## PhysicsBowl 2012

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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.
**Treat***g*= 10 m/s$^2$ for all questions

Which one of the following lengths is the largest?

$\textbf{(A) }$ one centimeter $ \qquad\newline$ $\textbf{(B) }$ one kilometer$ \qquad\newline$ $\textbf{(C) }$ one millimeter$ \qquad\newline$ $\textbf{(D) }$ one meter$ \qquad\newline$ $\textbf{(E) }$ one nanometer

$\textbf{B}$

The lengths in the problem are (A) 0.01 m, (B) 1000 m, (C) 0.001 m, (D) 1.0 m, (E) $1\times10^{-9}\ \text{m}$

A ray of light passes straight downward through the point labeled B in the diagram shown. The ray reaches a flat mirror placed at an angle $\theta$ to the horizontal as shown. Which one of the locations labeled in the figure best represents the point through which the ray reflected from the mirror will pass?

$\textbf{(A) }$ A$ \qquad$ $\textbf{(B) }$ B$ \qquad$ $\textbf{(C) }$ C$ \qquad$ $\textbf{(D) }$ D$ \qquad$ $\textbf{(E) }$ E

$\textbf{C}$

The ray of light will bounce off the mirror with the same reflected angle as incident angle measured from the normal of the mirror’s surface. The figure demonstrates the ray’s path.

Which one of the following choices best represents the value of the speed of light using units of $\dfrac{\text{miles}}{\text{week}}$?

$\textbf{(A) }$ $5.90\times10^{12} \qquad\newline$ $\textbf{(B) }$ $1.13\times10^{11} \qquad\newline$ $\textbf{(C) }$ $1.61\times10^{10} \qquad\newline$ $\textbf{(D) }$ $6.75\times10^8 \qquad\newline$ $\textbf{(E) }$ $1.13\times10^8$

$\textbf{B}$

The speed of light $3\times10^8\ \dfrac{\text{m}}{\text{s}}\times\dfrac{1\ \text{mile}}{1600\ \text{m}}\times\dfrac{7\times24\times60\times60\ \text{s}}{1\ \text{week}}=1.13\times10^{11}\ \dfrac{\text{mile}}{\text{week}}$.

The wave speed is 20.0 m/s for waves traveling on a string tied at both ends. If sinusoidal waves with a frequency of 2.00 Hz are traveling on this string, which one of the following choices best represents the period of these waves?

$\textbf{(A) }$ 40.0 s$ \qquad$ $\textbf{(B) }$ 10.0 s$ \qquad$ $\textbf{(C) }$ 3.16 s$ \qquad$ $\textbf{(D) }$ 0.50 s$ \qquad$ $\textbf{(E) }$ 0.10 s

$\textbf{D}$

The period $T=\dfrac1f=0.5\ \text{s}$.

A solid rectangular box is measured to have length $L=13.34\ \text{cm}$, width $W=8.45\ \text{cm}$, and height $H=3.36\ \text{cm}$. Which one of the following choices best represents the volume of the box using proper significant digits?

$\textbf{(A) }$ $4\times10^2\ \text{cm}^3 \qquad\newline$ $\textbf{(B) }$ $3.8\times10^2\ \text{cm}^3 \qquad\newline$ $\textbf{(C) }$ $3.79\times10^2\ \text{cm}^3 \qquad\newline$ $\textbf{(D) }$ $3.787\times10^2\ \text{cm}^3 \qquad\newline$ $\textbf{(E) }$ $3.7875\times10^2\ \text{cm}^3$

$\textbf{C}$

From rules of significant digits when multiplying, the result has an equal number of sig figs as the value with the fewest sig figs. Here, the values have 4,3, and 3 sig figs meaning that the result also must have 3 significant digits.

Electromagnetic radiation travels through vacuum with a wavelength of 400 nm. Which one of the following choices best describes this type of radiation?

$\textbf{(A) }$ X-rays$ \qquad\newline$ $\textbf{(B) }$ Radio Waves$ \qquad\newline$ $\textbf{(C) }$ Microwaves$ \qquad\newline$ $\textbf{(D) }$ Red Light$ \qquad\newline$ $\textbf{(E) }$ Violet Light

$\textbf{E}$

Visible light is in the wavelength range of 400 nm - 700 nm with violet at the 400 nm end.

**Questions 7 and 8 deal with the following information:**

An object of mass 5.00 kg moves only to the right along the +x-axis. During some time interval, the object’s speed is increased from 4.00 m/s to 8.00 m/s with a constant acceleration of 2.00 m/s$^2$.

What is the net force acting on the object during the time interval of the acceleration?

$\textbf{(A) }$ 10.0 N$ \qquad\newline$ $\textbf{(B) }$ 20.0 N$ \qquad\newline$ $\textbf{(C) }$ 30.0 N$ \qquad\newline$ $\textbf{(D) }$ 40.0 N$ \qquad\newline$

$\textbf{(E) }$ The answer cannot be determined without more information about the forces involved.

$\textbf{A}$

By Newton’s Second Law, $F=ma=10.0\ \text{N}$.

Through what distance does the object move during the time interval of the acceleration?

$\textbf{(A) }$ 2.00 m$ \qquad$ $\textbf{(B) }$ 4.00 m$ \qquad$ $\textbf{(C) }$ 8.00 m$ \qquad$ $\textbf{(D) }$ 12.0 m$ \qquad$ $\textbf{(E) }$ 24.0 m

$\textbf{D}$

The average speed $\overline{v}=\dfrac{v_i+v_f}2=6\ \text{m/s}$. The time for it to travel is $t=\dfrac{v_f-v_i}a=2\ \text{s}$. So the total distance $x=\overline{v}t=12\ \text{m}$.

A constant current of 4.00 A through a light bulb results in a power of 24.0 W associated with the bulb. Which one of the following choices best represents the resistance of the light bulb?

