## PhysicsBowl 2008

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**Instruction**

- Questions: The test is composed of 50 questions; however, students answer only 40 questions.
**Division 1 students will answer only questions 1 – 40. Do not answer questions 41 – 50.**

Division 2 students will answer only questions 11 – 50. Do not answer questions 1 – 10. - 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.

Of the following, which quantity is a vector?

$\textbf{(A) }$ Energy$ \qquad\newline$ $\textbf{(B) }$ Mass$ \qquad\newline$ $\textbf{(C) }$ Average speed$ \qquad\newline$ $\textbf{(D) }$ Temperature$ \qquad\newline$ $\textbf{(E) }$ Linear Momentum

$\textbf{E}$

Linear Momentum has both magnitude and direction, so it is a vector.

In one year, there are approximately $31.5×10^6\ \text{s}$. Which of the following representations using metric prefixes is equivalent to this value?

$\textbf{(A) }$ 31.5 ks$ \qquad$ $\textbf{(B) }$ 31.5 Ms$ \qquad$ $\textbf{(C) }$ 31.5 Gs$ \qquad$ $\textbf{(D) }$ 31.5 ms$ \qquad$ $\textbf{(E) }$ 31.5 ps

$\textbf{B}$

$10^3=\text{kilo(k)}$; $10^6=\text{Mega(M)}$; $10^9=\text{Giga(G )}$; $10^{-3}=\text{milli(m)}$; $10^{-12}=\text{pico(p)}$.

A dog starts from rest and runs in a straight line with a constant acceleration of $2.5\ \text{m/s}^2$. How much time does it take for the dog to run a distance of 10.0 m?

$\textbf{(A) }$ 8.0 s$ \qquad$ $\textbf{(B) }$ 4.0 s$ \qquad$ $\textbf{(C) }$ 2.8 s$ \qquad$ $\textbf{(D) }$ 2.0 s$ \qquad$ $\textbf{(E) }$ 1.4 s

$\textbf{C}$

Using $x=\dfrac12at^2$, we have $t=\sqrt{\dfrac{2x}{a}}=\sqrt{\dfrac{2(10)}{2.5}}=2.8\ \text{s}$.

One mole of an ideal gas has a temperature of $100\ ^\circ\text{C}$. If this gas fills the $10.0\ \text{m}^3$ volume of a closed container, what is the pressure of the gas?

$\textbf{(A) }$ 0.821 Pa$ \qquad$ $\textbf{(B) }$ 3.06 Pa$ \qquad$ $\textbf{(C) }$ 83.1 Pa$ \qquad$ $\textbf{(D) }$ 310 Pa$ \qquad$ $\textbf{(E) }$ $1.84\times10^{24}\ \text{Pa}$

$\textbf{D}$

Using the ideal gas equation $PV=nRT$, we have $P=\dfrac{nRT}{V}=\dfrac{(1)(8.31)(100+273)}{10}=310\ \text{Pa}$.

Approximately how much would it cost to keep a 100 W light bulb lit continuously for 1 year at a rate of $\$0.10/\text{kW}\cdot\text{hr}$?

$\textbf{(A) }$ $\$1 \qquad$ $\textbf{(B) }$ $\$10 \qquad$ $\textbf{(C) }$ $\$100 \qquad$ $\textbf{(D) }$ $\$1000 \qquad$ $\textbf{(E) }$ $\$100000$

$\textbf{C}$

The total cost is $\dfrac{\$0.10}{\text{kW}\cdot\text{hr}}\times100\ \text{W}\times\dfrac{1\ \text{kW}}{1000\ \text{W}}\times1\ \text{yr}\times\dfrac{365\ \text{dy}}{1\ \text{yr}}\times\dfrac{24\ \text{hr}}{1\ \text{dy}}=\$87.60$, which is closest to $\$100$.

A positive point charge exerts a force of magnitude $F$ on a negative point charge placed a distance $x$ away. If the distance between the two point charges is halved, what is the magnitude of the new force that the positive point charge exerts on the negative point charge?

$\textbf{(A) }$ $4F \qquad$ $\textbf{(B) }$ $2F \qquad$ $\textbf{(C) }$ $F \qquad$ $\textbf{(D) }$ $F/2 \qquad$ $\textbf{(E) }$ $F/4$

$\textbf{A}$

According to Coulomb’s Law $F=k\dfrac{Q_1Q_2}{r^2}$, when the distance $r$ is halved, the force will be 4 times greater.

Which of the following types of electromagnetic radiation has the greatest energy per photon?

$\textbf{(A) }$ infrared$ \qquad$ $\textbf{(B) }$ microwave$ \qquad$ $\textbf{(C) }$ FM radio$ \qquad$ $\textbf{(D) }$ AM radio$ \qquad$ $\textbf{(E) }$ violet light

$\textbf{E}$

The energy of a photon is related to the frequency as $E=h\nu$, so the electromagnetic wave with the greatest frequency has the greatest energy per photon.

An object on an inclined plane has a gravitational force of magnitude 10 N acting on it from the Earth. Which of the following gives the correct components of this gravitational force for the coordinate axes shown in the figure? The y-axis is perpendicular to the incline’s surface while the x-axis is parallel to the inclined surface.

