## PhysicsBowl 2019

**If you like our free learning materials, please click the advertisement below or anywhere. No paying, no donation, just a simple click. The advertising revenues will be used to provide more and better learning materials. Thank you!**

**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

Select the smallest value from the choices below

$\textbf{(A) }$ $15\times10^{-3} \qquad\newline$ $\textbf{(B) }$ $0.15\times10^0 \qquad\newline$ $\textbf{(C) }$ $0.00015\times10^3 \qquad\newline$ $\textbf{(D) }$ $150\times10^{-3} \qquad\newline$ $\textbf{(E) }$ $0.00000015\times10^6$

$\textbf{A}$

Zeroes come and zeroes go, leave the gay one alone 🙂

Related to the historical development of understanding gravity, which is the proper chronological order for the work of these three scientists, from earliest to latest?

$\textbf{(A) }$ Cavendish, Galileo, Newton$ \qquad\newline$

$\textbf{(B) }$ Galileo, Cavendish, Newton$ \qquad\newline$

$\textbf{(C) }$ Galileo, Newton, Cavendish$ \qquad\newline$

$\textbf{(D) }$ Newton, Galileo, Cavendish$ \qquad\newline$

$\textbf{(E) }$ Newton, Cavendish, Galileo

$\textbf{C}$

You know it, don't you?

An isolated solid metal sphere with a radius $R$ is given a positive charge $Q$. The electric potential at the surface of the sphere is $V$. What is the electric potential at a distance of 0.5R from the center of the sphere?

$\textbf{(A) }$ zero$ \qquad$ $\textbf{(B) }$ $0.5V \qquad$ $\textbf{(C) }$ $V \qquad$ $\textbf{(D) }$ $2V \qquad$ $\textbf{(E) }$ $4V$

$\textbf{C}$

Metal sphere is equipotential inside and on the surface.

Consider a situation where the acceleration of an object is always directed perpendicular to its velocity. This means that

$\textbf{(A) }$ the object is increasing speed.$ \qquad\newline$

$\textbf{(B) }$ the object is decreasing speed.$ \qquad\newline$

$\textbf{(C) }$ the object is not moving.$ \qquad\newline$

$\textbf{(D) }$ the object is turning.$ \qquad\newline$

$\textbf{(E) }$ this situation would not be physically

$\textbf{D}$

A perpendicular acceleration will change the direction but not the magnitude of the velocity.

The acceleration due to gravity on the Moon is less than the acceleration due to gravity on the Earth. Which of the following is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth?

$\textbf{(A) }$ Mass is less, weight is the same.$ \qquad\newline$

$\textbf{(B) }$ Mass is the same, weight is less.$ \qquad\newline$

$\textbf{(C) }$ Both mass and weight are less.$ \qquad\newline$

$\textbf{(D) }$ Both mass and weight are the same.$ \qquad\newline$

$\textbf{(E) }$ Mass is more, weight is less

$\textbf{B}$

The mass will always be the same, but the weight will change due to the change of gravity.

Which one of the following is not equivalent to 2.50 miles?$\newline$

(1.00 mi = 1.61 km = 5280 ft, 1.00 yd = 3.00 ft = 36.0 in.)

$\textbf{(A) }$ $1.32\times10^4\ \textrm{ft} \qquad\newline$ $\textbf{(B) }$ $1.58\times10^5\ \textrm{in} \qquad\newline$ $\textbf{(C) }$ $4.02\times10^3\ \textrm{km} \qquad\newline$ $\textbf{(D) }$ $4.40\times10^3\ \textrm{yd} \qquad\newline$ $\textbf{(E) }$ $4.02\times10^5\ \textrm{cm}$

$\textbf{BC}$

The original question take 1.00 yd as 12.0 in, which is a mistake (Actually 1 yd = 36 in).

A 100 kg person travels from sea level to an altitude of 5000 m. By how many Newtons does their weight change?

$\textbf{(A) }$ 0.8 N$ \qquad$ $\textbf{(B) }$ 1.2 N$ \qquad$ $\textbf{(C) }$ 1.6 N$ \qquad$ $\textbf{(D) }$ 2.0 N$ \qquad$ $\textbf{(E) }$ 2.4 N

$\textbf{C}$

The gravitational acceleration at sea level is $g=\dfrac{GM}{R^2}$, where $R=6400$ km is the radius of earth. at $h=5$ km, the gravitational acceleration is $g_h=\dfrac{GM}{(R+h)^2}=\dfrac{R^2}{(R+h)^2}g$.

A candle, a converging lens and a white screen are placed in a line with the lens between the candle and the screen. A distance of 72 cm separates the candle and screen. As the lens is moved to all points between the candle and the screen, only one focused image of the candle can be made on the screen. What is the focal length of the converging lens?

$\textbf{(A) }$ 12 cm$ \qquad$ $\textbf{(B) }$ 18 cm$ \qquad$ $\textbf{(C) }$ 24 cm$ \qquad$ $\textbf{(D) }$ 36 cm$ \qquad\newline$

$\textbf{(E) }$ It cannot be determined without knowing the location of the lens when the focused image is produced.

