PhysicsBowl 2014

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

Question 1

An FM radio station sends a signal with a frequency of $99.99\times10^6\ \text{Hz}$. Which one of the following choices best represents this frequency expressed using metric prefixes?

$\textbf{(A) }$ 99.99 kHz$ \qquad$ $\textbf{(B) }$ 99.99 MHz$ \qquad$ $\textbf{(C) }$ 99.99 GHz$ \qquad$ $\textbf{(D) }$ 99.99 THz$ \qquad$ $\textbf{(E) }$ 99.99 nHz

Solution

$\textbf{B}$

$10^6$ is represented by Mega which has a symbol of M.

Question 2

In the laboratory, a student makes the following six measurements for the length of an object: 5.05 cm, 5.06 cm, 5.07 cm, 5.06 cm, 5.07 cm, and 5.09 cm. Using the rules of significant digits, which one of the following choices correctly represents how she should express the average length of the object?

$\textbf{(A) }$ 5 cm$ \qquad$ $\textbf{(B) }$ 5.06 cm$ \qquad$ $\textbf{(C) }$ 5.06$\overline{6}$ cm$ \qquad$ $\textbf{(D) }$ 5.07 cm$ \qquad$ $\textbf{(E) }$ 5.1 cm

Solution

$\textbf{D}$

Averaging the values gives 5.06666 cm. Using the rules of significant digits, we can only record the answer out to the hundredths place. Hence, we round to obtain 5.07 cm.

Question 3

An object is dropped into free fall. Through how many meters does the object fall during the first 3.00 seconds of flight?

$\textbf{(A) }$ 10.0 m$ \qquad$ $\textbf{(B) }$ 15.0 m$ \qquad$ $\textbf{(C) }$ 30.0 m$ \qquad$ $\textbf{(D) }$ 45.0 m$ \qquad$ $\textbf{(E) }$ 90.0 m

Solution

$\textbf{D}$

$h=\dfrac12gt^2=45\ \text{m}$

Question 4

Three equal masses are suspended from a classroom ceiling by a series of strings as shown in the figure. Which string has the greatest tension?

$\textbf{(A) }$ Only String A$ \qquad\newline$
$\textbf{(B) }$ Only String B$ \qquad\newline$
$\textbf{(C) }$ Only String C$ \qquad\newline$
$\textbf{(D) }$ Strings A, B, and C have the same non-zero tension.$ \qquad\newline$
$\textbf{(E) }$ Strings A, B, and C all have no tension.

Solution

$\textbf{A}$

Taking three masses as a system, string A will carry the whole system. So it has the greatest tension.

Question 5

A simple pendulum consists of a massive bob connected to the end of a very light string. Which one of the following changes should be made in order to increase the period of the pendulum? Ignore air resistance.

$\textbf{(A) }$ Increase the mass of the bob$ \qquad\newline$
$\textbf{(B) }$ Decrease the mass of the bob$ \qquad\newline$
$\textbf{(C) }$ Increase the length of the string$ \qquad\newline$
$\textbf{(D) }$ Decrease the length of the string$ \qquad\newline$
$\textbf{(E) }$ Decrease the maximum angle of the pendulum’s oscillation

Solution

$\textbf{C}$

The period $T=2\pi\sqrt{\dfrac{l}{g}}$. So increasing the length of the string will increase the period.

Question 6

MRI is a commonly used acronym for a medical diagnostic technique. MRI stands for which one of the following choices?

$\textbf{(A) }$ Medical Radio Imaging$ \qquad\newline$
$\textbf{(B) }$ Minimally Radioactive Intervention$ \qquad\newline$
$\textbf{(C) }$ Multiaxis Radar Injection$ \qquad\newline$
$\textbf{(D) }$ Magnetic Radioisotope Injection$ \qquad\newline$
$\textbf{(E) }$ Magnetic Resonance Imaging

Solution

$\textbf{E}$

Bonus question: What is MAGA 🙂

Question 7

Starting from rest, a cart uniformly accelerates to a speed of 7.60 m/s in a time of 3.00 s. Through what distance does the cart move in this time?

$\textbf{(A) }$ 5.7 m$ \qquad$ $\textbf{(B) }$ 8.1 m$ \qquad$ $\textbf{(C) }$ 11.4 m$ \qquad$ $\textbf{(D) }$ 16.1 m$ \qquad$ $\textbf{(E) }$ 22.8 m

Solution

$\textbf{C}$

The average speed is the speed at 1.5 s, so $\overline{v}=3.8\ \text{m/s}$. The distance is $x=3.8\ \text{m/s}\times3\ \text{s}=11.4\ \text{m}$.

Question 8

A 2.50 kg mass connected to the end of an ideal spring oscillates in simple harmonic motion. The mass’s position is described as a function of time by $x(t)=0.20\cos(8.00t+0.50)$ where all quantities are in base SI units. Which one of the following choices gives the numerical value of the oscillation’s amplitude in base SI units?

$\textbf{(A) }$ 8.00$ \qquad$ $\textbf{(B) }$ 4.00$ \qquad$ $\textbf{(C) }$ 1.60$ \qquad$ $\textbf{(D) }$ 0.50$ \qquad$ $\textbf{(E) }$ 0.20

Solution

$\textbf{E}$

The amplitude of motion for a simple harmonic oscillator is found from the coefficient in front of the sinusoidal function. This means that the value 0.20 gives the amplitude of the motion.

Question 9

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

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

Solution

$\textbf{A}$

Force is a vector because it has both magnitude and direction.

Question 10

Two charges, $-Q$ and $+Q$, are fixed in place on the x-axis, each a distance $a$ from the origin as shown. At the point labeled P, a distance $d$ along the y-axis from the origin, what is the direction of the electric field from the given charges?