$\textbf{A}$

$F_x=mg\sin36.9^\circ=6.0\ \text{N}$, $F_y=-10\cos36.9^\circ=-8.0\ \text{N}$.

When a beam of white light passes through a prism, the exiting light is seen as a spectrum of visible colors. This phenomenon is known as

$\textbf{(A) }$ diffraction. $ \qquad\newline$ $\textbf{(B) }$ dispersion.$ \qquad\newline$ $\textbf{(C) }$ interference. $ \qquad\newline$ $\textbf{(D) }$ polarization.$ \qquad\newline$ $\textbf{(E) }$ reflection.

$\textbf{B}$

When a medium is dispersive, the different wavelengths of light experience different indices of refraction, resulting in different refraction angles, leading to the separation of the colors.

Which of the following could be a correct unit for pressure?

$\textbf{(A) }$ $\text{kg/m}^2 \qquad$ $\textbf{(B) }$ $\text{kg/m}\cdot\text{s} \qquad$ $\textbf{(C) }$ $\text{kg/s}^2 \qquad$ $\textbf{(D) }$ $\text{kg/m}\cdot\text{s}^2 \qquad$ $\textbf{(E) }$ $\text{m}\cdot\text{s/kg}$

$\textbf{D}$

Pressure is force divided by area, so the unit of pressure is $\dfrac{\text{N}}{\text{m}^2}=\dfrac{\text{kg}\cdot\text{m/s}^2}{\text{m}^2}=\text{kg/m}\cdot\text{s}^2$.

Which person has won a Nobel Prize in physics?

$\textbf{(A) }$ Marie Curie$ \qquad\newline$

$\textbf{(B) }$ Isaac Newton$ \qquad\newline$

$\textbf{(C) }$ Aristotle$ \qquad\newline$

$\textbf{(D) }$ Johannes Kepler$ \qquad\newline$

$\textbf{(E) }$ Stephen Hawking

$\textbf{A}$

Marie Curie won the Nobel Prize in physics in 1903 for work on radiation.

What is the average angular speed of the second hand on a clock (in rad/s)?

$\textbf{(A) }$ $6.28 \qquad$ $\textbf{(B) }$ $0.105 \qquad$ $\textbf{(C) }$ $0.0167 \qquad$ $\textbf{(D) }$ $1.75\times10^{-3} \qquad$ $\textbf{(E) }$ $2.778\times10^{-4}$

$\textbf{B}$

The second hand of the clock takes 60 s to rotate for a circle, so the angular speed is $$\omega=\dfrac{\Delta\theta}{\Delta t}=\dfrac{2\pi\ \text{rad}}{60\ \text{s}}=0.105\ \text{rad/s}$$

What is the ideal mechanical advantage for the pulley system shown in the figure?

$\textbf{(A) }$ $F/Mg \qquad$ $\textbf{(B) }$ $Mg/F \qquad$ $\textbf{(C) }$ $3 \qquad$ $\textbf{(D) }$ $4 \qquad$ $\textbf{(E) }$ $5$

$\textbf{D}$

The ideal mechanical advantage of a pulley system is computed in relation to the number of strings effectively acting to pull upward on a load. Here, the 4 inner strings provide a force acting upward on the bottom pulley system connected to the mass.

**Questions 14 and 15 refer to the following scenario:**

A particle continuously moves in a circular path at constant speed in a counterclockwise direction. Consider a time interval during which the particle moves along this circular path from point P to point Q. Point Q is exactly half-way around the circle from Point P.

What is the direction of the average velocity during this time interval?

$\textbf{(A) }$ $\rightarrow \qquad$ $\textbf{(B) }$ $\leftarrow \qquad$ $\textbf{(C) }$ $\uparrow \qquad$ $\textbf{(D) }$ $\downarrow \qquad$ $\textbf{(E) }$ The average velocity is zero.

$\textbf{B}$

Average velocity is found as displacement divided by time. The displacement points from P to Q, so the average velocity points to the left.

What is the direction of the average acceleration during this time interval?

$\textbf{(A) }$ $\rightarrow \qquad$ $\textbf{(B) }$ $\leftarrow \qquad$ $\textbf{(C) }$ $\uparrow \qquad$ $\textbf{(D) }$ $\downarrow \qquad$ $\textbf{(E) }$ The average acceleration is zero.

$\textbf{D}$

Average acceleration is computed as the change in velocity divided by the time. At point P the velocity points upwards, and final points downwards at point Q. So the change of velocity, and also the average acceleration, point downwards.

A block is connected to a light string attached to the bottom of a large container of water. The tension in the string is 3.0 N. The gravitational force from the earth on the block is 5.0 N. What is the block’s volume?

$\textbf{(A) }$ $2.0\times10^{-4}\ \text{m}^3 \qquad\newline$ $\textbf{(B) }$ $3.0\times10^{-4}\ \text{m}^3 \qquad\newline$ $\textbf{(C) }$ $5.0\times10^{-4}\ \text{m}^3 \qquad\newline$ $\textbf{(D) }$ $8.0\times10^{-4}\ \text{m}^3 \qquad\newline$ $\textbf{(E) }$ $1.0\times10^{-3}\ \text{m}^3$

$\textbf{D}$

Using the free body diagram, the buoyant force $F=mg+T=5+3=8.0\ \text{N}$. The buoyant force is equal to the weight of the water displaced, so $F=\rho_{water}Vg$. Thus we get the volume $V=\dfrac{F}{\rho_{water}g}=\dfrac{8}{\left(1.0\times10^3\right)(10)}=8.0\times10^{-4}\ \text{m}^3$.

