June 2024 Physics Regents — Answer Key & Worked Solutions

Every question from the June 2024 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 2024 Physics New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

Which quantity is a vector?

(A) electric field ✓
(B) electric potential difference
(C) electric charge
(D) electric power

Worked solution

$\vec{E} \text{ has both magnitude and direction}$

$\text{Potential difference, charge, and power are scalars}$

Answer: electric field

Electric field has both magnitude and direction (force per unit charge), making it a vector; the other three quantities are scalars.

Question 2 · 2 pts

What is the magnitude of the eastward component of the velocity of an airplane flying at 612 kilometers per hour in a direction 40.0° north of east?

(A) 393 km/h
(B) 469 km/h ✓
(C) 799 km/h
(D) 952 km/h

Worked solution

$v_x = v\cos\theta$

$v_x = 612\cos(40.0^\circ)$

$v_x = 612(0.766) = 469\ \text{km/h}$

Answer: 469 km/h

The eastward component is found by multiplying the resultant velocity by the cosine of the angle from the east axis.

Question 3 · 2 pts

A car, initially traveling at 25 meters per second, is uniformly brought to rest as the brakes are applied over a distance of 40. meters. The magnitude of the average acceleration of the car while braking is

(A) 0.31 m/s²
(B) 0.63 m/s²
(C) 7.8 m/s² ✓
(D) 16 m/s²

Worked solution

$v_f^2 = v_i^2 + 2ad$

$0 = (25)^2 + 2a(40.)$

$a = \frac{-625}{80} = -7.8\ \text{m/s}^2$

Answer: 7.8 m/s²

Using the kinematics equation relating final and initial velocity to acceleration and distance, the magnitude of acceleration is 7.8 m/s².

Question 4 · 2 pts

A brick starts from rest and falls freely from the top of a building to the ground. As the brick falls, its acceleration

(A) increases and its speed increases
(B) increases and its speed is constant
(C) is constant and its speed increases ✓
(D) is constant and its speed is constant

Worked solution

$a = g = 9.81\ \text{m/s}^2 = \text{constant}$

$v = gt \text{ increases with time}$

Answer: is constant and its speed increases

In free fall, the acceleration due to gravity is constant, so the speed of the falling brick continually increases.

Question 5 · 2 pts

Which object has the greatest inertia?

(A) a 0.10-kg baseball traveling at 30. m/s
(B) a 70-kg sprinter traveling at 10. m/s
(C) a 1000-kg car traveling at 50. m/s
(D) a 2000-kg truck traveling at 20. m/s ✓

Worked solution

$\text{Inertia depends only on mass, not velocity}$

$2000\ \text{kg} > 1000\ \text{kg} > 70\ \text{kg} > 0.10\ \text{kg}$

Answer: a 2000-kg truck traveling at 20. m/s

Inertia is a measure of an object's mass alone, so the object with the greatest mass, the 2000-kg truck, has the greatest inertia.

Question 6 · 2 pts

An unbalanced force is always necessary to

(A) keep a body at rest
(B) keep a body moving with constant velocity
(C) change the speed of a body ✓
(D) change the position of a body

Worked solution

$\Sigma F = ma$

$a \neq 0 \implies \text{net (unbalanced) force required}$

Answer: change the speed of a body

By Newton's second law, a net (unbalanced) force is required to produce an acceleration, which changes an object's speed.

Question 7 · 2 pts

Space probes launched from Earth send information back to Earth from space in the form of

(A) mechanical waves
(B) sound waves
(C) longitudinal waves
(D) electromagnetic waves ✓

Worked solution

$\text{Space is a vacuum; mechanical/sound waves need a medium}$

$\text{Only electromagnetic waves travel through vacuum}$

Answer: electromagnetic waves

Because space is a vacuum, only electromagnetic waves, which require no medium, can carry information from space probes back to Earth.

Question 8 · 2 pts

A ball is thrown from level ground at an angle of 55° above the horizontal and lands on level ground. Neglecting friction, if the ball is thrown again at the same angle but with a larger initial speed, the ball will travel

(A) higher and the same distance horizontally
(B) to the same maximum height and farther horizontally
(C) both higher and farther horizontally ✓
(D) to the same maximum height and the same distance horizontally

Worked solution

$h_{max} = \frac{(v\sin\theta)^2}{2g}$

$R = \frac{v^2\sin(2\theta)}{g}$

$\text{Both } h_{max} \text{ and } R \text{ increase with } v$

Answer: both higher and farther horizontally

Both the maximum height and range of projectile motion depend on the square of the initial speed, so increasing speed at the same angle increases both height and horizontal distance.

Question 9 · 2 pts

A photon collides with an electron, as represented in the diagram below. Incident photon approaches a stationary electron (e⁻); after the collision, the electron recoils and a scattered photon leaves in another direction. After the collision, the electron recoils and the photon is scattered in another direction with a longer wavelength than the incident photon. The increase in photon wavelength results because, during the collision, the photon

(A) loses energy to the electron ✓
(B) gains momentum from the electron
(C) loses some speed
(D) generates a magnetic field

Worked solution

$E = \frac{hc}{\lambda}$

$\lambda \uparrow \implies E \downarrow$

$\text{Energy lost by photon transferred to electron (Compton scattering)}$

Answer: loses energy to the electron

This is Compton scattering: the photon transfers some of its energy to the electron, and since photon energy is inversely proportional to wavelength, the wavelength increases.

Question 10 · 2 pts

What is the weight of a 60.0-kilogram student on the surface of Earth?

(A) 0.164 N
(B) 6.12 N
(C) 60.0 N
(D) 589 N ✓

Worked solution

$F_g = mg$

$F_g = (60.0)(9.81) = 589\ \text{N}$

Answer: 589 N

Weight equals mass times the acceleration due to gravity, giving 589 N.

Question 11 · 2 pts

A 120-newton box is pulled by a 48-newton horizontal force across a horizontal surface at constant velocity. The coefficient of kinetic friction between the box and the horizontal surface is

(A) 0.041
(B) 0.40 ✓
(C) 0.67
(D) 2.5

Worked solution

$F_f = F_{applied} = 48\ \text{N (constant velocity)}$

$\mu_k = \frac{F_f}{F_N} = \frac{48}{120}$

$\mu_k = 0.40$

Answer: 0.40

Since the box moves at constant velocity, the applied force equals the friction force; dividing friction force by normal force (weight) gives the coefficient of kinetic friction.

Question 12 · 2 pts

Box A has a mass of 10. kilograms and is at rest on a shelf that is 1.5 meters above the floor. Box B has a mass of 20. kilograms and is at rest on a shelf that is 3.0 meters above the floor. Compared to box A, box B has a gravitational potential energy relative to the floor that is

(A) one fourth as great
(B) the same
(C) twice as great
(D) four times as great ✓

Worked solution

$PE_A = m_Agh_A = (10.)(9.81)(1.5) = 147\ \text{J}$

$PE_B = m_Bgh_B = (20.)(9.81)(3.0) = 589\ \text{J}$

$\frac{PE_B}{PE_A} = \frac{589}{147} \approx 4$

Answer: four times as great

Box B has both twice the mass and twice the height of box A, so its gravitational potential energy is 2×2 = 4 times as great.

