PhysicsBowl 2016
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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 represents the smallest amount of time?
$\textbf{(A) }$ 1 day$ \qquad$ $\textbf{(B) }$ 1 minute$ \qquad$ $\textbf{(C) }$ 1 second$ \qquad$ $\textbf{(D) }$ 1 week$ \qquad$ $\textbf{(E) }$ 1 year
$\textbf{C}$
Bridegroom: I am yours and you are mine from this day until the end of my days, from this day until my last day...$\newline$
Bride: Your last DAY ? Why not your last minute? last second? You liar!
The position vs. time graph of an object moving on a horizontal line is shown. At what time(s) is the object moving with its greatest speed?
$\textbf{(A) }$ At time $t=3.0$ s only$ \qquad\newline$
$\textbf{(B) }$ At time $t=3.5$ s only$ \qquad\newline$
$\textbf{(C) }$ At times $t=3.5$ s and $t=9.0$ s$ \qquad\newline$
$\textbf{(D) }$ At time $t=7.0$ s only$ \qquad\newline$
$\textbf{(E) }$ At time $t=9.0$ s only
$\textbf{A}$
The velocity of the object is found from the slope of the line tangent to the point on the position vs. time graph. To have the greatest speed, we are looking for the largest magnitude of slope to the line tangent. The steepest curve on the position vs. time graph shown occurs at time $t=3.0\ \text{s}$.
A car accelerates uniformly from 0 km/hr to 60 km/hr in 4.50 s. Which one of the following choices best represents the acceleration of the car?
$\textbf{(A) }$ 13.3 m/s$^2 \qquad$ $\textbf{(B) }$ 9.8 m/s$^2 \qquad$ $\textbf{(C) }$ 4.8 m/s$^2 \qquad$ $\textbf{(D) }$ 3.7 m/s$^2 \qquad$ $\textbf{(E) }$ 0.37 m/s$^2$
$\textbf{D}$
First, we need to convert the unit of speed from km/hr to m/s, then the acceleration $a=\dfrac{\Delta v}{\Delta t}$.
The following four length measurements are recorded: $5.4\times10^{-2}\ \text{m}$, $5.4\times10^0\ \text{m}$, $5.40\times10^1\ \text{m}$, and $5.400\times10^2\ \text{m}$. Which one of the following choices best represents the sum of these values using the rules of significant digits?
$\textbf{(A) }$ $6\times10^2\ \text{m} \qquad\newline$
$\textbf{(B) }$ $6.0\times10^2\ \text{m} \qquad\newline$
$\textbf{(C) }$ $6.00\times10^2\ \text{m} \qquad\newline$
$\textbf{(D) }$ $5.999\times10^2\ \text{m} \qquad\newline$
$\textbf{(E) }$ $5.9994\times10^2\ \text{m}$
$\textbf{D}$
The simplest way to look at the addition of quantities is to line up the decimal place. That is, we are computing $0.5\textbf{4}+5.4+54.0+540.0=599.9\textbf{4}$. The 4 is bolded as it is the only digit known in the hundredths place since three of the quantities stop at the tenths. Hence, the final value is only kept to the tenths place making $599.9=5.999\times10^2\ \text{m}$ the correct answer.
In the figure shown, the mass $m=6.0$ kg and the mass $M=14.0$ kg are stationary. The mass M rests on the floor. Which one of the following choices represents the tension in the string connecting the masses?
$\textbf{(A) }$ 40 N$ \qquad$ $\textbf{(B) }$ 60 N$ \qquad$ $\textbf{(C) }$ 80 N$ \qquad$ $\textbf{(D) }$ 140 N$ \qquad$ $\textbf{(E) }$ 200 N
$\textbf{B}$
The forces exerted on $m$ is $T-mg=0$, so $T=mg=60\ \text{N}$.
An object, thrown straight downward with a speed of 20.0 m/s, takes 2.00 s to reach the ground below. From what height above the ground was the object thrown? Ignore air resistance.
$\textbf{(A) }$ 20.0 m$ \qquad$ $\textbf{(B) }$ 40.0 m$ \qquad$ $\textbf{(C) }$ 50.0 m$ \qquad$ $\textbf{(D) }$ 60.0 m$ \qquad$ $\textbf{(E) }$ 80.0 m
$\textbf{D}$
The height $H=v_0t+\dfrac12gt^2=60$ m.
Three cylindrical resistors are made of the same material. The length and radius of each resistor is:$\newline$
Resistor 1: Length $L$, radius $r\newline$
Resistor 2: Length $L/2$, radius $r/2\newline$
Resistor 3: Length $L/4$, radius $r/2\newline$
Which one of the following choices correctly ranks the resistance of these resistors ($R_1$,$R_2$,$R_3$)?
$\textbf{(A) }$ $R_1=R_2<R_3 \qquad\newline$
$\textbf{(B) }$ $R_2<R_1<R_3 \qquad\newline$
$\textbf{(C) }$ $R_3<R_2<R_1 \qquad\newline$
$\textbf{(D) }$ $R_1<R_2=R_3 \qquad\newline$
$\textbf{(E) }$ $R_1=R_3<R_2$
$\textbf{E}$
The expression for the resistance of a resistor is given as $R=\rho\dfrac{L}{A}$, So $R_1=R_3<R_2$.