$\textbf{(A) }$ Up the plane of the page$ \qquad\newline$
$\textbf{(B) }$ Down the plane of the page$ \qquad\newline$
$\textbf{(C) }$ To the right$ \qquad\newline$
$\textbf{(D) }$ To the left$ \qquad\newline$
$\textbf{(E) }$ There is no electric field

Solution 10

$\textbf{D}$

The electric field by $-Q$ at P points toward $-Q$, while the electric field by $+Q$ points away from $+Q$. Due to symmetry, the vertical components of electrics fields cancel each other. So the total field points left.

Question 11

A crate gains 36.0 J of kinetic energy while its speed is increased from 2.00 m/s to 4.00 m/s. Which one of the following choices best represents the mass of the crate?

$\textbf{(A) }$ 36.0 kg$ \qquad$ $\textbf{(B) }$ 18.0 kg$ \qquad$ $\textbf{(C) }$ 6.0 kg$ \qquad$ $\textbf{(D) }$ 3.0 kg$ \qquad$ $\textbf{(E) }$ 1.5 kg

Solution

$\textbf{C}$

The change of kinetic energy $\Delta KE=\dfrac12mv_f^2-\dfrac12m_i^2$. By solving the equation we get $m=6\ \text{kg}$.

Question 12

A box of mass 5.0 kg is being pushed to the right across a horizontal surface while a constant frictional force of 8.0 N acts on the box. At some instant of time, the box has a speed of 4.0 m/s and an acceleration of 3.0 m/s$^2$. What is the magnitude of the net force acting on the box at this instant?

$\textbf{(A) }$ 7.0 N$ \qquad$ $\textbf{(B) }$ 12.0 N$ \qquad$ $\textbf{(C) }$ 15.0 N$ \qquad$ $\textbf{(D) }$ 20.0 N$ \qquad$ $\textbf{(E) }$ 23.0 N

Solution

$\textbf{C}$

The net force $F_{net}=ma=15$ N.

Question 13

Which one of the following scientists is most associated with the following statement: “For small displacements of an object from equilibrium, there is a restoring force that is proportional to the displacement.”?

$\textbf{(A) }$ Einstein$ \qquad$ $\textbf{(B) }$ Hooke$ \qquad$ $\textbf{(C) }$ Huygens$ \qquad$ $\textbf{(D) }$ Kepler$ \qquad$ $\textbf{(E) }$ Lenz

Solution

$\textbf{B}$

Hooke: It is called Hooke’s Law.$\newline$
Newton: Eww...That's what you called a “law”... t( -$\_$- t )

Question 14

A box rests on the floor of an elevator. The elevator is accelerating upward. Which one of the following choices best represents the Newton’s Third Law pair force to the gravitational force acting on the box by the Earth?

$\textbf{(A) }$ There is no Newton’s Third Law pair force in this scenario.$ \qquad\newline$
$\textbf{(B) }$ The entire normal force of contact from the floor on the box.$ \qquad\newline$
$\textbf{(C) }$ Only a portion of the normal force of contact from the floor on the box.$ \qquad\newline$
$\textbf{(D) }$ The force of the cables pulling upward on the elevator.$ \qquad\newline$
$\textbf{(E) }$ The gravitational force acting on the Earth by the box.

Solution

$\textbf{E}$

When the earth acts a force on the box, the box also acts a force on the earth.

Question 15

Two objects, A and B, move in space and then collide.$\newline$
Before collision: Object A, of mass 5.0 kg, moves to the right with a speed of 25.0 m/s. Object B, of mass 10.0 kg, moves to the left with a speed of 20.0 m/s.$\newline$
After collision: Object A moves to the left with a speed of 25.0 m/s.$\newline$
What is the velocity of object B after the collision?

$\textbf{(A) }$ 30.0 m/s to the right$ \qquad\newline$
$\textbf{(B) }$ 20.0 m/s to the right$ \qquad\newline$
$\textbf{(C) }$ 20.0 m/s to the left$ \qquad\newline$
$\textbf{(D) }$ 5.0 m/s to the right$ \qquad\newline$
$\textbf{(E) }$ 5.0 m/s to the left

Solution

$\textbf{D}$

The linear momentum is conserved. If we take “to the right” as positive, then $v_{Ai}=25\ \text{m/s}$, $v_{Bi}=-20\ \text{m/s}$, $v_{Af}=-25\ \text{m/s}$. By $m_Av_{Ai}+m_Bv_{Bi}=m_Av_{Af}+m_Bv_{Bf}$, we get $v_{Bf}=5\ \text{m/s}$. Note that the sign is positive indicating motion to the right.

Questions 16 and 17 refer to the following information:

A toy car initially moves to the right at 60.0 cm/s. Five seconds later, the car is moving at 40.0 cm/s to the left. The total displacement of the car during this time is 10.0 cm to the left of where it started.

Question 16

Which one of the following choices best represents the magnitude of the average velocity of the car during the five second motion?

$\textbf{(A) }$ 50.0 cm/s$ \qquad$ $\textbf{(B) }$ 10.0 cm/s$ \qquad$ $\textbf{(C) }$ 4.0 cm/s$ \qquad$ $\textbf{(D) }$ 2.0 cm/s$ \qquad$ $\textbf{(E) }$ 0.40 cm/s

Solution

$\textbf{D}$

The average velocity $\overline{v}=\dfrac{\Delta x}{\Delta t}=2.0\ \text{cm/s}$.

Question 17

Which one of the following choices best represents the magnitude of the average acceleration of the car during the five second motion?

$\textbf{(A) }$ 20.0 cm/s$^2 \qquad$ $\textbf{(B) }$ 4.0 cm/s$^2 \qquad$ $\textbf{(C) }$ 2.00 cm/s$^2 \qquad$ $\textbf{(D) }$ 0.80 cm/s$^2 \qquad$ $\textbf{(E) }$ 0.40 cm/s$^2$

Solution

$\textbf{A}$

The average acceleration $\overline{a}=\dfrac{\Delta v}{\Delta t}=\dfrac{60-(-40)}{5}=20\ \text{cm/s}^2$.

