PhysicsBowl 2017
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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 scientist computes a speed as $\dfrac{12\ \text{mm}}{3\ \mu\text{s}}$. Which one of the following choices represents this same speed?
$\textbf{(A) }$ $4\dfrac{\text{km}}{\text{s}} \qquad$ $\textbf{(B) }$ $4\dfrac{\text{nm}}{\text{s}} \qquad$ $\textbf{(C) }$ $4\dfrac{\text{m}}{\text{s}} \qquad$ $\textbf{(D) }$ $4\dfrac{\text{Gm}}{\text{s}} \qquad$ $\textbf{(E) }$ $4\dfrac{\text{pm}}{\text{s}}$
$\textbf{A}$
$1\ \text{mm}=10^{-3}\ \text{m}\qquad 1\ \mu\text{s}=10^{-6}\ \text{s}$
A small object is released from rest and falls freely for 4.00 s until reaching the ground. What was the height above the ground from which the object was released?
$\textbf{(A) }$ 10.0 m$ \qquad$ $\textbf{(B) }$ 20.0 m$ \qquad$ $\textbf{(C) }$ 40.0 m$ \qquad$ $\textbf{(D) }$ 80.0 m$ \qquad$ $\textbf{(E) }$ 160.0 m
$\textbf{D}$
$h=\dfrac12gt^2$
Which one of the following choices is a vector quantity?
$\textbf{(A) }$ average speed$ \qquad$ $\textbf{(B) }$ kinetic energy$ \qquad$ $\textbf{(C) }$ mass$ \qquad$ $\textbf{(D) }$ time$ \qquad$ $\textbf{(E) }$ acceleration
$\textbf{E}$
A vector has both magnitude and direction.
At which one of the following temperatures does water undergo a phase change into a solid at a pressure of 1.0 atm?
$\textbf{(A) }$ 273 K$ \qquad$ $\textbf{(B) }$ $0\ ^\circ\text{F} \qquad$ $\textbf{(C) }$ $32\ ^\circ\text{C} \qquad$ $\textbf{(D) }$ $100\ ^\circ\text{F} \qquad$ $\textbf{(E) }$ 0 K
$\textbf{A}$
Water undergoes a phase change to a solid at $0\ ^\circ\text{C}$, which is equivalent to $32\ ^\circ\text{F}$ and 273 K.
Which one of the following choices is most closely associated with the following observation about an isolated object: The linear momentum of the object is constant.
$\textbf{(A) }$ Huygens’ Principle$ \qquad\newline$
$\textbf{(B) }$ Kepler’s First Law$ \qquad\newline$
$\textbf{(C) }$ Hooke’s Law$ \qquad\newline$
$\textbf{(D) }$ Pascal’s Principle$ \qquad\newline$
$\textbf{(E) }$ Newton’s First Law
$\textbf{E}$
An object with constant linear momentum will have a constant velocity. This relates most directly to Newton’s First Law (a body in motion continues in uniform motion).
A mass connected to an ideal spring oscillates with a period $T$ when it is released from rest at a position $A$ from equilibrium. Which one of the following choices correctly identifies the period of oscillation for the mass connected to the same spring when it is released from rest at a position $2A$ from equilibrium?
$\textbf{(A) }$ $\dfrac14T \qquad$ $\textbf{(B) }$ $\dfrac12T \qquad$ $\textbf{(C) }$ $T \qquad$ $\textbf{(D) }$ $2T \qquad$ $\textbf{(E) }$ $4T$
$\textbf{C}$
The period of an ideal mass-spring system does not depend on the amplitude of the oscillation (it depends on spring constant and mass), meaning that the period is unchanged.
A traveling sine wave of frequency 3.0 Hz is known to move at 6.0 m/s on a string. Which one of the following figures best represents what a one-meter section of the string could look like at an instant of time?
$\textbf{B}$
The wavelength $\lambda=\dfrac{v}{f}=2\ \text{m}$, so only $\dfrac12\lambda$ is shown in a one-meter section.
A mass moves along the x-axis following the position vs. time graph provided. During what time interval(s) does the mass have an acceleration of 0 m/s$^2$? Time is represented as $t$.
$\textbf{(A) }$ only from $0\ \text{s}<t<3\ \text{s} \qquad\newline$
$\textbf{(B) }$ only at $t=5\ \text{s} \qquad\newline$
$\textbf{(C) }$ only from $7\ \text{s}<t<10\ \text{s} \qquad\newline$
$\textbf{(D) }$ only from $0\ \text{s}<t<3\ \text{s}$ and $7\ \text{s}<t<10\ \text{s} \qquad\newline$
$\textbf{(E) }$ only from $0\ \text{s}<t<3\ \text{s}$, $t=5\ \text{s}$, and $7\ \text{s}<t<10\ \text{s}$.
$\textbf{D}$
The acceleration $a=0$ means the velocity is a constant, so the graph must be a straight line with a constant slope in the x-t graph.
A 4.00 kg object moves to the right with speed 5.00 m/s. A 5.00 kg object moves to the left with speed 3.00 m/s. After colliding, the 4.00 kg object moves to the left with speed 1.00 m/s. What is the speed of the 5.00 kg object after the collision?
$\textbf{(A) }$ 1.00 m/s$ \qquad$ $\textbf{(B) }$ 1.80 m/s$ \qquad$ $\textbf{(C) }$ 3.00 m/s$ \qquad$ $\textbf{(D) }$ 4.11 m/s$ \qquad$ $\textbf{(E) }$ 7.80 m/s
$\textbf{B}$
The linear momentum is conserved, so $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$.
