PHYSICS 221 SPRING 2012

FINAL EXAM: May 1, 2012 4:30pm - 6:30pm Name (printed): ______________________________________________ Recitation Instructor: _________________________ Section #_______ INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions, each worth 4 points. Choose one answer only for each question. Choose the best answer to each question. Answer all questions. Allowed material: Before turning over this page, put away all materials except for pens, pencils, erasers, rulers and your calculator. There is a formula sheet attached at the end of the exam. Other copies of the formula sheet are not allowed. Calculator: In general, any calculator, including calculators that perform graphing, is permitted. Electronic devices that can store large amounts of text, data or equations (like laptops, palmtops, pocket computers, PDA or e-book readers) are NOT permitted. Wireless devices are NOT permitted. If you are unsure whether or not your calculator is allowed for the exam, ask your TA. How to fill in the bubble sheet:

Use a number 2 pencil. Do NOT use ink. If you did not bring a pencil, ask for one. You will continue to use the same bubble sheet that you already used for the first midterm exam. Bubble answers 55-81 on the bubble sheet for this exam.

Please turn over your bubble sheet when you are not writing on it. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over.

Best of luck,

Drs. Soeren Prell and Kai-Ming Ho

55) On Earth, when an astronaut throws a 0.250-kg stone vertically upward, it

returns to his hand a time T later. On planet X he finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is

A) 2g B) g/√2 C) g/4 D) g/2 E) √2g Solution The y position of the stone above the astronaut’s hand as a function of time after the throw is given by y = y0 + v0t !

12 gt2 . At times t = 0 and t = T the y position is

zero: y(T ) = y0 = 0 and thus 12 gT = v0 ! g = 2v0 / T . If the time doubles on planet

X, the gravitational acceleration has to be half as strong there than on Earth. 56) The Ferris wheel in the figure rotates counter-

clockwise with uniform angular speed. What is the direction of the average acceleration of a gondola as it goes from the top to the bottom of its trajectory?

A) Up B) Down C) To the left D) To the right E) The acceleration is 0 because the motion is uniform. Solution Since the Ferris wheel in the figure rotates counter-clockwise, a gondola at the top has a velocity pointing to the left, whereas it has a velocity pointing to the right when it is at the bottom. The speed of the gondola is the same at the top and at the bottom. Thus,

!vbottom = !

!vtop and the average acceleration is given by

!a =

!!v!t

=

!vbottom "!vtop

!t=

2!vbottom

!t,

where Δt is the time interval it takes the gondola to move from top to bottom.

57) In the figure, a person pulls horizontally on block B and

both blocks A and B move horizontally as one unit. There is friction between block B and the horizontal table. Air resistance is negligible. If the two blocks are moving to the right at constant velocity,

A) … the horizontal force that B exerts on A points to the left. B) … the horizontal force that B exerts on A points to the right. C) … B exerts no horizontal force on A. D) … the pull force is greater than the friction force E) … the pull force is less than the friction force Solution The two blocks are moving with constant velocity à acceleration = 0à net force = 0 on each block. For block A, the only force exerted in the horizontal direction comes from B. Therefore, the horizontal force exerted by B on A = mass of A x acceleration of A = 0. 58) A ball sits at rest on a horizontal tabletop. The weight of the ball is one half of

a Newton’s third law force pair. Which force is the other half? A) the force of the Earth’s gravity on the ball B) the upward force that the table top exerts on the ball C) the upward force that the ball exerts on the Earth D) the downward force that the ball exerts on the table top E) the frictional force between the ball and the table top Solution Action and reaction are equal and opposite and act on different bodies. Weight acts on the ball, the reaction must act on what is causing the action i.e. the earth.

59) You push a block of weight mg against a wall with force F perpendicular to

the wall such that the static friction between the wall and the block barely keeps the block from falling. Consider the friction between your finger and the block negligible and the coefficient of static friction between the wall and the block is µS. When you push with a force 2F, the magnitude of the friction force will be _______.