Question 13 · 2 pts

A 0.10-kilogram yo-yo is whirled at the end of a length of string in a horizontal circular path of radius 0.80 meter at a speed of 6.0 meters per second. The magnitude of the centripetal acceleration of the yo-yo is

(A) 4.5 m/s²
(B) 7.5 m/s²
(C) 23 m/s²
(D) 45 m/s² ✓

Worked solution

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

$a_c = \frac{(6.0)^2}{0.80} = \frac{36}{0.80}$

$a_c = 45\ \text{m/s}^2$

Answer: 45 m/s²

Centripetal acceleration equals the speed squared divided by the radius of the circular path.

Question 14 · 2 pts

A 4.0-kilogram mass is initially at rest on a horizontal, frictionless surface. A constant 2.0-newton force to the east is applied to the mass for a 5.0-second interval. As a result of this action, the mass acquires a

(A) velocity of 10. m/s, east
(B) velocity of 10. m/s, west
(C) momentum of 10. kg·m/s, east ✓
(D) momentum of 10. kg·m/s, west

Worked solution

$\Delta p = F\Delta t$

$\Delta p = (2.0\ \text{N})(5.0\ \text{s}) = 10.\ \text{kg·m/s, east}$

Answer: momentum of 10. kg·m/s, east

Impulse equals the product of force and time interval, and equals the change in momentum, which is 10. kg·m/s to the east (the direction of the applied force).

Question 15 · 2 pts

A motor lifts a 1.2 × 10⁴-newton elevator 9.0 meters in 15 seconds. The minimum power output of the motor is

(A) 8.0 × 10² W
(B) 7.2 × 10³ W ✓
(C) 1.0 × 10⁵ W
(D) 1.6 × 10⁶ W

Worked solution

$P = \frac{W}{t} = \frac{Fd}{t}$

$P = \frac{(1.2\times10^4)(9.0)}{15}$

$P = 7.2\times10^3\ \text{W}$

Answer: 7.2 × 10³ W

Power is the work done (force times distance) divided by the time interval.

Question 16 · 2 pts

A train blows its horn, which emits a uniform sound as the train approaches a stationary observer. The observer hears a sound that has a

(A) lower frequency than the emitted sound and is decreasing in amplitude
(B) lower frequency than the emitted sound and is increasing in amplitude
(C) higher frequency than the emitted sound and is decreasing in amplitude
(D) higher frequency than the emitted sound and is increasing in amplitude ✓

Worked solution

$\text{Source approaching} \implies f_{observed} > f_{emitted} \text{ (Doppler effect)}$

$\text{Decreasing distance} \implies \text{increasing amplitude (louder)}$

Answer: higher frequency than the emitted sound and is increasing in amplitude

As the train approaches, the Doppler effect causes the observer to hear a higher frequency, and the sound grows louder (increasing amplitude) as the source gets closer.

Question 17 · 2 pts

A wood block is pulled at constant velocity across a horizontal wood floor. Which type of energy increases in this block-floor system as the block moves?

(A) gravitational potential
(B) kinetic
(C) mechanical
(D) thermal ✓

Worked solution

$\text{Constant velocity} \implies KE, PE_{grav}\ \text{unchanged}$

$\text{Friction does work} \implies \text{thermal energy increases}$

Answer: thermal

Since the block moves at constant velocity, kinetic and gravitational potential energy stay the same, but friction between the block and floor converts mechanical energy into thermal energy.

Question 18 · 2 pts

A total energy of 5.0 joules is used to move an electron from position A to position B in a uniform electric field. What is the potential difference between positions A and B?

(A) 3.1 × 10¹⁹ V ✓
(B) 8.0 × 10⁻¹⁹ V
(C) 3.2 × 10⁻²⁰ V
(D) 3.1 × 10¹⁸ V

Worked solution

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

$V = \frac{5.0\ \text{J}}{1.6\times10^{-19}\ \text{C}}$

$V = 3.1\times10^{19}\ \text{V}$

Answer: 3.1 × 10¹⁹ V

Potential difference equals the energy used divided by the charge of the electron.

Question 19 · 2 pts

A 0.14-kilogram lacrosse ball, traveling west at 17 meters per second, is brought to rest with a 0.21-kilogram lacrosse stick. If the force applied by the lacrosse stick on the ball is 220 newtons east, the force applied by the ball on the stick is

(A) 150 N east
(B) 150 N west
(C) 220 N east
(D) 220 N west ✓

Worked solution

$\text{By Newton's 3rd law: } F_{ball\ on\ stick} = -F_{stick\ on\ ball}$

$F_{ball\ on\ stick} = 220\ \text{N, west}$

Answer: 220 N west

By Newton's third law, the force the ball exerts on the stick is equal in magnitude and opposite in direction to the force the stick exerts on the ball, so it is 220 N west.

Question 20 · 2 pts

Four wires are tested for electrical conductivity. All the wires have the same length and the same cross-sectional area, but are made of different metals. Which wire has the highest conductivity at 20°C?

(A) aluminum
(B) copper
(C) gold
(D) silver ✓

Worked solution

$\text{Conductivity} \propto \frac{1}{\rho}$

$\text{Reference Table lists silver with lowest resistivity of the four}$

Answer: silver

Using the resistivity values in the reference tables, silver has the lowest resistivity of the four metals and therefore the highest conductivity.

Question 21 · 2 pts

The angle of incidence for a ray of light striking a plane mirror is 20º. What is the angle between the incident ray and the reflected ray?

(A) 20º
(B) 40º ✓
(C) 70º
(D) 90º

Worked solution

$\theta_i = \theta_r = 20^\circ \text{ (law of reflection)}$

$\text{Angle between rays} = \theta_i + \theta_r = 20^\circ+20^\circ = 40^\circ$

Answer: 40º

By the law of reflection, the angle of reflection equals the angle of incidence, so the total angle between the incident and reflected rays, both measured from the normal, is 40°.

Question 22 · 2 pts

As shown in the diagram below, mass M slides across a level, frictionless surface with speed vi. The mass strikes a spring at position A, causing the spring to compress. When the mass is at position B, it is moving at a slower speed, vf. Which statement best describes the energy conversion as the mass moves from position A to position B?

(A) Some of mass M's kinetic energy is converted to elastic potential energy. ✓
(B) All of mass M's kinetic energy is converted to elastic potential energy.
(C) Some of mass M's kinetic energy is converted to gravitational potential energy.
(D) All of mass M's kinetic energy is converted to internal energy.

Worked solution

$v_f < v_i,\ v_f \neq 0 \implies \text{mass still has kinetic energy at B}$

$\text{Frictionless surface} \implies \text{no energy lost to heat}$

$\text{Lost KE} \to \text{elastic PE of spring}$

Answer: Some of mass M's kinetic energy is converted to elastic potential energy.

Because the mass is still moving (though slower) at position B and the surface is frictionless, only part of its kinetic energy has been converted into the spring's elastic potential energy.

Question 23 · 2 pts

The diagram below shows a magnetic compass placed between unlike magnetic poles. The north pole of the compass needle will point toward

(A) A
(B) B
(C) C ✓
(D) D

Worked solution

$\text{Magnetic field lines point away from the N pole and toward the S pole.}$

$\vec{B}_{\text{net}} \text{ at the compass's location points from the N magnet (left) toward the S magnet (right), i.e., toward point } C.$

$\text{A compass needle's N-pole aligns with the local field, so it points toward } C.$

Answer: C

Between unlike poles, the net magnetic field points from the N pole toward the S pole. Since the compass sits directly between them, the field at its location points horizontally toward the S magnet, so the needle's north pole points toward point C.

