## PhysicsBowl 2013

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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

Red light from a laser is noted to have a wavelength of 632.8 nanometers. Which one of the following choices best represents the meaning of the prefix nano?

$\textbf{(A) }$ $1\times10^{-3} \qquad$ $\textbf{(B) }$ $1\times10^{-6} \qquad$ $\textbf{(C) }$ $1\times10^{-9} \qquad$ $\textbf{(D) }$ $1\times10^{-12} \qquad$ $\textbf{(E) }$ $1\times10^{-15}$

$\textbf{A}$

The prefixes for the answers given are A) milli, B) micro, C) nano, D) pico, E) femto

**Questions 2 and 3 deal with the following information:**

A small object is released from rest and reaches the ground in a time of 2.50 s. Neglect air resistance.

With what speed does the object reach the ground?

$\textbf{(A) }$ 31.3 m/s$ \qquad$ $\textbf{(B) }$ 25.0 m/s$ \qquad$ $\textbf{(C) }$ 12.5 m/s$ \qquad$ $\textbf{(D) }$ 10.0 m/s$ \qquad$ $\textbf{(E) }$ 2.50 m/s

$\textbf{B}$

The final speed $v=gt=25.0$ m/s.

From what height above the ground was the object released?

$\textbf{(A) }$ 6.25 m$ \qquad$ $\textbf{(B) }$ 12.5 m$ \qquad$ $\textbf{(C) }$ 25.0 m$ \qquad$ $\textbf{(D) }$ 31.3 m$ \qquad$ $\textbf{(E) }$ 62.5 m

$\textbf{D}$

The height $h=\dfrac12gt^2=31.25$ m.

A scientist calculated a quantity that was equal to one light-year. Which one of the following choices represents the type of quantity that the scientist calculated?

$\textbf{(A) }$ Time$ \qquad$ $\textbf{(B) }$ Mass$ \qquad$ $\textbf{(C) }$ Speed$ \qquad$ $\textbf{(D) }$ Force$ \qquad$ $\textbf{(E) }$ Distance

$\textbf{E}$

A light-year is the distance that light travels in one year.

At an instant of time , a point object of mass $M$ moves with velocity $\vec{V}$, has acceleration $\vec{A}$, and is at position $(x,y)$. In what direction must the linear momentum of the object be directed at this instant?

$\textbf{(A) }$ Along the direction of the velocity of the mass$ \qquad\newline$

$\textbf{(B) }$ Along the direction of the net force acting on the mass$ \qquad\newline$

$\textbf{(C) }$ Along the direction of the vector from the origin $(0,0)$ to $(x,y)$ $ \qquad\newline$

$\textbf{(D) }$ Along the direction of the acceleration of the mass$ \qquad\newline$

$\textbf{(E) }$ Along the direction perpendicular to the object’s acceleration

$\textbf{A}$

The linear momentum $\vec{P}=m\vec{V}$ has the same direction as the velocity.

A 1.50 m-long string clamped at both ends is vibrating at its second harmonic. What is the wavelength associated with the string for this scenario?

$\textbf{(A) }$ 3.00 m$ \qquad$ $\textbf{(B) }$ 2.25 m$ \qquad$ $\textbf{(C) }$ 1.50 m$ \qquad$ $\textbf{(D) }$ 1.00 m$ \qquad$ $\textbf{(E) }$ 0.75 m

$\textbf{C}$

For $L=\dfrac{n}{2}\lambda$ and $n=2$, $\lambda=L=1.50\ \text{m}$.

Two identical particles are fixed in place a distance of 0.5 m apart. The electric force that one particle exerts on the other has a magnitude of 3.00 N. Which one of the following choices best represents the magnitude of each particle’s charge?

$\textbf{(A) }$ $4.17\times10^{-11}\ \text{C} \qquad\newline$

$\textbf{(B) }$ $8.33\times10^{-11}\ \text{C} \qquad\newline$

$\textbf{(C) }$ $1.67\times10^{-10}\ \text{C} \qquad\newline$

$\textbf{(D) }$ $9.13\times10^{-6}\ \text{C} \qquad\newline$

$\textbf{(E) }$ $1.29\times10^{-5}\ \text{C}$

$\textbf{D}$

By Coulomb’s Law $F=k\dfrac{Q^2}{r^2}$, we have $Q=r\sqrt{\dfrac{F}{k}}=9.13\times10^{-6}\ \text{C}$.

Some standard household lights are being replaced with LEDs. LED is the acronym for which one of the following choices?

$\textbf{(A) }$ Low Emission Dial$ \qquad\newline$

$\textbf{(B) }$ Light Emitting Diode$ \qquad\newline$

$\textbf{(C) }$ Light Energy Divider$ \qquad\newline$

$\textbf{(D) }$ Lower Edge Disc$ \qquad\newline$

$\textbf{(E) }$ Limiting Emission Diode

$\textbf{B}$

LED is Light Emitting Diodes.

A simple pendulum oscillates with a period of 2.0 s. If the maximum oscillation of the pendulum is $4.0^\circ$ from equilibrium, what is the length of the string for this pendulum?

$\textbf{(A) }$ 6.4 m$ \qquad$ $\textbf{(B) }$ 3.2 m$ \qquad$ $\textbf{(C) }$ 1.6 m$ \qquad$ $\textbf{(D) }$ 1.0 m$ \qquad$ $\textbf{(E) }$ 0.5 m

$\textbf{D}$

The expression for the period of a simple pendulum at small angles is $T=2\pi\sqrt{\dfrac{l}{g}}$, so the length of the string $l=g\left(\dfrac{T}{2\pi}\right)^2=1.0\ \text{m}$.

