## PhysicsBowl 2022

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**Instruction**

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

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

A 20,000 kg truck is traveling at 25 km/hr. At what speed does a 1000 kg car need to travel to have the same kinetic energy as the truck?

$\textbf{(A) }$ 112 km/hr$ \qquad$ $\textbf{(B) }$ 132 km/hr$ \qquad$ $\textbf{(C) }$ 102 km/hr$ \qquad$ $\textbf{(D) }$ 79.0 km/hr$ \qquad$ $\textbf{(E) }$ 89.0 km/hr

$\textbf{A}$

When the kinetic energy is the same, we have $\dfrac12m_1v_1^2=\dfrac12m_2v_2^2\Rightarrow v_2=\sqrt{\dfrac{m_1}{m_2}}v_1=\sqrt{\dfrac{20000}{1000}}(25)=112\ \text{km/hr}$.

Who was the $\textit{last}$ person to walk on the Moon?

$\textbf{(A) }$ Buzz Aldrin$ \qquad\newline$ $\textbf{(B) }$ Eugene Cernan$ \qquad\newline$ $\textbf{(C) }$ Neil Armstrong$ \qquad\newline$ $\textbf{(D) }$ Sally Ride$ \qquad\newline$ $\textbf{(E) }$ Gus Grissom

$\textbf{B}$

Eugene Cernan was the last man to walk on the moon (till now, maybe the Chinese will land on the moon in the next decade. Shame on you, NASA). See: https://www.space.com/20790-eugene-cernan-astronaut-biography.html

Two bodies of equal mass are moving in circular paths at equal speed. The first body moves in a circle whose radius is twice that of the second. The ratio of the radial acceleration of the first body to that of the second is:

$\textbf{(A) }$ 1 to 4$ \qquad$ $\textbf{(B) }$ 1 to 3$ \qquad$ $\textbf{(C) }$ 1 to 2$ \qquad$ $\textbf{(D) }$ 1 to 1$ \qquad$ $\textbf{(E) }$ 1 to 0.5

$\textbf{C}$

The radial acceleration is $a=m\dfrac{v^2}{r}$. Since the masses and speeds of the two bodies are equal, and the radius $r_1=2r_2$, we have $\dfrac{a_1}{a_2}=\dfrac{r_2}{r_1}=\dfrac12$.

A bullet with a mass of 5.0 g is fired horizontally into a 2.0 kg wooden block which is resting on a horizontal table. The bullet stops in the block and the block and bullet combination move 2.0 m. The coefficient of kinetic friction between the block and surface of the table is 0.2. Find the initial speed of the bullet.

$\textbf{(A) }$ 1123 m/s$ \qquad$ $\textbf{(B) }$ 1134 m/s$ \qquad$ $\textbf{(C) }$ 1132 m/s$ \qquad$ $\textbf{(D) }$ 113.2 m/s$ \qquad$ $\textbf{(E) }$ 113.4 m/s

$\textbf{B}$

The mass of the bullet is $m_1=0.005\ \text{kg}$, and the mass the block is $m_2=2.0\ \text{kg}$. When the block and bullet combination decelerates on the surface, the acceleration is $a=-\mu g=-(0.2)(10)=-2\ \text{m/s}^2$. Hence the initial speed of the combination is $v=\sqrt{2|a|x}=\sqrt{2(2)(2)}=2\sqrt2\ \text{m/s}$. When the bullet collides with the block, the linear momentum is conserved. Supposing the initial speed of the bullet is $v_1$, then we have $$m_1v_1+m_2(0)=(m_1+m_2)v$$ so $$v_1=\dfrac{m_1+m_2}{m_1}v=\dfrac{2+0.005}{0.005}\cdot2\sqrt2=1134\ \text{m/s}$$

A car is moving along a straight horizontal road at a speed of 20 m/s. The brakes are applied and a constant force of 5000 N brings the car to a stop in 10 s. What is the mass of the car?

$\textbf{(A) }$ 1250 kg$ \qquad$ $\textbf{(B) }$ 2500 kg$ \qquad$ $\textbf{(C) }$ 5000 kg$ \qquad$ $\textbf{(D) }$ 7500 kg$ \qquad$ $\textbf{(E) }$ 10,000 kg

$\textbf{B}$

The acceleration of the car is $a=\dfrac{\Delta v}{\Delta t}=\dfrac{0-20}{10}=-2\ \text{m/s}^2$ (we take the absolute value as $a=2\ \text{m/s}^2$). According to Newton's second law, $F=ma$, so the mass of the car is $m=\dfrac{F}{a}=\dfrac{5000}{2}=2500\ \text{kg}$.

Two forces have magnitudes of 11.0 N and 5.0 N. The magnitude of their sum could NOT be equal to which of the following values?

$\textbf{(A) }$ 16.0 N$ \qquad$ $\textbf{(B) }$ 9.0 N$ \qquad$ $\textbf{(C) }$ 7.0 N$ \qquad$ $\textbf{(D) }$ 5.0 N$ \qquad$ $\textbf{(E) }$ 6.0 N

$\textbf{D}$

The two forces and their sum will form three sides of a triangle unless the two forces are in the same or opposite direction. Hence, the sum should be no more than $11+5=16\ \text{N}$, and no less than $11-5=6\ \text{N}$.

Which of the following colors of visible light has the longest wavelength?