Question 24 · 2 pts

Which circuit diagram contains a lamp that will not have current passing through it until switch S is closed?

(A) Diagram (1): the lamp (top), switch S (left branch), and battery (right branch) are connected in a single series loop, with no other path between the battery and the lamp. ✓
(B) Diagram (2): the lamp (top) is connected to a loop in which the battery and switch S are on separate parallel branches at the bottom, giving the lamp a closed path through the battery independent of S.
(C) Diagram (3): the lamp (top), switch S, and battery are arranged in a loop similar to (1) but with an additional branch that keeps the lamp's circuit closed regardless of the switch position.
(D) Diagram (4): the lamp (top), battery (left branch), and switch S (right branch) are arranged so that a path to the lamp exists independent of the switch.

Worked solution

$\text{For the lamp to be dark until S closes, the lamp must lie on a path that is broken (open) at S.}$

$\text{In diagram (1), the lamp, S, and battery form one single series loop, with no parallel bypass.}$

$\text{Closing S is therefore required to complete the only current path to the lamp.}$

Answer: Diagram (1)

A lamp will stay dark until a switch closes only if the switch is the sole connection completing the lamp's circuit. In diagram (1), the lamp, switch, and battery are all in one series loop with no alternate path, so current cannot reach the lamp until S is closed; in the other diagrams a parallel branch keeps a current path available regardless of the switch position.

Question 25 · 2 pts

The diagram below represents an electric circuit. A $3.0\ \Omega$ resistor, a $6.0\ \Omega$ resistor, and a $9.0\ \Omega$ resistor are connected in parallel with each other across a battery. The equivalent resistance of the circuit is

(A) $1.6\ \Omega$ ✓
(B) $2.0\ \Omega$
(C) $6.0\ \Omega$
(D) $18\ \Omega$

Worked solution

$\frac{1}{R_{eq}} = \frac{1}{3.0} + \frac{1}{6.0} + \frac{1}{9.0}$

$\frac{1}{R_{eq}} = \frac{6}{18} + \frac{3}{18} + \frac{2}{18} = \frac{11}{18}$

$R_{eq} = \frac{18}{11} \approx 1.6\ \Omega$

Answer: 1.6 \Omega

For resistors in parallel, the reciprocals of the resistances add; solving for the equivalent resistance of 3.0, 6.0, and 9.0 ohm resistors gives approximately 1.6 ohms, which is less than the smallest individual resistor as expected for parallel combinations.

Question 26 · 2 pts

An object is thrown vertically upward with an initial velocity of 9.81 meters per second. What is the maximum height reached by the object? [Neglect friction.]

(A) 1.00 m
(B) 4.91 m ✓
(C) 9.81 m
(D) 19.6 m

Worked solution

$v_f^2 = v_i^2 - 2g h \Rightarrow 0 = (9.81)^2 - 2(9.81)h$

$h = \frac{(9.81)^2}{2(9.81)} = \frac{9.81}{2}$

$h \approx 4.91\ \text{m}$

Answer: 4.91 m

At maximum height the vertical velocity is zero, so applying v_f^2 = v_i^2 - 2gh and solving for h with v_i = 9.81 m/s and g = 9.81 m/s^2 gives h = 4.91 m.

Question 27 · 2 pts

Which type of photon has the least amount of energy?

(A) ultraviolet
(B) visible light
(C) infrared
(D) radio ✓

Worked solution

$E = hf$

$\text{Radio waves have the lowest frequency of the choices listed on the EM spectrum.}$

$\text{Lowest } f \Rightarrow \text{lowest } E.$

Answer: radio

Photon energy is directly proportional to frequency (E = hf). Of ultraviolet, visible, infrared, and radio, radio waves have the lowest frequency and therefore the least photon energy.

Question 28 · 2 pts

A 7.5-kilogram object moving at 20. meters per second strikes a 60.-kilogram object initially at rest on a horizontal, frictionless surface. The two objects stick together and move off at a speed of

(A) 0.33 m/s
(B) 2.2 m/s ✓
(C) 2.5 m/s
(D) 18 m/s

Worked solution

$m_1 v_1 = (m_1+m_2)v_f$

$(7.5)(20.) = (7.5+60.)v_f$

$150 = 67.5\, v_f \Rightarrow v_f \approx 2.2\ \text{m/s}$

Answer: 2.2 m/s

Conservation of momentum for a perfectly inelastic collision gives the final combined speed of the two objects as approximately 2.2 m/s.

Question 29 · 2 pts

How is the electrostatic force between two positive charges affected as the charges are brought closer together?

(A) The force of attraction between them increases.
(B) The force of repulsion between them increases. ✓
(C) The force of attraction between them decreases.
(D) The force of repulsion between them decreases.

Worked solution

$F = \frac{k q_1 q_2}{r^2}$

$\text{As } r \text{ decreases, } F \text{ increases (inverse-square law).}$

$\text{Like charges repel, so the repulsive force increases.}$

Answer: The force of repulsion between them increases.

By Coulomb's law, the electrostatic force is inversely proportional to the square of the separation distance, so decreasing the distance between two like (positive) charges increases the repulsive force between them.

Question 30 · 2 pts

What is one difference between magnetic forces and gravitational forces?

(A) Magnetic forces are always attractive, whereas gravitational forces are always repulsive.
(B) Magnetic forces are always repulsive, whereas gravitational forces are always attractive.
(C) Magnetic forces may be attractive or repulsive, whereas gravitational forces are always attractive. ✓
(D) Magnetic forces may be attractive or repulsive, whereas gravitational forces are always repulsive.

Worked solution

$\text{Like magnetic poles repel; unlike poles attract} \Rightarrow \text{magnetic force can be either.}$

$\text{Gravity between masses is always attractive.}$

Answer: Magnetic forces may be attractive or repulsive, whereas gravitational forces are always attractive.

Magnetic forces can be either attractive (unlike poles) or repulsive (like poles), while gravitational forces between any two masses are always attractive, never repulsive.

Question 31 · 2 pts

A sound wave passes through an opening in a brick wall. An observer standing behind the wall is able to hear the sound. This spreading out of the sound wave as it passes through the opening is an example of

(A) interference
(B) reflection
(C) refraction
(D) diffraction ✓

Worked solution

$\text{Bending/spreading of a wave around obstacles or through openings} = \text{diffraction.}$

Answer: diffraction

The spreading of a wave as it passes through an opening or around a barrier, allowing it to be detected outside the direct line of sight, is the defining characteristic of diffraction.

Question 32 · 2 pts

A sound wave is emitted by a vibrating tuning fork. What is transferred as the sound wave travels to a student's ear?

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

Worked solution

$\text{Mechanical waves transfer energy through a medium without net transfer of the medium's mass.}$

Answer: energy, only

Sound is a mechanical wave; it transfers energy through the vibration of air particles, but the particles themselves (and thus mass) do not travel with the wave.