$\textbf{(A) }$ $96.0\ \Omega \qquad$ $\textbf{(B) }$ $6.00\ \Omega \qquad$ $\textbf{(C) }$ $2.45\ \Omega \qquad$ $\textbf{(D) }$ $1.50\ \Omega \qquad$ $\textbf{(E) }$ $0.67\ \Omega$

$\textbf{D}$

The resistance of the light bulb $R=\dfrac{P}{I^2}=1.5\ \text{W}$.

A string of negligible mass connects an object of mass $M=10\ \text{kg}$ to the ceiling of an elevator. The elevator experiences a constant downward acceleration of magnitude $6.0\ \text{m/s}^2$. Let $T$ represent the magnitude of the force by the string (tension) acting on the mass $M$, let $G$ represent the magnitude of the gravitational force by the Earth acting on the mass $M$, and let $F$ represent the magnitude of the net force acting on the mass $M$. Which one of the following choices describes the relationships between these forces?

$\textbf{(A) }$ $T=G=F \qquad\newline$

$\textbf{(B) }$ $F<G<T \qquad\newline$

$\textbf{(C) }$ $T=F<G \qquad\newline$

$\textbf{(D) }$ $F<T=G \qquad\newline$

$\textbf{(E) }$ $T<F<G$

$\textbf{E}$

The gravitational force $G=Mg=100\ \text{N}$. The net force $F=Ma=60\ \text{N}$. The tension $T=G-F=40\ \text{N}$. So $T<F<G$.

A girl twirls a small mass connected to the end of a string counterclockwise in a horizontal circle above her head. The figure shows an outline of the mass’s path viewed from above the twirling mass. If the girl needs the mass to pass through the point labeled P in the figure, at which lettered point on the path should she let go of the string?

$\textbf{(A) }$ A$ \qquad$ $\textbf{(B) }$ B$ \qquad$ $\textbf{(C) }$ C$ \qquad$ $\textbf{(D) }$ D$ \qquad$ $\textbf{(E) }$ E

$\textbf{B}$

By Newton’s First Law, the mass will continue to move horizontally with constant speed. For the object, the trajectories of the object as it twirls counterclockwise are shown for all choices.

Which one of the following quantities is a scalar quantity?

$\textbf{(A) }$ Impulse$ \qquad\newline$ $\textbf{(B) }$ Linear Momentum$ \qquad\newline$ $\textbf{(C) }$ Acceleration$ \qquad\newline$ $\textbf{(D) }$ Speed$ \qquad\newline$ $\textbf{(E) }$ Displacement

$\textbf{D}$

Speed is the magnitude of velocity.

Approximately how many electrons must be removed from an electrically neutral object to give it a net charge of $Q=+1.00\ \text{C}$?

$\textbf{(A) }$ $1.60\times10^{-19} \qquad\newline$ $\textbf{(B) }$ $1 \qquad\newline$ $\textbf{(C) }$ $6.25\times10^{18} \qquad\newline$ $\textbf{(D) }$ $6.02\times10^{23} \qquad\newline$ $\textbf{(E) }$ $1.10\times10^{30}$

$\textbf{C}$

The charge of an electron is $e=-1.60\times10^{-19}\ \text{C}$. For a net charge of $Q=+1.00\ \text{C}$, the number of electrons removed is $N=\dfrac{Q}{|e|}=6.25\times10^{-18}$.

For the circuit shown to the right, what is the equivalent resistance? Assume that all wires are ideal, the battery has no internal resistance, and all three resistors have identical resistance $R$.

$\textbf{(A) }$ $3R \qquad$ $\textbf{(B) }$ $\dfrac32R \qquad$ $\textbf{(C) }$ $\dfrac23R \qquad$ $\textbf{(D) }$ $\dfrac13R \qquad$ $\textbf{(E) }$ $R$

$\textbf{B}$

The two resistors on the right is in parallel, and then is series with the left one, so the equivalent resistance is $R_{eq}=\dfrac12R+R=\dfrac32R$.

Which one of the following phases of the lunar cycle immediately follows “First Quarter”?

$\textbf{(A) }$ Waxing Gibbous$ \qquad\newline$ $\textbf{(B) }$ Waning Gibbous$ \qquad\newline$ $\textbf{(C) }$ Waxing Crescent$ \qquad\newline$ $\textbf{(D) }$ Waning Crescent$ \qquad\newline$ $\textbf{(E) }$ New Moon

$\textbf{A}$

The phases of the Moon are : New Moon – Waxing Crescent – First Quarter – Waxing Gibbous – Full Moon – Waning Gibbous – Last (or Third) Quarter – Waning Crescent – New Moon ….

A constant force $F=50\ \text{N}$ (as shown in the figure) is applied for the 6.0 meter motion of the box upward along the incline. The mass of the block is $M=15\ \text{kg}$. Which one of the following choices best represents the work done by the force $F$ on the box for the motion?

$\textbf{(A) }$ 300 J$ \qquad$ $\textbf{(B) }$ 282 J$ \qquad$ $\textbf{(C) }$ 260 J$ \qquad$ $\textbf{(D) }$ 193 J$ \qquad$ $\textbf{(E) }$ 150 J

$\textbf{B}$

The work done by the force on the box is $W=\vec{F}\cdot\vec{d}=Fd\cos20^\circ=(50)(6)\cos20^\circ=282\ \text{J}$.

A mass moves according to the graph of position as a function of time shown below. Which one of the following choices correctly represents the time instants or time interval for which the instantaneous velocity of the mass is considered always to be negative? Let $t$ represent time.

$\textbf{(A) }$ $t=0.0\ \text{s}, t=5.0\ \text{s and}\ t=10.0\ \text{s} \qquad\newline$

$\textbf{(B) }$ $0.0\ \text{s}<t<2.5\ \text{s} \qquad\newline$

$\textbf{(C) }$ $2.5\ \text{s}<t<7.5\ \text{s} \qquad\newline$

$\textbf{(D) }$ $5.0\ \text{s}<t<10.0\ \text{s} \qquad\newline$

$\textbf{(E) }$ $2.5\ \text{s}<t<10.0\ \text{s}$

$\textbf{C}$

The slope at a given time of the position vs. time graph gives the velocity at that time. Negative velocity means negative local slope. This occurs everywhere from $2.5\ \text{s}<t<7.5\ \text{s}$.