In the history of physics, the names Nicolaus Copernicus, Galileo Galilei, James Clerk Maxwell, and Isaac Newton are very recognizable. Which one of the following choices correctly orders these names chronologically by the approximate dates of each person’s scientific work (ending with the most recent)?
$\textbf{(A) }$ Copernicus $\rightarrow$ Galileo $\rightarrow$ Newton $\rightarrow$ Maxwell$ \qquad\newline$
$\textbf{(B) }$ Copernicus $\rightarrow$ Galileo $\rightarrow$ Maxwell $\rightarrow$ Newton$ \qquad\newline$
$\textbf{(C) }$ Galileo $\rightarrow$ Copernicus $\rightarrow$ Newton $\rightarrow$ Maxwell$ \qquad\newline$
$\textbf{(D) }$ Galileo $\rightarrow$ Newton $\rightarrow$ Copernicus $\rightarrow$ Maxwell$ \qquad\newline$
$\textbf{(E) }$ Newton $\rightarrow$ Copernicus $\rightarrow$ Maxwell $\rightarrow$ Galileo
$\textbf{A}$
Copernicus published the heliocentric theory in 1512 while Galileo did a lot of work related to this theory about a century later. Isaac Newton started producing work in the 1660's while Maxwell's contributions to physics had to wait until the 1860's.
In the circuit shown, an electric current exists in the metal resistor connected to the battery. Which one of the following choices best describes the direction of electron flow through the resistor and the direction of the electric field through the interior of the resistor?
$\textbf{B}$
Since the conventional current in the circuit is directed to the right through the resistor, the electrons are flowing in the opposite direction (to the left). The electric field in the interior of the resistor would be in the direction of conventional current and hence points to the right.
A person wishes to accelerate an object upward uniformly. Which one of the following choices must be true of the force, $F$, provided by the person onto the object? $W$ represents the magnitude of the gravitational force acting on the object. Ignore any effects of the air or of the Earth’s rotation.
$\textbf{(A) }$ $F=W \qquad$ $\textbf{(B) }$ $F\ge W \qquad$ $\textbf{(C) }$ $F>W \qquad$ $\textbf{(D) }$ $F\ge 2W \qquad$ $\textbf{(E) }$ $F>2W$
$\textbf{C}$
By Newton’s Second Law, in order to accelerate an object, there must be an imbalance in the forces. To accelerate an object upward, a force must exist on the object that exceeds the gravitational force acting on it. Hence, the required force is $F>W$.
Questions 11 – 12 deal with the following information:
A 12.0 N net force directed to the right is exerted on a 3.0 kg block for 5.0 s. The block was initially at rest and slides on a horizontal surface.
What is the speed of the block at the end of the 5.0 s interval?
$\textbf{(A) }$ 12 m/s$ \qquad$ $\textbf{(B) }$ 15 m/s$ \qquad$ $\textbf{(C) }$ 20 m/s$ \qquad$ $\textbf{(D) }$ 36 m/s$ \qquad$ $\textbf{(E) }$ 60 m/s
$\textbf{C}$
By $F\Delta t=m\Delta v$, we get $v_f=\Delta v=\dfrac{F\Delta t}{m}=20\ \text{m/s}$.
What is the kinetic energy associated with the block after the 5.0 s interval?
$\textbf{(A) }$ 180 J$ \qquad$ $\textbf{(B) }$ 600 J$ \qquad$ $\textbf{(C) }$ 800 J$ \qquad$ $\textbf{(D) }$ 1200 J$ \qquad$ $\textbf{(E) }$ 1600 J
$\textbf{B}$
The kinetic energy $KR=\dfrac12mv^2=600$ J.
For the inclined plane shown, which one of the following choices best represents the value of the incline’s ideal mechanical advantage?
$\textbf{(A) }$ 1.67$ \qquad$ $\textbf{(B) }$ 1.33$ \qquad$ $\textbf{(C) }$ 1.25$ \qquad$ $\textbf{(D) }$ 0.80$ \qquad$ $\textbf{(E) }$ 0.75
$\textbf{A}$
The ideal mechanical advantage for an inclined plane is found as the length of the hypotenuse divided by the height. Hence, in this situation, we find $IMA=\dfrac53=1.67$.
A scientist obtains an answer that has units equivalent to those from computing the square root of the ratio of the Universal Gravitational constant to Coulomb's constant. Which one of the following choices has these same units?
$\textbf{(A) }$ energy divided by time$ \qquad\newline$
$\textbf{(B) }$ length divided by time$ \qquad\newline$
$\textbf{(C) }$ mass divided by energy$ \qquad\newline$
$\textbf{(D) }$ electric field strength divided by magnetic field strength$ \qquad\newline$
$\textbf{(E) }$ charge divided by mass
$\textbf{E}$
By $F=G\dfrac{mM}{R^2}$ and $F=k\dfrac{qQ}{R^2}$, the unit of universal gravitational constant is $\dfrac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$, and the unit of Coulomb's constant is $\dfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$, so the square root the of the ratio is $\sqrt{\dfrac{\text{C}^2}{\text{kg}^2}}=\dfrac{\text{C}}{\text{kg}}$, which is the unit of charge divided by mass.
A radio station broadcasts its signal at a frequency of 100.0 MHz. What is the wavelength of the station's signal?