$\textbf{B}$

For distance $d=15$ cm, $\dfrac{1}{u}+\dfrac{1}{d-u}=\dfrac1f$, thus $f=\dfrac{-u^2+du}{d}$. The quadratic function $g(u)=-u^2+du$ has only one real root at $u=d/2$, which means only one focused image. So $f=d/4$.

Standby power (sometimes called vampire power) is the power used by a device that is off but plugged in and in a standby mode. Regulations typically limit this power to 1 Watt. If electricity costs $\$$0.10 per kilowatt hour, then to the nearest order of magnitude, and assuming 1 Watt, how much does it cost to leave a device in standby mode for one year?

$\textbf{(A) }$ $\$0.01 \qquad$ $\textbf{(B) }$ $\$0.10 \qquad$ $\textbf{(C) }$ $\$1.00 \qquad$ $\textbf{(D) }$ $\$10.00 \qquad$ $\textbf{(E) }$ $\$100.00$

$\textbf{C}$

(1 W)$\times$(365 days)$\times$(24 hours/day)$\times$($\$$0.1 /kilowatt hour) = $\$$0.876

A student generates a transverse periodic wave on a string. The wave travels away from the student at a constant speed $v$. Which of the following changes by itself will increase the speed at which the wave travels away from the student?

$\textbf{(A) }$ The student could use the same string but increase the frequency at which they generate the wave.$ \qquad\newline$

$\textbf{(B) }$ The student could use the same string but increase the wavelength of the waves they generate.$ \qquad\newline$

$\textbf{(C) }$ The student could use the same string but increase the amplitude of the waves they generate.$ \qquad\newline$

$\textbf{(D) }$ The student could use a string with the same length and tension, but greater linear density.$ \qquad\newline$

$\textbf{(E) }$ The student could use the same string, but placed under greater tension.

$\textbf{E}$

Wave speed on a string is $v=\sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension and $\mu$ is linear density.

You throw a ball at angle $\theta$ and measure its horizontal range, $R$. You now throw the same ball again at the same angle, but with twice the original speed. Its horizontal range, compared to your first throw (in both cases, the ball lands at the same height from which it was thrown) would be

$\textbf{(A) }$ $1.4R \qquad$ $\textbf{(B) }$ $0.5R \qquad$ $\textbf{(C) }$ $2R \qquad$ $\textbf{(D) }$ $4R \qquad$ $\textbf{(E) }$ the same

$\textbf{D}$

The horizontal range $x=\dfrac{v^2\sin2\theta}{g}$.

A horizontal disk has a spring and a mass attached to a rod passing through its center. The spring has a spring constant of $k$ and the attached mass is $m$. When the disk is not rotating, the equilibrium length of the spring is $d$. The disk is now angularly accelerated until it reaches a constant angular velocity of $\omega$. At this angular velocity the length of the spring is now $2d$. What is the angular velocity of the disk?

$\textbf{(A) }$ $\sqrt{\dfrac{k}{d}} \qquad$ $\textbf{(B) }$ $\sqrt{\dfrac{k}{2m}} \qquad$ $\textbf{(C) }$ $\sqrt{\dfrac{2k}{m}} \qquad$ $\textbf{(D) }$ $\sqrt{\dfrac{2d}{k}} \qquad$ $\textbf{(E) }$ $\sqrt{\dfrac{d}{2k}}$

$\textbf{B}$

The stretched spring provides the centripetal force, thus $kd=m\omega^2\cdot2d$, so $\omega=\sqrt{\dfrac{k}{2m}}$.

You measure that 4000 kcal of heat conducts through a window in your house in one hour when the house is kept at 20 $^\circ$C. The window is a 4.0 m$^2$ pane of 0.30 cm thick glass ($k=2.0\times10^{-4}\ \textrm{kcal/s}\cdot\textrm{m}\cdot^\circ$C). What is the outside temperature?

$\textbf{(A) }$ 4 $^\circ$C$ \qquad$ $\textbf{(B) }$ 8 $^\circ$C$ \qquad$ $\textbf{(C) }$ 13 $^\circ$C$ \qquad$ $\textbf{(D) }$ 16 $^\circ$C$ \qquad$ $\textbf{(E) }$ 18 $^\circ$C

$\textbf{D}$

Based on the units of constant $k$, for area $A=4\ \textrm{m}^2$, thickness $d=3\times10^{-3}\ \textrm{m}$ and $t=3600\ \textrm{s}$, $\Delta Q=kAt\Delta T/d$.

The red glow in the neon tube of an advertising sign is a result of

$\textbf{(A) }$ fluorescence$ \qquad\newline$ $\textbf{(B) }$ incandescence$ \qquad\newline$ $\textbf{(C) }$ iridescence $ \qquad\newline$ $\textbf{(D) }$ coherence$ \qquad\newline$ $\textbf{(E) }$ de-excitation

$\textbf{E}$

After being excited and moving to a higher energy level, the electrons return to their original energy level and release the excess energy as light.

Which of the following people was a recipient of the 2018 Nobel Prize in Physics?