A 1.00 kg object is released from rest near the surface of the Earth. The gravitational force acting on the 1.00 kg object by the Earth does 10.0 J of work on the object as it falls 1.00 m to the ground. Which one of the following choices best represents the amount of work done by the gravitational force acting on the Earth by the 1.00 kg object during the fall?

$\textbf{(A) }$ 0.0 J$ \qquad\newline$ $\textbf{(B) }$ -10.0 J$ \qquad\newline$ $\textbf{(C) }$ 10.0 J$ \qquad\newline$ $\textbf{(D) }$ $-5.86\times10^{25}\ \text{J} \qquad\newline$ $\textbf{(E) }$ $5.86\times10^{25}\ \text{J}$

$\textbf{A}$

The force of gravity acting on the Earth from the object is 10 N as it is the same as the gravitational force acting on the object from the Earth. However, while the object falls a distance of 1.0 m, the Earth moves infinitesimally because of its large mass. Hence, the work $W=Fd\approx0\ \text{J}$.

The magnitude of the linear momentum of a 4.00 kg point mass is changed from $6.00\ \text{kg}\cdot\text{m/s}$ to $14.0\ \text{kg}\cdot\text{m/s}$ in a time interval of 6.00 s. What is the change in the kinetic energy of the mass during this time interval?

$\textbf{(A) }$ 8.0 J$ \qquad$ $\textbf{(B) }$ 10.0 J$ \qquad$ $\textbf{(C) }$ 16.0 J$ \qquad$ $\textbf{(D) }$ 20.0 J$ \qquad$ $\textbf{(E) }$ 32.0 J

$\textbf{D}$

The kinetic energy $KE=\dfrac{p^2}{2m}$. The change of kinetic energy $\Delta KE=\dfrac{p_f^2}{2m}-\dfrac{p_i^2}{2m}=\dfrac{14^2}{2(4)}-\dfrac{6^2}{2(4)}=20\ \text{J}$.

“Both the position and momentum of an electron cannot be known exactly at the same instant of time.” To whom is this concept attributed?

$\textbf{(A) }$ Pauli$ \qquad$ $\textbf{(B) }$ de Broglie$ \qquad$ $\textbf{(C) }$ Einstein$ \qquad$ $\textbf{(D) }$ Dirac$ \qquad$ $\textbf{(E) }$ Heisenberg

$\textbf{E}$

This is a statement of the Heisenberg Uncertainty Principle.

A child’s balloon is filled with pure Xenon gas. This balloon then is released from rest two meters above the ground on Earth. Which one of the following choices best describes the response of the balloon?

$\textbf{(A) }$ The balloon immediately falls toward the ground.$ \qquad\newline$

$\textbf{(B) }$ The balloon floats gently in the air, finally reaching the ground after several minutes.$ \qquad\newline$

$\textbf{(C) }$ The balloon floats gently in the air, essentially hovering at the same height for at least a day.$ \qquad\newline$

$\textbf{(D) }$ The balloon very, very slowly and gently rises upward.$ \qquad\newline$

$\textbf{(E) }$ The balloon rapidly rises into the sky.

$\textbf{A}$

Xenon gas is much denser than the surrounding air. As a result, when released, the balloon sinks immediately to the ground.

A girl swings a 4.0 kg mass with a constant speed of 3.24 m/s in a vertically-oriented circle of radius 0.75 m. What is the net force acting on the mass when it is at the lowest point of the circle?

$\textbf{(A) }$ 96 N$ \qquad$ $\textbf{(B) }$ 56 N$ \qquad$ $\textbf{(C) }$ 40 N$ \qquad$ $\textbf{(D) }$ 16 N$ \qquad$ $\textbf{(E) }$ 0 N

$\textbf{B}$

The circular motion is drove by the net force as the centripetal force, thus $F_{net}=m\dfrac{v^2}{R}=4\cdot\dfrac{3.24^2}{0.75}=56\ \text{N}$.

**Questions 23 and 24 deal with the following information:**

A collision of two blocks takes place along a horizontal surface without friction. A block with mass $M_1=3.00\ \text{kg}$ initially moves to the left with speed $V_1=5.00\ \text{m/s}$ when it hits a block with mass $M_2=5.00\ \text{kg}$ initially moving to the right with speed $V_2=2.00\ \text{m/s}$. After colliding, the block with mass $M_1$ is moving to the right with speed 1.00 m/s.

Which of the blocks underwent a larger magnitude of acceleration during the collision?

$\textbf{(A) }$ The block with mass $M_1$.$ \qquad\newline$

$\textbf{(B) }$ The block with mass $M_2$.$ \qquad\newline$

$\textbf{(C) }$ The magnitude of the acceleration was the same for both blocks.$ \qquad\newline$

$\textbf{(D) }$ The answer depends on how much kinetic energy was transferred out of the two-block system.$ \qquad\newline$

$\textbf{(E) }$ More information about the time of the collision is required to answer the question.

$\textbf{A}$

From Newton’s Third Law, the force that $M_1$ exerts on $M_2$ has the same magnitude of the force that $M_2$ exerts on $M_1$. By Newton’s Second Law, $a=\dfrac{F}{m}$, the less massive object undergoes a greater acceleration.

What is the speed of the block with mass $M_2$ after the collision?