A) mg B) 2mg C) µS F D) 2(µS F − mg) E) µS mg Solution In the free body diagram of the block there are 4 forces acting on the block. In the vertical direction the weight mg is pulling the block down and the friction between block and wall is holding the block in place. In the horizontal direction, your finger is pushing to the left, while the normal force by the wall is pushing to the right. As long as the block stays in place the two forces in the vertical direction balance each other. Therefore, fS = mg regardless how hard you push. Since the block was barely held in place originally when the force between the block and the wall was F, the static friction force was at its maximum of µS F=mg. 60) A small hockey puck slides without friction over the icy hill shown in the figure

and lands 6.20 m from the foot of the cliff with no air resistance. What was its speed v0 at the bottom of the hill?

A) 4.71 m/s B) 20.8 m/s C) 13.7 m/s D) 14.4 m/s E) 17.4 m/s

Solution The height above the ground after the puck shoots over the cliff as a function of time is given by

y = h + v0, yt !

12 gt2 . With y = 0 and v0,y = 0, we calculate the time it

takes the puck to hit the ground as

0 = h ! 12 gt2 " t = 2h / g = 2 # 8.5m / 9.81 ms!2 = 1.32s The x velocity at the top of

the cliff is given by vx = x / t = 4.70 m/s, where is x = 6.20 m is the distance between the foot of the hill and the point where the puck hits the ground. From energy conservation, we get

12 mv0

2 = 12 mvx

2 + mgh! v0 = vx2 + 2gh = 13.7m/s

 

       

   

 

61) A 58.0-gram tennis ball is dropped from a height of 2.54m above ground.

It hits the floor and bounces back up to a height of 1.40m. Neglecting air resistance, when the ball bounces it imparts on the floor an impulse of magnitude _______.

A) 1.4 kg m/s B) 0.11 kg m/s C) 2.2 kg m/s D) 0.67 kg m/s E) 0.71 kg m/s Solution The impulse is given by

!J = !!p = !pup "

!pdown = m !vup "!vdown( ) with

!vdown = 2g(2.54m) = 7.06 m/s and

!vup = 2g(1.40m) = 5.24 m/s such that

!J = (0.058 gram) !vup !

!vdown = (0.058 gram) vup + vdown( ) = 0.71 kg m/s

62) Two objects of the same mass move along the same line in opposite

directions. The first object is moving with speed v. The objects collide, stick together, and move with speed 0.100v in the direction of the velocity of the first object before the collision. What was the speed of the second object before the collision?

A) 0.900v B) 1.20v C) 0.800v D) Zero E) 1.10v Solution With v’ being the velocity of the second object, momentum conservation gives mv ! mv ' = 2 m (0.1v)" v ' = v ! 0.2v = 0.8v

63) Starting from rest, a small pebble slides down the frictionless inner surface of

a spherical bowl starting from a height of 0.2 m above the bottom of the bowl. The radius of the bowl is 1.0 m. Calculate the magnitude of the linear acceleration (in terms of g) of the pebble at the bottom of its trajectory.

A) 0.2 g B) 0.4 g C) 0.6 g D) 1.0 g E) Zero Solution

2 212

2

Conservation of energy 2At bottom of bowl, all forces are vertical no horizontal acceleration

acceleration is vertical and is purely centripetal2 0.4

mv mgh v gh

a v R gh R g

→ = → =

→

→

= = =

64) A planet is moving around the Sun in an elliptical orbit. As the planet moves

from aphelion (farthest point from the Sun) to perihelion (closest point to Sun), the Sun’s gravitational force does _____.

A) positive work on the planet at all points between aphelion and perihelion. B) negative work on the planet at all points between aphelion and perihelion. C) positive work on the planet during part of the motion and negative work during the other part with the total work being positive. D) positive work on the planet during part of the motion and negative work during the other part with the total work being negative. E) positive work on the planet during part of the motion and negative work during the other part with the total work being zero. Solution As the planet moves from aphelion to perihelion, the distance between the sun and the planet keeps decreasing. The force is attractive in the radial direction, if the displacement of the planet decreases the distance, the displacement always has a component in the direction of the force à work done is positive.

65) Three objects are connected as shown in

the figure. The strings and frictionless pulleys have negligible masses, and the coefficient of kinetic friction between the 2.0-kg block and the table is 0.22. What is the acceleration of the 2.0-kg block?

