## PhysicsBowl 2007

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**Instruction**

- Questions: The test is composed of 50 questions; however, students answer only 40 questions.
**Division 1 students will answer only questions 1 – 40. Do not answer questions 41 – 50.**

Division 2 students will answer only questions 11 – 50. Do not answer questions 1 – 10. - Calculator: A hand-held calculator may be used. Any memory must be cleared of data and programs. Calculators may not be shared.
- Formulas and constants: Only the formulas and constants provided with the contest may be used.
- Time limit: 45 minutes.

A standard centimeter ruler is shown. Which recorded value is the most correct for the location of the shaded object’s right end?

$\textbf{(A) }$ 10 cm$ \qquad$ $\textbf{(B) }$ 10. cm$ \qquad$ $\textbf{(C) }$ 10.0 cm$ \qquad$ $\textbf{(D) }$ 10.00 cm$ \qquad$ $\textbf{(E) }$ 10.$\bar{0}$ cm

$\textbf{D}$

Proper measurement technique means that one must estimate the NEXT digit past what is read. Here, we can see that the end of the object is at 10.0 cm and we estimate the hundredths place to be 0 as the object looks to line up exactly with the mark.

How thick is the average page of a physics textbook in micrometers?

$\textbf{(A) }$ 0.1$ \qquad$ $\textbf{(B) }$ 1$ \qquad$ $\textbf{(C) }$ 10$ \qquad$ $\textbf{(D) }$ 100$ \qquad$ $\textbf{(E) }$ 1000

$\textbf{D}$

A book that is an inch thick is about 500 pages, which means 250 sheets of paper. There are about 2.5 cm in an inch. So, about 100 sheets per cm, or $100\ \mu\text{m/sheet}$.

Two automobiles are 150 kilometers apart and traveling toward each other. One automobile is moving at 60 km/h and the other is moving at 40 km/h. In how many hours will they meet?

$\textbf{(A) }$ 1.5$ \qquad$ $\textbf{(B) }$ 1.75$ \qquad$ $\textbf{(C) }$ 2.0$ \qquad$ $\textbf{(D) }$ 2.5$ \qquad$ $\textbf{(E) }$ 3.0

$\textbf{A}$

From the point of view of one of the vehicles, the other looks to move at 100 km/h. Hence, it takes 150/100=1.5 h for the cars to meet.

A particle moves on the x-axis. When the particle’s acceleration is positive and increasing

$\textbf{(A) }$ its velocity must be positive.$ \qquad\newline$

$\textbf{(B) }$ its velocity must be negative.$ \qquad\newline$

$\textbf{(C) }$ it must be slowing down.$ \qquad\newline$

$\textbf{(D) }$ it must be speeding up.$ \qquad\newline$

$\textbf{(E) }$ none of the above must be true.

$\textbf{E}$

If the initial velocity is positive, with a positive acceleration it will speed up. However, if the initial velocity is negative, it will slow down.

The position-time, $y$ vs. $t$, graph for the motion of an object is shown. What would be a reasonable equation for the acceleration $a$ that would account for this motion?

$\textbf{(A) }$ $a=0 \qquad\newline$

$\textbf{(B) }$ $a=\text{positive constant} \qquad\newline$

$\textbf{(C) }$ $a=\text{negative constant} \qquad\newline$

$\textbf{(D) }$ $a=\text{positive constant times} t \qquad\newline$

$\textbf{(E) }$ $a=\text{negative constant times} t$

$\textbf{C}$

Since the position moves upward and then downward, this graph is reminiscent of a particle in free fall as it is parabolic… hence, the acceleration is a negative constant.

A 500-kg car is moving at 28 m/s. The driver sees a barrier ahead. If the car takes 95 meters to come to rest, what is the magnitude of the minimum average net force necessary to stop?

$\textbf{(A) }$ 47.5 N$ \qquad$ $\textbf{(B) }$ 1400 N$ \qquad$ $\textbf{(C) }$ 2060 N$ \qquad$ $\textbf{(D) }$ 19600 N$ \qquad$ $\textbf{(E) }$ 133000 N

$\textbf{C}$

The acceleration is given by the equation $v^2=2ax$, which is $a=\dfrac{v^2}{2x}=\dfrac{28^2}{2(95)}=4.13\ \text{m/s}^2$. Using Newton’s Second Law, the net force $F=ma=(500)(4.13)=2060\ \text{N}$.

A mass connected to a string swings back and forth as a pendulum with snapshots of the motion seen in the figure. Ignore the friction in the system. Which of the following statements about the pendulum-Earth system is correct?

$\textbf{(A) }$ The total mechanical energy in the system is constant.$ \qquad\newline$

$\textbf{(B) }$ The total mechanical energy in the system is maximum at B.$ \qquad\newline$

$\textbf{(C) }$ The potential energies at A and C are equal.$ \qquad\newline$

$\textbf{(D) }$ The kinetic energies at C and D are equal.$ \qquad\newline$

$\textbf{(E) }$ The kinetic energies at E equals the kinetic energy at C.

