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学霸学习网 这下你爽了

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PHYSICSBOWL 2016

March 30 – April 15, 2016

40 QUESTIONS – 45 MINUTES

The sponsors of the 2016 PhysicsBowl, including the American Association of Physics Teachers, are providing some of the prizes to recognize outstanding high school physics students and their teachers through their performance on this year’s contest. two divisions, each with nineteen regions. o Division 1 is for students taking physics for the first time (even if that first course is AP Physics). o Division 2 is for students taking a second (or more) course in physics or anyone wishing a challenge. ● A school's team score in each division is the sum of the five highest student scores in that division. ● A school may compete in either or both divisions. INSTRUCTIONS Answer sheet: Write and bubble-in the following REQUIRED information on your answer sheet: ● Your Name ● Your Teacher’s AAPT Teacher code (given to you by your teacher – only one code per school!) ●Your Region (given to you by your teacher) ●Your Division (1 for first-year physics students, 2 for students in a second physics course) If this information is not properly bubbled, you will be disqualified as your official score will be a zero. Your School’s CEEB code (given to you by your teacher), though not required, IS helpful in the event of a disqualification for identifying your school. Your answer sheet will be machine graded. Be sure to use a #2 pencil, fill the bubbles completely, and make no stray marks on the answer sheet. Questions: The test is composed of 50 questions; however, students answer only 40 questions. Answers should be marked on the answer sheet next to the number corresponding to the question number on the test. Division 1 students will answer only questions 1 – 40. Numbers 41 – 100 on the answer

sheet should remain blank for all Division 1 students. Division 2 students will answer only questions 11 – 50. Numbers 1 – 10 and 51 – 100 on the answer sheet should remain blank for all Division 2 students. Calculator: A hand-held calculator may be used. Any memory must be cleared of data and programs. Calculators may not be shared. Formulas and constants: Only the formulas and constants provided with the contest may be used. Time limit: 45 minutes. Score: Your score is equal to the number of correct answers (no deduction for incorrect answers). If there are tie scores, the entries will be compared from the end of the test forward until the tie is resolved. Thus, the answers to the last few questions may be important in determining the winner and you should consider them carefully. Good Luck! Copyright ? 2016, AAPT

ATTENTION:

All Division 1 students – START HERE. All Division 2 students – Begin on question #11. *** Treat =. ? for ALL questions #1 – #50. 2016 PhysicsBowl Solutions

1. Which one of the following choices represents the smallest amount of time? (A) 1 day (B) 1 minute (C) 1 second (D) 1 week (E) 1 year

Solution：C… 1 second < 1 minute < 1 day < 1 week < 1 year 2. The position vs. time graph of an object moving on a horizontal line is shown. At what time(s) is the object moving with its greatest speed? (A) At time =3.0 only (B) At time =3.5 only (C) At times =3.5 and =9.0 (D) At time =7.0 only (E) At time =9.0 only Solution：A… The velocity of the object is found from the slope of the line tangent to the point on the position vs. time graph. To have the greatest speed, we are looking for the largest magnitude of slope to the line tangent. The steepest curve on the position vs. time graph shown occurs at time =3.0 . 3. A car accelerates uniformly from 0

km hr

to 60

km hr

in 4.50 . Which one of the following

choices best represents the acceleration of the car? (A) 13.3 s?

m

(B) 9.8 ?

(C) 4.8 ?

(D) 3.7 ?

(E) 0.37 ?

Solution：D… First, we convert units to obtain 60 constant acceleration kinematics then yields =

1000m 1hr 1min × × × 1km 60min 60s

= 16.7 . Using

m s

?0 t

=

16.7?0 = 4.5

3.7 .

?

4. The following four length measurements are recorded: 5.4× 10?1 ,5.4× 100 ,5.40× 101 , and 5.400× 102 . Which one of the following choices best represents the sum of these values using the rules of significant digits? (A) 6× 102 (B) 6.0 × 102 (C) 6.00 × 102 (D) 5.999× 102 (E) 5.9994× 102 Solution：D… The simplest way to look at the addition of quantities is to line up the decimal place. That is, we are computing 0.5+5.4+54.0+540.0=599.9. The 4 is bolded as it is the only digit known in the hundredths place since three of the quantities stop at the tenths. Hence, the final value is only kept to the tenths place making 599.9=5.999× 102 the correct answer. 5. In the figure shown, the mass =6.0 and the mass =14.0 are stationary. The mass rests on the floor. Which one of the following choices represents the tension in the string connecting the masses? (A) 40 (B) 60 (C) 80 (D) 140 (E) 200