Question 33 · 2 pts

An electric current passing through a copper wire at constant temperature would result in

(A) an increase in the resistivity of the wire
(B) a decrease in the resistivity of the wire
(C) the emission of protons from the wire
(D) the production of a magnetic field around the wire ✓

Worked solution

$\text{Moving charge (current) always produces a magnetic field around the conductor.}$

$\text{Resistivity depends on material and temperature, not on whether current is flowing.}$

Answer: the production of a magnetic field around the wire

Resistivity is a temperature-dependent material property and is unaffected by the mere presence of current, so at constant temperature it stays the same; however, any current-carrying wire produces a magnetic field around it.

Question 34 · 2 pts

The velocity of an object in uniform circular motion has a

(A) constant magnitude and changing direction ✓
(B) constant magnitude and constant direction
(C) changing magnitude and constant direction
(D) changing magnitude and changing direction

Worked solution

$\text{Uniform circular motion: speed } |v| \text{ is constant, but direction of } \vec{v} \text{ continuously changes.}$

Answer: constant magnitude and changing direction

In uniform circular motion the speed stays constant while the direction of the velocity vector continuously changes as the object moves around the circle, which is why there is a centripetal acceleration.

Question 35 · 2 pts

Tuning fork A starts to vibrate at 320 hertz when it is held near tuning fork B, already vibrating at 320 hertz. Which phenomenon is exemplified by the action of tuning fork A?

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

Worked solution

$\text{Fork B's sound waves match fork A's natural frequency (320 Hz), driving A into vibration.}$

Answer: resonance

When an object begins to vibrate because it is driven at its own natural frequency by an external source (here, fork B's sound at 320 Hz matching fork A's natural frequency), this is resonance.

Question 36 · 2 pts

A light ray with a frequency of $5.09 \times 10^{14}$ hertz has a wavelength of $2.44 \times 10^{-7}$ meter in diamond. The wavelength of this light ray in sodium chloride is

(A) $1.55 \times 10^{-7}$ m
(B) $2.44 \times 10^{-7}$ m
(C) $3.83 \times 10^{-7}$ m ✓
(D) $5.89 \times 10^{-7}$ m

Worked solution

$v_{diamond} = f\lambda = (5.09\times10^{14})(2.44\times10^{-7}) \approx 1.24\times10^{8}\ \text{m/s}$

$n_{NaCl} = 1.54\ \text{(from reference table)}, \quad v_{NaCl} = \frac{c}{n_{NaCl}} = \frac{3.00\times10^{8}}{1.54} \approx 1.95\times10^{8}\ \text{m/s}$

$\lambda_{NaCl} = \frac{v_{NaCl}}{f} = \frac{1.95\times10^{8}}{5.09\times10^{14}} \approx 3.83\times10^{-7}\ \text{m}$

Answer: $3.83 \times 10^{-7}$ m

Frequency does not change when light passes between media; using the reference table's absolute index of refraction for sodium chloride (1.54) to find the speed of light in NaCl, then dividing by the (unchanged) frequency gives the new wavelength.

Question 37 · 2 pts

The diagram below shows resistors $R_1$, $R_2$, and $R_3$ connected to a 12-volt source. $R_1 = 15\ \Omega$ is connected in series along the top wire, $R_3 = 24\ \Omega$ is connected in series along the bottom wire, and both the top and bottom wires connect to $R_2 = 21\ \Omega$, which is connected between them on the right side, forming a single series loop with the 12-volt source. The current flowing through resistor $R_3$ is

(A) 5.0 A
(B) 2.0 A
(C) 0.50 A
(D) 0.20 A ✓

Worked solution

$R_{total} = R_1+R_2+R_3 = 15+21+24 = 60\ \Omega$

$I = \frac{V}{R_{total}} = \frac{12}{60} = 0.20\ \text{A}$

$\text{Since it is a single series loop, the same current flows through } R_3.$

Answer: 0.20 A

R1, R2, and R3 form a single series loop with the 12-V source, so the same current flows through all three resistors; dividing the total voltage by the total resistance (60 ohms) gives 0.20 A.

Question 38 · 2 pts

A charge of 25 coulombs moves past a point in a circuit in 2.5 seconds. What is the current at that point in the circuit?

(A) 0.10 A
(B) 10. A ✓
(C) 50. A
(D) 63 A

Worked solution

$I = \frac{q}{t} = \frac{25}{2.5}$

$I = 10.\ \text{A}$

Answer: 10. A

Current is charge per unit time; dividing 25 coulombs by 2.5 seconds gives 10 amperes.

Question 39 · 2 pts

Which statement describes an object with constant kinetic energy?

(A) A car accelerates along a straight road.
(B) A runner decreases her speed along a curved path.
(C) A bicycle travels around a curve at constant speed. ✓
(D) A sled travels down a frictionless, steep, straight hill.

Worked solution

$KE = \frac{1}{2}mv^2$

$KE \text{ depends only on speed } |v|, \text{ not on direction.}$

$\text{Constant speed around a curve} \Rightarrow \text{constant KE, even though direction (velocity) changes.}$

Answer: A bicycle travels around a curve at constant speed.

Kinetic energy depends only on the magnitude of velocity (speed), not its direction, so an object moving at constant speed along a curved path (like the bicycle) has constant kinetic energy even though its velocity direction is changing.

Question 40 · 2 pts

Which phrase describes a box in equilibrium?

(A) box in an elevator slowing down as it rises vertically
(B) box at rest on a stationary table ✓
(C) box sliding down a frictionless ramp
(D) box in free fall

Worked solution

$\text{Equilibrium} \Rightarrow \sum F = 0 \text{ (net force zero, no acceleration).}$

$\text{A box at rest on a stationary table has zero net force and zero acceleration.}$

Answer: box at rest on a stationary table

An object is in equilibrium when the net force acting on it is zero, meaning zero acceleration; a box at rest on a stationary table satisfies this, unlike the accelerating, sliding, or free-falling boxes in the other choices.

Question 41 · 2 pts

Quarks may combine to produce a meson of charge

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

Worked solution

$\text{A meson consists of one quark and one antiquark, each with charge } \pm\frac{1}{3}e \text{ or } \pm\frac{2}{3}e.$

$\text{Combining a quark with its own antiquark (e.g., up + anti-up) gives a net charge of } +\frac{2}{3}e - \frac{2}{3}e = 0e.$

Answer: 0 e

Mesons are made of a quark-antiquark pair. Combinations such as an up quark with an anti-up quark, or a down quark with an anti-down quark, produce a net charge of 0e, which is one of the allowed meson charges.

Question 42 · 2 pts

The diagram below shows waves A and B as they travel through a region in the same medium. Which wave best represents the superposition of waves A and B?

(A) Graph (1): a wave of somewhat reduced amplitude relative to A and B, roughly in phase with them.
(B) Graph (2): a flat horizontal line (zero amplitude everywhere), representing complete cancellation.
(C) Graph (3): a wave with amplitude and shape similar to A and B but shifted in phase. ✓
(D) Graph (4): a wave with a visibly larger amplitude than A and B, reflecting their point-by-point addition.