$\textbf{A}$

By ignoring friction, there is no loss or gain of energy to the system as potential energy and kinetic energy are constantly transformed into each other.

What does one obtain by dividing the distance of 12 Mm by the time of 4 Ts?

$\textbf{(A) }$ 3 nm/s$ \qquad$ $\textbf{(B) }$ 3 $\mu$m/s$ \qquad$ $\textbf{(C) }$ 3 mm/s$ \qquad$ $\textbf{(D) }$ 3 km/s$ \qquad$ $\textbf{(E) }$ 3 Gm/s

$\textbf{B}$

$\dfrac{12\ \text{Mm}}{4\ \text{Ts}}=\dfrac{12\times10^6\ \text{m}}{4\times10^{12}\ \text{s}}=3\times10^{-6}\ \text{m/s}=3\ \mu\text{m/s}$

A block rests on an incline that makes the angle $\phi$ with the horizontal. The block remains at rest as $\phi$ is slowly increased. The magnitudes of the normal force and the static frictional force of the incline on the block

$\textbf{(A) }$ both increase.$ \qquad\newline$

$\textbf{(B) }$ both decrease.$ \qquad\newline$

$\textbf{(C) }$ both remain the same.$ \qquad\newline$

$\textbf{(D) }$ increase and decrease, respectively.$ \qquad\newline$

$\textbf{(E) }$ decrease and increase, respectively.

$\textbf{E}$

When the angle $\phi$ increases, the normal force $N=mg\cos\phi$ decreases while the frictional force $f=mg\sin\phi$ increases.

Contact forces are examples of which of the fundamental forces?

$\textbf{(A) }$ Strong$ \qquad$ $\textbf{(B) }$ Electromagnetic$ \qquad$ $\textbf{(C) }$ Weak$ \qquad$ $\textbf{(D) }$ Gravitational $ \qquad$ $\textbf{(E) }$ None of these

$\textbf{B}$

Contact forces arise from the interaction of molecules between the surfaces… these arise from the positive and negative charges… hence the forces are electromagnetic.

A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?

$\textbf{(A) }$ 1 m$ \qquad$ $\textbf{(B) }$ 2 m$ \qquad$ $\textbf{(C) }$ 3 m$ \qquad$ $\textbf{(D) }$ 4 m$ \qquad$ $\textbf{(E) }$ 8 m

$\textbf{D}$

Using $x=\dfrac{v^2}{2a}$, when the initial velocity doubled, the distance $x$ will be 4 times greater.

The definition of average velocity is

$\textbf{(A) }$ the average acceleration multiplied by the time.$ \qquad\newline$

$\textbf{(B) }$ distance traveled divided by the time.$ \qquad\newline$

$\textbf{(C) }$ $\dfrac12\left(v_f+v_0\right). \qquad\newline$

$\textbf{(D) }$ radius multiplied by angular velocity.$ \qquad\newline$

$\textbf{(E) }$ displacement divided by the time.

$\textbf{E}$

Average velocity is defined as the displacement divided by the time.

A student weighing 500 N stands on a bathroom scale in the school’s elevator. When the scale reads 520 N, the elevator must be

$\textbf{(A) }$ accelerating upward.$ \qquad\newline$

$\textbf{(B) }$ accelerating downward.$ \qquad\newline$

$\textbf{(C) }$ moving upward at a constant speed.$ \qquad\newline$

$\textbf{(D) }$ moving downward at a constant speed.$ \qquad\newline$

$\textbf{(E) }$ at rest.

$\textbf{A}$

Since the scale reads a higher number than the gravitational force on the student… this means that there is a net force UPWARD on the student meaning that the student accelerates upward.

An object moves to the East across a frictionless surface with constant speed. A person then applies a constant force to the North on the object. What is the resulting path that the object takes?

$\textbf{(A) }$ A straight line path partly Eastward, partly Northward$ \qquad\newline$

$\textbf{(B) }$ A straight line path totally to the North$ \qquad\newline$

$\textbf{(C) }$ A parabolic path opening toward the North$ \qquad\newline$

$\textbf{(D) }$ A parabolic path opening toward the East$ \qquad\newline$

$\textbf{(E) }$ An exponential path opening upward toward the North

$\textbf{C}$

With a constant force applied to the North, the path will be parabolic opening to the North. This is akin to an object moving to the right and then falling off of a table. The path would be a parabola opening downward in the direction of the gravitational force.

**Questions 15 and 16 refer to the following scenario:**

Two identical mass objects are launched with the same speed from the same starting location. Object 1 is launched at an angle of $30^\circ$ above the horizontal while Object 2 is launched at an angle of $60^\circ$ above the horizontal. Ignore air resistance and consider the flight of each object from launch until it returns to the same launch height above the ground.

Which object returns to the starting height with the greatest speed?