A 2000 kg car moves over a hill at a constant speed of 12.0 m/s. At the very top of the hill, the shape of the road can be approximated as a circle with radius 25 m, as indicated in the figure. Which one of the following choices best represents the magnitude of the upward force exerted by the road on the car?
$\textbf{(A) }$ 0 kN$ \qquad$ $\textbf{(B) }$ 8.5 kN$ \qquad$ $\textbf{(C) }$ 11.5 kN$ \qquad$ $\textbf{(D) }$ 20.0 kN$ \qquad$ $\textbf{(E) }$ 31.5 kN
$\textbf{B}$
The centripetal force is provided by $mg-N=m\dfrac{v^2}{R}$, so $N=mg-m\dfrac{v^2}{R}$.
Questions 11 and 12 refer to the following information:
A 5.00 kg block slides with uniform acceleration on a horizontal surface for 5.00 s. The block’s kinetic energy strictly decreases from 50.0 J to 20.0 J in this time.
What is the average speed of the block during the motion described?
$\textbf{(A) }$ 7.00 m/s$ \qquad$ $\textbf{(B) }$ 3.74 m/s$ \qquad$ $\textbf{(C) }$ 3.65 m/s$ \qquad$ $\textbf{(D) }$ 2.65 m/s$ \qquad$ $\textbf{(E) }$ 2.58 m/s
$\textbf{C}$
According to the initial and final kinetic energy, we get the initial speed $v_i=2\sqrt5\ \text{m/s}$ and final speed $v_f=2\sqrt2\ \text{m/s}$. Since the acceleration is a constant, the average speed $\overline{v}=\dfrac{v_i+v_f}{2}$.
Which one of the following choices best represents the magnitude of the block's acceleration during the motion described?
$\textbf{(A) }$ 3.43 m/s$^2 \qquad$ $\textbf{(B) }$ 2.40 m/s$^2 \qquad$ $\textbf{(C) }$ 1.20 m/s$^2 \qquad$ $\textbf{(D) }$ 0.329 m/s$^2 \qquad$ $\textbf{(E) }$ 0.232 m/s$^2$
$\textbf{D}$
The acceleration $a=\lvert\dfrac{v_f-v_i}{\Delta t}\rvert$
According to currently accepted theory, dark matter makes up approximately what percent of the Universe?
$\textbf{(A) }$ $5\% \qquad$ $\textbf{(B) }$ $25\% \qquad$ $\textbf{(C) }$ $50\% \qquad$ $\textbf{(D) }$ $80\% \qquad$ $\textbf{(E) }$ $99\%$
$\textbf{B}$
According to current theory, the majority of the Universe consists of dark matter and dark energy with about $5\%$ “normal matter”, ~$25\%$ dark matter and ~$70\%$ dark energy.
A block is suspended in the air at the end of a string held by a person. The person pulls on the string uniformly accelerating the block upward for a short time. During this time, which one of the following choices correctly identifies the Newton’s Third Law pair force to the force of the string on the block?
$\textbf{(A) }$ the gravitational force from the Earth on the block$ \qquad\newline$
$\textbf{(B) }$ the force from the string on the person$ \qquad\newline$
$\textbf{(C) }$ the force from the person on the string$ \qquad\newline$
$\textbf{(D) }$ the force from the block on the string$ \qquad\newline$
$\textbf{(E) }$ the force from the person on the block
$\textbf{D}$
Newton’s Third Law deals with an interaction between two bodies. Here, the string exerts a force on the block with the block exerting an equal sized force on the string.
Two blocks X and Y are in contact on a frictionless horizontal
surface. A constant 35 N force is applied to the right onto block X. What is the magnitude of the force that block Y exerts onto block X?
$\textbf{(A) }$ 35 N$ \qquad$ $\textbf{(B) }$ 28 N$ \qquad$ $\textbf{(C) }$ 25 N$ \qquad$ $\textbf{(D) }$ 14 N$ \qquad$ $\textbf{(E) }$ 10 N
$\textbf{C}$
The acceleration $a=\dfrac{F}{M_X+M_Y}$, so the force X exerted on Y is $F_{XY}=M_Ya$, the force Y exerted on X is $F_{YX}=F_{XY}$.
Which one of the following choices best represents the type of electromagnetic wave associated with a frequency of $10^{18}\ \text{Hz}$?
$\textbf{(A) }$ AM radio$ \qquad$ $\textbf{(B) }$ microwaves$ \qquad$ $\textbf{(C) }$ ultraviolet light$ \qquad$ $\textbf{(D) }$ X ray$ \qquad$ $\textbf{(E) }$ red light
$\textbf{D}$
From $v=\lambda f$, we determine the wavelength of the light to be $\lambda=v/f=30.3\ \text{nm}$. This wavelength is about 1000 times smaller than what one might expect for ultraviolet light (violet light’s wavelength is approximately 400 nm). So, we have light much more energetic (higher frequency) than ultraviolet light. Of the choices given, this puts the light in the X-ray range.
The following lengths are added together: $L_1=8.36\times10^1\ \text{m}$, $L_2=9.20\times10^2\ \text{m}$, $L_3=1.39\times10^{-1}\ \text{m}$, and $L_4=2.1\times10^0\ \text{m}$. How many significant digits are in the resulting sum?
$\textbf{(A) }$ 2$ \qquad$ $\textbf{(B) }$ 3$ \qquad$ $\textbf{(C) }$ 4$ \qquad$ $\textbf{(D) }$ 5$ \qquad$ $\textbf{(E) }$ 7
$\textbf{C}$
Adding the values together gives 1005.839. Because 920. only goes as far as the nearest whole number and the other lengths have values after the decimal, we record the answer as a whole number. Hence, we would write the sum as 1006 with proper significant digits.