An ideal red pigment is mixed with an ideal blue pigment. After mixing, what color pigment results?

$\textbf{(A) }$ cyan$ \qquad$ $\textbf{(B) }$ magenta$ \qquad$ $\textbf{(C) }$ yellow$ \qquad$ $\textbf{(D) }$ green$ \qquad$ $\textbf{(E) }$ black

$\textbf{E}$

The red pigment reflects red light and absorbs both green and blue light. The blue pigment reflects blue light and absorbs both red and green light. By mixing them, the new pigment reflects nothing, so it looks black.

A car is moving to the left at 20 m/s while a truck is moving to the right at 25 m/s. If the truck emits a sound of frequency 5000 Hz, what is the perceived frequency of the sound by the driver of the car? Assume the vehicles are moving directly away from each other on a $20.0\ ^\circ\text{C}$ day.

$\textbf{(A) }$ 4384 Hz$ \qquad$ $\textbf{(B) }$ 4706 Hz$ \qquad$ $\textbf{(C) }$ 4932 Hz$ \qquad$ $\textbf{(D) }$ 5079 Hz$ \qquad$ $\textbf{(E) }$ 5714 Hz

$\textbf{A}$

Using the Doppler Shift equation, and that the speed of sound is $v_s=340\ \text{m/s}$, we have $$f_{observer}=f_{source}\dfrac{v_s-v_{observer}}{v_s+v_{source}}=5000\cdot\dfrac{340-20}{340+25}=4384\ \text{Hz}$$ The minus sign in the numerator arises since the observer is moving away from the source and the positive sign in the denominator arises since the source is moving away from the observer.

Kepler’s Second Law about “sweeping out equal areas in equal time” can be derived most directly from which conservation law?

$\textbf{(A) }$ energy$ \qquad\newline$ $\textbf{(B) }$ angular momentum$ \qquad\newline$ $\textbf{(C) }$ linear momentum$ \qquad\newline$ $\textbf{(D) }$ mechanical energy$ \qquad\newline$$\textbf{(E) }$ mass

$\textbf{B}$

Kepler’s Second Law is most directly related to the law of conservation of angular momentum.

A person pushes a block of mass $M=6.0\ \text{kg}$ with a constant speed of 5.0 m/s straight up a flat surface inclined $30.0^\circ$ above the horizontal. The coefficient of kinetic friction between the block and the surface is $\mu=0.40$. What is the net force acting on the block?

$\textbf{(A) }$ 0 N$ \qquad$ $\textbf{(B) }$ 21 N$ \qquad$ $\textbf{(C) }$ 30 N$ \qquad$ $\textbf{(D) }$ 51 N$ \qquad$ $\textbf{(E) }$ 76 N

$\textbf{A}$

The block moves with a constant velocity, so the acceleration is zero. According to Newton’s second law, the net force $F=ma=0$.

Modern telescopes use mirrors, rather than lenses, to form images. One advantage of mirrors over lenses is that the images formed by mirrors are not affected by:

$\textbf{(A) }$ destructive interference$ \qquad\newline$

$\textbf{(B) }$ constructive interference$ \qquad\newline$

$\textbf{(C) }$ chromatic aberration$ \qquad\newline$

$\textbf{(D) }$ spherical aberration$ \qquad\newline$

$\textbf{(E) }$ atmospheric refraction

$\textbf{C}$

Lenses are made of material which will be subject to dispersion. This means that different colors of visible light focus to different places because they experience different indices of refraction within the lens material.

In a calorimeter, 20 grams of liquid water at $100\ ^\circ\text{C}$ is mixed with 50 grams of water vapor at $100\ ^\circ\text{C}$. The system is allowed to come to equilibrium. Assuming that the calorimeter and the surroundings can be ignored, which of the following best describes the net energy exchange between the vapor and the liquid during the process of coming to equilibrium?

$\textbf{(A) }$ There is no net energy exchange.$ \qquad\newline$

$\textbf{(B) }$ Energy is transferred from the vapor to the liquid, vaporizing some of the liquid.$ \qquad\newline$

$\textbf{(C) }$ Energy is transferred from the vapor to the liquid, increasing the liquid’s temperature.$ \qquad\newline$

$\textbf{(D) }$ Energy is transferred from the vapor to the liquid until all of the liquid vaporizes.$ \qquad\newline$

$\textbf{(E) }$ Energy is transferred from the liquid to the vapor, condensing some vapor.

$\textbf{A}$

By the Zeroth Law of Thermodynamics, when materials are brought into contact at the same temperature, they are in thermal equilibrium. Hence, no net energy exchange occurs.

**Questions 23 and 24 refer to the following scenario:**

The velocity vs. time graph for the motion of a car on a straight track is shown in the diagram. The thick line represents the velocity. Assume that the car starts at the origin $x=0$.