An object is thrown straight upward. The object remains in free fall until it returns to its initial launch point. Which one of the following graphs could represent the velocity of the object as a function of time during its flight?

$\textbf{E}$

For an object launched upward and then returning to its initial position, the initial velocity is upward and the final velocity is downward. This means that half of the time, the velocity is positive with it being negative the other half of the time (by symmetry). Of the graphs shown, only (E) has this property (and the velocity has a constant slope indicating a constant acceleration).

There was great excitement in the physics community because of an announcement from the LHC during the summer of 2012. Which one of the following choices best represents the reason for the excitement?

$\textbf{(A) }$ The announcement that life was found on Mars.$ \qquad\newline$

$\textbf{(B) }$ The announcement that dark matter had been created and studied in the laboratory.$ \qquad\newline$

$\textbf{(C) }$ The announcement that the Hubble Telescope discovered a “spaceship-like” object near Alpha Centauri.$ \qquad\newline$

$\textbf{(D) }$ The announcement that the mass of a neutrino had been determined.$ \qquad\newline$

$\textbf{(E) }$ The announcement that there was experimental evidence of a particle consistent with a Higgs Boson.

$\textbf{E}$

The announcement of the discovery of a particle consistent with the Higgs Boson took place last summer from the LHC (Large Hadron Collider).

It takes the Earth one day to rotate about its axis. Which one of the following choices best represents the time that it takes the Moon to make one rotation about its axis?

$\textbf{(A) }$ One day$ \qquad\newline$ $\textbf{(B) }$ One week$ \qquad\newline$ $\textbf{(C) }$ One month$ \qquad\newline$ $\textbf{(D) }$ One year$ \qquad\newline$ $\textbf{(E) }$ It does not rotate at all

$\textbf{C}$

The Moon does rotate on its axis. Because of this rotation rate, there is a “dark side of the Moon” which is always directed away from the Earth.

Which one of the following choices represents the measurement with the most number of significant digits?

$\textbf{(A) }$ 6.75 m$ \qquad\newline$ $\textbf{(B) }$ $4.67\times10^9\ \text{m} \qquad\newline$ $\textbf{(C) }$ 0.000000012 m$ \qquad\newline$ $\textbf{(D) }$ 8100 m$ \qquad\newline$ $\textbf{(E) }$ 2.00005 m

$\textbf{E}$

The significant digits are not counted until the first non-zero value is encountered. For the choices given, the number of sig figs is A) 3, B) 3, C) 2, D) 2, E) 6

A small ball is thrown at an angle of $30.0^\circ$ above the horizontal ground with a speed of 20.0 m/s. To what maximum height above the launch point does the ball rise during its motion? Ignore air resistance.

$\textbf{(A) }$ 2.5 m$ \qquad$ $\textbf{(B) }$ 5.0 m$ \qquad$ $\textbf{(C) }$ 10.0 m$ \qquad$ $\textbf{(D) }$ 15.0 m$ \qquad$ $\textbf{(E) }$ 20.0 m

$\textbf{B}$

The vertical component of the initial speed is $v_\perp=v_0\sin\theta=10.0\ \text{m/s}$. The maximum height $h=\dfrac{v_\perp^2}{2g}=5.0\ \text{m}$.

In a circuit, the flow of electrons in a horizontal wire produces a constant current of 3.20 A for a time of 3.9 hr. Which one of the following choices best represents the number of electrons that pass through a vertical cross-section of the wire during this time?

$\textbf{(A) }$ $9.60 \qquad\newline$ $\textbf{(B) }$ $6.00\times10^{19} \qquad\newline$ $\textbf{(C) }$ $7.20\times10^{22} \qquad\newline$ $\textbf{(D) }$ $2.16\times10^{23} \qquad\newline$ $\textbf{(E) }$ $6.02\times10^{23}$

$\textbf{D}$

The total number of coulombs of charge in 3.0 hr to pass through the circuit is $Q=It=(3.20\ \text{A})(3.0\times3600\ \text{s})=3.456\times10^4\ \text{C}$. The charge of an electron has magnitude $e=1.6\times10^{-19}\ \text{C}$, so the total number of electrons $n=\dfrac{Q}{e}=2.16\times10^{23}$.

A block initially moving at 8.0 m/s accelerates uniformly to rest on a horizontal surface. The block travels a distance of 12.0 m during the slide. Which one of the following choices best represents the coefficient of kinetic friction between the surface and the block?

$\textbf{(A) }$ 1.20$ \qquad$ $\textbf{(B) }$ 0.667$ \qquad$ $\textbf{(C) }$ 0.533$ \qquad$ $\textbf{(D) }$ 0.267$ \qquad$ $\textbf{(E) }$ 0.133

$\textbf{D}$

The acceleration of the block (take it as a positive number) $a=\dfrac{v^2}{2x}=2.67\ \text{m/s}^2$. The acceleration comes from the frictional force, which is $f=\mu mg=ma$. So the coefficient of friction $\mu=\dfrac{a}{g}=0.267$.

A position vs. time graph of a particle moving along a horizontal axis is shown. What is the total distance traveled by the particle from $t=0\ \text{s}$ to $t=10\ \text{s}$ ?