$\textbf{(A) }$ 3.00 km$ \qquad$ $\textbf{(B) }$ 3.00 m$ \qquad$ $\textbf{(C) }$ 3.00 mm$ \qquad$ $\textbf{(D) }$ 0.33 m$ \qquad$ $\textbf{(E) }$
$\textbf{B}$
According to $c=\lambda f$, we have $f=\dfrac{c}{f}=3.00$ m.
A uniform stick is fixed to rotate about an axis through its center. The stick starts from rest and rotates through an angle of $90^\circ$ in a time of 1.0 s. If the angular acceleration of the stick is constant, what is the angular speed of the stick about its rotation axis after one full revolution?
$\textbf{(A) }$ $4\pi\ \text{rad/s} \qquad$ $\textbf{(B) }$ $2\sqrt2\pi\ \text{rad/s} \qquad$ $\textbf{(C) }$ $2\pi\ \text{rad/s} \qquad$ $\textbf{(D) }$ $\sqrt2\pi\ \text{rad/s} \qquad$ $\textbf{(E) }$ $\pi\ \text{rad/s}$
$\textbf{C}$
The stick rotates for $\theta=90^\circ=\dfrac12\pi\ \text{rad}$ in 1 s, According to $\theta=\dfrac12\beta t^2$, the angular acceleration can be found as $\beta=\dfrac{2\theta}{t^2}=\pi\ \text{rad/s}^2$. When rotating for $\theta_0=2\pi$, we have $\omega^2=2\beta\theta_0$, so the angular speed is $\omega=\sqrt{2\beta\theta_0}=2\pi\ \text{rad/s}$.
“All planets in the solar system follow elliptical orbits with the Sun located at one focus of the ellipse.” This statement is best summarized by which one of the following choices?
$\textbf{(A) }$ Kepler’s First Law$ \qquad\newline$
$\textbf{(B) }$ Kepler’s Second Law$ \qquad\newline$
$\textbf{(C) }$ Newton’s Universal Law of Gravitation$ \qquad\newline$
$\textbf{(D) }$ Einstein’s Law of Gravity$ \qquad\newline$
$\textbf{(E) }$ Schwarzchild’s Law of Orbits
$\textbf{A}$
Boy: I will always be with you, like the earth around the sun...$\newline$
Girl: You know the elliptical orbit has two focuses, all right?
For any motion in two dimensions, which one of the following choices identifies quantities that must have the same direction?
$\textbf{(A) }$ average velocity and displacement$ \qquad\newline$
$\textbf{(B) }$ average acceleration and average velocity$ \qquad\newline$
$\textbf{(C) }$ average acceleration and displacement$ \qquad\newline$
$\textbf{(D) }$ final velocity and average acceleration$ \qquad\newline$
$\textbf{(E) }$ final velocity and displacement
$\textbf{A}$
By definition, average velocity is displacement divided by time. Hence, these quantities must be directed the same way.
An ideal gas in a sealed container with $1.20\times10^{24}$ particles has a pressure of 1.0 atm at a temperature of $27.0^\circ C$. Which one of the following choices best represents the volume of the container?
$\textbf{(A) }$ 450 m$^3 \qquad$ $\textbf{(B) }$ 4.5 m$^3 \qquad$ $\textbf{(C) }$ 0.49 m$^3 \qquad$ $\textbf{(D) }$ 0.049 m$^3 \qquad$ $\textbf{(E) }$ 0.00049 m$^3$
$\textbf{D}$
The more number $n=\dfrac{N}{N_A}=2$ mol. By the ideal gas equation $pV=nRT$, the volume $V=\dfrac{nRT}{p}=0.049\ \text{m}^3$.
A uniform 10.0 m long plank is pivoted about its center of gravity. A mass $M_1$ then is placed at the left edge of the plank and a second mass $M_2$ is placed 1.0 m from the right end. This system is in static equilibrium (shown). If each mass now is moved 2.0 m closer to the center of the plank, which one of the following choices best describes the subsequent motion of the plank?
$\textbf{(A) }$ The plank remains in static equilibrium.$ \qquad\newline$
$\textbf{(B) }$ The plank rotates at constant angular speed until the right side reaches the ground.$ \qquad\newline$
$\textbf{(C) }$ The plank rotates at constant angular speed until the left side reaches the ground.$ \qquad\newline$
$\textbf{(D) }$ The plank angularly accelerates until the right side reaches the ground.$ \qquad\newline$
$\textbf{(E) }$ The plank angularly accelerates until the left side reaches the ground.
$\textbf{E}$
The torque exerted on the plank is balanced in the beginning, so $M_2>M_1$ since $M_2$ located closer to the pivot. When both objects moves 2.0 m closer to the pivot, the right side loses more torque because $M_2$ is heavier, so the plank will lean to the left.
At 1.94 s after launch, a projectile in free fall achieved its minimum speed during flight. If that minimum speed was 15.0 m/s, what was the projectile’s initial angle of launch from the horizontal?
