June 2023 Physics Regents — Answer Key & Worked Solutions

Every question from the June 2023 Physics Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then see how it's scored.

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These are real questions from the June 2023 Physics New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

Which pair of quantities are both scalar quantities?

(A) speed and mass ✓
(B) speed and momentum
(C) momentum and displacement
(D) mass and displacement

Worked solution

$\text{Speed and mass have magnitude only, no direction.}$

$\text{Momentum and displacement are vectors.}$

Answer: speed and mass

Scalars have magnitude only; speed and mass are both scalars, while momentum and displacement are vectors.

Question 2 · 2 pts

In an attempt to get ketchup out of a bottle, a student takes off the cap, turns the bottle upside down, accelerates it downward, and then suddenly stops. Ketchup is released from the open bottle because the ketchup doesn't stop moving when the bottle does. The ketchup leaving the bottle illustrates

(A) inertia ✓
(B) resistivity
(C) resonance
(D) mass being converted to energy

Worked solution

$\text{An object in motion tends to stay in motion (Newton's 1st Law).}$

$\text{The ketchup keeps moving downward when the bottle suddenly stops.}$

Answer: inertia

This is Newton's first law: the ketchup's inertia keeps it moving even after the bottle stops.

Question 3 · 2 pts

The same net force is applied to object A and object B. The mass of B is three times greater than the mass of A. Compared to the acceleration of A, the acceleration of B is

(A) the same
(B) one-third as great ✓
(C) three times as great
(D) one-ninth as great

Worked solution

$a=\frac{F_{net}}{m}$

$F_{net} \text{ is constant, } m_B=3m_A$

$a_B=\frac{F_{net}}{3m_A}=\frac{1}{3}a_A$

Answer: one-third as great

Since a = F/m and F is the same for both, tripling the mass reduces acceleration to one-third.

Question 4 · 2 pts

What is the mass equivalent of $3.37\times10^{-19}$ joule?

(A) $1.26\times10^{-54}$ kg
(B) $3.74\times10^{-36}$ kg ✓
(C) $1.12\times10^{-27}$ kg
(D) $5.08\times10^{14}$ kg

Worked solution

$E=mc^2 \Rightarrow m=\frac{E}{c^2}$

$m=\frac{3.37\times10^{-19}\ \text{J}}{(3.00\times10^8\ \text{m/s})^2}$

$m\approx 3.74\times10^{-36}\ \text{kg}$

Answer: $3.74\times10^{-36}$ kg

Using mass-energy equivalence, divide the energy by the speed of light squared to find the equivalent mass.

Question 5 · 2 pts

Which object is in equilibrium?

(A) Earth orbiting the Sun
(B) a thrown baseball at its highest point above the ground
(C) a car moving at a constant speed in a straight line ✓
(D) a bicycle skidding to a stop in a straight line

Worked solution

$\text{Equilibrium: } F_{net}=0, a=0$

$\text{Constant velocity in a straight line means zero net force.}$

Answer: a car moving at a constant speed in a straight line

An object in equilibrium has zero net force and zero acceleration, which describes constant velocity in a straight line, not circular motion, a point of zero velocity with gravity acting, or a decelerating object.

Question 6 · 2 pts

A race car travels around a flat circular track at constant speed. The net force on the car acts

(A) perpendicular to the car's velocity and toward the center of the circle ✓
(B) perpendicular to the car's velocity and away from the center of the circle
(C) parallel to the car's velocity and in the same direction as the velocity
(D) parallel to the car's velocity and in the opposite direction as the velocity

Worked solution

$\text{Centripetal force points toward the center of circular motion.}$

$\text{It is always perpendicular to instantaneous velocity.}$

Answer: perpendicular to the car's velocity and toward the center of the circle

For uniform circular motion, the net (centripetal) force is always directed toward the center, perpendicular to the velocity.

Question 7 · 2 pts

An object is traveling in a horizontal, circular path at a constant speed. If the radius of the path were doubled while the speed remained constant, the centripetal acceleration would be

(A) quartered
(B) doubled
(C) halved ✓
(D) quadrupled

Worked solution

$a_c=\frac{v^2}{r}$

$\text{Doubling } r \text{ with } v \text{ constant halves } a_c$

Answer: halved

Centripetal acceleration is inversely proportional to radius when speed is constant, so doubling the radius halves the acceleration.

Question 8 · 2 pts

A 600.-newton student pushes on a vertical wall for 20.0 seconds with a constant force having a magnitude of 100. newtons. What is the magnitude of the force that the wall exerts on the student?

(A) 0.00 N
(B) 5.00 N
(C) 100. N ✓
(D) 600. N

Worked solution

$\text{By Newton's 3rd Law, } F_{wall\ on\ student}=-F_{student\ on\ wall}$

$|F_{wall\ on\ student}|=100.\ \text{N}$

Answer: 100. N

Newton's third law states the wall exerts an equal and opposite reaction force of 100. N on the student, regardless of the student's weight or the time pushed.

Question 9 · 2 pts

An unbalanced force of 20. newtons is applied to a mass of $1.0\times10^3$ kilograms. After 10. seconds, the momentum of the mass will have changed by

(A) $2.0\times10^2\ \text{kg}\cdot\text{m/s}$ ✓
(B) $2.0\times10^3\ \text{kg}\cdot\text{m/s}$
(C) $1.0\times10^4\ \text{kg}\cdot\text{m/s}$
(D) $2.0\times10^4\ \text{kg}\cdot\text{m/s}$

Worked solution

$\Delta p = F\Delta t$

$\Delta p = (20.\ \text{N})(10.\ \text{s}) = 200\ \text{kg}\cdot\text{m/s}$

$\Delta p = 2.0\times10^2\ \text{kg}\cdot\text{m/s}$

Answer: $2.0\times10^2\ \text{kg}\cdot\text{m/s}$

The impulse-momentum theorem gives the change in momentum as the product of force and time, independent of the object's mass.

Question 10 · 2 pts

To lift a heavy block off the floor, a student pulls with force F on a rope that passes over a pulley, as shown in the diagram below. Which pair of vectors represents the perpendicular components of the force the student exerts on the rope?

(A) Vector diagram (1): a downward arrow and a shorter downward arrow
(B) Vector diagram (2): a leftward arrow and an upward-then-rightward arrow forming a right angle ✓
(C) Vector diagram (3): a vertical upward arrow and a horizontal rightward arrow forming an L shape at origin
(D) Vector diagram (4): two arrows, one upward and one leftward, both from the same point

Worked solution

$\text{Resolve force } F \text{ (directed up and to the right along the rope) into perpendicular components.}$

$\text{The correct pair shows one horizontal and one vertical component consistent with the rope's angle.}$

Answer: Vector diagram (2)

The perpendicular components of F must be one purely horizontal and one purely vertical vector whose vector sum reconstructs F along the rope's actual direction; this requires seeing the figure to confirm orientation.

Question 11 · 2 pts

Which graph best represents the relationship between the mass of an object and its distance from Earth's surface?