Solution：B… By performing the free body diagram on the mass , we have the tension in the string and the gravitational force. Since everything is at rest and not accelerating, we find from Newton’s Second Law that =→+(?)=0 → =60 . Note that for the mass on the ground, there is an additional normal force acting! 6. An object, thrown straight downward with a speed of 20.0 s , takes 2.00 to reach the ground below. From what height above the ground was the object thrown? Ignore air resistance. (A) 20.0 (B) 40.0 (C) 50.0 (D) 60.0 (E) 80.0 m

1 m

Solution：D… Using constant acceleration kinematics, we have =0+0+22 → 0=+(?20)(2)+ 2(?10)(2)2 → =40+20=60 . 7. Three cylindrical resistors are made of the same material. The length and radius of each resistor is: Resistor 1: Length , radius Resistor 2: LengthL?2, radiusr?2

1

Resistor 3: Length L?4, radiusr?2 Which one of the following choices correctly ranks the resistance of these resistors (1,2,3)? (A) 1=2<3 (B) 2<1<3 (C) 3<2<1 (D) 1<2=3 (E) 1=3<2

ρL .

Solution：E… The expression for the resistance of a resistor is given as = in turn, we find 1= π2 ; 2=

ρL

2 2 π( ) 2

So, computing

ρ( )

= 2π2 ; 3=

ρL

4 2 π( ) 2

ρ( )

= π2 . Hence, 1=3<2.

ρL

8. In the history of physics, the names Nicolaus Copernicus, Galileo Galilei, James Clerk Maxwell, and Isaac Newton are very recognizable. Which one of the following choices correctly orders these names chronologically by the approximate dates of each person’s scientific work (ending with the most recent)? (A) Copernicus → Galileo → Newton → Maxwell (B) Copernicus → Galileo → Maxwell → Newton (C) Galileo → Copernicus → Newton → Maxwell (D) Galileo → Newton → Copernicus → Maxwell (E) Newton → Copernicus → Maxwell → Galileo Solution： A… Copernicus published the heliocentric theory in 1512 while Galileo did a lot of work related to this theory about a century later. Isaac Newton started producing work in the 1660’s while Maxwell’s contributions to physics had to wait until the 1860’s. 9. In the circuit shown, an electric current exists in the metal resistor connected to the battery. Which one of the following choices best describes the direction of electron flow through the resistor and the direction of the electric field through the interior of the resistor? Direction of electron flow in resistor Direction of electric field in resistor

(A) → → (B) ← → (C) → ← (D) ← ← There is no electric field. (E) → Solution：B… Since the conventional current in the circuit is directed to the right through the resistor, the electrons are flowing in the opposite direction (to the left). The electric field in the interior of the resistor would be in the direction of conventional current and hence points to the right.

10. A person wishes to accelerate an object upward uniformly. Which one of the following choices must be true of the force, , provided by the person onto the object? represents the magnitude of the gravitational force acting on the object. Ignore any effects of the air or of the Earth’s rotation. (A) = (B) ≥ (C) > (D) ≥2 (E) >2 Solution：C… By Newton’s Second Law, in order to accelerate an object, there must be an imbalance in the forces. To accelerate an object upward, a force must exist on the object that exceeds the gravitational force acting on it. Hence, the required force is >.

ATTENTION:

All Division 1 students continue to question #40. All Division 2 students START HERE. Numbers 1 – 10 on your answer sheet should be blank. Your first answer should be for #11. *** Treat =. ? for ALL questions #1 – #50. Questions 11 – 12 deal with the following information:

A 12.0 net force directed to the right is exerted on a 3.0 block for 5.0 . The block was initially at rest and slides on a horizontal surface. 11. What is the speed of the block at the end of the 5.0 interval? (A) 12

(B) 15

(C) 20

(D) 36

(E) 60

m

Solution：C… From the information, we start by finding the acceleration as =

12.0 = 3.0 m s?

=

4.0 . Using constant acceleration kinematics, we then have =0+ →=0+(4)(5)=

20 .

12. What is the kinetic energy associated with the block after the 5.0 interval? (A) 180 (B) 600 (C) 800 (D) 1200

1

(E) 1600

1

Solution：B… Kinetic energy is computed as = 22 = 2 (3.0)(20)2 = 600 . 13. For the inclined plane shown, which one of the following choices best represents the value of the incline’s ideal mechanical advantage? (A) 1.67 (B) 1.33 (C) 1.25 (D) 0.80 (E) 0.75

Solution：A… The ideal mechanical advantage for an inclined plane is found as the length of the hypotenuse divided by the height. Hence, in this situation, we find = = 3.0 =1.67. 14. A scientist obtains an answer that has units equivalent to those from computing the square root of the ratio of the Universal Gravitational constant to Coulomb’s constant. Which one of the following choices has these same units? (A) energy divided by time (D) electric field strength divided by magnetic field strength

? 5.0

(B) length divided by time (C) mass divided by energy

(E) charge divided by mass

Solution：E… From the equation sheet, we find the units for these quantities. Doing this gives√

2 2 N2 C2

=√2 =

2

.