$\textbf{(A) }$ Object 1 since it keeps a lower trajectory.$ \qquad\newline$

$\textbf{(B) }$ Object 2 since it is in the air for a longer time.$ \qquad\newline$

$\textbf{(C) }$ Object 2 since there is more work done on the object during flight.$ \qquad\newline$

$\textbf{(D) }$ The speeds are the same.$ \qquad\newline$

$\textbf{(E) }$ It cannot be determined without more information.

$\textbf{D}$

The mechanical energy is conserved since there is no energy loss. When reaching the starting height, the potential energy does not change, so the kinetic energy and the velocity remain unchanged. Since the two objects are launched with the same speed, their speeds are the same when returning the starting height.

Which object experiences the greatest change in the linear momentum?

$\textbf{(A) }$ Object 1 since it has a higher final speed.$ \qquad\newline$

$\textbf{(B) }$ Object 2 since it has a higher final speed.$ \qquad\newline$

$\textbf{(C) }$ Object 2 since it is in the air for a longer time.$ \qquad\newline$

$\textbf{(D) }$ The change in momentum is the same for each.$ \qquad\newline$

$\textbf{(E) }$ It cannot be determined without more information.

$\textbf{C}$

Using the impulse-momentum theorem, the change in momentum equals the net force multiplied by the time. Since the force on each object is the same, the object in the air for a longer time experiences a larger momentum change. Since Object 2 is launched at a higher angle, it reaches a higher position above the ground and takes longer to return to the ground.

A toy car moves along the x-axis according to the velocity versus time curve as shown. When does the car have zero acceleration?

$\textbf{(A) }$ at 2 and 4 seconds$ \qquad\newline$

$\textbf{(B) }$ at approximately 3.0 seconds$ \qquad\newline$

$\textbf{(C) }$ at approximately 3.3 and 5.1 seconds$ \qquad\newline$

$\textbf{(D) }$ the acceleration is always zero$ \qquad\newline$

$\textbf{(E) }$ at no time

$\textbf{A}$

The acceleration is found from taking the slope of the line tangent to each point on the velocity-time graph. Zero acceleration occurs when the line tangent is horizontal. This occurs at times $t=2\ \text{s}$ and $t=4\ \text{s}$.

In which one of the following situations is the net force constantly zero on the object?

$\textbf{(A) }$ A mass attached to a string and swinging like a pendulum.$ \qquad\newline$

$\textbf{(B) }$ A stone falling freely in a gravitational field.$ \qquad\newline$

$\textbf{(C) }$ An astronaut floating in the International Space Station.$ \qquad\newline$

$\textbf{(D) }$ A snowboarder riding down a steep hill.$ \qquad\newline$

$\textbf{(E) }$ A skydiver who has reached terminal velocity.

$\textbf{E}$

For answer A, there is a net force on the mass on a pendulum since its speed and direction of motion keep changing. Answer B is not correct because the stone is in free fall. Answer C is not correct as the astronaut is accelerating toward the Earth (circular motion). Answer D is not correct as the person would increase speed as they ride down the hill. Terminal velocity (when the object achieves its maximum speed) means that there is no acceleration.

What net force is necessary to keep a 1.0 kg puck moving in a circle of radius 0.5 m on a horizontal frictionless surface with a speed of 2.0 m/s?

$\textbf{(A) }$ 0 N$ \qquad$ $\textbf{(B) }$ 2.0 N$ \qquad$ $\textbf{(C) }$ 4.0 N$ \qquad$ $\textbf{(D) }$ 8.0 N$ \qquad$ $\textbf{(E) }$ 16 N

$\textbf{D}$

The centripetal force $F=m\dfrac{v^2}{r}=(1.0)\dfrac{2.0^2}{0.5}=8.0\ \text{N}$.

A large wedge rests on a horizontal frictionless surface as shown. A block starts from rest and slides down the inclined surface of the wedge, which is rough. During the motion of the block, the center of mass of the block and wedge system

$\textbf{(A) }$ does not move.$ \qquad\newline$

$\textbf{(B) }$ moves vertically with increasing speed.$ \qquad\newline$

$\textbf{(C) }$ moves horizontally with constant speed.$ \qquad\newline$

$\textbf{(D) }$ moves horizontally with increasing speed.$ \qquad\newline$

$\textbf{(E) }$ moves both horizontally and vertically.

$\textbf{B}$

Since there is no force acting on the system in the horizontal direction, the center of mass does not move horizontally. As for the vertical component… since the wedge does not change its vertical position while the block moves downwards with a acceleration driven by gravity, the center of mass moves downwards with a increasing speed.

A box slides to the right across a horizontal floor. A person called Ted exerts a force $T$ to the right on the box. A person called Mario exerts a force $M$ to the left, which is half as large as the force $T$. Given that there is friction and the box accelerates to the right, rank the sizes of these three forces exerted on the box.

$\textbf{(A) }$ $f<M<T \qquad\newline$

$\textbf{(B) }$ $M<f<T \qquad\newline$

$\textbf{(C) }$ $M<T<f \qquad\newline$

$\textbf{(D) }$ $f=M<T \qquad\newline$

$\textbf{(E) }$ It cannot be determined.