$\textbf{(A) }$ $82.6^\circ \qquad$ $\textbf{(B) }$ $52.3^\circ \qquad$ $\textbf{(C) }$ $50.6^\circ \qquad$ $\textbf{(D) }$ $39.4^\circ \qquad$ $\textbf{(E) }$ $37.7^\circ$
$\textbf{B}$
The minimum speed achieved when the vertical component of the speed turned to zero, thus only horizontal component remained, so $v_\parallel=15.0\ \text{m/s}$. In the beginning the vertical component is $v_\perp=gt=19.4\ \text{m/s}$, so the initial angle $\theta=\arctan\dfrac{v_\perp}{v_\parallel}=\arctan\dfrac{19.4}{15}=52.3^\circ$.
For the following nuclear reaction, which one of the following choices correctly identifies the quantity labeled X? $$^{40}_{19}K+e^-\rightarrow X+^0_0\nu_e$$
$\textbf{(A) }$ $^{40}_{19}K \qquad$ $\textbf{(B) }$ $^{40}_{18}K \qquad$ $\textbf{(C) }$ $^{39}_{18}Ar \qquad$ $\textbf{(D) }$ $^{39}_{19}K \qquad$ $\textbf{(E) }$ $^{40}_{18}Ar$
$\textbf{E}$
In order to balance the reaction, we write $^{40}_{19}K+^{\ 0}_{-1}e^-\ \rightarrow\ ^A_ZX+^0_0\nu_e$. This means that we have $40+0=A+0\rightarrow A=40$. Further, $19+(-1)=Z+0\rightarrow Z=18$. By having $Z=18$, this means that the element is different from Potassium as that has 19 protons. In other words, from the choices provided, we see that $^A_ZX=^{18}_{40}Ar$.
At a party, two spherical balloons are expanded to have the same radius. Balloon 1 is filled with helium gas while balloon 2 is filled with xenon gas. Which one of the following statements about the buoyant forces on the balloons is correct?
$\textbf{(A) }$ The buoyant force is greater for the helium balloon.$ \qquad\newline$
$\textbf{(B) }$ The buoyant force is greater for the xenon balloon.$ \qquad\newline$
$\textbf{(C) }$ The buoyant force is the same for each balloon.$ \qquad\newline$
$\textbf{(D) }$ The balloon experiencing the greater buoyant force cannot be determined without knowing the radius.$ \qquad\newline$
$\textbf{(E) }$ The balloon experiencing the greater buoyant force cannot be determined without knowing the atmospheric pressure at the time.
$\textbf{C}$
Buoyant force is computed as the “weight of the fluid displaced.” The balloons have the same size and therefore displace the same amount of air. In other words, the buoyant forces are the same! The difference between the balloons would be seen on release when the helium balloon rises and the xenon balloon falls.
A proton moves with constant non-zero velocity in a region of space that has a uniform magnetic field directed into the plane of the page. The proton moves directly up the plane of the page. What is the direction of the electric field in this region of space? Ignore gravity.
$\textbf{(A) }$ To the left$ \qquad\newline$
$\textbf{(B) }$ To the right$ \qquad\newline$
$\textbf{(C) }$ Down the plane of the page$ \qquad\newline$
$\textbf{(D) }$ Up the plane of the page$ \qquad\newline$
$\textbf{(E) }$ There is no electric field necessary
$\textbf{C}$
For the proton to move with constant velocity, the net force on it must be zero. Since the proton moves at a right angle to the magnetic field, there is a magnetic force on the proton. By the right-hand rule, with the velocity up the plane of the page and the field into the plane, the magnetic force is directed to the left. In order to cancel this force, the electric force acting on the proton must be directed to the right, so the electric field is directed to the right.
The plates of a large parallel-plate capacitor are separated by a distance of 0.05 m. The potential difference between the plates is 24.0 V. A charge released from rest between the plates experiences an electric force of 1.00 N. What is the magnitude of the charge released between the plates?
$\textbf{(A) }$ 2.08 mC$ \qquad$ $\textbf{(B) }$ 4.17 mC$ \qquad$ $\textbf{(C) }$ 8.33 mC$ \qquad$ $\textbf{(D) }$ 16.6 mC$ \qquad$ $\textbf{(E) }$ 33.3 mC
$\textbf{A}$
The electric field between the plates $E=\dfrac{V}{d}=480\ \text{V/m}$, the force on the charge is $F=qE$, so we have $q=\frac{F}{E}=2.08\ \text{mC}$.
A 10.0 kg mass moves to the right at 8.00 m/s. A 5.0 kg mass moves to the left at 7.00 m/s. With what speed does the center of mass of this two-mass system move?
$\textbf{(A) }$ 7.67 m/s$ \qquad$ $\textbf{(B) }$ 7.50 m/s$ \qquad$ $\textbf{(C) }$ 3.00 m/s$ \qquad$ $\textbf{(D) }$ 2.00 m/s$ \qquad$ $\textbf{(E) }$ 0.50 m/s
$\textbf{C}$
The position of center of mass can be calculated as $$x_c=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$$
By differentiating both side of the equation we get the speed of center of mass $$v_c=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}=\dfrac{10\times8+5\times{-7}}{10+5}=3\ \text{m/s}$$
A tuning fork of frequency $f$ is placed over a long tube closed at one end producing the third lowest frequency standing wave for the tube. What frequency tuning fork would be needed to produce the fourth lowest frequency standing wave?
