PhysicsBowl 2010 – Ivy Learner (2024)

The very first PhysicsBowl took place in 1985. Approximately how many seconds ago did students compete in this first contest?

$\textbf{(A) }$ $10^2 \qquad$ $\textbf{(B) }$ $10^5 \qquad$ $\textbf{(C) }$ $10^7 \qquad$ $\textbf{(D) }$ $10^9 \qquad$ $\textbf{(E) }$ $10^{12}$

$\textbf{D}$

$25\ \text{yr}\times\dfrac{365\ \text{day}}{1\ \text{yr}}\times\dfrac{24\ \text{hr}}{1\ \text{dy}}\times\dfrac{60\ \text{min}}{1\ \text{hr}}\times\dfrac{60\ \text{s}}{1\ \text{min}}=7.88\times10^8\ \text{s}\approx10^9\ \text{s}$

Which of the following is $\textit{NOT}$ a vector quantity?

$\textbf{(A) }$ Acceleration$ \qquad\newline$ $\textbf{(B) }$ Average Velocity $ \qquad\newline$ $\textbf{(C) }$ Linear Momentum $ \qquad\newline$ $\textbf{(D) }$ Potential Energy $ \qquad\newline$ $\textbf{(E) }$ Force

$\textbf{D}$

Energy is a scalar quantity since it has no direction.

The following three measurements of length are given: $L_1=22.05\ \text{m}$, $L_2=6.1123\ \text{m}$, and $L_3=89.6\ \text{m}$. What is the sum of these measurements using rules of proper significant digits?

$\textbf{(A) }$ 120 m$ \qquad$ $\textbf{(B) }$ 118 m$ \qquad$ $\textbf{(C) }$ 117.8 m$ \qquad$ $\textbf{(D) }$ 117.76 m$ \qquad$ $\textbf{(E) }$ 117.7623 m

$\textbf{C}$

Addition rules of sig. figs means that the calculation stops at the “last column” for which each number is known. Hence, since $L_3$ is known only to the tenths place, the sum must be recorded to the tenths. Hence, 117.7623 m is rounded to 117.8 m.

Which of the following relationships correctly ranks the three given speeds from least to greatest? The speeds are given as $v_1=1.25\times10^{-4}\ \text{cm/}\mu\text{s}$, $v_2=0.076\ \text{Mm/week}$, and $v_3=9.50\ \text{kg/day}$.

$\textbf{(A) }$ $v_1<v_2<v_3 \qquad\newline$ $\textbf{(B) }$ $v_3<v_2<v_1 \qquad\newline$ $\textbf{(C) }$ $v_2<v_3<v_1 \qquad\newline$ $\textbf{(D) }$ $v_1<v_3<v_2 \qquad\newline$ $\textbf{(E) }$ $v_3<v_2=v_1$

$\textbf{B}$

Let’s change all quantities to . Changing units leads to $\newline$
$v_1=1.25\times10^{-4}\ \dfrac{\text{cm}}{\mu\text{s}}=1.25\times10^{-4}\ \dfrac{10^{-2}\ \text{m}}{10^{-6}\ \text{s}}=1.25\ \text{m/s}$;$\newline$
$v_2=0.076\ \dfrac{\text{Mm}}{\text{week}}=0.076\ \dfrac{10^6\ \text{m}}{(7)(24)(60)(60)\ \text{s}}=0.126\ \text{m/s}$;$\newline$
$v_3=9.50\ \dfrac{\text{km}}{\text{dy}}=9.50\ \dfrac{10^3\ \text{m}}{(24)(60)(60)\ \text{s}}=0.110\ \text{m/s}$;$\newline$
Hence we get $v_3<v_2<v_1$.

When computed with proper MKS units, the Universal Gas Constant divided by Boltzmann’s constant is equal to

$\textbf{(A) }$ the speed of light.$ \qquad\newline$
$\textbf{(B) }$ Planck’s constant.$ \qquad\newline$
$\textbf{(C) }$ Avogadro’s number.$ \qquad\newline$
$\textbf{(D) }$ the permittivity of free space.$ \qquad\newline$
$\textbf{(E) }$ the Universal Gravitational constant.

$\textbf{C}$

You can find out the conclusion by calculating $$\dfrac{R}{k_B}=\dfrac{8.31\ \text{J/mol}\cdot{K}}{1.38\times10^{-23}\ \text{J/K}}=6.02\times10^{23}\ \text{mol}^{-1}=N_A$$ Or just remember it $k_B=\dfrac{R}{N_A}$, which is quite useful in thermodynamics.

Which of the following statements is most closely associated with Newton’s Third Law of Motion?