$\textbf{A}$

Since the box accelerates to the right, the force $T$ must be greater than the sum of $M$ and $f$. As $M$ is half as large as $T$, $f$ must be less than $M$. So we get $f<M<T$.

A mass $m$ is pulled outward until the string of length $L$ to which it is attached makes a 90-degree angle with the vertical. The mass is released from rest and swings through a circular arc. What is the tension in the string when the mass swings through the bottom of the arc?

$\textbf{(A) }$ $0 \qquad$ $\textbf{(B) }$ $mg \qquad$ $\textbf{(C) }$ $2mg \qquad$ $\textbf{(D) }$ $3mg \qquad$ $\textbf{(E) }$ It cannot be determined.

$\textbf{D}$

Using the conservation of energy, the velocity at the bottom is given in $\dfrac12mv^2=mgL$, so $v=\sqrt{2gL}$. The tension provides the centripetal force in the circular motion, so $T-mg=m\dfrac{v^2}{L}$. Finally we get $T=3mg$.

The period of a mass-spring system undergoing simple harmonic oscillation is $T$. If the amplitude of the mass-spring system’s motion is doubled, the period will be

$\textbf{(A) }$ $\dfrac14T \qquad$ $\textbf{(B) }$ $\dfrac12T \qquad$ $\textbf{(C) }$ $T \qquad$ $\textbf{(D) }$ $2T \qquad$ $\textbf{(E) }$ $4T$

$\textbf{C}$

The period of the mass-spring system $T=2\pi\sqrt{\dfrac{m}{k}}$ depends on the spring constant and the mass, but not the amplitude.

A resonance occurs with a tuning fork and an air column of size 39 cm. The next highest resonance occurs with an air column of 65 cm. What is the frequency of the tuning fork? Assume that the speed of sound is 343 m/s.

$\textbf{(A) }$ 329.8 Hz$ \qquad$ $\textbf{(B) }$ 527.7 Hz$ \qquad$ $\textbf{(C) }$ 659.6 Hz$ \qquad$ $\textbf{(D) }$ 879.5 Hz$ \qquad$ $\textbf{(E) }$ 1319 Hz

$\textbf{C}$

The standing wave mode has a displacement antinode at the opening, and a displacement node at the water-air interface. By increasing the height of the air column, to go from one harmonic to the next, an additional length equal to 1/2 wavelength is required. Hence, $\dfrac12\lambda=(0.65-0.39)\ \text{m}\rightarrow\lambda=0.52\ \text{m}$. Using $v=\lambda f$, we get the frequency $f=\dfrac{v}{\lambda}=\dfrac{343}{0.52}=659.6\ \text{Hz}$. If one checks, this problem deals with the 3rd and 5th harmonics.

If two protons are spaced by a distance $R$, what is the ratio of the gravitational force that one proton exerts on the other to the electric force that one proton exerts on the other? That is, $F_{gravity}/F_{electric}=$?

$\textbf{(A) }$ $\approx10^{-8} \qquad$ $\textbf{(B) }$ $\approx10^{-16} \qquad$ $\textbf{(C) }$ $\approx10^{-20} \qquad$ $\textbf{(D) }$ $\approx10^{-36} \qquad$ $\textbf{(E) }$ $\approx10^{-43}$

$\textbf{D}$

$\dfrac{F_{gravity}}{F_{electric}}=\dfrac{Gm^2/r^2}{kq^2/r^2}=\dfrac{Gm^2}{kq^2}=\dfrac{\left(6.67\times10^{-11}\right)\left(1.67\times10^{-27}\right)^2}{\left(9\times10^9\right)\left(1.6\times10^{-19}\right)^2}=8\times10^{-37}\approx10^{-36}$

For the diagram shown below, what is the ratio of the charges $q_2/q_1$, where the diagram shown has a representation of the field lines in the space near the charges.

$\textbf{(A) }$ $-\dfrac32 \qquad$ $\textbf{(B) }$ $-\dfrac23 \qquad$ $\textbf{(C) }$ $\dfrac23 \qquad$ $\textbf{(D) }$ $\dfrac32 \qquad$ $\textbf{(E) }$ $1$

$\textbf{B}$

Field lines point into negative charges and out of positive charges. Hence, $q_2$ is negative. Further, the number of lines from the charge is a measure of the charge magnitude. This means that $\dfrac{q_2}{q_1}=-\dfrac69=-\dfrac23$.

A junior Thomas Edison wants to make a brighter light bulb. He decides to modify the filament. How should the filament of a light bulb be modified in order to make the light bulb produce more light at a given voltage?

$\textbf{(A) }$ Increase the resistivity only.$ \qquad\newline$

$\textbf{(B) }$ Increase the diameter only.$ \qquad\newline$

$\textbf{(C) }$ Decrease the diameter only.$ \qquad\newline$

$\textbf{(D) }$ Decrease the diameter and increase the resistivity.$ \qquad\newline$

$\textbf{(E) }$ Increase the length only.

