## PhysicsBowl 2015

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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 choices correctly represents a length of 3.00 mm?

$\textbf{(A) }$ $3.00\times10^{-6}\ \text{m} \qquad\newline$

$\textbf{(B) }$ $3.00\times10^{-3}\ \text{m} \qquad\newline$

$\textbf{(C) }$ $3.00\times10^{-2}\ \text{m} \qquad\newline$

$\textbf{(D) }$ $3.00\times10^{3}\ \text{m} \qquad\newline$

$\textbf{(E) }$ $3.00\times10^{6}\ \text{m}$

$\textbf{B}$

$1\ \text{mm}=10^{-3}\ \text{m}$

A box uniformly slides 7.50 m to rest across a flat surface in a time of 12.0 s. What was the initial speed of the box when it started its slide?

$\textbf{(A) }$ 0.313 m/s$ \qquad$ $\textbf{(B) }$ 0.625 m/s$ \qquad$ $\textbf{(C) }$ 1.25 m/s$ \qquad$ $\textbf{(D) }$ 2.50 m/s$ \qquad$ $\textbf{(E) }$ 5.00 m/s

$\textbf{C}$

The average speed is $\overline{v}=\dfrac{x}{t}=\dfrac{7.5}{12}=0.625\ \text{m/s}$. The acceleration is a constant, so the initial speed is $v_i=2\overline{v}=1.25\ \text{m/s}$.

Which one of the following quantities is not a vector quantity?

$\textbf{(A) }$ Average speed$ \qquad\newline$

$\textbf{(B) }$ Average velocity$ \qquad\newline$

$\textbf{(C) }$ Linear momentum$ \qquad\newline$

$\textbf{(D) }$ Acceleration $ \qquad\newline$

$\textbf{(E) }$ Average force

$\textbf{A}$

The velocity is a vector. The speed is a scalar because it is the quantity of velocity.

A standing wave on a string is produced. Which one of the following choices best describes the location on the string at which maximum constructive interference occurs?

$\textbf{(A) }$ node$ \qquad$ $\textbf{(B) }$ antinode$ \qquad$ $\textbf{(C) }$ harmonic$ \qquad$ $\textbf{(D) }$ overtone$ \qquad$ $\textbf{(E) }$ amplitude

$\textbf{B}$

Locations of complete constructive interference are known as antinodes on the standing wave.

An object initially is moving upward while in free fall. Which one of the following choices best represents the direction of the object’s acceleration during its flight?

$\textbf{E}$

The gravitational acceleration is always directed downward.

A car travels at 20.0 miles/hr. Which one of the following choices best represents the speed of the car in SI units of m/s?

$\textbf{(A) }$ 533 m/s$ \qquad$ $\textbf{(B) }$ 45.0 m/s$ \qquad$ $\textbf{(C) }$ 20.0 m/s$ \qquad$ $\textbf{(D) }$ 8.9 m/s$ \qquad$ $\textbf{(E) }$ 0.75 m/s

$\textbf{D}$

$\dfrac{20\ \text{miles}}{1\ \text{hr}}=\dfrac{20\cdot1600\ \text{m}}{3600\ \text{s}}=8.9\ \text{m/s}$

An object moves clockwise with constant speed around the vertical circle shown. Which arrow best indicates the direction of the object’s instantaneous acceleration at the point labeled X?

$\textbf{A}$

The centripetal acceleration points the center.

A negatively charged balloon remains at rest when placed on a vertical wall. Which one of the following terms is most closely associated with the electrical phenomenon allowing the balloon to remain on the wall?

$\textbf{(A) }$ Radiation$ \qquad$ $\textbf{(B) }$ Grounding$ \qquad$ $\textbf{(C) }$ Reduction$ \qquad$ $\textbf{(D) }$ Current$ \qquad$ $\textbf{(E) }$ Polarization

$\textbf{E}$

After the balloon obtains charge from being rubbed, it causes polarization in the wall allowing like charges to be closer to each other when they come in contact, providing an attractive force to prevent the balloon from falling to the ground.

Two cars are moving to the right on a horizontal track, each with constant acceleration. At an instant of time, the information about the cars is shown:$\newline$

Car 1: position = 125.0 m; velocity = 13.0 m/s; constant acceleration = 1.5 m/s$^2\newline$

Car 2: position = 80.0 m; velocity = 9.30 m/s; constant acceleration = 5.5 m/s$^2\newline$

During the next 1.0 s of motion, which one of the following choices best represents what happens to the distance between the cars?

$\textbf{(A) }$ It decreases during the entire 1.0 second of motion.$ \qquad\newline$

$\textbf{(B) }$ It increases during the entire 1.0 second of motion.$ \qquad\newline$

$\textbf{(C) }$ It initially increases and then decreases resulting in a greater distance between the cars after 1.0 s.$ \qquad\newline$

$\textbf{(D) }$ It initially increases and then decreases resulting in a smaller distance between the cars after 1.0 s.$ \qquad\newline$

$\textbf{(E) }$ It initially increases and then decreases resulting in the same distance between the cars after 1.0 s.

$\textbf{C}$

Car 1 is in front of car 2, and its velocity is greater, so the distance will increase in the beginning. However, the acceleration of car 2 is greater. One second later the speed of car 2 will exceed the speed of car 2 slightly. So the distance will decrease when speed of car 2 catch up with car 1, but the general distance will still increase in 1 second.

For the circuit shown, the three light bulbs have identical resistance $R$, the battery is ideal, and all wires have no resistance. Which one of the following choices correctly identifies the light bulbs that either become dimmer or go out completely when the switch, S, in the circuit is closed?