$\textbf{(A) }$ “What goes up must come down.”$ \qquad\newline$
$\textbf{(B) }$ “For every action force, there is an equal but opposite reaction force.”$ \qquad\newline$
$\textbf{(C) }$ “An object at rest remains at rest.”$ \qquad\newline$
$\textbf{(D) }$ “The acceleration of an object is directly proportional to the force acting on the object but inversely proportional to the mass of the object.”$ \qquad\newline$
$\textbf{(E) }$ “The Universe tends toward disorder.”

$\textbf{B}$

Anyone who chooses the right answer will be rewarded with a piece of Newton cookie!

A small object is thrown straight downward on Earth with an initial speed of 12.0 m/s from a position 10.0 m above the ground. Ignoring air resistance, the speed of the object when it reaches the ground is

$\textbf{(A) }$ 18.4 m/s$ \qquad$ $\textbf{(B) }$ 14.6 m/s$ \qquad$ $\textbf{(C) }$ 14.0 m/s$ \qquad$ $\textbf{(D) }$ 12.8 m/s$ \qquad$ $\textbf{(E) }$ 12.0 m/s

$\textbf{A}$

By the formula $v_f^2-v_i^2=2ax$, we have $v_f=\sqrt{v_i^2+2gh}=\sqrt{(12)^2+2(9.8)(10)}=18.4\ \text{m/s}$.

What is the period of the mass’s oscillation (in units of seconds)?

$\textbf{(A) }$ 2.0$ \qquad$ $\textbf{(B) }$ 4.0$ \qquad$ $\textbf{(C) }$ 5.0$ \qquad$ $\textbf{(D) }$ 8.0$ \qquad$ $\textbf{(E) }$ 10.0

$\textbf{D}$

The troughs of wave appear at $t=0\ \text{s}$ and $t=8\ \text{s}$, thus the period is $T=8\ \text{s}$.

What is the amplitude of the mass’s oscillation (in units of meters)?

$\textbf{(A) }$ 1.0$ \qquad\newline$
$\textbf{(B) }$ 2.0$ \qquad\newline$
$\textbf{(C) }$ 3.0$ \qquad\newline$
$\textbf{(D) }$ 4.0$ \qquad\newline$
$\textbf{(E) }$ 5.0

$\textbf{A}$

The height of the crest is 5.0 m, and the trough is 3.0 m. The amplitude is measured as half the distance between crest and trough, thus we have $A=1.0\ \text{m}$.

A particle travels at a constant speed around a circular path of radius $R$. If the particle makes one complete trip around the entire circle, what is the magnitude of the displacement for this trip?

$\textbf{(A) }$ $\pi R \qquad$ $\textbf{(B) }$ $2R \qquad$ $\textbf{(C) }$ $2\pi R \qquad$ $\textbf{(D) }$ $\pi R^2 \qquad$ $\textbf{(E) }$ 0

$\textbf{E}$

Displacement (a vector) is the change in position. For an object completing one trip around a circle, it ends at the same location at which it begins meaning that there is NO change in position.

Consider the motion of an object given by the velocity vs. time graph shown. For which time(s) is the acceleration of the object equal to $0\ \text{m/s}^2$?

$\textbf{B}$

Acceleration is related to the slope of the velocity vs. time graph. To find where the acceleration is zero, we need to find a place on the velocity graph which has a tangent line that is locally horizontal. This occurs only at time $t=5.0\ \text{s}$.

$\textbf{B}$

By Newton’s Third Law, the force of the proton on the nucleus is equal and opposite to the force on the helium nucleus by the proton. By Newton’s Second Law $F=ma$, since the force is the same on each object, the object with less mass (the proton) has a larger magnitude of acceleration.

The Large Hadron Collider (LHC) is the largest particle accelerator in the world. To which of the following countries would you have to travel in order to visit the LHC?

$\textbf{(A) }$ Germany$ \qquad$ $\textbf{(B) }$ Italy$ \qquad$ $\textbf{(C) }$ Switzerland $ \qquad$ $\textbf{(D) }$ United States$ \qquad$ $\textbf{(E) }$ Russia

$\textbf{C}$

The LHC is located along the border of France and Switzerland at CERN near Geneva.

What temperature change on the Kelvin scale is equivalent to a 27 degree change on the Celsius scale?

$\textbf{(A) }$ 300 K$ \qquad$ $\textbf{(B) }$ 273 K$ \qquad$ $\textbf{(C) }$ 246 K$ \qquad$ $\textbf{(D) }$ 27 K$ \qquad$ $\textbf{(E) }$ 9 K

$\textbf{D}$

The equation between Kelvin temperature and Celsius temperature is $T_K=T_C+273$. Hence we know the change in temperature has the same magnitude on both scales.

Three blocks, labeled A,B, and C, remain at rest on a table. The magnitude of the gravitational force on each block is indicated on the figure. What is the magnitude of the contact force on block C from block B?