(A) Graph (1): a horizontal line (mass constant with distance) ✓
(B) Graph (2): a straight line increasing from the origin
(C) Graph (3): an exponentially increasing curve
(D) Graph (4): a straight line decreasing

Worked solution

$\text{Mass is an intrinsic property of an object.}$

$\text{Mass does not change with distance from Earth's surface (weight does, mass does not).}$

Answer: Graph (1): horizontal line

Mass is independent of location, so a graph of mass versus distance from Earth's surface is a horizontal line.

Question 12 · 2 pts

A compass is placed near a strong bar magnet as represented in the diagram below (compass on the left, bar magnet to its right with N pole on the left side and S pole on the right side). Which diagram best represents the direction of the compass needle?

(A) Diamond needle with N (dot, filled diamond) on top pointing up, S on bottom
(B) Diamond needle with S on top, N on bottom
(C) Horizontal diamond needle with N pointing left, S pointing right ✓
(D) Horizontal diamond needle with S pointing left, N pointing right

Worked solution

$\text{The compass's N pole is attracted to the magnet's S pole.}$

$\text{Near the side of the bar magnet, the external field runs roughly parallel to the magnet's axis.}$

$\text{The needle aligns horizontally with its N pole pointing toward the magnet's N end (leftward).}$

Answer: Horizontal diamond needle with N pointing left, S pointing right

A compass needle aligns with the local magnetic field of the bar magnet; near the side of the magnet the field lines run roughly parallel to the magnet's axis, so the needle orients horizontally with its N pole pointing toward the magnet's N pole.

Question 13 · 2 pts

As Earth orbits the Sun in its elliptical orbit, the gravitational force between the Sun and Earth is

(A) always attractive ✓
(B) attractive as the Sun and Earth get closer together and repulsive as the Sun and Earth get farther apart
(C) repulsive as the Sun and Earth get closer together and attractive as the Sun and Earth get farther apart
(D) always repulsive

Worked solution

$\text{Gravitational force is always attractive between masses, regardless of distance.}$

Answer: always attractive

Gravity is always an attractive force between any two masses, so the Sun-Earth gravitational force remains attractive throughout the orbit.

Question 14 · 2 pts

Which unit is used to measure the work per unit charge required to move a charge in an electric field?

(A) ampere
(B) coulomb
(C) volt ✓
(D) watt

Worked solution

$V=\frac{W}{q}$

$\text{Units of } V = \text{J/C} = \text{volt}$

Answer: volt

Electric potential difference (voltage) is defined as work per unit charge, and its SI unit is the volt.

Question 15 · 2 pts

The diagram below shows a negatively charged sphere and a point, P, located within the electric field produced by the charge on the sphere. Points A, C, D, and B are arranged around P (A above, C below, D to the left, B to the right of P, with the sphere farther to the left of D). The direction of the electric field at point P is toward point

(A) A
(B) B
(C) C
(D) D ✓

Worked solution

$\text{Electric field lines point toward a negative charge (direction a positive test charge would move).}$

$\text{Since the sphere is negative, the field at P points toward the sphere, i.e., toward D.}$

Answer: D

For a negative charge, electric field vectors point toward the charge; since the sphere lies in the direction of point D from P, the field at P points toward D.

Question 16 · 2 pts

Two point charges, $q_1$ and $q_2$, are initially a distance, d, apart. Which change will cause the greatest increase in the electrostatic force between the two point charges, $q_1$ and $q_2$?

(A) double $q_1$
(B) halve $q_2$
(C) double $d$
(D) halve $d$ ✓

Worked solution

$F=k\frac{q_1q_2}{d^2}$

$\text{Doubling } q_1 \Rightarrow F \text{ doubles}$

$\text{Halving } d \Rightarrow F \text{ quadruples (since } d^2 \text{ decreases by factor of 4)}$

Answer: halve $d$

Because electrostatic force depends on the inverse square of distance, halving the distance quadruples the force, a greater increase than doubling a charge (which only doubles the force).

Question 17 · 2 pts

The diagram below represents two identical conducting spheres, A and B, each showing a mixture of + and - charge symbols distributed within them (sphere A shows charges scattered somewhat randomly; sphere B shows + charges concentrated on the left side and - charges concentrated on the right side near where it faces sphere A). Which statement could be the correct explanation for the charge distribution on the spheres?

(A) A small positively charged object is located between sphere A and sphere B.
(B) A small negatively charged object is located between sphere A and sphere B. ✓
(C) A small positively charged object is located to the right of sphere B.
(D) A small negatively charged object is located to the right of sphere B.

Worked solution

$\text{Induction: a nearby charge repels like charges and attracts opposite charges within a conductor.}$

$\text{A negative charge between A and B repels negative charges in each sphere away from it and attracts positive charges toward it, producing the charge separation shown at the facing surfaces.}$

Answer: A small negatively charged object is located between sphere A and sphere B.

The charge redistribution pattern between the two spheres is consistent with electrostatic induction caused by a small negative charge placed between them, repelling electrons in each sphere to the far sides.

Question 18 · 2 pts

The diagram below represents a circuit containing a battery, two operating lamps, A and B, and four closed switches, $S_1$, $S_2$, $S_3$, and $S_4$. Lamp A is in series with switch $S_2$ in one branch, and lamp B is in series with switch $S_3$ in a parallel branch; switch $S_4$ connects across the top joining both branches, and switch $S_1$ is in the main line from the battery. Which switch, when opened, causes both lamps to turn off?

(A) $S_1$ ✓
(B) $S_2$
(C) $S_3$
(D) $S_4$

Worked solution

$\text{Lamps A and B are in parallel branches, each individually controlled by } S_2 \text{ and } S_3.$

$S_4 \text{ is a cross-connection that does not solely control both lamps.}$

$S_1 \text{ is in the main line from the battery, so opening it cuts off current to the entire circuit, turning off both lamps.}$

Answer: $S_1$

Since S1 is located in the main circuit line before the parallel branches split, opening it disconnects the battery from both lamps simultaneously.

Question 19 · 2 pts

What is the resistance of a 100.-watt bulb when operating a potential difference of 120. volts?

(A) $1.20\ \Omega$
(B) $100.\ \Omega$
(C) $120.\ \Omega$
(D) $144\ \Omega$ ✓

Worked solution

$P=\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P}$

$R=\frac{(120.\ \text{V})^2}{100.\ \text{W}}=\frac{14400}{100}$

$R=144\ \Omega$

Answer: $144\ \Omega$

Using the power formula P = V^2/R and solving for R gives 144 ohms.

Question 20 · 2 pts

The interaction that is most responsible for binding three quarks together in a proton is the

(A) strong force ✓
(B) electromagnetic force
(C) weak force
(D) gravitational force

Worked solution

$\text{Quarks are bound together within nucleons by the strong nuclear force.}$

Answer: strong force

The strong force (mediated by gluons) is responsible for binding quarks together within protons and neutrons.