$\textbf{B}$

The power for a resistor is $P=\dfrac{V^2}{R}$, so to increase the power, the resistance must be reduced. As resistance is $R=\rho\dfrac{L}{A}$, increasing the length, decreasing the area or increasing the resistivity all INCREASE the resistance. The only option listed decreasing the resistance is found through increase the diameter of the wire.

Which statement about a system of point charges that are fixed in space is necessarily true?

$\textbf{(A) }$ If the potential energy of the system is negative, net positive work by an external agent is required to take the charges in the system back to infinity.$ \qquad\newline$

$\textbf{(B) }$ If the potential energy of the system is positive, net positive work is required to bring any new charge not part of the system in from infinity to its final resting location.$ \qquad\newline$

$\textbf{(C) }$ If the potential energy of the system is zero, no negative charges are in the configuration.$ \qquad\newline$

$\textbf{(D) }$ If the potential energy of the system is negative, net positive work by an external agent was required to assemble the system of charges.$ \newline$

$\textbf{(E) }$ If the potential energy of the system is zero, then there is no electric force anywhere in space on any other charged particle not part of the system.

$\textbf{A}$

If the potential energy is negative, then positive work is required to disassemble the system. If the potential energy of the system is positive, positive work was required to set the system up. The work done by the external agent goes into the potential energy of the system. The potential energy of a system of charges can be zero if constructed with the appropriate combination of positive and negative charges.

In the circuit diagram below, all of the bulbs are identical. Which bulb will be the brightest?

$\textbf{(A) }$ A$ \qquad$ $\textbf{(B) }$ B$ \qquad$ $\textbf{(C) }$ C$ \qquad$ $\textbf{(D) }$ D$ \qquad$ $\textbf{(E) }$ The bulbs all have the same brightness.

$\textbf{C}$

The currents passing through each branch will pass through bulb C. So bulb C get the most current, meaning that it is the brightest.

In the following circuit diagram, which one of the bulbs will not light?

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

$\textbf{E}$

A shorting wire across a bulb will short the bulb out. But it did not happen in the circuit shown.

James Clerk Maxwell's great contribution to electromagnetic theory was his idea that

$\textbf{(A) }$ work is required to move a magnetic pole through a closed path surrounding a current.$ \qquad\newline$

$\textbf{(B) }$ a time-changing electric field acts as a current and produces a magnetic field.$ \qquad\newline$

$\textbf{(C) }$ the speed of light could be determined from simple electrostatic and magnetostatic experiments and finding the values of $\mu_0$ and $\varepsilon_0$.$ \qquad\newline$

$\textbf{(D) }$ the magnetic force on a moving charge particle is perpendicular to both its velocity and the magnetic field.$ \qquad\newline$

$\textbf{(E) }$ magnetism could be explained in terms of circulating currents in atoms.

$\textbf{B}$

Maxwell’s contribution to Ampere’s Law was the realization that a time-changing electric field generated a magnetic field. We call the term related to the time-changing electric flux a displacement current.

What does LASER stand for?

$\textbf{(A) }$ Light Amplification by Simulated Emission of Radiation$ \qquad\newline$

$\textbf{(B) }$ Light Amplification by Stimulated Emission of Radiation$ \qquad\newline$

$\textbf{(C) }$ Light Amplification by Simultaneous Emission of Radiation$ \qquad\newline$

$\textbf{(D) }$ Light Amplification by Systematic Emission of Radiation$ \qquad\newline$

$\textbf{(E) }$ Light Amplification by Serendipitous Emission of Radiation

$\textbf{B}$

Laser is cool, isn't it?

For the circuit shown, the ammeter reading is initially $I$. The switch in the circuit then is closed. Consequently:

$\textbf{(A) }$ The ammeter reading decreases.$ \qquad\newline$

$\textbf{(B) }$ The potential difference between E and F increases.$ \qquad\newline$

$\textbf{(C) }$ The potential difference between E and F stays the same.$ \qquad\newline$

$\textbf{(D) }$ Bulb $\#3$ lights up more brightly.$ \qquad\newline$

$\textbf{(E) }$ The power supplied by the battery decreases.

$\textbf{C}$

Since FCD shorts bulb $\#3$, the circuit is equivalently wires BC and EF in parallel. As the voltage across each branch must be equal to the voltage of the battery, the voltage from E to F is unchanged by closing the switch. Initially, there is voltage across the switch and after it is closed, the voltage is across the resistor. By closing the switch, the current in the circuit increases (a new branch with current) and the power from the battery must increase.

For the solenoids shown in the diagram (which are assumed to be close to each other), the resistance of the left-hand circuit is slowly increased. In which direction does the galvanometer needle in the right-hand circuit move in response to this change?

$\textbf{(A) }$ The needle deflects to the left.$ \qquad\newline$

$\textbf{(B) }$ The needle deflects to the right.$ \qquad\newline$

$\textbf{(C) }$ The needle oscillates back and forth.$ \qquad\newline$

$\textbf{(D) }$ The needle rotates in counterclockwise circles.$ \qquad\newline$

$\textbf{(E) }$ The needle never moves.