$\textbf{(A) }$ 2 m$ \qquad$ $\textbf{(B) }$ 18 m$ \qquad$ $\textbf{(C) }$ 26 m$ \qquad$ $\textbf{(D) }$ 34 m$ \qquad$ $\textbf{(E) }$ 42 m

$\textbf{D}$

Given the position vs. time graph, we see that the particle starts at position 8.0 m and stays there for 3 seconds. Then it moves at a constant rate to the 0 m mark traveling 8.0 m. After stopping for 1 second, it walks at a constant rate 8 more meters to -8 m. Finally, it travels from -8 m to 10 m during the last second for a total of 18 m. Hence, for the entire trip, it moved $8+8+18=34$ meters.

Which one of the following choices correctly identifies all of the listed situations for which there is a non-zero acceleration?$\newline$

Situation I : A point object moves in a straight line with increasing speed.$\newline$

Situation II : A point object moves in a circular path with constant speed.$\newline$

Situation III: A point objects moves in a circular path with decreasing speed.$\newline$

$\textbf{(A) }$ Situations I, II, $\&$ III$ \qquad\newline$

$\textbf{(B) }$ Only Situations I $\&$ III$ \qquad\newline$

$\textbf{(C) }$ Only Situations II $\&$ III$ \qquad\newline$

$\textbf{(D) }$ Only Situation III$ \qquad\newline$

$\textbf{(E) }$ Only Situation I

$\textbf{A}$

Any time the velocity changes (magnitude or direction), there is acceleration. Consequently, since the speed changes in I and III, there is tangential acceleration. As the direction of motion changes in II and III, there is radial (centripetal) acceleration here. Hence, there is acceleration in all cases presented.

Which physicist won the Nobel Prize in physics partly for the explanation of the photoelectric effect?

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

$\textbf{(B) }$ Steven Hawking$ \qquad\newline$

$\textbf{(C) }$ Albert Einstein$ \qquad\newline$

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

$\textbf{(E) }$ Neil deGrasse Tyson

$\textbf{C}$

Albert Einstein won his Nobel Prize partly for his work explaining the photoelectric effect.

A sample of ideal gas at a temperature of $40.0\ ^\circ\text{C}$ is in a container of volume $3.50\times10^{-2}\ \text{m}^3$. If the pressure of the gas is 0.50 atm, how many molecules of the gas are in the container?

$\textbf{(A) }$ $4.05\times10^{18} \qquad\newline$ $\textbf{(B) }$ $4.10\times10^{20} \qquad\newline$ $\textbf{(C) }$ $4.05\times10^{21} \qquad\newline$ $\textbf{(D) }$ $4.10\times10^{23} \qquad\newline$ $\textbf{(E) }$ $4.10\times10^{26}$

$\textbf{D}$

By the ideal gas equation $PV=nRT$, we get the mole number $$n=\dfrac{PV}{RT}=\dfrac{(0.5\times1.013\times10^5)(3.50\times10^{-2})}{(8.31)(40+273)}=0.68\ \text{mol}$$The number of molecules $N=nN_A=(0.68)(6.02\times10^{23})=4.10\times10^{23}$.

The following nuclear reaction occurs: $_{\ 53}^{131}I\ \rightarrow\ _{\ 54}^{131}Xe\ +\ ^{A}_{Z}X$. What is $_Z^AX$?

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

$\textbf{E}$

By $131=131+A$ and $53=54+Z$ we get $A=0$ and $Z=-1$. So $^{\ 0}_{-1}X$ is an electron $^{\ 0}_{-1}e$.

Four resistors, each of resistance $R$, are connected to a battery in the following way: “Two resistors are connected in series. This combination of two resistors is connected in parallel to a third resistor. This set of three resistors is connected in series to a fourth resistor.” What is the equivalent resistance of this arrangement of four resistors?

$\textbf{(A) }$ $\dfrac52R \qquad$ $\textbf{(B) }$ $\dfrac53R \qquad$ $\textbf{(C) }$ $\dfrac43R \qquad$ $\textbf{(D) }$ $\dfrac35R \qquad$ $\textbf{(E) }$ $\dfrac25R$

$\textbf{B}$

Two resistors in series have a equivalent resistance of $R_{12}=R+R=2R$. Putting this in parallel with a resistor gives a total resistance of $\dfrac{1}{R_{123}}=\dfrac1R+\dfrac1{2R}\Rightarrow R_{123}=\dfrac23R$. Finally, putting another resistor in series gives $R_{1234}=\dfrac23R+R=\dfrac53R$.

A small object of mass 11.0 g is at rest 30.0 cm from a horizontal disk’s center. The disk starts to rotate from rest about its center with a constant angular acceleration of 4.50 rad/s$^2$. What is the magnitude of the net force acting on the object after a time of $t=\dfrac13\ \text{s}$ if the object remains at rest with respect to the disk?

$\textbf{(A) }$ 0 N$ \qquad\newline$ $\textbf{(B) }$ $7.43\times10^{-3}\ \text{N} \qquad\newline$ $\textbf{(C) }$ $1.49\times10^{-2}\ \text{N} \qquad\newline$ $\textbf{(D) }$ $1.66\times10^{-2}\ \text{N} \qquad\newline$ $\textbf{(E) }$ $2.33\times20^{-2}\ \text{N}$

$\textbf{D}$

At $t=\dfrac13\ \text{s}$ the angular speed of the disk is $\omega=\beta t=1.5\ \text{rad/s}$. As the object is moving in a circle with increasing angular speed, there are both tangential and radial accelerations. The tangential acceleration $a_\tau=\beta r=(4.5)(0.3)=1.35\ \text{m/s}^2$. The radial acceleration is $a_r=\omega^2r=(1.5)^2(0.3)=0.675\ \text{m/s}^2$. The total acceleration is computed using the Pythagorean Theorem (the tangential and radial accelerations are at right angles to each other) to give $a=\sqrt{a_\tau^2+a_r^2}=1.51\ \text{m/s}^2$. So the net force $F=ma=(0.011)(1.51)=1.66\times10^{-2}\ \text{N}$.

