04/19/23 Physics 201, Spring 2011 1

Physics 201: ReviewPhysics 201: Review

Final Exam: Wednesday, May 11, 10:05 am - 12:05 pm, BASCOM 272

The exam will cover chapters 1 – 14

The exam will have about 30 multiple choice questions

Consultations hours the same as before.

Another review sessions will be held by your TA’s

at the discussion session

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Problem SolvingProblem Solving

Read and understand the problem statement completely:Often this is helped by a diagram showing the relationships of the

objects.Be sure you understand what is wanted Be sure you understand what information is available to you (or

can be found from the available information)

Translate the situation described to physics concepts.Be alert for clues regarding the choice of relationships

(e.g. conservation of energy, conservation of momentum, rotational or linear motion, …….)

Be alert for detail that would qualify the use of some concepts (e.g. friction affecting conservation of energy.)

If there are other “unknowns” involved how can you find them (or eliminate them).

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Problem solving…..Problem solving…..

After choosing the appropriate relationship between the concepts (equation), find the target quantities -- at first algebraically, then substitute numbers at the end. Check

Does the answer make sense?Are the units consistent?Often with multiple choice questions you can round the numerical quantities and check the final choice without the calculator.

Techniques and HintsBe clear and organized -- neat in your solution.You must be able to read and understand your own notes!Go through the exam completely at first and complete those questions that you are confident in solving. Then return to the others.

Kinematics:Kinematics:

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Chapters 2, 3 (linear) 9 (rotational)

Dynamics:Dynamics:

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Chapters 4, 5, (the 2nd law) 6, 7, (energy and work) 10, 11 (rotational, gravity)

No net force, No net torque:No net force, No net torque:

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Chapters 8 (linear: consequence of the 2nd law) 10 (rotational)

Statics: Stationary balanceStatics: Stationary balance

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Chapters 12 (static equilibrium, elasticity)

FluidsFluidsArchimedes’ Principle :A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid.

Oscillations:Oscillations:

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Resonance frequency:

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Question (Chapt 2)

An European sports car dealer claims that his product will accelerate at a constant rate from rest to a speed of 100 km/hr in 8s. What is the speed after first 5 s of acceleration?     

17.4 m/s     

53.2 m/s     

44.4 m/s    

34.7 m/s     

28.7 m/sv =v0 + at (for constant acceleration)

a=v−v0

t=100km/ hr −0

8s=100000 / 3600

8m/ s2

After 5 seconds: v=0 +100000 / 3600

8m/ s2 ×5s=17.4m/ s

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Two gliders of unequal mass mA<mB are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right.

Let FhA represent the magnitude of the force of the hand on the glider A. Let FBA represent the magnitude of the force exerted by the glider A on the glider B.

Which one of the following is true?     

FhA < FBA  

FhA = FBA      

FhA > FBA     

Question (Chapt 4)

Newton’s Second Law:

Net external, FhA-FBA, is causing block A to accelerate to the right.

FBA = FAB < FhA

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Two gliders of unequal mass mA<mB are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right.

Let FhA represent the magnitude of the force of the hand on the glider A. Let FBA represent the magnitude of the force exerted by the glider A on the glider B.

Which one of the following is true?     

FBA < FAB  

FBA = FAB      

FBA > FAB     

Question , Continued

Newton’s Third Law

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Two gliders of unequal mass mA<mB are placed on a frictionless air track. Glider A is pushed horizontally as shown so that the gliders accelerate together to the right.

How does the net force on glider B (FB) compare to the magnitude of the net force on glider A (FA)?     

FB < FA  

FB = FA      

FB > FA

Question, Continued

€

Given : mA < mB and aA = aBSecond Law : FA = mAaA and FB = mBaB

∴ FBFA

=mBmA

⇒ FB > FA

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Question (Chapt. 6)

How much power is needed to lift a 75-kg student vertically upward at a constant speed of 0.33 m/s?    

 25 W     

12.5 W     

243 W     

115 W    

  230 W

€

Power =Work

Time=

Force × Distance

Time= Force ×Speed

Power = mgv = 75kg × 9.8m/s2 × 0.33m/s = 243W

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A moving object collides with an object initially at rest.

Question (Chapt 8)

Is it possible for both objects to be at rest after the collision?YesNo

Can one of them be at rest after the collision?YesNo

If the objects stick together after the collision, is the kinetic energy conserved?YesNo

Realize that momentum is conserved

m1v1i = m1v1 f + m2v2 f

For this to be true both speeds cannot be zero

m1v1i =(m1 + m2 )vf ⇒ vf =m1

m1 + m2

v1i

KEi =12

m1v1i2

KE f =12(m1 + m2 )vf

2 =12

m12

m1 + m2

v1i2 =

m1

m1 + m2

KEi

Kinetic energy is reduced - an inelastic collision

Energy is lost as heat

Elastic collision (Q4)Elastic collision (Q4)

A block of mass m moving at to the right with speed v hits a block of mass M that is at rest. If the surface is frictionless and the collision is elastic, what are the final velocities of the two blocks?