$\textbf{(A) }$ All 3 bulbs$ \qquad\newline$ $\textbf{(B) }$ Bulbs 1 and 2 only$ \qquad\newline$ $\textbf{(C) }$ Bulb 3 only$ \qquad\newline$ $\textbf{(D) }$ Bulb 1 only$ \qquad\newline$ $\textbf{(E) }$ None of the bulbs

$\textbf{E}$

When the switch is closed, another wire is added to the circuit. The new wire is functionally parallel to the left wire in the circuit, so it can be regarded as to widen the left wire. However, the wire is of no resistance, so broadening the wire means nothing to the circuit.

Two length measurements are made and recorded as $L_1=84.55\ \text{cm}$ and $L_2=33.55\ \text{cm}$. Two other length measurements are made and recorded as $L_3=1.750\ \text{cm}$ and $L_4=1.250\ \text{cm}$. These measurements are used to compute the quantity $(L_1+L_2)-(L_3+L_4)$ using the rules of significant figures. Which one of the following choices best represents the correct result to this calculation?

$\textbf{(A) }$ 115.100 cm$ \qquad$ $\textbf{(B) }$ 115.10 cm$ \qquad$ $\textbf{(C) }$ 115.1 cm$ \qquad$ $\textbf{(D) }$ 115 cm$ \qquad$ $\textbf{(E) }$ 120 cm

$\textbf{B}$

$L_1$ and $L_2$ are correct to two decimal places while $L_3$ and $L_4$ are to three decimal places. So the final result should be correct to only two decimal places.

A box of mass 12.0 kg is being pushed to the right across a horizontal surface. When the box has 12.0 J of kinetic energy, a 12.0 N net force acts on it. Which one of the following choices best represents the magnitude of the linear momentum of the box at this instant?

$\textbf{(A) }$ 6.00 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(B) }$ 8.50 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(C) }$ 12.0 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(D) }$ 17.0 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(E) }$ 24.0 kg$\cdot$m/s

$\textbf{D}$

The formula for kinetic energy and linear momentum is $E=\dfrac{p^2}{2m}$, so $p=\sqrt{2mE}=12\sqrt2\ \text{kg}\cdot\text{m/s}$.

An object is being pushed at constant speed on an inclined plane. The free body diagram of the object is shown with the gravitational force represented by $W$, the friction force by $f$, the applied external push parallel to the incline by $F$, and the normal force with the surface by $n$. Which one of the following choices represents correct relationships between the forces?

$\textbf{(A) }$ $n>W$ and $F<f$ $ \qquad\newline$

$\textbf{(B) }$ $n<W$ and $F=f$ $ \qquad\newline$

$\textbf{(C) }$ $n<W$ and $F<f$ $ \qquad\newline$

$\textbf{(D) }$ $n=W$ and $F>f$ $ \qquad\newline$

$\textbf{(E) }$ $n=W$ and $F=f$

$\textbf{C}$

Assuming the angle of the inclined plane to the ground is $\theta$, we have $n=W\cos\theta$ and $F+W\sin\theta=f$. So $n<W$, $F<f$.

A particle has a position, $x$, as a function of time, $t$, given as $x(t)=-15-25t+10t^2$. Which one of the following choices represents the magnitude of the particle’s acceleration? All quantities are expressed in base SI units.

$\textbf{(A) }$ 5 m/s$^2 \qquad$ $\textbf{(B) }$ 10 m/s$^2 \qquad$ $\textbf{(C) }$ 15 m/s$^2 \qquad$ $\textbf{(D) }$ 20 m/s$^2 \qquad$ $\textbf{(E) }$ 40 m/s$^2$

$\textbf{D}$

$v=\dot{x}=-25+20t\newline$

$a=\dot{v}=20$

A skydiver falls downward through the air with constant speed. Which one of the following choices correctly describes the Newton’s Third Law pair force to the air resistance acting on the skydiver during the fall?

$\textbf{(A) }$ There is no Third Law pair force for this kind of situation.$ \qquad\newline$

$\textbf{(B) }$ The gravitational force acting on the skydiver by the Earth.$ \qquad\newline$

$\textbf{(C) }$ The force that molecules in the air exert on neighboring molecules in the air.$ \qquad\newline$

$\textbf{(D) }$ The force exerted on molecules in the air by the ground.$ \qquad\newline$

$\textbf{(E) }$ The force exerted on molecules in the air by the skydiver.

$\textbf{E}$

The molecules in the air exert a force upward on the skydiver. The skydiver exerts a force downward on the molecules in the air.

“Particles of matter also have associated wavelengths and can behave as waves.” To which scientist is this concept attributed?

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

$\textbf{A}$

The concept of matter wave

An object starts at the origin and its velocity along a line vs. time is graphed. Which one of the following choices best gives the proper interval(s) of time for which the object is moving away from the origin?

$\textbf{(A) }$ Only for times $0\ \text{s}<t<3\ \text{s} \qquad\newline$

$\textbf{(B) }$ Only for times $0\ \text{s}<t<5\ \text{s} \qquad\newline$

$\textbf{(C) }$ Only for times $3\ \text{s}<t<5\ \text{s} \qquad\newline$

$\textbf{(D) }$ Only for times $0\ \text{s}<t<7\ \text{s} \qquad\newline$

$\textbf{(E) }$ For times $0\ \text{s}<t<3\ \text{s}$ and $5\ \text{s}<t<9\ \text{s}$

$\textbf{D}$

The velocity is positive for time $0\ \text{s}<t<7\ \text{s}$, so the object is moving away from the origin. For $7\ \text{s}<t<10\ \text{s}$ the velocity is negative. The position at $t=10\ \text{s}$ is $x=10\ \text{m}$, so for $7\ \text{s}<t<10\ \text{s}$ the object is moving toward the origin.

A 680 Hz tuning fork is placed over a tube open at both ends that is filled with air. As a result, a standing wave in the 3rd harmonic is produced. The speed of sound in air is 340 m/s. What is the length of the tube?