$\textbf{(A) }$ Violet$ \qquad$ $\textbf{(B) }$ Yellow$ \qquad$ $\textbf{(C) }$ Blue$ \qquad$ $\textbf{(D) }$ Green$ \qquad$ $\textbf{(E) }$ Red

$\textbf{E}$

The red and violet light are at opposite ends of the spectrum of visible light. The red light has the longest wavelength and the lowest frequency, while the violet has the shortest wavelength and the highest frequency.

n increase in the translational motion of the molecules of a gas confined in a steel tank will be observed in one of the following ways. It will produce an increase in:

$\textbf{(A) }$ the temperature of the gas only.$ \qquad\newline$

$\textbf{(B) }$ the pressure of the gas only.$ \qquad\newline$

$\textbf{(C) }$ both the temperature and pressure of the gas.$ \qquad\newline$

$\textbf{(D) }$ the temperature of the gas and a decrease in its pressure.$ \qquad\newline$

$\textbf{(E) }$ the volume of the gas.

$\textbf{C}$

An increase in the translational motion of the gas molecules results in a more frequent collision with container wall, so the pressure of the gas will increase. Since the volume of the steel tank will not change, the volume of the gas is a constant. Using ideal gas equation $PV=nRT$, the temperature $T$ will increase.

When descending mountain roads, large trucks pulling a heavy load can burn up the brakes. Once the brakes are no longer useful, the driver may need to guide the truck up a “runaway truck lane” on the side of the road. The runaway truck lane is directed uphill and often has a thick layer or sand or gravel or both on the surface. Which of the following is $\textbf{one}$ of the reasons the truck will stop?

$\textbf{(A) }$ An increase in kinetic energy$ \qquad\newline$

$\textbf{(B) }$ A decrease in potential energy$ \qquad\newline$

$\textbf{(C) }$ A decrease in fuel$ \qquad\newline$

$\textbf{(D) }$ A transfer of energy to the gravel on the track of the runaway truck lane$ \qquad\newline$

$\textbf{(E) }$ The change in temperature of the engine

$\textbf{D}$

The runaway truck lane with a thick layer of sand or gravel has a greater coefficient of friction. More kinetic energy will be transferred to the lane and therefore the truck stops.

A 4.0 kg mass at the end of a spring moves with simple harmonic motion on a horizontal frictionless table with a period of 2.0 s and an amplitude of 2.0 m. Determine the maximum force exerted on the spring.

$\textbf{(A) }$ 25.1 N$ \qquad$ $\textbf{(B) }$ 158 N$ \qquad$ $\textbf{(C) }$ 39.5 N$ \qquad$ $\textbf{(D) }$ 63.0 N$ \qquad$ $\textbf{(E) }$ 79.0 N

$\textbf{E}$

The angular frequency $\omega=2\pi f=\dfrac{2\pi}{T}=\dfrac{2\pi}{2}=\pi\ \text{rad/s}$. The spring constant $k=m\omega^2$. The force reaches its maximum when the displacement $x=A=2.0\ \text{m}$. Hence, $F_{\max}=kA=m\omega^2A=4\pi^2(2)=79.0\ \text{N}$.

**The following information is to be used for Questions 11-12:**

The James Webb Space Telescope was launched on December 25, 2021 and arrived at its destination, LaGrange Point 2, on January 24, 2022. This point is approximately 1,500,000 km from the Earth and it is where the telescope will orbit the sun.

What was James Webb’s former occupation?

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

$\textbf{(B) }$ Telescope designer/engineer$ \qquad\newline$

$\textbf{(C) }$ Silicon valley entrepreneur and financier of the telescope project$ \qquad\newline$

$\textbf{(D) }$ Space shuttle astronaut/pilot$ \qquad\newline$

$\textbf{(E) }$ NASA Administrator

Well...now you know it.

How long will it take the radio signal sent by the James Webb Space Telescope to reach the Earth?

$\textbf{(A) }$ 0.005 s$ \qquad$ $\textbf{(B) }$ 0.05 s$ \qquad$ $\textbf{(C) }$ 5 s$ \qquad$ $\textbf{(D) }$ 50 s$ \qquad$ $\textbf{(E) }$ 5000 s

$\textbf{C}$

The time for electromagnetic wave to travel is $t=\dfrac{d}{c}=\dfrac{1.5\times10^5\cdot10^3\ \text{m}}{3\times10^8\ \text{m/s}}=5\ \text{s}$

car engine moves a piston with a circular cross-section of 7.500 cm in diameter a distance of 3.250 cm to compress the gas in the cylinder. By what amount is the gas decreased in volume?

$\textbf{(A) }$ $143.6\ \text{cm}^3 \qquad$ $\textbf{(B) }$ $153.6\ \text{cm}^3 \qquad$ $\textbf{(C) }$ $662.7\ \text{cm}^3 \qquad$ $\textbf{(D) }$ $682.7\ \text{cm}^3 \qquad$ $\textbf{(E) }$ $88.36\ \text{cm}^3$

$\textbf{A}$

The volume $V=\pi R^2x=\pi\left(\dfrac{D}2\right)^2x=\pi\left(\dfrac{7.5}2\right)^2(3.25)=143.6\ \text{cm}^3$.

A fisherman watches a dolphin leap out of the water at an angle of $35^\circ$ above the horizontal. The horizontal component of the dolphin’s velocity is 7.7 m/s. Find the magnitude of the vertical component of the velocity.

$\textbf{(A) }$ 4.4 m/s$ \qquad$ $\textbf{(B) }$ 6.3 m/s$ \qquad$ $\textbf{(C) }$ 11 m/s$ \qquad$ $\textbf{(D) }$ 5.4 m/s$ \qquad$ $\textbf{(E) }$ 3.2 m/s

$\textbf{D}$

The vertical component $v_y=v_x\tan\theta=7.7\tan35^\circ=5.4\ \text{m/s}$.

Two simple pendula are 60 cm and 63 cm in length. They hang vertically, one in front of the other. If they are set in motion simultaneously, find the time taken for one to gain a complete cycle of oscillation on the other.