$\textbf{D}$

Considering the blocks A+B to be a system, they have a 12.0 N gravitational force. Consequently, from the free body diagram of the A+B system, there is a contact force of $n=12\ \text{N}$ directed upward to keep the system from accelerating. By Newton’s Third Law, this 12.0 N force exerted on the surface of block B by block C is the same in magnitude of the force provided onto block C from the surface of block B.

A mass $M_1=40\ \text{kg}$ is at the very edge of a 6.0 m long plank which is pivoted about its center of mass located directly at the center of its length. How far from the center of the plank $(X)$ should the mass $M_2=80\ \text{kg}$ be placed so that the plank remains in static equilibrium in a horizontal position?

$\textbf{C}$

By the lever rule, $M_1\cdot(3\ \text{m})=M_2x$. Hence we get $x=3\dfrac{M_1}{M_2}=1.5\ \text{m}$.

A tube of air open at only one end is vibrating in the 5th harmonic from a tuning fork of frequency 120 Hz placed at the open end. If this tube of air were to vibrate at its fundamental frequency, what frequency tuning fork would be needed at the open end?

$\textbf{(A) }$ 24 Hz$ \qquad$ $\textbf{(B) }$ 30 Hz$ \qquad$ $\textbf{(C) }$ 40 Hz$ \qquad$ $\textbf{(D) }$ 125 Hz$ \qquad$ $\textbf{(E) }$ 600 Hz

$\textbf{A}$

The frequency for the $n$th harmonic of a tube is equal to $n$ times the fundamental frequency $f_1$. This means that we have (regardless of whether the tube was open at both ends or not) that $f_n=nf_1\rightarrow f_1=\dfrac{f_n}{n}=\dfrac{125\ \text{Hz}}{5}=24\ \text{Hz}$.

Of the following choices, which best describes a “rapidly rotating neutron star emitting radiation” ?

$\textbf{(A) }$ Supernova$ \qquad$ $\textbf{(B) }$ Quasar$ \qquad$ $\textbf{(C) }$ Nebula$ \qquad$ $\textbf{(D) }$ White Dwarf$ \qquad$ $\textbf{(E) }$ Pulsar

$\textbf{E}$

Rapidly rotating neutron stars emitting radiation are most closely associated with pulsars.

Which is the correct relationship for the speed $(v)$ of the following three electromagnetic waves in vacuum? The waves are : X-rays, UV rays, microwaves.

$\textbf{(A) }$ $v_{X ray}<v_{microwaves}<v_{UV rays} \qquad\newline$
$\textbf{(B) }$ $v_{microwaves}<v_{UV rays}<v_{X ray} \qquad\newline$
$\textbf{(C) }$ $v_{X ray}=v_{microwaves}=v_{UV rays} \qquad\newline$
$\textbf{(D) }$ $v_{UV rays}<v_{microwaves}<v_{X ray} \qquad\newline$
$\textbf{(E) }$ $v_{X ray}<v_{UV rays}<v_{microwaves}$

$\textbf{C}$

All electromagnetic waves travel in vacuum with the speed of light $c=3.0\times10^8\ \text{m/s}$.

A sample of ideal gas is in a container at a temperature of $100\ ^\circ\text{C}$ and a pressure of 2.50 atm . If the volume of the container is 25 L , approximately how many molecules of gas are in the container?

$\textbf{(A) }$ $4.58\times10^{24} \qquad$ $\textbf{(B) }$ $1.23\times10^{24} \qquad$ $\textbf{(C) }$ $6.25\times10^{23} \qquad$ $\textbf{(D) }$ $4.53\times10^{22} \qquad$ $\textbf{(E) }$ $1.21\times10^{22}$

$\textbf{B}$

By the ideal gas equation, the mole number of molecules is $$n=\dfrac{PV}{RT}=\dfrac{(2.50\times1.013\times10^5)(25\times10^{-3})}{(8.31)(100+273)}=2.04\ \text{mol}$$ So the number of molecules is $N=nN_A=(2.04)(6.02\times10^{23})=1.23\times10^{24}$.

Which terminology is best associated with the amount of energy required to change the phase of a material per unit mass?

$\textbf{(A) }$ Thermal conductivity$ \qquad\newline$ $\textbf{(B) }$ Specific heat$ \qquad\newline$ $\textbf{(C) }$ Work$ \qquad\newline$ $\textbf{(D) }$ Latent heat$ \qquad\newline$ $\textbf{(E) }$ Entropy

$\textbf{D}$

To change the phase of a material, the quantity most associated with this is called the latent heat. Specific heat refers to the energy associated with the change in temperature of a material per unit mass.

