2211_Test3_Key
-
Upload
amber-lucas -
Category
Documents
-
view
5 -
download
2
description
Transcript of 2211_Test3_Key
PHYS 2211 Test 3
Spring 2015
Name(print) Lab Section
Greco (K or M), Schatz(N)Day 12-3pm 3-6pm 6-9pmMonday K01 K02Tuesday M01 N01 M02 N02 M03 N03Tuesday K03 K05 K04 K07 K06 K08Thursday M04 N04 M05 N05 M06 N06
Instructions
• Read all problems carefully before attempting to solve them.
• Your work must be legible, and the organization must be clear.
• You must show all work, including correct vector notation.
• Correct answers without adequate explanation will be counted wrong.
• Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anythingyou do not want us to grade
• Make explanations correct but brief. You do not need to write a lot of prose.
• Include diagrams!
• Show what goes into a calculation, not just the final number, e.g.: a·bc·d = (8×10−3)(5×106)
(2×10−5)(4×104) =
5× 104
• Give standard SI units with your results.
Unless specifically asked to derive a result, you may start from the formulas given on theformula sheet, including equations corresponding to the fundamental concepts. If a formulayou need is not given, you must derive it.
If you cannot do some portion of a problem, invent a symbol for the quantity you can notcalculate (explain that you are doing this), and use it to do the rest of the problem.
Honor Pledge
“In accordance with the Georgia Tech Honor Code, I have neither givennor received unauthorized aid on this test.”
Sign your name on the line above
- Test - key - # '
Sakura kinomoto
PHYS 2211Please do not write on this page
Problem Score GraderProblem 1 (25 pts)Problem 2 (25 pts)Problem 3 (25 pts)Problem 4 (25 pts)
Problem 1 (25 Points)
An object of mass m and velocity v is acted upon by a net force Fnet at time t, as indicated in figure below.
(a 6pts) Using the numbered direction arrows shown, indicate (by number) which arrow best represents
the direction of the quantities listed below. If the quantity has zero magnitude or if more information isneeded to determine the direction, indicate using the corresponding number listed below.
• p, the object’s momentum.
• (Fnet)∥, the component of Fnet that is parallel to the object’s velocity.
• (Fnet)⊥, the component of Fnet that is perpendicular to the object’s velocity.
• dpdt , the time rate of change of the object’s momentum.
• (dpdt )∥, the component of dpdt that is parallel to the object’s velocity.
• (dpdt )⊥, the component of dpdt that is perpendicular to the object’s velocity.
(b 4pts)At time t, the object’s speed is (circle one):
• increasing.
• decreasing.
• constant.
• unable to be determined with the given information.
5
I
3'Ey 2
1
3
*
(dpdt )∥ and (dpdt )⊥ for an object of mass m are shown at time t in the figure below.
(c 10pts) Using the numbered direction arrows shown, indicate (by number) which arrow best represents
the direction of the quantities listed below. If the quantity has zero magnitude or if more information isneeded to determine the direction, indicate using the corresponding number listed below.
• p, the object’s momentum.
• Fnet, the net force acting on the object
• (Fnet)∥, the component of Fnet that is parallel to the object’s velocity.
• (Fnet)⊥, the component of Fnet that is perpendicular to the object’s velocity.
• dpdt , the time rate of change of the object’s momentum.
(d 5pts)At time t, the object’s speed is (circle one):
• increasing.
• decreasing.
• constant.
• unable to be determined with the given information.
10 - or - ( 6 or2)7
Metty 6
8
7
AY
Problem 2 (25 Points)
A person rides in an elevator above the surface of the Earth. The elevator descends a distance h towardsthe Earth at a constant speed.
(a 5pts) The work done on the person by the Earth is (circle one) Positive Negative Zero
Insufficient Information to Answer
(b 5pts) The change in gravitational potential energy of the person+Earth system is (circle one) Positive
Negative Zero Insufficient Information to Answer
(c 5pts) The elevator cable breaks and the person+elevator fall an additional distance h. During thissecond phase, the work done on the person by the Earth is compared to that done in part (a). The workdone now is (circle one) More Than Less Than The Same As Insufficient Information to
Answer
A
:= F. DF = I -
mg)
th) ⇒ W > 0
A :U = mg Bh = mg ( hf - hi ) ⇒ hf < hi ⇒ DU < 0
A := ( - mg) th ) = same as in Part (a)
No#: F = mg close to the surface of the Earth . If the elevator
moved at much greater heights ,then you need to consider
F = Gmt ,so the force is larger at smaller heights ,
and
in that Case,
the correct answer here would be ' ' More than'!
