final exam: saturday, december 19, 5:05-7:05pm,...

12/14/09 Physics 201, Fall 2009, UW-Madison 1

Final Exam: Saturday, December 19, 5:05-7:05pm, 3650 Humanities

The exam will cover chapters 1 – 14 The exam will have about 28 questions

Practice questions are posted − follow link to “practice questions for exams” from course home pagehttp://www.physics.wisc.edu/undergrads/courses/current/201/

Review sessions: Wed, Dec 16 1 - 3 pm, 2104 Chamberlin (Jialu) Wed, Dec 16 4 - 6 pm, 2116 Chamberlin (Andrew) Thu, Dec 17 2:45-4:45 pm, 2104 Chamberlin (Jared) Thu, Dec 17 5:05-7:05 pm, 2104 Chamberlin (Eunsong) Fri, Dec 18 2:45-4:45 pm, 2104 Chamberlin (Dan)


  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).


  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 Hints   Be 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.


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

SIGN

wire

θ = 30o

1 m


Is there a net force acting on the system?   Yes   No

Is there a net torque acting on the system?   Yes   No

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

  2   3   4   5 What is the direction of the contact force at the

hinge between the wall and the beam ?   Vertical   Horizontal   It has both vertical and horizontal components

SIGN

wire

θ = 30o 1 m

mg, Mg, tension, force from hinge


F∑ = 0 ⇒ Fx = −T cos 300

Fy + T sin 300 = mg + Mg

mg Mg

T Fy

Fx 300

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

2m

3m

τ∑ = 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


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.

Please do not try to solve problems like this in your head. Please follow the procedure, although this does not look like a traditional problem.


Is there a net force acting on the system?   Yes   No

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).


F1 = mv12

r

F2 = mv22

r

v1 = rω1

F2F1

=v22

v12 =

ω22

ω12 =

22

12= 4

Tension goes up by a factor of 4!

Centripetal force for the two situations:

Need to change to angular velocity because only that is specified

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


A moving object collides with an object initially at rest.

Question 3

Is it possible for both objects to be at rest after the collision?   Yes   No

Can one of them be at rest after the collision?   Yes   No

If the objects stick together after the collision, is the kinetic energy conserved?   Yes   No

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 =12m1v1i

2

KEf =12

(m1 + m2 )vf2 =

12

m12

m1 + m2

v1i2 =

m1

m1 + m2

KEi

Kinetic energy is reduced - an inelastic collisionEnergy is lost as heat

  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?


m M v

  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


m M v

m M v-vCM -vCM

m M -(v-vCM) vCM

vCM = mv/(m+M)

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)


m M v

vCM = mv/(m+M)


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?


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 =FY A

L =mgLY π r2

= 0.15m

F = force A = area of rod L = length of rod ΔL = change of length of the rod


Question 6 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/s

v = v0 + at (for constant acceleration)

a =v − v0

t=

100km / hr − 08s

=100000 / 3600

8m / s2

After 5 seconds: v = 0 +100000 / 3600

8m / s2 × 5s = 17.4m / s


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

Newton’s Second Law: Net external, FhA-FBA, is causing block A to accelerate to the right.

FBA = FAB < FhA



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 Newton’s Third Law



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

�

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

∴ FBFA

= mB

mA

⇒ FB > FA


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 m B. 118 m C. 135 m D.  114 m E.  None of the above

�

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

ρg

h = 0.015 ×105Pa1.29kg /m3 × 9.8m /s2 = 118m


Question 9 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.)?


Question 9 A venturi tube may be used as the inlet to an automobile


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


•  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


Question 9 (continued) A venturi tube may be used as the inlet to an automobile


70 Pa 85 Pa 100 Pa 115 Pa 81 Pa

Constant of volume flow rate resultedcontinuity 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 d1

4

d24 −1

⎛⎝⎜

⎞⎠⎟v1

2


Question 10 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


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 N B. 4 N C. 80 N D.  8 N E.  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 ≈ 78NWeight: W = MAlg = ρAlVg = 3041× 8 ×10−3 × 9.8 = 238NTension: T =W − FB = 160N

TFb

W


Question 12 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 bladesMagnitude of the force = Pressure x Blade Area

Power = WorkTime

=Force x Distance

Time= Force x Velocity

Power = 12ρvi

2 30%⎛⎝⎜

⎞⎠⎟πR2( ) v( )


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 cm B. 6 cm C. 4 cm D.  2 cm E.  None of the above Flow is constant in hose ⇒ A1v1 = A2v2

⇒πr12v1 = πr2

2v2 ⇒ r2 = r1v1

v2

⇒ d2 = d1v1

v2

10m

Point 1 Point 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 =2ρP2 − P1 + 1

2 ρv22( )

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


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.


Question 15 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 =12kx2 +

12mv2

at x = A,v = 0

U =12kA2

final exam: saturday, december 19, 5:05-7:05pm,...

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