A small and uniform 10.0 kg object floats at rest with $75\ \%$ of the object submerged in a cubical tank containing $50.0\ \text{m}^3$ of water. What is the buoyant force acting on the object? Treat $g=10\ \text{m/s}^2$.

$\textbf{(A) }$ 500000 N$ \qquad$ $\textbf{(B) }$ 375000 N$ \qquad$ $\textbf{(C) }$ 100 N$ \qquad$ $\textbf{(D) }$ 75 N$ \qquad$ $\textbf{(E) }$ 25 N

$\textbf{C}$

The object floats at rest, thus the buoyant force is equal to the gravitational force $F_{buoyant}=G=mg=(10)(10)=100\ \text{N}$.

A point object of mass $M$ is connected to the end of a long string of negligible mass and the system swings as a simple pendulum with period $T$ . The point object of mass $M$ is now replaced with a point object of mass $4M$. When this new system swings as a simple pendulum, what is its period?

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

$\textbf{C}$

The period of a simple pendulum depends only on the string length and the acceleration due to gravity $T=2\pi\sqrt{\dfrac{l}{g}}$, not the mass connected to the end of the pendulum.

By computing the area under the acceleration vs. time graph for a fixed time interval of an object’s motion, what quantity has been determined for that object?

$\textbf{(A) }$ The average velocity during the time interval$ \qquad\newline$
$\textbf{(B) }$ The velocity at the end of the time interval$ \qquad\newline$
$\textbf{(C) }$ The average speed during the time interval$ \qquad\newline$
$\textbf{(D) }$ The change in velocity during the time interval$ \qquad\newline$
$\textbf{(E) }$ The velocity at the time midway through the time interval

$\textbf{D}$

The acceleration is defined as $a=\dfrac{\text{d}v}{\text{d}t}$. It can be rewrote in integral form as $\Delta v=\int a\ \text{d}t$, which means the area under the acceleration vs. time graph is the change in velocity during the time interval.

Two point objects are launched straight upward with identical linear momentum. Object 1 has mass $M$ and reaches a maximum height $H$ above the launch point. If object 2 has mass $2M$, what is its maximum attained height above the launch point in terms of $H$?

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

$\textbf{A}$

Since the two objects have the same linear momentum but the mass of object 2 is twice the mass of object 1, the initial speed of object 2 is half of object 1. By $H=\dfrac{v^2}{2g}$, the maximum height object 2 can reach is 1/4 of height $H$.

$\textbf{E}$

METHOD 1: Using the right-hand rule. Right fingers point in the direction of the electron’s velocity and the fingers are curled into the direction of the magnetic field. Here, that results in the right thumb pointing into the plane of the page. However, since the electron is negatively charged, one must flip the hand over to find the final direction of the force. This means that the force when entering the field region points directly out of the plane of the page.

METHOD 2: Only motion perpendicular to a magnetic field results in a magnetic force on a charged particle. For this scenario, only the motion of the electron directed up the page gives rise to a magnetic force (the component of velocity directed to the right is parallel to the magnetic field and produces no force). Therefore, using the right-hand rule, the right fingers are pointed upward, the fingers are curled to the right, leaving the right thumb pointing into the plane of the page. Since the charge in question is negative, the hand is flipped over to find the direction of the magnetic force (out of the plane of the page).

The circuit shown contains a battery (of emf $V_{bat}$ ) with an internal resistance $r$ connected to a rheostat (variable resistor). When the resistance of the rheostat is increased, which of the following statements is true?

$\textbf{A}$

When the resistance of the variable resistor increased, the total resistance increased. Thus the current decreased, leading to the decrease of the voltage across resistor $r$. The total voltage $V_{bat}$ is a constant, so the terminal $(V_a-V_b)$ increases.

Which of the following $\textit{could}$ produce an enlarged but inverted image of a real object?

$\textbf{(A) }$ Place a converging lens at a distance greater than its focal length from the object.$ \qquad\newline$
$\textbf{(B) }$ Place a converging lens at a distance less than its focal length from the object.$ \qquad\newline$
$\textbf{(C) }$ Place a diverging lens at a distance less than the magnitude of its focal length from the object.$ \qquad\newline$
$\textbf{(D) }$ Place a diverging lens at a distance greater than the magnitude of its focal length from the object.$ \qquad\newline$
$\textbf{(E) }$ It is not possible to create the type of image desired.

$\textbf{A}$

If one places a converging lens at a distance between the focal length and twice the focal length, the resulting image will be real (inverted) and it would be larger. If the lens is placed at a distance equal to twice the focal length, the image is the same size as the object… placing the object outside of twice the focal length results in a smaller inverted image. The diverging lens always produces a smaller virtual image and by placing the object inside the focal length of the converging lens produces a larger but virtual image.

A ball initially at rest falls without air resistance from a height $h$ above the ground. If the ball falls the first distance $h/2$ in a time $t$, how much time is required to fall the remaining distance of $h/2$?