A disk of radius 75.0 cm makes 3.5 revolutions when coming to rest in 2.50 s. What was the initial angular speed of the disk?
$\textbf{(A) }$ 1.40 rad/s$ \qquad$ $\textbf{(B) }$ 6.60 rad/s$ \qquad$ $\textbf{(C) }$ 8.80 rad/s$ \qquad$ $\textbf{(D) }$ 13.2 rad/s$ \qquad$ $\textbf{(E) }$ 17.6 rad/s
$\textbf{E}$
Assuming the disk is decelerated by a constant deceleration, the average angular speed is $$\overline{\omega}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{3.5\cdot2\pi\ \text{rad}}{2.5\ \text{s}}=2.8\pi\ \text{rad/s}$$ So the initial angular speed is $\omega_i=2\overline{\omega}$.
An ideal gas at a temperature of $100^\circ\text{C}$ has 3.0 moles of particles enclosed in a 5.0 L container. Which one of the following choices best represents the pressure of the gas?
$\textbf{(A) }$ $4.99\times10^5\ \text{Pa} \qquad\newline$
$\textbf{(B) }$ $1.86\times10^6\ \text{Pa} \qquad\newline$
$\textbf{(C) }$ $5.02\times10^6\ \text{Pa} \qquad\newline$
$\textbf{(D) }$ $5.05\times10^7\ \text{Pa} \qquad\newline$
$\textbf{(E) }$ $1.88\times10^8\ \text{Pa}$
$\textbf{B}$
The temperature can be calculated by $pV=nRT$.
An object appears green when ideal cyan light shines on it. If ideal yellow light were to shine on the object, which one of the following choices identifies the color(s) that the object could appear?
$\textbf{(A) }$ only red$ \qquad\newline$
$\textbf{(B) }$ only yellow$ \qquad\newline$
$\textbf{(C) }$ only green or black$ \qquad\newline$
$\textbf{(D) }$ only yellow or green$ \qquad\newline$
$\textbf{(E) }$ only yellow, green, or red
$\textbf{D}$
Cyan light is a combination of blue and green, which means that the object reflects green light and absorbs blue light. Since yellow light is made of green and red, we know that green light will be reflected and the red light will either be reflected (resulting in the object appearing yellow) or be absorbed (resulting in the object appearing green).
The change in linear momentum of an object is always in the same direction as the
$\textbf{(A) }$ velocity of the object.$ \qquad\newline$
$\textbf{(B) }$ displacement of the object.$ \qquad\newline$
$\textbf{(C) }$ average speed of the object.$ \qquad\newline$
$\textbf{(D) }$ average velocity of the object.$ \qquad\newline$
$\textbf{(E) }$ average acceleration of the object.
$\textbf{E}$
By the impulse-momentum theorem, the change in linear momentum is the force multiplied by the time. Hence, since the force is mass multiplied by acceleration, the change in linear momentum is in the same direction as the acceleration.
A large swimming pool is filled with a liquid having a density of $5.00\times10^2\ \text{kg/m}^3$. A person wants to see what happens if she jumps into the deep end of the pool (3.0 meters deep). Which one of the following choices correctly identifies what will happen to the person? Assume that the person does not try to move their arms or legs until discovering what happens after equilibrium is established.
$\textbf{(A) }$ The person will float with much more than 50$\%$ of their body’s volume above the liquid’s surface.$ \qquad\newline$
$\textbf{(B) }$ The person will float with approximately 50$\%$ of their body’s volume above the liquid’s surface.$ \qquad\newline$
$\textbf{(C) }$ The person will float with much less than 50$\%$ of their body’s volume above the liquid’s surface.$ \qquad\newline$
$\textbf{(D) }$ The person will sink but remain off the bottom of the pool.$ \qquad\newline$
$\textbf{(E) }$ The person will sink to the pool’s bottom.
$\textbf{E}$
The density of the fluid in the pool is less than the density of water, which is very close to the density of a human being. This means that when the person jumps into the pool, they will sink all the way to the bottom.
A 1500 kg car and a 6000 kg truck have a perfectly inelastic collision on a street. Which one of the following choices correctly identifies the object(s) that lose kinetic energy from the impact?
$\textbf{(A) }$ Neither object loses kinetic energy.$ \qquad\newline$
$\textbf{(B) }$ Only the car loses kinetic energy.$ \qquad\newline$
$\textbf{(C) }$ Only the truck loses kinetic energy.$ \qquad\newline$
$\textbf{(D) }$ Both the car and truck lose kinetic energy.$ \qquad\newline$
$\textbf{(E) }$ The answer cannot be determined without more information.
$\textbf{E}$
The overall kinetic energy associated with the system of the two objects colliding will decrease as a result of the collision, but the lighter one may not.
What condition must be met for the angular momentum of a system to be constant?
$\textbf{(A) }$ The mechanical energy is constant.$ \qquad\newline$
$\textbf{(B) }$ There is no net external force acting on the system.$ \qquad\newline$
$\textbf{(C) }$ The system is in static equilibrium.$ \qquad\newline$
$\textbf{(D) }$ There is no net external torque acting on the system.$ \qquad\newline$
$\textbf{(E) }$ Only the gravitational force is acting on the system.
$\textbf{D}$
The angular momentum is conserved when no external torque applied.
A box slides with uniform acceleration up an incline. The box has an
initial speed of 9.0 m/s and rises vertically 2.60 m before coming to rest. If the angle of the incline is $30^\circ$, what is the coefficient of kinetic friction between the box and the incline?