(d 5pts) You push down on a positive charge Q1 and your friend pushes up on a positive charge Q2. ChargeQ1 moves down through a distance ∆r1 and Q2 moves up through a distance ∆r2, as shown in the diagram.For the system of the two charges, the net work done by the external forces F1 and F2 is (circle one):
Positive Negative Zero Insufficient Information to Answer
(e 5pts) You push to the left on a positive charge Q1 and your friend pushes up on a positive charge Q2.Charge Q1 moves through a distance ∆r1 to the left and Q2 moves through a distance ∆r2 to the right,as shown in the diagram. For the system of the two charges, the net work done by the external forces F1
and F2 is (circle one):
Positive Negative Zero Insufficient Information to Answer
AX
FTSDF,
>0 and Es
.sFz>o
,so Whet >0
AX
FT. of >0 and FiDFz=0 , so Wnet >0
Problem 3 (25 Points)
Consider the child’s toy shown in the diagram. This toy is made by attaching a ball of mass m to a woodenstick by two identical strings. The stick is then rotated between the palms of your hand so that the balltravels around a circle of radius R at constant velocity v. In the diagram, gravity points down (i.e. parallelto the stick)
(a 5pts) In the space below draw a free bodydiagram for the ball.
.
Ip :i.→
•.⇐
÷¥lpt g Npt
(b 20pts) Determine the tension in each string if both strings make an angle of θ with the stick.
Ball moves in a circle at constant speed ,so Frett to
Circular motion, so /Fnet±|= m÷
, pointing to the left
Trajectory is Parallel to the ground ( no motion in y - direction)
⇒ CEEI ⇒ T,
coso- Tzuso -
mg= 0 [ Eg . I ]
vertical
⇒K¥t⇒T
,
sina.it#non=nxn.=s.zB%9aEYI
SdueEq1f#
UseEq4inEq3T, coso - Tz Gso - mg
=°
T,
= z + IT =
( T,
- Tz) cos o = mgmv
2
T - Tz = It = # .
- II. + I÷=
T,
= Tz + and [E8i3]myT = #,→ +mom cost
PlugEq3intoEg÷sinotzsino
= mtf [EqS]
(T, +Tz ) Sino =T÷ * The tensions in the two
( It # +Tfs ; no = my strings are given by
ZTZ + ¥ = myEq 4 and Eq .
S .
Rsino
⇐n÷o . ¥ ElliffTz=zk¥o - F÷o [E£4]
Problem 4 (25 Points)
A rock of mass m is released from rest very far from the Earth (i.e. r = ∞). The only force acting onthe rock is the gravitational force of the Earth. In the following questions you can assume the mass of therock is much less than the mass of the Earth and neglect air resistance.
(a 5pts) Determine the velocity of the rock the instant it reaches the surface of the Earth. Your answershould not be numeric.
(b 5pts) For the case considered above, sketch the: Kinetic, Potential, and Total energy for the Earth+Rocksystem.
←
.ee#eiitux::ogYi:FE*
Kf t Uf =° direction :
= GM⇐ towards center
MVZ +'
GMT = O of Earth
fM = mass of Earth
lz m/v2=GMI W¥ :R= radius of Earth
-:t←UR
lphto-
A rock with mass m is released from rest at the Earth’s surface and drops through a hole all the way tothe center of the Earth. The magnitude of the gravitational force on the rock is mgr/R, where r is therock’s distance from the center of the Earth and R is the Earth’s radius. This force points towards thecenter of the Earth and is the only force acting on the rock. In the following questions you can assume themass of the rock is much less than the mass of the Earth and neglect air resistance.
(c 10pts) Determine the velocity of the rock the instant it reaches the center of the Earth. Your answershould not be numeric.