In other words, this has units of charge divided by mass.

15. A radio station broadcasts its signal at a frequency of 100.0 . What is the wavelength of the station’s signal? (A) 3.00 (B) 3.00 (C) 3.00 (D) 0.33 (E) 0.33

Solution：B… Since all EM waves travel at the speed of light, we have = → = =100×106=3.0 16. A uniform stick is fixed to rotate about an axis through its center. The stick starts from rest and rotates through an angle of 90°in a time of 1.0 . If the angular acceleration of the stick is constant, what is the angular speed of the stick about its rotation axis after one full revolution? (A) 4

3.0×108

(B) 2√2

(C) 2

(D) √2

(E)

Solution：C… From the first quarter rotation, we determine the angular acceleration as Δ=0+ 2 → =0+ (1)2 → =

1 2 π 2 1 2 rad . s?

And so, for one full revolution, we have 2=02+2Δ

→ 2=02+2()(2)=(2)2. Hence, =(2)

.

17. “All planets in the solar system follow elliptical orbits with the Sun located at one focus of the ellipse.” This statement is best summarized by which one of the following choices? (A) Kepler’s First Law (B) Kepler’s Second Law (C) Newton’s Universal Law of Gravitation (D) Einstein’s Law of Gravity (E) Schwarzchild’s Law of Orbits Solution：A… This is a basic statement of Kepler’s First Law. 18. For any motion in two dimensions, which one of the following choices identifies quantities that must have the same direction?

(A) average velocity and displacement (B) average acceleration and average velocity (C) average acceleration and displacement (D) final velocity and average acceleration (E) final velocity and displacement Solution：A… By definition, average velocity is displacement divided by time. Hence, these quantities must be directed the same way. 19. An ideal gas in a sealed container with 1.20× 1024 particles has a pressure of 1.0 at a temperature of 27.0 ℃. Which one of the following choices best represents the volume of the container? (A) 450 3 (B) 4.5 3 (C) 0.49 3 (D) 0.049 3 (E) 0.00049 3

Solution：D… Using the ideal gas equation in the form =, we compute (1.0 )()=(1.20× 1024)(1.38× 10?23)(27+273)=4968 . So, we need to convert the pressure as 1.0 =1.01× 105. Hence, =

4968 =0.049 1.01×105

3.

20. A uniform 10.0 long plank is pivoted about its center of gravity. A mass 1 then is placed at the left edge of the plank and a second mass 2 is placed 1.0 from the right end. This system is in static equilibrium (shown). If each mass now is moved 2.0 closer to the center of the plank, which one of the following choices best describes the subsequent motion of the plank? (A) The plank remains in static equilibrium. (B) The plank rotates at constant angular speed until the right side reaches the ground. (C) The plank rotates at constant angular speed until the left side reaches the ground. (D) The plank angularly accelerates until the right side reaches the ground. (E) The plank angularly accelerates until the left side reaches the ground. Solution：E… From the original locations, we can balance the torques about the center of the plank to find =0 → (1)(5)?(2)(4)=0 → 2=41 where the distance of the mass 2 to the rotation axis is 4.0 . So, by moving the masses, we have the following torque condition: = (1)(3)?(2)(2). Replacing the mass 2=41 into the new expression gives = (1)(3)?( 41 )(2)= 2>0. As we assumed counterclockwise torques about the center are

5 1 5 5

positive, the net torque being positive means that the right side will rise and the left side will fall. Since there is a torque imbalance, this motion occurs with a non-zero angular acceleration! 21. At 1.94 after launch, a projectile in free fall achieved its minimum speed during flight. If that minimum speed was 15.0, what was the projectile’s initial angle of launch from the horizontal? (A) 82.6° (B) 52.3° (C) 50.6° (D) 39.4° (E) 37.7°

Solution：B… The minimum speed after launch for a projectile is when it reaches its apex (highest point) when the y-component of the projectile’s velocity is 0 . From constant acceleration kinematics, we find =0+ →0=0+(?10)(1.94)→0=19.4 . At the highest point, the total velocity of the object is the x-component of the velocity (which is constant). Making a right triangle using the initial velocity components gives the angle above the horizontal at launch to be tan=

m s m s

→ =tan?1(

19.4 )=52.3° . 15.0

22. For the following nuclear reaction, which one of the following choices correctly identifies the quantity labeled ? 40 0 +? →+ 19 0 (A) 40 19 (B) 40 18 (C) 39 18 (D) 39 19 (E) 40 18