Question 18

A rectangular block of wood has a mass of 17.8 g and dimensions of length: 3.00 cm, width: 4.00 cm, and height: 2.00 cm. Which one of the following choices correctly gives the average density of the block?

$\textbf{(A) }$ $7.42\times10^5\ \text{kg/m}^3 \qquad\newline$ $\textbf{(B) }$ $7.42\times10^2\ \text{kg/m}^3 \qquad\newline$ $\textbf{(C) }$ $7.42\ \text{kg/m}^3 \qquad\newline$ $\textbf{(D) }$ $7.42\times10^{-1}\ \text{kg/m}^3 \qquad\newline$ $\textbf{(E) }$ $7.42\times10^{-4}\ \text{kg/m}^3$

Solution

$\textbf{B}$

The density $\rho=\dfrac{m}{V}=\dfrac{17.8}{4\times3\times2}\ \text{g/cm}^3=0.742\ \text{g/cm}^3=7.42\times10^2\ \text{kg/m}^3$

Question 19

In 2013, the Nobel Prize in physics was awarded "for the theoretical discovery of a mechanism that contributes to our understanding of the origin of mass of subatomic particles, and which recently was confirmed through the discovery of the predicted fundamental particle, by the ATLAS and CMS experiments at CERN's Large Hadron Collider". The winners were François Englert and

$\textbf{(A) }$ Stephen Hawking$ \qquad\newline$ $\textbf{(B) }$ Edward Witten$ \qquad\newline$ $\textbf{(C) }$ Edwin Hubble$ \qquad\newline$ $\textbf{(D) }$ Brian Greene$ \qquad\newline$ $\textbf{(E) }$ Peter Higgs

Solution

$\textbf{E}$

The Nobel Prize was awarded for the experimental verification of the Higgs Boson, named after Peter Higgs.

Question 20

A uniform block of mass $m$ is placed on an inclined plane of angle $\theta$. When released, the block does not move. It is determined that the normal force acting from the incline on the block has a magnitude of 62 N while the force of static friction acting on the block has a magnitude of 38 N. The coefficient of static friction between the block and inclined plane is $\mu_s=0.92$. Which one of the following choices best represents the mass, $m$, of the block?

$\textbf{(A) }$ 2.4 kg$ \qquad$ $\textbf{(B) }$ 6.2 kg$ \qquad$ $\textbf{(C) }$ 7.3 kg$ \qquad$ $\textbf{(D) }$ 8.4 kg$ \qquad$ $\textbf{(E) }$ 10.0 kg

Solution

$\textbf{C}$

Noting that the object is in static equilibrium, the sum of the vector forces is equal to zero. This means that the sum of the normal and friction forces must be equal and opposite to the gravitational force. The size of the force can be found from the Pythagorean Theorem, since the normal and friction forces are at right angles to each other, to give $mg=\sqrt{f_s^2+n^2}=72.7\ \text{N}$. Consequently, the mass is 7.27 kg.

Question 21

A tuning fork placed over a 57 cm column of air closed only at its bottom end produces a standing wave in the 3rd harmonic. The speed of sound in air is 342 m/s. What is the frequency of the tuning fork?

$\textbf{(A) }$ 150 Hz$ \qquad$ $\textbf{(B) }$ 300 Hz$ \qquad$ $\textbf{(C) }$ 450 Hz$ \qquad$ $\textbf{(D) }$ 600 Hz$ \qquad$ $\textbf{(E) }$ 900 Hz

Solution

$\textbf{C}$

For a tube closed at the bottom, only odd harmonics exist. The formula for standing waves in a closed tube is $L=n\dfrac{\lambda}{4}$. By $v=\lambda f$, when $n=3$, $f=\dfrac{3v}{2L}=450\ \text{Hz}$.

Question 22

For the circuit shown, all three light bulbs have the same resistance. The battery and wires have no resistance. What is the proper ranking of the bulbs’ brightness?

$\textbf{(A) }$ Bulb 1 = Bulb 2 = Bulb 3$ \qquad\newline$
$\textbf{(B) }$ Bulb 3 $<$ Bulb 2 = Bulb 1$ \qquad\newline$
$\textbf{(C) }$ Bulb 2 $<$ Bulb 1 $<$ Bulb 3$ \qquad\newline$
$\textbf{(D) }$ Bulb 1 = Bulb 2 $<$ Bulb 3$ \qquad\newline$
$\textbf{(E) }$ Bulb 1 = Bulb 3 $<$ Bulb 2

Solution

$\textbf{D}$

Bulb 1 and 2 are in parallel, so the currents passing through them are the same since they have the same resistance. All currents will pass through Bulb 3, so Bulb 1 = Bulb 2 $<$ Bulb 3.

Question 23

Two small identical coins (labeled X and Y) are at rest on a horizontal disk rotating at a constant rate about an axis perpendicular to the plane of the disk and through its center. The distance of the coins from the center of disk is related by $d_X=\dfrac12d_Y$. Which one of the following choices correctly identifies the relationship between $f_X$ and $f_Y$, the frictional force on coin X and on coin Y, respectively?

$\textbf{(A) }$ $f_X=\dfrac14f_Y \qquad$ $\textbf{(B) }$ $f_X=\dfrac12f_Y \qquad$ $\textbf{(C) }$ $f_X=f_Y \qquad$ $\textbf{(D) }$ $f_X=2f_Y \qquad$ $\textbf{(E) }$ $f_X=4f_Y$

Solution

$\textbf{B}$

The frictional force provides the centripetal force as $f=ma=m\omega^2r$. By $d_X=\dfrac12d_Y$, $f_X=\dfrac12f_Y$.

Question 24

For the bar magnet shown in the figure, which choice best describes the direction of the magnetic field at the point P located directly above the center of the magnet?

$\textbf{(A) }$ Up the plane of the page$ \qquad\newline$
$\textbf{(B) }$ To the right$ \qquad\newline$
$\textbf{(C) }$ Down the plane of the page$ \qquad\newline$
$\textbf{(D) }$ To the left$ \qquad\newline$
$\textbf{(E) }$ Out of the plane of the page

Solution

$\textbf{D}$

Magnetic field lines point from North to South poles outside the magnet. As a result, the field points directly to the left above the middle of the bar magnet shown.