$\textbf{(A) }$ Frances Arnold$ \qquad\newline$ $\textbf{(B) }$ Donna Strickland$ \qquad\newline$ $\textbf{(C) }$ Tasuku Honjo$ \qquad\newline$ $\textbf{(D) }$ Kip Thorne$ \qquad\newline$ $\textbf{(E) }$ Joachim Frank

$\textbf{B}$

The 2018 Nobel Prize in Physics was awarded to Arthur Ashkin, Gerard Mourou, and Donna Strickland “for groundbreaking inventions in the field of laser physics”.

Water flows out of a horizontal drainpipe at the rate of 120 kg per minute. Its initial vertical velocity is zero and it falls 3.20 m to the ground. What is the average force it exerts when it hits the ground?

$\textbf{(A) }$ 6.0 N$ \qquad$ $\textbf{(B) }$ 10.0 N$ \qquad$ $\textbf{(C) }$ 12.0 N$ \qquad$ $\textbf{(D) }$ 16.0 N$ \qquad$ $\textbf{(E) }$ 20.0 N

$\textbf{D}$

In the formula of impulse-momentum theorem $F\Delta t=m\Delta v$, the change of speed $\Delta v$ is the speed hitting the ground, so $\Delta v=\sqrt{2gh}=8\ \text{m/s}$. The rate of water flow is $k=120\ \textrm{kg/min}=2\ \textrm{kg/s}$, the mass $m=k\Delta t$, so $F=k\Delta v=16\ \textrm{N}$.

A sphere has a radius of $1.96\pm0.01\ \text{m}$. What is the approximate uncertainty in its volume?

$\textbf{(A) }$ $31.5\pm0.2\ \text{m}^3 \qquad\newline$ $\textbf{(B) }$ $31.5\pm0.3\ \text{m}^3 \qquad\newline$ $\textbf{(C) }$ $31.5\pm0.4\ \text{m}^3 \qquad\newline$ $\textbf{(D) }$ $31.5\pm0.5\ \text{m}^3 \qquad\newline$ $\textbf{(E) }$ $31.5\pm0.6\ \text{m}^3$

$\textbf{D}$

Calculate the volume of a sphere with radium $r_1=1.95\ \text{m}$ and $r_2=1.97\ \text{m}$ individually.

A block slides up and back down a rough incline. Which of the following graphs could represent the velocity of the block as a function of time? All graphs use the same scale and uphill as the positive direction.

$\textbf{B}$

When sliding up the velocity is positive until decreasing to zero, then sliding down with the a negative velocity, but the magnitude of velocity can not reach its initial velocity due to frictional force.

A ball with a mass of 0.500 kg traveling at 4.80 m/s strikes a wall and rebounds in the opposite direction at 3.60 m/s. What is the magnitude of the impulse that acted on the ball during the collision with the wall?

$\textbf{(A) }$ 0.600 N$\cdot$s$ \qquad$ $\textbf{(B) }$ 1.20 N$\cdot$s$ \qquad$ $\textbf{(C) }$ 2.70 N$\cdot$s$ \qquad$ $\textbf{(D) }$ 4.20 N$\cdot$s$ \qquad$ $\textbf{(E) }$ 16.8 N$\cdot$s

$\textbf{D}$

For $v_1=-4.8\ \text{m/s}$ and $v_2=3.6\ \text{m/s}$, the impulse $I=m\Delta v=m(v_2-v_1)$.

A string, fixed at both ends, vibrates at a frequency of 12 Hz with a standing transverse wave pattern containing 3 loops (antinodes). What frequency is needed if the standing wave pattern is to contain 4 loops (antinodes)?

$\textbf{(A) }$ 12 Hz$ \qquad$ $\textbf{(B) }$ 16 Hz$ \qquad$ $\textbf{(C) }$ 36 Hz$ \qquad$ $\textbf{(D) }$ 48 Hz$ \qquad$ $\textbf{(E) }$ 60 Hz

$\textbf{B}$

For the string of length $l$, the first standing transverse wave pattern containing 3 antinodes means $\lambda_1=\dfrac13l$. For the second wave of 4 antinodes, $\lambda_2=\dfrac14l$. According to $v=\lambda f$, $v$ is a constant for the same string, so $\lambda_1f_1=\lambda_2f_2$, $f_2=\dfrac{\lambda_1}{\lambda_2}f_1=16\ \text{Hz}$.

Let $M$ represent the magnification of an image. For which of the following arrangements of an object and an optical device would $-1 <M<0$?

$\textbf{(A) }$ The object is placed less than one focal length in front of a converging mirror.$ \qquad\newline$

$\textbf{(B) }$ The object is placed between one focal length and two focal lengths in front of a diverging mirror.$ \qquad\newline$

$\textbf{(C) }$ The object is placed less than one focal length in front of a diverging lens.$ \qquad\newline$

$\textbf{(D) }$ The object is placed more than two focal lengths in front of a converging lens.$ \qquad\newline$

$\textbf{(E) }$ The object is placed between one focal length and two focal lengths in front of a plane mirror.