$\textbf{(A) }$ 0.38 m$ \qquad$ $\textbf{(B) }$ 0.67 m$ \qquad$ $\textbf{(C) }$ 0.75 m$ \qquad$ $\textbf{(D) }$ 1.33 m$ \qquad$ $\textbf{(E) }$ 1.50 m

$\textbf{C}$

The formula for a standing wave in an open tube is $L=\dfrac{n}{2}\lambda$. For $n=3$ and $v=\lambda f$, we have $L=\dfrac{3v}{2f}=0.75\ \text{m}$.

Ten moles of helium gas are enclosed in a container at a pressure of 1.00 atm and at a temperature of 400 K. Which one of the following choices best represents the density of this gas sample?

$\textbf{(A) }$ 0.012 kg/m$^3 \qquad\newline$ $\textbf{(B) }$ 0.12 kg/m$^3 \qquad\newline$ $\textbf{(C) }$ 1.2 kg/m$^3 \qquad\newline$ $\textbf{(D) }$ 120 kg/m$^3 \qquad\newline$ $\textbf{(E) }$ 1.2$\times10^4$ kg/m$^3$

$\textbf{B}$

The volume of the container can be found in $PV=nRT$. The density of the gas $\rho=\dfrac{m}{V}=\dfrac{nM}{V}$, where the molecular weight of helium gas is $M=4\ \text{g/mol}=0.004\ \text{kg/mol}$. So the density $\rho=\dfrac{nM}{V}=\dfrac{MP}{RT}=0.12\ \text{kg/m}^3$.

Which one of the following choices best represents the work for which the 2014 Nobel Prize in Physics was awarded?

$\textbf{(A) }$ Landing the Rosetta Philae Lander on the surface of Comet 67P/Churyumov-Gerasimenko$ \qquad\newline$

$\textbf{(B) }$ The detection of dark matter$ \qquad\newline$

$\textbf{(C) }$ The invention of the blue LED$ \qquad\newline$

$\textbf{(D) }$ The creation of a tractor beam using sound$ \qquad\newline$

$\textbf{(E) }$ The detection of neutrinos from the Sun which agreed with theory

$\textbf{C}$

The Nobel Prize was awarded to Shuji Nakamura, Hiroshi Amano, and Isamu Akasaki for the invention of the blue LED.

An object is launched from the ground at an angle of 60$^\circ$ above the horizontal with a speed of 20.0 m/s. What is the magnitude of the average velocity of the object from just after launch until it reaches its highest vertical position during flight?

$\textbf{(A) }$ 13.7 m/s$ \qquad$ $\textbf{(B) }$ 13.2 m/s$ \qquad$ $\textbf{(C) }$ 10.0 m/s$ \qquad$ $\textbf{(D) }$ 9.3 m/s$ \qquad$ $\textbf{(E) }$ 8.7 m/s

$\textbf{B}$

The vertical component of the initial velocity is $v_{i\perp}=v_i\sin\theta$, and the horizontal component is $v_{i\parallel}=v_i\cos\theta$. The time for reaching the highest point is $t=\dfrac{v_{i\perp}}{g}=\dfrac{v_i\sin\theta}{g}$. When reaching the highest point, the displacement in the horizontal direction is $x=v_{i\parallel}t=\dfrac{v_i^2\sin\theta\cos\theta}{g}$, and the displacement in the vertical direction is $y=\dfrac{v_{i\perp}^2}{2g}=\dfrac{v_i^2\sin^2\theta}{2g}$. So the total displacement is $r=\sqrt{x^2+y^2}$, thus the average velocity is $\overline{v}=\dfrac{r}{t}=v_i\sqrt{\cos^2\theta+\dfrac14\sin^2\theta}=5\sqrt7\ \text{m/s}=13.2\ \text{m/s}$.

Satellite 1 makes a circular orbit around the Earth with a radius $r_1=R.\newline$

Satellite 2 makes a circular orbit around the Earth with a radius $r_2=2R.\newline$

We let $v$ represent the speed of a satellite and $a$ represent the magnitude of a satellite’s acceleration. Which one of the following choices gives the correct relation between the speeds and accelerations of the satellites?

$\textbf{(A) }$ $v_2=\dfrac{1}{\sqrt2}v_1\quad a_2=\dfrac14a_1 \qquad\newline$

$\textbf{(B) }$ $v_2=\dfrac{1}{2}v_1\quad a_2=\dfrac14a_1 \qquad\newline$

$\textbf{(C) }$ $v_2=\dfrac{1}{\sqrt2}v_1\quad a_2=\dfrac12a_1 \qquad\newline$

$\textbf{(D) }$ $v_2=\dfrac{1}{2}v_1\quad a_2=\dfrac12a_1 \qquad\newline$

$\textbf{(E) }$ $v_2=v_1\quad a_2=\dfrac12a_1$

$\textbf{A}$

The gravitational force provides the centripetal force, so $\dfrac{GMm}{r^2}=ma=n\dfrac{v^2}{r}$. Hence we have $v=\sqrt{\dfrac{GM}{r}}$ and $a=\dfrac{GM}{r^2}$. The ratio $\dfrac{v_2}{v_1}=\sqrt{\dfrac{r_1}{r_2}}=\dfrac{1}{\sqrt2}$, $\dfrac{a_2}{a_1}=\dfrac{r_1^2}{r_2^2}=\dfrac14$.

A car moves with constant speed around a horseshoe-shaped path as shown with the arrows in the figure. Which one of the following choices best describes the direction of the average acceleration of the car in traveling from W to X?

$\textbf{(A) }$ $\swarrow \qquad$ $\textbf{(B) }$ $\nwarrow \qquad$ $\textbf{(C) }$ $\nearrow \qquad$ $\textbf{(D) }$ $\searrow \qquad$ $\textbf{(E) }$ There is no average acceleration

$\textbf{A}$

The average acceleration has the same direction as the change of velocity. At point W the velocity points right, while at point X the velocity points downward. So the change of velocity points bottom left.