$\textbf{(A) }$ 15.4 s$ \qquad$ $\textbf{(B) }$ 15.7 s$ \qquad$ $\textbf{(C) }$ 31.5 s$ \qquad$ $\textbf{(D) }$ 62.4 s$ \qquad$ $\textbf{(E) }$ 79.7 s

$\textbf{D}$

The periods of the two pendula are $T_1=2\pi\sqrt{\dfrac{l_1}{g}}$ and $T_2=2\pi\sqrt{\dfrac{l_2}{g}}$. Since $l_1<l_2$, the first pendulum oscillates faster than the second one. When the second pendulum finished $n$ cycles, the first did $n+1$ cycles. Hence, we get $$(n+1)T_1=nT_2\rightarrow n=\dfrac{T_1}{T_2-T_1}=\dfrac{\sqrt{l_1}}{\sqrt{l_2}-\sqrt{l_1}}=\dfrac{\sqrt{60}}{\sqrt{63}-\sqrt{60}}=40.5$$ So the total time is $$T=nT_2=40.5\cdot2\pi\sqrt{\dfrac{0.63}{10}}=63.9\ \text{s}$$

If a copper wire is stretched to make it $0.1\%$ longer, what is the percentage change in its resistance?

$\textbf{(A) }$ $0.01\% \qquad$ $\textbf{(B) }$ $0.1\% \qquad$ $\textbf{(C) }$ $0.2\% \qquad$ $\textbf{(D) }$ $0.4\% \qquad$ $\textbf{(E) }$ $0.5\%$

$\textbf{C}$

Since the volume of the copper wire $V=L\cdot A$ is a constant, when the length $l_2=(1+0.1\%)l_1$, the sectional area $A_2=(1-0.1\%)A_1$. The resistance $R=\rho\dfrac{L}{A}$, so the change of resistance is $$\dfrac{R_2}{R_1}-1=\dfrac{l_2}{l_1}\cdot\dfrac{A_1}{A_2}-1=(1+0.1\%)^2-1=0.2\%$$

A block of wood initially at rest slides down an inclined plane. Neglecting friction, the kinetic energy of the block at the bottom of the plane is:

$\textbf{(A) }$ all converted into heat.$ \qquad\newline$

$\textbf{(B) }$ less than its kinetic energy at the top of the plane.$ \qquad\newline$

$\textbf{(C) }$ dependent on the materials of which the block is made.$ \qquad\newline$

$\textbf{(D) }$ dependent on the materials of which the inclined plane is made.$ \qquad\newline$

$\textbf{(E) }$ equal to its potential energy (with respect to the bottom of the plane) when it was at the top of the plane.

$\textbf{E}$

In the absence of friction, all gravitational potential energy will be converted to kinetic energy at the bottom of the plane.

In the last one second of a free fall, an apple traveled three-fourths of its total path. From what height did the apple fall?

$\textbf{(A) }$ 10 m$ \qquad$ $\textbf{(B) }$ 15 m$ \qquad$ $\textbf{(C) }$ 20 m$ \qquad$ $\textbf{(D) }$ 25 m$ \qquad$ $\textbf{(E) }$ 30 m

$\textbf{C}$

Let the total time for the free fall be $t$. Since the apple traveled 3/4 of the total path in the last second, we get $$\dfrac12g(t-1)^2=\dfrac14\cdot\dfrac12gt^2\rightarrow t=2\ \text{s}$$ So the height $$h=\dfrac12gt^2=\dfrac12(10)(2)^2=20\ \text{m}$$

A helicopter with a mass of 700 kg will hover when its rotating blades move through an area of $50\ \text{m}^2$. Find the average speed imparted to the air ($\text{density of air}=1.3\ \text{kg/m}^3$)

$\textbf{(A) }$ 7.66 m/s$ \qquad$ $\textbf{(B) }$ 10.4 m/s$ \qquad$ $\textbf{(C) }$ 16.5 m/s$ \qquad$ $\textbf{(D) }$ 38.0 m/s$ \qquad$ $\textbf{(E) }$ 44.1 m/s

$\textbf{B}$

By the free body diagram, the force from the air $$F=\rho v^2A=mg\rightarrow v=\sqrt{\dfrac{mg}{\rho A}}=\sqrt{\dfrac{(700)(10)}{(1.3)(50)}}=10.4\ \text{m/s}$$

A refrigerator has a mass of 150 kg and rests in the open back end of a delivery truck. If the truck accelerates from rest at $1.5\ \text{m/s}^2$, what is the minimum coefficient of static friction between the refrigerator and the bed of the truck that is required to prevent the refrigerator from sliding off the back of the truck?

$\textbf{(A) }$ 0.08$ \qquad$ $\textbf{(B) }$ 0.10$ \qquad$ $\textbf{(C) }$ 0.12$ \qquad$ $\textbf{(D) }$ 0.15$ \qquad$ $\textbf{(E) }$ 0.18

$\textbf{D}$

In preventing the refrigerator from sliding, the friction must provide it the same acceleration as the truck to keep the refrigerator and the truck relatively static. So we have $$\mu mg=ma\rightarrow\mu=\dfrac{a}{g}=\dfrac{1.5}{10}=0.15$$

Three blocks ($m_1=1\ \text{kg}$, $m_2=2\ \text{kg}$, $m_3=3\ \text{kg}$) connected by cords are pulled by a constant force, $F$, of 18 N on a frictionless horizontal table. $T_2$ is the tension in the rope between $m_2$ and $m_3$. What is $T_2$?

$\textbf{(A) }$ 3 N$ \qquad$ $\textbf{(B) }$ 6 N$ \qquad$ $\textbf{(C) }$ 9 N$ \qquad$ $\textbf{(D) }$ 12 N$ \qquad$ $\textbf{(E) }$ 15 N

$\textbf{C}$

The acceleration of the three-block system is $a=\dfrac{F}{m_1+m_2+m_3}$. However, for the $m_1+m_2$ system, the acceleration is given by $T_2=(m_1+m_2)a=\dfrac{m_1+m_2}{m_1+m_2+m_3}F=\dfrac{1+2}{1+2+3}\cdot18=9\ \text{N}$.