$\textbf{(A) }$ $0.25t \qquad$ $\textbf{(B) }$ $0.41t \qquad$ $\textbf{(C) }$ $0.50t \qquad$ $\textbf{(D) }$ $0.71t \qquad$ $\textbf{(E) }$ $1.00t$

$\textbf{B}$

For the first $h/2$, we have $\dfrac{h}2=\dfrac12gt^2$, so the time $t=\sqrt{\dfrac{h}{g}}$. For the whole height $h$, the time to travel $T$ can be calculated from $h=\dfrac12gT^2$, thus $T=\sqrt{\dfrac{2h}{g}}=\sqrt2t$. Finally we get the time to fall the second $h/2$ as $T-t=(\sqrt2-1)t=0.41t$.

Chromatic aberration from a lens is a consequence of

$\textbf{(A) }$ polarization.$ \qquad\newline$ $\textbf{(B) }$ interference.$ \qquad\newline$ $\textbf{(C) }$ total internal reflection.$ \qquad\newline$ $\textbf{(D) }$ diffraction.$ \qquad\newline$ $\textbf{(E) }$ dispersion.

$\textbf{E}$

Chromatic aberration is a phenomenon that results when light shines through a lens and different colors focus at different places because the wavelengths of light experience different indices of refraction. This is called dispersion.

An object of mass $M$ starts from rest at the bottom of a fixed incline of height $H$. A person decides to push the object up the incline in one of two ways with an applied force shown in the diagram. In each of the trials, the object reaches the top of the incline with speed $V$. How would the work done by the person on the block compare for the two trials? Assume the same constant non-zero coefficient of kinetic friction between the incline and the object for both trials.

$\textbf{A}$

The work $W=\vec{F}\cdot\vec{s}$. In both diagrams $\vec{s}$ is the displacement from the bottom of the inclined plane to the top. $F_2$ is parallel to $\vec{s}$, so the work been done in Trial 2 is $W=F_2s$. $F_1$ has a component $F_\perp$ perpendicular to the incline and a component $F_\parallel$ parallel to the incline, so the work $W_1=F_\parallel s$. To compare $W_1$ and $W_2$, we need to compare $F_\parallel$ and $F_2$ first. The perpendicular component $F_\perp$ leads to a larger supporting force by the incline to the object, which results a larger frictional force. To balance the greater frictional force and keep the object moving with a constant speed, $F_\parallel$ must be greater than $F_2$, so $W_1>W_2$.

What is the magnitude of the average acceleration (in units of $\text{m/s}^2$) of the ball while it is in contact with the ground?

$\textbf{(A) }$ 1.2$ \qquad$ $\textbf{(B) }$ 6.0$ \qquad$ $\textbf{(C) }$ 10.0$ \qquad$ $\textbf{(D) }$ 13.0$ \qquad$ $\textbf{(E) }$ 16.0

$\textbf{B}$

The average acceleration is the change in velocity divided by the time, so $\bar{a}=\dfrac{\Delta v}{\Delta t}=\dfrac{0.6-(-0.9)}{0.25}=6.0\ \text{m/s}^2$

What is the magnitude of the average force (in units of Newtons) exerted by the ground on the ball while they are in contact?

$\textbf{(A) }$ 2.4$ \qquad$ $\textbf{(B) }$ 12.0$ \qquad$ $\textbf{(C) }$ 20.0$ \qquad$ $\textbf{(D) }$ 26.0$ \qquad$ $\textbf{(E) }$ 32.0

$\textbf{E}$

Both gravitational and contact force exerts on the object in the collision process, which leads to an average acceleration $\bar{a}=6.0\ \text{m/s}^2$. By Newton’s Second Law, we have $N-mg=ma$. So the contact force by the ground is $N=m(g+a)=(2)(10+6)=32.0\ \text{N}$.

A 4.0 kg object in deep space moves at a constant velocity of 10 m/s along the $+x$ axis. The object suddenly explodes into 2 equal mass pieces. Immediately after the explosion, one piece is now moving directly along the $–y$ axis with a speed of 8.0 m/s. What was the size of the impulse (in proper MKS units) provided to this piece now moving along the $–y$ axis from the explosion?

$\textbf{C}$

The 2.0 kg half piece was initially moving along the $+x$ axis with speed 10 m/s (linear momentum $p_i=m_{half}v_i=(2)(10)=20\ \text{kg}\cdot\text{m/s}$ ). Finally it moved along the $-y$ axis with speed 8.0 m/s (linear momentum $p_f=m_{half}v_f=(2)(8)=16\ \text{kg}\cdot\text{m/s}$ ). By the impulse-momentum theorem and Pythagorean Theorem, the impulse $I=\Delta p=\sqrt{p_i^2+p_f^2}=\sqrt{(20)^2+(16)^2}=25.6\ \text{kg}\cdot\text{m/s}$.