$\textbf{A}$

Increasing the resistance causes a decrease in the current on the left. This reduces the field strength for the solenoid. By the right-hand rule, the field through the solenoid is directed to the RIGHT. Hence, in the right-hand circuit, since there will be fewer field lines directed to the right, there is an induced electric field that will produce a current in the wires to “replace” the lost field lines. This means that current will be directed with the same orientation as the current in the left circuit … the ammeter needle will deflect to the LEFT.

Two objects labeled K and L have equal mass but densities $0.95D_0$ and $D_0$, respectively. Each of these objects floats after being thrown into a deep swimming pool. Which is true about the buoyant forces acting on these objects?

$\textbf{(A) }$ The buoyant force is greater on Object K since it has a lower density and displaces more water.$ \qquad\newline$

$\textbf{(B) }$ The buoyant force is greater on Object K since it has lower density and lower density objects always float “higher” in the fluid.$ \qquad\newline$

$\textbf{(C) }$ The buoyant force is greater on Object L since it is denser than K and therefore “heavier.”$ \qquad\newline$

$\textbf{(D) }$ The buoyant forces are equal on the objects since they have equal mass.$ \qquad\newline$

$\textbf{(E) }$ Without knowing the specific gravity of the objects, nothing can be determined.

$\textbf{D}$

By drawing a free body diagram, there are two forces acting: a gravitational force and a buoyant force. Since neither object accelerates and the gravitational forces are equal, this means that there are identical buoyant forces.

A driveway is 22.0 m long and 5.0 m wide. If the atmospheric pressure is $1.0\times10^5\ \text{Pa}$, what force does the atmosphere exert on the driveway?

$\textbf{(A) }$ $9.09\times10^{-8}\ \text{N} \qquad$ $\textbf{(B) }$ $1.1\times10^{-3}\ \text{N} \qquad$ $\textbf{(C) }$ $909\ \text{N} \qquad$ $\textbf{(D) }$ $4545\ \text{N} \qquad$ $\textbf{(E) }$ $1.1\times10^7\ \text{N}$

$\textbf{E}$

$F=PA=\left(1.0\times10^5\ \text{Pa}\right)\left(22\ \text{m}\times5\ \text{m}\right)=1.1\times10^7\ \text{N}$

A place of zero displacement on a standing wave is called

$\textbf{(A) }$ an antinode.$ \qquad\newline$ $\textbf{(B) }$ a node.$ \qquad\newline$ $\textbf{(C) }$ the amplitude.$ \qquad\newline$ $\textbf{(D) }$ the wavenumber.$ \qquad\newline$ $\textbf{(E) }$ the harmonic.

$\textbf{B}$

A place of zero displacement on a standing wave is called zero-displacement point.

Absolute zero is best described as that temperature at which

$\textbf{(A) }$ water freezes at standard pressure.$ \qquad\newline$

$\textbf{(B) }$ water is at its triple point.$ \qquad\newline$

$\textbf{(C) }$ the molecules of a substance have a maximum kinetic energy.$ \qquad\newline$

$\textbf{(D) }$ the molecules of a substance have a maximum potential energy.$ \qquad\newline$

$\textbf{(E) }$ the molecules of a substance have minimum kinetic energy.

$\textbf{E}$

The molecules are absolutely frozen at absolute zero that they can not even tremble.

A mass of material exists in its solid form at its melting temperature $10^\circ\text{C}$. The following processes then occur to the material:$\newline$

Process 1: An amount of thermal energy $Q$ is added to the material and 3/4 of the material melts.$\newline$

Process 2: An identical additional amount of thermal energy $Q$ is added to the material and the material is now a liquid at $50^\circ\text{C}$.$\newline$

What is the ratio of the latent heat of fusion to the specific heat of the liquid for this material?

$\textbf{(A) }$ $80^\circ\text{C} \qquad$ $\textbf{(B) }$ $60^\circ\text{C} \qquad$ $\textbf{(C) }$ $40^\circ\text{C} \qquad$ $\textbf{(D) }$ $20^\circ\text{C} \qquad\newline$

$\textbf{(E) }$ More information is needed to answer this question.

$\textbf{A}$

In process 1 $Q=\left(\dfrac34m\right)L$. A additional $Q$ in process 2 melts the remaining material and then increases the temperature $Q=\left(\dfrac14m\right)L+cm\left(40^\circ\text{C}\right)$. So we get $$\left(\dfrac34m\right)L=\left(\dfrac14m\right)L+cm\left(40^\circ\text{C}\right)\rightarrow\dfrac{L}{c}=80^\circ\text{C}$$

Which is not true of an isochoric process on an enclosed ideal gas in which the pressure decreases?