A mass on a frictionless incline has a gravitational force, a normal force from the incline, and a force applied by a person that all are equal in magnitude. The mass remains at rest and the incline makes an angle $\theta$ counterclockwise from the horizontal. Which one of the following choices best describes the orientation of the applied force by the person? The +x-axis is directed upward, parallel to the incline’s surface as shown in the figure.

$\textbf{(A) }$ The applied force is oriented directly along the +x axis.$ \qquad\newline$

$\textbf{(B) }$ The applied force is oriented at an angle $\theta$ clockwise from the +x axis.$ \qquad\newline$

$\textbf{(C) }$ The applied force is oriented at an angle 90$^\circ-\theta$ clockwise from the +x axis.$ \qquad\newline$

$\textbf{(D) }$ The applied force is oriented at an angle 90$^\circ-\theta$ counterclockwise from the +x axis.$ \qquad\newline$

$\textbf{(E) }$ This is a completely impossible situation that never can be realized physically.

$\textbf{C}$

In order for the sum of 3 equal-sized vectors to be zero, the angle between each vector with any other needs to $120^\circ$. Since the gravitational force is straight downward and the normal force is perpendicular to the surface, $\theta$ must be $60^\circ$. So the applied force would have to be $30^\circ$ below the incline’s surface.

A gas undergoes the unusual process M → N in the pressure vs. volume graph shown. Which one of the following choices properly represents the signs of the internal energy change of the gas, $\Delta U$, the total energy transferred as heat to the gas, $Q$, and the total work done on the gas by the surroundings, $W$, for this process?

$\textbf{E}$

Recall the equation $\Delta U=Q+W$ (be careful! The work is done by the surroundings to the gas). Since the volume decreased from M to N, the gas has been compressed. So the work has been done by the surroundings, thus $W>0$. From M to N both volume and pressure decreased. By $PV=nRT$ we know the temperature also decreased, so the change of internal energy $\Delta U<0$. Now $Q=\Delta U-W$ must be negative.

The position of a mass connected to a spring obeys $x(t)=A\cos(\omega t)$. What is the average speed of the mass for one full oscillation in terms of the mass’s maximum speed during oscillation, $v_{\max}$?

$\textbf{(A) }$ $\dfrac{2}{\pi}v_{\max} \qquad$ $\textbf{(B) }$ $\dfrac{1}{\sqrt2}v_{\max} \qquad$ $\textbf{(C) }$ $\dfrac12v_{\max} \qquad$ $\textbf{(D) }$ $\dfrac{\sqrt2}{\pi}v_{\max} \qquad$ $\textbf{(E) }$ $\dfrac{1}{2\pi\sqrt2}v_{\max}$

$\textbf{A}$

The maximum speed $v_{\max}=\omega A$. In one full oscillation the object moves for $4A$ in $T=\dfrac{2\pi}{\omega}$, so the average speed $\overline{v}=\dfrac{4A}{T}=\dfrac{2a\omega}{\pi}=\dfrac{2}{\pi}v_{\max}$.

**Questions 24 and 25 deal with the following information:**

On a frictionless horizontal surface, two bodies make a head-on collision and stick together. Body 1 has a mass of 3.50 kg and initially moves to the right with speed 7.0 m/s. Body 2 initially is at rest. A graph of the force exerted onto Body 2 from Body 1 during the collision is shown.

What is the mass of Body 2?

$\textbf{(A) }$ 2.81 kg$ \qquad$ $\textbf{(B) }$ 3.50 kg$ \qquad$ $\textbf{(C) }$ 4.59 kg$ \qquad$ $\textbf{(D) }$ 5.53 kg$ \qquad$ $\textbf{(E) }$ 7.50 kg

$\textbf{D}$

The area under the force-time curve gives the impulse on mass $I=15\ \text{N}\cdot\text{s}$. By the impulse-momentum theorem, the change of linear momentum of body 1 is $\Delta p_1=m_1\Delta v_1=I$. so $\Delta v_1=\dfrac{I}{m_1}=4.3\ \text{m/s}$. Finally the common speed of two bodies is $\Delta v_2=7.0-4.3=2.7\ \text{m/s}$. The change of linear momentum of body 2 is also $\Delta p_2=m_2\Delta v_2=I$, so $m_2=\dfrac{I}{\Delta v_2}=5.54\ \text{kg}$.

How much kinetic energy was transformed to other kinds of energy from the collision?

$\textbf{(A) }$ 67.1 J$ \qquad$ $\textbf{(B) }$ 52.6 J$ \qquad$ $\textbf{(C) }$ 42.9 J$ \qquad$ $\textbf{(D) }$ 38.2 J$ \qquad$ $\textbf{(E) }$ 30.0 J

$\textbf{B}$

The initial kinetic energy of the system is $KE_i=\dfrac12m_1v_1^2=\dfrac12(3.5)(7.0)^2=85.75\ \text{J}$. The final kinetic energy is $KE_f=\dfrac12(m_1+m_2)v_f^2=\dfrac12(3.50+5.53)(2.7)^2=33.16\ \text{J}$. So the change of kinetic energy is $\Delta KE=KE_f-KE_i=-52.6\ \text{J}$.

An electron moves at constant non-zero velocity directly between two long straight wires. The conventional current in each wire has the same magnitude, but the currents are in opposite directions as shown in the figure. Ignoring gravity, which choice best reflects the direction of the magnetic field and the direction of the electric field that exist at the location of the electron? Any electric field in the region originates from an unseen external source.

$\textbf{E}$

The magnetic fields induced by both currents point out of the plane, so the magnetic force points to the left(be careful! the electron is negatively charged). To cancel the magnetic force, the electric field must be toward the right.

A spring scale reads 2.50 N when a small solid mass hangs from it in air. The spring scale reads 1.58 N when the mass at the end of the spring is completely submerged in a container of water. Which one of the following choices best represents the density of the solid mass?