Question 25

It is observed that a light ray changes direction when it enters a new material. Which one of the following choices is the term best associated with this phenomenon?

$\textbf{(A) }$ Doppler Effect $ \qquad\newline$ $\textbf{(B) }$ Interference$ \qquad\newline$ $\textbf{(C) }$ Polarization$ \qquad\newline$ $\textbf{(D) }$ Diffraction$ \qquad\newline$ $\textbf{(E) }$ Refraction

Solution

$\textbf{E}$

The bending of light occurs because of a change in the index of refraction… the phenomenon is refraction.

Question 26

A solid cube of iron and a solid cube of aluminum have equal mass. The cubes are placed into the same large pool of water so that each is completely submerged and resting on the pool’s bottom. Which object experiences the greater buoyant force from the water?

$\textbf{(A) }$ The iron cube$ \qquad\newline$
$\textbf{(B) }$ The aluminum cube$ \qquad\newline$
$\textbf{(C) }$ The buoyant forces are equal$ \qquad\newline$
$\textbf{(D) }$ The mass of the cubes is needed to answer the question$ \qquad\newline$
$\textbf{(E) }$ The answer depends on whether the pool is filled with fresh water or salt water

Solution

$\textbf{B}$

The buoyant force $F=\rho_{water}Vg$, so the only variable here is the volume of the cube. Since two cubes have the same mass but the density of iron is greater than aluminum, the volume of aluminum cube is greater. So the aluminum cube has a greater buoyant force.

Question 27

Two spherical, non-rotating planets, X and Y, have the same uniform density $\rho$. Planet X has twice the radius of Planet Y. Let $g_X$ and $g_Y$ represent the accelerations due to gravity at the surfaces of Planet X and Planet Y, respectively. What is the ratio of $g_X:g_Y$?

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

Solution

$\textbf{A}$

The gravity on a planet is $mg=\dfrac{GMm}{r^2}$, so the gravitational acceleration $g=\dfrac{GM}{r^2}$. The mass of the planet $M=\rho V=\dfrac43\pi r^3\rho$, so $g=\dfrac{GM}{r^2}=\dfrac43G\pi r$. As we know, Planet X has twice the radius of Planet Y, thus the ratio $g_X:g_Y=r_X:r_Y=2:1$.

Question 28

A waterproof speaker placed at the bottom of a swimming pool emits a sound wave that travels toward the surface of the water. In the water, the sound wave has a frequency $f_{water}$, wavelength $\lambda_{water}$, and wave speed $v_{water}$. When the sound wave enters the air it has a frequency $f_{air}$, wavelength $\lambda_{air}$, and wave speed $v_{air}$. Which one of the following relationships correctly compares the frequencies, wavelengths, and wave speeds of the waves in the air and water?

$\textbf{(A) }$ $f_{water}=f_{air}\ ;\ \lambda_{water}=\lambda_{air}\ ;\ v_{water}=v_{air} \qquad\newline$
$\textbf{(B) }$ $f_{water}=f_{air}\ ;\ \lambda_{water}>\lambda_{air}\ ;\ v_{water}>v_{air} \qquad\newline$
$\textbf{(C) }$ $f_{water}<f_{air}\ ;\ \lambda_{water}>\lambda_{air}\ ;\ v_{water}=v_{air} \qquad\newline$
$\textbf{(D) }$ $f_{water}<f_{air}\ ;\ \lambda_{water}=\lambda_{air}\ ;\ v_{water}<v_{air} \qquad\newline$
$\textbf{(E) }$ $f_{water}=f_{air}\ ;\ \lambda_{water}<\lambda_{air}\ ;\ v_{water}>v_{air}$

Solution

$\textbf{B}$

The frequency does not change when a wave travels from one medium to another, but the wavelength and speed does due to the change of refractive index. The refractive index of air is 1 (the same value for vacuum), and the refractive index of water is greater than one (actually $n=1.33$). The wave speed in water $v_{water}=nv_{air}$. The wavelength in water $\lambda_{water}=n\lambda_{water}$.

Question 29

An electron with charge $-e$ and mass $m$ travels at a speed $v$ in a plane perpendicular to a magnetic field of magnitude $B$. The electron follows a circular path of radius $R$. In a time $t$, the electron travels halfway around the circle. What is the amount of work done by the magnetic force on the electron in this time?

$\textbf{(A) }$ zero$ \qquad$ $\textbf{(B) }$ $-\pi evBR \qquad$ $\textbf{(C) }$ $\pi evBR \qquad$ $\textbf{(D) }$ $-2evBR \qquad$ $\textbf{(E) }$ $\dfrac{-\pi mv}{eB}$

Solution

$\textbf{A}$

The magnetic force is always perpendicular to the direction of velocity. So no work has been done by magnetic force.

Question 30

Considering only the Moon-Earth system (ignore any influence of the Sun), which one of the following best expresses the magnitude of the Moon’s acceleration about the Earth?

$\textbf{(A) }$ $3\times10^{-1}\ \text{m/s}^2 \qquad\newline$ $\textbf{(B) }$ $3\times10^{-2}\ \text{m/s}^2 \qquad\newline$ $\textbf{(C) }$ $3\times10^{-3}\ \text{m/s}^2 \qquad\newline$ $\textbf{(D) }$ $3\times10^{-4}\ \text{m/s}^2 \qquad\newline$ $\textbf{(E) }$ $3\times10^{-5}\ \text{m/s}^2$

Solution

$\textbf{C}$

The mass of the earth $M_E=6.0\times10^{24}\ \text{kg}$. It takes about 1 second for light from the Moon to reach the earth, which means that the Earth-Moon distance is approximately $3.0\times10^8\ \text{m}$. The moon’s acceleration $a=\dfrac{GM_E}{r^2}=\dfrac{\left(6,7\times10^{-11}\right)\left(6\times10^{24}\right)}{\left(3\times10^8\right)^2}=4,47\times10^{-3}\ \text{m/s}^2$.