$\textbf{(A) }$ 0.40 m/s$ \qquad$ $\textbf{(B) }$ 1.00 m/s$ \qquad$ $\textbf{(C) }$ 1.60 m/s$ \qquad$ $\textbf{(D) }$ 4.29 m/s$ \qquad$ $\textbf{(E) }$ 4.40 m/s

$\textbf{C}$

If we take the direction to the right as the positive direction, then the linear momentum before collision is $p_{1i}=M_1V_1=(3)(-5)=-15\ \text{kg}\cdot\text{m/s}$, $p_{2i}=M_2V_2=(5)(2)=10\ \text{kg}\cdot\text{m/s}$. After collision, $p_{1f}=(3)(1)=3\ \text{kg}\cdot\text{m/s}$. Using linear momentum conservation, we have $p_{2f}=p_{1i}+p_{2i}-p_{1f}=-8\ \text{kg}\cdot\text{m/s}$. So the speed of $M_2$ after collision is $v=p_{2f}/M_2=-1.6\ \text{kg}\cdot\text{m/s}$. Negative speed means $M_2$ moves to the left.

Which one of the following choices best represents the average angular speed of the hour hand on a standard clock (in units of rad/s)?

$\textbf{(A) }$ $5.24\times10^{-1} \qquad\newline$ $\textbf{(B) }$ $2.62\times10^{-1} \qquad\newline$ $\textbf{(C) }$ $1.75\times10^{-3} \qquad\newline$ $\textbf{(D) }$ $1.45\times10^{-4} \qquad\newline$ $\textbf{(E) }$ $7.27\times10^{-5}$

$\textbf{D}$

The hour hand on a standard clock makes a complete circle every 12 hours. Hence, $$\omega=\dfrac{\Delta\theta}{\Delta t}=\dfrac{2\pi\ \text{rad}}{12\ \text{hr}}=\dfrac{2\pi\ \text{rad}}{12\times60\times60\ \text{s}}=1.45\times10^{-4}\ \text{rad/s}$$

An object is thrown horizontally with speed 10.0 m/s from a height $H$ above the ground. The object reaches the ground with a speed of 20.0 m/s. Which one of the following choices best represents the time of the object’s flight to the ground? Ignore air resistance.

$\textbf{(A) }$ 1.00 s$ \qquad$ $\textbf{(B) }$ 1.22 s$ \qquad$ $\textbf{(C) }$ 1.41 s$ \qquad$ $\textbf{(D) }$ 1.50 s$ \qquad$ $\textbf{(E) }$ 1.73 s

$\textbf{E}$

The horizontal component of the velocity does not change during flight, but not the vertical component. By the Pythagorean Theorem, $v_f=\sqrt{v_\parallel^2+v_\perp^2}$, so the vertical component of velocity $v_\perp=\sqrt{v_f^2-v_\parallel^2}=\sqrt{20^2-10^2}=17.3\ \text{m/s}$. The time of dropping $t=\dfrac{v_\perp}{g}=1.73\ \text{s}$.

The pressure inside a container with two moles of an ideal gas is 0.75 atm. The temperature of the gas is $100\ ^\circ\text{C}$. The container maintains constant volume as the pressure is tripled. Which one of the following choices best represents the temperature of the gas after the pressure is tripled?

$\textbf{(A) }$ $33\ ^\circ\text{C} \qquad$ $\textbf{(B) }$ $300\ ^\circ\text{C} \qquad$ $\textbf{(C) }$ $573\ ^\circ\text{C} \qquad$ $\textbf{(D) }$ $846\ ^\circ\text{C} \qquad$ $\textbf{(E) }$ $1119\ ^\circ\text{C}$

$\textbf{D}$

By the ideal gas equation $PV=nRT$, when the volume is a constant and pressure tripled, the temperature also tripled. However, the temperature is measured in the unit of Kelvin. The initial temperature $100\ ^\circ\text{C}=(100+273)\ \text{K}=373\ \text{K}$. The final temperature is $3(373)\ \text{K}=1119\ \text{K}=(1119-273)\ ^\circ\text{C}=846\ ^\circ\text{C}$.

A long straight wire has a conventional current directed into the plane of the page as shown in the figure. Which one of the arrows shown best indicates the direction of the magnetic field associated with this wire at the location labeled P?

$\textbf{(A) }$ A$ \qquad$ $\textbf{(B) }$ B$ \qquad$ $\textbf{(C) }$ C$ \qquad$ $\textbf{(D) }$ D$ \qquad$ $\textbf{(E) }$ E

$\textbf{E}$

The magnetic field lines form concentric circles around the wire with the right-hand rule of the right thumb along the current and the fingers wrapping in the sense of the field. At a given location, the field direction is tangent to the circle. For this configuration, the field lines form clockwise circles and therefore at P, the field is directed straight downward.

A vehicle completes one lap around a circular track at an average speed of 50 m/s and then completes a second lap at an average speed of $V$. The average speed of the vehicle for the completion of both laps was 80 m/s. What was the average speed $V$ of the second lap?

$\textbf{(A) }$ 100 m/s$ \qquad$ $\textbf{(B) }$ 110 m/s$ \qquad$ $\textbf{(C) }$ 125 m/s$ \qquad$ $\textbf{(D) }$ 150 m/s$ \qquad$ $\textbf{(E) }$ 200 m/s

$\textbf{E}$

For a general track length $L$, the first lap takes $t_1=\dfrac{L}{50}$ seconds, the second lap takes $t_2=\dfrac{L}{V}$ seconds. The average speed for the total two laps is $$80=\dfrac{2L}{t_1+t_2}=\dfrac{2L}{\dfrac{L}{50}+\dfrac{L}{V}}=\dfrac{2}{\dfrac1{50}+\dfrac1V}$$ By solving the equation we have $V=200\ \text{m/s}$.

Coherent light of wavelength 550 nm shines on a double slit apparatus that has point slits spaced by a distance of $42.4\ \mu\text{m}$. In theory, what is the maximum order bright fringe that can be viewed?

$\textbf{(A) }$ 1297$ \qquad$ $\textbf{(B) }$ 77$ \qquad$ $\textbf{(C) }$ 12$ \qquad$ $\textbf{(D) }$ 8$ \qquad$ $\textbf{(E) }$ 1

$\textbf{B}$

From the double slit experiment, we have the bright fringes $d\sin\theta=m\lambda$ where $m$ is the order of the bright fringe. The maximum value for $m$ occurs when $\theta=90^\circ$ leading to $$m_{max}=\dfrac{d}{\lambda}=\dfrac{42.4\times10^{-6}\ \text{m}}{550\times10^{-9}\ \text{m}}=77.1$$ Since there can only be integer orders viewed, the maximum value is the 77th order bright fringe.