Solution： E… In order to balance the reaction, we write

40 0 0 + ?→ + . This means 19 ?1 0

that we have 40+0=+0 → =40. Further, 19+(?1)=+0 →=18. By having =18, this means that the element is different from Potassium as that has 19 protons. In other words, 40 from the choices provided, we see that = . 18 23. At a party, two spherical balloons are expanded to have the same radius. Balloon #1 is filled with helium gas while balloon #2 is filled with xenon gas. Which one of the following statements about the buoyant forces on the balloons is correct? (A) The buoyant force is greater for the helium balloon. (B) The buoyant force is greater for the xenon balloon. (C) The buoyant force is the same for each balloon. (D) The balloon experiencing the greater buoyant force cannot be determined without knowing the radius. (E) The balloon experiencing the greater buoyant force cannot be determined without knowing the atmospheric pressure at the time.

Solution：C… Buoyant force is computed as the “weight of the fluid displaced.” The balloons have the same size and therefore displace the same amount of air. In other words, the buoyant forces are the same! The difference between the balloons would be seen on release when the helium balloon rises and the xenon balloon falls. 24. A proton moves with constant non-zero velocity in a region of space that has a uniform magnetic field directed into the plane of the page. The proton moves directly up the plane of the page. What is the direction of the electric field in this region of space? Ignore gravity. (A) To the left (B) To the right (C) Down the plane of the page (D) Up the plane of the page (E) There is no electric field necessary Solution：B… For the proton to move with constant velocity, the net force on it must be zero. Since the proton moves at a right angle to the magnetic field, there is a magnetic force on the proton. By the right-hand rule, with the velocity up the plane of the page and the field into the place, the magnetic force is directed to the left. In order to cancel this force, the electric force acting on the proton must be directed to the right. By = , the field and force are in the same direction for a positive charge. The field is directed to the right. 25. The plates of a large parallel-plate capacitor are separated by a distance of 0.05 . The potential difference between the plates is 24.0 . A charge released from rest between the plates experiences an electric force of 1.00 . What is the magnitude of the charge released between the plates? (A) 2.08 (B) 4.17 (C) 8.33 (D) 16.6 (E) 33.3

Solution：A… The force on the charged particle is computed as = . The electric field between the plates of a capacitor can be found from = d =0.05 m=480=480 . So, we find the charge between the plates as == 480 =0.00208 =2.08 .

1.00 ΔV 24.0 V

26. A 10.0 mass moves to the right at 8.00 . A 5.0 mass moves to the left at 7.00 . With what speed does the center of mass of this two-mass system move? (A) 7.67

(B) 7.50

(C) 3.00

(D) 2.00

(E) 0.50

Solution：C… By adding the momenta of the masses together, we have the total momentum

of the center of mass of the system. That is, 1 1+2 2=(1+2) →(10)(8)+(5)(?7)=(10+5) →45=15

→ =3.00 .

27. A tuning fork of frequency is placed over a long tube closed at one end producing the third lowest frequency standing wave for the tube. What frequency tuning fork would be needed to produce the fourth lowest frequency standing wave? (A) 43 (B) 53 (C) 65 (D) 75 (E) 54

Solution：D… For the initial situation, the third lowest frequency standing wave produced by the tube would be the fifth harmonic. Tubes closed at one end only have odd harmonics. This means that we have = . Since there are only odd harmonics, the next harmonic for the tube would be the seventh which has frequency =4L=5 (4L)= 5. 28. Recent excitement in the physics community came from the February, 2016 announcement from LIGO that (A) alien life had been discovered on a distant planet. (B) dark matter had been discovered. (C) there was proof of dark energy’s existence. (D) gravitational waves had been detected. (E) objects had been observed to travel faster than the speed of light. Solution：D… It was recently announced that gravitational waves have been detected! 29. A ray of light is incident onto a thin glass lens as shown. Which one of the following arrows best indicates the path of the refracted ray? (A) A (B) B (C) C (D) D (E) E Solution：E… The lens shown would be converging because of its shape. With the incident ray coming from the right side, the light will refract through the focus on the opposite side of the lens. This refracted ray

7v 7 5v 7 5v 4L

is shown in the figure. 30. A scientist wishes to compute the amount of energy necessary to melt a known mass of a solid sample. Which quantity associated with the sample would the scientist be most interested in knowing? (A) Latent heat of fusion (B) Specific heat (C) Coefficient of thermal expansion (D) Latent heat of vaporization (E) Thermal conductivity Solution：A… In order to melt a material, the property desired is the latent heat of fusion.