Which one of the following statements best describes Huygens’s Principle?

$\textbf{(A) }$ An additional pressure is transmitted undiminished to all points in the fluid and to the walls of the container.$ \qquad\newline$

$\textbf{(B) }$ Each point on a wavefront acts as a source of secondary spherical wavelets (new waves).$ \qquad\newline$

$\textbf{(C) }$ For every action force, there is an equal but opposite reaction force.$ \qquad\newline$

$\textbf{(D) }$ It is impossible to have a process which has the sole result of transferring energy from a low temperature reservoir to a high temperature reservoir.$ \qquad\newline$

$\textbf{(E) }$ A time-changing magnetic field has an associated induced electric field.

$\textbf{B}$

Huygens’s Principle is related to wave phenomena and that one can treat each point on a wave front to be the sources of new spherical wave fronts (sometimes called wavelets).

An ideal fluid completely fills a small horizontal tube that has a narrowing cross-sectional area as seen in the figure. Which one of the following choices best describes what has happened to the fluid’s speed and its associated pressure in the narrower region as compared to the wider region?

$\textbf{(A) }$ The fluid speed increased and the fluid pressure decreased.$ \qquad\newline$

$\textbf{(B) }$ The fluid speed increased and the fluid pressure increased.$ \qquad\newline$

$\textbf{(C) }$ The fluid speed increased and the fluid pressure remained the same.$ \qquad\newline$

$\textbf{(D) }$ The fluid speed decreased and the fluid pressure increases.$ \qquad\newline$

$\textbf{(E) }$ The fluid speed decreased and the fluid pressure decreases.

$\textbf{A}$

Using Bernoulli’s Principle, we have $p+\dfrac12\rho v^2+\rho gh=costant$. Since the tube is horizontal, there is no change in $h$ as the fluid enters the narrower region. However, from continuity, as the size of the opening decreases, the fluid speed must increase (what goes in, must come out). Using the Bernoulli expression now with the second term increasing and the third term unchanged, this means that the pressure of the fluid has to decrease as it enters the narrow region.

An object moving only to the right completes a 20.0 second trip in two stages, I and II. The average speed of the entire 20.0 second trip is 10.0 m/s. For stage I, the object moves with a constant velocity of 6.0 m/s for 12.0 seconds. What constant acceleration must the object have during the 8.0 seconds of stage II?

$\textbf{(A) }$ $2.25\ \text{m/s}^2 \qquad$ $\textbf{(B) }$ $2.50\ \text{m/s}^2 \qquad$ $\textbf{(C) }$ $4.00\ \text{m/s}^2 \qquad$ $\textbf{(D) }$ $6.25\ \text{m/s}^2 \qquad$ $\textbf{(E) }$ $8.50\ \text{m/s}^2$

$\textbf{B}$

The average speed of the 20.0 second trip is 10.0 m/s, so the total distance is $x=\overline{v}t=200\ \text{m}$. For stage I, the object moves with a constant velocity of 6.0 m/s for 12.0 seconds, so the distance $x_1=(6)(12)=72\ \text{m}$. The distance for stage II is $x_2=x-x_1=128\ \text{m}$, so the average speed is $\overline{v_2}=128/8=16\ \text{m/s}$. Since the time of stage II is 8.0 second, 16 m/s is the speed at the middle of stage II, so the acceleration is $a_2=\dfrac{16-6}{4}=2.50\ \text{m/s}$.

Light of wavelength 600 nm is transmitted from air into a piece of glass. Which one of the labeled arrows best indicates the path of the light ray after it enters the glass?

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

$\textbf{C}$

As the light enters a medium with a higher index of refraction, it bends toward the normal. We need to look at the figure to determine which way that is directed. Since the angles are measured from normal to the surface, that angle must decrease in the glass, meaning that the ray travels along C.

An object is in free fall close to the ground. A person intervenes and slows the object uniformly to rest. Which one of the following statements must be true about the magnitude of the acceleration of the object as it is being stopped by the person? The magnitude of the object’s acceleration is $a_{obj}$ and the magnitude of the acceleration from gravity is $g$.

$\textbf{(A) }$ $a_{obj}=g \qquad\newline$

$\textbf{(B) }$ $a_{obj}>g \qquad\newline$

$\textbf{(C) }$ $a_{obj}<g \qquad\newline$

$\textbf{(D) }$ $a_{obj}\geq g \qquad\newline$

$\textbf{(E) }$ None of the previous relations must be true.

$\textbf{E}$

Imagine an object dropped from rest and let it attain a final downward speed of 10 m/s. $\newline$

If the object is stopped in 2.0 seconds, the magnitude of the acceleration is $|a|=\left|\dfrac{\Delta v}{\Delta t}\right|=\left|-\dfrac{10}{2}\right|=5\ \text{m/s}^2<g$. $\newline$

If the object is stopped in time 1.0 s, the acceleration is $|a|=\left|\dfrac{\Delta v}{\Delta t}\right|=\left|-\dfrac{10}{1}\right|=10\ \text{m/s}^2=g$. $\newline$

If the object is stopped in time 0.5 s, the acceleration is $|a|=\left|\dfrac{\Delta v}{\Delta t}\right|=\left|-\dfrac{10}{0.5}\right|=20\ \text{m/s}^2>g$.