A point object with mass $M=2.0\ \text{kg}$ is attached a distance $R=1.75\ \text{m}$ from the fixed center of a disk as shown in the figure. The disk starts rotating from rest with constant angular acceleration $\alpha=5.00\ \text{rad/s}^2$ about an axis perpendicular to the plane of the page through the disk’s center. After how much time $T$ (in seconds) is the tangential component of the point object’s acceleration equal in magnitude to the centripetal component of the point object’s acceleration?

$\textbf{D}$

The tangential acceleration $a_\tau=\alpha R$. At time $T$, the speed of the object is $v=a_\tau T=\alpha RT$, the centripetal acceleration $a_r=\dfrac{v^2}{R}$. When $a_\tau=a_r$, we get $T=\dfrac1{\sqrt{\alpha}}=\dfrac1{\sqrt5}=0.447\ \text{s}$.

In terms of the seven fundamental SI units in the MKS system, the unit for capacitance is written as which of the following?

$\textbf{(A) }$ $\dfrac{\text{A}^2}{\text{kg}\cdot\text{m}^2} \qquad$ $\textbf{(B) }$ $\dfrac{\text{A}\cdot\text{s}}{\text{kg}\cdot\text{m}^2} \qquad$ $\textbf{(C) }$ $\dfrac{\text{A}\cdot\text{s}^2}{\text{kg}\cdot\text{m}} \qquad$ $\textbf{(D) }$ $\dfrac{\text{A}^2\cdot\text{s}^3}{\text{kg}\cdot\text{m}^2} \qquad$ $\textbf{(E) }$ $\dfrac{\text{A}^2\cdot\text{s}^4}{\text{kg}\cdot\text{m}^2}$

$\textbf{E}$

By $C=\dfrac{Q}{U}$, we have $\text{F}=\dfrac{\text{C}}{\text{V}}=\dfrac{\text{C}}{\text{J/C}}=\dfrac{\text{C}^2}{\text{J}}=\dfrac{\text{(As)}^2}{\text{Nm}}=\dfrac{\text{A}^2\text{s}^2}{(\text{kg}\cdot\text{m/s}^2)\cdot\text{m}}=\dfrac{\text{A}^2\cdot\text{s}^4}{\text{kg}\cdot\text{m}^2}$.

A positively charged particle is moved in a straight line to the right from position $x=-4.0\ \text{m}$ to position $x=-2.0\ \text{m}$ by an external agent. The electric potential at these positions in space are $V(x=-4\ \text{m})=-4.0\ \text{volts}$ and $V(x=-2\ \text{m})=-2.0\ \text{volts}$. Which statement is true about the work done by the external agent moving the charge between these positions and the electric field component parallel to the direction of motion that the moving charged particle experiences? Assume the particle is moved at constant speed.

$\textbf{(A) }$ The external agent does no work since the speed was constant.$ \qquad\newline$
$\textbf{(B) }$ The external agent does negative work and the electric field is directed to the right.$ \qquad\newline$
$\textbf{(C) }$ The external agent does positive work and the electric field is directed to the right.$ \qquad\newline$
$\textbf{(D) }$ The external agent does negative work and the electric field is directed to the left.$ \qquad\newline$
$\textbf{(E) }$ The external agent does positive work and the electric field is directed to the left.

$\textbf{E}$

The electric potential decreases along the direction of electric field. As $V(x=-4\ \text{m})=-4.0\ \text{volts}$ and $V(x=-2\ \text{m})=-2.0\ \text{volts}$, the electric field is directed to the left. Since the particle is positively charged, the electric force points toward left. To balance the the electric force and keep the particle moving to the right, the external agent must exert a force to the right. Thus the external agent does a positive work.

Two objects both move and uniformly accelerate to the right. At time $t=0$, the objects are at the same initial position but:$\newline$
Object 1 has initial speed twice that of Object 2.$\newline$
Object 1 has one-half the acceleration of Object 2.$\newline$
After some time $T$, the velocity of the two objects is the same.$\newline$
What is the ratio of the distance traveled in this time $T$ by Object 2 to that traveled by Object 1?