31. The circuit shown has 4 identical light bulbs connected to an ideal battery. If bulb #2 were removed leaving an open circuit there, what happens to the voltage across both bulb #1 and from Q to X?

Voltage across bulb #1 Remains the same Decreases Increases Increases Decreases Voltage from Q to X Decreases Remains the same Decreases Increases Increases

(A) (B) (C) (D) (E)

Solution：E… By removing bulb #2, the current for this branch becomes zero Amps meaning that the effective resistance of the circuit has risen. By increasing the effective resistance of the circuit means that the total current in the circuit will decrease. Since bulb #1 always receives the total current in the circuit, it now has a smaller potential difference from Ohm’s aw = as the resistance of bulb #1 is unchanged. By Kirchhoff’s Loop Rule, if the battery stays the same (which it did) and there is less potential difference across bulb #1, then there is more potential difference in the rest of the circuit (from Q to X)!

32. A mass connected to a string forms a simple pendulum. The mass is released from rest and undergoes simple harmonic motion. Which one of the following choices correctly ranks the magnitude of the tension in the string (), the gravitational force on the mass (), and the net force acting on the mass () when the mass swings through its lowest point?

(A) <= (B) << (C) <= (D) << (E) More information is required to answer the question.

Solution：D… Since the mass is in simple harmonic motion, the angle from the vertical upon first release would be fairly small (<15° ) . When the mass swings through its lowest location,

the acceleration of the mass would be straight up to the pivot of the pendulum. This means that > to have more upward force compared to downward force. To check on the size of the net force, let us assume that it is equal to the gravitational force and determine the angle at which the mass is released. In other words, ==

mv ? = →? = where is the L

length of the pendulum. From mechanical energy conservation during the fall of the mass to the lowest point from the highest, we write Δ+Δ=0→(

1 mv ? ? 0 )+(?(1?cos))=0 2

where is the maximum angle that the string makes with the vertical. Using our substitution, we have

1 1 mgL ?((1?cos))=0 →cos= → =60° . In other words, since we are 2 2

nowhere near that initial angle, < and <<.

33. Given the configuration of three charges,+,+, and ?2, which one of the following choices best represents the direction of the electric field at the point and the sign of the electric potential at from these charges?

Field Direction at P ↘ ↖ ↘ ↖ Zero Sign of Electric Potential at P Positive Positive Negative Negative Zero

(A) (B) (C) (D) (E)

Solution： B… The electric potential is computed as =

kQ for a point charge. As a result, at r

the point P, we have =

k Q k Q k (?2Q) k kQ + + =( )( 2Q ? 2Q )=(2? 2 ) ( )>0 a a a a 2a

As for the field, we need to vector add. By the symmetry, the fields from the positive charges are directed at a 45 ? upward from the left. In other words, it is in the direction opposite to that from the negative charge. The total field from the positive charges comes to

√( ) + ( 2 ) = 2 k Q . The field from the negative charge is 2 2 2 (2) (√2 )

2

a?

= k Q . Since the field

a?

from the negative charge has a smaller magnitude that the fields from the positive charges, the resultant field points up and to the left.

34. The graph shows the magnitude of the linear momentum of two solid objects colliding on the x-axis. Which one of the following statements must be true based on the information provided?

(A) The objects remain stuck together after the collision. (B) Object 2 experiences the greater force from the collision. (C) Object 2 cannot have more mass than object 1. (D) The kinetic energy loss was maximized with no external forces present in the two-object system. (E) The kinetic energy of the two-object system is the same both before and after the collision. Solution：C… Object 1 must be catching up to object 2 from behind because 1) the instantaneous momentum of neither object ever reached 0 kg

m s

and 2) object 1 loses

momentum while object 2 gains momentum. Because object 2 is in front of object 1 with 11= 22 after collision then 2≥1. This means that 2≤1. In the limiting case of 2=1, then answers (A) and (D) would be true and (E) would be false. Answer B is incorrect from Newton’s Third Law.

35. Which one of the following choices best describes what it means for an object to be iridescent?

(A) The color of the object appears to change with a change in the angle of view. (B) The object appears to have two images formed from it because of differences in index of refraction. (C) The color of the object appears to be different based on its temperature. (D) The object remains illuminated after turning off all surrounding light sources. (E) The object is alive and producing light biochemically. Solution：A… Iridescence has to do with the angle at which one is looking at an object and that the color of the object appears to change with it such as what happens with a peacock!