Three particles with charges $Q$, $Q$, and $Q_2$ are initially at rest very (infinitely) far apart from one another. The particles are moved to the locations shown in the figure where they are fixed in place. The particles on the left and right have charge $Q$ and are separated by a distance $R$. The particle with charge $Q_2$ is located at the midpoint directly between the other charged particles. The total work required to configure this three-particle arrangement is 0 Joules. Ignore any self-energies for the particles. What is the value of the charge $Q_2$?

$\textbf{(A) }$ $-\dfrac14Q \qquad$ $\textbf{(B) }$ $-\dfrac{\sqrt2}4Q \qquad$ $\textbf{(C) }$ $-\dfrac12Q \qquad$ $\textbf{(D) }$ $-\dfrac18Q \qquad\newline$

$\textbf{(E) }$ It is not possible to accomplish what is required.

$\textbf{A}$

In the beginning three particles are far away from each other, so the potential energy is zero. When they are moved to the location as shown in the figure, then work been done is zero, so the potential energy is still zero. Thus we have $$\dfrac12\left(\dfrac{kQ^2}{R}+\dfrac{kQQ_2}{R/2}\right)+\dfrac12\left(\dfrac{kQQ_2}{R/2}+\dfrac{kQQ_2}{R/2}\right)+\dfrac12\left(\dfrac{kQ^2}{R}+\dfrac{kQQ_2}{R/2}\right)=0$$ By solving the equation we have $Q_2=-\dfrac14Q$.

To whom was the first Nobel Prize in physics awarded?

$\textbf{(A) }$ Isaac Newton for his contributions to physics and calculus.$ \qquad\newline$

$\textbf{(B) }$ James Chadwick for the discovery of the neutron.$ \qquad\newline$

$\textbf{(C) }$ Wilhelm Röntgen for the discovery of X-rays.$ \qquad\newline$

$\textbf{(D) }$ Marie Curie for her work in radioactivity.$ \qquad\newline$

$\textbf{(E) }$ Albert Einstein for his explanation of the photoelectric effect and for the theories of relativity.

$\textbf{C}$

The first Nobel Prize in physics went to Wilhelm Röntgen in 1901 for his work with X-rays. Isaac Newton never won a Nobel Prize and Albert Einstein did win a Nobel Prize in 1921. Marie Curie was the first woman to win the Nobel Prize in physics in 1903. James Chadwick won the Nobel Prize in 1935.

A solid, uniform sphere rolls without slipping on a floor along the $+x$-axis (to the right). The rotational kinetic energy associated with the sphere about an axis of rotation through its center of mass along the $+z$-axis (out of the plane of the page) is 20 Joules. What is the translational kinetic energy associated with the sphere?

$\textbf{(A) }$ 8 J$ \qquad$ $\textbf{(B) }$ 10 J$ \qquad$ $\textbf{(C) }$ 20 J$ \qquad$ $\textbf{(D) }$ 40 J$ \qquad$ $\textbf{(E) }$ 50 J

$\textbf{E}$

The rotational kinetic energy is $KE_r=\dfrac12I\omega^2=\dfrac12\left(\dfrac25mR^2\right)\omega^2=\dfrac15mr^2\omega^2$. The translational kinetic energy $KE_t=\dfrac12mv^2$. As we know, the sphere is rolling without slipping, so $v=\omega r$. So $KE_t=\dfrac12mv^2=\dfrac12m\omega^2r^2=\dfrac52KE_r=50\ \text{J}$.

A mass $m$ attached to a light string of length $L$ is located at an angle $\theta$ below the horizontal as shown in the figure. The mass then is released from rest. Calculated from an axis perpendicular to the plane of the page through the pivot, which one of the following choices represents the magnitude of the torque produced by the gravitational force acting on the mass at this instant?

$\textbf{(A) }$ $mgL \qquad\newline$ $\textbf{(B) }$ $mgL\sin\theta \qquad\newline$ $\textbf{(C) }$ $mgL\cos\theta \qquad\newline$ $\textbf{(D) }$ $mgL(1-\sin\theta) \qquad\newline$ $\textbf{(E) }$ $mgL(1-\cos\theta)$

$\textbf{C}$

The gravitational force is $mg$ directing downward. The distance between the pivot and the line of gravitational force is $L\cos\theta$. So the torque is $\tau=mgL\cos\theta$.

A student wants to set up an experiment with a thin convex lens of focal length $f$ such that a thin real object produces a focused real image on a movable screen. At how many locations along the optical axis (principal axis) can the object be placed so that the distance between the object and the focused image on the screen is equal to $3f$?

$\textbf{(A) }$ There is no location.$ \qquad\newline$

$\textbf{(B) }$ There is exactly one location.$ \qquad\newline$

$\textbf{(C) }$ There are exactly two locations.$ \qquad\newline$

$\textbf{(D) }$ There are exactly four locations.$ \qquad\newline$

$\textbf{(E) }$ There are an infinite number of locations.

$\textbf{A}$

By $u+v=3f$ and the lens equation $\dfrac1u+\dfrac1v=\dfrac1f$, we get an expression for the variable $u$ as $u^2-3fu+3f^2=0$. The discriminant of the quadratic equation is $\Delta=(-3f)^2-4\cdot3f=-3f^2<0$, which means there are no real solutions for the equation. So it is impossible to have a distance $3f$ between the object and the real image.

A positive charge moves with constant velocity through a region of space containing both an electric field and a magnetic field. The electric field is directed out of the plane of the page. Ignoring any gravitational field, which one of the following choices represents possible directions of both the particle’s velocity and the total magnetic field in the region of space?