An object starting from rest moves on a circular path with a radius 40 cm and a constant tangential acceleration of $10\ \text{cm/s}^2$. How much time is needed after the motion begins for the centripetal acceleration of the object to be equal to the tangential acceleration?

$\textbf{(A) }$ 0.2 s$ \qquad$ $\textbf{(B) }$ 1.0 s$ \qquad$ $\textbf{(C) }$ 1.2 s$ \qquad$ $\textbf{(D) }$ 1.8 s$ \qquad$ $\textbf{(E) }$ 2.0 s

$\textbf{E}$

Given that the tangential acceleration $a_{\tau}=0.1\ \text{m/s}^2$ is a constant, the speed of the object at time $t$ is $v=a_{\tau}t$. Then we get the centripetal acceleration $a_r=\dfrac{v^2}{r}=\dfrac{a_{\tau}^2t^2}{r}$. When the centripetal acceleration is equal to the tangential acceleration, we have $$a_r=a_{\tau}\rightarrow t=\sqrt{\dfrac{r}{a_{\tau}}}=\sqrt{\dfrac{0.4}{0.1}}=2.0\ \text{s}$$

The distance between the electron and the proton in the hydrogen atom is about $0.53\times10^{-10}\ \text{m}$. By what factor is the electrical force between the electron and proton stronger than the gravitational force between them?

$\textbf{(A) }$ $1.3\times10^{39} \qquad$ $\textbf{(B) }$ $2.3\times10^{39} \qquad$ $\textbf{(C) }$ $3.3\times10^{39} \qquad$ $\textbf{(D) }$ $4.3\times10^{39} \qquad$ $\textbf{(E) }$ $5.3\times10^{39}$

$\textbf{B}$

Let the factor be $x$, then we have $$ k\dfrac{e^2}{r^2}=x\cdot G\dfrac{m_em_p}{r^2}\rightarrow x=\dfrac{ke^2}{Gm_em_p}=\dfrac{\left(9\times10^9\right)\left(1.6\times10^{-19}\right)^2}{\left(6.67\times10^{-11}\right)\left(9.11\times10^{-31}\right)\left(1.67\times10^{-27}\right)}=2.3\times10^{39}$$

A 3.00 kg bucket of water is raised with an upward acceleration of $2.20\ \text{m/s}^2$ from a well by means of an attached rope. What is the tension in the rope?

$\textbf{(A) }$ 30.2 N$ \qquad$ $\textbf{(B) }$ 33.3 N$ \qquad$ $\textbf{(C) }$ 36.6 N$ \qquad$ $\textbf{(D) }$ 39.0 N$ \qquad$ $\textbf{(E) }$ 43.2 N

$\textbf{C}$

By the free body diagram, we $T-mg=ma\rightarrow T=m(g+a)=3(10+2.2)=36.6\ \text{N}$.

Two light waves, initially emitted in phase, will interfere constructively with maximum amplitude if the path-length difference between them is:

$\textbf{(A) }$ 1.5 wavelengths$ \qquad\newline$

$\textbf{(B) }$ one wavelength$ \qquad\newline$

$\textbf{(C) }$ one-half wavelength$ \qquad\newline$

$\textbf{(D) }$ one-quarter wavelength$ \qquad\newline$

$\textbf{(E) }$ one-eighth wavelength

$\textbf{B}$

To interfere constructively, the phase difference should be an integer multiple of the wavelength.

For an object moving in uniform circular motion, the magnitude of the centripetal acceleration is given by $a_c=v^2/r$, where $v$ is the speed of the object, and $r$ is the radius of the circle. The “jerk” is the rate of change of acceleration. For uniform circular motion, the magnitude of the jerk is given by:

$\textbf{(A) }$ zero$ \qquad$ $\textbf{(B) }$ $v^2/r \qquad$ $\textbf{(C) }$ $v^2r \qquad$ $\textbf{(D) }$ $v^3/r \qquad$ $\textbf{(E) }$ $v^3/r^2$

$\textbf{E}$

Let the object rotates by a very small angle $\Delta\theta$ in the very short amount of time $\Delta t$. The acceleration rotates for the same angle $\Delta\theta$, where the change of acceleration is $\Delta a_c=a_c\Delta\theta$. Therefore, the rate of change of acceleration is $\dfrac{\Delta a_c}{\Delta t}=a_c\dfrac{\Delta\theta}{\Delta t}=a_c\omega=a_c\dfrac{v}{r}=\dfrac{v^3}{r^2}$.

A machine gun fires 100 g bullets at a speed of 1000 m/s. The person holding the machine gun in their hands can exert an average force of 150 N against the gun. If the gun is to remain stationary, what is the maximum number of bullets that can be fired per minute?

$\textbf{(A) }$ 10$ \qquad$ $\textbf{(B) }$ 15$ \qquad$ $\textbf{(C) }$ 30$ \qquad$ $\textbf{(D) }$ 60$ \qquad$ $\textbf{(E) }$ 90

$\textbf{E}$

The impulse for the force $F=150\ \text{N}$ in a minute is $I=Ft=(150)(60)=9000\ \text{N}\cdot\text{s}$. The momentum of a bullet is $p=mv=(0.1)(1000)=100\ \text{kg}\cdot\text{m/s}$. By the impulse-momentum theorem, the number of bullets is given as $n=\dfrac{I}{p}=90$.

A spring with spring constant $k=20\ \text{kN/m}$, is used to stop a 50 kg box that is sliding on a horizontal surface. The spring is initially in its equilibrium state. At position A, shown in the top diagram, the box has a speed of 3.0 m/s. The compression of the spring when the box is instantaneously at rest (position B in the bottom diagram) is 120 mm. Determine the coefficient of kinetic friction between the box and the surface.