$\textbf{(A) }$ 0.298$ \qquad$ $\textbf{(B) }$ 0.322$ \qquad$ $\textbf{(C) }$ 0.372$ \qquad$ $\textbf{(D) }$ 0.483$ \qquad$ $\textbf{(E) }$ 0.557
$\textbf{B}$
The supporting force by the incline is $$N=mg\cos\theta$$ the acceleration can be derived from $ma=mg\sin\theta+\mu N$, so $$a=g\sin\theta+\mu g\cos\theta$$ The box slides for $l=h/\sin\theta=2h$, so the acceleration can also been found as $a=\dfrac{v^2}{2l}$, now we have $$g\sin\theta+\mu g\cos\theta=\dfrac{v^2}{2l}$$ By solving the equation we get $\mu=0.322$.
A 4.00 kg mass is in uniform circular motion as shown in the figure. The string to which the mass is attached has a length of $L=3.00\ \text{m}$ and forms an angle of $\theta=50.0^\circ$ with the vertical. What is the speed of the mass?
$\textbf{(A) }$ 5.23 m/s$ \qquad$ $\textbf{(B) }$ 5.54 m/s$ \qquad$ $\textbf{(C) }$ 5.98 m/s$ \qquad$ $\textbf{(D) }$ 6.62 m/s$ \qquad$ $\textbf{(E) }$ 6.90 m/s
$\textbf{A}$
In the vertical direction $$mg=F\cos\theta$$ in the horizontal direction $$F\sin\theta=m\dfrac{v^2}{L\sin\theta}$$ thus we get $$v=\sqrt{gL\sin\theta\tan\theta}$$
A 2.0 kg mass moves at 10.0 m/s when it has a collision in space. The mass's speed was unchanged, but the direction of its velocity was altered by 80°. If the collision lasted 0.25 s, what was the magnitude of the average force exerted on the mass?
$\textbf{(A) }$ 0 N$ \qquad$ $\textbf{(B) }$ 54.7 N$ \qquad$ $\textbf{(C) }$ 102.8 N$ \qquad$ $\textbf{(D) }$ 122.5 N$ \qquad$ $\textbf{(E) }$ 150.3 N
$\textbf{C}$
By the impulse-momentum theorem $\vec{F}\cdot\Delta t=m\cdot\Delta\vec{v}$, where $\Delta v=2v\sin\dfrac{\theta}{2}$, we get $F=m\dfrac{\Delta v}{\Delta t}=102.8\ \text{N}$.
A shipment of parts is labeled in units of ohm$\cdot$second. Which one of the following choices represents an equivalent unit?
$\textbf{(A) }$ Henry$ \qquad$ $\textbf{(B) }$ Farad$ \qquad$ $\textbf{(C) }$ Joule$ \qquad$ $\textbf{(D) }$ Volt$ \qquad$ $\textbf{(E) }$ Weber
$\textbf{A}$
According to $\mathcal{E}=L\dfrac{\text{d}I}{\text{d}t}$, the unit $$\text{Volt}=\text{Henry}\dfrac{\text{Ampere}}{\text{second}}$$ it can be rewrote as $$\text{Henry}=\dfrac{\text{Volt}}{\text{Ampere}}\cdot\text{second}=\text{Ohm}\cdot\text{second}$$
Two lightbulbs (X and Y) are connected in series to a battery. Bulb X is brighter than bulb Y. Assume all circuit elements are ideal. If bulbs X and Y are connected in parallel with each other to the same battery, which one of the following choices best represents what is observed?
$\textbf{E}$
Bulbs in series have the same current, so the resistor with higher resistance will be brighter, which means $R_X>R_Y$. Bulbs connected in parallel have the same potential difference, so the resistor with smaller resistance will be brighter. Further, in the parallel configuration, both bulbs get the potential difference associated with the battery, whereas in series, each bulb only gets a fraction of the potential difference. This means that both bulbs now are brighter than before.
Two cars, P and Q, each start from rest at the origin and will move along the x-axis. The acceleration vs. time graph for each car is shown. Which one of the following choices correctly identifies the relationship for both the speeds ($v_P$ and $v_Q$) and the positions ($x_P$ and $x_Q$) of the cars at time $T$?
$\textbf{A}$
The area under an acceleration vs time curve gives the change in velocity. For the curves given, the area under each is equal. This means that at time $T$, the speeds are the same. Since the acceleration of X is much greater at the start, it gains speed more quickly than Y. This means that X travels further than Y. As Y only matches the speed of X at the end of the interval, X moves faster than Y at all other times and therefore is always moving further ahead of Y.
A solid cylinder rolls without slipping on a rough inclined plane. Which one of the following choices best represents the type and direction of friction (if any) acting on the cylinder as it rolls up the incline?
$\textbf{(A) }$ Static friction directed up the incline$ \qquad\newline$
$\textbf{(B) }$ Kinetic friction directed up the incline$ \qquad\newline$
$\textbf{(C) }$ There is no friction$ \qquad\newline$
$\textbf{(D) }$ Kinetic friction directed down the incline$ \qquad\newline$
$\textbf{(E) }$ Static friction directed down the incline
$\textbf{A}$
Since the object is rolling without slipping, any friction would have to be static on the incline (there is no relative motion between cylinder and surface). As for the direction, one notes that if the friction acts down the incline (as one might suspect to be opposite to the motion), this would cause the speed of the object to decrease, but it also would cause an increase in the rotational speed of the object. The torque from the center of the cylinder would make the object spin faster and faster which cannot happen. Consequently, in order to slow the rotation, the friction must be pointing up the incline!