A) 1.8 m/s2 B) 4.1 m/s2 C) 3.3 m/s2 D) 2.6 m/s2 E) 5.2 m/s2 Solution Taking the positive direction to be along the string with the center block moving to the right, from Newton’s second law we get

(3kg)g ! µk (2kg)g ! (1kg)g = (6kg)a

" a = (3kg ! 0.22(2kg) !1kg)/(6kg)g = 2.6 m/s2

66) The angle that a swinging simple pendulum makes with the vertical obeys the

equation !(t) = (0.240 rad) cos 2.85 rad/s( ) t + 1.66!" #$ . What is the length of the pendulum?

A) 0.83 m B) 1.21 m C) 2.02 m D) 0.15 m E) It cannot be determined from the information given. Solution The equation is given in the standard form !(t) =!0 cos "t + ![ ] and we read off

ω = 2.85 rad/s. For a simple pendulum ! = g / L and therefore L = g /! 2 = (9.81 m/s2 ) / (2.85 rad/s)2 =1.21m

67) A sewing machine needle moves up and down in simple harmonic motion.

The distance between the highest and lowest point of the needle tip is 2.6 cm and the frequency of the motion is 3.2 Hz. What is the maximum speed of the needle?

A) 8.2 cm/s B) 21 cm/s C) 4.1 cm/s D) 13 cm/s E) 26 cm/s Solution The maximum speed of an object in one-dimensional simple harmonic motion is given by vmax = Aω, where A is the amplitude and ω = 2 π f is the angular frequency. With A = 2.6/2 cm = 1.3 cm and ω = 2 π f = (2 π)(3.2 Hz) = 20 s−1 we have vmax = 26 cm/s. 68) In designing buildings to be erected in an area prone to earthquakes, what

relationship should the designer try to achieve between the natural frequency of the building and the typical earthquake frequencies?

A) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly lower. B) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly higher. C) The natural frequency of the building should be very different from typical earthquake frequencies. D) The natural frequency of the building should be exactly the same as typical earthquake frequencies. E) The designer does not have to worry about typical earthquake frequencies. Solution If the natural frequency of a building is close to the range of typical earthquake frequencies, the building will experience resonance during an earthquake where even small forces can cause a large amplitude of forced oscillation leading to damage or the destruction of the building.

69) The sound from a single source can reach point P by two different paths. One

path is 20.0 m long and the second path is 21.0 m long. The sound destructively interferes at point P. What is the minimum frequency of the source if the speed of sound is 343 m/s?

A) 343 Hz B) 515 Hz C) 686 Hz D) 6860 Hz E) 172 Hz Solution When the path length difference is an odd integer multiple of half a wavelength destructive interference occurs. The largest wavelength corresponds to the lowest frequency. Thus, ΔL = λ/2 = 1.0m and f = v / λ = 343 m/s / 2.0m = 172 Hz. 70) A brass rod is 69.5 cm long and an

aluminum rod is 49.0 cm long when both rods are at an initial temperature of 0°C. The rods are placed in line with a gap between them, as shown in the figure. The distance between the far ends of the rods is maintained at 120.0 cm throughout. The temperature of both rods is raised until the two rods are barely in contact. The coefficients of linear expansion of brass and aluminum are 2.0 x 10−5 K−1 and 2.4 x 10−5 K−1, respectively. The temperature at which contact of the rods barely occurs is closest to

A) 560 oC B) 585 oC C) 605 oC D) 630 oC E) 660 oC Solution The width of the gap at 0°C is equal to the combined expansion of the two rods: (120.0!69.5! 49.0)cm = [(69.5cm)(2.0"10!5K!1)+ (49.0cm)(2.4"10!5K!1)]#T!"T = (1.5 / 2.57!10"3)oC = 585oC

71) A substance has a melting point of 20°C and a heat of fusion of 1.9 x 104

J/kg. The boiling point is 150°C and the heat of vaporization is 3.8 x 104 J/kg at a pressure of one atmosphere. The specific heats for the solid, liquid, and gaseous phases are 600, 1000, and 400 J/(kg K), respectively. The quantity of heat required to raise the temperature of 1.00 kg of the substance from −5°C to 129°C is closest to:

