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Unformatted text preview: Physics 1116 : Prelim #2
7:30  9:00 PM, April 2, 2009 Name : C/ ‘9 (Tc/M7???“ 0 Don’t forget to write down your name
0 Breathe deeply and relax. Nothing can save you now.
0 Calculators are permitted, but laptops or communication devices are not. 0 There are 6 problems (+1 bonus) on the exam and 14 pages, not including
the equation sheets. Please check that you have all of them. Feel free to
remove the equation sheets from the back of the exam. 0 Please show your reasoning clearly on all problems. Partial credit will not be
awarded if we do not understand what you are trying to do. Feel free to use
the back of pages for your working (but let us know that you are doing so). 0 Don’t panic if you have difficulty with a problem, and don’t get hung up on
any one particular problem. This test is designed to be sufﬁciently challenging
that you do not have to be perfect in order to get a good score. 1 3 30 pts 25 pts
Bonus 5 pts
TOTAL DO NOT GO TO THE NEXT PAGE UNTIL INSTRUCTED TO DO SO Short Questions Question 1 10 pts Batman (mass M B) hangs from a building ledge at a height 3h above the ground. A BatPlat—
form consists of a massless ideal spring with spring constant k attached vertically between the
ground and a platform of mass M p. The BatPlatform is initially a height h above the ground and
is directly below Bruce Wayne. Batman loses his grip on the ledge and falls onto the platform, causing the spring to compress
initially. The spring extends again, and when the platform once again reaches h, Batman leaps off
the BatPlatform. If Batman wants to reach the ledge again, what is the minimum velocity that
Batman needs to leap off the platform with? Please give this velocity relative to the platform. ' J / e7Z m
#7514 “MM ene7/ ,& mmm74m/ 7m 4514 a// ‘
54““ 4? you vfové Via/«1‘ reAt'Ki/erg jrewm/ MAUI 3474‘” ’5" C“? éq ’Vﬂén &‘”‘b )dA' ”15 +Wllo Minimum velocity : #— M“ 2 Question 2 : True or False (no partial credit) 2 pts each, 10 pts Each of these true / false questions is only worth 2 points. If you get stuck, you may want to
move on, and come back to some of these questions later. 0 If the external forces acting on a system are conservative, the momentum of the system is
conserved. o In an elastic collision, the speed of an object can change. False 0 To ﬁnd the kinetic energy of an object in the center of mass frame KCM one can add the
kinetic energy in the lab frame KM to %mvg~M where ’UCM is the center of mass velocity. 0 The speed of a rocket can never exceed the speed of its exhaust (exhaust speed taken relative
to rocket) —a o F = A6 9 is a conservative force, where A is a constant. . Question 3 10 pts An observer on a train is moving at a constant velocity :10 forward relative to the ground.
Looking out the window, she sees an accelerating car of mass m on the ground. At t = to, she
sees the car moving a velocity —v0 (backwards) relative to the train. At t 2 151, she sees the car
stationary relative to the train, and at £2, she sees the car moving at +210 forwards relative to the
train. Given that the car burns C’ liters of fuel between to and t1, how much fuel does the car burn
between t1 and t2, assuming that we can neglect air resistance and internal friction? What is the
average power delivered by the engine between t1 and t2? Velalib; Jfo ammo“ :1 r:::I—> [:3———> 06f Wﬁho “U053 , no W12) . 9110‘
i=6, 7.. 7' . .
AKEéfl—tu)= ém (rub '" gimCO) = Jimﬂpl 4—"? C lulu; (E rpuj n. ’L a. ‘
Am; (were): Jilv'W) ~5Emtvo) = 333ml, ,_,., 36, Has GEM rpm = AK‘E “gm’lél —,__.————
.. A15 Liters of fuel burned : 30
3mm Average Power : 2 £131: 3, ) END OF SHORT QUESTIONS 7.. Question 4 20 pts Brainiac has trapped the last Kryptonian (mass M3) in a pair of ﬁelds which have the form
U1: A62” and U2: Ae‘” , such that UTOT — U1 + U2, as shown below. ( 3.) Find the position of S Jperman meg, assuming that he IS at an equilibrium point. {3119 where d—Lo—s “(A9, +Ae,x—) 4,43," er‘j so .—\ (b) Given UToT, ﬁnd the resulting force from this potential, FTO’I‘ and sketch it below.
_. ch!