A projectile is launched at an angle of $40^\circ$ above the horizontal with a speed of 30 m/s. How much time passes before the position of the projectile makes an angle of $20^\circ$ above the horizontal from the original launch point?
$\textbf{(A) }$ 3.02 s$ \qquad$ $\textbf{(B) }$ 2.38 s$ \qquad$ $\textbf{(C) }$ 2.18 s$ \qquad$ $\textbf{(D) }$ 1.93 s$ \qquad$ $\textbf{(E) }$ 1.64 s
$\textbf{C}$
The vertical component of the speed at $40^\circ$ is $$v_{\perp1}=v_0\sin\theta_1$$ and the horizontal component is $$v_{\parallel1}=v_0\cos\theta_1$$ The horizontal component do not change in the process, so the vertical component at $20^\circ$ is $$v_{\perp2}=v_{\parallel1}\tan\theta_2$$. The time for the change of vertical speed is $$t=\dfrac{v_{\perp1}-v_{\perp2}}{g}$$
Two long wires are fixed in space so that the conventional current in the left wire (2A) comes out of the plane of the page and the conventional current in the right wire (3A) goes into the plane of the page. In which Region(s) is there a place on the x-axis (aside from infinity) at which the magnetic field is equal to zero from these currents?
$\textbf{(A) }$ Only in Region I$ \qquad\newline$
$\textbf{(B) }$ In both Regions I and II$ \qquad\newline$
$\textbf{(C) }$ Only in Region II$ \qquad\newline$
$\textbf{(D) }$ In both Regions I and III$ \qquad\newline$
$\textbf{(E) }$ In both Regions II and III
$\textbf{A}$
Assuming the position of the left wire on the x-axis is the zero point and the position of right wire is $x_2$, for any point of coordinate $x$ on the x-axis, the magnetic field by $I_1=2\ \text{A}$ is $$B_1=\dfrac{\mu_0I_1}{2\pi x}$$ the magnetic field by $I_2=3\ \text{A}$ is $$B_2=-\dfrac{\mu_0I_2}{2\pi(x-x_2)}$$ When $B=B_1+B_2=0$, by solving the equation we get $$x=-2x_2$$ which is in region I.
A 9.20 m long uniform plank rests on a frictionless ice pond. A 52 kg box rests on the plank's left end while a 71 kg person stands at the plank’s right end. After the person walks to the left on the plank and stands at the same location as the box, the plank has slid 3.84 m to the right relative to the pond's shore. Which one of the following choices best represents the mass of the plank?
$\textbf{(A) }$ 123 kg$ \qquad$ $\textbf{(B) }$ 61.5 kg$ \qquad$ $\textbf{(C) }$ 47.1 kg$ \qquad$ $\textbf{(D) }$ 36.5 kg$ \qquad$ $\textbf{(E) }$ 31.2 kg
$\textbf{C}$
Considering that the ice pond is frictionless, there is no force in the horizontal direction, thus the center of mass of the whole system (box $m_1$, person $m_2$ and plank $M$) do not move. In the beginning the position of the center of mass is $$x_c=\dfrac{M\cdot\frac{1}{2}l+m_2l}{m_1+m_2+M}$$ After the person moving left and the plank sliding right for $x_0=3.84\ \text{m}$, the position of the center of mass is $$x_c=\dfrac{(m_1+m_2)x_0+M(x_0+\frac12l)}{m_1+m_2+M}$$ Since the center of mass do not move, we have $$M\cdot\frac{1}{2}l+m_2l=(m_1+m_2)x_0+M(x_0+\frac12l)$$ so $$M=\dfrac{m_2l}{x_0}-m_1-m_2=47.1\ \text{kg}$$
A metal bar is moving to the left across a set of frictionless conducting rails as seen in the figure. Throughout the region between the rails, there is a uniform magnetic field directed into the plane of the page. The resistors labeled $X$ and $Y$ are identical. Which one of the following choices correctly indicates the direction of the conventional current in the resistors and the relation between the magnitude of the currents through each resistor at the instant shown?
$\textbf{B}$
Supposing there is a positive charge in the metal bar. When the bar is moving to the left, the force exerted on the charge $\vec{F}=q\vec{v}\times\vec{B}$ pointing downward, so the bottom side of the metal bar is the positive terminal, thus both resistors experience upward currents. Since the two resistors are identical and connected to the same battery (the moving metal bar), the currents passing through them are equal.
A scientist performs an experiment in which she determines the shortest length of a gas column needed to create resonance for a vibrating tuning fork over a tube closed at one end. She plots the gas column length against the inverse of the frequency for a set of tuning forks and finds that she has a straight line fit through the data. Representing the slope of the line as $m$, which one of the following choices correctly identifies the speed of waves through the gas in the experiment?
$\textbf{(A) }$ $\dfrac14m \qquad$ $\textbf{(B) }$ $m \qquad$ $\textbf{(C) }$ $2m \qquad$ $\textbf{(D) }$ $\dfrac43m \qquad$ $\textbf{(E) }$ $4m$
$\textbf{E}$
From the expression for a tube closed at one end, $f_n=n\left(\dfrac{v}{4L}\right)$. In the experiment, only first harmonics are being sought, so $n=1$. Rearranging this expression, we write it as $L=\left(\dfrac{v}{4}\right)\dfrac1f$. Plotting L vs. $\dfrac1f$ will give a straight line of slope $m=\dfrac{v}{4}$. So, the speed of the waves is $v=4m$.
Which one of the following choices best approximates the magnitude of the Earth's angular momentum (expressed in base MKS units) associated with its orbit around the Sun?