Question 31

A toy crane exerts an upward force and delivers a useful power output of 0.10 W to raise a block vertically at a constant speed. At what constant speed will this crane raise a 0.20 kg block?

$\textbf{(A) }$ 0.01 m/s$ \qquad$ $\textbf{(B) }$ 0.02 m/s$ \qquad$ $\textbf{(C) }$ 0.05 m/s$ \qquad$ $\textbf{(D) }$ 0.20 m/s$ \qquad$ $\textbf{(E) }$ 0.50 m/s

Solution

$\textbf{C}$

The power $P=Fv$. Since the block moves in a constant speed, the force $F$ is equal to the gravitational force $F=mg=2\ \text{N}$. So the speed $v=\dfrac{P}{F}=0.05\ \text{m/s}$.

Question 32

Planck’s constant is multiplied by the speed of light. The resulting value then is divided by three meters. This final value has the units of which one of the following quantities?

$\textbf{(A) }$ Force$ \qquad\newline$ $\textbf{(B) }$ Linear Momentum$ \qquad\newline$ $\textbf{(C) }$ Speed$ \qquad\newline$ $\textbf{(D) }$ Frequency $ \qquad\newline$ $\textbf{(E) }$ Energy

Solution

$\textbf{E}$

The value can be wrote as $\dfrac{hc}{l}=\dfrac{h\nu\lambda}{l}$. The unit of wavelength $\lambda$ is the same as the denominator, so their units cancel each other. As we know, the energy of a photon is $E=h\nu$, so the final value has the unit of the energy.

Question 33

An object moves with constant acceleration starting with velocity $v_0=5.00\ \text{m/s}$ and ending with a velocity of $v=-1.00\ \text{m/s}$ in a time of 3.00 s. For this motion, what is the average speed associated with the object?

$\textbf{(A) }$ 2.00 m/s$ \qquad$ $\textbf{(B) }$ 2.17 m/s$ \qquad$ $\textbf{(C) }$ 2.50 m/s$ \qquad$ $\textbf{(D) }$ 2.83 m/s$ \qquad$ $\textbf{(E) }$ 3.00 m/s

Solution

$\textbf{B}$

The acceleration $a=\dfrac{v_f-v_0}{t}=\dfrac{-1-5}{3}=-2\ \text{m/s}^2$ (we take the absolute value as $a=2\ \text{m/s}^2$). In the deceleration process from $v_0=5\ \text{m/s}$ to 0 m/s the distance it moved is $x_1=\dfrac{v_0^2}{2a}=6.25\ \text{m}$. Then the object will accelerate from 0 to $v=1\ \text{m/s}$, so the distance is $x_2=\dfrac{v^2}{2a}=0.25\ \text{m}$. The total distance $x=x_1+x_2=6.5\ \text{m}$. The average speed $\overline{v}=\dfrac{x}{t}=2.17\ \text{m/s}$.

Question 34

Two circular loops of resistive wire are placed next to each other as in the figure. The circular loop on the left is connected to a constant voltage source V. The resistance of this loop is increasing linearly with time. As the resistance is changing, what is the direction of the magnetic field at point P (the center of the left-hand loop) and what is the orientation of the conventional current in the right-hand loop (as viewed in the figure from above)?

Solution

$\textbf{A}$

By a right-hand rule, the fingers curl in the direction of the conventional current (clockwise) and the right thumb points in the direction of the magnetic field at the center of the loop (into the plane). Because magnetic field lines form closed loops, the field is directed out of the plane of the page inside the loop on the right. Because the resistance of the left-hand circuit is increasing, the current in that loop is decreasing, thereby decreasing the magnetic field strength. In the right-hand loop, since the field through the loop is decreasing in magnitude, induction occurs to reinforce the lost field. This means from the right-hand rule that the fingers will curl counterclockwise around the loop to produce a magnetic field out of the plane, trying to replace the field lines that are being lost.

Question 35

There are several statements presented below that attempt to describe physical phenomena. Which one of the following statements is correct?

$\textbf{(A) }$ The coefficient of friction is a value always less than or equal to one, but greater than or equal to zero.$ \qquad\newline$
$\textbf{(B) }$ For horizontal surfaces, the normal force acting on an object always cancels the gravitational force.$ \qquad\newline$
$\textbf{(C) }$ An ideal gas’s temperature must change if both work is done and energy is exchanged as heat with it.$ \qquad\newline$
$\textbf{(D) }$ Increasing the spacing between slits in the Young’s double slit experiment results in an increase in the spacing between the dark regions on a distant viewing screen.$ \qquad\newline$
$\textbf{(E) }$ In electrostatic equilibrium, an electric field is perpendicular to the surface of a charged conductor.

Solution

$\textbf{E}$

A: The coefficient of friction can exceed the value of 1$\newline$
B: The normal force is not equal to gravitational force when the object is accelerating in the vertical direction$\newline$
C: The internal energy may not change when work is done while energy exchanged (isothermal process), thus the temperature may not change either$\newline$
D: By $\Delta x=\dfrac{D}{d}\lambda$, increasing the spacing $d$ between slits results in a decrease in the spacing $\Delta x$ $\newline$
The last statement is a property of conductors in electrostatic equilibrium.

Question 36

A concave mirror with focal length $f$ is shown in the figure. A real object now is placed to the left of the mirror. In theory, which one of the following choices best describes everywhere that it is $\textbf{impossible}$ for an image to form from the mirror?