36. On a straight road, a self-driving car races with a robot. The robot accelerates at 8.50s?

until reaching its maximum constant speed of 15.0 s . The car accelerates at 5.60s? until reaching its maximum constant speed of 25.0 s . The car and robot start from rest from the same location. What is the maximum distance that the robot leads the car during the race? (A) 3.3 (B) 6.9 (C) 10.2 (D) 13.2 (E) 17.3

m m m

m

Solution：B… METHOD #1: One approach to the problem is to graph the velocity of each racer as a function of time. This is shown in the figure. The robot leads by its maximum amount when the speeds of the objects are the same. This is indicated at time 2 whereas time 1 indicates when the robot reaches its maximum speed. From the figure, we can find the change in position of each object with the area under the curve. This means that the distance that the robot leads the car is equal to the beige-colored triangle in the graph. All that is needed is to find the times of interest. This is done as follows: robot=0+robot1→ 1=arobot=5.60=1.76 and

robot

v

15.0

robot = 0+2→ 2=

vrobot 15.0 =5.60=2.68 acar 1

.

1 1 2

The area of the triangle is found as =2?=2(2?1)=

(2.68?1.76)(15)=6.9 .

METHOD #2: Alternatively, we can approach the problem from an algebraic point of view. In order to determine when the robot leads by the greatest distance, we need to find where the objects are when they have the same speed! In other words, it takes the car =0+→ =

vrobot 15.0 =5.60=2.68 . acar

By looking at constant acceleration kinematics for the car, we have its

1 2 1 2

position as =0+0+ ?→=0+0+ (5.60)(2.68)?=20.1 . The motion for the robot occurs in two stages as it accelerates at 8.50s? for some time and then travels at a constant rate thereafter. The robot reaches its maximum speed after =0+→ =

vrobot 15.0 = =1.76 arobot 5.60 m

. So, we compute the position of the robot after 2.68 as Δ0 1.76+Δ1.76 2.68.

For the first part, we write Δ0 1.76=0+1 ?=0+1 (8.50)(1.76)?=13.2 and Δ1.76 2 2 1 2.68=0+ ?=(15)(2.68?1.76)+0=13.8 . Hence, the robot is at a position of 2 13.2+13.8=27.0 … a full 6.9 ahead of the car. Questions 37 – 38 deal with the following information: A string is connected to a mechanical oscillator on one end and to a cube-shaped mass, =8.0 , at the other end as shown. The oscillator vibrates the string with a frequency of producing the standing wave in the figure on the left. When the mass is submerged completely in water, the string vibrates in the standing wave pattern shown on the right.

37. What is the value of the ratio for the tension when the mass is submerged in water to the tension when the mass is hanging in air?

(A) 4 (B) 2 (C)

√

(D) ?

(E) ?

Solution：E… From the figure, we see that the string length from the oscillator to the pulley went from being one full wavelength on the left to being two full wavelengths on the right. This means that the wavelength is now ? as large with the mass submerged. Since the oscillator is not changing the frequency of vibration, this means that the wave speed for the string is now ? as large from =. The wave speed on a string is determined as =√?μ which means to cut the speed in half resulted from a tension now ? as large.

38. What is the length of one side of the cube of mass ?

(A) 0.006 (B) 0.091 (C) 0.126 (D) 0.182 (E) 0.200

Solution：D… From the free body diagram of the mass hanging on the string in the figure on the left, we find =→?=0 →=80 . Since we determined in the previous question that the tension is now 20 with the mass submerged. This means that there is a buoyant force of 60 acting on the mass from the water. So, = →60=(1000)(10) →=0.006 ? where is the volume of displaced water (which is the same as the volume of the mass). Since the mass is a cube, ?=0.006 ? → =0.182

Questions 39 – 40 deal with the following information: An object moves in a circle, starting at the top, with initial speed 17.0 s . The object’s speed increases uniformly until it has moved counterclockwise through an angle of 55°. The average acceleration for this motion is 9.8s2 directed straight downward.

m m

39. What is the final speed of the object?

(A) 29.6 s

m

(B) 26.8 s

m

(C) 20.8 s

m

(D) 18.4 s

→

m

(E) 17.1 s

m

Solution：A… METHOD #1: By definition, ? ?=Δv . Call the final speed of the object to be Δt and so we can write =?17 +0 indicating it is moving to the left at the start and then =?cos55° ?sin55°. Also, we have ? ?=0 ?9.8 . And so, by looking at the

x-component, we see =

(?Vcos55°)?（?17） Vfx?Vix 17 m →0= → = =29.6 T T cos55° s

METHOD #2: One can construct a vector picture representing the velocities as shown →=→+→ . Noting that the resultant change in velocity is straight

v vi Δv

downward (direction of acceleration), we see from the construction that =

17 m =29.6 . cos55° s

40. What is the time for this motion?

(A) 1.54 (B) 1.80 (C) 2.26 (D) 2.48 (E) 2.93

Solution：D… From the previous answer, we now consider the y-component of the acceleration to find the time. That is, we have =

17

?