$\textbf{(A) }$ $10^{36} \qquad$ $\textbf{(B) }$ $10^{40} \qquad$ $\textbf{(C) }$ $10^{44} \qquad$ $\textbf{(D) }$ $10^{48} \qquad$ $\textbf{(E) }$ $10^{52}$
$\textbf{B}$
To compute the angular momentum we are going to approximate the Earth as point-like since its radius is small compared to its orbit radius. So $L=I\omega=MR^2\omega$. The distance from the Earth to the Sun can be approximated as the distance that light travels in about 10 minutes, that is, $$R=vt=(3\times10^8\ \text{m/s})(10\ \text{min}\times\dfrac{60\ \text{s}}{1\ \text{min}})=1.8\times10^{11}\ \text{m}$$ The mass is on the constants sheet and the angular speed of the Earth around the Sun is approximately $$\omega=\dfrac{2\pi\ \text{rad}}{1\ \text{year}}\times\dfrac{1\ \text{year}}{365\ \text{day}}\times\dfrac{1\ \text{day}}{24\ \text{hour}}\times\dfrac{1\ \text{hour}}{60\ \text{min}}\times\dfrac{1\ \text{min}}{60\ \text{s}}=1.99\times10^{-7}\ \text{rad/s}$$ Putting this together gives the angular momentum as $$L=(6.0\times10^{24})(1.8\times10^{11})^2(1.99\times10^{-7})=3.87\times10^{40}\ \text{kg}\cdot\text{m}^2/\text{s}^2$$ Hence, the answer is $10^40$. The angular momentum associated with the Earth’s rotation about its axis is several orders of magnitude smaller than this number.
A small 2.0 kg block rests at the bottom of a bucket. The bucket is spun in a vertical circle of radius L by a rope. When the bucket reaches the highest point in its motion, it moves just fast enough for the block to remain in place in the bucket. When the bucket is at an angle $\theta=30^\circ$ from the vertical, as seen in the figure, what is the magnitude of the normal force (perpendicular to the surface) provided by the bucket onto the block? Note that the direction of the gravitational field is indicated in the diagram by $\vec{g}$ and that the block does not touch any sides of the bucket aside from the bottom of it.
$\textbf{(A) }$ 8.0 N$ \qquad$ $\textbf{(B) }$ 10.0 N$ \qquad$ $\textbf{(C) }$ 15.4 N$ \qquad$ $\textbf{(D) }$ 18.7 N$ \qquad$ $\textbf{(E) }$ 37.3 N
$\textbf{A}$
When the bucket reaches the highest point in its motion, it moves just fast enough for the block to remain in place in the bucket, so the normal force here is 0, thus the centripetal force is only provided by gravity $$mg=m\dfrac{v_1^2}{L}$$ When the bucket moves to angle $\theta$, the kinetic energy is conserved in the moving process, thus part the of potential energy is converted into kinetic energy, so $$\dfrac12mv_2^2=\dfrac12mv_1^2+mgL(1-\cos\theta)$$ At angle $\theta$ the centripetal force is provided by both gravity and normal force, so $$mg\cos\theta+N=m\dfrac{v_2^2}{L}$$ Finally we get $$N=3mg(1-\cos\theta)=8.0\ \text{N}$$
A point wave source travels along the x-axis at constant speed $v_s$. Stationary observers on the x-axis measure the wavelength of the waves that they receive. The ratio of the wavelength measured at a location behind the source to the wavelength measured at a location in front of the source is 1.50. If the wave speed is $v$, what is the source speed $v_s$?
$\textbf{(A) }$ $\dfrac15v \qquad$ $\textbf{(B) }$ $\dfrac14v \qquad$ $\textbf{(C) }$ $\dfrac13v \qquad$ $\textbf{(D) }$ $\dfrac12v \qquad$ $\textbf{(E) }$ $\dfrac23v$
$\textbf{A}$
Since the ratio of the wavelengths is 3:2, this makes the ratio of the corresponding frequencies to be 2:3 as $v=\lambda f$. Using the Doppler Shift expression, we have $f_0=f_s\left(\dfrac{v}{v\pm v_s}\right)$ where the sign in the denominator relates to whether the source is approaching or receding from the observer. For our ratio, we have $$\dfrac{f_b}{f_f}=\dfrac23=f_s\left(\dfrac{v}{v+v_s}\right)/f_s\left(\dfrac{v}{v-v_s}\right)=\dfrac{v-v_s}{v+v_s}$$ Solving for the source speed gives $v_s=\dfrac15v$.
A small object of mass M is released from rest at the top of a frictionless incline. The incline has a mass M and makes an angle $\theta$ with the horizontal. The incline remains at rest on a table as the small object slides. During the slide, what is the magnitude of the normal force from the table on the incline?
$\textbf{(A) }$ $2Mg(1-\tan\theta) \qquad\newline$ $\textbf{(B) }$ $2Mg(1-\sin\theta) \qquad\newline$ $\textbf{(C) }$ $2Mg \qquad\newline$ $\textbf{(D) }$ $Mg(2-\sin^2\theta) \qquad\newline$ $\textbf{(E) }$ $Mg(2-\sin\theta)$
$\textbf{D}$
Taking the two objects as a system, the force analysis in the vertical direction can be wrote as $$2Mg-N=Ma\sin\theta$$ where the acceleration $a=g\sin\theta$, so $$N=2Mg-Ma\sin\theta=Mg(2-\sin^2\theta)$$
Pure red light shines through a diffraction grating of 1200 lines/cm and produces an interference pattern on a screen a distance 5.0 m away. Which one of the following choices best represents the distance between the first and third principal bright spots on the screen?
