Physics 120

Exam #2

May 4, 2011

Name___________________________________

Multiple Choice /16 Problem #1 /28 Problem #2 /28 Problem #3 /28

Total /100

PartI:Multiple­Choice:Circlethebestanswertoeachquestion.Anyothermarkswillnotbegivencredit.Eachmultiple­choicequestionisworth4pointsforatotalof16points. 1. A person hoists a bucket of water from a

well and holds the rope, keeping the bucket at rest as in the left photo. Call this situation A. A short time later the person ties the rope to the bucket so that the rope holds the bucket in place as in the right photo. Call this situation B.

a. The tension in situation B is greater than in situation A.

b. The tension in situation B is equal to that in situation A. c. The tension in situation B is less than that in situation A. d. Three is no way that the individual tensions in the ropes can be determined. 2. A 2.0 x 103 kg car travels at a constant speed of 12 m/s around a circular curve of radius

30 meters. As the car goes around the curve, the time rate of change of the car’s momentum is directed

a. toward the center of the circular curve. b. away from the circular curve. c. tangent to the curve in the direction of motion. d.tangenttothecurveoppositethedirectionofmotion.

3. If the Earth orbit the sun in a nearly circular orbit of radius r = 1.5x1011m and the orbital period of the Earth in its orbit is T = 31.5x106s (1 year), the mass of the sun is most nearly given by the expression

a.

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MSun =2πr2

GT 2 b.

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MSun =4π 2r3

GT 2 c.

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MSun =2π 2Gr3

T 2 d.

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MSun =4πr3

GT 2

4. A spring of stiffness k is suspended from the ceiling an allowed to hang vertically.

The spring has a relaxed length (measured from the ceiling) of Lo. A mass m is suspended from the spring and the spring stretches by an amount L (measured from the ceiling.) The stiffness of the spring can be approximated using

a.

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k =m

L − Lo b.

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k =L − Lomg

c.

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k =mgL

d.

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k =mg

L − Lo

PartII:FreeResponseProblems:Thethreeproblemsbelowareworth84pointstotalandeachsubpartisworth7pointseach.Pleaseshowallworkinordertoreceivepartialcredit.Ifyoursolutionsareillegibleorillogicalnocreditwillbegiven.Anumberwithnoworkshown(evenifcorrect)willbegivennocredit.Pleaseusethebackofthepageifnecessary,butnumbertheproblemyouareworkingon.1. Suppose that you are going to make a 1m long tungsten wire that has a square cross-

sectional area of 1cm2. The density of tungsten is

€

ρ =19250 kgm 3 and its molecular

mass is

€

183.8 gmole .

a. What is the interatomic bond spacing, d?

Assuming a cube with a1m3 volume we have

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m = ρV =19250 kgm 3 ×1m3 =19250kg. The

number of atoms in this volume is given by

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# volume =19250kg× 1mol0.1838kg

×6.02 ×1023atoms

1mol= 6.3×1028 . The number of atoms on a side is

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# side = 6.3×10283 = 3.98 ×109 . Therefore in a side of length 1m, we have

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d =1m# side

=1m

3.98 ×109= 2.51×10−10m

b. What is the interatomic bond stiffness, kiab? (Hint: Assume that the wire has a

stiffness kwire = 4.11x107 N/m.) In 1m of wire there are

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Nseries =1m

2.51×10−10m= 3.98 ×109 springs in series. The stiffness of

an interatiomic bond is

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k iab= Nserieskchainwhere

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kchain =kwireNparallel

and the number of parallel

chains of atoms is

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Nparallel =

1cm2 ×1m2

100cm( )2

2.51×10−10m( )2=1.58 ×1015 . Therefore,

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k iab= Nserieskchain =NserieskwireNparallel

=3.98 ×109

1.58 ×1015× 4.11×107 N

m =103.5 Nm

.

c. What is Young’s modulus for tungsten?

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Y =kiabd

=103.5 N

m

2.51×10−10m= 4.12 ×1011 N

m 2

d. If a 5kg mass were hung from this wire when the wire is held vertical, by how

much would the wire stretch?

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Stress =Y × strain→ FA

=Y ΔLL

ΔL =FLYA

=5kg × 9.8 m

s2×1m

4.12 ×1011 Nm 2 × 1cm2 ×

1m2

100cm( )2

=1.2 ×10−6m =1.2µm

2. An amusement park ride shown below, consists of a rotating circular platform 8.0m in diameter from which 10kg seats are suspended from a chain with mass 4kg and assume that the mass of the chain is located at the center of the chain. When the system rotates it does so at constant speed the chain makes and angle θ = 28o with respect to the vertical and a 40kg child is riding in a chair.

a. Starting with the momentum principle, what

is the speed of the chair when it is empty (no child is in the seat)?