Two electrons enter a region between charged capacitor plates with equal speed $v$. Electron A is directed horizontally to the left while electron B is directed at $30^\circ$ below the horizontal. Each electron makes it to the left-hand plate. Which one of the following choices best compares the speeds of the charges $(v_A,v_B)$ upon arrival at the left plate? Consider only the electrons A and B’s interactions with the constant electric field between the plates, ignoring any relativistic effects.

$\textbf{(A) }$ $v_A>v_B \qquad\newline$

$\textbf{(B) }$ $v_A=v_B \qquad\newline$

$\textbf{(C) }$ $v_A<v_B \qquad\newline$

$\textbf{(D) }$ The answer depends on the size of the plate separation, $d. \qquad\newline$

$\textbf{(E) }$ The answer depends on the magnitude of the charge, $Q$, on each plate.

$\textbf{B}$

Using mechanical energy conservation for the particle-field system as the particle crosses the region between the plates, we write $\Delta KE+\Delta PE=0$ where $\Delta PE=q\Delta V$. Since the potential difference between the plates will be the same for each charge, the change in PE of the chargefield system is the same in each case, meaning that the change in KE is the same for each charge. Hence, each charge reaches the other side with the same speed.

Electrons flow to the left in a wire as shown. For the proton moving toward the top of the page at the instant shown, what is the direction of the magnetic force on the proton?

$\textbf{(A) }$ To the left$ \qquad\newline$

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

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

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

$\textbf{(E) }$ There is no force

$\textbf{B}$

Conventional current is in the direction of positive charge flow, so that is directed to the right in the wire. Using the right-hand rule, the direction of the magnetic field associated with this current is out of the plane of the page at the location of the proton (thumb to the right, fingers curl around the thumb). Now, we use the right-hand rule that the force on a charged particle can be found by pointing the right fingers along the velocity (up the plane) and curling them into the direction of the field (out of the plane), resulting in the thumb pointing to the right, in the direction of the force.

A monatomic ideal gas undergoes a reversible isothermal expansion in an enclosed container. Which one of the following quantities associated with the gas has a value of zero?

$\textbf{(A) }$ Heat$ \qquad\newline$

$\textbf{(B) }$ Entropy change$ \qquad\newline$

$\textbf{(C) }$ Work done$ \qquad\newline$

$\textbf{(D) }$ Internal energy change$ \qquad\newline$

$\textbf{(E) }$ Pressure change

$\textbf{D}$

An ideal gas during an isothermal process has a constant temperature (that is the meaning of isothermal). Since internal energy depends on the temperature, there is no internal energy change.

An acceleration vs. time graph for an object moving along a line is shown. The object starts from rest at time $t=0\ \text{s}$. At what time(s) does the object attain a maximum displacement from its starting position?

$\textbf{(A) }$ At times $t=2.5\ \text{s}$ and $t=7.5\ \text{s}$ only$ \qquad\newline$

$\textbf{(B) }$ At times $t=5.0\ \text{s}$ and $t=10\ \text{s}$ only$ \qquad\newline$

$\textbf{(C) }$ At times $t=1.25\ \text{s}$, $t=3.75\ \text{s}$, $t=6.25\ \text{s}$, and $t=8.75\ \text{s}$ only$ \qquad\newline$

$\textbf{(D) }$ At times $t=2.5\ \text{s}$, $t=5.0\ \text{s}$, $t=7.5\ \text{s}$, and $t=10\ \text{s}$ only$ \qquad\newline$

$\textbf{(E) }$ At times $t=10\ \text{s}$ only

$\textbf{E}$

While it looks like a simple harmonic oscillation, it is not because it is a acceleration vs. time graph but not displacement vs. time graph. Instead, we need to find the area under the curve to give the velocity as a function of time. For the first 2.5 seconds, the acceleration is positive, so the velocity is becoming more positive. For the next 2.5 seconds, the acceleration is negative, indicating that the object is now slowing down, but still moving forward (velocity is still positive) until it comes back to rest at time t = 5 seconds. The process begins again for the next 5 seconds with the object again having positive velocity. Hence, the object is getting further from its starting position during this motion and achieves its greatest displacement from the origin at $t=10\ \text{s}$.

At the top of a high cliff, a small rock is dropped from rest. A ball is launched straight downward with an initial speed of 36.0 m/s at a time 2.10 s after the rock was dropped. When the ball has fallen 28.0 m further than the initially dropped rock, what is the speed of the ball relative to the rock?

$\textbf{(A) }$ 15.0 m/s$ \qquad$ $\textbf{(B) }$ 16.0 m/s$ \qquad$ $\textbf{(C) }$ 20.0 m/s$ \qquad$ $\textbf{(D) }$ 21.0 m/s$ \qquad$ $\textbf{(E) }$ 36.0 m/s

$\textbf{A}$

At $t=2.10\ \text{s}$ the speed of the rock is $v_{rock}=gt=21\ \text{m/s}$. Since both objects are accelerated by gravity, the relative speed between them $v_r=36-21=15\ \text{m/s}$ do not change in the whole process.

Which one of the following choices represents the base MKS units for sound intensity?