$\textbf{D}$

Here, we look at the Lorentz Force Law $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. Since the electric force is out of the plane of the page, in order for the positive charge to experience no net force (moves with constant velocity), there must be a magnetic force directed into the plane of the page. By the right hand rule, the right fingers point in the direction of the velocity and curl into the direction of the magnetic field, leaving the right thumb to point in the direction of the force. Doing this, we find that only the result of (D) gives the right thumb pointing into the plane of the page.

The cylindrical head of an aluminum nail has a diameter of 1.00 cm. For the top layer of atoms in the nail’s head, which one of the following choices best represents the number of aluminum atoms in that layer?

$\textbf{(A) }$ $10^{10} \qquad$ $\textbf{(B) }$ $10^{15} \qquad$ $\textbf{(C) }$ $10^{20} \qquad$ $\textbf{(D) }$ $10^{25} \qquad$ $\textbf{(E) }$ $10^{30}$

$\textbf{B}$

An aluminum atom has an approximate radius of $r=10^{-10}\ \text{m}$, so it has an approximate area of $A=\pi r^2=3.14\times10^{-20}\ \text{m}$. The area of the nail’s head is $S=\pi(0.5\times10^{-2})^2$. Taking the ratio, we have $S/A=2.5\times10^{15}$ atoms in the top layer.

A square, conducting wire loop sits in a plane perpendicular to a spatially uniform magnetic field pointing into the plane of the page as shown. The magnetic field strength steadily increases with time. Which one of the following effects best describes the result of this field increase?

$\textbf{(A) }$ The entire loop moves up the plane of the page.$ \qquad\newline$

$\textbf{(B) }$ The loop rotates with the top edge of the loop initially moving out of the plane of the page and the bottom edge moving into the plane of the page.$ \qquad\newline$

$\textbf{(C) }$ The loop rotates with the top edge of the loop initially moving into the plane of the page and the bottom edge moving out of the plane of the page.$ \qquad\newline$

$\textbf{(D) }$ The legs of the loop attempt to increase the area enclosed by the loop.$ \qquad\newline$

$\textbf{(E) }$ The legs of the loop attempt to decrease the area enclosed by the loop.

$\textbf{E}$

With the time-changing magnetic field, there is an induction by Lenz’s Law to oppose the change. This increasing field strength into the plane of the loop results in a counterclockwise-oriented electric field that produces a conventional current in the wire. Using the right-hand rule for the force on a wire in a magnetic field for the right-most wire, we have the right fingers point along the current curling into the page results in the right thumb pointing to the left. Performing the same analysis on each wire element results in forces on each leg directed toward the center, thereby trying to decrease the size of the region receiving magnetic field. The legs of the loop are trying to decrease the area enclosed by the loop.

For the circuit shown, all wires have no resistance, the battery has a constant internal resistance of $r=8.0\ \Omega$ and the two light bulbs ($\#1$ and $\#2$) are identical, each with resistance $R_{bulb}$. The variable resistor is initially set to $R=26.0\ \Omega$. The switch $S$ in the circuit now is closed. To what resistance must the variable resistor be set if bulb $\#1$ is to have the same brightness after the switch is closed as it did with the switch open?

$\textbf{(A) }$ $9.0\ \Omega \qquad$ $\textbf{(B) }$ $13.0\ \Omega \qquad$ $\textbf{(C) }$ $16.0\ \Omega \qquad$ $\textbf{(D) }$ $22.0\ \Omega \qquad\newline$

$\textbf{(E) }$ The answer can be computed only if the bulbs’ resistance $R_{bulb}$ is known

$\textbf{A}$

By closing the switch in the circuit, the light bulbs have been put into parallel, thereby reducing the resistance of the grouping. In order to maintain the same brightness, there must be the same ratio of potential difference for the light bulbs compared to the two resistors in the circuits so that the potential differences do not change. In other words, we write $$\dfrac{R_{bulb}}{r+R}=\dfrac{\frac12R_{bulb}}{r+R^\prime}$$ This expression leads to $$\left(\dfrac{1}{8+26}\right)=\dfrac12\left(\dfrac{1}{8+R^\prime}\right)$$ Finally we get $R^\prime=9\ \Omega$.

Using the kinetic theory of gases, which one of the following choices best represents the rms (root mean square) speed of 58 grams of a monatomic ideal gas at a pressure of 3.0 atm in an enclosed container of volume 6.0 L?

$\textbf{(A) }$ 0.557 m/s$ \qquad$ $\textbf{(B) }$ 9.71 m/s$ \qquad$ $\textbf{(C) }$ 177 m/s$ \qquad$ $\textbf{(D) }$ 307 m/s$ \qquad$ $\textbf{(E) }$ 3150 m/s

$\textbf{D}$

From the kinetic theory of gases, the rms speed is found as $v=\sqrt{\dfrac{3RT}{\mu}}$ where $\mu$ is the molar mass of the gas. Using the ideal gas equation, we have $PV=nRT=\dfrac{m}{\mu}RT\Rightarrow\dfrac{RT}{\mu}=\dfrac{PV}{m}$. So we rewrite the rms speed as $v=\sqrt{\dfrac{3RT}{\mu}}=\sqrt{\dfrac{3PV}{m}}=\sqrt{\dfrac{3(3\times1.013\times10^5)(6\times10^{-3})}{58\times10^{-3}}}=307\ \text{m/s}$.

**Questions 41 and 42 deal with the following information:**

A hypothetical radioactive substance Aaptinium decays via alpha-emission into Physicsbowlium. The decay constant for this alpha-emission is $20\ \text{s}^{-1}$.

Which one of the following statements correctly compares Physicsbowlium to Aaptinium?

$\textbf{(A) }$ Physicsbowlium has 4 fewer protons and 2 fewer neutrons than Aaptinium.$ \qquad\newline$

$\textbf{(B) }$ Physicsbowlium has 4 fewer neutrons and 2 fewer protons than Aaptinium.$ \qquad\newline$

$\textbf{(C) }$ Physicsbowlium has 2 fewer protons and 2 fewer neutrons than Aaptinium.$ \qquad\newline$

$\textbf{(D) }$ Physicsbowlium has 4 fewer protons than, and the same number of neutrons as, Aaptinium.$ \qquad\newline$

$\textbf{(E) }$ Physicsbowlium has 2 fewer protons than, and the same number of neutrons as, Aaptinium.