$\textbf{D}$

For object distance $d_o$ and image distance $d_i$, $\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac1f$, the magnification $$M=-\dfrac{d_i}{d_o}=\dfrac{f}{f-d_o}$$

For choice A, $f>0$, $0<d_o<f$, the magnitude $M>0$;$\newline$

For choice B, $f<0$, $\lvert f\rvert<d_o<2\lvert f\rvert$, the the magnitude $M>0$;$\newline$

For choice C, $f<0$, $0<d_o<\lvert f\rvert$, the the magnitude $M>0$;$\newline$

For choice D, $f>0$, $d_o>2f$, so $-1 <M<0$;$\newline$

For choice E, $f=\infty$, $M=1$.

An Atwood machine is shown in the diagram with $m_1=0.60\ \text{kg}$ and $m_2=0.40\ \text{kg}$. What is the magnitude of the acceleration of $m_2$? Ignore friction and the mass of the pulley.

$\textbf{(A) }$ 4.2 m/s$^2 \qquad$ $\textbf{(B) }$ 3.3 m/s$^2 \qquad$ $\textbf{(C) }$ 2.0 m/s$^2 \qquad$ $\textbf{(D) }$ 5.0 m/s$^2 \qquad$ $\textbf{(E) }$ 1.0 m/s$^2$

$\textbf{C}$

$F_1-F_2=(m_1-m_2)g=(m_1+m_2)a$, so $a=\dfrac{m_1-m_2}{m_1+m_2}g=2\ \text{m/s}^2$.

A 30.0 N block falls straight down from a height of 10.0 m, and strikes the ground with a velocity of 7.00 m/s. What average force of air friction acts on it as it falls?

$\textbf{(A) }$ 22.65 N$ \qquad$ $\textbf{(B) }$ 45.45 N$ \qquad$ $\textbf{(C) }$ 75,0 N$ \qquad$ $\textbf{(D) }$ 206.5 N$ \qquad$ $\textbf{(E) }$ 293 N

$\textbf{A}$

The acceleration in the falling process $a=\dfrac{G-f}{m}$ satisfies the equation of velocity $v^2=2ah$, thus we get $f=G-\dfrac{mv^2}{2h}=22.65\ \text{N}$.

**Questions 24 and 25 refer to the circuit below:**

The capacitor is initially uncharged. Which of the following choices best represents the current at Point X immediately after the switch is closed?

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

$\textbf{(B) }$ 24 mA directed to the right$ \qquad\newline$

$\textbf{(C) }$ 36 mA directed to the left$ \qquad\newline$

$\textbf{(D) }$ 48 mA directed to the left$ \qquad\newline$

$\textbf{(E) }$ 72 mA directed to the right

$\textbf{B}$

At the moment of closing the switch, the capacitor will be charged and current seems passing through the capacitor. Thus at this moment the capacitor can be regarded as wire. $100\ \Omega$ is in parallel with $400\ \Omega$, $300\ \Omega$ is in parallel with $200\ \Omega$, so the resistance of the circuit is $200\ \Omega$. The total current is 60 mA, and the distribution of current passing through $100\ \Omega$ and $400\ \Omega$ is 4:1 due to the ratio of resistance, so the current passing through $100\ \Omega$ is 48 mA. For the same reason, the current passing through $300\ \Omega$ is 24 mA. The difference between 48 mA and 24 mA means the remaining 24 mA passing through the capacitor, from the left to the right.

If the capacitor has a capacitance of $15\ \mu \text{F}$ and is uncharged prior to closing the switch, what is the steady-state charge on the capacitor?

$\textbf{(A) }$ $3\ \mu \text{C} \qquad$ $\textbf{(B) }$ $45\ \mu \text{C} \qquad$ $\textbf{(C) }$ $75\ \mu \text{C} \qquad$ $\textbf{(D) }$ $90\ \mu \text{C} \qquad$ $\textbf{(E) }$ $120\ \mu \text{C}$

$\textbf{C}$

At steady-state, the voltage is 9 V on the left side of the capacitor and 4 V on the right side, so the charge with be $Q=CU=15\ \mu\text{F}\cdot5 V=75\ \mu\text{C}$.

A 15.0 mW laser emits a beam that is 2.00 mm in diameter. What is the intensity of this beam?

$\textbf{(A) }$ $1.19\times10^3\ \text{W/m}^2 \qquad\newline$

$\textbf{(B) }$ $1.19\times10^4\ \text{W/m}^2 \qquad\newline$

$\textbf{(C) }$ $3.59\times10^4\ \text{W/m}^2 \qquad\newline$

$\textbf{(D) }$ $4.77\times10^4\ \text{W/m}^2 \qquad\newline$

$\textbf{(E) }$ $2.98\times10^5\ \text{W/m}^2$

$\textbf{D}$

The intensity $I=P/A=4.77\times10^4\ \text{W/m}^2$.

The unit that is used to measure magnetic flux is:

$\textbf{(A) }$ Coulomb (C)$ \qquad\newline$ $\textbf{(B) }$ Farad (F)$ \qquad\newline$ $\textbf{(C) }$ Tesla (T) $ \qquad\newline$ $\textbf{(D) }$ Weber (Wb) $ \qquad\newline$ $\textbf{(E) }$ Henry (H)

$\textbf{D}$

Coulomb (C) measures electric charge, Farad (F) measures capacitance, Tesla (T) measures magnetic field, and Henry (H) measures mutual inductance. So tell me who can measure the intelligence of the webmaster 🙂

A curve on a racetrack has a radius of 80 m and is banked at $45^\circ$. Suppose that the road surface on the curve somehow became frictionless (perhaps caused by an ice storm or an oil spill) and a car needs to navigate this curve. What is the safe speed with which to take the curve without either sliding up or down the bank of the curve?

