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) .

This page is for extra work, if needed.

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