$\textbf{(A) }$ The work done is zero.$ \qquad\newline$

$\textbf{(B) }$ The internal energy of the gas decreases.$ \qquad\newline$

$\textbf{(C) }$ The heat is zero.$ \qquad\newline$

$\textbf{(D) }$ The rms speed of the gas molecules decreases.$ \qquad\newline$

$\textbf{(E) }$ The gas temperature decreases.

$\textbf{C}$

An isochore is a process of constant volume. So, with the pressure decreasing, the temperature must decrease from the ideal gas law. By the First Law of Thermodynamics $\Delta U=Q+W$, there is no work done (no change in volume) and the internal energy decreases (internal energy of an ideal gas is dependent on temperature). Hence, there is a transfer of energy through heat with the system, making C incorrect. The rms speed of the gas molecules is proportional to the square root of the temperature (which decreased).

For the diagram shown, what is the magnitude of the torque from the applied force as measured from the center of the disk?

$\textbf{(A) }$ $Fd\sin30^\circ \qquad$ $\textbf{(B) }$ $Fd\tan30^\circ \qquad$ $\textbf{(C) }$ $Fd\sin90^\circ \qquad$ $\textbf{(D) }$ $Fd\sin120^\circ \qquad$ $\textbf{(E) }$ $Fd\cos120^\circ$

$\textbf{D}$

The distance from the center to the line of force is $d\sin60^\circ$, which is equal to $d\sin120^\circ$. So the torque is $Fd\sin120^\circ$.

A solid spherical conductor has charge $+Q$ and radius $R$. It is surrounded by a solid spherical shell with charge $-Q$, inner radius $2R$, and outer radius $3R$. Which of the following statements is true for the labeled points in the diagram? Assume these objects are isolated in space and that the electric potential is zero as the distance from the spheres approaches infinity. Point A is at the center of the inner sphere, Point B is located at $r=R+\delta$, Point D is located at $r=3R+\delta$ and Point C is located at $r=2R-\delta$ here $\delta$ is an infinitesimal amount and all distances are from the center of the inner sphere.

$\textbf{(A) }$ The electric potential has a maximum magnitude at C and the electric field has a maximum magnitude at A.$ \qquad\newline$

$\textbf{(B) }$ The electric potential has a maximum magnitude at D and the electric field has a maximum magnitude at B.$ \qquad\newline$

$\textbf{(C) }$ The electric potential at A is zero and the electric field has a maximum magnitude at D.$ \qquad\newline$

$\textbf{(D) }$ The electric potential at A is zero and the electric field has a maximum magnitude at B.$ \qquad\newline$

$\textbf{(E) }$ Both the electric potential and electric field achieve a maximum magnitude at B.

$\textbf{E}$

The electric field in the interior of conductors in equilibrium is zero. Since the charge of $+Q$ was on the surface of radius $R$, then all of the charge $-Q$ on the outer conductor will be on the inner surface of radius $2R$. Consequently, there is no electric field from $2R$ to infinity and from 0 to $R$. Hence, the electric field only exists between $R$ and $2R$ with the field strongest at B (the field drops off like that of a point charge). In addition, the electric potential is zero at all points from infinity to C, at which point the magnitude increases up to point B (in the region of field). From 0 to $R$, there is no electric field and hence no change in potential. Answer E is correct.

Which of the following best represents the ray diagram construction for finding the image formed for the virtual object shown? The solid dots on either side of the lens locate the equal magnitude foci of the lens.

$\textbf{B}$

By careful! It is a VIRTUAL OBJECT. A virtual object with a diverging lens will act like a real object with a converging lens. So the ray diagram in answer B is correct. We can also solve the problem in a mathematical way. The focal length of the diverging lens is $-f$, and the object distance of the virtual distance is $d_o=-2f$. Using $\dfrac1{d_o}+\dfrac1{d_i}=\dfrac1{-f}$, we get the image distance $d_i=-2f$. Both the object distance and the image distance are negative, so they are on the different sides of the diverging lens. The magnification $M=-d_i/d_o=-1$, so the object and the image are of equal size but upside down. Hence, the image in answer B is correct.

A person vibrates the end of a string sending transverse waves down the string. If the person then doubles the rate at which he vibrates the string, the speed of the waves

$\textbf{(A) }$ doubles and the wavelength is unchanged$ \qquad\newline$

$\textbf{(B) }$ doubles and the wavelength doubled$ \qquad\newline$

$\textbf{(C) }$ doubles while the wavelength is halved$ \qquad\newline$

$\textbf{(D) }$ is unchanged while the wavelength is doubled$ \qquad\newline$

$\textbf{(E) }$ is unchanged while the wavelength is halved.

$\textbf{E}$

The wave speed depends on the tension in the string and its composition, which are both unchanged. Hence, since the wave speed is unchanged, with an increased frequency, by $v=\lambda f$, there is a decreased wavelength.

If the temperature of a material doubles on the Kelvin scale, by how much does the time-rate at which energy is radiated from the material change?