Question 21 · 2 pts

One example of a force doing work is the force exerted by

(A) Earth on a high diver falling toward a pool from a platform above ✓
(B) a hook on an engine held stationary above a car
(C) a frictionless horizontal air hockey table on a puck moving across the table at a constant velocity
(D) a weightlifter on a barbell he holds motionless over his head

Worked solution

$W=Fd\cos\theta$

$\text{Work requires displacement with a component along the force.}$

$\text{Gravity acting on the falling diver has displacement parallel to the force (downward), so work is done.}$

$\text{The other cases involve zero displacement or force perpendicular to displacement, so no work is done.}$

Answer: Earth on a high diver falling toward a pool from a platform above

Work is done only when a force causes displacement along its direction; gravity does work on the falling diver, whereas the other examples involve either no displacement or a force perpendicular to the motion.

Question 22 · 2 pts

A ball falls freely from a rooftop to the street below. The ball starts from rest with 20. joules of gravitational potential energy with respect to the street. The total mechanical energy of the ball just before it hits the street is

(A) 0 J
(B) 5.0 J
(C) 10. J
(D) 20. J ✓

Worked solution

$\text{By conservation of mechanical energy (no air resistance):}$

$ME_{initial}=PE_i+KE_i=20.\ \text{J}+0=20.\ \text{J}$

$ME_{final}=ME_{initial}=20.\ \text{J}$

Answer: 20. J

In free fall with no friction, total mechanical energy is conserved, so the total mechanical energy just before impact equals the initial 20. J of potential energy.

Question 23 · 2 pts

Which statement describes the gravitational potential energy (PE), kinetic energy (KE), and internal (thermal) energy (Q), of a wooden crate as it is pushed across a level classroom floor at constant speed?

(A) The PE decreases, KE remains the same, and Q decreases.
(B) The PE increases, KE increases, and Q decreases.
(C) The PE remains the same, KE decreases, and Q increases.
(D) The PE remains the same, KE remains the same, and Q increases. ✓

Worked solution

$\text{Level floor} \Rightarrow \Delta PE = 0$ height does not change

$\text{Constant speed} \Rightarrow \Delta KE = 0$

$\text{Friction converts input work into heat} \Rightarrow Q \text{ increases}$

Answer: The PE remains the same, KE remains the same, and Q increases.

On a level floor at constant speed there is no change in height or speed, so PE and KE stay constant; the work done against friction is converted into internal thermal energy.

Question 24 · 2 pts

The oscillation of electrons up and down a metal antenna produces waves. These waves are

(A) mechanical and longitudinal
(B) mechanical and transverse
(C) electromagnetic and transverse ✓
(D) electromagnetic and longitudinal

Worked solution

$\text{Accelerating charges radiate electromagnetic waves}$

$\text{Electromagnetic waves are always transverse}$

Answer: electromagnetic and transverse

Accelerating electrons in an antenna radiate electromagnetic radiation, and all electromagnetic waves are transverse.

Question 25 · 2 pts

The diagram below shows an incident light ray reflecting from a plane mirror. The incident ray and the reflected ray meet at the mirror surface, forming an angle of 84° between the incident ray and the reflected ray. What is the angle of reflection?

(A) 96°
(B) 84°
(C) 48°
(D) 42° ✓

Worked solution

$\theta_i = \theta_r \text{ (law of reflection, measured from the normal)}$

$\theta_i + \theta_r = 84^\circ$

$\theta_r = \dfrac{84^\circ}{2} = 42^\circ$

Answer: 42°

The 84° angle shown is the angle between the incident and reflected rays; since these angles are equal and measured from the normal, each equals half of 84°.

Question 26 · 2 pts

A characteristic common to both sound waves and x rays is that they both

(A) travel fastest in a vacuum
(B) cause particles to vibrate in a direction parallel to the wave's direction of motion
(C) transmit energy without transmitting matter ✓
(D) are mechanical

Worked solution

$\text{Sound: mechanical, longitudinal}$

$\text{X-rays: electromagnetic, transverse}$

$\text{Shared property: both transport energy without net transport of matter}$

Answer: transmit energy without transmitting matter

Sound waves are mechanical and longitudinal while x rays are electromagnetic and transverse, so the only property they share from the choices is that both transmit energy without transmitting matter.

Question 27 · 2 pts

The diagram below represents a wave moving to the right through a rope. Point P is marked on the rope. Point P in the rope is moving toward the

(A) top of the page ✓
(B) bottom of the page
(C) right
(D) left

Worked solution

$\text{Transverse wave particles move perpendicular to wave travel direction}$

$\text{As the waveform shifts right, the shape currently to the left of P will arrive at P next}$

$\text{That upcoming shape is higher, so P is moving upward (toward top of page)}$

Answer: top of the page

For a transverse wave moving right, a point's future position matches the waveform currently just to its left; since that portion of the curve at P's location is rising, P is momentarily moving toward the top of the page.

Question 28 · 2 pts

What is the wavelength of a 300.-hertz sound wave in air at STP?

(A) 0.906 m
(B) 1.10 m ✓
(C) $3.00 \times 10^2$ m
(D) $1.00 \times 10^6$ m

Worked solution

$v_{sound,STP} = 331\ \text{m/s (reference table)}$

$v = f\lambda \Rightarrow \lambda = \frac{v}{f}$

$\lambda = \frac{331}{300.} \approx 1.10\ \text{m}$

Answer: 1.10 m

Using the speed of sound in air at STP (331 m/s) and the wave equation v = fλ, dividing gives a wavelength of about 1.10 m.

Question 29 · 2 pts

In the diagram below, a remote control is aimed at a television. When the signal from the remote reaches the sensor on the television, the signal will most likely be

(A) neither reflected nor absorbed
(B) partially absorbed and partially reflected ✓
(C) completely reflected
(D) completely absorbed

Worked solution

$\text{Real sensor surfaces are neither perfect absorbers nor perfect reflectors}$

$\text{Most of the infrared signal is absorbed to be detected, but some is reflected}$

Answer: partially absorbed and partially reflected

No real surface completely absorbs or completely reflects all incident light, so the infrared signal striking the sensor is partially absorbed (to be detected) and partially reflected.

Question 30 · 2 pts

Compared to waves of blue light traveling in a vacuum, waves of red light traveling in a vacuum have

(A) a lower frequency and a lower speed
(B) a lower frequency and the same speed ✓
(C) the same frequency and a lower speed
(D) the same frequency and the same speed

Worked solution

$\text{All light travels at } c = 3.00\times10^8\ \text{m/s in a vacuum, regardless of color}$

$\text{Red light has a longer wavelength and thus lower frequency than blue light}$

Answer: a lower frequency and the same speed

In a vacuum all electromagnetic waves travel at the same speed c, but red light has a lower frequency than blue light because it has a longer wavelength.

Question 31 · 2 pts

Earthquakes often cause buildings between twelve and forty stories high to vibrate at an amplitude high enough to be destructive. Buildings are often designed to absorb this vibrational energy that might cause them to vibrate at their natural frequency. This tendency for an earthquake to cause a building to vibrate at a large amplitude is an example of

(A) resonance ✓
(B) the Doppler effect
(C) refraction
(D) diffraction

Worked solution

$\text{Driving frequency} \approx \text{natural frequency} \Rightarrow \text{large amplitude response}$

Answer: resonance

When a periodic driving force (the earthquake) matches an object's natural frequency, the resulting large-amplitude vibration is called resonance.