$\textbf{(A) }$ $3.68\times10^3\ \text{kg/m}^3 \qquad\newline$

$\textbf{(B) }$ $2.72\times10^3\ \text{kg/m}^3 \qquad\newline$

$\textbf{(C) }$ $1.58\times10^3\ \text{kg/m}^3 \qquad\newline$

$\textbf{(D) }$ $9.20\times10^2\ \text{kg/m}^3 \qquad\newline$

$\textbf{(E) }$ $1.58\times10^2\ \text{kg/m}^3$

$\textbf{B}$

The gravitational force of the mass is $G=mg=\rho Vg=2.5\ \text{N}$. The buoyancy force $F=\rho_{water}Vg=2.5-1.58=0.92\ \text{N}$. So $\dfrac{G}{F}=\dfrac{\rho}{\rho_{water}}=\dfrac{2.5}{0.92}=2.72$. We get the density of the solid mass $\rho=2.72\rho_{water}=2.72\times10^3\ \text{kg/m}^3$.

Approximately how many hydrogen atoms are there in the liquid water of Earth’s oceans?

$\textbf{(A) }$ $10^{62} \qquad$ $\textbf{(B) }$ $10^{57} \qquad$ $\textbf{(C) }$ $10^{52} \qquad$ $\textbf{(D) }$ $10^{47} \qquad$ $\textbf{(E) }$ $10^{42}$

$\textbf{D}$

Approximately 70$\%$ of the Earth’s surface covered by water. Assuming the average depth of ocean is 5000 m, the volume of water is $V=0.7\cdot4\pi r^2\cdot5000=1.8\times10^{18}\ \text{m}^3$. The mass of water in the oceans is $(1.8\times10^{18}\ \text{m}^3)\cdot(1.0\times10^3\ \text{kg/m}^3)=1.8\times10^{21}\ \text{kg}$. The molar mass of water is 18 g/mol, so the mole number of water molecules is $n=1.8\times10^{21}/0.0018=1.0\times10^{23}\ \text{mol}$. There are 2 hydrogen atoms per water molecule, so the number of hydrogen atoms is $2\cdot(1.0\times10^{23})\cdot(6.02\times10^{23})=1.2\times10^{47}$.

An object moving along a line completes a 20.0-second trip with an average speed of 10.0 m/s in two stages. During stage 1, the object moves with a constant velocity of 6.0 m/s to the right for 12.0 seconds. What constant magnitude acceleration directed to the left must the object have during the 8.0 seconds of stage 2?

$\textbf{(A) }$ 2.5 m/s$^2 \qquad$ $\textbf{(B) }$ 2.7 m/s$^2 \qquad$ $\textbf{(C) }$ 4.0 m/s$^2 \qquad$ $\textbf{(D) }$ 5.3 m/s$^2 \qquad$ $\textbf{(E) }$ 6.3 m/s$^2$

$\textbf{D}$

The object moves with an average speed of 10.0 m/s for 20 s, so the total distance is 200 m. In the first stage it moves with a constant speed of 6.0 m/s for 12 s, so the distance it moves is 72 m, which means the distance for stage 2 is 128 m. To move 128 m in 8 s, the average speed for stage 2 is 16 m/s. As we know, the object moves with a constant acceleration in stage 2, so the average speed is the speed 4 s after accelerated. To accelerate the object from 6 m/s to 16 m/s in 4 s, the acceleration is 2.5 m/s$^2$.

Two spherical speakers separated by 30.0 m each emit a constant frequency signal of 57.0 Hz in phase with each other. The speed of sound is 342 m/s. How many locations of complete destructive interference of the incoming signals are there on the line between the speakers?

$\textbf{(A) }$ 12$ \qquad$ $\textbf{(B) }$ 11$ \qquad$ $\textbf{(C) }$ 10$ \qquad$ $\textbf{(D) }$ 9$ \qquad$ $\textbf{(E) }$ 6

$\textbf{C}$

The wavelength of the sound is $\lambda=\dfrac{v}{f}=6\ \text{m}$. The distance between two speakers, 30 m, contains 5 wavelengths. As the two waves are in phase but moves in different direction, every crest will meet a trough, and every trough will meet a crest. Each wave has 1 crest and 1 trough, so destructive interference will appear for 10 times.

An upward-pointing object is placed 15.0 cm to the left of a lens system. The first lens is convex with focal length 10.0 cm. The second lens is convex with focal length 10 cm and its location from the first lens is varied from 10 cm away to 110 cm away. Which one of the following choices best represents the description of the final image formed as the second lens is moved from $x=10\ \text{cm}$ to $x=110\ \text{cm}$ from the first lens?

$\textbf{A}$

For lens 1, by $\dfrac1{d_{o1}}+\dfrac1{d_{i1}}=\dfrac1{f_1}$, the image distance is $d_{i1}=30\ \text{cm}$. The object distance for lens 2 is $-20\ \text{cm}\leq d_{o2}\leq80\ \text{cm}$. When $-20\ \text{cm}\leq d_{o2}\leq0\ \text{cm}$, $d_{i2}$ varies from 6.7 cm to 0 cm, here we get a real and downward image (The magnification after 2 times of converging is negative). When $0\ \text{cm}\leq d_{o2}\leq10\ \text{cm}$, $d_{i2}$ varies from 0 cm to negative infinity, here we get a virtual and downward image. When $10\ \text{cm}\leq d_{o2}\leq80\ \text{cm}$, $d_{i2}$ varies from positive infinity to 11.4 cm, here we get a real and upward image.

Two uniform disks, X and Y, have equal masses, $M$, but different radii such that $r_X<r_Y$. Both disks initially are at rest. A force $F$ is applied tangent to each disk at its right edge for the same amount of time. As a result, each disk rotates counterclockwise in the plane of the page about a fixed frictionless axis through its center. Which one of the following choices correctly compares the magnitudes of angular momentum $L$ about the center axis and total kinetic energy $K$ of disk X and disk Y?