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m Mv

Elastic collisionElastic collision

A block of mass m moving at to the right with speed v hits a block of mass M that is at rest. If the surface is frictionless and the collision is elastic, what are the final velocities of the two blocks?

In the center of mass frame, velocities reverse after an elastic collision

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m Mv

m Mv-vCM

-vCM

m M-(v-vCM) vCM

vCM = mv/(m+M)

Elastic collisionElastic collision

Now find velocity of each block in lab frame:

Velocity of m = vCM - (v-vCM) = 2vCM – v = (m-M)v/(m+M)

Velocity of M = 2vCM = 2mv/(m+M)

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m Mv

vCM = mv/(m+M)

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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the string?     

The tension increases to four times its original value.    

The tension increases to two times its original value.

The tension is unchanged.

The tension reduces to one half its original value.

The tension reduces to one fourth its original value.

Question (Chapt 9)

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ConceptsConcepts

Is there a net force acting on the system?YesNo

Yes, the direction of velocity is changing.

Centripetal acceleration is provided by the tension in the string.

The centripetal acceleration is different in the two cases presented, therefore, the tension will be different

Note that the radius has not changed in the two conditions

Note also that angular velocity is given (in words).

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SolutionSolution

F1 =mv12

r

F2 =mv22

r

v1 =rω1

F2

F1

=v22

v12 =

ω22

ω12 =

22

12=4

Tension goes up by a factor of 4!

Centripetal force for the two situations:

Need to write in terms of change to angular velocity because that is what is specified

Some of the quantities are not given but we are comparing situations, I.e., take ratios and cancel common factors!

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The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.

In which case is the torque on the nut the biggest?     

Case 1 

Case 2      

Case 3

Question (Chapt. 9)

= F d sin θ Longest lever arm, dat 90o angle

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Question (Chapt 12)Question (Chapt 12)A sign of mass M is hung 1 m from the end of a 4 m long uniform beam of mass m, as shown in the diagram. The beam is hinged at the wall. What is the tension in the guy wire? Determine the tension T, and the contact force F at the hinge.

SIGN

wire

θ 30

1 m

3 m

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What are the concepts involved?What are the concepts involved?

Is there a net force acting on the system?YesNo

Is there a net torque acting on the system?YesNo

Draw the free body diagram.How many forces are acting on the system?

2345 What is the direction of the contact force at the

hinge between the wall and the beam ?VerticalHorizontalIt has both vertical and horizontal components

SIGN

wire

θ 30

1 m

mg, Mg, tension, force from hinge

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SolutionSolution

rF∑ =0 ⇒ Fx −T cos300 =0

Fy +T sin300 =mg+ Mg

mg Mg

TFy

Fx

300

Hint: Choose axis of rotation at support because Fx & Fy are not known

2m

3m

r∑ =0 ⇒ 2mg + 3Mg = 4T sin 300

∴T = g 2m + 3M( ) / 2 (T can now be computed)

Substitute T in Force equations to get Fx and Fy

Forces

Torques

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A rock is thrown straight up from the Earth’s surface. Which one of the following statements concerning the net force acting on the rock at the top of its path is true?     

It is equal to zero for an instant.    

It is equal to the force used to throw it up but in opposite direction

It is equal to the weight of the rock

Its direction changes from up to down

Its magnitude is equal to the sum of the force used to throw it up and its weight    

Motion in Gravity (Chapt. 11)

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Question (Chapt 12)Question (Chapt 12)

A mass of 100 tons (105 kg) is lifted on a steel rod two cm in diameter and 10 m in length.

(Young’s modulus of steel is 210 109 N/m2)

(a) How long does the rod stretch?

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A mass of 100 tons (105 kg) is lifted on a steel rod two cm in diameter and 10 m in length.

(Young’s modulus of steel is 210 109 N/m2)

(a) How long does the rod stretch?

Y =stressstrain

=F / AΔL / L

ΔL =F

Y AL =

mgLYπ r2 =0.15m

F = forceA = area of rodL = length of rodΔL = change of length of the rod

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1) The pressure on the roof of a tall building is 0.985 × 105 Pa and the pressure on the ground is 1.000 × 105 Pa. The density of air is 1.29 kg/m3. What is the height of the building?

A. 100 mB. 118 mC. 135 m• 114 m• None of the above

€

P1 = ρh1g; P2 = ρh2g⇒ h = h2 − h1 =P2 − P1

ρg

h =0.015 ×105Pa

1.29kg /m3 × 9.8m /s2=118m

Question (Chapt 13)Question (Chapt 13)

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Question (Chapt 13)

A venturi tube may be used as the inlet to an automobile carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm diameter, what is the pressure drop in the constricted section for an airflow of 3.0 m/s in the 2.0-cm section? (fuel density = 1.2 kg/m3.)?     

Velocity is faster in constricted section because mass flow is conserved (mass that flows into constriction must also flow out).