At which time is the car the greatest distance from the origin?

$\textbf{(A) }$ $t=10\ \text{s} \qquad$ $\textbf{(B) }$ $t=6\ \text{s} \qquad$ $\textbf{(C) }$ $t=5\ \text{s} \qquad$ $\textbf{(D) }$ $t=3\ \text{s} \qquad$ $\textbf{(E) }$ $t=0\ \text{s}$

$\textbf{C}$

The car moves away from the origin initially and then slows to a stop at 5.0 seconds. One notes, though, that while the car is slowing down from $t=3.0\ \text{s}$ to $t=5.0\ \text{s}$… since the velocity is still positive, the car continues to move away from the origin. After $t=5.0\ \text{s}$, the car reverses direction as the velocity is negative and heads back toward the origin.

What is the average speed of the car for the 10 second interval?

$\textbf{(A) }$ 1.20 m/s$ \qquad$ $\textbf{(B) }$ 1.40 m/s$ \qquad$ $\textbf{(C) }$ 3.30 m/s$ \qquad$ $\textbf{(D) }$ 5.00 m/s$ \qquad$ $\textbf{(E) }$ 5.40 m/s

$\textbf{D}$

Average speed is distance divided by time. The total distance traveled by the car is found as the magnitude of the area under the velocity vs. time curve. Breaking this computation into pieces:$\newline$

$t=0\rightarrow3\ \text{s}$: $A=(8)(3)=24\ \text{m}\newline$

$t=3\rightarrow5\ \text{s}$: $A=\dfrac12(8)(2)=8\ \text{m}\newline$

$t=5\rightarrow6\ \text{s}$: $A=\dfrac12(4)(1)=2\ \text{m}\newline$

$t=6\rightarrow10\ \text{s}$: $A=(4)(5)=16\ \text{m}\newline$

Adding the magnitudes of the position changes together gives a total distance traveled of $24+8+2+16=50\ \text{m}$ . Hence, the average speed $\bar{v}=\dfrac{50\ \text{m}}{10\ \text{s}}=5.0\ \text{m/s}$.

The following nuclear reaction occurs: $_2^4\text{He}+_4^9\text{Be}\rightarrow^{12}_{\ 6}\text{C}+_Z^A\text{X}$. What is $_Z^A\text{X}$?

$\textbf{(A) }$ a proton$ \qquad$ $\textbf{(B) }$ an electron$ \qquad$ $\textbf{(C) }$ a positron$ \qquad$ $\textbf{(D) }$ an alpha particle$ \qquad$ $\textbf{(E) }$ a neutron

$\textbf{E}$

By $4+9=12+A$ and $2+4=6+Z$, we get $A=-1$ and $Z=0$. So X is a particle with 1 nucleon but no proton. This is a neutron. $_Z^A\text{X}=^1_0\text{n}$.

If the principal quantum number of an electron is $n=4$, how many possible values of the orbital magnetic quantum number $m_i$ are there for this electron?

$\textbf{(A) }$ 3$ \qquad$ $\textbf{(B) }$ 4$ \qquad$ $\textbf{(C) }$ 7$ \qquad$ $\textbf{(D) }$ 9$ \qquad$ $\textbf{(E) }$ 16

$\textbf{C}$

The principal quantum number allows one to find the azimuthal quantum number, $l$, as $l=3,2,1,0$. The magnetic quantum number is then found as $m_i=-l,-l+1,...,0,...,l-1,l$. Since $l=3$, the values of $m_i$ are $m_i=-3,-2,-1,0,1,2,3$, which is a total of 7 possible values.

A tube of length $L_1$ is open at both ends. A second tube of length $L_2$ is closed at one end and open at the other end. This second tube resonates at the same fundamental frequency as the first tube. What is the value of $L_2$?

$\textbf{(A) }$ $4L_1 \qquad$ $\textbf{(B) }$ $2L_1 \qquad$ $\textbf{(C) }$ $L_1 \qquad$ $\textbf{(D) }$ $\dfrac12L_1 \qquad$ $\textbf{(E) }$ $\dfrac14L_1$

$\textbf{D}$

For a tube open at both ends, there is 1/2 wavelength within the tube. For the tube closed at one end, there is only 1/4 wavelength contained in the tube. Hence, the tube closed at one end is half as long as the tube open at both ends when the fundamental frequency is the same.

A diverging lens produces an image of a real object. This image is

$\textbf{(A) }$ virtual, larger than the object, and upright.$ \qquad\newline$

$\textbf{(B) }$ virtual, smaller than the object, and upright.$ \qquad\newline$

$\textbf{(C) }$ virtual, smaller than the object, and inverted.$ \qquad\newline$

$\textbf{(D) }$ real, smaller than the object, and inverted.$ \qquad\newline$

$\textbf{(E) }$ real, larger than the object, and inverted.

$\textbf{B}$

For a real object $d_o>0$ and diverging lens $f<0$, by $\dfrac1{d_o}+\dfrac1{d_i}=\dfrac1f$, we get the image distance $d_i=\dfrac{fd_o}{d_o-f}<0$, so it is a virtual image. The magnification $M=-\dfrac{d_i}{d_o}=-\dfrac{f}{d_o-f}$ is greater than 0 bu less than 1, so it is a upright and smaller image.