Question 32 · 2 pts

During the radioactive decay of a uranium-238 atom, a thorium-234 atom and an alpha particle are produced. During this process, there is conservation of

(A) charge, only
(B) mass-energy, only
(C) both charge and mass-energy ✓
(D) neither charge nor mass-energy

Worked solution

$Z_U = Z_{Th} + Z_{\alpha} \Rightarrow 92 = 90 + 2$ charge conserved

$\text{Total mass-energy before} = \text{Total mass-energy after (including released energy)}$

Answer: both charge and mass-energy

In all nuclear reactions, including alpha decay, both total charge and total mass-energy are conserved.

Question 33 · 2 pts

An antiproton has a charge of

(A) +1e
(B) $+\frac{2}{3}e$
(C) 0
(D) -1e ✓

Worked solution

$\text{Antiproton = antiparticle of the proton}$

$\text{Proton charge} = +1e \Rightarrow \text{Antiproton charge} = -1e$

Answer: -1e

An antiparticle has the same mass as its corresponding particle but opposite charge, so since the proton has charge +1e, the antiproton has charge -1e.

Question 34 · 2 pts

Two pulses approach each other in the same medium. Which pair of pulses will result in the largest magnitude displacement of the medium as the pulses pass through each other?

(A) Pulse pair 1 (as shown in the diagram)
(B) Pulse pair 2 (as shown in the diagram) ✓
(C) Pulse pair 3 (as shown in the diagram)
(D) Pulse pair 4 (as shown in the diagram)

Worked solution

$\text{Superposition principle: displacements add algebraically when pulses overlap}$

$\text{Largest resultant magnitude occurs when two pulses of the largest same-sign amplitudes overlap}$

Answer: Pulse pair 2 (as shown in the diagram)

By the principle of superposition, the greatest combined displacement occurs for the pair whose two pulses have the largest amplitudes displaced in the same direction, so their heights add constructively; pulse pair 2 shows this combination.

Question 35 · 2 pts

Two waves of the same wavelength interfere to form a standing wave pattern as represented in the diagram below. Which point on the diagram represents a node?

(A) A
(B) B
(C) C
(D) D ✓

Worked solution

$\text{Nodes occur where the medium has zero displacement at all times}$

$\text{Antinodes occur at the points of maximum displacement (loop crests/troughs)}$

$\text{Point D lies where adjacent loops cross the central axis} \Rightarrow \text{node}$

Answer: D

A node is a point of the standing wave pattern that stays at zero displacement; point D is located where the standing wave pattern crosses the central axis between loops, identifying it as a node, while A and C sit at the crest/trough of a loop (antinodes).

Question 36 · 2 pts

A round dinner plate has a diameter closest to

(A) $2\times10^{-2}$ m
(B) $2\times10^{-1}$ m ✓
(C) $2\times10^{0}$ m
(D) $2\times10^{1}$ m

Worked solution

$\text{A typical dinner plate is about 20 cm across}$

$20\ \text{cm} = 0.20\ \text{m} = 2\times10^{-1}\ \text{m}$

Answer: $2\times10^{-1}$ m

A typical dinner plate is roughly 20 centimeters, or 2×10⁻¹ meters, in diameter.

Question 37 · 2 pts

Several springs are lying on frictionless tabletops with one end attached to a wall and a variable force F applied to the free end of each spring. The springs have different spring constants, k. The diagram below shows the setup for one of the springs. The elongation of the springs produced by force F depends

(A) directly on both F and k
(B) directly on F and inversely on k ✓
(C) inversely on F and directly on k
(D) inversely on both F and k

Worked solution

$F = kx \Rightarrow x = \frac{F}{k}$

Answer: directly on F and inversely on k

By Hooke's law, F = kx, so elongation x = F/k, which increases directly with F and decreases (inversely) as k increases.

Question 38 · 2 pts

The table below shows the weight of four athletes (A, B, C, and D) and the time required for each athlete to run from the base of a hill to its top. Which athlete ran up the hill with the greatest average power?

(A) A
(B) B
(C) C ✓
(D) D

Worked solution

$P = \frac{W}{t} = \frac{(\text{weight})(\text{height})}{t}\ \text{, height same for all athletes}$

$\frac{\text{weight}}{t}:\ A=\frac{600}{11.8}\approx50.8,\ B=\frac{650}{10.7}\approx60.7,\ C=\frac{700}{10.8}\approx64.8,\ D=\frac{750}{11.9}\approx63.0$

$\text{Athlete C has the largest weight/time ratio} \Rightarrow \text{greatest average power}$

Answer: C

Since all athletes climb the same hill (same height gain), average power is proportional to weight divided by time; computing this ratio for each athlete shows athlete C has the highest value.

Question 39 · 2 pts

A copper wire carries 2.8 amperes of current. The total amount of charge that passes a point in the wire in 1.3 milliseconds is

(A) $4.6\times10^{-4}$ C
(B) $3.6\times10^{-3}$ C ✓
(C) 3.6 C
(D) $2.2\times10^3$ C

Worked solution

$I = \frac{q}{t} \Rightarrow q = It$

$q = (2.8\ \text{A})(1.3\times10^{-3}\ \text{s}) = 3.64\times10^{-3}\ \text{C} \approx 3.6\times10^{-3}\ \text{C}$

Answer: $3.6\times10^{-3}$ C

Using q = It with the current in amperes and time converted to seconds gives a charge of about 3.6×10⁻³ C.

Question 40 · 2 pts

The diagram below represents four solid copper wire segments at 20°C with different lengths (L) and cross-sectional areas (A): Segment A has A = 2 cm² and L = 2 cm; segment B has A = 2 cm² and L = 1 cm; segment C has A = 1 cm² and L = 2 cm; segment D has A = 1 cm² and L = 1 cm. Which two segments have the same resistance?

(A) A and B
(B) B and D
(C) B and C
(D) A and D ✓

Worked solution

$R = \rho\frac{L}{A}\ \text{, same material (copper) so } \rho \text{ is constant}$

$A:\ \frac{L}{A}=\frac{2}{2}=1,\ B:\ \frac{1}{2}=0.5,\ C:\ \frac{2}{1}=2,\ D:\ \frac{1}{1}=1$

$\text{A and D both have } \frac{L}{A}=1 \Rightarrow \text{equal resistance}$

Answer: A and D

Since resistance is proportional to L/A for wires of the same material, computing L/A for each segment shows that segments A and D have the same ratio (1) and therefore the same resistance.

Question 41 · 2 pts

In substance X, a ray of light with a frequency of $5.09\times10^{14}$ hertz has a speed of $2.04\times10^{8}$ meters per second. Substance X could be

(A) diamond
(B) water
(C) zircon
(D) glycerol ✓

Worked solution

$n = \frac{c}{v} = \frac{3.00\times10^8}{2.04\times10^8} \approx 1.47$

$\text{Reference table value for glycerol: } n \approx 1.47$

Answer: glycerol

Dividing the speed of light in a vacuum by the given speed yields an index of refraction of about 1.47, which matches the reference-table value for glycerol.