→?9.8=

(?Vsin55°)?(0) T

→

V=cos55° 9.8

sin55°

=

17tan55° =2.48 . 9.8

IMPORTANT: All Division 1 students STOP HERE. Your last answer should be for #40. Numbers 41-50 should remain blank for Division 1 students. All Division 2 students continue to Questions 41 – 50. ATTENTION: All Division 1 students – STOP HERE. All Division 2 students – continue to question #50.

41. A small object is launched 30°above the horizontal from a height of 10.0 above the

ground. If the initial speed of the object is 15.0

m s

, at what angle below the horizontal is the

object moving when it reaches the ground? Ignore air resistance. (A) 36.1° (B) 39.1° (C) 43.3° (D) 50.9° (E) 68.5°

Solution: D… METHOD #1: Let’s find things using kinematics. In the vertical direction, we

2 2 2 write =0 +2Δ → =(15sin30°)2+2(?10) (?10) → =?16.0

. The

x-component of the velocity is a constant =15cos30°=13.0 . Consequently, upon landing, the angle made by the object with the horizontal is computed as tan= →

=tan?1(? )=?50.9° METHOD #2: Using mechanical energy conservation, we can find the speed of the object

2 2 2 when it reaches the ground as Δ+Δ=0 → ( ? 2 )+Δ=0 → =0 ?2Δ 2 Hence, =(15)2?2(10)(?10) →=20.6 . Since the x-component of the velocity is m s 1 2 1 2

16 13

=15cos30° =13.0 , we can find the angle below the horizontal as cos= =cos?1(

13 )=50.9° . 20.6

m s

→

42. An object (which points upward) is placed 21.0 from a converging lens of focal length 18.0 . Which one of the following choices is true about the image formed by the lens?

(A) The image is larger than the object, virtual, and points upward. (B) The image is larger than the object, real, and points downward. (C) The image is smaller than the object, virtual, and points upward. (D) The image is smaller than the object, real, and points upward. (E) The image is smaller than the object, real, and points downward. Solution: B… Using the lens equation, we have = +→18=21+ →=126 . The magnification is found as =? =? 21 =?6. As the magnification is negative, this makes the image real and inverted. Since ||>1, the image is also larger than the object.

126 1 1 1 1 1 1

43. A uniform beam connected to a wall at the hinge is in static equilibrium as shown. The cable connected to the beam is massless. What are the directions of the horizontal and vertical components of force acting at the pivot on the beam?

Horizontal component To the right To the right To the left To the left It cannot be determined Vertical component Upward Downward Upward Downward It cannot be determined

(A) (B) (C) (D) (E)

Solution: C… As the object is in static equilibrium, the sum of forces and the sum of torques must be zero. The force from the cable is upward and rightward. Because the gravitational force is downward, the pivot must be providing a force to the left in order to cancel the rightward force from the cable. As for the vertical component, by choosing an axis of rotation

through the direction of the gravitational force at the vertical location of the pivot, the only non-zero torques are from the cable and the vertical component of the pivot force. As the torque from the cable would be oriented counterclockwise from this axis of rotation, the vertical force at the pivot needs to provide a clockwise torque. This is possible only if that vertical component is upward.

44. A proton moves in a circular orbit in a uniform magnetic field, . A helium nucleus moves in a circular orbit in the same magnetic field. If each particle experiences the same magnitude magnetic force during its motion, what is the ratio of the speed of the helium to the speed of the proton? (A) 4:1 (B) 1:4 (C) 1:1 (D) 2:1 (E) 1:2

Solution: E… As the objects move in circular paths, it is because of the magnetic force which has a magnitude of . Taking a ratio of the forces, we find

1 →1=(2)( )(1) →( )= . 2

=

45. Light of wavelength 250 shines onto a metallic surface. It is known that the electrons

ejected from the surface range in speed as 0≤≤ 4.85× 105 . What is the work function of the surface? (A) 4.97 (B) 1.60 (C) 3.63 (D) 4.30 (E) 1.34

1 1 m s

Solution: D… The kinetic energy of the electrons emitted range from 0 up to 22=2 (9.1×10?31)(4.85×105)2=1.07×10?19 =0.67 . The energy of the incoming photons is determined to be = λ =

hc (4.14×10?15 )(3.0×108 ) 250×10?9

=4.97 . The work function is determined by

taking the difference between the maximum kinetic energy and the energy of the incoming photon, meaning =4.97?0.67=4.30 .