$\textbf{(A) }$ $\dfrac43f \qquad$ $\textbf{(B) }$ $\dfrac53f \qquad$ $\textbf{(C) }$ $\dfrac65f \qquad$ $\textbf{(D) }$ $\dfrac75f \qquad$ $\textbf{(E) }$ $\dfrac54f$
$\textbf{D}$
The standing wave exists when $L=(2n-1)\dfrac{\lambda}{4}$. So the third lowest frequency happens when $n=3$ and $\lambda_3=\dfrac45L$, the fourth lowest frequency happens when $n=4$ and $\lambda_4=\dfrac47L$, its frequency is $f_4=\dfrac{\lambda_3}{\lambda_4}f_3=\dfrac75f$.
Recent excitement in the physics community came from the February, 2016 announcement from LIGO that
$\textbf{(A) }$ alien life had been discovered on a distant planet.$ \qquad\newline$
$\textbf{(B) }$ dark matter had been discovered.$ \qquad\newline$
$\textbf{(C) }$ there was proof of dark energy’s existence.$ \qquad\newline$
$\textbf{(D) }$ gravitational waves had been detected.$ \qquad\newline$
$\textbf{(E) }$ objects had been observed to travel faster than the speed of light.
$\textbf{D}$
Call me when the other four are found.
A ray of light is incident onto a thin glass lens as shown. Which one of the following arrows best indicates the path of the refracted ray?
$\textbf{(A) }$ A$ \qquad$ $\textbf{(B) }$ B$ \qquad$ $\textbf{(C) }$ C$ \qquad$ $\textbf{(D) }$ D$ \qquad$ $\textbf{(E) }$ E
$\textbf{E}$
The light parallel to the principal axis of a converging lens will be refracted through the focus on the opposite side of the lens.
A scientist wishes to compute the amount of energy necessary to melt a known mass of a solid sample. Which quantity associated with the sample would the scientist be most interested in knowing?
$\textbf{(A) }$ Latent heat of fusion$ \qquad\newline$
$\textbf{(B) }$ Specific heat$ \qquad\newline$
$\textbf{(C) }$ Coefficient of thermal expansion$ \qquad\newline$
$\textbf{(D) }$ Latent heat of vaporization$ \qquad\newline$
$\textbf{(E) }$ Thermal conductivity
$\textbf{A}$
In order to melt a material, the property desired is the latent heat of fusion.
The circuit shown has 4 identical light bulbs connected to an ideal battery. If bulb $\#$2 were removed leaving an open circuit there, what happens to the voltage across both bulb $\#$1 and from Q to X?
$\textbf{E}$
Bulb 2 and 3 are in parallel with bulb 4 in the beginning. By removing bulb 2, the resistance of 2+3+4 increases, so the voltage across bulb 1 decreases and the voltage across bulb 4 increases.
A mass connected to a string forms a simple pendulum. The mass is released from rest and undergoes simple harmonic motion. Which one of the following choices correctly ranks the magnitude of the tension in the string (T), the gravitational force on the mass (W), and the net force acting on the mass (F) when the mass swings through its lowest point?
$\textbf{(A) }$ $F<W=T \qquad$ $\textbf{(B) }$ $W<T<F \qquad$ $\textbf{(C) }$ $W<F=T \qquad$ $\textbf{(D) }$ $F<W<T \qquad\newline$
$\textbf{(E) }$ More information is required to answer the question.
$\textbf{D}$
Since the mass is in simple harmonic motion, the angle from the vertical upon first release would be fairly small ($<15^\circ$). When the mass swings through its lowest location, the acceleration of the mass would be straight up to the pivot of the pendulum. This means that $W<T$ to have more upward force compared to downward force. To check on the size of the net force, let us assume that it is equal to the gravitational force and determine the angle at which the mass is released. In other words, $$F=W=mg=m\dfrac{v^2}{L}$$ where $L$ is the length of the pendulum. From mechanical energy conservation during the fall of the mass to the lowest point from the highest, we write $$mgL(1-\cos\theta)=\dfrac12mv^2$$ where $\theta$ is the maximum angle that the string makes with the vertical. By solving the two equations we get $\cos\theta=\dfrac12$ or $\theta=60^\circ$, which is much greater than the small angle assumption. In other words, since we are nowhere near that initial angle, $F<W$ and $F<W<T$.
Given the configuration of three charges, $+Q$, $+Q$, and $-2Q$, which one of the following choices best represents the direction of the electric field at the point P and the sign of the electric potential at P from these charges?
$\textbf{B}$
The electric field of $+Q$ on point P is $\dfrac{kQ}{a^2}$, so the sum of electric fields of two $+Q$ charges is $\sqrt2\dfrac{kQ}{a^2}$ pointing top left. The electric field of $-2Q$ at P is $\dfrac{2kQ}{(\sqrt2a)^2}=\dfrac{kQ}{a^2}$ pointing bottom right, which is less than the electric field of the two positive charges, so the direction of the total electric field at P is pointing top left. The potential at P is $\dfrac{kQ}{a}+\dfrac{kQ}{a}-\dfrac{2kQ}{\sqrt2a}=(2-\sqrt2)\dfrac{kQ}{a}>0$, so the sign of electric potential at P is positive.
The graph shows the magnitude of the linear momentum of two solid objects colliding on the x-axis. Which one of the following statements must be true based on the information provided?