Question 42 · 2 pts

Which could not be the charge on a particle?

(A) $3.2\times10^{-19}$ C
(B) $4.5\times10^{-19}$ C ✓
(C) $4.8\times10^{-19}$ C
(D) $6.4\times10^{-19}$ C

Worked solution

$\text{Charge must be an integer multiple of } e = 1.6\times10^{-19}\ \text{C}$

$\frac{3.2\times10^{-19}}{1.6\times10^{-19}}=2,\ \frac{4.5\times10^{-19}}{1.6\times10^{-19}}=2.81,\ \frac{4.8\times10^{-19}}{1.6\times10^{-19}}=3,\ \frac{6.4\times10^{-19}}{1.6\times10^{-19}}=4$

$\text{Only } 4.5\times10^{-19}\ \text{C is not an integer multiple of } e$

Answer: $4.5\times10^{-19}$ C

Electric charge is quantized in integer multiples of the elementary charge e; dividing each choice by e shows only 4.5×10⁻¹⁹ C is not a whole-number multiple, so it cannot be a particle's charge.

Question 43 · 2 pts

Which graph best represents the relationship between photon energy ($E_{photon}$) and wavelength?

(A) Graph 1: a steeply rising, concave-up curve of $E_{photon}$ vs. Wavelength
(B) Graph 2: a straight line rising linearly with Wavelength
(C) Graph 3: a horizontal, constant line of $E_{photon}$ vs. Wavelength
(D) Graph 4: a decreasing, concave-up curve (hyperbola-like) of $E_{photon}$ vs. Wavelength ✓

Worked solution

$E_{photon} = \frac{hc}{\lambda}$

$E_{photon} \propto \frac{1}{\lambda} \Rightarrow \text{inverse relationship, a decreasing hyperbola-shaped curve}$

Answer: Graph 4: a decreasing, concave-up curve (hyperbola-like) of $E_{photon}$ vs. Wavelength

Since photon energy equals hc/λ, energy is inversely proportional to wavelength, which is represented by a decreasing, concave-up curve like graph 4, not a straight line or constant value.

Question 44 · 2 pts

Light travels from air into another medium with an index of refraction of n. The light has a wavelength of $6.0\times10^{-7}$ meter in the new medium. Which expression represents the wavelength of this light in air?

(A) $n(6.0\times10^{-7}\text{m})$ ✓
(B) $\dfrac{6.0\times10^{-7}\text{m}}{n}$
(C) $\dfrac{n}{6.0\times10^{-7}\text{m}}$
(D) $(6.0\times10^{-7}\text{m}) - n$

Worked solution

$n = \frac{\lambda_{air}}{\lambda_{medium}}$

$\lambda_{air} = n \cdot \lambda_{medium} = n(6.0\times10^{-7}\ \text{m})$

Answer: $n(6.0\times10^{-7}\text{m})$

The index of refraction relates the wavelength in air to the wavelength in the medium by n = λ_air/λ_medium, so solving for λ_air gives n times the wavelength in the medium.

Question 45 · 2 pts

If 80. joules of electrical energy is dissipated by a 10.-ohm resistor in 2.0 seconds, the current in the resistor is

(A) 5.0 A
(B) 2.0 A ✓
(C) 8.0 A
(D) 4.0 A

Worked solution

$P = \frac{W}{t} = \frac{80.\ \text{J}}{2.0\ \text{s}} = 40\ \text{W}$ Find power dissipated

$P = I^2R \implies I = \sqrt{\frac{P}{R}} = \sqrt{\frac{40\ \text{W}}{10.\ \Omega}} = \sqrt{4.0} = 2.0\ \text{A}$

Answer: 2.0 A

Power is energy per unit time, and once power and resistance are known the current follows directly from P = I^2R.

Question 46 · 2 pts

Base your answer on the diagrams below, and on your knowledge of physics. The diagrams represent waves W, A, B, C, and D traveling through the same uniform medium. Which wave has a period that is twice that of wave W?

(A) A
(B) B
(C) C ✓
(D) D

Worked solution

$T \propto \frac{1}{f}$

$\text{Wave } W \text{ completes 2 full cycles over the shown distance; Wave } C \text{ completes only 1 full cycle over the same distance/time}$

$\implies T_C = 2T_W$

Answer: C

Since all waves travel through the same medium at the same speed, the wave that completes half as many full cycles over the same horizontal extent as wave W has twice the period of wave W.

Question 47 · 2 pts

Which wave is always 180 degrees out of phase with wave W?

(A) A ✓
(B) B
(C) C
(D) D

Worked solution

$\text{Two waves are 180}^{\circ}\text{ out of phase when one is at a crest wherever the other is at a trough, at every point along the medium}$

$\text{Wave } A \text{ has the same period/wavelength as } W \text{ but its crests align with } W\text{'s troughs everywhere}$

Answer: A

A constant 180 degree phase difference requires the same period and an inverted displacement pattern at every point, which matches wave A relative to wave W.

Question 48 · 2 pts

Which graph best represents the relationship between acceleration and time for a freely falling object as the object falls near the surface of Earth?

(A) Graph (1): Acceleration increasing linearly with time (straight line through origin, positive slope)
(B) Graph (2): Acceleration decreasing linearly with time (straight line, negative slope)
(C) Graph (3): Acceleration constant with time (horizontal line) ✓
(D) Graph (4): Acceleration increasing exponentially/curving upward with time

Worked solution

$a_{\text{freefall}} = g = 9.81\ \text{m/s}^2 = \text{constant, independent of } t$

Answer: Graph (3): a constant horizontal line

Near Earth's surface a freely falling object (neglecting air resistance) has a constant acceleration equal to g, so acceleration versus time is a horizontal line.

Question 49 · 2 pts

The diagram below represents the forces acting on a skydiver with his parachute. An arrow labeled "Force of air friction" points upward above the skydiver, and an arrow labeled "Gravitational force" points downward below the skydiver. The total mass of the skydiver with his parachute is 85.0 kilograms. If the magnitude of the gravitational force is 834 newtons, and the magnitude of the force of air friction is 652 newtons, the acceleration of the skydiver at the time shown is

(A) 2.14 m/s$^2$ up
(B) 2.14 m/s$^2$ down ✓
(C) 7.67 m/s$^2$ up
(D) 9.81 m/s$^2$ down

Worked solution

$F_{net} = F_{gravity} - F_{friction} = 834\ \text{N} - 652\ \text{N} = 182\ \text{N (down)}$

$a = \frac{F_{net}}{m} = \frac{182\ \text{N}}{85.0\ \text{kg}} = 2.14\ \text{m/s}^2\ \text{down}$

Answer: 2.14 m/s² down

The gravitational force is larger than the upward air friction force, so the net force and resulting acceleration both point downward.