$\textbf{(A) }$ Region II only$ \qquad\newline$
$\textbf{(B) }$ Regions II and III only$ \qquad\newline$
$\textbf{(C) }$ Regions III and IV only$ \qquad\newline$
$\textbf{(D) }$ Regions II and IV only$ \qquad\newline$
$\textbf{(E) }$ Regions I, II, and IV only

Solution

$\textbf{A}$

By $\dfrac1p+\dfrac1q=\dfrac1f$, once placing the object in Region I ($p>f$), the image formed will be real and located in Region I. Once inside the focal length of the mirror (in Region II where $0<p<f$), the image formed will become virtual and appear on the right-hand side of the mirror as $q=-\dfrac{p}{f-p}f$. For $0<\dfrac{p}{f-p}<1$ or $0<p<\dfrac{f}{2}$, the image will be in Region III. For $\dfrac{p}{f-p}>1$ or $f>p>\dfrac{f}{2}$, the image will be in Region IV. So it is $\textbf{impossible}$ to form an image anywhere in Region II.

Prime minister Cao: Impossible! Absolutely impossible!$\newline$
Prince Liu: Impossible! My brother is invincible!

Question 37

A new element is discovered and named PhysicsBowlium (atomic symbol $Phys$). Its entry onto the standard periodic table of elements appears as in the figure (with Helium shown as well). Given a sample of $Phys$ which acts as a perfect monatomic ideal gas, what is the root-mean-square speed of the atoms of the gas if the sample is at $20^\circ$C?

$\textbf{(A) }$ 1.04 m/s$ \qquad$ $\textbf{(B) }$ 4.00 m/s$ \qquad$ $\textbf{(C) }$ 33.0 m/s$ \qquad$ $\textbf{(D) }$ 126 m/s$ \qquad$ $\textbf{(E) }$ 191 m/s

Solution

$\textbf{D}$

The root-mean-square speed is computed as $v_{rms}=\sqrt{\dfrac{3RT}{M}}$. From the periodic table entry, we have the molar mass $M=458.7\ \text{g/mol}$. As a result, we substitute our values to obtain $v_{rms}=\sqrt{\dfrac{3(8.31)(293)}{0.4587}}=126\ \text{m/s}$ as we have converted the temperature into Kelvin and the molar mass into kg/mol.

Question 38

In the circuit shown, the switch S has been left open for a very long time. All circuits elements are considered to be ideal. Which one of the following statements best describes the behavior of the current through the switch S once it is closed?

$\textbf{(A) }$ The current initially is 12 mA and decreases to a steady 3 mA.$ \qquad\newline$
$\textbf{(B) }$ The current initially is 3 mA and increases to a steady 12 mA.$ \qquad\newline$
$\textbf{(C) }$ The current initially is 9 mA and decreases to a steady 3 mA.$ \qquad\newline$
$\textbf{(D) }$ The current initially is 6 mA and decreases to a steady 3 mA.$ \qquad\newline$
$\textbf{(E) }$ The current is a steady 3 mA.

Solution

$\textbf{A}$

After a long time has passed, the current through the circuit is zero Amps. Using a Kirchhoff Loop Rule (KLR) around the outside of the circuit reveals that the potential difference across the capacitor is 9 V since there is no current after a long time. Once the switch is closed, there is a loop that encloses the 9 V battery and 3 k$\Omega$ resistor resulting in a 3 mA current through the resistor. For the other branch on the right, there is a capacitor initially charged at 9 V and a 1 k$\Omega$ resistor. So the current passing through the resistor is 9 mA. As a result, when the switch is initially closed, from Kirchhoff’s Junction Rule, we have 3 mA from the left and 9 mA from the right resulting in 12 mA through the switch. Eventually, the capacitor is completely discharged and there is no more current in the right half of the circuit, meaning that only the 3 mA through the 3 k$\Omega$ resistor exists in the circuit.

Question 39

Two uniform disks, X and Y, have masses $m_X<m_Y$, equal radii, and equal initial non-zero kinetic energies. Each disk rotates counterclockwise in the plane of the page about a fixed frictionless axis through its center. As shown in the figure, a force $F$ is applied tangent to each disk at its right edge for the same amount of time. After the forces are applied, let $L$ represent the magnitude of the angular momentum about the center of a disk and $K$ represent the kinetic energy of a disk. Which one of the following choices correctly
compares these quantities for disk X and disk Y?

$\textbf{(A) }$ $L_X>L_Y\ ;\ K_X<K_Y \qquad\newline$
$\textbf{(B) }$ $L_X>L_Y\ ;\ K_X>K_Y \qquad\newline$
$\textbf{(C) }$ $L_X=L_Y\ ;\ K_X=K_Y \qquad\newline$
$\textbf{(D) }$ $L_X<L_Y\ ;\ K_X<K_Y \qquad\newline$
$\textbf{(E) }$ $L_X<L_Y\ ;\ K_X>K_Y$

Solution

$\textbf{E}$

The change of angular momentum $\Delta L=\tau t=Frt$. As the two disks have the same radii, the changes of angular momentum are the same. To compare the final angular momentum $L_X$ and $L_Y$, we need to find the initial angular momentum $L_{0X}$ and $L_{0Y}$. The initial kinetic energy $K_0=\dfrac12I\omega^2=\dfrac{L_0^2}{2I}$. The momentum of inertia $I=\dfrac12mr^2$. As $m_X<m_Y$, $I_X<I_Y$. So the angular momentum $L_{0X}=\sqrt{2I_XK_0}<L_{0Y}$, the final angular momentum $L_X<L_Y$. $\newline$

As for the final kinetic energy $K_X$ and $K_Y$, we know that the initial kinetic energies are the same. The change of kinetic energy depends on the distance the disk rotated. By $K_0=\dfrac12I\omega^2=\dfrac12mr^2\omega^2$, disk X has a lesser mass, so it has a greater angular velocity, which means it moves a longer distance. The work done by the force $F$ is the change of kinetic energy, so disk X gains more energy from the force and thus $K_X>K_Y$.

Question 40

A small 1.35 kg mass is launched from the top of a cliff at an angle of $15.9^\circ$ above the horizontal. When the mass reaches the ground 4.33 seconds later, its velocity is directed at $34.4^\circ$ below the horizontal. What is the speed of the mass when it reaches the ground? Ignore air resistance.