Worked solution

$\text{Superposition: resultant displacement at each point} = y_A + y_B$

$\text{A and B are similar in size but offset in phase, so their sum has an amplitude similar to A and B, with crests and troughs shifted from both}$

Answer: a wave with amplitude and shape similar to A and B but shifted in phase

Wave A (solid) has a clearly larger amplitude than wave B (dashed), and although both complete the same two cycles over the region shown, B's labeled crest sits at a different point along the axis than A's crest, not aligned with it. Because the two waves are neither perfectly in phase, which would require matching crests and add up to a much bigger wave, nor equal in size and exactly opposite, which would cancel completely to a flat line, their point-by-point sum comes out similar in size to the originals, with its own crests and troughs falling at positions offset from both A and B rather than matching either one.

Question 43 · 2 pts

Four diagrams each show a wave traveling to the right, differing in the direction in which the particles of the medium vibrate. Which diagram best represents the characteristics of a sound wave?

(A) Diagram (1): particle vibration parallel to (left-right along) the direction of wave travel ✓
(B) Diagram (2): particle vibration perpendicular (up-down) to the direction of wave travel
(C) Diagram (3): particle vibration shown as circular motion
(D) Diagram (4): particle vibration shown as a curved, wavy motion

Worked solution

$\text{Sound is a longitudinal wave.}$

$\text{In a longitudinal wave, particle vibration is parallel to the direction of wave travel (compressions and rarefactions).}$

Answer: Diagram (1)

Sound waves are longitudinal, meaning the medium's particles vibrate back and forth parallel to the direction the wave travels, which matches diagram (1); the other diagrams show perpendicular (transverse), circular, or other non-longitudinal vibration patterns.

Question 44 · 2 pts

In a sample of gas, many excited hydrogen atoms are in the $n = 4$ energy level. What is the maximum number of different photon energies that can be emitted by these atoms as they return to the ground state?

(A) 6 ✓
(B) 5
(C) 3
(D) 4

Worked solution

$\text{Possible downward transitions from } n=4: 4\to3, 4\to2, 4\to1, 3\to2, 3\to1, 2\to1.$

$\text{Number of distinct transitions} = \binom{4}{2} = \frac{4!}{2!2!} = 6.$

Answer: 6

With electrons able to fall from n=4 down through each lower level in various single or multi-step paths, the number of distinct possible transitions (and thus distinct photon energies) between 4 levels is the combination of 4 levels taken 2 at a time, which equals 6.

Question 45 · 2 pts

Base your answers to questions 45 and 46 on the information below and on your knowledge of physics. Two students did an experiment to measure the acceleration of a freely falling object. One student dropped an object from rest. The other student measured the distance fallen by the object and the corresponding time of fall. The data for the dropped object are shown below. distance fallen by object = 2.4 meters time of fall = 0.71 second Based on the data for the dropped object, the experimental value calculated for the object's acceleration is

(A) 11 m/s$^2$
(B) 9.5 m/s$^2$ ✓
(C) 6.8 m/s$^2$
(D) 4.8 m/s$^2$

Worked solution

$d = \frac{1}{2}at^2$ object starts from rest

$a = \frac{2d}{t^2} = \frac{2(2.4\ \text{m})}{(0.71\ \text{s})^2}$

$a \approx 9.5\ \text{m/s}^2$

Answer: 9.5 m/s$^2$

Since the object starts from rest, the distance-time relation d = (1/2)at^2 can be solved for a using the measured distance and time.

Question 46 · 2 pts

The ideal value for the acceleration differs from the one obtained experimentally by the students. What is one possible cause of this discrepancy?

Two students did an experiment to measure the acceleration of a freely falling object. One student dropped an object from rest. The other student measured the distance fallen by the object and the corresponding time of fall. The data for the dropped object are shown below. distance fallen by object = 2.4 meters time of fall = 0.71 second

(A) The object was given some initial horizontal velocity.
(B) The force of gravity was much stronger outside the building than inside.
(C) Motion formulas should not be used in an experimental setting.
(D) There may have been errors in the measurement of distance and/or time. ✓

Worked solution

$\text{Experimental error in measured quantities is the most common source of a discrepancy between measured and ideal (9.81 m/s}^2\text{) values.}$

Answer: There may have been errors in the measurement of distance and/or time.

Small errors in timing and distance measurement during a real experiment naturally cause the calculated acceleration to differ from the accepted value of g.

Question 47 · 2 pts

Which graph represents the motion of an object falling freely from rest near the surface of the Moon?

(A) Graph (1): Distance vs. Time, a straight line with constant positive slope through the origin
(B) Graph (2): Distance vs. Time, a straight line with constant negative slope
(C) Graph (3): Distance vs. Time, a curve starting at the origin that gets steeper with time (concave up) ✓
(D) Graph (4): Distance vs. Time, a curve starting at the origin that levels off with time (concave down)

Worked solution

$d = \frac{1}{2}at^2$

$\text{Since } d \propto t^2, \text{ the distance-time graph must be a parabola that curves upward (steepens) with time.}$

Answer: Graph (3): the concave-up curve, since distance increases as d = (1/2)at^2 for constant acceleration starting from rest

Free fall from rest is uniformly accelerated motion, so distance grows with the square of time, giving a concave-up curve rather than a straight line or a leveling-off curve.

Question 48 · 2 pts

The diagrams below represent the initial velocities, $v_i$, of four identical projectiles launched from level ground at various angles above the horizontal. Which projectile will have the longest time of flight? [Neglect friction.]

(A) Projectile (1): $v_i = 10.$ m/s at 30°
(B) Projectile (2): $v_i = 10.$ m/s at 45°
(C) Projectile (3): $v_i = 9.0$ m/s at 45°
(D) Projectile (4): $v_i = 9.0$ m/s at 60° ✓

Worked solution

$v_{iy} = v_i \sin\theta$

$(1):\ 10\sin30^\circ = 5.0\ \text{m/s} \quad (2):\ 10\sin45^\circ \approx 7.1\ \text{m/s}$

$(3):\ 9.0\sin45^\circ \approx 6.4\ \text{m/s} \quad (4):\ 9.0\sin60^\circ \approx 7.8\ \text{m/s}$

$t = \frac{2v_{iy}}{g}, \text{ so the largest } v_{iy} \text{ (projectile 4) gives the longest time of flight.}$

Answer: Projectile (4): $v_i = 9.0$ m/s at 60°

Time of flight depends only on the vertical component of initial velocity; projectile (4) has the greatest vertical component despite not having the greatest speed.

Question 49 · 2 pts

Four forces act on a crate on a level floor, as shown in the diagram below: a normal force $F_N = 150$ N upward, a friction force $F_f = 50$ N to the left, an applied rope force $F_{rope} = 100$ N to the right, and gravity $F_g = 150$ N downward. At the instant shown, the crate is

(A) accelerating to the right ✓
(B) accelerating to the left
(C) moving at constant velocity to the right
(D) remaining at rest

Worked solution

$F_{net,vertical} = F_N - F_g = 150\ \text{N} - 150\ \text{N} = 0\ \text{N}$

$F_{net,horizontal} = F_{rope} - F_f = 100\ \text{N} - 50\ \text{N} = 50\ \text{N, to the right}$

$\text{A nonzero net horizontal force means the crate is accelerating to the right.}$

Answer: accelerating to the right

Vertical forces balance, but the horizontal forces do not, leaving a net 50 N force to the right, so the crate accelerates rather than moving at constant velocity or remaining at rest.