Pressure drops because of Bernoulli principle: (applies to incompressible, frictionless fluid)

P +12ρv2 + ρgh=constant

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Calculate velocity in constrictionCalculate velocity in constriction

• Volume flow rate: ΔV/Δt = A Δx/Δt = Av (m3/s)

• Continuity: A1 v1 = A2 v2

i.e., mass that flows in must then flow out

Fluid flow without friction

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Question, continued

A venturi tube may be used as the inlet to an automobile carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm diameter, what is the pressure drop in the constricted section for an airflow of 3.0 m/s in the 2.0-cm section? (fuel density = 1.2 kg/m3.)?     

70 Pa     

85 Pa     

100 Pa    

115 Pa     

81 Pa

Constant of volume flow rate resulted

continuity equation:

A1v1 =A2v2 ⇒ v2 =A1

A2

v1 =πr1

2

πr22 v1 =

d12

d22 v1

Bernoulli Equation (same height):

P1 +12ρv1

2 =P2 +12ρv2

2 ⇒ ΔP =12ρ v2

2 −v12( )

Combining with above: ΔP =12ρ

d14

d24 −1

⎛

⎝⎜⎞

⎠⎟v12

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Question (Chapt 13)

The water level in identical bowls, A and B, is exactly the same. A contains only water; B contains ice as well as water. When we weigh the bowls, we find that     

WA < WB     

WA = WB      

WA > WB     

WA < WB if the volume of the ice cubes is greater than one-ninths the volume of the water.    

WA < WB if the volume of the ice cubes is greater than one-ninths the volume of the water.

Eureka! Archimedes Principle.

Weight of the water displaced = Bouyant Force

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1) A block of aluminum (density 3041 kg/m3) is lifted very slowly but at constant speed from the bottom of a tank filled with water. If it is a cube 20 cm on each side, the tension in the cord is:

A. 160 NB. 4 NC. 80 ND. 8 NE. None of the above

Volume of fluid displaced: V =VAl = 20 ×10−2( )3=8×10−3m3

Buoyant force: FB =ρwVg=1000 ×8 ×10−3 ×9.8 ≈78N

Weight: W =MAlg=ρAlVg=3041×8 ×10−3 ×9.8 =238NTension: T =W−FB =160N

TFb

W

Question (Chapt 13)Question (Chapt 13)

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Question (Chapt 13)

A wind with velocity 10 m/s is blowing through a wind generator with blade radius 5.0 meters. What is the maximum power output if 30% of the wind’s energy can be extracted? (air density = 1.25 kg/m3.)     

7.2 kW     

14.7 kW     

21.3 kW    

29.4 kW    

39.6 kW

Bernoulli Equation (same height):

P1 +12ρv1

2 =P2 +12ρv2

2 ⇒ ΔP =12ρ v2

2 −v12( )

Pressure difference results in net force on the blades

Magnitude of the force = Pressure x Blade Area

Power = Work

Time=

Force x DistanceTime

=Force x Velocity

Power=12ρvi

230%⎛⎝⎜

⎞⎠⎟πR2( ) v( )

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Firemen connect a hose (8 cm in diameter) to a fire hydrant. When the nozzle is open, the pressure in the hose is 2.35 atm. (1 atm. = 105 Pa). The firemen hold the nozzle at the same height of the hydrant and at 45o to the horizontal. The stream of water just barely reaches a window 10 m above them. The diameter of the nozzle is about:

A. 8 cmB. 6 cmC. 4 cmD. 2 cmE. None of the above Flow is constant in hose ⇒ A1v1 =A2v2

⇒ πr12v1 =πr2

2v2 ⇒ r2 =r1v1v2

⇒ d2 =d1

v1v2

10m

Point 1Point 2

Point 3

P + 12 ρv2

2 + ρgh=constant (Bernoulli's Equation).

But, P and 12ρv2x

2 are constant, so

12 ρv2y

2 =ρgh3 =1000 ×9.8 ×10 =98000Pa⇒

v2y = 2gh3 = 196 =14m/ s

P1 +12 ρv1

2 =P2 +12 ρv2

2 ⇒ v1 =2ρ

P2 −P1 +12 ρv2

2( )

12 ρv2

2 =196000Pa

v1 =2

1000100000 −235000 +196000( ) =11.045m/ s

45o angle ⇒ v2 = v2x2 + v2y

2 = 2v2y =19.799m/ s

d2 =811.04519.799

=5.975 ≈6cm

Question (Chapt 13)Question (Chapt 13)

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Question (Chapt 14)Question (Chapt 14)

At t=0, a 795-g mass at rest on the end of a horizontal spring (k=127 N/m) is struck by a hammer, giving it an initial speed of 2.76 m/s. The position of the mass is described by , with

What is period of the motion? period = 2π/ω

What is the frequency of the motion?

What is the maximum acceleration?

What is the total energy?

0.497 s

2.01 Hz

34.9 m/s2

3.03 J

x(t) =Acos ωt+π2

⎛⎝⎜

⎞⎠⎟

ω = k / m = 12.64 rad/s.

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Question (Chapt 14)

The amplitude of a system moving with simple harmonic motion is doubled. The total energy will then be     

4 times larger     

2 times larger

the same as it was

half as much    

quarter as much

U =12

kx2 +12

mv2

at x=A,v=0

U =12

kA2