Question 50 · 2 pts

A 15.0-gram air hockey puck sliding on a horizontal surface at a velocity of 7.00 meters per second north collides with a 15.0-gram air hockey puck traveling at a velocity of 8.00 meters per second south. The momentum of the system of pucks after the collision is

(A) 0.0150 kg•m/s north
(B) 0.0150 kg•m/s south ✓
(C) 0.225 kg•m/s north
(D) 0.225 kg•m/s south

Worked solution

$m = 15.0\ \text{g} = 0.0150\ \text{kg (each puck)}$

$p_1 = (0.0150\ \text{kg})(7.00\ \text{m/s north}) = 0.105\ \text{kg·m/s north}$

$p_2 = (0.0150\ \text{kg})(8.00\ \text{m/s south}) = 0.120\ \text{kg·m/s south}$

$p_{total} = 0.120 - 0.105 = 0.0150\ \text{kg·m/s south}$

Answer: 0.0150 kg·m/s south

Total momentum is conserved in the collision, so the system's momentum after collision equals its momentum before collision, found by subtracting the two oppositely-directed momenta.

Question 51 · 1 pt · constructed response

Base your answer on the information and circuit diagram below and on your knowledge of physics. An electric circuit contains a battery, a 3.0-ohm resistor, a 6.0-ohm resistor, and three meters. Ammeter A1 reads 6.0 amperes, ammeter A2 reads 4.0 amperes, and the voltmeter reads 12.0 volts. Determine the potential difference across the battery.

Model answer: 12.0 V

Worked solution

$\text{The 3.0-}\Omega\text{ and 6.0-}\Omega\text{ resistors are wired in parallel directly across the battery}$

$\text{Check: } I_{3\Omega}=\frac{12.0\text{ V}}{3.0\ \Omega}=4.0\text{ A (matches }A_2\text{); } I_{6\Omega}=\frac{12.0\text{ V}}{6.0\ \Omega}=2.0\text{ A; total } = 6.0\text{ A (matches }A_1\text{)}$

$V_{battery} = V_{parallel\ network} = 12.0\ \text{V}$

Answer: 12.0 V

In a parallel circuit with no other series resistance, the voltage across the battery equals the voltage across the parallel combination, which is given directly by the voltmeter reading and is confirmed by checking both ammeter readings.

Question 52 · 1 pt · constructed response

Determine the current through the 6.0-ohm resistor.

Model answer: 2.0 A

Worked solution

$I = \frac{V}{R} = \frac{12.0\ \text{V}}{6.0\ \Omega} = 2.0\ \text{A}$

Answer: 2.0 A

Ohm's law applied to the 6.0-ohm resistor using the 12.0 V reading from the voltmeter across it gives the branch current directly.

Question 53 · 2 pts · constructed response

Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]

Model answer: 2.0 Ω

Worked solution

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3.0\ \Omega} + \frac{1}{6.0\ \Omega} = \frac{2}{6.0\ \Omega} + \frac{1}{6.0\ \Omega} = \frac{3}{6.0\ \Omega}$

$R_{eq} = \frac{6.0\ \Omega}{3} = 2.0\ \Omega$

$\text{Check: } R_{eq} = \frac{V_{battery}}{I_{total}} = \frac{12.0\ \text{V}}{6.0\ \text{A}} = 2.0\ \Omega$

Answer: 2.0 Ω

For resistors wired in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of each resistor; the result is confirmed by dividing total battery voltage by total current.

Question 55 · 1 pt · constructed response

The diagram below shows parallel straight wave fronts approaching a rigid barrier that has a small opening in its center. On the diagram in your answer booklet, draw three wave fronts after they have passed through the opening.

Model answer: Three semicircular (circular arc) wave fronts, centered on the opening, spreading outward on the far side of the barrier, evenly spaced by one wavelength.

Worked solution

$\text{When a wave passes through an opening comparable in size to its wavelength, it diffracts}$

$\text{The transmitted wave fronts spread out as circular arcs centered at the opening rather than continuing as straight lines}$

Answer: Three semicircular (circular arc) wave fronts, centered on the opening, spreading outward on the far side of the barrier, evenly spaced by one wavelength.

Diffraction causes the straight wave fronts to bend around the edges of the small opening, producing circular (semicircular) wave fronts that spread out in all directions beyond the barrier.

Question 56 · 2 pts · constructed response

Base your answer on the information below and on your knowledge of physics. A projectile is launched from level ground with an initial vertical speed of 26 meters per second upward and an initial horizontal speed of 15 meters per second. [Neglect friction.] Calculate the total time the projectile is in the air before the projectile returns to the level ground. [Show all work, including the equation and substitution with units.]

Model answer: 5.3 s

Worked solution

$t_{total} = \frac{2v_{iy}}{g} = \frac{2(26\ \text{m/s})}{9.81\ \text{m/s}^2}$

$t_{total} = \frac{52\ \text{m/s}}{9.81\ \text{m/s}^2} = 5.3\ \text{s}$

Answer: 5.3 s

For a projectile launched from and landing at the same height, the total flight time is twice the time to reach maximum height, found from the vertical initial speed and gravitational acceleration.

Question 58 · 1 pt · constructed response

Determine the total horizontal distance the projectile travels before the projectile returns to the ground.

Model answer: ≈80. m

Worked solution

$\Delta x = v_x \cdot t = (15\ \text{m/s})(5.3\ \text{s})$

$\Delta x \approx 80.\ \text{m}$

Answer: ≈80. m

Since horizontal velocity is constant (no friction), the horizontal range is simply the horizontal speed multiplied by the total flight time found in question 56-57.

Question 59 · 1 pt · constructed response

Determine the angle above the horizontal at which the projectile is launched.

Model answer: 60.° above the horizontal

Worked solution

$\theta = \tan^{-1}\left(\frac{v_{iy}}{v_{ix}}\right) = \tan^{-1}\left(\frac{26\ \text{m/s}}{15\ \text{m/s}}\right)$

$\theta = \tan^{-1}(1.73) \approx 60.^{\circ}$

Answer: 60.° above the horizontal

The launch angle is found from the inverse tangent of the ratio of the vertical to horizontal initial speed components.

Question 60 · 1 pt · constructed response

Base your answer on the information below and on your knowledge of physics. A 20.-newton force due east and a 25-newton force due south act concurrently at point P. The diagram shows a vector drawn to scale representing the 20.-newton force due east. Using a ruler, determine the scale used in the diagram.

Model answer: 1.0 cm = 5.0 N (scale read from the printed diagram)

Worked solution

$\text{Measure the printed length of the 20. N arrow with a metric ruler (e.g., 4.0 cm)}$

$\text{scale} = \frac{20.\ \text{N}}{4.0\ \text{cm}} = 5.0\ \text{N/cm} \implies 1.0\ \text{cm} = 5.0\ \text{N}$

Answer: 1.0 cm = 5.0 N (scale read from the printed diagram)

The scale is found by physically measuring the printed length of the 20. N vector with a ruler and dividing the known force by that length; this requires the actual printed diagram rather than the transcribed text.

Question 61 · 1 pt · constructed response

On the diagram in your answer booklet, use a metric ruler and the scale shown in the diagram to draw a vector to represent the 25 newton force due south acting at point P. Begin the vector at point P and label its magnitude in newtons.