$\textbf{(A) }$ $L_X<L_Y\ ;\ K_X<K_Y \qquad\newline$

$\textbf{(B) }$ $L_X<L_Y\ ;\ K_X>K_Y \qquad\newline$

$\textbf{(C) }$ $L_X=L_Y\ ;\ K_X=K_Y \qquad\newline$

$\textbf{(D) }$ $L_X=L_Y\ ;\ K_X<K_Y \qquad\newline$

$\textbf{(E) }$ $L_X<L_Y\ ;\ K_X=K_Y$

$\textbf{E}$

The angular momentum $L=\tau t=Frt$. As $r_X<r_Y$, $L_X<L_Y$. The kinetic energy $$K=\dfrac{L^2}{2I}=\dfrac{F^2r^2t^2}{2\left(\dfrac12Mr^2\right)}=\dfrac{F^2t^2}{M}$$ So $K_X=K_Y$.

Rain falls vertically at 12.0 m/s with respect to a stationary observer. A car is moving at an angle of $40^\circ$ below the horizontal with respect to the observer. A passenger sitting in the car notes that the rain makes an angle of $29.0^\circ$ with the vertical. What is the car's speed with respect to the observer?

$\textbf{(A) }$ 2.29 m/s$ \qquad$ $\textbf{(B) }$ 5.93 m/s$ \qquad$ $\textbf{(C) }$ 9.03 m/s$ \qquad$ $\textbf{(D) }$ 11.8 m/s$ \qquad$ $\textbf{(E) }$ 16.2 m/s

$\textbf{B}$

By the law if sines, $\dfrac{12}{\sin101^\circ}=\dfrac{v_{car}}{\sin29^\circ}$, we get $v_{car}=5.93\ \text{m/s}$.

Which one of the following choices represents the base SI units of inductance?

$\textbf{(A) }$ $\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}^2\text{s}^2} \qquad$ $\textbf{(B) }$ $\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}\cdot\text{s}} \qquad$ $\textbf{(C) }$ $\dfrac{\text{kg}\cdot\text{m}}{\text{A}^2\text{s}^2} \qquad$ $\textbf{(D) }$ $\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}^2\text{s}^3} \qquad$ $\textbf{(E) }$ $\dfrac{\text{kg}\cdot\text{m}}{\text{A}^2\text{s}^3}$

$\textbf{A}$

The inductance can be found in $\mathcal{E}=L\dfrac{\text{d}I}{\text{d}t}$, so the unit of inductance is $$\text{H}=\dfrac{\text{V}\cdot\text{s}}{\text{A}}=\dfrac{\text{W}\cdot\text{s}}{\text{A}^2}=\dfrac{\text{J}}{\text{A}^2}=\dfrac{\text{N}\cdot\text{m}}{\text{A}^2}=\dfrac{\left(\text{kg}\cdot\text{m/s}^2\right)\text{m}}{\text{A}^2}=\dfrac{\text{kg}\cdot\text{m}^2}{\text{A}^2\text{s}^2}$$

A thin film of alcohol ($n_{alcohol}=1.35$) lies on a flat glass surface ($n_{glass}=1.60$). When light of wavelength 540 nm is incident normal to the alcohol surface from air, the light is strongly reflected, but when light of wavelength 432 nm is incident normal to the surface from air, the reflected light is minimized. Which one of the following choices could represent the thickness, $t$, of the alcohol film?

$\textbf{(A) }$ 216 nm$ \qquad$ $\textbf{(B) }$ 320 nm$ \qquad$ $\textbf{(C) }$ 324 nm$ \qquad$ $\textbf{(D) }$ 400 nm$ \qquad$ $\textbf{(E) }$ 486 nm

$\textbf{D}$

To create constructive interference with the 540 nm light, we need to be sure that the interfering rays from the reflection off the alcohol and the ray that passes into the alcohol and is reflected back at the glass surface are in phase. Since the light is traveling from air to alcohol to glass, the index of refraction increases at each interface, meaning that the reflected light is phase-shifted by $\lambda/2$ for each reflection and canceled out each other. This means that for light traveling down through the alcohol a distance $t$ to the glass surface and then traveling an

additional distance $t$ back to the air, we need this extra path length to be an integer number of wavelengths to put the two waves in phase, that is, $2t=p\dfrac{\lambda_1}{n_{alcohol}}$. So $t=200p$, where $p$ is an integer. Likewise, destructive interference for the 432 nm light satisfies $2t=\left(q+\dfrac12\right)\dfrac{\lambda_2}{n_{alcohol}}$. So $t=160\left(q+\dfrac12\right)$, where $q$ is also an integer. $(p,q)=(2,2)$ is one pair of solution in the solution set, here $t=400\ \text{nm}$.

For the circuit shown, the four light bulbs have identical resistance, the battery is ideal and all wires have no resistance. After the switch, S, in the circuit is closed, which one of the following choices correctly describes what happens to the magnitude of the current at the point labeled P and to the magnitude of the potential difference from W to X?

$\textbf{E}$

Before the switch is closed, the resistance for the circuit is $\dfrac23R$ (assuming the resistance for each bulb is $R$), so the current passing through point P is $\dfrac{3\mathcal{E}}{2R}$. The potential difference from W to X is the voltage across the top bulb, so the voltage is $\dfrac{\mathcal{E}}{2}$. After the switch is closed, the bulb in the switch branch is in parallel with the top bulb. So the resistance for the total circuit is $\dfrac35R$, the current passing through point P is $\dfrac{5\mathcal{E}}{3R}$ (increased). The potential difference is $\dfrac13\mathcal{E}$ (decreased).

A 2.0 kg mass is connected to the end of string and moves about the string’s fixed end in a conical motion with a constant speed of 4.0 m/s. The string has a length of 2.50 m and forms an angle of $\theta$ with the vertical. What is the tension in the string?