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dpdt

=mtotalv

2

R,0,0 = Fnet = FT sinθ,FT cosθ −mchairg −mchaing,0

y − dir : FT cosθ −mchairg −mchaing = 0⇒ FT =mchair + mchain( )g

cosθ=

14kg× 9.8 ms2

cos28=155.4N

x − dir : mtotalv2

R= FT sinθ ⇒ v =

FTRsinθmtotal

=155.4N × 4m + 2.5msin28( )sin28

14kg= 5.2 m

s

b. What is the tension in the chain when there is no child in the chair?

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FT =155.4N

c. Again, using the momentum principle, what is the speed of the chair when the child is in the seat?

€

dpdt

=mtotalv

2

R,0,0 = Fnet = FT sinθ,FT cosθ −mchairg −mchaing −mchild g,0

y − dir : FT cosθ −mchairg −mchaing −mchild g = 0⇒ FT =mchair + mchain + mchild( )g

cosθ=

54kg × 9.8 ms2

cos28= 599.4N

x − dir : mtotalv2

R= FT sinθ ⇒ v =

FTRsinθmtotal

=599.4N × 4m + 2.5msin28( )sin28

54kg= 5.2 m

s

d. What is the tension in the chain when a child is in the chair?

€

FT = 599.4N

3. A block of mass m = 5kg is moving along a horizontal frictionless surface at a constant velocity of

€

v i = 2,0,0 ms . At a location taken as the origin the block

encounters a region in which there is friction present and the coefficient of kinetic friction is µk = 0.08.

a. Starting from the momentum principle

€

d p dt

= F net , what is the velocity of the block

as a function of time? (Hint: You will need the integral

€

xndxxi

x f∫ =xn+1

n +1 xi

x f

.)

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d p dt

=dpx

dt,0,0 =

F net = −µkFN ,FN −mg,0

y − dir : FN −mg = 0⇒ FN = mg

x − dir : dpx

dt= m dvx

dt= −µkmg⇒ v fx − vix = dv

vix

v fx∫ = −µkg dt0

t∫ = −µkgt

v f = v i + −µkgt,0,0 = vix −µkgt,0,0

b. How long will it take before the block comes to rest? (Hint: Take time ti = 0

when the block first encounters the region with friction.)

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v fx = 0 = vix −µkgt→ t =vixµkg

=2 m

s

0.08 × 9.8 ms2

= 2.55s

c. What is the final location of the block when it comes to rest?

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r f = r i + v avgΔt = 0,0,0 m +

2,0,0 ms + 0,0,0 m

s

2

× 2.55s = 2.55,0,0 m

d. Suppose that instead of friction bringing the block to rest, the block encounters a

region of space (the beginning of which is taken again as the origin) over which the block is subject to a variable force given by

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F res = −ρv 2,0,0 N , where ρ is a

constant with units of Ns2/m2. In this case, what is the velocity of the block as a function of time, taking ti = 0 when the block first encounters this variable force?

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d p dt

=dpx

dt,0,0 =

F net = −ρv 2,FN −mg,0

y − dir : FN −mg = 0⇒ FN = mg

x − dir : dpx

dt= m dvx

dt= −ρv 2 ⇒

dvv 2vix

v fx∫ = −1

v fx

−1

vix

−

ρm

dt0

t∫ = −

ρm

t

v fx =mvix

ρvixt + m⇒ v f =

mvix

ρvixt + m,0,0

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g = 9.8 ms2

G = 6.67 ×10−11 Nm 2

kg 2

1e =1.6 ×10−19C

k =14πεo

= 9 ×109 C 2

Nm 2

εo = 8.85 ×10−12 Nm 2

C 2

1eV =1.6 ×10−19Jµo = 4π ×10−7 TmAc = 3×108 m

s

h = 6.63×10−34 Js

me = 9.11×10−31kg =0.511MeV

c 2

mp =1.67 ×10−27kg =937.1MeV

c 2

mn =1.69 ×10−27kg =948.3MeV

c 2

1amu =1.66 ×10−27kg =931.5MeV

c 2

NA = 6.02 ×1023

Ax 2 + Bx + C = 0→ x =−B ± B2 − 4AC

2A

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Circles : C = 2πr = πD A = πr2

Triangles : A = 12 bh

Spheres : A = 4πr2 V = 43 πr

3

Usefulformulas:

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p = γm v keff ,parallel = nparallelkindividual

γ =1

1− v 2

c 2

keff ,series =kindividual

nseries

v avg = v i + v f

2 stress = Ystrain → F

A= Y ΔL

L; Y =

kiab

d F g = m g F gravity =

GM1M2

r122

ˆ r 12

F spring = −k s ; s = L − Lo( )ˆ s

ρ =MV

MomentumPrinciple:

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p f = p i + F netΔt

d p dt

= F net

Position‐update:

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r f = r i + v avgΔt =

r i + p

m 1+p2

m2c 2

Δt

Vectors

Constants Geometry

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magnitude of a vector : a = ax2 + ay

2 + az2

writing a vector : a = ax,ay,az = a ˆ a = ax

ˆ i + ayˆ j + az

ˆ k