Model answer: A vector starting at point P, pointing straight down (south), drawn to a length consistent with the diagram's scale (e.g., 5.0 cm long if the scale is 1.0 cm = 5.0 N), labeled "25 N".

Worked solution

$\text{length} = \frac{25\ \text{N}}{\text{scale (N/cm)}}$

$\text{Draw the vector starting at P, pointing due south, with that length, and label it 25 N}$

Answer: A vector starting at point P, pointing straight down (south), drawn to a length consistent with the diagram's scale (e.g., 5.0 cm long if the scale is 1.0 cm = 5.0 N), labeled "25 N".

The vector's direction (south) is given, and its drawn length is set by dividing the 25 N magnitude by the diagram's established scale from question 60.

Question 62 · 1 pt · constructed response

Determine the magnitude of the resultant force acting at point P.

At point P, a 20. N force acts toward the east and a 25 N force acts toward the south (the two forces are perpendicular).

Model answer: 32 N

Worked solution

$F_R = \sqrt{F_1^2+F_2^2} = \sqrt{(20.\ \text{N})^2+(25\ \text{N})^2}$

$F_R = \sqrt{400+625} = \sqrt{1025} \approx 32\ \text{N}$

Answer: 32 N

Because the 20 N and 25 N forces act perpendicular to each other (east and south), the Pythagorean theorem gives the magnitude of their resultant.

Question 63 · 1 pt · constructed response

Determine the amount of energy, in electronvolts, that must be absorbed by a hydrogen atom to excite the atom from the n = 1 energy level directly to the n = 4 energy level.

Model answer: 12.75 eV

Worked solution

$E_1 = -13.6\ \text{eV}, \quad E_4 = -0.85\ \text{eV (from the hydrogen energy level reference table)}$

$\Delta E = E_4 - E_1 = -0.85\ \text{eV} - (-13.6\ \text{eV}) = 12.75\ \text{eV}$

Answer: 12.75 eV

The energy absorbed to move an electron between energy levels equals the difference between the final and initial energy level values listed in the reference tables.

Question 64 · 1 pt · constructed response

Determine the sign and magnitude of the net charge, in coulombs, of a particle that has a quark composition of bbb.

Model answer: -1.6 × 10⁻¹⁹ C (negative)

Worked solution

$\text{charge of a bottom quark } b = -\frac{1}{3}e$

$q_{particle} = 3\left(-\frac{1}{3}e\right) = -1e = -(1.6\times10^{-19}\ \text{C})$

Answer: -1.6 × 10⁻¹⁹ C (negative)

Each bottom quark carries a charge of negative one-third the elementary charge, so three bottom quarks combine to give a net charge of exactly minus one elementary charge.

Question 65 · 1 pt · constructed response

What type of electromagnetic radiation would contain photons each having an energy of 2.65 × 10⁻²⁰ joule?

Model answer: Infrared radiation

Worked solution

$E = hf \implies f = \frac{E}{h} = \frac{2.65\times10^{-20}\ \text{J}}{6.63\times10^{-34}\ \text{J·s}}$

$f \approx 4.00\times10^{13}\ \text{Hz}$

$\text{This frequency falls within the infrared region of the electromagnetic spectrum reference table}$

Answer: Infrared radiation

Dividing the photon energy by Planck's constant gives the photon frequency, which falls in the infrared band of the electromagnetic spectrum chart in the reference tables.

Question 66 · 2 pts · constructed response

Base your answers to questions 66 through 70 on the information below and on your knowledge of physics. A car, initially traveling at a velocity of 25 meters per second west, accelerates uniformly at a rate of 1.5 meters per second squared east for 6.0 seconds. Calculate the magnitude of the car's velocity after the 6.0 seconds. [Show all work, including the equation and substitution with units.]

Model answer: 16 m/s

Worked solution

$\text{Let west be positive: } v_i = +25\ \text{m/s}, \quad a = -1.5\ \text{m/s}^2\ \text{(east)}$

$v_f = v_i + at = 25\ \text{m/s} + (-1.5\ \text{m/s}^2)(6.0\ \text{s})$

$v_f = 25\ \text{m/s} - 9.0\ \text{m/s} = 16\ \text{m/s}$

Answer: 16 m/s

Choosing west as the positive direction, the eastward acceleration is negative; applying v_f = v_i + at over the given time gives the final velocity's magnitude.

Question 68 · 1 pt · constructed response

Determine the direction of the car's velocity after the 6.0 seconds.

Model answer: West

Worked solution

$v_f = -25 \text{ m/s} + 9.0 \text{ m/s} = -16 \text{ m/s}$

$\text{Negative sign (west convention) means the velocity remains directed west}$

Answer: West

Since v_f is still negative in the west-negative convention, the car (though it has slowed down) is still moving west after 6.0 seconds.

Question 69 · 2 pts · constructed response

Calculate the total distance traveled by the car during the 6.0 seconds. [Show all work, including the equation and substitution with units.]

Model answer: 123 m

Worked solution

$d = v_i t + \tfrac{1}{2} a t^2$

$d = (-25 \text{ m/s})(6.0 \text{ s}) + \tfrac{1}{2}(1.5 \text{ m/s}^2)(6.0 \text{ s})^2$

$d = -150 \text{ m} + 27 \text{ m} = -123 \text{ m}$

$\text{Total distance} = |d| = 123 \text{ m}$

Answer: 123 m

Because the velocity never reaches zero during the 6.0 s interval (it would take about 16.7 s to stop), the car never reverses direction, so the magnitude of displacement equals the total distance traveled.

Question 71 · 2 pts · constructed response

Base your answer to questions 71 through 76 on the information below and on your knowledge of physics. Starting from rest at the top of the slide, a 40.-kilogram child accelerates uniformly along the surface of a 4.2-meter-long slide at a rate of 0.30 meter per second squared until she reaches the bottom of the slide. Calculate the speed of the child at the instant she reaches the bottom of the slide. [Show all work, including the equation and substitution with units.]

Model answer: 1.6 m/s

Worked solution

$v_f^2 = v_i^2 + 2 a d$

$v_f^2 = 0 + 2(0.30 \text{ m/s}^2)(4.2 \text{ m}) = 2.52 \text{ m}^2/\text{s}^2$

$v_f = \sqrt{2.52} = 1.6 \text{ m/s}$

Answer: 1.6 m/s

Using kinematics with initial speed zero, the child's speed at the bottom of the slide is found from v_f squared equals 2ad, using the 4.2 m slide length and 0.30 m/s squared acceleration given in the text.

Question 73 · 2 pts · constructed response

Calculate the child's kinetic energy at the bottom of the slide. [Show all work, including the equation and substitution with units.]

Model answer: 50.4 J

Worked solution

$KE = \tfrac{1}{2} m v^2$

$KE = \tfrac{1}{2}(40. \text{ kg})(2.52 \text{ m}^2/\text{s}^2)$

$KE = 50.4 \text{ J}$

Answer: 50.4 J

Using the mass of 40. kg and the speed squared (2.52 m^2/s^2) found from the previous kinematics calculation, the kinetic energy at the bottom of the slide is 50.4 J.