$\textbf{(A) }$ 0.125$ \qquad$ $\textbf{(B) }$ 0.230$ \qquad$ $\textbf{(C) }$ 0.245$ \qquad$ $\textbf{(D) }$ 0.280$ \qquad$ $\textbf{(E) }$ 0.315

$\textbf{B}$

The kinetic energy of the box is converted into the heat by friction and the potential energy of the spring. Hence, we have $$\dfrac12mv^2=\mu mg(x_1+x_2)+\dfrac12kx_2^2$$ The coefficient of kinetic friction $$\mu=\dfrac{\dfrac12mv^2-\dfrac12kx_2^2}{mg(x_1+x_2)}=\dfrac{\dfrac12(50)(3)^2-\dfrac12(20000)(0.12)^2}{(50)(10)(0.6+0.12)}=0.225$$

A cyclist, using a power meter while on a training ride, checks and sees that she is doing work at the rate of 500 W. How much average force does her foot exert on the pedals when she is traveling at 8.0 m/s?

$\textbf{(A) }$ 31 N$ \qquad$ $\textbf{(B) }$ 63 N$ \qquad$ $\textbf{(C) }$ 80 N$ \qquad$ $\textbf{(D) }$ 320 N$ \qquad$ $\textbf{(E) }$ 710 N

$\textbf{B}$

Using $P=Fv$, we get the force $F=\dfrac{P}{v}=\dfrac{500}{8}=63\ \text{N}$.

A ticker tape timer is operating at 60 Hz. It was used to analyze the motion of a battery-powered car. The following displacements were measured for five intervals on the ticker tape.

What is the average speed of the car?

$\textbf{(A) }$ 0.612 m/s$ \qquad$ $\textbf{(B) }$ 3.66 m/s$ \qquad$ $\textbf{(C) }$ 0.366 m/s$ \qquad$ $\textbf{(D) }$ 0.947 m/s$ \qquad$ $\textbf{(E) }$ 6.31 m/s

$\textbf{B}$

The average displacement of an interval is $\bar{x}=\dfrac{x_1+x_2+x_3+x_4+x_5}{5}=0.061\ \text{m}$. The time for each interval is $t=\dfrac1{60}\ \text{s}$. So the average speed is $\bar{v}=\dfrac{\bar{x}}{t}=3.66\ \text{m/s}$.

Water drips from the nozzle of a shower onto the floor 2.45 m below. The drops fall at regular intervals of time, the first drop striking the floor at the instant the third drop begins to fall. Measured from the shower head, what is the location of the second drop when the first drop strikes the floor.

$\textbf{(A) }$ 0.313 m$ \qquad$ $\textbf{(B) }$ 0.613 m$ \qquad$ $\textbf{(C) }$ 0.938 m$ \qquad$ $\textbf{(D) }$ 1.25 m$ \qquad$ $\textbf{(E) }$ 1.563 m

$\textbf{B}$

The time of the free fall from $h=2.45\ \text{m}$ is $t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2(2.45)}{10}}=0.7\ \text{s}$. Since the drops fall at regular intervals of time, and the first drop striking the floor at the instant the third drop begins to fall, we get the time interval $t_0=\dfrac12t=0.35\ \text{s}$. When the first drop strikes the floor, the second drop has traveled for $t_0=0.35\ \text{s}$. Therefore, the location of the second drop from the shower head is $h_0=\dfrac12gt_0^2=\dfrac12(10)(0.35)^2=0.613\ \text{s}$.

A star initially has a radius of $6\times10^8\ \text{m}$ and a period of rotation about its axis of 30 days. Eventually, it collapses to form a neutron star with a radius of only $1\times10^4\ \text{m}$ and a period of 0.1 s. Assuming that the mass has not changed, find the ratio of initial and final angular momentum of the star.

$\textbf{(A) }$ 15.92$ \qquad$ $\textbf{(B) }$ 214.6$ \qquad$ $\textbf{(C) }$ 103.1$ \qquad$ $\textbf{(D) }$ 442.1$ \qquad$ $\textbf{(E) }$ 138.9

$\textbf{E}$

A solid ball of mass $m$ and radius $r$ rotating about its axis has a momentum of inertia $I=\dfrac25mr^2$. The angular momentum of the solid ball is $L=I\omega=I\cdot\dfrac{2\pi}{T}=\dfrac{4\pi mr^2}{5T}$. Hence, the ratio of angular momentum is $$\dfrac{L_1}{L_2}=\left(\dfrac{r_1}{r_2}\right)^2\dfrac{T_2}{T_1}=\left(\dfrac{6\times10^8}{1\times10^4}\right)^2\dfrac{0.1}{30\times24\times60\times60}=138.9$$

For ordinary conversation, a sound level meter reads 60 dB. What is the intensity of this sound wave?

$\textbf{(A) }$ $6.0\times10^{-6}\ \text{W/m}^2 \qquad\newline$

$\textbf{(B) }$ $1.0\times10^{-4}\ \text{W/m}^2 \qquad\newline$

$\textbf{(C) }$ $6.0\times10^{-4}\ \text{W/m}^2 \qquad\newline$

$\textbf{(D) }$ $1.0\times10^{-6}\ \text{W/m}^2 \qquad\newline$

$\textbf{(E) }$ $1.0\times10^{-8}\ \text{W/m}^2$

$\textbf{D}$

The intensity level of sound wave $60\ \text{dB}=10\log\left(\dfrac{I}{I_0}\right)$ where $I_0=10^{-12}\ \text{W/m}^2$. Hence, we get the intensity of the sound wave $I=1.0\times10^{-6}\ \text{W/m}^2$.

A tiny ball with a mass of 0.6 g carries a charge of magnitude $8\ \mu\text{C}$. It is suspended by a thread in a downward directed electric field of intensity 300 N/C. What is the tension in the thread if the charge on the ball is positive?