$\textbf{(A) }$ 9 m/s$ \qquad$ $\textbf{(B) }$ 14 m/s$ \qquad$ $\textbf{(C) }$ 21 m/s$ \qquad$ $\textbf{(D) }$ 28 m/s$ \qquad$ $\textbf{(E) }$ 33 m/s

$\textbf{D}$

On a banked curve the normal force on the car is no longer vertical and has a horizontal component. This component then acts as the centripetal force $mg\tan\theta=m\dfrac{v^2}{R}$, so $R=mg\tan\theta=20\sqrt{2}\ \text{m/s}$.

A convex lens has a focal length of 50 mm. How far from the lens must an object be placed if it is to form a virtual image magnified in size by a factor of three?

$\textbf{(A) }$ 33 mm$ \qquad$ $\textbf{(B) }$ 38 mm$ \qquad$ $\textbf{(C) }$ 43 mm$ \qquad$ $\textbf{(D) }$ 48 mm$ \qquad$ $\textbf{(E) }$ 53 mm

$\textbf{A}$

Virtual image magnified in size by a factor of three means $d_i=-3d_o$. According to $\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac1f$, we get $d_o=\dfrac23f=33\ \text{mm}$.

Consider the bicycle wheel shown at the right to be a ring with a 60 cm diameter and a mass of 1.5 kg. Attached to the wheel is a gear with a radius of 4.0 cm and negligible mass. A force of 20 N is applied tangentially to the gear for 4.0 s. Starting from rest, what linear speed does the wheel achieve, assuming it rolls without slipping?

$\textbf{(A) }$ 3.0 m/s$ \qquad$ $\textbf{(B) }$ 5.9 m/s$ \qquad$ $\textbf{(C) }$ 7.1 m/s$ \qquad$ $\textbf{(D) }$ 16.4 m/s$ \qquad$ $\textbf{(E) }$ 24 m/s

$\textbf{C}$

The torque applied on the gear is $\tau=Fr=I\beta$, where $r=4\ \text{cm}$ is the radius of the gear, $I=mR^2$ is the moment of inertia of the wheel, and $\beta$ is the angular acceleration. By the torque equation we get $\beta=\dfrac{Fr}{mR^2}$, the angular speed after $t=4\ \text{s}$ is $\omega=\beta t$, and the linear speed of the wheel is $v=\omega R$.

The intensity level is measured to be 60 dB at a distance of 15 m from a speaker. What is the intensity level at a point 2.0 m from the speaker? Assume that the speaker radiates equally in all directions

$\textbf{(A) }$ 55.7 dB$ \qquad$ $\textbf{(B) }$ 57.5 dB$ \qquad$ $\textbf{(C) }$ 67.0 dB$ \qquad$ $\textbf{(D) }$ 75.5 dB$ \qquad$ $\textbf{(E) }$ 77.5 dB

$\textbf{E}$

Supposing the intensity at $r_1=15\ \text{m}$ is $I_1$ and the intensity at $r_2=2\ \text{m}$ is $I_2$. According to $I=\dfrac PA$ and $A=4\pi r^2$, $I_1\cdot4\pi r_1^2=I_2\cdot2\pi r_2^2$, we have $$I_2=\dfrac{r_1^2}{r_2^2}I_1$$ For $dB=10\log\dfrac{I}{I_0}$, $60=10\log\dfrac{I_1}{I_0}$, what we need is $$10\log\dfrac{I_2}{I_0}=10\log\dfrac{r_1^2}{r_2^2}\dfrac{I_1}{I_0}=10(\log r_1^2-\log r_2^2+\log\dfrac{I_1}{I_0}=60+20(\log r_1-\log r_2)$$

A pine wood block is floating in a small pool. There is a second pine wood block that sits on top of the first and does not touch the water. If the top block is taken off and placed in the water, how does the new water level in the pool compare to the original water level?

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

$\textbf{(B) }$ Lowers$ \qquad\newline$

$\textbf{(C) }$ Does not change$ \qquad\newline$

$\textbf{(D) }$ It depends upon the size of the blocks$ \qquad\newline$

$\textbf{(E) }$ It depends upon the amount of water in the pool

$\textbf{C}$

Archimedes' principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. For the floating block, the buoyant force is equal to the weight of the block. Because the mass remains the same in either arrangement of the blocks, the water level remains unchanged.

Two cars are being tested on a track. Car 1 accelerates from rest on this straight track at $a_1=3.0\ \text{m/s}^2$. Two seconds later, Car 2 accelerates from rest at $a_2=12.0\ \text{m/s}^2$. How much time after Car 1 starts will Car 2 pass Car 1?