1:: """ =A(é:x .— 2329
F , _. ~75 2.96
FTOT 2 44(6 “'26 (c) Using his heat vision, Superman is able to displace himself slightly by an amount +Am
away from meg. However, drained of his solar reserves, he is no longer able to use his heat vision,
and slides back towards equilibrium Find Supermahs period of oscillation, assuming that he
utilizes none of his remaining superpowers, and 13 subject only to FTOT keﬁlé $= A("‘°1 ear1) ——a M's fem. answer is 4“” if:
a, “'1 at? T._.. yin: 7'43
x 2:: ‘1» ‘1 Period: if A (6 "' ‘le, ) (cl) Now, accounting for air resistance ( fair oc ——v' ), Superman’s oscillations will be very
gradually damped. We do not know the damping constant, b, but we do know the quality factor
of the system, Q. Find the amount of time, T, it takes for the magnitude of Superman’s oscillations
to be reduced to 1/3 of his initial displacement, Act. You may assume that the air resistance is
very small, and hence the damping is very light. _:‘Q__: Xel’ X Lu ﬁg amFlilude. —‘, X'Ct)= X0 CL
251“ X69 .1. 1 .3115. Question 5 25 pts Iron Man is ﬁghting Crimson Dynamo (both of mass M), and both are ﬂying on a collision course towards one another. Iron Man has an initial velocity vector of 27 = v0[2, 0, 2] and Crimson
Dynamo has an initial velocity vector of 23' = v0[2, 4, 2]. (a) Find the Avelocity of each combatant 1n the center— of—mass frame.
—‘ ‘Pa 2“ o[2°9]+E1421 au [.2 a a]
CM :2. o I I
”M131 lpa’
.3 I,“ = “D: in z wad—2 oil [.2 2 ﬂ) = v, [or—4,0]
W)c_ = “1—7me = RAID 41]” [111]) = 'Uo[0,l,o] .1
'CM
vIr/LCM I ‘1)‘[0, ~11 0] ’UCD,CMI en” 0 9‘ D] (b) If the two collide in a. completely inelastic collision, is it possible given only this information,
to determine their resultant velocity vector immediately after the coliision? If yes, give the velocity;
if no, explain why or why not. —9 41nd, [III Ill A ..——\
(P1311 . = Re I“ I Tl ﬁ‘M‘ as}? Collision ,dm‘vw ’ Since. ‘fl‘e7 are, STUCK $39M
(”[7 3': unkmms). (c) Now assume that the collision was completely elastic, and after the collision, Crimson
Dynamo is observed to be traveling completely in the 5: — y plane (1);, = O) in the lab frame at an
unknown Speed. Find the ﬁnal velocity of Iron Man (in the lab frame). rﬂméa') Veg: 1) ”1)“ ——)'Vo[€l’a,0] = U°[c,d 21+ v,b,;1,;71
a = C..+ Q
a = d"? Q
0 = 64' Q ———9 €=“‘Q.
hr! VLJ‘ m [at 41c.— CJOM
New ms mm; c
I; e=wﬂr ‘ltm (944:0, if waned Wag“; l
c4
.' ’Uco’: MED: 01’2“]
7131“? “Mama?
._.. .4 .._3
‘rUIMI: IMI+FUCM= Wotoforaj+fuotg( 9‘,91 (d) Now assume a third possible collision, where after the collision, Crimson Dynamo is ob— served to be traveling with a ﬁnal velocity of [2%, 2'00, 2%]. How much kinetic energy was lost in
this collision? ,4 ___‘ / ‘Marnznium artworks“: flab. =E/ +191” ‘U W
'F/Vo 'Vq, hi0 VIM fl).
. 9— 7. '1' 1r ‘7—
.. k]; m , Ell—m'vm +—;m’U a“.31((o+21)+ (.2 +412) ,4 14, M190
7' ’L (‘L ? ‘1‘!
l1.
Kg; 3. Jimmie[P1, + "lilting, tang (9.1‘11+911>+C3~ 4; +32) : ll mVD
AK‘E "‘ KEDE ’ Ken AKEzQ: M Question 6 25 pts A block of mass m rests on a frictionless surface. Two hoses are pointed at the block — one
from the left and one from the right. Assume that both hoses spray water at a Speed 110, and both
hoses spray water a constant rate ff? = c. However, the water collides elastically with the left side
of the block, but inelastically with the right side of the block (after colliding, the water just drips
down the right side of the block). Assume that the block is initially at rest, and the hoses turn on right at t : 0. 4K
»——9 V0 (a) Find the force FL that the water exerts on the left side of the block, as well as the force
FR that the water exerts on the right side of the block at t = 0. APTOT =0 9 Milan MM +13»: v0 + A AV 93m. , 2A 1/ MA____\_/__
do 41’9”" AV m a = 'E
—_""' At
'FIJ 0') ‘H "‘ At
Am 9 5m PL: QCVO
before.