Question 50 · 2 pts

The graph below represents the relationship between the potential energy stored in a spring, $PE_s$, and the elongation of the spring, $x$. The graph is a curve starting at the origin passing through approximately (0.0200 m, 0.0500 J) and (0.0400 m, 0.200 J). The value of the spring constant is

(A) 5.00 N/m
(B) 10.0 N/m
(C) 125 N/m
(D) 250. N/m ✓

Worked solution

$PE_s = \frac{1}{2}kx^2$

$k = \frac{2PE_s}{x^2} = \frac{2(0.200\ \text{J})}{(0.0400\ \text{m})^2}$

$k = 250.\ \text{N/m}$

Answer: 250. N/m

Using the point (0.0400 m, 0.200 J) from the parabolic PE_s vs. x graph and the elastic potential energy formula, the spring constant is found to be 250. N/m.

Question 51 · 2 pts · constructed response

Planet Nede has a mass of $6.50 \times 10^{23}$ kilograms and a radius of $2.96 \times 10^{6}$ meters. The magnitude of the gravitational force of attraction exerted on a 5.00-kilogram object by planet Nede is 7.15 newtons when the object is located at a position $5.50 \times 10^{6}$ meters from the center of Nede. Calculate the gravitational field strength of planet Nede at the position of the object. [Show all work, including the equation and substitution with units.]

Model answer: 1.43 N/kg

Worked solution

$g = \frac{F_g}{m}$ gravitational field strength equals force per unit mass

$g = \frac{7.15\ \text{N}}{5.00\ \text{kg}}$

$g = 1.43\ \text{N/kg}$

Answer: 1.43 N/kg

The gravitational field strength at a point equals the gravitational force per unit mass experienced by an object at that point, so dividing the given force by the given mass gives g directly (mass and radius of Nede are extra/unneeded data).

Question 53 · 2 pts · constructed response

A toy car, initially traveling at 1.0 meter per second, uniformly accelerates to a speed of 4.0 meters per second as it travels down a 5.0-meter-long slope. Calculate the time required for the toy car to travel the 5.0 meters from the top of the slope to the bottom. [Show all work, including the equation and substitution with units.]

Model answer: 2.0 s

Worked solution

$d = \left(\frac{v_i + v_f}{2}\right)t$

$5.0\ \text{m} = \left(\frac{1.0\ \text{m/s} + 4.0\ \text{m/s}}{2}\right)t = (2.5\ \text{m/s})t$

$t = 2.0\ \text{s}$

Answer: 2.0 s

For uniform acceleration, the average velocity is the mean of the initial and final velocities, so distance divided by average velocity gives the time of travel.

Question 55 · 2 pts · constructed response

Calculate the resistance of an incandescent lightbulb that operates at 20. watts of power when connected to a 12-volt battery. [Show all work, including the equation and substitution with units.]

Model answer: 7.2 ohms

Worked solution

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

$R = \frac{V^2}{P} = \frac{(12\ \text{V})^2}{20.\ \text{W}}$

$R = 7.2\ \Omega$

Answer: 7.2 ohms

Rearranging the power formula P = V^2/R for resistance and substituting the given voltage and power gives the bulb's resistance.

Question 57 · 1 pt · constructed response

Base your answers to questions 57 through 59 on the information and graph below and on your knowledge of physics. The motion of a car traveling along a straight road is represented in the graph below (Speed (m/s) on the vertical axis from 0 to 12.0, Time (s) on the horizontal axis from 0 to 10.0). The graph is a straight line starting at a speed of 4.0 m/s at t = 0 and rising steadily to about 11.5 m/s at t = 10.0 s. Determine the average speed of the car from t = 0 to t = 8.0 seconds.

Model answer: 7.0 m/s

Worked solution

$\text{From the graph: } v(0) = 4.0\ \text{m/s}, \quad v(8.0\ \text{s}) = 10.0\ \text{m/s}$

$\bar{v} = \frac{v_i + v_f}{2} = \frac{4.0 + 10.0}{2}\ \text{m/s}$

$\bar{v} = 7.0\ \text{m/s}$

Answer: 7.0 m/s

Because the speed-time graph is a straight line (constant acceleration), the average speed over any interval equals the arithmetic mean of the initial and final speeds read from the graph.

Question 58 · 2 pts · constructed response

The graph below shows the speed of a car traveling along a straight road as a function of time. Calculate the magnitude of the acceleration of the car from t = 0 to t = 8.0 seconds. [Show all work, including the equation and substitution with units.]

Model answer: 0.75 m/s^2

Worked solution

$\text{From the graph: } v_i = 4.0\ \text{m/s at } t=0,\quad v_f = 10.0\ \text{m/s at } t = 8.0\ \text{s}$

$a = \frac{v_f - v_i}{t} = \frac{10.0\ \text{m/s} - 4.0\ \text{m/s}}{8.0\ \text{s}}$

$a = 0.75\ \text{m/s}^2$

Answer: 0.75 m/s^2

The acceleration equals the slope of the speed-time graph, found from the change in speed over the change in time between the two given instants.

Question 60 · 1 pt · constructed response

A light ray is traveling through a transparent medium at a speed of $1.75 \times 10^{8}$ meters per second. Determine the absolute index of refraction of this medium for this light ray.

Model answer: 1.71

Worked solution

$n = \frac{c}{v}$

$n = \frac{3.00 \times 10^{8}\ \text{m/s}}{1.75 \times 10^{8}\ \text{m/s}}$

$n \approx 1.71$

Answer: 1.71

The absolute index of refraction is the ratio of the speed of light in a vacuum to the speed of light in the medium.

Question 61 · 1 pt · constructed response

As shown in the diagram below, a ray of light traveling through medium X is incident upon the surface of medium Y. The absolute index of refraction of medium Y is greater than the absolute index of refraction of medium X. On the diagram in your answer booklet, use a straightedge to draw a ray that could represent the path of the light in medium Y.

Model answer: The refracted ray in medium Y bends toward the normal, continuing into medium Y at an angle smaller than the angle of incidence in medium X.

Worked solution

$n_X \sin\theta_X = n_Y \sin\theta_Y$

$\text{Since } n_Y > n_X, \text{ it follows that } \sin\theta_Y < \sin\theta_X, \text{ so } \theta_Y < \theta_X$

$\text{The ray drawn in medium Y must therefore bend toward the normal line relative to the incident ray.}$

Answer: The refracted ray in medium Y bends toward the normal, continuing into medium Y at an angle smaller than the angle of incidence in medium X.

Light slows down when entering a medium with a higher index of refraction, which by Snell's Law causes the ray to bend toward the normal rather than away from it.

Question 62 · 2 pts · constructed response

A tungsten wire has a cross-sectional area of $5.03 \times 10^{-7}$ meter squared and has a resistance of $1.21 \times 10^{-2}$ ohm when operated at 20°C. Calculate the length of this wire. [Show all work, including the equation and substitution with units.]

Model answer: approximately 0.109 m

Worked solution

$R = \frac{\rho L}{A}$

$L = \frac{RA}{\rho} = \frac{(1.21 \times 10^{-2}\ \Omega)(5.03 \times 10^{-7}\ \text{m}^2)}{5.6 \times 10^{-8}\ \Omega\cdot\text{m}}$ resistivity of tungsten at 20 C, 5.6 x 10^-8 ohm m, from the Reference Tables for Physical Setting/Physics

$L \approx 0.109\ \text{m}$

Answer: approximately 0.109 m

Rearranging the resistivity formula R = ρL/A for length and substituting the given resistance, area, and the tabulated resistivity of tungsten gives the wire's length.