$\textbf{C}$

An alpha particle is a helium nucleus, comprised of 2 neutrons and 2 protons.

What is the half-life of Aaptinium?

$\textbf{(A) }$ 0.035 s$ \qquad$ $\textbf{(B) }$ 0.050 s$ \qquad$ $\textbf{(C) }$ 0.100 s$ \qquad$ $\textbf{(D) }$ 0.297 s$ \qquad$ $\textbf{(E) }$ 0.693 s

$\textbf{A}$

For radioactive decay, the mathematical expression is given as $N(t)=N_0e^{-\lambda t}$. The half-life is the time for 1/2 of the particles to decay. Hence, we have from the decay equation $\dfrac12N_0=N_0e^{-\lambda t}\Rightarrow t=\dfrac{\ln2}{\lambda}=\dfrac{\ln2}{20}=0.035\ \text{s}$.

Several forces act on a rigid body. If the resultant (net) force on the body is zero, which one of the following statements must be true?

$\textbf{(A) }$ The object is in translational equilibrium and rotational equilibrium.$ \qquad\newline$

$\textbf{(B) }$ The object is in translational, but not necessarily rotational, equilibrium.$ \qquad\newline$

$\textbf{(C) }$ The object is in rotational, but not necessarily translational, equilibrium.$ \qquad\newline$

$\textbf{(D) }$ The object is in static equilibrium.$ \qquad\newline$

$\textbf{(E) }$ The object is in neither translational nor rotational equilibrium.

$\textbf{B}$

Having the net force equaling zero means that there is no linear acceleration of the body. This puts it into translational equilibrium. However, the net force equaling zero is no guarantee that the net torque is also equal to zero, so nothing can be concluded about the rotational motion of the system. The only statement that MUST be true is that the body is in translational equilibrium.

A monatomic ideal gas is the working substance for a refrigerator that undergoes the cyclic process (ABCDA) shown in the PV diagram. The processes are all isochoric or isobaric with pressures between $P_0$ and $2P_0$ and volumes between $V_0$ and $2V_0$. What is the coefficient of performance for this refrigerator?

$\textbf{(A) }$ 1/4$ \qquad$ $\textbf{(B) }$ 1/3$ \qquad$ $\textbf{(C) }$ 4/3$ \qquad$ $\textbf{(D) }$ 11/2$ \qquad$ $\textbf{(E) }$ 13/2

$\textbf{D}$

The coefficient of performance is computed as “what you get divided by what you pay for”. For the refrigerator, you are hoping for energy to be removed from the cold temperature reservoir to the high temperature reservoir and paying for the work done to remove the energy. In equations, the coefficient is $e=\dfrac{Q}{W}$. For the PV diagram given, the magnitude of the work done is equal to magnitude of the area under $PV$ curve which here is $W=P_0V_0$ where $T_0$ is the temperature at point A in the cycle. For the processes shown, there is heat lost from the cold surroundings in processes AB and BC. For a monatomic ideal gas, the heat associated with these processes is computed as $Q_{AB}=nc_P\Delta T_{AB}$ and $Q_{BC}=nc_V\Delta T_{BC}$. Here, $c_V=\dfrac32R$, $C_P=\dfrac52R$, $\Delta T_{AB}=T_0$ and $\Delta T_{BC}=2T_0$. Using the ideal gas equation, the temperatures at A,B,C, and D are found as $T_0$, $2T_0$, $4T_0$ and $2T_0$, respectively, allowing us to find the temperature changes. So, $Q=n\left(\dfrac52R\right)(T_0)+n\left(\dfrac32R\right)(2T_0)=\dfrac{11}{2}nRT_0$. This means that the coefficient of performance is $e=\dfrac{Q}{W}=\dfrac{11}{2}$.

A stationary atom of mass $4.00\times10^{-26}\ \text{kg}$ spontaneously emits a photon of energy 10.0 eV. Which one of the following choices best represents the speed, in units of m/s, of the atom after emitting the photon?

$\textbf{(A) }$ $4.00\times10^7 \qquad$ $\textbf{(B) }$ $8.94\times10^3 \qquad$ $\textbf{(C) }$ $1.33\times10^{-1} \qquad$ $\textbf{(D) }$ $1.58\times10^{-4} \qquad$ $\textbf{(E) }$ $2.50\times10^{-8}$

$\textbf{C}$

The linear momentum of the photon is $p=\dfrac{E}{c}=\dfrac{10\times1.6\times10^{-19}}{3\times10^{8}}=5.33\times10^{-27}\ \text{kg}\cdot\text{m/s}$. Since the linear momentum is conserved, the atom gains the same momentum but in opposite direction. The speed of atom is $v=\dfrac{p}{m}=\dfrac{5.33\times10^{-27}}{4.00\times10^{-26}}=0.133\ \text{m/s}$. The calculated speed result is far away from the speed of light, so no relativistic effect need to be considered.

There is a quantity called the Planck time, $t_{planck}$, which is computed in terms of constants as $t_{planck}^2=\dfrac{\hbar G}{c^n}$ where $\hbar$ is Planck’s constant divided by $2\pi$, $G$ is the Universal Gravitational Constant, and $c$ is the speed of light. In order for this expression for time to be consistent, what is the numerical value of $n$, the power to which the speed of light is raised?