A strong bar magnet is held very close to the opening of a solenoid as shown in the diagram. As the magnet is moved away from the solenoid at constant speed, what is the direction of conventional current through the resistor shown and what is the direction of the force on the magnet because of the induced current?

$\textbf{C}$

As the magnet moves away from the circuit, the field directed to the left through the solenoid (toward the South Pole of the magnet) is decreasing. This means that there is an induced electric field in the vicinity of the solenoid to produce a magnetic field directed to the left. Using the right-hand rule for a solenoid, the current through the solenoid must be directed upward “in front” of the solenoid… meaning that the current is directed from A to B through the resistor (the induced current and magnetic field are shown in the figure). The induction takes place to “fight the change” of the moved magnet, meaning that there is a force on the magnet trying to pull it back. We can model the magnet of the solenoid as having a North Pole on the left side, thereby attracting the magnet to the right.

A light beam passes through the air and strikes the surface of a plastic block. Which pair of statements correctly describes the phase changes for the reflected wave and the transmitted wave?

$\textbf{E}$

There is no phase change for transmitted waves, but when light bounces off a material which has a higher index of refraction, there is a phase shift of $180^\circ$.

For the circuit shown, $\xi=6.0\ \text{V}$, $R_1=7.0\Omega$, $R_2=3.0\Omega$, $R_3=6.0\Omega$, and $R_4=12.0\Omega$. After operating for a long time, equilibrium is established. What is the voltage across the capacitor at equilibrium?

$\textbf{(A) }$ 6.0 V$ \qquad$ $\textbf{(B) }$ 4.2 V$ \qquad$ $\textbf{(C) }$ 3.0 V$ \qquad$ $\textbf{(D) }$ 2.2 V$ \qquad$ $\textbf{(E) }$ 0.2 V

$\textbf{D}$

Once equilibrium is established in the circuit, there is no current to the capacitor. This means that the only current is through the branches with resistors. The voltage of the left side of the capacitor is $V_{left}=\dfrac{R_2}{R_1+R_2}\xi=\dfrac{3}{7+3}6=1.8\ \text{V}$. The voltage of the right side is $V_{right}=\dfrac{R_4}{R_3+R_4}\xi=\dfrac{12}{6+12}6=4\ \text{V}$. So the voltage across the capacitor is $4-1.8=2.2\ \text{V}$.

On February 20, 2008, there was a total lunar eclipse. What was the phase of the Moon during the eclipse?

$\textbf{(A) }$ New Moon$ \qquad\newline$ $\textbf{(B) }$ Full Moon$ \qquad\newline$ $\textbf{(C) }$ Dark Moon$ \qquad\newline$ $\textbf{(D) }$ Last Quarter$ \qquad\newline$ $\textbf{(E) }$ First Quarter

$\textbf{B}$

When there is a total lunar eclipse, the Moon passes into the shadow of the Earth… which means that the Moon is Full.

**Questions 33 and 34 refer to the following scenario:**

Two point charges are fixed on the x-axis in otherwise empty space as shown below.

In which Region(s) is there a place on the x-axis (aside from infinity) at which the electric potential is equal to zero?

$\textbf{(A) }$ Only in Region II$ \qquad\newline$

$\textbf{(B) }$ Only in Region III$ \qquad\newline$

$\textbf{(C) }$ In both Regions I and II$ \qquad\newline$

$\textbf{(D) }$ In both Regions I and III$ \qquad\newline$

$\textbf{(E) }$ In both Regions II and III

$\textbf{E}$

The electric potential of a point charge is $kQ/r$. In order for the potential to be zero in this configuration, one needs to be closer to the smaller charge. As a result, there is no place in Region I where the potential is zero as all points in that region are closer to the larger charge. There are locations, however, in both Regions II and III for which the total potential is zero.

In which Region(s) is there a place on the x-axis (aside from infinity) at which the electric field is equal to zero?

$\textbf{(A) }$ Only in Region II$ \qquad\newline$

$\textbf{(B) }$ Only in Region III$ \qquad\newline$

$\textbf{(C) }$ In both Regions I and II$ \qquad\newline$

$\textbf{(D) }$ In both Regions I and III$ \qquad\newline$

$\textbf{(E) }$ In both Regions II and III

$\textbf{B}$

Again, one must be closer to the smaller charge in order for the fields to cancel. This rules out Region I. Here, though, there is a direction associated with the field. In region II, the field from the $+3Q$ charge is to the right, whereas the field from the $–Q$ charge is also to the right. These fields cannot cancel. In Region III, the fields are in opposite directions and there is a place on the axis at which the fields from the two charges have equal magnitude.

Astronauts on the Moon perform an experiment with a simple pendulum that is released from the horizontal position at rest. At the moment shown in the diagram with $0^\circ<\theta<90^\circ$, the total acceleration of the mass may be directed in which of the following ways?