$\textbf{(A) }$ 3.0 s$ \qquad$ $\textbf{(B) }$ 4.0 s$ \qquad$ $\textbf{(C) }$ 5.0 s$ \qquad$ $\textbf{(D) }$ 6.0 s$ \qquad$ $\textbf{(E) }$ 7.0 s

$\textbf{B}$

For car 1, $x_1=\dfrac12a_1t^2$;$\newline$

For car 2, $x_2=\dfrac12a_2(t-2)^2\ (\text{only for } t>2\ \text{s})$;$\newline$

Car 2 catches up with car 1 when $x_1=x_2$, so $$\dfrac12a_1t^2=\dfrac12a_2(t-2)^2$$by solving the equation we get $$t=\dfrac{2\sqrt{\frac{a_2}{a_1}}}{\sqrt{\frac{a_2}{a_1}}-1}=4\ \text{s}$$

The table below lists the finish times for the 2008 Men’s Olympic 100 m butterfly swim final in Beijing. From the data, what is the best estimate of the distance the 2nd place finisher, Cavic, was behind the 1st place finisher, Phelps?

$\textbf{(A) }$ 0.2 cm$ \qquad$ $\textbf{(B) }$ 2.0 cm$ \qquad$ $\textbf{(C) }$ 5.0 cm$ \qquad$ $\textbf{(D) }$ 10 cm$ \qquad$ $\textbf{(E) }$ 20 cm

$\textbf{B}$

Phelps takes $t_1=50.58\ \text{s}$ to finish $x=100\ \text{m}$, so $v_1=x/t_1$; Cavic takes $t_2=50.59\ \text{s}$ for the same distance, so $v_2=x/t_2$. When Phelps reaches the finish, Cavic swims for $v_2t_1$, which is $x-v_2t_1=1.96\ \text{cm}$ behind.

In a mixture of hydrogen, oxygen, and nitrogen gases at room temperature, the molecules having the greatest average speed are those of

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

$\textbf{(B) }$ Oxygen$ \qquad\newline$

$\textbf{(C) }$ Nitrogen$ \qquad\newline$

$\textbf{(D) }$ All have the same speed$ \qquad\newline$

$\textbf{(E) }$ It depends upon the composition of the mixture

$\textbf{A}$

The average speed $\overline{v}=\sqrt{\dfrac{8kT}{\pi m}}$ is greater for lighter atom.

**Questions 36 and 37 refer to the circuit below. The battery is assumed ideal with an emf of 3.0 V**

The resistor that dissipates the most power is:

$\textbf{(A) }$ $R_1 \qquad$ $\textbf{(B) }$ $R_2 \qquad$ $\textbf{(C) }$ $R_3 \qquad$ $\textbf{(D) }$ $R_4 \qquad$ $\textbf{(E) }$ $R_5$

$\textbf{A}$

$R_3$ and $R_4$ are in parallel so $R_{34}=20\ \Omega$;$\newline$

$R_{34}$ and $R_5$ are in series so $R_{345}=50\ \Omega$; $\newline$

$R_{345}$ and $R_2$ are in parallel, so $R_{2345}=25\ \Omega$;$\newline$

The voltage on each resistor is $V_1=2\ \text{V}$, $V_2=1\ \text{V}$, $V_3=V_4=0.4\ \text{V}$, $V_5=0.6\ \text{V}$. By $P=V^2/R$ we can get the power of each resistor.

The voltage across resistor $R_4$ is:

$\textbf{(A) }$ 0.4 V$ \qquad$ $\textbf{(B) }$ 0.6 V$ \qquad$ $\textbf{(C) }$ 1.2 V$ \qquad$ $\textbf{(D) }$ 1.5 V$ \qquad$ $\textbf{(E) }$ 3.0 V

$\textbf{A}$

Problem solved in Question 36.

Shown here is accelerometer data for an elevator ride in a skyscraper in Chicago. The spikes at roughly 10 s and 72 s are caused by the handling of the accelerometer which rested on the floor of the elevator and occurred before the elevator started moving up, and after it reached the top, respectively. What is the best estimate for the peak vertical speed for this elevator

$\textbf{(A) }$ 4 m/s$ \qquad$ $\textbf{(B) }$ 8 m/s$ \qquad$ $\textbf{(C) }$ 12 m/s$ \qquad$ $\textbf{(D) }$ 16 m/s$ \qquad$ $\textbf{(E) }$ 20 m/s

$\textbf{B}$

The largest acceleration value from the graph is $-10.4\ \text{m/s}^2$, which is $0.6\ \text{m/s}^2$ deeper than the base line of $-9.8\ \text{m/s}^2$ (the gravitational acceleration), and it lasts for about 13 s. So the peak vertical velocity is $0.6\ \text{m/s}^2\cdot13\ \text{s}=7.8\ \text{m/s}$.

The picture shows two speakers, $A$ and $B$, which emit the same 680 Hz tones in phase. Point P is located 1.000 m directly in front of Speaker $A$. A sound sensor at Point P records a minimum sound intensity. Speaker $B$ is slowly moved away from Speaker $A$ along the line joining them. How far must Speaker $B$ be moved further from Speaker $A$ until the sound sensor at Point P first records a maximum sound intensity?