A) 34 kJ B) 120 kJ C) 130 kJ D) 140 kJ E) 160 kJ Solution Heat required to raise the temperature of the solid from −5°C to the melting point = 600 x 25 J = 1.5 x 104 J Latent heat of fusion = 1.9 x 104 J Heat required to raise temperature of the liquid from melting point to 129°C = 1000 x 109 = 10.9 x104 J Total heat required=14.3 x104 J 72) When a fixed amount of a monatomic ideal gas is expanded in volume at

constant pressure, the average kinetic energy of the gas molecules ______. A) increases. B) decreases. C) does not change. D) may either increase or decrease, depending on whether or not the process is carried out adiabatically. E) may or may not change, but insufficient information is given at make such a determination. Solution The average kinetic energy of a gas molecule of an ideal gas is proportional to the temperature of the gas

Ktrans = 3

2 kT . From T = pV/(nR), we see that for constant pressure, fixed amount of gas we have T = constant x V. Therefore, if the volume increases, the temperature and the average kinetic energy also increase.

73) Consider two pistons filled with ideal gases. One piston has a volume of 10

liters and contains 0.5 mol of Xenon gas at atmospheric pressure. The other piston has a volume of 16 liters and contains 1.0 mol of Argon gas at a temperature of 400K. The atomic mass of Argon is 39.9 gram/mol and the atomic mass of Xenon is 131 gram/mol. The ratio of rms speeds vrms(Xenon)/vrms(Argon) of the gas atoms is closest to

A) 0.43 B) 5.2 C) 0.81 D) 8.4 E) 0.12 Solution The rms speed of an ideal gas is given by vrms = 3RT / M . The temperature of the Argon gas is TArgon = 400K. The temperature of the Xenon gas is

TXenon =

pXenonVXenon

nR=

(1.013!105 Pa)(0.01m3)0.5! 8.31 J/(mol K)

= 244K . The ratio of rms speeds is

then

vrms(Xenon)vrms(Argon)

=3RTXenon

M Xenon

M Argon

3RTArgon

=TXenon

M Xenon

M Argon

TArgon

=244K

131g/mol39.9g/mol

400K= 0.43

74) The figure shows a pV diagram for 0.47 mol of

gas that undergoes the process 1 → 2. The gas then undergoes an isochoric heating from point 2 until the pressure is restored to the value it had at point 1. What is the final temperature of the gas?

A) −40 oC B) 2 oC C) 130 oC D) 230 oC E) 510 oC Solution

We have n = 0.47 mol, p final = 3atm !1.013!105 Pa/atm = 3.04 !105 Pa

V final = 3000cm3 !10"6 m3 /cm3 = 3.00 !10"3m3 and thus

Tfinal =

p finalV final

nR=

(3.039 !105 Pa)(0.003m3)0.47 ! 8.31 J/(mol K)

= 233K= " 40oC

75) A glass window pane is 2.6 m high, 2.3 m wide, and 3.0 mm thick. The

temperature at the inner surface of the glass is 24°C and at the outer surface 4.0°C. How much heat is lost each hour through the window? (properties of glass: density: 2300 kg/m3, specific heat: 840 J/kg·C°, coefficient of linear thermal expansion: 8.5 × 10-6 (C°)-1, thermal conductivity: 0.80 W/(m·C°))

A) 1.1 x 108 J B) 1.1 x 106 J C) 1.1 x 105 J D) 32 J E) 320 J Solution The heat lost each hour is given by

t !

dQdt

"

#$%

&'= tkA

TH ( TC

L= 3600s ! 0.80 W/(m°C) ! 2.6 ! 2.3m2( ) 20°C

3!10(3m= 1.1!108 J

76) A Carnot refrigerator operates between a hot reservoir at 600K and a cold

reservoir at 200K. The refrigerator consumes 50 W of power. How much heat is removed from the interior of the refrigerator in 1 hour?