__ Am AV Am A'PT¢T=O ) —AmU = —MA\I+ M
v, ' Arm}. r: _.{_N M Ti; Al M
1+: c'U, —a"1’e d‘w FL I 4' QCTUO FRI —CV (b) Assume that at some moment :51, the block is traveling with a known velocity 121. Find the force F L at time 151 as well as the force FR at t1
, _ ’ = P. ' 1’ +  ~
[email protected])‘  '1). (QtM13411) 1).+AV 0 A TT9AM ° M1)! " WI+®+M(QVI “WED a 01—5 66.0% MVoWWMN~AmV°+JAmﬂ+W
sew = Ivei! W
:. v of An = av wow SPQA 4 aﬁnaok L‘M—V‘l I
'1} 1) —'vo—) Mao /
kW“? A’m Wm'ﬂdﬁd’l’oa a": “PL—LEI ' A
I E I at run ¥ (”1 a 0 I bmac/ /
'1. H
" .. 'L
50 9%(ﬁ°~V)QC (”03:10 ~— Mfﬂ 10—? F1." «26 (Va—Wu) Same, ‘03}, ’91“ R APTW=0§ MilMm = (MiAmﬂl—A’i fl}. ‘1), ’IJFH’XJ y—Amti = M+Mﬂlr MAW) +W
5 0 ﬁg mﬁw) = Mm 7..
4. $22. (”0"“)
mm) : 'Vo
— CL CV0 +1]! )1
FR(tl) I “V0 (d) Find the “terminal” (1.6. equilibrium) velocity of the block as t —> 00. m4 mm “0:47 “st Eor=o 20¢ “—11: JFK! 2/5614, WVr zo/(V +'V.)L M 24/ )5 " 'S f ﬁ—iﬁm (U0!
(1)434?) slim m% s a“ U(t~—>oo): (U0 Bﬂﬁ 11 (e) Find the rate of change of kinetic energy of the entire block / water system as a function of the velocity of the block, 1). For simplicity, do not worry about what happens to the water well
after it collides with the block. 0rd Yawn7 am‘l (R143 4 Hog, 13““ LHS (MM e/(axh‘c.
COH'SIW, So no KE €14”!ij an ‘J‘Ltq‘l‘ Suite (PM ”gm PH3 —9 862.0 {3'0 (SW BlocJL dqm__g AKE . At RHJ‘ ' f1) 9. q 1
MW Jm “Ill"D = 3% “C; 14:32" )nt) on, .1;
RH:
AKrHs= LCUJV‘, )(1‘ +1) )("Uo ~19
At ave
'3. v 3
.. c .00 __
AKToT = Tquo‘fvD (fl) '1‘ a .. java Bonus Problem 5 pts Superman is able to “leap tall buildings in a single bound”. Assume that Superman’s vastly
superior physiology is purely a result of the planet Krypton’s much stronger gravitational ﬁeld
(and not the different spectral properties of the Kryptonian sun). Provide a rough estimate of how
many more times denser Krypton is than the Earth, assuming that it has a comparable radius to EZE'ERZQSMWL ‘SLWM C—w ()qu 0L I’m/1614+ lxﬁan M'l‘k. lru‘s [M'h‘al Welou'l7: fuzz: 231m. W' ’2
blunt (S _l1~°~ rm Elba} Superman's [288 can Merl“
_. 191E __ MS'vg—O)
S _ At At 6 I S albeit $5M (La/W ( 0... (IR——
7:10:01: (Lfc :Qal +3 a Far—“U? S'lﬂngf‘t‘37W3kf‘E m g )
a 7 6f 3 P: w
(1)323le ah ~05,“ s. 720 ~ m/s , “was;
4/1000 N 80 1' Su mom can affly 1:: wdumrm ,"l'Lus IS «a 7
eziuaf~lw ms Wa‘ahl‘ (magK7lylm. a T N [0'0 £2 (2.2u300m) N /‘/000 N.
3 039: /u// [cu/X147 M 300m «vino 414,50,
If mm a... am 3:: ~M‘ 74% mil a7! ewe. ghoégqgiﬁlr—e alrdfk.
Nilf For 1 point, Superman’s Kryptonian name is : Kﬂ/‘Eﬂ, Sow 070 def—E1 pKrypton 3 PS. : assam' he. has 79%” Jam 7% weﬂu' 0—1.) ...
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 Spring '05
 ELSER, V
 mechanics, Special Relativity

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