$\textbf{(A) }$ The objects remain stuck together after the collision.$ \qquad\newline$
$\textbf{(B) }$ Object 2 experiences the greater force from the collision.$ \qquad\newline$
$\textbf{(C) }$ Object 2 cannot have more mass than object 1.$ \qquad\newline$
$\textbf{(D) }$ The kinetic energy loss was maximized with no external forces present in the two-object system.$ \qquad\newline$
$\textbf{(E) }$ The kinetic energy of the two-object system is the same both before and after the collision.
$\textbf{C}$
Object 1 loses part of its linear momentum after collision, its speed becomes lower, while the speed of object 2 increases. So object 1 must be catching up to object 2 before collision, thus the speed of object 1 after collision should be no greater than object 2. Since the linear momentum after collision is the same for both objects, the mass of object 1 is no less than object 2.
Which one of the following choices best describes what it means for an object to be iridescent?
$\textbf{(A) }$ The color of the object appears to change with a change in the angle of view.$ \qquad\newline$
$\textbf{(B) }$ The object appears to have two images formed from it because of differences in index of refraction.$ \qquad\newline$
$\textbf{(C) }$ The color of the object appears to be different based on its temperature.$ \qquad\newline$
$\textbf{(D) }$ The object remains illuminated after turning off all surrounding light sources.$ \qquad\newline$
$\textbf{(E) }$ The object is alive and producing light biochemically.
$\textbf{A}$
Iridescence has to do with the angle at which one is looking at an object and that the color of the object appears to change with it.
On a straight road, a self-driving car races with a robot. The robot accelerates at $8.50\ \text{m/s}^2$ until reaching its maximum constant speed of $15.0\ \text{m/s}$. The car accelerates at $5.60\ \text{m/s}^2$ until reaching its maximum constant speed of $25.0\ \text{m/s}$. The car and robot start from rest from the same location. What is the maximum distance that the robot leads the car during the race?
$\textbf{(A) }$ 3.3 m$ \qquad$ $\textbf{(B) }$ 6.9 m$ \qquad$ $\textbf{(C) }$ 10.2 m$ \qquad$ $\textbf{(D) }$ 13.2 m$ \qquad$ $\textbf{(E) }$ 17.3 m
$\textbf{B}$
The maximum distance is achieved when the car accelerated to the same velocity $15.0\ \text{m/s}$ as the robot can do. The robot needs $t_1=\dfrac{v_1}{a_1}=\dfrac{15}{8.5}\ \text{m/s}^2$ to reach $15\ \text{m/s}$, while the car needs $t_2=\dfrac{v_1}{a_2}=\dfrac{15}{5.6}\ \text{m/s}^2$. At $t_2$, the robot moved for $x_1=\dfrac12a_1t_1^2+v_1(t_2-t_1)$, the car drove for $x_2=\dfrac12a_2t_2^2$. So the maximum distance is $\Delta x=x_2-x_1$.
Questions 37 and 38 deal with the following information:
A string is connected to a mechanical oscillator on one end and to a cube-shaped mass, $M=8.0\ \text{kg}$, at the other end as shown. The oscillator vibrates the string with a frequency of $f$ producing the standing wave in the figure on the left. When the mass is submerged completely in water, the string vibrates in the standing wave pattern shown on the right.
What is the value of the ratio for the tension when the mass is submerged in water to the tension when the mass is hanging in air?
$\textbf{(A) }$ $4 \qquad$ $\textbf{(B) }$ $2 \qquad$ $\textbf{(C) }$ $\dfrac{1}{\sqrt2} \qquad$ $\textbf{(D) }$ $\dfrac12 \qquad$ $\textbf{(E) }$ $\dfrac14$
$\textbf{E}$
The wavelength on the right figure is half the wavelength on the left, thus $\lambda_2=\dfrac12\lambda_1$. According to $v=\lambda f$, since the frequency does not change, we have $v_2=\dfrac12 v_1$. The speed of wave on a stretched string is determined by $v=\sqrt{\dfrac{T}{\mu}}$, so the ratio of tension is $T_2=\dfrac14T_1$.
What is the length of one side of the cube of mass $M$?
$\textbf{(A) }$ 0.006 m$ \qquad$ $\textbf{(B) }$ 0.091 m$ \qquad$ $\textbf{(C) }$ 0.126 m$ \qquad$ $\textbf{(D) }$ 0.182 m$ \qquad$ $\textbf{(E) }$ 0.200 m
$\textbf{D}$
The tension $T_2=\dfrac14T_1=\dfrac14Mg$ when the cube is immersed in water, so the buoyant force $\rho_wVg=\dfrac34Mg=\dfrac34\rho Vg$, thus we get the density of the cube $\rho=\dfrac43\rho_w=1.33\times10^3\ \text{kg/m}^3$. The mass of the cube $M=\rho V=\rho L^3$, so the length of the cube is $L=\sqrt[3]{\dfrac{M}{\rho}}=0.182\ \text{m}$.
Questions 39 and 40 deal with the following information:
An object moves in a circle, starting at the top, with initial speed $17.0\ \text{m/s}$. The object’s speed increases uniformly until it has moved counterclockwise through an angle of $55^\circ$. The average acceleration for this motion is $9.8\ \text{m/s}^2$ directed straight downward.
What is the final speed of the object?