$\textbf{(A) }$ 20.0 N$ \qquad$ $\textbf{(B) }$ 23.7 N$ \qquad$ $\textbf{(C) }$ 27.4 N$ \qquad$ $\textbf{(D) }$ 29.8 N$ \qquad$ $\textbf{(E) }$ 32.5 N

$\textbf{C}$

From the free body diagram we have $T\cos\theta=mg$ and $T\sin\theta=m\dfrac{v^2}{l\sin\theta}$. So $\cos\theta=\dfrac{mg}{T}$, $\sin\theta=v\sqrt{\dfrac{m}{lT}}$. By $\sin^2\theta+\cos^2\theta=1$, we get an equation for $T$ as $T^2-\dfrac{mv^2}{l}T-m^2g^2=0$. By solving the equation we get $T=27.4\ \text{N}$ or $T=-14.6\ \text{N}$. The reasonable answer is $T=27.4\ \text{N}$.

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

A car and a truck are moving on a horizontal track. The position vs. time graph for the two vehicles is shown.

For the entire time shown in the graph, which one of the following choices correctly describes the relationship between the average speed of the truck to that of the car?

$\textbf{(A) }$ The truck’s average speed is less than the average speed of the car.$ \qquad\newline$

$\textbf{(B) }$ The truck’s average speed is the same as the average speed of the car.$ \qquad\newline$

$\textbf{(C) }$ The truck’s average speed is greater than the average speed of the car.$ \qquad\newline$

$\textbf{(D) }$ The truck’s average speed is positive while the car’s average speed is negative but of the same magnitude.$ \qquad\newline$

$\textbf{(E) }$ A relationship cannot be determined without more information.

$\textbf{A}$

The average speed is distance divided by time. The truck moves at constant rate from the start to end position while the car initially moves “backward” and then forward to the same ending location of the truck. Consequently, the car travels a greater distance than the truck and has a higher average speed.

Which one of the following choices best describes the instants of time, $t$, at which the car and truck travel with the same speed?

$\textbf{(A) }$ Only at times $t=0$, $t=T$ and $t=2T$.$ \qquad\newline$

$\textbf{(B) }$ At one instant during the interval $0<t<T$ and at one instant during the interval $T<t<2T$.$ \qquad\newline$

$\textbf{(C) }$ At two instants during the interval $0<t<T$ and at one instant during the interval $T<t<2T$$. \qquad\newline$

$\textbf{(D) }$ At one instant during the interval $0<t<T$ and at two instants during the interval $T<t<2T$$. \qquad\newline$

$\textbf{(E) }$ At two instants during the interval $0<t<T$ and at two instants during the interval $T<t<2T$.

$\textbf{C}$

The slope of the position vs, time graph is the speed. The car moves faster than the truck initially and eventually come to rest, which means that at some point, the speed of the car and truck are the same. From rest, the car accelerates and catches up to the truck by eventually moving faster. This means that the car and truck have the same speed again before time $T$. Finally, for times $T<t<2T$, the car slows down close to rest and the truck catches up. Hence, the car’s speed is again equal to that of the truck at some time.

An object of mass 4.0 kg has a total kinetic energy of 100.0 J and an x-component of linear momentum equal to 24.0 kg$\cdot$m/s. The object is moving in the x-y plane. What is the y-component of the object’s linear momentum?

$\textbf{(A) }$ 8.00 kg$\cdot$m/s $ \qquad\newline$ $\textbf{(B) }$ 15.0 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(C) }$ 26.0 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(D) }$ 32.0 kg$\cdot$m/s$ \qquad\newline$ $\textbf{(E) }$ 97.0 kg$\cdot$m/s

$\textbf{B}$

By $K=\dfrac{p^2}{2m}$, the total kinetic energy $p=\sqrt{2mK}=20\sqrt2\ \text{kg}\cdot\text{m/s}$. As $p=\sqrt{p_x^2+p_y^2}$, the y-component of the linear momentum is $p_y=\sqrt{224}\ \text{kg}\cdot\text{m/s}\approx15.0\ \text{kg}\cdot\text{m/s}$.

Which one of the following choices is most associated with the following statement: “When the pressure of a gas is held constant, the volume of the gas is directly proportional to the temperature.” ?

$\textbf{(A) }$ Newton’s Law$ \qquad\newline$ $\textbf{(B) }$ Boyle’s Law$ \qquad\newline$ $\textbf{(C) }$ Avogadro’s Law$ \qquad\newline$ $\textbf{(D) }$ Graham’s Law $ \qquad\newline$ $\textbf{(E) }$ Charles’s Law

$\textbf{E}$

Jacques Charles is the scientist associated with the observation that volume is proportional to temperature at constant pressure.

A 6.00 $\mu$F parallel-plate capacitor is disconnected from a 12 volt battery after being fully charged. A person now carefully inserts a dielectric material of constant $\kappa=2$ so that it fills one-half of the space between the plates as shown. How much work was done by the person while inserting the dielectric?

$\textbf{(A) }$ -81 $\mu$J$ \qquad$ $\textbf{(B) }$ -108 $\mu$J$ \qquad$ $\textbf{(C) }$ -144 $\mu$J$ \qquad$ $\textbf{(D) }$ -216 $\mu$J$ \qquad$ $\textbf{(E) }$ -288 $\mu$J

$\textbf{B}$

The work done by the person inserting the dielectric would be equal to the change in the potential energy for the capacitor. Since the battery was disconnected, the charge on the plates remains constant and $W_{person}=\Delta U=\dfrac12Q\Delta V$. The charge on the plates would be $Q=CV=(6.00\ \mu\text{F})(12.0\ \text{V})=72.0\ \mu\text{C}$. The final capacitance can be found by moving the dielectric to any location in the space between plates and treating the system as 2 capacitors in series. The potential difference through the air-filled portion would be 6 volts since the field strength is unchanged and we are crossing half the original capacitor (which had 12 volts). For the dielectric portion, the dielectric decreases the field strength by a factor of 2, which means that the potential difference across the dielectric would be 3 volts for a grand total of 9 volts across the entire capacitor. Hence, $W_{person}=\Delta U=\dfrac12Q\Delta V=\dfrac12(72\ \mu\text{C})(-3\ \text{V})=-108\ \mu\text{J}$.