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

$\textbf{A}$

Intensity is power (in Watts) divided by area (in $\text{m}^2$). So, we have for the analysis: $$\dfrac{\text{W}}{\text{m}^2}=\dfrac{\text{J/s}}{\text{m}^2}=\dfrac{\text{J}}{\text{s}\cdot\text{m}^2}=\dfrac{\text{N}\cdot\text{m}}{\text{s}\cdot\text{m}^2}=\dfrac{\text{N}}{\text{s}\cdot\text{m}}=\dfrac{\text{kg}\cdot\text{m/s}^2}{\text{s}\cdot\text{m}}=\dfrac{\text{kg}}{\text{s}^3}$$

For the figure shown, the variable resistance of the inner circuit, $R_{Inner}$, is increasing at a constant rate. While this is occurring, in which direction is the magnetic field associated with the inner circuit at the point P in the plane of the circuit and in which direction is the flow of electrons through the resistor labeled $R_{Outer}$?

$\textbf{C}$

The current of the inner circuit is oriented counterclockwise. By the right hand rule, this means that the field is directed out of the plane in the interior of the circuit, but since field lines for closed loops, they would enter into the plane of the page exterior to the loop (for example, point P). Since not all of the field lines coming back into the plane are interior to the outer loop, the net magnetic flux is directed out of the plane of the page for the outer loop, but decreasing in strength. By Lenz’s Law, this means that there is an induced current directed counterclockwise around the outer loop. Hence, this is generated by pushing electrons from left to right from X to Y in the upper branch.

The principal quantum number of an electron is $n=5$. How many possible values of the orbital magnetic quantum number are there for this electron?

$\textbf{(A) }$ 25$ \qquad$ $\textbf{(B) }$ 11$ \qquad$ $\textbf{(C) }$ 9$ \qquad$ $\textbf{(D) }$ 5$ \qquad$ $\textbf{(E) }$ 4

$\textbf{C}$

The principal quantum number allows one to find the azimuthal quantum number, $l$, as $l=0,1,2,3,4$. The magnetic quantum number is then found as $m_l=-l,...-1,0,1,...,l$. Since $l=4$, the values of $m_l$ are $m_l=-4,-3,-2,-1,0,1,2,3,4$ which is a total of 9 possible values.

A real object in air is placed in front of a glass lens. The calculated image size is larger than the size of the original object. Which one of the following conclusions about the type of lens used and the type of image formed is correct?

$\textbf{A}$

By $\dfrac1u+\dfrac1v=\dfrac1f$, the image distance $v=\dfrac{uf}{u-f}$. When the image size is larger than the size of the object, we have $|v|=u\left|\dfrac{f}{u-f}\right|>u$. It can be rewrote as $|f|>|u-f|=|f-u|$. The relation of inequality is impossible if $f<0$, so the lens must be a convex. When $f>u>0$, the image is virtual since $v<0$. When $u>f>0.5u$, we get a real image as $v>0$. So the image could be virtual or real.

Five identical light bulbs are connected into a circuit as shown. All wires are ideal with no resistance, and the ideal battery has emf $\xi$. When the switch S in the circuit is closed, aside from bulb $\#5$, which of the other bulbs brighten?

$\textbf{(A) }$ Only Bulb $\#4$ $ \qquad\newline$

$\textbf{(B) }$ Only Bulbs $\#1$ and $\#3$ $ \qquad\newline$

$\textbf{(C) }$ Only Bulbs $\#3$ and $\#4$ $ \qquad\newline$

$\textbf{(D) }$ Only Bulbs $\#2$, $\#3$, and $\#4$ $ \qquad\newline$

$\textbf{(E) }$ Only Bulbs $\#1$, $\#3$, and $\#4$

$\textbf{B}$

The equivalent circuit before and after the switch is closed for the resistors is shown in the figure. In words, by closing the switch the resistance of the entire circuit goes down since the resistance of the bottom branch drops from $2R$ to $\dfrac32R$. Since there is less resistance in the circuit, there is more current, meaning that there is more current through bulb $\#1$ directly connected to the battery. Bulb $\#1$ gets brighter and has more potential difference. Consequently, there is less potential difference now for bulb $\#2$ from Kirchhoff’s Loop Rule with the battery and bulb $\#1$. Bulb $\#2$ is dimmer. Finally, since the resistance of the bottom branch decreased, it now gets a higher percentage of a slightly higher current. With the switch closed, bulbs $\#4$ and $\#5$ now share the current equally, resulting in less current through bulb $\#4$, thereby making it dimmer than before. At the same time, all of the slightly higher current is now through $\#3$, thereby making it brighter.

**Questions 39 – 40 deal with the following information:**

An ideal uniform solid disk and an ideal uniform ring each have mass $M$ and radius $R$. Each object begins purely rolling without slipping down a rough inclined plane. The coefficients of friction for the disk and ring with the incline are $\mu_{disk}>\mu_{ring}$.

As each object rolls down the incline, which statement is correct about the force of friction from the incline on the objects?

$\textbf{(A) }$ The ring experiences a greater force of friction than the disk.$ \qquad\newline$

$\textbf{(B) }$ The disk experiences a greater force of friction than the ring.$ \qquad\newline$

$\textbf{(C) }$ The force of friction is equal and non-zero for both objects.$ \qquad\newline$

$\textbf{(D) }$ The force of friction is equal to zero for both objects.$ \qquad\newline$

$\textbf{(E) }$ Nothing can be concluded about the force of friction without more information.

$\textbf{A}$

The object rolls without slipping, so $v=\omega R$. Along the inclined plane $Mg\sin\theta-f=Ma=MR\beta$ where $\beta$ is the angular acceleration. In the view of rotation, $fR=I\beta$, where $I_r=MR^2$ is the ring's momentum of inertia of to the center and $I_d=\dfrac12MR^2$ is the disk's. So $I_r>I_d$. As $\beta=\dfrac{fR}{I}$, we have $Mg\sin\theta-f=MR\beta=MR\dfrac{fR}{I}$, or it can be rewrote as $Mg\sin\theta=\left(\dfrac{MR^2}{I}+1\right)f$. As $I_r>I_d$, we get $f_r>f_d$, which means the ring experiences a greater force of friction than the disk.