$\textbf{(A) }$ 29.6 m/s$ \qquad$ $\textbf{(B) }$ 26.8 m/s$ \qquad$ $\textbf{(C) }$ 20.8 m/s$ \qquad$ $\textbf{(D) }$ 18.4 m/s$ \qquad$ $\textbf{(E) }$ 17.1 m/s
$\textbf{A}$
The initial speed $v_i=17\ \text{m/s}$ points left. The change of velocity has the same direction as the average acceleration, so the triangle shown in the figure is right-angled. The speed of the object $v=\dfrac{v_i}{\cos55^\circ}$.
What is the time for this motion?
$\textbf{(A) }$ 1.54 s$ \qquad$ $\textbf{(B) }$ 1.80 s$ \qquad$ $\textbf{(C) }$ 2.26 s$ \qquad$ $\textbf{(D) }$ 2.48 s$ \qquad$ $\textbf{(E) }$ 2.93 s
$\textbf{D}$
The change of velocity $\Delta v=v_i\tan55^\circ$, and we know the average acceleration $a$, so the time is $t=\dfrac{\Delta v}{a}$.
A small object is launched $30^\circ$ above the horizontal from a height of $10.0\ \text{m}$ above the ground. If the initial speed of the object is 15.0 m/s, at what angle below the horizontal is the object moving when it reaches the ground? Ignore air resistance.
$\textbf{(A) }$ $36.1^\circ \qquad$ $\textbf{(B) }$ $39.1^\circ \qquad$ $\textbf{(C) }$ $43.4^\circ \qquad$ $\textbf{(D) }$ $50.9^\circ \qquad$ $\textbf{(E) }$ $68.5^\circ$
$\textbf{D}$
The vertical component of velocity when reaching the ground can be calculated from $v_{f\perp}^2-v_{i\perp}^2=2gh$, where $v_{i\perp}=v_0\sin30^\circ$. The horizontal component remains the same, so $v_{f\parallel}=v_{i\parallel}=v_0\cos30^\circ$. The angle of the final velocity is $\theta=\arctan\dfrac{v_{f\perp}}{v_{f\parallel}}$.
An object (which points upward) is placed 21.0 cm from a converging lens of focal length 18.0 cm. Which one of the following choices is true about the image formed by the lens?
$\textbf{(A) }$ The image is larger than the object, virtual, and points upward.$ \qquad\newline$
$\textbf{(B) }$ The image is larger than the object, real, and points downward.$ \qquad\newline$
$\textbf{(C) }$ The image is smaller than the object, virtual, and points upward.$ \qquad\newline$
$\textbf{(D) }$ The image is smaller than the object, real, and points upward.$ \qquad\newline$
$\textbf{(E) }$ The image is smaller than the object, real, and points downward.
$\textbf{B}$
By the lens equation $\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac1f$, we get $d_i=126\ \text{cm}$, so the magnification $M=-\dfrac{d_i}{d_o}=-6$. As the magnification is negative, this makes the image real and inverted. Since $\lvert M\rvert>1$, the image is also larger than the object.
A uniform beam connected to a wall at the hinge is in static equilibrium as shown. The cable connected to the beam is massless. What are the directions of the horizontal and vertical components of force acting at the pivot on the beam?
$\textbf{C}$
As the object is in static equilibrium, the sum of forces and the sum of torques must be zero. The force from the cable is upward and rightward. Because the gravitational force is downward, the pivot must be providing a force to the left in order to cancel the rightward force from the cable. As for the vertical component, by choosing an axis of rotation through the direction of the gravitational force at the vertical location of the pivot, the only non-zero torques are from the cable and the vertical component of the pivot force. As the torque from the cable would be oriented counterclockwise from this axis of rotation, the vertical force at the pivot needs to provide a clockwise torque. This is possible only if that vertical component is upward.
A proton moves in a circular orbit in a uniform magnetic field, $B$. A helium nucleus moves in a circular orbit in the same magnetic field. If each particle experiences the same magnitude magnetic force during its motion, what is the ratio of the speed of the helium to the speed of the proton?
$\textbf{(A) }$ 4:1$ \qquad$ $\textbf{(B) }$ 1:4$ \qquad$ $\textbf{(C) }$ 1:1$ \qquad$ $\textbf{(D) }$ 2:1$ \qquad$ $\textbf{(E) }$ 1:2
$\textbf{E}$
The magnetic force is $F=qvB$. The ratio $\dfrac{F_{He}}{F_p}=\dfrac{q_{He}v_{He}B}{q_pv_pB}=1$, so $\dfrac{v_{He}}{v_p}=\dfrac{q_p}{q_{He}}=\dfrac12$.
Light of wavelength 250 nm shines onto a metallic surface. It is known that the electrons ejected from the surface range in speed as $0\leq v\leq4.85\times10^5\ \text{m/s}$. What is the work function of the surface?