Question 64 · 2 pts · constructed response

A block is hanging motionless from a vertical spring, having caused the spring to elongate from its unstretched length of 0.200 meter to a length of 0.500 meter. The spring has a spring constant of 250. newtons per meter. Calculate the magnitude of the force exerted by the block on the spring. [Show all work, including the equation and substitution with units.]

Model answer: 75 N

Worked solution

$x = L_{stretched} - L_{unstretched} = 0.500\ \text{m} - 0.200\ \text{m} = 0.300\ \text{m}$

$F_s = kx = (250.\ \text{N/m})(0.300\ \text{m})$

$F_s = 75\ \text{N}$

Answer: 75 N

By Newton's third law, the force the hanging block exerts on the spring is equal in magnitude to the spring force found from Hooke's Law using the spring's elongation.

Question 66 · 1 pt · constructed response

Base your answers to questions 66 through 70 on the information and data table below and on your knowledge of physics. In an experiment, a resistor is connected to a source of varying potential difference. A voltmeter and an ammeter are included in the circuit so that the current can be measured at different potential difference settings. The data table below shows the potential differences and currents measured. Data Table | Potential Difference (V) | Current (A) | | --- | --- | | 0.0 | 0.00 | | 2.0 | 0.26 | | 3.0 | 0.35 | | 5.0 | 0.63 | | 8.0 | 0.98 | Directions (66-67): Using the information in the data table, construct a graph on the grid in your answer booklet following the directions below. Plot the data points for current versus potential difference.

Model answer: Plot the five points (0.0, 0.00), (2.0, 0.26), (3.0, 0.35), (5.0, 0.63), and (8.0, 0.98) with Potential Difference (V) on the horizontal axis and Current (A) on the vertical axis.

Worked solution

$\text{Label the horizontal axis 'Potential Difference (V)' and the vertical axis 'Current (A)', choosing scales that use most of the grid.}$

$\text{Plot each ordered pair from the data table: } (0.0, 0.00),\ (2.0, 0.26),\ (3.0, 0.35),\ (5.0, 0.63),\ (8.0, 0.98).$

Answer: Plot the five points (0.0, 0.00), (2.0, 0.26), (3.0, 0.35), (5.0, 0.63), and (8.0, 0.98) with Potential Difference (V) on the horizontal axis and Current (A) on the vertical axis.

Full credit requires plotting all five given data points accurately on appropriately scaled and labeled axes, with potential difference as the independent (horizontal) variable and current as the dependent (vertical) variable.

Question 67 · 1 pt · constructed response

Base your answers to questions 67 through 70 on the information and data table below and on your knowledge of physics. In an experiment, a resistor is connected to a source of varying potential difference. A voltmeter and an ammeter are included in the circuit so that the current can be measured at different potential difference settings. The data table below shows the potential differences and currents measured. Data Table | Potential Difference (V) | Current (A) | | --- | --- | | 0.0 | 0.00 | | 2.0 | 0.26 | | 3.0 | 0.35 | | 5.0 | 0.63 | | 8.0 | 0.98 | Directions (66-67): Using the information in the data table, construct a graph on the grid in your answer booklet following the directions below. 67 Draw the line of best fit.

Model answer: A single straight line of best fit, passing near the origin (0,0) and through the general trend of the plotted points, with roughly equal numbers of points scattered above and below the line.

Worked solution

$V \propto I$ The data show current increasing steadily with potential difference, consistent with Ohm's Law (a linear, directly-proportional relationship).

$\text{Line of best fit: straight line through/near (0,0), extending to about (8.0, 0.98)}$

Answer: A single straight line of best fit, passing near the origin (0,0) and through the general trend of the plotted points, with roughly equal numbers of points scattered above and below the line.

Since the current-vs-potential-difference data is linear (Ohm's Law), the line of best fit is a single straight line drawn through the cluster of plotted points and close to the origin.

Question 68 · 2 pts · constructed response

Base your answer on the data table below and on your knowledge of physics. | Potential Difference (V) | Current (A) | | --- | --- | | 0.0 | 0.00 | | 2.0 | 0.26 | | 3.0 | 0.35 | | 5.0 | 0.63 | | 8.0 | 0.98 | Calculate the slope of the line of best fit on a graph of current versus potential difference for this data. [Show all work, including the equation and substitution with units.]

Model answer: slope ≈ 0.12 A/V

Worked solution

$\text{slope} = \frac{\Delta I}{\Delta V}$

$\text{slope} = \frac{0.98\ \text{A} - 0.00\ \text{A}}{8.0\ \text{V} - 0.0\ \text{V}}$

$\text{slope} = \frac{0.98\ \text{A}}{8.0\ \text{V}} \approx 0.12\ \text{A/V}$

Answer: slope ≈ 0.12 A/V

The slope of the current-versus-potential-difference line equals the change in current divided by the change in potential difference. Using the endpoints (0.0 V, 0.00 A) and (8.0 V, 0.98 A), slope = 0.98 A / 8.0 V, which is about 0.12 ampere per volt.

Question 70 · 1 pt · constructed response

Base your answers to questions 67 through 70 on the information and data table below and on your knowledge of physics. Data Table | Potential Difference (V) | Current (A) | | --- | --- | | 0.0 | 0.00 | | 2.0 | 0.26 | | 3.0 | 0.35 | | 5.0 | 0.63 | | 8.0 | 0.98 | 70 Determine the resistance of the resistor.

Model answer: R ≈ 8.2 Ω

Worked solution

$\text{slope} = \frac{I}{V} = \frac{1}{R} \implies R = \frac{1}{\text{slope}} = \frac{V}{I}$

$R = \frac{8.0\ \text{V}}{0.98\ \text{A}} \approx 8.2\ \Omega$

Answer: R ≈ 8.2 Ω

Because I versus V is linear through the origin (Ohm's Law), the resistance equals the reciprocal of the slope, or equivalently V/I evaluated at any point on the line.

Question 71 · 1 pt · constructed response

Model answer: q ≈ 3.2 × 10⁻¹⁹ C

Worked solution

$q = 2e$

$q = 2(1.6\times10^{-19}\ \text{C}) = 3.2\times10^{-19}\ \text{C}$

Answer: q ≈ 3.2 × 10⁻¹⁹ C

The alpha particle's charge is two elementary charges, so its magnitude is twice the fundamental charge of an electron/proton.

Question 72 · 1 pt · constructed response

Base your answers to questions 71 through 75 on the information and diagram below and on your knowledge of physics. An alpha particle with a charge of +2 elementary charges is moving toward two oppositely charged parallel plates, as shown in the diagram below. The magnitude of the electric field strength between the plates is 5.0 × 10^4 newtons per coulomb. (Diagram: the alpha particle moves downward toward the gap between two vertical parallel plates; the left plate is marked with minus signs (negative plate) and the right plate is marked with plus signs (positive plate).) 72 On the diagram in your answer booklet, draw at least three field lines to show the direction of the electric field in the space between the charged plates.