46. Two spherical, non-rotating planets, X and Y, have the same uniform density . Planet X has twice the radius of Planet Y. Let and represent the escape speed at the surfaces of Planet X and Planet Y, respectively. What is the ratio of : ? (A) 2:1 (B) 1:2 (C) 1:1 (D) 4:1 (E) 1:4

Solution: A… The escape speed from a planet is found using mechanical energy conservation.

2 That is, Δ+Δ=0 →(0?2 )+(0?(? 1 ))=0

→ =√

2 .

The mass of a planet

2( π3 )

4 3

would be computed as == ( 3). With this substitution, we have =√

4 3

=√ ρGm. This means that

8 3

2 = = . 1

47. Two small loops of wire are located close to a wire with conventional current directed up the page. A scientist then moves the loops away from the current as shown. What is the orientation of the conventional current in each loop while this occurs?

Solution: C… Lenz’s Law is required to answer this question. The magnetic field associated with the long wire is into the page to the right of the wire and out of the page to the left of the wire. As the loop on the right is moved away, there are weaker field lines penetrating into the loop. As a result, there is an induced current oriented to try to replace the missing field lines. This means that the current in the right loop is oriented clockwise to put field lines into the plane in the interior of the loop. Likewise, on the left side of the long wire, there are again weaker field lines coming out of the plane of the loop. Here, the induction would be for a current oriented counterclockwise to have additional field lines out of the plane through the interior of the loop.

48. From a given state of ,, and for an ideal gas, which one of the following reversible processes has the most heat associated with it?

(A) Isothermal expansion doubling the volume volume (B) Isobaric expansion doubling the volume (C) Isovolumic pressure doubling Solution: B… The heat associated with the adiabatic process is zero Joules (as that is what it means to be reversibly adiabatic). For isobaric processes, the heat is =c Δ and for isovolumic processes, the heat is = c Δ. Also, we need the ideal gas equation to determine what happens to the temperature. For (B), the temperature doubles resulting in =n( )(2?)= , for (C), we have =( )(2?)= , and for (E) we compute =(2)( 2 ?)=? 4. Lastly, because there is no internal energy change for an isothermal process, = for this process. For the isobar in (B) there is a doubling in the

5 5 5 2 5 2 3 2 3 2

(D) Adiabatic expansion doubling the (E) Isobaric compression halving the volume

internal energy with more negative work done than for the isotherm meaning that > from Δ=+. 49. John is at rest on a platform that he measures to be 500 long on the x-axis. He then sees a spaceship moving along the x-axis with a speed of 1.80× 108

m s

as it crosses the length of

the platform. Sitting in her chair, the spaceship's pilot measures the time it takes her to cross the platform. What type of time interval does she measure and what value does she obtain?

Type of time interval Proper time Improper time Proper time Improper time Neither proper nor improper Time 2.22 2.22 2.78 2.78 3.48

(A) (B) (C) (D) (E)

Solution: A… Because the spaceship pilot is in one location, she has a single resting clock. This means that the time she records to cross the platform will be a proper time. From the pilot’s perspective, the platform is length contracted to be of length =

1.8

=√1 ? ( )2 =√1 ? ( 3 )2 (500)=400 . So, her clock will read a time of Δ=

=

400 =2.22 1.80×108

50. A uniform stick of mass and length is fixed to rotate about a frictionless pivot located ?3 from one end. The stick is released from rest incrementally away from being perfectly vertical, resulting in the stick rotating clockwise about its pivot. When the stick is horizontal, what is the tangential speed of the center of mass about the pivot?

(A)

1 √15

√

(B)√ √

(C)

√

√ (D

√

√ (E)

√

√

Solution: E… We will use mechanical energy conservation to approach this problem, Δ+Δ=0 →(2 2?0)+(Δ)=0. To find the moment of inertia about the pivot, we need the parallel axis theorem to obtain =122 +2 where = 6 (the distance between the center of the mass and the pivot) leading to = 122 +m(6)2 =9 2 . Also, the center of mass will have Δ=? 6 when the stick is horizontal. Upon substitution, we obtain (2ω2

L 1 1 L 1 1 L 1

?0)+(Δ)=0→

1 1 2 ω2 2 9

? =0 → ω2 ?=0 → =√ . Finally, the speed of the

L 3 L 6

L 6

1 3

3

center of mass will be computed as = with = 6 leading to =√

=√ 36 =

3

1 √ √12