$\textbf{(A) }$ $2.40\times10^{-3}\ \text{N} \qquad\newline$

$\textbf{(B) }$ $6.00\times10^{-3}\ \text{N} \qquad\newline$

$\textbf{(C) }$ $8.40\times10^{-3}\ \text{N} \qquad\newline$

$\textbf{(D) }$ $6.00\times10^{-2}\ \text{N} \qquad\newline$

$\textbf{(E) }$ $6.24\times10^{-3}\ \text{N}$

$\textbf{C}$

By the free body diagram, the tension $$T=mg+qE=\left(0.6\times10^{-3}\right)(10)+\left(8\times10^{-6}\right)(300)=8.4\times10^{-3}\ \text{N}$$

A combination of two thin convex lenses is placed as shown. An object is placed 5 cm in front of $L_1$ which has a focal length of 10 cm. $L_2$ is 10 cm behind $L_1$ and has a focal length of 12 cm. How far from $L_2$ is the final image for this lens combination?

$\textbf{(A) }$ 8 cm$ \qquad$ $\textbf{(B) }$ 15 cm$ \qquad$ $\textbf{(C) }$ 22 cm$ \qquad$ $\textbf{(D) }$ 24 cm$ \qquad$ $\textbf{(E) }$ 30 cm

$\textbf{E}$

The object distance for $L_1$ is $d_{o1}=5\ \text{cm}$. Using $\dfrac1{d_o}+\dfrac1{d_i}=\dfrac1f$, we get $d_{i1}=\dfrac{f_1d_{o1}}{d_{o1}-f_1}=\dfrac{(10)(5)}{5-10}=-10\ \text{cm}$. This means the image are on the left side of $L_1$. Hence, the object distance for $L_2$ is $d_{o2}=10+10=20\ \text{cm}$. Therefore, the image distance is $d_{i2}=\dfrac{f_2d_{o2}}{d_{o2}-f_2}=\dfrac{(12)(20)}{20-12}=30\ \text{cm}$.

A thick glass plate has parallel sides. A beam of white light is incident on one side at an angle between $0^\circ$ and $90^\circ$ with the normal. Which color emerges from the other side first?

$\textbf{(A) }$ All of them$ \qquad$ $\textbf{(B) }$ red$ \qquad$ $\textbf{(C) }$ green$ \qquad$ $\textbf{(D) }$ violet$ \qquad$ $\textbf{(E) }$ None of them

$\textbf{A}$

There is no dispersion in a glass block with parallel sides.

A car and a truck are both traveling with a constant speed of 20 m/s. The car is 10 m behind the truck. The truck driver suddenly applies his brakes, causing the truck to slow to a stop at the constant rate of $2\ \text{m/s}^2$. Two seconds later, the driver of the car applies their brakes and just manages to avoid a rear-end collision. Determine the constant rate at which the car slowed.

$\textbf{(A) }$ $3.33\ \text{m/s}^2 \qquad$ $\textbf{(B) }$ $4.33\ \text{m/s}^2 \qquad$ $\textbf{(C) }$ $1.33\ \text{m/s}^2 \qquad$ $\textbf{(D) }$ $3.03\ \text{m/s}^2 \qquad$ $\textbf{(E) }$ $3.93\ \text{m/s}^2$

$\textbf{A}$

In the velocity vs. time graph, the motion of the truck is a straight line passing (0,20) and (10,0). The car keeps its speed at 20 m/s for 2 seconds and then decelerates. If the car stops at $t=10\ \text{s}$ either, the area enclosed by the truck line and the car line is the distance caught up by the car. However, the area of the triangle is $\dfrac12(2)(20)=20\ \text{m}$, which is greater than 10 m. Therefore, the car must stop before $t=10\ \text{s}$. Since the area of the triangle should be 10, given that the base of the triangle is 2, the height must be 10. Hence, the car line intersects the truck line at point (5,10). The acceleration of the car is $a=\dfrac{\Delta v}{\Delta t}=\dfrac{20-10}{5-2}=3.33\ \text{m/s}^2$.

In the circuit shown below, the voltage source, switch, capacitor and connecting wires have no resistance. Let $\Delta V_R$ and $\Delta V_C$ represent the potential differences across the resistor and the capacitor, respectively. The capacitor is initially uncharged. Which of the following choices correctly compares these potential differences immediately after the switch is closed and after the circuit has reached steady state.

$\textbf{A}$

The potential differences across the resistance and the capacitor are all equal to the potential difference of the voltage source.

A radioactive carbon-14 nucleus decays into a beta particle, an antineutrino, and a nitrogen-14 nucleus. In a particular decay, the beta particle has momentum, $p$, and the nitrogen nucleus has momentum of magnitude $4p/3$ at an angle of $90^\circ$ to $p$. At what angle (with respect to the beta particle) do you expect the antineutrino to be emitted?

$\textbf{(A) }$ $53^\circ \qquad$ $\textbf{(B) }$ $37^\circ \qquad$ $\textbf{(C) }$ $90^\circ \qquad$ $\textbf{(D) }$ $127^\circ \qquad$ $\textbf{(E) }$ $33^\circ$

$\textbf{D}$

The antineutrino is in the opposite direction of the resultant of $p$ and $4p/3$. The angle between the resultant and the direction of the beta particle is $\arctan\dfrac43=53^\circ$. Therefore, the angle between the direction of antineutrino and beta particle is $180^\circ-53^\circ=127^\circ$.

An Olympic ice skater in Beijing spins at $4\pi\ \text{rad/s}$ with her arms extended. If her moment of inertia with arms folded is $80\%$ of that with arms extended, what is her angular velocity when she folds her arms?