$\textbf{(A) }$ 4.97 eV$ \qquad$ $\textbf{(B) }$ 1.60 eV$ \qquad$ $\textbf{(C) }$ 3.63 eV$ \qquad$ $\textbf{(D) }$ 4.30 eV$ \qquad$ $\textbf{(E) }$ 1.34 eV
$\textbf{D}$
The maximum kinetic energy of emitted electrons $$KE=\dfrac12mv^2=\dfrac12(9.1\times10^{-31})(4.85\times10^5)^2=1.07\times10^{-19}\ \text{J}=0.67\ \text{eV}$$ The energy of incoming photons $$E=\dfrac{hc}{\lambda}=7.95\times10^{-19}\ \text{J}=4.97\ \text{eV}$$ So the working function of the surface $$W=E-KE=4.30\ \text{eV}$$
Two spherical, non-rotating planets, X and Y, have the same uniform density $\rho$. Planet X has twice the radius of Planet Y. Let $v_X$ and $v_Y$ represent the escape speed at the surfaces of Planet X and Planet Y, respectively. What is the ratio of $v_X:v_Y$?
$\textbf{(A) }$ 2:1$ \qquad$ $\textbf{(B) }$ 1:2$ \qquad$ $\textbf{(C) }$ 1:1$ \qquad$ $\textbf{(D) }$ 4:1$ \qquad$ $\textbf{(E) }$ 1:4
$\textbf{A}$
The escape speed is the minimum speed to escape from a planet. It can be calculated from $$\dfrac12mv^2-G\dfrac{Mm}{R}=0$$ so $v=\sqrt{\dfrac{2GM}{R}}$. The mass $M=\dfrac43\pi R^3\rho$. Since the densities for X and Y are the same, and $R_X=2R_Y$, so $M_X=8M_Y$, the ratio $v_X:v_Y=2:1$.
Two small loops of wire are located close to a wire with conventional current directed up the page. A scientist then moves the loops away from the current as shown. What is the orientation of the conventional current in each loop while this occurs?
$\textbf{C}$
Lenz’s Law is required to answer this question. The magnetic field associated with the long wire is into the page to the right of the wire and out of the page to the left of the wire. As the loop on the right is moved away, there are weaker field lines penetrating into the loop. As a result, there is an induced current oriented to try to replace the missing field lines. This means that the current in the right loop is oriented clockwise to put field lines into the plane in the interior of the loop. Likewise, on the left side of the long wire, there are again weaker field lines coming out of the plane of the loop. Here, the induction would be for a current oriented counterclockwise to have additional field lines out of the plane through the interior of the loop.
From a given state of $P$, $V$, and $T$ for an ideal gas, which one of the following reversible processes has the most heat associated with it?
$\textbf{(A) }$ Isothermal expansion doubling the volume$ \qquad\newline$
$\textbf{(B) }$ Isobaric expansion doubling the volume$ \qquad\newline$
$\textbf{(C) }$ Isovolumic pressure doubling$ \qquad\newline$
$\textbf{(D) }$ Adiabatic expansion doubling the volume$ \qquad\newline$
$\textbf{(E) }$ Isobaric compression halving the volume
$\textbf{B}$
$Q=\Delta U+W$, so the expansion with the increase of temperature need more heat.
John is at rest on a platform that he measures to be 500 m long on the x-axis. He then sees a spaceship moving along the x-axis with a speed of $1.80\times10^8\ \text{m/s}$ as it crosses the length of the platform. Sitting in her chair, the spaceship's pilot measures the time it takes her to cross the platform. What type of time interval does she measure and what value does she obtain?
$\textbf{A}$
Because the spaceship pilot is in one location, she has a single resting clock. This means that the time she records to cross the platform will be a proper time. From the pilot’s perspective, the platform is length contracted to be of length $L=L_0\sqrt{1-\dfrac{v^2}{c^2}}=400\ \text{m}$. So her clock will read a time of $\Delta t=\dfrac{L}{v}=2.22\ \mu\text{s}$.
A uniform stick of mass $M$ and length $L$ is fixed to rotate about a frictionless pivot located $L/3$ from one end. The stick is released from rest incrementally away from being perfectly vertical, resulting in the stick rotating clockwise about its pivot. When the stick is horizontal, what is the tangential speed of the center of mass about the pivot?
$\textbf{(A) }$ $\dfrac{1}{\sqrt{15}}\sqrt{gL} \qquad$ $\textbf{(B) }$ $\sqrt{\dfrac{3}{10}}\sqrt{gL} \qquad$ $\textbf{(C) }$ $\dfrac{1}{\sqrt{10}}\sqrt{gL} \qquad$ $\textbf{(D) }$ $\dfrac{3}{\sqrt{20}}\sqrt{gL} \qquad$ $\textbf{(E) }$ $\dfrac{1}{\sqrt{12}}\sqrt{gL}$
$\textbf{E}$
The center of mass is $L/6$ above the pivot in the beginning. When the stick is horizontal, the change of potential energy is $$\Delta PE=-Mg\cdot\dfrac{L}{6}$$ the change of kinetic energy is $$\Delta KE=\dfrac12Mv_c^2+\dfrac12I_c\omega^2$$ where the center of mass moves in speed $v_c=\omega\cdot\dfrac{L}{6}$, and the moment of inertia to the center is $I_0=\dfrac{1}{12}ML^2$. The mechanical energy is conserved, so $$\Delta KE+\Delta PE=0$$ here we get $\omega=\sqrt{\dfrac{3g}{L}}$. The speed of the center of mass is $v_c=\omega\cdot\dfrac{L}{6}=\dfrac{1}{\sqrt{12}}\sqrt{gL}$.