$\textbf{(A) }$ straight to the right$ \qquad\newline$

$\textbf{(B) }$ straight to the left$ \qquad\newline$

$\textbf{(C) }$ straight upward$ \qquad\newline$

$\textbf{(D) }$ straight downward$ \qquad\newline$

$\textbf{(E) }$ straight along the connecting string toward point P (the pivot)

$\textbf{A}$

The acceleration at $\theta=0^\circ$ is straight down whereas the acceleration at $\theta=90^\circ$ is straight upward. In transitioning between these angles, the sum of the two forces acting on the mass effectively rotates by $180^\circ$. As a result, the acceleration undergoes a transition like that shown in the figure (not quite drawn to scale) here (the red vector is the total acceleration). One notes that this result holds whether the angle $\theta$ increases (mass falling) or decreases (mass rising) and that the total acceleration is directed along the string only at $\theta=90^\circ$, outside the range of angles allowed in the problem.

An electron moves in the plane of the page through two regions of space along the dotted-line trajectory shown in the figure. There is a uniform electric field in Region I directed into the plane of the page (as shown). There is no electric field in Region II. What is a necessary direction of the magnetic field in regions I and II? Ignore gravitational forces.

$\textbf{C}$

Using the right-hand rule in Region II… fingers point along the velocity (to the right initially) and if the fingers are curled upward out of the page, then the right thumb points down the page… but since this is an electron, we flip our hand over to find the direction of the force. Hence, the field in Region II is out of the page. In region I, the electric force on the electron is directed OUT of the page with the electric field pointing into the page. Hence, we need a magnetic force INTO the page. In order to do this, we use the right-hand rule again and with the fingers pointing to the right, with the magnetic field UP the plane, then the right thumb points out of the page… but again, since this is an electron, the hand is flipped to find the force direction.

A uniform meter stick has a 45.0 g mass placed at the 20 cm mark as shown in the figure. If a pivot is placed at the 42.5cm mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick?

$\textbf{(A) }$ 18.0 g$ \qquad$ $\textbf{(B) }$ 45.0 g$ \qquad$ $\textbf{(C) }$ 72.0 g$ \qquad$ $\textbf{(D) }$ 120.0 g$ \qquad$ $\textbf{(E) }$ 135.0 g

$\textbf{E}$

The center of mass of the meter stick locates at the 50 cm mark, so the arm of force is $50-42.5=7.5\ \text{cm}$. The arm of force of the 45.0 g mass is $42.5-20=22.5\ \text{cm}$. According to Archimedes principle, $(45\ \text{g})\times(22.5\ \text{cm})=m\times(7.5\ \text{cm})$, so the mass of the meter stick is $m=\dfrac{(42)(22.5)}{7.5}=135\ \text{g}$.

A 1200 kg satellite orbits Planet X in a circular orbit with a constant speed of $5.00\times10^3\ \text{m/s}$. The radius of orbit is $7.50\times10^7\ \text{m}$. What is the magnitude of the gravitational force exerted on the satellite by Planet X?

$\textbf{(A) }$ 400 N$ \qquad$ $\textbf{(B) }$ 200 N$ \qquad$ $\textbf{(C) }$ 0.080 N$ \qquad$ $\textbf{(D) }$ 0.0127 N$ \qquad\newline$

$\textbf{(E) }$ More information is required to answer this question.

$\textbf{A}$

The gravitational force acts as the centripetal force of the circular motion, so $$F=\dfrac{mv^2}{r}=\dfrac{(1200)\left(5.00\times10^3\right)^2}{7.50\times10^7}=400\ \text{N}$$

An ideal gas is enclosed in a container. The volume of the container is reduced to half the original volume at constant temperature. According to kinetic theory, what is the best explanation for the increase in pressure created by the gas?

$\textbf{(A) }$ The average speed of the gas particles decreases, but they hit the container walls more frequently.$ \qquad\newline$

$\textbf{(B) }$ The average speed of the gas particles is unchanged, but they hit the container walls more frequently.$ \qquad\newline$

$\textbf{(C) }$ The average speed of the gas particles increases as does the frequency with which they hit the container walls.$ \qquad\newline$

$\textbf{(D) }$ The average speed of the gas particles increases, overcoming the decreased frequency that they hit the container walls.$ \qquad\newline$

$\textbf{(E) }$ The internal energy of the gas increases.

$\textbf{B}$

The average speed of molecules (rms-speed effectively) is dependent on temperature. With the temperature constant, there is no change in the speeds of the molecules. Hence, with a smaller volume, the distance that a particle has to travel in order to collide with the walls of the container is greatly decreased… hence there is an increase in the frequency with which particles hit the wall. This is reason for the pressure increase.

A point particle of mass $m$ collides with a thin rod pivoted at one end. The rod has mass $M=2m$, length $L$, and moment of inertia $I=\dfrac13ML^2$. The particle moves horizontally with speed $V$ when it hits the bottom of the rod and sticks to it. What is the speed of the particle immediately after collision?

$\textbf{(A) }$ $\dfrac13V \qquad$ $\textbf{(B) }$ $\dfrac1{\sqrt3}V \qquad$ $\textbf{(C) }$ $\dfrac35V \qquad$ $\textbf{(D) }$ $\dfrac34V \qquad$ $\textbf{(E) }$ $\dfrac{\sqrt3}2V$

$\textbf{C}$

The angular momentum about the pivot is conserved. The angular momentum before collision is $L_1=mVL$. Supposing the angular speed of the rod after collision is $\omega$, The angular momentum of the system is $L_2=\dfrac13ML^2\omega+m\omega L\cdot L=\dfrac53mL^2\omega$. By $L_1=L_2$, we get $\omega=\dfrac{3V}{5L}$. So the speed of particle after collision is $V_f=\omega L=\dfrac35V$.