As the objects roll, what is the ratio of the ring’s angular acceleration to the disk’s angular acceleration calculated about an axis perpendicular to the object’s face and through its center of mass?

$\textbf{(A) }$ 1:2$ \qquad$ $\textbf{(B) }$ 2:1$ \qquad$ $\textbf{(C) }$ $\mu_{disk}:\mu_{ring} \qquad$ $\textbf{(D) }$ 4:3$ \qquad$ $\textbf{(E) }$ 3:4

$\textbf{E}$

As $F=\dfrac{I\beta}{R}$, $Mg\sin\theta-f=Mg\sin\theta-\dfrac{I\beta}{R}=MR\beta$. So $\beta=\dfrac{MR\sin\theta}{MR+I/R}$. The ratio of angular acceleration $$\dfrac{\beta_r}{\beta_d}=\dfrac{MR+I_d/R}{MR+I_r/R}=\dfrac{MR+\frac12MR^2/R}{MR+MR^2/R}=\dfrac34$$

An engine operates between a low temperature of $273^\circ\text{C}$ and a high temperature of $546^\circ\text{C}$. What is the maximum theoretical efficiency of this engine?

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

$\textbf{C}$

The Carnot efficiency $\eta=1-\dfrac{T-2}{T_1}=1-\dfrac{(273+273)\text{K}}{(546+273)\text{K}}=\dfrac13$.

An object of mass $12M$ is at rest when it suddenly explodes into 3 pieces with masses $3M$, $4M$ and $5M$. The piece of mass $3M$ moves with speed $V$ in the direction shown in the diagram. What is the speed of the piece with mass $5M$ knowing that it is moving directly to the right?

$\textbf{(A) }$ $V \qquad$ $\textbf{(B) }$ $\dfrac{3\sqrt2}{20}V \qquad$ $\textbf{(C) }$ $\dfrac{3\sqrt2}{5}V \qquad$ $\textbf{(D) }$ $\dfrac{3}{5\sqrt2}V \qquad$ $\textbf{(E) }$ $\dfrac{7}{5\sqrt2}V$

$\textbf{C}$

Assuming the speed of each mass is $V_3=V$, $V_4$ and $V_5$. The object of $12M$ was initially at rest. After the explosion, the center of mass is still at rest in both x and y direction, so we have $$3MV\sin45^\circ=4MV_4\sin45^\circ$$ $$3MV\cos45^\circ+4MV_4\cos45^\circ=5MV_5$$ By solving equations we have $V_5=\dfrac{3\sqrt2}{5}V$.

A small 1.0 kg mass is launched from the top of a cliff with speed $V$ at an angle of $30^\circ$ above the horizontal. When the mass reaches the ground, its velocity is directed at $45^\circ$ below the horizontal. Which one of the following choices is the magnitude of the total impulse that was imparted to the mass during its flight? Ignore air resistance.

$\textbf{(A) }$ $\dfrac12\left(\sqrt3+1\right)V \qquad\newline$ $\textbf{(B) }$ $\sqrt{\dfrac32}\dfrac{\left(\sqrt2+1\right)}{2}V \qquad\newline$ $\textbf{(C) }$ $\dfrac12\left(\sqrt3-1\right)V \qquad\newline$ $\textbf{(D) }$ $\dfrac12\left(\sqrt{\dfrac32}+1\right)V \qquad\newline$ $\textbf{(E) }$ $\sqrt{\dfrac32}\dfrac{\left(\sqrt2+1\right)}{\left(\sqrt2-1\right)}V$

$\textbf{A}$

The impulse leads to the change of velocity in the vertical direction (the horizontal component do not change). The vertical component in the beginning is $V\sin30^\circ=\dfrac12V$ upwards and finally $V\cos30^\circ\tan40^\circ=\dfrac{\sqrt3}{2}V$ downwards. So the impulse $$I=m\Delta v=m\left(\dfrac12V+\dfrac{\sqrt3}{2}V\right)=\dfrac12\left(\sqrt3+1\right)V$$

Which one of the following terms/quantities is most closely associated with “the measure of the resistance of an object to length change under lengthwise tension or compression.”?

$\textbf{(A) }$ Bulk modulus$ \qquad\newline$

$\textbf{(B) }$ Plastic deformation $ \qquad\newline$

$\textbf{(C) }$ Shear modulus $ \qquad\newline$

$\textbf{(D) }$ Elastic limit$ \qquad\newline$

$\textbf{(E) }$ Young’s modulus

$\textbf{E}$

Young’s modulus deals with extension/compression of materials.

The switch S in the RC circuit shown is closed at time $t=0$. All circuit elements are ideal and $R=10.0\ \Omega$, $C=2.20\ \text{F}$ and $\xi=12.0\ \text{V}$. The capacitor is initially uncharged. How long after the switch is closed is the voltage across the capacitor three times as large as the voltage across the resistor?