$\textbf{(A) }$ 0.02 m$ \qquad$ $\textbf{(B) }$ 0.10 m$ \qquad$ $\textbf{(C) }$ 1.0 m$ \qquad$ $\textbf{(D) }$ 2.0 m$ \qquad$ $\textbf{(E) }$ 4.0 m
$\textbf{C}$
The diffraction grating formula $d\sin\theta=n\lambda$, where $d=1/1200\ \text{cm}$, and the wavelength for red light is about 650 - 750 nm (we take $\lambda=700\ \text{nm}$). The angle $\theta$ is pretty small, so $\sin\theta\approx\theta$, then we have $d\theta=n\lambda$, or $d\Delta\theta=\Delta n\lambda$. For the first and third bright spots $\Delta n=2$, the distance $\Delta x=L\Delta\theta=2\lambda L/d=0.84\ \text{m}$.
A long uniform rod, pivoted at one end affixed to the ground, is in static equilibrium. A horizontal rope acts 75$\%$ of the way up the rod while a vertical rope acts at the far end. The magnitude of each rope's force is the same as the gravitational force acting on the rod. What is the value of the angle $\theta$ in the figure?
$\textbf{(A) }$ 18.4$^\circ \qquad$ $\textbf{(B) }$ 21.8$^\circ \qquad$ $\textbf{(C) }$ 33.7$^\circ \qquad$ $\textbf{(D) }$ 36.9$^\circ \qquad$ $\textbf{(E) }$ 45.0$^\circ$
$\textbf{C}$
The torque from the pivot at the ground is $$mg\cdot\dfrac12l\cos\theta+T_1\cdot\dfrac34l\sin\theta-T_2\cdot l\cos\theta=0$$ As $T_1=T_2=mg$, we get $\theta=\arctan\dfrac23=33.7^\circ$.
A semi-circle has radius R, total charge Q, and a charge per unit length given as $\lambda=\lambda_0\cos\theta$ with $\theta$ defined in the figure and $\lambda_0$ a positive constant. Which one of the following choices gives the electric field strength at the point labeled P?
$\textbf{(A) }$ $0 \qquad$ $\textbf{(B) }$ $\dfrac{2k_e\lambda_0}{R^2} \qquad$ $\textbf{(C) }$ $\dfrac{k_eQ\lambda_0}{R} \qquad$ $\textbf{(D) }$ $\dfrac{2k_eQ\lambda_0}{\pi R^2} \qquad$ $\textbf{(E) }$ $\dfrac{k_e\lambda_0\pi}{2R}$
$\textbf{E}$
For the charge at angle $\theta$ and $-\theta$, the horizontal component of their electric fields at point P canceled out, so only the vertical component remained. The charge at angle $\theta$ is $$\text{d}q=\lambda\cdot R\text{d}\theta=\lambda_0R\cos\theta\text{d}\theta$$ the horizontal component of its electric field is $$\text{d}E=k_e\dfrac{\text{d}q}{R^2}\cos\theta=\dfrac{k_e\lambda_0}{R}\cos^2\theta\text{d}\theta$$ so the electric field $$E=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{k_e\lambda_0}{R}\cos^2\theta\text{d}\theta=\dfrac{k_e\lambda_0}{R}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos2\theta+1}{2}\text{d}\theta=\dfrac{k_e\lambda_0}{2R}\left(\dfrac12\sin2\theta+\theta\right)\biggr\rvert_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\dfrac{k_e\lambda_0\pi}{2R}$$
A ball is launched vertically upward. The vertical position of the ball 0.70 s after release is the same as its vertical position 4.10 s after its release. At what time is the object first at a vertical position that is one-half of the maximum height it obtains during its flight?
$\textbf{(A) }$ 0.35 s$ \qquad$ $\textbf{(B) }$ 0.70 s$ \qquad$ $\textbf{(C) }$ 0.85 s$ \qquad$ $\textbf{(D) }$ 1.20 s$ \qquad$ $\textbf{(E) }$ 1.70 s
$\textbf{B}$
By the symmetry, the ball reach the maximum height at $t_3=\dfrac12(t_1+t_2)=2.4\ \text{s}$, the maximum height is $H=\dfrac12gt_3^2$. To fall half the maximum height $\dfrac12gt^2=\dfrac12H=\dfrac14gt_3^2$, we get $t=\dfrac{t_3}{\sqrt2}=1.70\ \text{s}$, so the time the object first reach half the maximum height is $t_3-t=0.70\ \text{s}$.
Which one of the following magnifications cannot be produced using a single converging lens?
$\textbf{(A) }$ $\dfrac12 \qquad$ $\textbf{(B) }$ $2 \qquad$ $\textbf{(C) }$ $-\dfrac12 \qquad$ $\textbf{(D) }$ $-1 \qquad$ $\textbf{(E) }$ $-2$
$\textbf{A}$
The magnification $M=-\dfrac{d_i}{d_o}$, where $d_o$ and $f$ are positive, and $d_i$ can be found in $\dfrac1{d_o}+\dfrac1{d_i}=\dfrac1f$. So $d_i=\dfrac{d_of}{d_o-f}$, $M=\dfrac{f}{f-d_o}$.$\newline$
For choice A, $M=\dfrac{f}{f-d_o}=\dfrac12$, we get $f=-d_o$, which is impossible because both $f$ and $d_o$ are positive.
Two copper spheres, X and Y, with different radii ($R_X>R_Y$) and equal excess charge +Q, are placed apart from each other. A scientist using insulating gloves connects the spheres with a copper wire. Which one of the following choices best describes what happens after the connection is made?