$\textbf{(A) }$ 5:6$ \qquad$ $\textbf{(B) }$ 4:5$ \qquad$ $\textbf{(C) }$ 3:4$ \qquad$ $\textbf{(D) }$ 2:3$ \qquad$ $\textbf{(E) }$ 1:2

$\textbf{B}$

Object 1 moves with speed $2v$ and acceleration $a$, while Object 2 moves with speed $v$ and acceleration $2a$. At time $T$, the speeds of both objects are the same, which means that $2v+aT=v+2aT$, or simplified as $v=aT$. The ration of distance is $$\dfrac{x_2}{x_1}=\dfrac{vT+\frac12(2a)T^2}{(2v)T+\frac12aT^2}=\dfrac{aT^2+aT^2}{2aT^2+\frac12aT^2}=\dfrac45$$

$\textbf{A}$

The induced current in the solenoid wraps from behind the solenoid to in front of the solenoid. This means that the induced magnetic field (from the right hand rule) is directed to the right in the solenoid (the right end of the solenoid is modeled with a N pole). By Lenz’s law, the induced current is trying to stop the movement of the magnet to the left. This means that the force on the magnet points toward the right, and the ?? is a North Pole since the two approaching North poles repel each other.

A uniform, solid cylinder with a mass $M$ and radius $R$ is pulled by a horizontal force $F$ acting through the center as shown. The cylinder rolls to the right without slipping. What is the magnitude of the force of friction between the cylinder and the ground?

$\textbf{B}$

The acceleration of the cylinder is $a=\dfrac{F-f}{M}$ where $f$ is the frictional force. The torque from the center drives the rotation with angular acceleration $\beta$, which is $\tau=fR=I\beta=\left(\dfrac12MR^2\right)\beta$. To roll without slipping, the relationship between acceleration and angular acceleration is $a=\beta R$, thus we have $$\dfrac{F-f}{M}=\dfrac{fR}{1/2MR^2}R$$ By solving the equation we get $f=\dfrac13F$.

Two spheres are heated to the same temperature and allowed to radiate energy to identical surroundings. The spheres have the same emissivity, but one sphere has twice the diameter of the other. If the smaller sphere radiates energy at a rate $P$, at what rate will the larger sphere radiate energy?

$\textbf{(A) }$ $P \qquad$ $\textbf{(B) }$ $2P \qquad$ $\textbf{(C) }$ $4P \qquad$ $\textbf{(D) }$ $8P \qquad$ $\textbf{(E) }$ $16P$

$\textbf{C}$

The rate at which energy is radiated from a sample is given by the expression $P=\sigma T^4A$ where the surface area is given as $A=4\pi R^2$. When the diameter is doubled, the surface area is 4 times as before, thus the power will be $4P$.

$\textbf{B}$

The angular momentum of the system is constant since there is no net external torque acting on the comet-sun system. As the comet comes closer to the sun, its speed $v=\dfrac{L}{mR}$ will increase and the gravitational potential energy $E_p=-G\dfrac{Mm}R$ will decrease.

Two ions travel perpendicular to the same uniform magnetic field. The ions carry the same charge and have the same path radius in the magnetic field. Which of the following quantities must be the same for the two ions?

$\textbf{(A) }$ Mass$ \qquad\newline$
$\textbf{(B) }$ Speed$ \qquad\newline$
$\textbf{(C) }$ Charge-to-mass ratio$ \qquad\newline$
$\textbf{(D) }$ Kinetic energy$ \qquad\newline$
$\textbf{(E) }$ Magnitude of linear momentum

$\textbf{E}$

The Lorentz force provides the centripetal force of the circular motion as $qvB=m\dfrac{v^2}{R}$. So we get the the radius $R=\dfrac{mv}{qB}$. Since the two ions carry the same charge $q$ and the magnetic field $B$ is uniform, the same radius means the same linear momentum $mv$.

A radioactive sample of gas has a half-life of 100 seconds. If there are initially 10000 of these gas molecules in a closed container, approximately how many of the molecules remain after a time of 250 seconds elapses?

$\textbf{(A) }$ 2500$ \qquad$ $\textbf{(B) }$ 2190$ \qquad$ $\textbf{(C) }$ 1770$ \qquad$ $\textbf{(D) }$ 1560$ \qquad$ $\textbf{(E) }$ 1250

$\textbf{C}$

For half life $\tau=100\ \text{s}$, the expression for radioactive decay is $N=N_02^{-t/\tau}=N_02^{-t/1000}$. When $N_0=10000$ and $t=250\ \text{s}$, the final number of molecules is $N=10000\cdot2^{-250/100}=1768$.

For the ideal $RL$ circuit shown, the resistance is $R=10.0\ \Omega$, the inductance is $L=5.0\ \text{H}$ and the battery has voltage $\xi_{bat}=12\ \text{volts}$. Some time after the switch S in the circuit is closed, the ammeter in the circuit reads 0.40 A. At what rate is energy being stored by the inductor at this instant (in Watts)?

$\textbf{D}$

At the moment of $I=0.4\ \text{A}$, the power of the battery is $P_{bat}=\xi_{bat}I=(12)(0.4)=4.8\ \text{W}$, the power of the resistor is $P_r=I^2R=(0.4)^2(10)=1.6\ \text{W}$. So the power of the inductance is $P_L=P_{bat}-P_r=4.9-1.6=3.2\ \text{W}$.