The circuit shown has been operating for a long time. The instant after the switch in the circuit labeled S is $\textit{opened}$, what is the voltage across the inductor $V_L$ and which labeled point (A or B) of the inductor is at a higher potential? Take $R_1=4.0\Omega$, $R_2=8.0\Omega$, and $L=2.5\ \text{H}$.

$\textbf{(A) }$ $V_L=30\ \text{V}$; Points A and B are at equal potentials.$ \qquad\newline$

$\textbf{(B) }$ $V_L=12\ \text{V}$; Point A is at the higher potential.$ \qquad\newline$

$\textbf{(C) }$ $V_L=12\ \text{V}$; Point B is at the higher potential.$ \qquad\newline$

$\textbf{(D) }$ $V_L=6\ \text{V}$; Point A is at the higher potential.$ \qquad\newline$

$\textbf{(E) }$ $V_L=6\ \text{V}$; Point B is at the higher potential.

$\textbf{E}$

When the switch is closed, the resistance of the circuit is $\dfrac83\Omega$, so the current passing through the inductor is 4.5 A. At the moment when the switch is opened, the current passing through the inductor can not change abruptly. So the voltage across $R_1$ is $4.0\Omega\times4.5\ \text{A}=18\ \text{V}$. The voltage across the inductor is $12-18=-6\ \text{V}$. Negative sign means point B is at the higher potential.

A parallel-plate capacitor is connected to a battery. Without disconnecting the capacitor, a student pulls the capacitor’s plates apart so that the plate separation doubles. As a result of this action, what happens to the voltage across the capacitor and the energy stored by the capacitor?

$\textbf{(A) }$ the voltage doubles; the energy stays the same$ \qquad\newline$

$\textbf{(B) }$ the voltage halves; the energy doubles$ \qquad\newline$

$\textbf{(C) }$ the voltage doubles; the energy halves$ \qquad\newline$

$\textbf{(D) }$ the voltage stays the same; the energy halves$ \qquad\newline$

$\textbf{(E) }$ the voltage stays the same; the energy doubles

$\textbf{D}$

Since the battery remains connected, the voltage across the capacitor is unchanged. Since the geometry of the capacitor changes, so does the capacitance. The capacitance $C=\dfrac{\epsilon_0A}{d}$, doubling the distance between the plates halves the capacitance. Consequently, the total energy stored by the capacitor, given by $U=\dfrac12CV^2$, is also cut in half.

Unpolarized light of intensity $I_0$ enters a polarizer-analyzer system in which the angle between the transmission axes of the polarizer and analyzer is $30^\circ$. What is the intensity of the light leaving the analyzer?

$\textbf{(A) }$ $\dfrac38I_0 \qquad$ $\textbf{(B) }$ $\dfrac18I_0 \qquad$ $\textbf{(C) }$ $\dfrac34I_0 \qquad$ $\textbf{(D) }$ $\dfrac14I_0 \qquad$ $\textbf{(E) }$ $\dfrac12I_0$

$\textbf{A}$

By Malus’ Law, the intensity of the light out of the analyzer has intensity $I\cos^230^\circ=\dfrac34I$. The polarizer through which the unpolarized light initially shines reduces the intensity by a factor of 2, meaning that the total intensity of light out of the analyzer is given as $\dfrac34\left(\dfrac12I_0\right)=\dfrac38I_0$.

A mole of a monatomic ideal gas has pressure $P$, volume $V$, and temperature $T$. Which of the following processes would result in the greatest amount of energy added to the gas from heat?

$\textbf{(A) }$ A process doubling the temperature at constant pressure.$ \qquad\newline$

$\textbf{(B) }$ An adiabatic free expansion doubling the volume.$ \qquad\newline$

$\textbf{(C) }$ A process doubling the pressure at constant volume.$ \qquad\newline$

$\textbf{(D) }$ An adiabatic expansion doubling the volume.$ \qquad\newline$

$\textbf{(E) }$ A process doubling the volume at constant temperature.

$\textbf{A}$

For choice B and D, the adiabatic processes have no heat exchanged with the surroundings. Of the remaining processes, no matter doubling the temperature (choice A), the pressure (choice C) or the volume (choice E), by the ideal gas equation $PV=nRT$, the temperature is doubled, so the internal energy is double. So the process did the most work to the surrounding absorbs the greatest amount of energy. As shown in the figure, process A did the most work.

Electron $\#1$ moves with speed $0.30c$ where $c$ is the speed of light. Electron $\#2$ moves with speed $0.60c$. What is the ratio of the kinetic energy of electron $\#2$ to electron $\#1$?