$\textbf{(A) }$ 60.7 m/s$ \qquad$ $\textbf{(B) }$ 54.1 m/s$ \qquad$ $\textbf{(C) }$ 46.4 m/s$ \qquad$ $\textbf{(D) }$ 43.3 m/s$ \qquad$ $\textbf{(E) }$ 38.8 m/s

Solution

$\textbf{B}$

The horizontal component of the velocity does not change. The change of velocity in the vertical direction is $\Delta v=gt=43.3\ \text{m/s}$. By the law of sins we get $\dfrac{v_f}{\sin74.1^\circ}=\dfrac{\Delta v}{\sin50.3^\circ}$, so the final speed $v_f=\Delta v\dfrac{\sin74.1^\circ}{\sin50.3^\circ}=54.1\ \text{m/s}$.

Question 41

“No two electrons in an atom can have an identical set of the four quantum numbers.” is a statement most closely associated with which one of the following scientists?

$\textbf{(A) }$ Albert Einstein$ \qquad\newline$ $\textbf{(B) }$ Enrico Fermi$ \qquad\newline$ $\textbf{(C) }$ Sheldon Cooper$ \qquad\newline$ $\textbf{(D) }$ Wolfgang Pauli $ \qquad\newline$ $\textbf{(E) }$ Isaac Newton

Solution

$\textbf{D}$

The sentence is usually described as the Pauli Exclusion Principle. Even though, I choose Sheldon Copper (a cute guy in the show The Big Bang Theory).

Question 42

Plane-polarized light with intensity $I$ is incident on a single polarizing sheet. If the intensity of the light become $\dfrac14I$ after passing through the polarizer, what is the angle between the transmission axis of the polarizer and the polarization plane of the incident light?

$\textbf{(A) }$ $75^\circ \qquad$ $\textbf{(B) }$ $67.5^\circ \qquad$ $\textbf{(C) }$ $60^\circ \qquad$ $\textbf{(D) }$ $30^\circ \qquad$ $\textbf{(E) }$ $22.5^\circ$

Solution

$\textbf{C}$

By Malus’ Law $I=I_0\cos^2\theta$, we get $\dfrac14I=I\cos^2\theta$. So $\cos\theta=\dfrac12$, $\theta=60^\circ$.

Question 43

A long rod of length $L$ is pivoted about its left end. It is released from an angle $\theta$ above the horizontal. What is the magnitude of the angular acceleration of the rod about the pivot when the rod is released?

$\textbf{(A) }$ $\dfrac{6g}{L}\cos\theta \qquad$ $\textbf{(B) }$ $\dfrac{6g}{L}\sin\theta \qquad$ $\textbf{(C) }$ $\dfrac{3g}{2L}\sin\theta \qquad$ $\textbf{(D) }$ $\dfrac{3g}{2L}\cos\theta \qquad$ $\textbf{(E) }$ $\dfrac{3g}{L}\sin\theta$

Solution

$\textbf{D}$

The torque about the pivot $\tau=I\beta=mg\cdot\dfrac{L}{2}\cos\theta$, where the moment of inertia about the end $I=\dfrac13mL^2$. So the angular acceleration $\beta=\dfrac{\tau}{I}=\dfrac{3g\cos\theta}{2L}$.

Question 44

A ray of monochromatic light enters the right-triangular glass as shown. The glass has an index of refraction of 2.00 and it is surrounded by air. Which one of the lettered rays shows the path of the light after it exits the glass?

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

Solution

$\textbf{E}$

Light coming in perpendicular to a surface travels undeflected. When it reaches the back surface, we apply Snel’s Law to obtain the angle of refraction as $n_g\sin\theta_g=n_a\sin\theta_a$, thus we get $\sin\theta_a=1.44$. Since the mathematics is impossible, we have total internal reflection of the light, which bounces it straight downward in the prism. The same analysis now will occur when this ray reaches the glass/air interface. Consequently, internal reflection occurs again and the ray now bounces to the left, where it exits the glass.

Question 45

A long thin rod of mass $M$ and length $L$ is pivoted at one end so that it swings as a pendulum. The rod is set into simple harmonic oscillation and has a period of motion $T$. A second thin rod with mass $2M$ and length $2L$ also is pivoted at one end to swing as a pendulum. When this second rod is set into simple harmonic oscillation, what is its period?

$\textbf{(A) }$ $2T \qquad$ $\textbf{(B) }$ $\sqrt2T \qquad$ $\textbf{(C) }$ $T \qquad$ $\textbf{(D) }$ $\dfrac1{\sqrt2}T \qquad$ $\textbf{(E) }$ $\dfrac12T$

Solution

$\textbf{B}$

For a physical pendulum, the period $$T=2\pi\sqrt{\dfrac{I}{mgd}}=2\pi\sqrt{\dfrac{\frac13ML^2}{Mg\frac{L}2}}=2\pi\sqrt{\dfrac{2L}{3g}}$$ By doubling the mass, that has no effect on the period, whereas by doubling the distance, we have $$T_2=2\pi\sqrt{\dfrac{2(2L)}{3g}}=\sqrt2T$$

Question 46

A monatomic, ideal gas undergoes an isobaric process. During the process, the gas performs 80 joules of work on the surroundings. Which one of the following statements best describes the heat exchange during this process?

$\textbf{(A) }$ 200 joules of energy was added to the gas.$ \qquad\newline$
$\textbf{(B) }$ 200 joules of energy was removed from the gas.$ \qquad\newline$
$\textbf{(C) }$ 80 joules of energy was added to the gas.$ \qquad\newline$
$\textbf{(D) }$ 80 joules of energy was removed from the gas.$ \qquad\newline$
$\textbf{(E) }$ It cannot be determined without knowing the change in temperature for the gas.