$\textbf{(A) }$ $\pi\ \text{rad/s} \qquad$ $\textbf{(B) }$ $2\pi\ \text{rad/s} \qquad$ $\textbf{(C) }$ $3\pi\ \text{rad/s} \qquad$ $\textbf{(D) }$ $5\pi\ \text{rad/s} \qquad$ $\textbf{(E) }$ $6\pi\ \text{rad/s}$

$\textbf{D}$

Let the momentum of inertia of the ice skater be $I_1$ when her arms extended. Then the momentum of inertia with arms folded is $I_2=0.8I_1$. Using the conservation of angular momentum, we have $$I_1\omega_1=I_2\omega_2\rightarrow\omega_2=\dfrac{I_1}{I_2}\omega_1=1.25(4\pi)=5\pi\ \text{rad/s}$$

A weather balloon is loosely filled with $2.0\ \text{m}^3$ of helium at 1.0 atm and $27\ ^\circ\text{C}$. The balloon is then released. When it has reached an elevation of 7000 m, the pressure has dropped to 0.41 atm and the balloon has expanded. If the temperature at this elevation is $-31\ ^\circ\text{C}$, what is the new volume of the balloon?

$\textbf{(A) }$ $2.7\ \text{m}^3 \qquad$ $\textbf{(B) }$ $3.2\ \text{m}^3 \qquad$ $\textbf{(C) }$ $3.9\ \text{m}^3 \qquad$ $\textbf{(D) }$ $4.6\ \text{m}^3 \qquad$ $\textbf{(E) }$ $5.1\ \text{m}^3$

$\textbf{C}$

When $P_1=1.0\ \text{atm}$, $V_1=2.0\ \text{m}^3$ and $T_1=273+27=300\ \text{K}$, using the ideal gas equation, we get $$P_1V_1=nRT_1$$ Similarly, when $P_2=0.41\ \text{atm}$ and $T_2=273-31=242\ \text{K}$, we have $$P_2V_2=nRT_2$$ So the new volume of the balloon is $$V_2=\dfrac{P_1T_2}{P_2T_1}V_1=\dfrac{(1)(242)}{(0.41)(300)}(2)=3.9\ \text{m}^3$$

Three resistors of $4\Omega$, $6\Omega$, and $12\Omega$ are connected in parallel. This parallel arrangement is then connected in series with a 1-$\Omega$ and a 2-$\Omega$ resistor. A potential difference of 120 V is applied across the ends of the circuit. What will be the potential drop across the part of the circuit connected in parallel?

$\textbf{(A) }$ 12 V$ \qquad$ $\textbf{(B) }$ 48 V$ \qquad$ $\textbf{(C) }$ 24 V$ \qquad$ $\textbf{(D) }$ 36 V$ \qquad$ $\textbf{(E) }$ 72 V

$\textbf{B}$

Using $\dfrac1{R}=\dfrac1{R_1}+\dfrac1{R_2}+\dfrac1{R_3}$, the equivalent resistance of the three resistors in parallel is $R=2\Omega$, which takes $\dfrac{2}{2+1+2}=\dfrac25$ of the total potential difference. So the potential drop across in-parallel part is $\dfrac25\times120=48\ \text{V}$.

A triangular glass prism ($n=1.6$) is immersed in a liquid ($n=1.1$) as shown. A laser light is incident as shown on face AB making an angle of $20^\circ$ with the normal (N). Calculate the angle that the ray emerging from AC makes with the ground when it leaves AC and strikes the ground.

$\textbf{(A) }$ $28.1^\circ \qquad$ $\textbf{(B) }$ $30.3^\circ \qquad$ $\textbf{(C) }$ $33.8^\circ \qquad$ $\textbf{(D) }$ $36.1^\circ \qquad$ $\textbf{(E) }$ $18.9^\circ$

$\textbf{B}$

The incident angle $\theta_1=20^\circ$. Using $n_1\sin\theta_1=n_2\sin\theta_2$, we get the refraction angle $$\theta_2=\arcsin\left(\dfrac{n_1}{n_2}\sin\theta_1\right)=\arcsin\left(\dfrac{1.1}{1.6}\sin20^\circ\right)13.6^\circ$$ Let the points where the laser ray intersects AB be D, the ray intersects AC be E, and the ray intersects the ground be F. Then we get $\angle ADE=90-13.6=76.4^\circ$, $\angle AED=180-40-76.4=63.6^\circ$. The incident angle on face AC is $\theta_3=90-63.6=26.4^\circ$. By $n_2\sin\theta_3=n_1\sin\theta_4$, we get the refraction angle $\theta_4=40.3^\circ$, $\angle CEF=90-40.3=49.7^\circ$. Considering that $\angle ACB=180-40-60=80^\circ$, the angle between the refraction ray and the ground is $80-49.7=30.3^\circ$.

How long would it take $4.0\times10^{20}$ nuclei to decay to $1.0\times10^{19}$ atoms if their half-life was 14.7 years?

$\textbf{(A) }$ 29.4 years$ \qquad$ $\textbf{(B) }$ 58.8 years$ \qquad$ $\textbf{(C) }$ 78.2 years$ \qquad$ $\textbf{(D) }$ 147 years$ \qquad$ $\textbf{(E) }$ 161 years

$\textbf{C}$

Given the half-life $\tau=14.7\ \text{years}$, we have $$N=N_0\left(\dfrac12\right)^{t/\tau}\rightarrow 1.0\times10^{19}=4.0\times10^{20}\left(\dfrac12\right)^{t/14.7}$$ Hence, we get $t=78.2\ \text{years}$.

An electric field of 1500 V/m and a magnetic field act on an electron moving with a speed of 3000 m/s. If the resultant force is to be zero, what should be the strength of the magnetic field?