A) 180 kJ B) 72 kJ C) 720 kJ D) 7.5 kJ E) 90 kJ Solution The coefficient of performance of a Carnot refrigerator depends on the temperatures between it is operating:

Krefrigerator =

TC

TH ! TC

=200K

600K ! 200K= 0.5

Thus,

Krefrigerator =

QC

W!QC = Krefrigerator W = Krefrigerator P " t = 0.5" 50J/s " 3600s = 90kJ

77) A hot piece of iron is thrown into the ocean and its temperature eventually

stabilizes. Which of the following statements concerning this process is correct?

A) The entropy lost by the iron is equal to the entropy gained by the ocean. B) The ocean gains more entropy than the iron loses. C) The entropy gained by the iron is equal to the entropy lost by the ocean. D) The change in the entropy of the iron-ocean system is zero. E) The ocean gains less entropy than the iron loses.

Solution

The entropy gained by the ocean is given by S =

dQTocean! and the entropy lost by

the iron is S =

dQTiron! . Since the temperature of the ocean is always less than the

temperature of the iron, the magnitude of the second integral is smaller than the one of the first. 78) The figure below shows the measured displacement of a mass oscillating at

the end of a spring as a function of time. What is the frequency of the oscillations?

A) 0.47 Hz B) 0.94 Hz C) 1.2 Hz D) 1.9 Hz E) 2.6 Hz Solution There are about 18.75 periods in 20 seconds. Thus, T = (20 sec)/ 18.75 = 1.07 s. The oscillation frequency is f = 1 / T = 0.94 Hz.

79) In the collisions experiment, you obtained traces of the motion of a puck like

the one shown below:

This trace is produced by the sparks generated at a rate of 60 sparks per second. The distance between the first and last point in the trace shown above is measured with a standard metric ruler and found to be 3.40 cm. Due to the limitations of the ruler, we can estimate the uncertainty in the reading to be ± 0.5 mm. You may neglect the uncertainty in the spark generator frequency. What is the uncertainty in the speed of the puck?

A) ± 0.5 mm/s B) ± 1 mm/s C) ± 3 mm/s D) ± 5 mm/s E) ± 3 cm/s Solution There are 6 gaps between sparks. Since they come at a rate of 60 sparks per second the time interval between the first and the last spark is 0.1 seconds. The puck moves at a speed of v = 3.4 cm / 0.10 s = 34.0 cm/s. The uncertainty in the distance between the first and the last spark is ± 0.5 mm. Therefore, the uncertainty in the speed is given by Δv = ± 0.5 mm / 0.10 s = ± 5 mm/s.

80) A 40-cm3 closed container filled with air is warmed up while probes are

measuring the temperature and pressure inside the container. The results are shown below in a pressure versus temperature graph. As expected, a linear behavior is observed, so a linear fit is performed.

Assuming that the air inside the container is an ideal gas, how many moles of air are trapped inside the container?

A) 0.0021 moles B) 0.056 moles C) 0.24 moles D) 0.88 moles E) 1.3 moles

Solution Reading the p/T slope from the fit, we can calculate n with the ideal gas equation

n = V

R!

pT=

(40.0 !10"6 m3)8.31 J/(mol K)

! 0.0044 atm/K( ) = 0.0021mol

81) In the Forces and Motion experiment, a force probe is fixed to a cart that can

roll with or without friction on a track. The whole system (probe + cart) is pulled through a hook screwed onto the probe by a light string that goes through a pulley and has a cylindrical weight attached to the other end (see figure).

Which of the following most accurately describes what the probe measures? Note that in some situations, two or more of these forces may have the same magnitude, but it is important to know which force you are really measuring.

A) The net force on the cart + probe system B) The net force on the probe alone C) The tension in the string D) The weight of the cylinder that pulls on the system E) The weight of the cart + probe system Solution The probe measures the tension in the string. To see that A) and B) are incorrect, consider a situation with a large kinetic friction force. The cart would almost not move, but the probe would still measure the tension in the string, which in this case would be close to the weight of the cylinder. On the other hand the acceleration and thus the net force on the cart or cart-and-probe system would be almost zero. Only, if the car and probe system is not being accelerated, the tension in the string is equal to the mass of the cylinder. So, in general D) is incorrect. In no situation, does the probe directly measure the weight of the cart and probe system.

Cart    

Pulley  Force  probe  

Cylinder