Question 75 · 1 pt · constructed response

Determine the total gravitational potential energy gained by the child when climbing 1.5 meters vertically from the bottom to the top of the slide.

Model answer: 589 J (5.9 x 10^2 J)

Worked solution

$PE = mgh$

$PE = (40. \text{ kg})(9.81 \text{ m/s}^2)(1.5 \text{ m})$

$PE = 588.6 \text{ J} \approx 589 \text{ J}$

Answer: 589 J (5.9 x 10^2 J)

The gravitational potential energy gained equals the weight of the child times the vertical height climbed, using the 1.5 m height shown in the diagram.

Question 76 · 1 pt · constructed response

Determine the total amount of mechanical energy lost by the child-slide system as the child slides from the top to the bottom of the slide.

Model answer: About 538 J

Worked solution

$PE_{top} = mgh = (40. \text{ kg})(9.81 \text{ m/s}^2)(1.5 \text{ m}) = 588.6 \text{ J}$

$KE_{bottom} = 50.4 \text{ J}$

$\Delta E_{lost} = PE_{top} - KE_{bottom} = 588.6 \text{ J} - 50.4 \text{ J} \approx 538 \text{ J}$

Answer: About 538 J

Since the child starts at rest at the top with only gravitational potential energy and ends at the bottom with only kinetic energy, the mechanical energy lost to friction is the difference between the initial potential energy and the final kinetic energy.

Question 77 · 2 pts · constructed response

Base your answer to questions 77 through 80 on the information below and on your knowledge of physics. A ray of light ($f = 5.09 \times 10^{14}$ Hz) in flint glass is incident on air at an angle of incidence of 30.$^\circ$. Calculate the angle of refraction of the light ray in the air. [Show all work, including the equation and substitution with units.]

Model answer: About 56 degrees

Worked solution

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

$n_{flint glass} = 1.66, \ n_{air} = 1.00 \text{ (from reference table)}$

$(1.66)(\sin 30^\circ) = (1.00)(\sin\theta_2)$

$\sin\theta_2 = 1.66 (0.500) = 0.830$

$\theta_2 = \sin^{-1}(0.830) \approx 56^\circ$

Answer: About 56 degrees

Applying Snell's Law with the absolute index of refraction of flint glass (1.66) from the reference tables and air (1.00), the angle of refraction in air is about 56 degrees, bending away from the normal since light travels from a denser to a less dense medium.

Question 79 · 1 pt · constructed response

Using a protractor and a straightedge, draw the refracted light ray on the diagram in your answer booklet.

Model answer: A ray drawn in the air region starting at the point of incidence, making an angle of approximately 56 degrees with the normal, bent away from the normal (on the same side of the normal as the incident ray) since the light travels from a denser medium (flint glass) into a less dense medium (air).

Worked solution

$\theta_2 \approx 56^\circ \text{ from the normal, on the air side, same side as the incident ray}$

Answer: A ray drawn in the air region starting at the point of incidence, making an angle of approximately 56 degrees with the normal, bent away from the normal (on the same side of the normal as the incident ray) since the light travels from a denser medium (flint glass) into a less dense medium (air).

The refracted ray must be drawn at approximately 56 degrees from the normal on the air side of the boundary, bending away from the normal because the light is moving from a more optically dense medium into a less dense one.

Question 80 · 1 pt · constructed response

State what happens to the wavelength of the light ray as the light ray passes into the air from the flint glass.

Model answer: The wavelength increases.

Worked solution

$v = f\lambda, \ f \text{ stays constant when changing media}$

$\text{Speed increases in air (lower index of refraction), so } \lambda \text{ must increase}$

Answer: The wavelength increases.

Frequency does not change when light crosses a boundary between media, but since air has a lower index of refraction than flint glass, the speed of light increases, so the wavelength must increase proportionally.

Question 81 · 1 pt · constructed response

Base your answers to questions 81 through 85 on the information, graph, and diagram below and on your knowledge of physics. A student pulls a block weighing 2.0 newtons, resting on a horizontal laboratory table, with a string and force sensor. The sensor measures the force applied to the block when the block is at rest and when the block is moving. The force applied to the block is plotted against time on the graph. (Diagram: a force sensor connected by a string to a block on a horizontal table. Graph: Force (N) on the y-axis from 0 to 1.2 N, Time (s) on the x-axis from 0 to 4.0 s. The force rises roughly linearly (slightly curved) from 0 N at t=0 to about 1.0 N at t=2.0 s, labeled "Block started in motion" at that peak; then the force drops and levels off at about 0.9 N from t=2.0 s to t=4.0 s.) Based on the graph, determine the force of kinetic friction acting on the block as it is pulled at constant velocity across the laboratory table.

Model answer: 0.9 N

Worked solution

$\text{At constant velocity, applied force} = \text{friction force (Newton's 2nd Law, } a=0\text{)}$

$\text{Graph shows the force leveling off at } 0.9 \text{ N for } t > 2.0 \text{ s}$

Answer: 0.9 N

Once the block is moving at constant velocity, the net force is zero, so the applied force read from the flat portion of the graph (about 0.9 N) equals the kinetic friction force.

Question 82 · 2 pts · constructed response

Calculate the coefficient of kinetic friction between the block and the laboratory table. [Show all work, including the equation and substitution with units.]

Model answer: 0.45

Worked solution

$\mu_k = \frac{F_f}{F_N}$

$F_f = 0.9 \text{ N (from graph)}, \ F_N = 2.0 \text{ N (weight of block)}$

$\mu_k = \frac{0.9 \text{ N}}{2.0 \text{ N}} = 0.45$

Answer: 0.45

The coefficient of kinetic friction is the ratio of the kinetic friction force (read from the graph) to the normal force, which equals the block's weight of 2.0 N on a horizontal table.

Question 84 · 1 pt · constructed response

At the instant the block begins to move, what happens to the value of the coefficient of friction?

Model answer: It decreases.

Worked solution

$\mu_s > \mu_k \text{ for the same pair of surfaces}$

$\text{At the instant of motion, friction changes from static to kinetic, so } \mu \text{ decreases}$

Answer: It decreases.

The coefficient of static friction is always greater than or equal to the coefficient of kinetic friction, so as the block transitions from rest to sliding, the coefficient of friction decreases (matching the drop in force seen on the graph at t = 2.0 s).

Question 85 · 1 pt · constructed response

How would the coefficient of kinetic friction ($\mu$) between the block and the laboratory table be affected if a 0.50-kilogram mass were placed on the block?

Model answer: The coefficient of kinetic friction would not be affected (it would stay the same).

Worked solution

$\mu_k = \frac{F_f}{F_N}$

$\text{Adding mass increases both } F_f \text{ and } F_N \text{ proportionally, so their ratio } \mu_k \text{ is unchanged}$

Answer: The coefficient of kinetic friction would not be affected (it would stay the same).

The coefficient of kinetic friction depends only on the nature of the two surfaces in contact, not on the normal force or mass, so adding a 0.50 kg mass to the block would not change the value of mu.

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