$\textbf{(A) }$ 0.35 T$ \qquad$ $\textbf{(B) }$ 0.50 T$ \qquad$ $\textbf{(C) }$ 0.72 T$ \qquad$ $\textbf{(D) }$ 0.81 T$ \qquad$ $\textbf{(E) }$ 0.96 T

$\textbf{B}$

By $qE=qvB$, we get the magnetic field $B=\dfrac{E}{v}=\dfrac{1500}{3000}=0.5\ \text{T}$.

Each of the two curved rods shown in the picture form one-quarter of a circle with a radius $R$. Both rods carry a uniformly-distributed electric charge $+Q$. Which of the following choices correctly expresses the net electric field and net electric potential at the origin? Assume $V\rightarrow0$ as $r\rightarrow\infty$.

$\textbf{B}$

For any point charge A on the top right rod, we can find a point charge B on the bottom left rod where AB passes through the origin O. So the electric field of these pair at the origin cancel each other out. Hence, the electric field at the origin is zero. The total charge $+Q$ on a rod is the sum of point charges. Each point charge is distance $R$ away from the origin. So the electric potential by charge $+Q$ at the origin is $\dfrac{kQ}{R}$. Considering that there are two rods of charge $+Q$, the electric potential is $\dfrac{2kQ}{R}$.

A shipping box in a warehouse has a mass of 2.00 kg and slides down an inclined plane that makes an angle of $30^\circ$ with the horizontal. The coefficient of kinetic friction between the box and the plane surface is 0.866. How much force (parallel to the incline) should be applied to the box so that it moves down the plane at constant speed?

$\textbf{(A) }$ 5.00 N$ \qquad$ $\textbf{(B) }$ 6.50 N$ \qquad$ $\textbf{(C) }$ 3.80 N$ \qquad$ $\textbf{(D) }$ 4.60 N$ \qquad$ $\textbf{(E) }$ 11.0 N

$\textbf{A}$

The component of gravity paralleling to the incline is $mg\sin\theta=(2)(10)\sin30^\circ=10\ \text{N}$. The friction when the box slides is $\mu mg\cos\theta=(0.866)(2)(10)\cos30^\circ=15\ \text{N}$. To make the box slide down with a constant speed, we need to push the box. The force $F=15-10=5\ \text{N}$.

The diagram below shows combinations X, Y, and Z of three identical resistors. What is the correct order of the $\textbf{total}$ resistance of the combinations going from $\textbf{lowest}$ resistance to $\textbf{highest}$ resistance?

$\textbf{(A) }$ Y, X, Z$ \qquad$ $\textbf{(B) }$ Z, X, Y$ \qquad$ $\textbf{(C) }$ X, Y, Z$ \qquad$ $\textbf{(D) }$ Z, Y, X$ \qquad$ $\textbf{(E) }$ Y, Z, X

$\textbf{D}$

Let the resistance of each resistor be $R$. Then we have $R_X=\dfrac32R$, $R_Y=\dfrac23R$, $R_Z=\dfrac13R$. So the order is $R_Z<R_Y<R_X$.

If the Earth suddenly stopped in its orbit, and the orbit is assumed to be circular, how much time would elapse before it falls into the Sun?

$\textbf{(A) }$ 147 minutes$ \qquad$ $\textbf{(B) }$ 1.22 days$ \qquad$ $\textbf{(C) }$ 64.5 days$ \qquad$ $\textbf{(D) }$ 228 days$ \qquad$ $\textbf{(E) }$ 2.34 days

$\textbf{C}$

The earth falls to the sun in a straight line. One way to simplify the calculation is to take the line as a thin and long ellipse where the focal points, for example, the location of the sun, is at the edge of the long axis. Hence, the semimajor axis $a$ is half the distance $d$ between the sun and the earth. According to Kepler's third law, $$\dfrac{a^3}{T^2}=\dfrac{GM_S}{4\pi^2}$$ The period $$T=\sqrt{\dfrac{4\pi^2a^3}{GM_S}}=\sqrt{\dfrac{\pi^2d^3}{2GM_S}}=\sqrt{\dfrac{\pi^2\left(1.5\times10^{11}\right)^3}{2\left(6.67\times10^{-11}\right)\left(2\times10^{30}\right)}}=1.1\times10^7\ \text{s}=129\ \text{days}$$ The time for the earth to falls to the sun is half the period, which is 64.5 days.

An ideal gas system, with an initial volume of $1.0\ \text{m}^3$ at standard temperature and pressure, undergoes the following three-stage cycle: Stage 1 – an isothermal expansion to twice its original volume. Stage 2 – a process by which its volume remains constant, its pressure returns to its original value and 104 J of heat is added to the system. Stage 3 – an isobaric compression to its original volume, with $3\times10^4\ \text{J}$ of heat being removed from the system. Calculate the work done on the system during Stage 3.

$\textbf{(A) }$ $-7.005\times10^4\ \text{J} \qquad\newline$

$\textbf{(B) }$ $-1.013\times10^4\ \text{J} \qquad\newline$

$\textbf{(C) }$ $-1.013\times10^5\ \text{J} \qquad\newline$

$\textbf{(D) }$ $-7.050\times10^5\ \text{J} \qquad\newline$

$\textbf{(E) }$ $-2.030\times10^5\ \text{J}$

$\textbf{C}$

The initial state of the gas is $V_0=1.0\ \text{m}^3$, $P_0=1.013\times10^5\ \text{pa}$, and $T_0=273.15\ \text{K}$.$\newline$

After stage 1, $T_1=T_0$, $V_1=2V_0$, $P_2=\dfrac12P_0$ due to the ideal gas equation.$\newline$

After stage 2, $V_2=V_1=2V_0$, $P_2=P_0$, $T_2=2T_0.\newline$

After stage 3, $P_3=P_2=P_0$, $V_3=V_0$. The work done on the system is $P_3(V_3-V_2)=-P_0V_0=-1.013\times10^5\ \text{J}$.