¥ system : onlythe rock
or> Initial : rock @ surface ( v = R )r -
Final rock @ center ( r = o )
W = SF . dr. = f÷T±dr= TI ,[ rdr =
=T÷E÷ti=tYIt*⇒= met
BE= Dk = W
- os
"
EYE:# Iscv2= GR
V = GRM→ direction :
Away from the
center of the Earth
(d 5pts) For the case of the rock falling through the Earth, the potential energy of the Earth+Rocksystem is given by mgr2/(2R). On the graph below, sketch the Kinetic, Potential and Total energy for theEarth+Rock system.
(Extra credit 5pts) How does your answer to part (c) compare to your answer in part (a)? Hint: the ratioof the two velocities should not depend on any of the parameters in the problem.
- -
k+uftp.2pt⇒i
2
A :what is g ? ⇒ rxg=Gyre ⇒ g= Gary
P¥A: Va = 2*17 Part: ve =gRr = Greth = off
'
⇒ nano ÷=¥TnHfTe= £ or "÷ a
* The speed at crashing Mart A) is MZ times larger
than the speed at the center of Earth LPart C) .
Things you must have memorized
The Momentum Principle The Energy Principle The Angular Momentum PrincipleDefinition of Momentum Definition of Velocity Definition of Angular Momentum
Definitions of angular velocity, particle energy, kinetic energy, and work
Other potentially useful relationships and quantities
γ ≡1
!
1−
"
|v|
c
#2E2 − (pc)2 =
$
mc2%2
dp
dt=
d|p|
dtp+ |p|
dp
dtF∥ =
d|p|
dtp and F⊥ = |p|
dp
dt= |p|
|v|
Rn
Fgrav = −Gm1m2
|r|2r Ugrav = −G
m1m2
|r|&
&
&Fgrav
&
&
&≈ mg near Earth’s surface ∆Ugrav ≈ mg∆y near Earth’s surface
Felec =1
4πϵ0
q1q2
|r|2r Uelec =
1
4πϵ0
q1q2|r|
&
&
&Fspring
&
&
&= kss Uspring =
1
2kss2
Ui ≈1
2ksis2 −EM ∆Ethermal = mC∆T
rcm =m1r1 +m2r2 + . . .
m1 +m2 + . . .I = m1r21⊥ +m2r22⊥ + . . .
Ktot = Ktrans +Krel Krel = Krot +Kvib
Krot =L2rot
2IKrot ==
1
2Iω2
LA = Ltrans,A + Lrot Lrot = Iω
ω =
'
ksm
v = d
'
ksima
Y =F/A
∆L/L(macro) Y =
ksid
(micro)
Ω =(q +N − 1)!
q! (N − 1)!S ≡ k lnΩ
1
T≡
∂S
∂E∆S =
Q
T(small Q)
prob(E) ∝ Ω (E) e−E
kT EN = −13.6eV
N2where N = 1, 2, 3 . . .
EN = N!ω0 + E0 where N = 0, 1, 2 . . . and ω0 =
'
ksima
(Quantized oscillator energy levels)
Moment of intertia for rotation about indicated axis
The cross productA× B = ⟨AyBz −AzBy, AzBx −AxBz, AxBy −AyBx⟩
Constant Symbol Approximate ValueSpeed of light c 3× 108 m/sGravitational constant G 6.7× 10−11 N · m2/kg2
Approx. grav field near Earth’s surface g 9.8 N/kgElectron mass me 9× 10−31 kgProton mass mp 1.7× 10−27 kgNeutron mass mn 1.7× 10−27 kg
Electric constant1
4πϵ09× 109 N · m2/C2
Proton charge e 1.6× 10−19 CElectron volt 1 eV 1.6× 10−19 JAvogadro’s number NA 6.02 × 1023 atoms/molPlank’s constant h 6.6× 10−34 joule · second
hbar =h
2π! 1.05 × 10−34 joule · second
specific heat capacity of water C 4.2 J/g/KBoltzmann constant k 1.38 × 10−23 J/K
milli m 1× 10−3 kilo K 1× 103
micro µ 1× 10−6 mega M 1× 106
nano n 1× 10−9 giga G 1× 109
pico p 1× 10−12 tera T 1× 1012