$\textbf{(A) }$ 0.368 m$ \qquad$ $\textbf{(B) }$ 0.438 m$ \qquad$ $\textbf{(C) }$ 0.500 m$ \qquad$ $\textbf{(D) }$ 0.686 m$ \qquad$ $\textbf{(E) }$ 0.982 m

$\textbf{A}$

The wavelength $\lambda=v/f=0.5\ \text{m}$. $AP=2\lambda$, $BP=2.5\lambda$, so the two waves cancel each other at point P. When we move speak $B$ to $B_2$, to records a maximum sound intensity at P, we need $B_2P=3\lambda=1.5\ \text{m}$, so $AB_2=\sqrt{1.5^2-1^2}\ \text{m}=\dfrac{\sqrt5}{2}\ \text{m}$, $BB_2=(\dfrac{\sqrt5}{2}-0.75)\ \text{m}$.

The bulb of a mercury thermometer has a volume of $0.100\ \text{cm}^3$ at $10^\circ$C and contains $0.100\ \text{cm}^3$ of mercury. The capillary tube above the bulb has a cross-sectional area of $0.012\ \text{mm}^2$. The volume thermal expansion coefficient of mercury is $1.8\times10^{-4}(^\circ\text{C})^{-1}$. How much will the mercury rise when the temperature rises by $30\ ^\circ\text{C}$ (the expansion of the glass is negligible)?

$\textbf{(A) }$ 0.045 mm$ \qquad$ $\textbf{(B) }$ 0.45 mm$ \qquad$ $\textbf{(C) }$ 4.5 mm$ \qquad$ $\textbf{(D) }$ 45 mm$ \qquad$ $\textbf{(E) }$ 45 cm

$\textbf{D}$

For the change of temperature $\Delta T=20\ ^\circ\text{C}$ and thermal expansion coefficient $k=1.8\times10^{-4}(^\circ\text{C})^{-1}$, the change of volume is $\Delta V=k\Delta T$, so the height is $h=\Delta V/A$.

To monitor the breathing of a hospital patient, a thin belt is wrapped around a patient’s chest and back. The belt is a 200 turn coil of conducting wire. When the patient inhales, the area of the coil increases by $39.0\ \text{cm}^2$. The magnitude of the Earth’s magnetic field is $50.0\ \mu\text{T}$ and makes an angle of $28.0^\circ$ to the plane of the coil. Assuming the patient takes 1.80 s to inhale, how much emf is induced in the coil?

$\textbf{(A) }$ $1.91\times10^{-5}\ \text{V} \qquad\newline$ $\textbf{(B) }$ $9.57\times10^{-8}\ \text{V} \qquad\newline$ $\textbf{(C) }$ $9.57\times10^{-4}\ \text{V} \qquad\newline$ $\textbf{(D) }$ $1.02\times10^{-5}\ \text{V} \qquad\newline$ $\textbf{(E) }$ $3.44\times10^{-5}\ \text{V}$

$\textbf{D}$

The induced emf $\mathcal{E}=NB\dfrac{\Delta A}{\Delta t}\cos\theta$.

Three identical samples of a monatomic ideal gas are taken from a temperature $T$ to a temperature $2T$. Sample A undergoes an adiabatic process. Sample B undergoes an isobaric process and Sample C undergoes an isochoric process. Which of the following correctly ranks the heats added to the samples during the three processes?

$\textbf{(A) }$ $Q_A<Q_B<Q_C \qquad\newline$ $\textbf{(B) }$ $Q_A<Q_C<Q_B \qquad\newline$ $\textbf{(C) }$ $Q_B<Q_A<Q_C \qquad\newline$ $\textbf{(D) }$ $Q_C<Q_A<Q_B \qquad\newline$ $\textbf{(E) }$ $Q_C<Q_B<Q_A$

$\textbf{B}$

In an adiabatic process, no heat is allowed to flow into or out of the system, so $Q_A=0$;$\newline$

In an isobaric process, pressure is held constant so $Q_B=\Delta U+W$;$\newline$

In an isochoric process, $V$ is constant so $Q_C=\Delta U$.

Two boxes are stacked on a table as shown below. The mass of box 1 is

$m$ and the mass of box 2 is $3m$. The surface between box 2 and the table is smooth and the surface between the two boxes is rough. When a force, $F$, is applied, box 1 does not slide on box 2. What is the minimum coefficient of static friction between the boxes?

$\textbf{(A) }$ $\dfrac{F}{4mg} \qquad$ $\textbf{(B) }$ $\dfrac{F}{3mg} \qquad$ $\textbf{(C) }$ $\dfrac{F}{2mg} \qquad$ $\textbf{(D) }$ $\dfrac{F}{mg} \qquad$ $\textbf{(E) }$ $\dfrac{2mg}{F}$

$\textbf{A}$

Two boxes are moving with the same acceleration $a=\dfrac{F}{4m}$, where box 1 is drove by frictional force, so $\mu mg=ma$, thus we get $\mu=\dfrac{F}{4mg}$.

Consider a traveling wave on a string of length $L$, mass $M$, and tension $T$. A standing wave is set up. Which of the following is true?

$\textbf{(A) }$ The wave velocity depends on $M$, $L$, $T$. $ \qquad\newline$

$\textbf{(B) }$ The wavelength of the wave is proportional to the frequency.$ \qquad\newline$

$\textbf{(C) }$ The velocity of a given particle in the string is equal to the wave velocity.$ \qquad\newline$

$\textbf{(D) }$ The wavelength is proportional to $T$.$ \qquad\newline$

$\textbf{(E) }$ The frequency depends upon $L$.