Solution

$\textbf{A}$

For an isobaric process, the work done by the gas on the surroundings is computed as $W=P\Delta V=80\ \text{J}$. The heat associated with this kind of process is $$Q=nc_p\Delta T=n\left(\dfrac52R\right)\Delta T=\dfrac52P\Delta V=\dfrac52(80) = 200\ \text{J}$$ since $P\Delta V=nR\Delta T$ for an isobaric process on an ideal gas.

Question 47

An air-filled parallel-plate capacitor of capacitance $C$ is fully charged after being connected to a battery of voltage $V$. The battery then is disconnected and an insulating dielectric slab of constant $\kappa$ is inserted between the capacitor’s plates, fully filling the region. What is the voltage between the plates once equilibrium is established with the dielectric in place?

$\textbf{(A) }$ $\kappa^2V \qquad$ $\textbf{(B) }$ $\kappa V \qquad$ $\textbf{(C) }$ $V \qquad$ $\textbf{(D) }$ $\dfrac{V}{\kappa} \qquad$ $\textbf{(E) }$ $\dfrac{V}{\kappa^2}$

Solution

$\textbf{D}$

By disconnecting the battery, the charge on the plates of the capacitor is fixed. Because the capacitance would increase to $\kappa C_0$. Using for a capacitor that $Q=C\Delta V$, since $Q$ is unchanged and the capacitance went up, this means that the voltage is now $V⁄\kappa$. Alternatively, since the electric field would be decreased by a factor of $\kappa$ with the introduction of the dielectric, and because $\Delta V=Ed$, then the voltage also decreases by a factor of $\kappa$.

Question 48

A laser beam with wavelength 632.8 nm shines onto a newly fabricated single slit. As a result, the width of the principal bright region on a viewing screen 1.25 m away is 1.00 m. Which one of the following best represents the size of the single slit opening?

$\textbf{(A) }$ 0.79 $\mu$m $ \qquad$ $\textbf{(B) }$ 0.85 $\mu$m$ \qquad$ $\textbf{(C) }$ 1.01 $\mu$m$ \qquad$ $\textbf{(D) }$ 1.58 $\mu$m$ \qquad$ $\textbf{(E) }$ 1.70 $\mu$m

Solution

$\textbf{E}$

From the single slit relation, we have $a\sin\theta=n\lambda$. The angle from the slit to the dark region is computed as $\tan\theta=\dfrac{0.5\ \text{m}}{1.25\ \text{m}}\Rightarrow\theta=21.8^\circ$. The 0.50 m is the distance from the center of the bright region to the first dark spot. So, we now write $a=\dfrac{n\lambda}{\sin\theta}=\dfrac{632.8\times10^{-9}}{\sin21.8^\circ}=1.70\times10^{-6}\ \text{m}=1.70\ \mu\text{m}$.

Question 49

A solid, uniform disk of mass $M$ and radius $R$ rotates clockwise about its center with an angular speed $\omega_0$. The disk then is placed onto a horizontal surface and begins moving only to the right, slipping as it rolls. The coefficient of friction between the floor and the disk is $\mu$ and the frictional force is considered constant throughout the motion. What is the angular speed of the disk when the disk starts rolling without slipping?

$\textbf{(A) }$ $\dfrac{\mu}{2}\omega_0 \qquad$ $\textbf{(B) }$ $\dfrac12\omega_0 \qquad$ $\textbf{(C) }$ $\dfrac13\omega_0 \qquad$ $\textbf{(D) }$ $\dfrac{2\mu}{3}\omega_0 \qquad$ $\textbf{(E) }$ $\dfrac35\omega_0$

Solution

$\textbf{C}$

In the beginning the contact point on the ground has a relative movement trend to the left, which means that the frictional force $f=\mu mg$ points to the right. The frictional force provides translation velocity while decelerates the rotation. For the translation, the acceleration $a=\dfrac{f}{M}=\mu g$, so the velocity of the center at time $t$ is $v=at=\mu gt$. For the rotation, the angular deceleration is $\beta=\dfrac{\tau}{I}=\dfrac{fR}{\frac12MR^2}=\dfrac{2\mu g}{R}$, so the angular speed at time $t$ is $\omega=\omega_0-\beta t=\omega_0-\dfrac{2\mu g}{R}t$. Finally the disk rolls without slipping, now we get $v=\omega R$. It can be rewrote as $\mu gt=\omega_0R-2\mu gt$. The time $t=\dfrac{\omega_0R}{3\mu g}$. The final angular speed $\omega=\omega_0-\dfrac{2\mu g}{R}t=\dfrac13\omega_0$.

Question 50

Two clocks, A and B, are synchronized on Earth. Clock A is placed onto a space ship that leaves Earth in a straight line with a speed of $2.40\times10^8\ \text{m/s}$. On Earth, a scientist with clock B has her telescope fixed directly on clock A. If each clock started at $t=0\ \text{s}$, what time does the scientist observe on clock A when
clock B reads $t=90\ \text{s}$? Assume the time of acceleration for the ship leaving the Earth was negligible.

$\textbf{(A) }$ 24 s$ \qquad$ $\textbf{(B) }$ 30 s$ \qquad$ $\textbf{(C) }$ 50 s$ \qquad$ $\textbf{(D) }$ 54 s$ \qquad$ $\textbf{(E) }$ 72 s

Solution

$\textbf{B}$

The key to this problem is recognizing that what is seen in the telescope is light that has made it all the way back to Earth. This means that for the Earth, the time that passes will be the time it takes the ship to move out some distance $l_0$ and allow the light to return back to Earth. That is, $$90\ \text{s}=t_{out}+t_{bacl}=\dfrac{l_0}{0.8c}+\dfrac{l_0}{c}\Rightarrow l_0=40c=1.2\times10^10\ \text{m}$$ From length contraction, the spaceship claims that a distance of $l=l_0\sqrt{1-0.8^2}=7.2\times10^9\ \text{m}$ has been traversed. Hence, at a speed of 0.8c, a time of $t=\dfrac{l}{0.8c}=30\ \text{s}$ would be reading on the clock.