$\textbf{(A) }$ 2$ \qquad$ $\textbf{(B) }$ 3$ \qquad$ $\textbf{(C) }$ 4$ \qquad$ $\textbf{(D) }$ 5$ \qquad$ $\textbf{(E) }$ 6

$\textbf{D}$

Using units analysis, we have $t_{planck}^2=\dfrac{\hbar G}{c^n}\rightarrow \text{s}^2=(\text{Js})\left(\dfrac{\text{Nm}^2}{\text{kg}^2}\right)\dfrac{\text{s}^n}{\text{m}^n}$. The Newton is $\text{N}=\text{kg}\cdot\text{m/s}^2$ while a Joule $\text{J}=\text{Nm}=\text{kg}\cdot\text{m}^2/\text{s}^2$. These substitutions lead to $\text{s}^2=\left(\text{kg}\dfrac{\text{m}^2}{\text{s}^2}\text{s}\right)\left(\dfrac{\text{kg}\frac{\text{m}}{\text{s}^2}\text{m}^2}{\text{kg}^2}\right)\dfrac{\text{s}^n}{\text{m}^n}$. Rearranging the equation leads to $\dfrac{\text{s}^n}{\text{m}^n}=\dfrac{\text{s}^5}{\text{m}^5}$ meaning that $n=5$.

Water flows ideally through a cylindrically-shaped pipe. At the lower end, the pipe’s cross-sectional area is $30.0\ \text{cm}^2$ whereas in the upper portion, the pipe’s cross-sectional area is $10.0\ \text{cm}^2$ and fluid is moving at 9.0 m/s. Which one of the following choices best represents the difference in pressure between the lower section of the pipe and the upper section if the vertical distance between the centers of the pipe sections is 2.0 m?

$\textbf{(A) }$ 5.6 Pa$ \qquad$ $\textbf{(B) }$ 6.05 Pa$ \qquad$ $\textbf{(C) }$ 56 kPa$ \qquad$ $\textbf{(D) }$ 60.5 kPa$ \qquad$ $\textbf{(E) }$ 65 kPa

$\textbf{C}$

To solve this problem, we need a combination of Bernoulli’s Equation along with the Equation of Continuity. The mass flow rate is the same at the top and bottom portion of the pipe. Since the cross-sectional area at the bottom is 3 times as the top, the speed of flow is one third as $v_b=\dfrac13v_t=3\ \text{m/s}$. Now employing Bernoulli’s equation for two points along a streamline at the center of the tube, we write $P_b+\dfrac12\rho v_b^2+\rho gy_b=P_t+\dfrac12\rho v_t^2+\rho gy_t$. Choosing the location of the lower pipe to be $y_b=0$, we have $P_b+\dfrac12(1000)(3)^3=P_t+\dfrac12(1000)(9)^2+(1000)(10)(2)$. So $P_b-P_t=56000\ \text{Pa}$.

An object of mass $M$ is dropped from a height $H$ above the ground. The object bounces off of a horizontal surface in a collision lasting time $T$. The object then rises upward to a maximum height $H/2$. What was the magnitude of the average net force acting on the mass during the collision with the surface?

$\textbf{(A) }$ $\left(2-\sqrt2\right)\dfrac{M\sqrt{gH}}{T} \qquad\newline$

$\textbf{(B) }$ $\left(\dfrac1{\sqrt2}+1\right)\dfrac{M\sqrt{gH}}{T} \qquad\newline$

$\textbf{(C) }$ $\left(\sqrt3\right)\dfrac{M\sqrt{gH}}{T} \qquad\newline$

$\textbf{(D) }$ $\left(2\sqrt2-1\right)\dfrac{M\sqrt{gH}}{T} \qquad\newline$

$\textbf{(E) }$ $\left(\sqrt2+1\right)\dfrac{M\sqrt{gH}}{T}$

$\textbf{E}$

The speed of object reaching the ground is $v_1=-\sqrt{2gH}$. The bouncing-off speed is $v_2=\sqrt{2g\cdot\dfrac{H}{2}}=\sqrt{gH}$. By the impulse-momentum theorem, the change of linear momentum $m(v_2-v_1)=FT$ where $F$ is the net force acting on the object during the collision. Finally we get $F=\left(\sqrt2+1\right)\dfrac{M\sqrt{gH}}{T}$.

A beam of unpolarized light traveling in air strikes a piece of optically flat glass at an angle of incidence of $58^\circ$. Some of the light is reflected while the remainder is transmitted into the glass. The reflected beam is $100\%$ polarized parallel to the surface of the glass. What is the index of refraction for the glass?

$\textbf{(A) }$ 1.60$ \qquad$ $\textbf{(B) }$ 1.53$ \qquad$ $\textbf{(C) }$ 1.47$ \qquad$ $\textbf{(D) }$ 1.38$ \qquad$ $\textbf{(E) }$ 1.18

$\textbf{A}$

When the reflected light is $100\%$ polarized, it means that there is a right angle between reflected and refracted rays. Since the reflected angle is $\theta_1=58^\circ$, the refraction angle is $\theta_2=180^\circ-90^\circ-58^\circ=32^\circ$. By Snel’s Law $n_1\sin\theta_1=n_2\theta_2$, the index of refraction for the air is $n_1=1$, so the index of refraction for the glass is $n_2=\dfrac{\sin58^\circ}{\sin32^\circ}=1.60$.

A spatially uniform electric field is constrained within the circular region of radius $R$ as shown. The field is directed out of the plane of the page and its strength is decreasing uniformly with time. Which one of the following choices best represents the direction of the Lorentz force on the electron at the instant shown in the figure when the electron is moving up the plane of the page? Ignore gravity.

$\textbf{(A) }$ No Force$ \qquad\newline$

$\textbf{(B) }$ Into the plane of the page$ \qquad\newline$

$\textbf{(C) }$ Out of the plane of the page$ \qquad\newline$

$\textbf{(D) }$ To the right$ \qquad\newline$

$\textbf{(E) }$ To the left

$\textbf{B}$

A time-changing electric field has an associated magnetic field. Outside the region of the electric field, this magnetic field is oriented in clockwise circles due to Lenz’s Law. So, at the location of the moving electron, there is a magnetic field directed to the left at the instant shown. Using the right-hand rule, with right fingers pointed along the velocity of the electron, the fingers are curled to the left. Since the charge of interest is negative, the hand is flipped 180 degrees resulting in the right thumb pointing into the plane of the page.