$\textbf{A}$

The speed of a wave on a stretched string is $v=\sqrt{\dfrac{T}{\mu}}$, where the linear density $\mu=\dfrac{M}{L}$.

A heat engine is operating between $40^\circ\text{C}$ and $380^\circ\text{C}$ and has an efficiency that is $60\%$ of a Carnot engine that is operating between the same temperatures. If the engine absorbs heat at a rate of 60 kW, at what rate does it exhaust heat?

$\textbf{(A) }$ 36 kW$ \qquad$ $\textbf{(B) }$ 41 kW$ \qquad$ $\textbf{(C) }$ 48 kW$ \qquad$ $\textbf{(D) }$ 57 kW$ \qquad$ $\textbf{(E) }$ 60 kW

$\textbf{B}$

The efficiency is $\eta=60\%\times(1-\dfrac{T_2}{T_1})=0.31$, so the exhausted heat is $Q_2=(1-\eta)Q=41\ \text{kW}$.

In physics experiments located deep underground, the two types of cosmic rays that most commonly reach the experimental apparatus are:

$\textbf{(A) }$ alpha particles and neutrons$ \qquad\newline$

$\textbf{(B) }$ protons and electrons$ \qquad\newline$

$\textbf{(C) }$ iron nuclei and carbon nuclei$ \qquad\newline$

$\textbf{(D) }$ muons and neutrinos$ \qquad\newline$

$\textbf{(E) }$ positrons and electrons

$\textbf{D}$

The two most common types of cosmic rays that reach the experimental apparatus are muons and neutrinos.

When two identical resistors are connected in series to an ideal voltage source, the current supplied by the source is 2.0 Amperes. When these two resistors are connected in parallel to the same ideal voltage source, what is the current supplied by the source?

$\textbf{(A) }$ 0.5 Ampere$ \qquad\newline$ $\textbf{(B) }$ 1.0 Ampere$ \qquad\newline$ $\textbf{(C) }$ 2.8 Amperes$ \qquad\newline$ $\textbf{(D) }$ 4.0 Amperes$ \qquad\newline$ $\textbf{(E) }$ 8.0 Amperes

$\textbf{E}$

For two identical resistors of resistance $R$, $I_{series}=\dfrac{U}{2R}=2\ \text{A}$, $I_{parallel}=\dfrac{U}{0.5R}=8\ \text{A}$.

It is observed that a charged particle moves through a region of space and experiences no magnetic force. From this we can conclude that

$\textbf{(A) }$ no magnetic field exists in that region of space.$ \qquad\newline$

$\textbf{(B) }$ the particle is moving parallel to the magnetic field.$ \qquad\newline$

$\textbf{(C) }$ the particle is moving perpendicular to the magnetic field.$ \qquad\newline$

$\textbf{(D) }$ either no magnetic field exists or the particle is moving parallel to the magnetic field.$ \qquad\newline$

$\textbf{(E) }$ either no magnetic field exists or the particle is moving perpendicular to the magnetic field.

$\textbf{D}$

$\vec{F}=q\vec{v}\times\vec{B}=0$, the magnetic field could be 0 or parallel to the velocity.

A lump of clay whose rest mass is 4.0 kg is traveling at three-fifths of the speed of light when it collides head-on with an identical lump going in the opposite direction at the same speed. If the two lumps stick together and no energy is radiated away, what is the mass of the composite lump?

$\textbf{(A) }$ 4.0 kg$ \qquad$ $\textbf{(B) }$ 6.4 kg$ \qquad$ $\textbf{(C) }$ 8.0 kg$ \qquad$ $\textbf{(D) }$ 10.0 kg$ \qquad$ $\textbf{(E) }$ 13.3 kg

$\textbf{D}$

The total energy $$E=2\cdot\dfrac{m_0c^2}{\sqrt{1-v^2/c^2}}=M_0c^2$$ so the mass of the composite lump $M_0=\dfrac52m_0=10\ \text{kg}$.

One mole of ideal gas initially at temperature $T_0$ and volume $V_0$ undergoes a reversible isothermal expansion to $V_1$. If the ratio of specific heats is $c_p/c_v=\gamma$ and if $R$ is the gas constant, the work done by the gas is:

$\textbf{(A) }$ Zero$ \qquad\newline$ $\textbf{(B) }$ $RT_0\left(\dfrac{V_1}{V_0}\right)^\gamma \qquad\newline$ $\textbf{(C) }$ $RT_0\left(\dfrac{V_1}{V_0}-1\right) \qquad\newline$ $\textbf{(D) }$ $c_vT_0\left[1-\left(\dfrac{V_0}{V_1}\right)^{\gamma-1}\right] \qquad\newline$ $\textbf{(E) }$ $RT_0\ln\left(\dfrac{V_1}{V_0}\right)$

$\textbf{E}$

The work $W=\int p\ \text{d}V=\int_{V_0}^{V_1}\dfrac{RT_0}{V}\ \text{d}V=RT_0\ln(\dfrac{V_1}{V_0})$