$\textbf{(A) }$ 1.19$ \qquad$ $\textbf{(B) }$ 1.32$ \qquad$ $\textbf{(C) }$ 2.00$ \qquad$ $\textbf{(D) }$ 4.00$ \qquad$ $\textbf{(E) }$ 5.18

$\textbf{E}$

The kinetic energy $KE=mc^2-m_0c^2=\left(\dfrac{1}{\sqrt{1-v^2/c^2}}-1\right)m_0c^2$. For $v_1=0.3c$ and $v_2=0.6c$, we get $KE_1=0.048m_0c^2$ and $KE_2=0.25m_0c^2$. So the ratio $KE_2/KE_1=0.25/0.048=5.18$.

A traveling wave has the form $y(x,t)=3.0\sin(2.5x-5.0t)$ where all quantities given are in MKS units, $x$ is position, and $t$ represents time. What is the period of the wave (in seconds)?

$\textbf{(A) }$ 2.00$ \qquad$ $\textbf{(B) }$ 1.26$ \qquad$ $\textbf{(C) }$ 1.00$ \qquad$ $\textbf{(D) }$ 0.63$ \qquad$ $\textbf{(E) }$ 0.20

$\textbf{B}$

For a traveling wave, the general form is $y(x,t)=A\sin(kx-\omega t)$. So the period of the wave is found as $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{5}=1.26\ \text{s}$.

A radioactive sample decays with a half-life of 2.0 yr . Approximately how much time must pass so that only 1/3 of the original sample remains?

$\textbf{(A) }$ 6.0 yr$ \qquad$ $\textbf{(B) }$ 3.4 yr$ \qquad$ $\textbf{(C) }$ 3.2 yr$ \qquad$ $\textbf{(D) }$ 3.0 yr$ \qquad$ $\textbf{(E) }$ 2.8 yr

$\textbf{C}$

The radioactive decay can be expressed as $N=N_0\left(\dfrac12\right)^{t/\tau}$ where the half life $\tau=2.0\ \text{yr}$. When $N=\dfrac13N_0$, we have $\dfrac13=\left(\dfrac12\right)^{t/2}$, so $t=2\dfrac{\ln3}{\ln2}=3.2\ \text{yr}$.

A block of mass $M$ on a horizontal surface is connected to the end of a massless spring of spring constant $k$. The block is pulled a distance $x$ from equilibrium and when released from rest, the block moves toward equilibrium. What minimum coefficient of kinetic friction between the surface and the block would prevent the block from returning to equilibrium with non-zero speed?

$\textbf{(A) }$ $\dfrac{kx^2}{2Mg} \qquad$ $\textbf{(B) }$ $\dfrac{kx}{Mg} \qquad$ $\textbf{(C) }$ $\dfrac{kx}{2Mg} \qquad$ $\textbf{(D) }$ $\dfrac{Mg}{2kx} \qquad$ $\textbf{(E) }$ $\dfrac{k}{4Mgx}$

$\textbf{C}$

The coefficient of kinetic friction should be large enough to transfer all the potential energy of the spring into frictional loss before reaching the equilibrium point. The critical value is at which $\dfrac12kx^2=\mu Mgx$, so $\mu_{\min}=\dfrac{kx}{2Mg}$.

A circuit consists of a resistor, capacitor, and inductor connected in series to an AC source. As the source frequency increases, the current in the circuit decreases. Which statement about the circuit is $\textbf{NOT}$ correct as the source frequency increases?

$\textbf{(A) }$ The impedance of the circuit increases.$ \qquad\newline$

$\textbf{(B) }$ The circuit is said to become more capacitive than inductive.$ \qquad\newline$

$\textbf{(C) }$ The phase angle for the circuit becomes more positive.$ \qquad\newline$

$\textbf{(D) }$ The inductive reactance increases.$ \qquad\newline$

$\textbf{(E) }$ The total power from the source decreases.

$\textbf{B}$

The impedance $Z=R+j\left(\omega L-\dfrac1{\omega C}\right)$. As the frequency increases, the current decreases, so the impedance must increase (choice A). Impedance increase means $\omega L-\dfrac1{\omega C}>0$, so the circuit is more inductive (choice B), the phase angle for the circuit becomes more positive when the frequency increases (choice C). The inductive reactance $\omega L$ increases when the frequency increases (choice D). When the impedance increases and voltage remains unchanged, the power from the source decreases (choice E).

An infinitely long solenoid passes through the circuit as shown. The magnetic field of the solenoid, directed into the plane of the page, is weakening which produces a constant emf of magnitude $\xi$ for a closed loop around the outside of the solenoid. Once equilibrium is established in this circuit, what is the voltage across the switch S ?

$\textbf{(A) }$ $0 \qquad$ $\textbf{(B) }$ $\dfrac12\xi \qquad$ $\textbf{(C) }$ $\dfrac23\xi \qquad$ $\textbf{(D) }$ $\xi \qquad$ $\textbf{(E) }$ $\dfrac43\xi$

$\textbf{E}$

Since the field is directed into the plane and is decreasing, there is an induced electric field oriented clockwise in space from the solenoid. The switch is open in the circuit, so there is no current in the branch of the circuit with the battery once equilibrium is established. Considering the loop around the solenoid with the three resistors in a clockwise manner, one has $\xi=3IR$. In looking at a loop around the circuit with the battery, the bulb down the middle branch, and the switch, one has $\Delta V_S=IR+\xi=\dfrac43\xi$.