$\textbf{(A) }$ It is unchanged$ \qquad\newline$

$\textbf{(B) }$ It is doubled$ \qquad\newline$

$\textbf{(C) }$ It is 4 times greater$ \qquad\newline$

$\textbf{(D) }$ It is 8 times greater$ \qquad\newline$

$\textbf{(E) }$ It is 16 times greater

$\textbf{E}$

The rate of black body radiation is $P=\sigma AT^4$. When the temperature $T$ is doubled, the radiation power is $2^4=16$ times greater.

An ideal gas undergoes a reversible isothermal expansion at $T=300\ \text{K}$. The total change in entropy of the gas is 2.5 J/K. How much work was done by the environment on the gas during this process?

$\textbf{(A) }$ -750 J$ \qquad$ $\textbf{(B) }$ -120 J$ \qquad$ $\textbf{(C) }$ 120 J$ \qquad$ $\textbf{(D) }$ 750 J$ \qquad\newline$

$\textbf{(E) }$ More information is required to answer this question.

$\textbf{A}$

The entropy $\Delta S=\dfrac{Q}{T}=2.5\ \text{J/K}$, so the energy absorbed by the gas is $Q=T\Delta S=(300)(2.5)=750\ \text{J}$. By the First Law of Thermodynamics and the fact that there is no change in temperature (and hence no change in the internal energy of the ideal gas) $\delta U=Q+W=0$, so the work done by the surrounding on the gas is $W=-Q=-750\ \text{J}$.

Two spaceships travel along paths that are at right angles to each other. Each ship travels at $0.60c$ where $c$ is the speed of light in a vacuum according to a stationary observer. If one of the ships turns on a green laser and aims it at a right angle to the direction of its travel, with what speed does the other speed record the speed of the green light?

$\textbf{(A) }$ $0.40c \qquad$ $\textbf{(B) }$ $0.85c \qquad$ $\textbf{(C) }$ $1.00c \qquad$ $\textbf{(D) }$ $1.17c \qquad\newline$

$\textbf{(E) }$ More information is required about the direction that the light is traveling in order to answer the question.

$\textbf{C}$

Einstein’s Second Postulate of Special Relativity states that the speed of light is the same for all observers in an inertial frame. Hence, regardless of which way the green light is directed, the observer records that the light travels at speed $c$.

How fast must an observer move so that a stationary object appears to be one-half of its proper length?

$\textbf{(A) }$ $0.50c \qquad$ $\textbf{(B) }$ $0.67c \qquad$ $\textbf{(C) }$ $0.75c \qquad$ $\textbf{(D) }$ $0.87c \qquad$ $\textbf{(E) }$ $0.93c$

$\textbf{D}$

Using the formula of length contraction effect $l=l_0\sqrt{1-\dfrac{v^2}{c^2}}$, when $l=\dfrac12l_0$, we get $\dfrac{v^2}{c^2}=\dfrac12\rightarrow v=\sqrt{\dfrac34}c=0.87c$.

The ratio $\lambda_1/\lambda_2$ of the deBroglie wavelengths of two non-relativistic particles with masses $m_1$ and $m_2$ and the same kinetic energy, is equal to

$\textbf{(A) }$ $\dfrac{m_2}{m_1} \qquad$ $\textbf{(B) }$ $\dfrac{m_1}{m_2} \qquad$ $\textbf{(C) }$ $\sqrt{\dfrac{m_2}{m_1}} \qquad$ $\textbf{(D) }$ $\sqrt{\dfrac{m_1}{m_2}} \qquad$ $\textbf{(E) }$ $1$

$\textbf{C}$

The deBroglie wavelength is $\lambda=\dfrac{h}{p}$, where $p$ is the linear momentum. The kinetic energy $KE$ is related to the linear momentum in the equation $KE=\dfrac{p^2}{2m}\rightarrow p=\sqrt{2mK}$. Since the kinetic energy is the same, we have $\dfrac{p_1}{p_2}=\dfrac{\sqrt{2m_1K}}{\sqrt{2m_2K}}=\sqrt{\dfrac{m_1}{m_2}}$. The ratio of wavelength $\dfrac{\lambda_1}{\lambda_2}=\dfrac{p_2}{p_1}=\sqrt{\dfrac{m_2}{m_1}}$.

A gas undergoes radioactive decay with time constant $\tau$. A sample of 10000 particles is put into a container. After one time constant has passed, the experimenter places another 10000 particles into the original container. How much time passes from the addition of the particles until the container of gas reaches 10000 total particles again?

$\textbf{(A) }$ $0.405\tau \qquad$ $\textbf{(B) }$ $0.500\tau \qquad$ $\textbf{(C) }$ $0.693\tau \qquad$ $\textbf{(D) }$ $\tau \qquad$ $\textbf{(E) }$ $\tau$

$\textbf{A}$

The first 10000 particles after one time constant will be $N=N_0e^{-t/\tau}=10000e^{-1}=3678$ particles remained. Adding 10000 particles gives a total of 13678 particles, the time needed to decay to 10000 particles is given in $10000=13678e^{-t/\tau}\rightarrow t=-\tau\ln\dfrac{10000}{13678}=0.313\tau$.