$\textbf{(A) }$ There is a net movement of electrons from sphere Y to sphere X until the electric field just outside the surface of each sphere has the same magnitude.$ \qquad\newline$
$\textbf{(B) }$ There is a net movement of electrons from sphere Y to sphere X until the electric potential just outside the surface of each sphere has the same magnitude.$ \qquad\newline$
$\textbf{(C) }$ There is a net movement of electrons from sphere X to sphere Y until the electric field just outside the surface of each sphere has the same magnitude.$ \qquad\newline$
$\textbf{(D) }$ There is a net movement of electrons from sphere X to sphere Y until the electric potential just outside the surface of each sphere has the same magnitude.$ \qquad\newline$
$\textbf{(E) }$ There is no net movement of electrons from one sphere to the other because the spheres already have equal charge.
$\textbf{D}$
With different radii, the potentials $V=k\dfrac{Q}{R}$ are not the same, so the current flows from high potential (Y) to low potential (X), and the electrons move in the opposite direction.
Once the circuit shown reaches equilibrium, what is the magnitude of the potential difference across the capacitor?
$\textbf{(A) }$ $0 \qquad$ $\textbf{(B) }$ $\dfrac19\mathcal{E} \qquad$ $\textbf{(C) }$ $\dfrac29\mathcal{E} \qquad$ $\textbf{(D) }$ $\dfrac59\mathcal{E} \qquad$ $\textbf{(E) }$ $\dfrac23\mathcal{E}$
$\textbf{D}$
The current $I=\dfrac{\mathcal{E}-\frac13\mathcal{E}}{R+2R}=\dfrac{2\mathcal{E}}{9R}$ flows anticlockwise, so the potential difference across the capacitor is $\Delta V=IR+\dfrac13\mathcal{E}=\dfrac59\mathcal{E}$.
In the binary star system shown, the two stars follow circular orbits about the system's center of mass. The stars are separated by a distance D that is large compared to their size and are subject only to their mutual gravitational attraction. The orbital period of the star of mass $M$ is $T$. Which one of the following choices represents the total mass of the binary star system?
$\textbf{(A) }$ $\dfrac{\pi^2D^3}{5GT^2} \qquad$ $\textbf{(B) }$ $\dfrac{4\pi^2D^3}{5GT^2} \qquad$ $\textbf{(C) }$ $\dfrac{4\pi^2D^3}{GT^2} \qquad$ $\textbf{(D) }$ $\dfrac{16\pi^2D^3}{5GT^2} \qquad$ $\textbf{(E) }$ $\dfrac{25\pi^2D^3}{4GT^2}$
$\textbf{C}$
For two stars of mass $m_1$ and $m_2$ $$G\dfrac{m_1m_2}{D^2}=m_1\left(\dfrac{2\pi}{T}\right)^2r$$ the position of the center of mass $$r=\dfrac{m_2}{m_1+m_2}D$$ so we get $$m_1+m_2=\dfrac{4\pi^2D^3}{GT^2}$$
Two identical particles travel to the right. The particle traveling at $2.40\times10^8\ \text{m/s}$ collides with and sticks to the other particle traveling at $1.80\times10^8\ \text{m/s}$. Which one of the following choices best represents the speed of the resulting object after collision?
$\textbf{(A) }$ $2.20\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(B) }$ $2.14\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(C) }$ $2.10\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(D) }$ $2.06\times10^8\ \text{m/s} \qquad\newline$ $\textbf{(E) }$ $2.00\times10^8\ \text{m/s}$
$\textbf{B}$
According to the conservation of linear momentum $$\dfrac{m_0}{\sqrt{1-v_1^2/c^2}}v_1+\dfrac{m_0}{\sqrt{1-v_2^2/c^2}}v_2=\dfrac{2m_0}{\sqrt{1-v^2/c^2}}v$$ $v_1=0.8c$, $v_2=0.6c$, so $$v=\dfrac{25}{\sqrt{24^2+25^2}}c=2.16\times10^8\ \text{m/s}$$
A monatomic ideal gas undergoes the reversible cyclic process (ABCA) shown in the PV diagram. Process A$\rightarrow$B is adiabatic. What is the efficiency of this engine?
$\textbf{(A) }$ 0.15$ \qquad$ $\textbf{(B) }$ 0.22$ \qquad$ $\textbf{(C) }$ 0.33$ \qquad$ $\textbf{(D) }$ 0.47$ \qquad$ $\textbf{(E) }$ 0.67
$\textbf{B}$
B$\rightarrow$C is isobaric, so the heat released $$Q_1=nc_p(T_B-T_C)$$ C$\rightarrow$A is isochoric, so the heat absorbed $$Q_2=nc_v(T_A-T_C)$$ and $$T_A=3T_C$$ A$\rightarrow$B is adiabatic, so no heat absorbed or released, and $$T_A^\gamma p_A^{1-\gamma}=T_B^\gamma p_B^{1-\gamma}$$ for $p_A=3p_B$, it can be rewrote as $$T_B=3^{\frac{1-\gamma}{\gamma}}T_A$$ for monatomic gas $$\dfrac{c_p}{c_v}=\gamma=\dfrac{i+2}{i}=\dfrac53$$ the efficiency $$\eta=1-\dfrac{Q_1}{Q_2}=1-\dfrac{nc_p(T_B-T_C)}{nc_v(T_A-T_C)}=1-\gamma\dfrac{3^{\frac{1-\gamma}{\gamma}}-1}{1-\frac13}=0.223$$