For the following nuclear reaction, what is the unknown labeled by X ?$$^{22}_{11}Na+X\rightarrow^{22}_{10}Ne+\nu_e$$

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

$\textbf{B}$

For the unknown particle $^A_ZX$, we need $22+A=22$ and $11+Z=10$ to balance the equation. So we get $A=0$ and $Z=-1$. Particle $^{\ 0}_{-1}X=^{\ 0}_{-1}e$ is an electron.

An open cylindrical container with very large radius is at rest a distance $H$ above the ground at the edge of a platform. A tiny hole develops at the bottom of the container and water from the container squirts out horizontally landing a distance $H$ from the edge of the platform. For the water to land at this location, what is the depth of the water $L$ in the container? The figure is not drawn to scale and air resistance is ignored.

$\textbf{A}$

The time for a free fall of height $H$ is $t=\sqrt{\dfrac{2H}{g}}$. To land a distance $H$ in the horizontal direction, the initial speed of the water drop is $v=\dfrac{H}{t}=\sqrt{\dfrac12gH}$. By Bernoulli’s Equation, $\rho gL=\dfrac12\rho v^2$, thus we get the depth of water $L=\dfrac{v^2}{2g}=\dfrac{H}{4}$.

Which of the following wavelengths (in nm) of electromagnetic radiation will produce photoelectrons of the least kinetic energy if the radiation is incident on a material with a work function of 4.80 eV?

$\textbf{(A) }$ 992$ \qquad$ $\textbf{(B) }$ 496$ \qquad$ $\textbf{(C) }$ 248$ \qquad$ $\textbf{(D) }$ 124$ \qquad$ $\textbf{(E) }$ 62

$\textbf{C}$

To produce photoelectrons, the energy of the photon should be no less than the work function of the metal. So we have $h\nu=h\dfrac{c}{\lambda}=W$. The wavelength $\lambda=\dfrac{hc}{W}=\dfrac{(6.62\times10^{-34})(3\times10^8)}{(4.8)(1.6\times10^{-19})}=2.59\times10^{-7}\ \text{m}=259\ \text{nm}$.

A monatomic ideal gas is the working substance for an engine that undergoes the cyclic process (ABCDA) shown in the PV diagram. The processes are all isochoric or isobaric with pressures between $P_0$ and $2P_0$ and volumes between $V_0$ and $\dfrac32V_0$. What is the efficiency of this engine?

$\textbf{A}$

The efficiency of the engine is the work done divided by the heat absorbed. The work done is the area surrounded by the curve $W=(2P_0-P_0)\left(\dfrac32V_0-V_0\right)=\dfrac12P_0V_0=\dfrac12nRT_0$. The heat absorbed in $A\rightarrow B$ process is $Q_{AB}=nc_V\Delta T=n\left(\dfrac32R\right)(2T_0-T_0)=\dfrac32nRT_0$. The heat absorbed in $B\rightarrow C$ process is $Q_{BC}=nc_P\Delta T=n\left(\dfrac52R\right)(3T_0-T_0)=\dfrac nRT_0$. The total heat absorbed is $Q=Q_{AB}+Q_{BC}=4nRT_0$. So the efficiency of the engine is $\eta=\dfrac{W}{Q}=\dfrac18$.

An object of mass $m$ is initially at rest. After this object is accelerated to a speed of $2.40\times10^8\ \text{m/s}$, it collides with and sticks to a second object of mass $m$ at rest. Immediately after the collision, what is the common speed of the two masses?

$\textbf{(A) }$ $2.25\times10^8\ \text{m/s} \qquad\newline$
$\textbf{(B) }$ $1.80\times10^8\ \text{m/s} \qquad\newline$
$\textbf{(C) }$ $1.66\times10^8\ \text{m/s} \qquad\newline$
$\textbf{(D) }$ $1.50\times10^8\ \text{m/s} \qquad\newline$
$\textbf{(E) }$ $1.20\times10^8\ \text{m/s}$

$\textbf{D}$

The initial speed $v_0=2.40\times10^8\ \text{m/s}=0.8c$. Assuming the speed after collision is $v$ and the rest mass is $M$. By the conservation of linear momentum and energy, we have $$\dfrac{m}{\sqrt{1-v_0^2/c^2}}v_0=\dfrac{M}{\sqrt{1-v^2/c^2}}v$$ $$mc^2+\dfrac{m}{\sqrt{1-v_0^2/c^2}}c^2=\dfrac{M}{\sqrt{1-v^2/c^2}}c^2$$ By solving the equation we have $v=0.5c=1.50\times10^8\ \text{m/s}$.