) to ﬁnd Fmax 2
2J/At 2 1440 N Appropriate units of force Total Max
20 5 30 Score Exam #2 Problem Steps 2 Find Break Angle Apply conservation of energy Change in potential energy is AUg =
—mgL sin6' Change in kinetic energy is AK = §mv2 Apply circular motion Partial credit for accurate FBD Net force in radial direction mv2 / L = T—
mg sin 0 Algebraically solve for 0
mgL sin 0 = %mv2 T — mgsinﬂ = mug/L
sing = T/(3mg) '2 43° Total PHYS 125
31 October 2006 Max Score 30 30 l  HI PHYS 125 Exam #2 31 October 2006
“EH Max Score Problem Steps 3a 3b 3c FBDS Spike Weight must point down Tension must point up There must be no other forces present
One end of the vertical axis must be 1a» beled Angelus
Weight must point down Tension must point up
There must be no other forces present One end of the vertical axis must be la
beled Find Acceleration Application of 2‘“ Law to Spike msg~T =
msa,3 Application of 2“d Law to Angelus T —
mag = maaa Application of acceleration constraint a =
as = :lzaa, depending on labeling of axes
from part (a) ' _ m —a.
Solve for acceleration a —— my '2
0.61 m/s2
Tensions The tensions are equalzTa = T3
Substitute acceleration into one of the
2‘“ Law equations to ﬁnd a tension T = ma(a +g) = ms(g — a) e: 781 N Total 10
5 1
1
1
2 MHI—It—AQ‘ 30 {C ...
View
Full
Document
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

Click to edit the document details