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

$\textbf{(A) }$ $\dfrac16g \qquad$ $\textbf{(B) }$ $\dfrac12g \qquad$ $\textbf{(C) }$ $\dfrac43g \qquad$ $\textbf{(D) }$ $\dfrac23g \qquad$ $\textbf{(E) }$ $\dfrac14g$

$\textbf{E}$

The moment of inertia to the pivot is $I=I_0+Md^2=\dfrac1{12}ML^2+M\left(\dfrac{L}6\right)^2=\dfrac19ML^2$. The torque to the pivot is $\tau=Mg\cdot\dfrac16L=\dfrac16MgL$. By $\tau=I\beta$, the angular acceleration $\beta=\dfrac{\tau}{I}=\dfrac{3g}{2L}$. So the tangential acceleration of the center of mass is $a_\tau=\beta\cdot\dfrac{L}{6}=\dfrac14g$.

One mole of a diatomic ideal gas undergoes a reversible adiabatic process. The pressure and volume initially are given as $P=2.0\ \text{atm}$ and $V=30\ \text{L}$. If the volume is halved during the adiabatic process, how much work was done on the gas sample by the surroundings?

$\textbf{(A) }$ 6790 J$ \qquad$ $\textbf{(B) }$ 5530 J$ \qquad$ $\textbf{(C) }$ 4850 J$ \qquad$ $\textbf{(D) }$ 4200 J$ \qquad$ $\textbf{(E) }$ 3040 J

$\textbf{C}$

For a diatomic gas, $\gamma=\dfrac{i+2}{i}=\dfrac{5+2}{5}=\dfrac75$. For an adiabatic process, $PV^\gamma=Costant$, so $PV^\gamma=P_0V_0^\gamma$. The word done by the gas is $$W=\int P\ \text{d}V=\int_{30\text{L}}^{15\text{L}}\dfrac{P_0V_0^\gamma}{V^\gamma}\ \text{d}V=P_0V_0^\gamma\cdot\dfrac{1}{-\gamma+1}V^{-\gamma+1}\bigg|_{15\text{L}}^{30\text{L}}=-4850\ \text{J}$$ So the work done by the surroundings is 4850 J.

Two concentric charged conducting shells are in free space. The outer shell has inner radius $2a$ and outer radius $3a$. The inner shell has radius $a$. It is known that the electric potential at $r=3a$ is $V_{3a}=\dfrac{kQ}{3a}$. If the electric potential $V_a$ at $r=a$ is 0 volts, what is the charge on the inner spherical shell, $Q_{in}$?

$\textbf{(A) }$ $Q_{in}=-\dfrac32Q \qquad\newline$ $\textbf{(B) }$ $Q_{in}=-\dfrac23Q \qquad\newline$ $\textbf{(C) }$ $Q_{in}=-\dfrac13Q \qquad\newline$ $\textbf{(D) }$ $Q_{in}=-2Q \qquad\newline$ $\textbf{(E) }$ $Q_{in}=-\dfrac12Q$

$\textbf{B}$

The electric field at $a<r<2a$ is $E=\dfrac{kQ_{in}}{r^2}$. The potential difference from $r=a$ to $r=2a$ is $\Delta V=\int E\ \text{d}r=\int_a^{2a}\dfrac{kQ_{in}}{r^2}\ \text{d}r=\dfrac{kQ_{in}}{2a}$. As we know, the electric potential at $r=a$ is 0 and at $r=3a$ is $\dfrac{kQ}{3a}$. Between $2a$ and $3a$, there is no field interior to the conductor. So the potential difference between is $-\dfrac{kQ}{3a}=\dfrac{kQ_{in}}{2a}$, here we get $Q_{in}=-\dfrac23Q$.

The kinetic energy associated with an electron is twice its rest energy. At what speed is the electron traveling?

$\textbf{(A) }$ $2.83\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(B) }$ $2.67\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(C) }$ $2.60\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(D) }$ $2.25\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(E) }$ $2.12\times10^8\ \text{m/s}$

$\textbf{A}$

When the kinetic energy is twice the rest energy, the total energy $mc^2=\dfrac{m_0}{\sqrt{1-v^2/c^2}}c^2=3m_0c^2$, so $\dfrac{1}{\sqrt{1-v^2/c^2}}=3$. Here we get $v=\dfrac23\sqrt2c=2.83\times10^8\ \text{m/s}$.

A magnetic field directed into the plane of the page is decreasing in time. A constant emf $\mathcal{E}$ is produced for the square loop enclosing the field in the figure. The square loop has three identical light bulbs of resistance $R$ in it and an ideal voltmeter connected to the corners through the center of the loop. What is the magnitude of the voltmeter’s reading?

$\textbf{(A) }$ $0\ \mathcal{E} \qquad$ $\textbf{(B) }$ $\dfrac12\ \mathcal{E} \qquad$ $\textbf{(C) }$ $\dfrac13\ \mathcal{E} \qquad$ $\textbf{(D) }$ $\dfrac16\ \mathcal{E} \qquad$ $\textbf{(E) }$ $\dfrac23\ \mathcal{E}$

$\textbf{D}$

The square loop is divided into two parts by the wire of voltmeter. Each part has the same area, thus the emf in each part is $\mathcal{E}/2$. Assuming the current passing through each resistor is $I$ and the voltage across the voltmeter is $V$, here we have $2IR-V=\dfrac12\mathcal{E}$ and $IR+V=\dfrac12\mathcal{E}$. So $V=\dfrac16\mathcal{E}$.