$\textbf{(A) }$ 22.0 s$ \qquad$ $\textbf{(B) }$ 24.2 s$ \qquad$ $\textbf{(C) }$ 30.5 s$ \qquad$ $\textbf{(D) }$ 36.0 s$ \qquad$ $\textbf{(E) }$ 54.7 s

$\textbf{C}$

When the voltage across the capacitor $V_C=3V_R$ along with $V_C+V_R=12\ \text{V}$, we get $V_C=9\ \text{V}$ and $V_R=3\ \text{V}$. So the current at this moment is $I=V_R/R=0.3\ \text{A}$. The current follows the formula $I=I_0e^{-t/\tau}$ where the maximum current at $t=0$ is $I_0=\xi/R=1.2\ \text{A}$ and the time constant $\tau=RC=22\ \text{s}$. So we get $0.3=1.2e^{-t/22}$, $t=22\ln4=30.5\ \text{s}$.

Which one of the following quarks was the last to be confirmed experimentally?

$\textbf{(A) }$ Charmed$ \qquad$ $\textbf{(B) }$ Up$ \qquad$ $\textbf{(C) }$ Strange$ \qquad$ $\textbf{(D) }$ Down$ \qquad$ $\textbf{(E) }$ Top

$\textbf{E}$

The top quark was confirmed experimentally in 1995.

Which one of the following choices best represents the magnitude of the angular momentum of the Earth (expressed in base MKS units) associated with its rotation about its axis?

$\textbf{(A) }$ $10^{38} \qquad$ $\textbf{(B) }$ $10^{34} \qquad$ $\textbf{(C) }$ $10^{30} \qquad$ $\textbf{(D) }$ $10^{26} \qquad$ $\textbf{(E) }$ $10^{22}$

$\textbf{B}$

The angular momentum of a solid ball on its axis is $L=I\omega=\dfrac25MR^2\omega$. The angular speed $$\omega=\dfrac{\Delta\theta}{\Delta t}=\dfrac{2\pi\ \text{rad}}{1\ \text{dy}}\times\dfrac{1\ \text{dy}}{24\ \text{hr}}\times\dfrac{1\ \text{hr}}{3600\ \text{s}}=7.27\times10^{-5}\ \text{rad/s}$$ So we find $$L=\dfrac25\left(6.0\times10^{24}\right)\left(6.4\times10^6\right)^2\left(7.27\times10^{-5}\right)=7\times10^{33}\ \text{kg}\cdot\text{m}^2/\text{s}\approx10^{34}\ \text{kg}\cdot\text{m}^2/\text{s}$$

Two identical samples of a monatomic ideal gas are to undergo reversible processes. Which one of the following choices is a correct statement about the heat associated with the processes?$\newline$

Process 1: An isochoric pressure doubling$\newline$

Process 2: An isobaric volume doubling

$\textbf{(A) }$ There is less heat associated with Process 1 than Process 2.$ \qquad\newline$

$\textbf{(B) }$ The heat is the same non-zero value for Processes 1 and 2.$ \qquad\newline$

$\textbf{(C) }$ There is more heat associated with Process 1 than Process 2.$ \qquad\newline$

$\textbf{(D) }$ The heat is zero for Processes 1 and 2.$ \qquad\newline$

$\textbf{(E) }$ More information is required to determine the relationship for the heats.

$\textbf{A}$

By $PV=nRT$, both processes double the temperature, thus the change of internal energy is the same. However, work has been done to the surrounding in process 2, while no work has been done in process since the volume did not change. The heat absorbed $Q=\Delta U+W$, so $Q_1<Q_2$.

Two electrons move with the magnitude of their linear momentum having a ratio of 2:1. If the slower electron moves with a speed of $1.20\times10^8\ \text{m/s}$, what is the speed of the faster moving electron?

$\textbf{(A) }$ $2.67\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(B) }$ $2.40\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(C) }$ $2.24\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(D) }$ $1.97\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(E) }$ $1.56\times10^8\ \text{m/s}$

$\textbf{D}$

The linear momentum $p=\dfrac{m_0v}{\sqrt{1-v^2/c^2}}$. The speed of the slower electron $v_1=0.4c$. The ratio $$\dfrac{p_2}{p_1}=\dfrac{v_2}{v_1}\dfrac{\sqrt{1-v_1^2/c^2}}{\sqrt{1-v_2^2/c^2}}=2$$ By solving the equation we get $v_2=0.66c=1.97\times10^8\ \text{m/s}$.

A U-tube is filled with mercury (density $13.6\ \text{g/cm}^3$) as shown in the left-most figure. Water of mass 800 grams is added to the lefthand side of the tube. When equilibrium is re-established, the tube appears as shown in the right-most figure. The cross-sectional area of the left tube is $6.50\ \text{cm}^2$ while the right tube has cross-sectional area $15.0\ \text{cm}^2$. Which one of the following choices best represents the height $x$ above the original equilibrium that the mercury rises in the right tube? The drawings are not to scale.

$\textbf{(A) }$ 1.96 cm$ \qquad$ $\textbf{(B) }$ 2.74 cm$ \qquad$ $\textbf{(C) }$ 3.92 cm$ \qquad$ $\textbf{(D) }$ 4.92 cm$ \qquad$ $\textbf{(E) }$ 9.05 cm

$\textbf{B}$

The cross-sectional area of the right tube is $15/6.5=30/13$ times than the left tube. When the mercury level rises $x$ cm in the right tube, $\dfrac{30}{13}x$ cm declined in the left. The volume of water is $V=\dfrac{m}{\rho_w}=800\ \text{cm}^3$, so the height of water in the left tube is $h=800/6.5=\dfrac{1600}{13}\ \text{cm}$. The pressure at the bottom of water is $\rho_wgh=\rho_mg\cdot\dfrac{30}{13}x+\rho_mgx$. Finally we get $x=2.74\ \text{cm}$.