Model answer: At least three straight, evenly spaced, horizontal field lines drawn pointing from the positive plate (right) toward the negative plate (left).

Worked solution

$\vec{E}: \text{points from + plate toward - plate}$

Answer: At least three straight, evenly spaced, horizontal field lines drawn pointing from the positive plate (right) toward the negative plate (left).

Electric field lines always point away from positive charge and toward negative charge, so between the plates they run horizontally from the positive plate to the negative plate.

Question 73 · 1 pt · constructed response

Base your answers to questions 71 through 75 on the information and diagram below and on your knowledge of physics. An alpha particle with a charge of +2 elementary charges is moving toward two oppositely charged parallel plates, as shown in the diagram below. The magnitude of the electric field strength between the plates is 5.0 × 10^4 newtons per coulomb. (Diagram: the alpha particle moves downward toward the gap between two vertical parallel plates; the left plate is marked with minus signs (negative plate) and the right plate is marked with plus signs (positive plate).) 73 What is the direction of the electrostatic force exerted on the alpha particle as it passes between the plates?

Model answer: The force is directed horizontally from the positive plate toward the negative plate (toward the left, i.e., in the same direction as the electric field).

Worked solution

$\vec{F} = q\vec{E}, \ q>0 \implies \vec{F} \parallel \vec{E}$

Answer: The force is directed horizontally from the positive plate toward the negative plate (toward the left, i.e., in the same direction as the electric field).

Since the alpha particle is positively charged, the electrostatic force on it points in the same direction as the electric field, which runs from the positive plate toward the negative plate.

Question 74 · 2 pts · constructed response

An alpha particle with a charge of +2 elementary charges is moving toward two oppositely charged parallel plates. The magnitude of the electric field strength between the plates is $5.0 \times 10^{4}$ newtons per coulomb. Calculate the magnitude of the electrostatic force exerted on the alpha particle by the electric field as the alpha particle passes through the electric field. [Show all work, including the equation and substitution with units.]

Model answer: F ≈ 1.6 × 10⁻¹⁴ N

Worked solution

$F = qE$

$F = (3.2\times10^{-19}\ \text{C})(5.0\times10^{4}\ \text{N/C})$

$F = (3.2\times10^{-19})(5.0\times10^{4}) = 1.6\times10^{-14}\ \text{N}$

Answer: F ≈ 1.6 × 10⁻¹⁴ N

The electrostatic force on the alpha particle is F = qE. Substituting the charge q = 3.2 x 10^-19 C and the field strength E = 5.0 x 10^4 N/C gives F = 1.6 x 10^-14 N.

Question 76 · 2 pts · constructed response

Base your answers to questions 76 through 80 on the information below and on your knowledge of physics. Two students investigate transverse waves using a long, stretched spring. One student holds one end of the spring stationary, while the other student produces 12 waves every 4.0 seconds. The waves have a uniform amplitude of 0.20 meter and the distance between two adjacent crests is 0.40 meter. 76-77 On the diagram in your answer booklet, draw at least one complete wave produced in the spring.

Model answer: A smooth transverse waveform showing at least one complete cycle, drawn with a wavelength (crest-to-crest distance) of 0.40 m and an amplitude of 0.20 m (each crest 0.20 m above and each trough 0.20 m below the rest position).

Worked solution

$\lambda = 0.40\ \text{m (distance between adjacent crests)}$

Answer: A smooth transverse waveform showing at least one complete cycle, drawn with a wavelength (crest-to-crest distance) of 0.40 m and an amplitude of 0.20 m (each crest 0.20 m above and each trough 0.20 m below the rest position).

Full credit requires a smooth transverse wave of at least one complete cycle with both the correct wavelength (0.40 m crest to crest) and the correct amplitude (0.20 m above and below the rest axis).

Question 78 · 1 pt · constructed response

Model answer: f = 3.0 Hz

Worked solution

$f = \frac{\text{number of waves}}{\text{time}}$

$f = \frac{12}{4.0\ \text{s}} = 3.0\ \text{Hz}$

Answer: f = 3.0 Hz

Frequency is the number of complete waves produced per unit time.

Question 79 · 2 pts · constructed response

Two students investigate transverse waves using a long, stretched spring. One student holds one end of the spring stationary, while the other student produces 12 waves every 4.0 seconds. The waves have a uniform amplitude of 0.20 meter and the distance between two adjacent crests is 0.40 meter. Calculate the speed of the waves. [Show all work, including the equation and substitution with units.]

Model answer: v = 1.2 m/s

Worked solution

$v = f\lambda$

$v = (3.0\ \text{Hz})(0.40\ \text{m})$

$v = (3.0\ \text{Hz})(0.40\ \text{m}) = 1.2\ \text{m/s}$

Answer: v = 1.2 m/s

Wave speed is the product of frequency and wavelength, v = f times lambda. With f = 3.0 Hz and lambda = 0.40 m, v = 1.2 m/s.

Question 81 · 1 pt · constructed response

Base your answers to questions 81 through 85 on the information below and on your knowledge of physics. Alpha particles are stable decay products of many nuclear disintegrations. An alpha particle consists of two protons and two neutrons and has a mass of 6.644 × 10⁻²⁷ kilogram. When the individual masses of two protons and two neutrons are added, the sum is 6.695 × 10⁻²⁷ kilogram. The difference between the mass of an alpha particle and the total mass of its four individual component particles is called the mass defect. 81 Determine the mass defect in kilograms when two protons and two neutrons combine to form an alpha particle.

Model answer: Δm ≈ 5.1 × 10⁻²⁹ kg

Worked solution

$\Delta m = m_{\text{components}} - m_{\alpha}$

$\Delta m = 6.695\times10^{-27}\ \text{kg} - 6.644\times10^{-27}\ \text{kg} = 5.1\times10^{-29}\ \text{kg}$

Answer: Δm ≈ 5.1 × 10⁻²⁹ kg

The mass defect is the difference between the summed mass of the four separate nucleons and the actual mass of the bound alpha particle.

Question 82 · 2 pts · constructed response

Base your answers to questions 81 through 85 on the information below and on your knowledge of physics. Alpha particles are stable decay products of many nuclear disintegrations. An alpha particle consists of two protons and two neutrons and has a mass of 6.644 × 10⁻²⁷ kilogram. When the individual masses of two protons and two neutrons are added, the sum is 6.695 × 10⁻²⁷ kilogram. The difference between the mass of an alpha particle and the total mass of its four individual component particles is called the mass defect. Calculate the total amount of energy in joules that would result from the complete conversion of this mass defect to energy. [Show all work, including the equation and substitution with units.]

Model answer: E ≈ 4.6 × 10⁻¹² J

Worked solution

$E = \Delta m\, c^2$

$E = (5.1\times10^{-29}\ \text{kg})(3.00\times10^{8}\ \text{m/s})^2$

$E = (5.1\times10^{-29})(9.00\times10^{16}) \approx 4.6\times10^{-12}\ \text{J}$

Answer: E ≈ 4.6 × 10⁻¹² J

The energy equivalent of the mass defect is E = (mass defect) times c^2. Substituting a mass defect of 5.1 x 10^-29 kg and c = 3.00 x 10^8 m/s gives E of about 4.6 x 10^-12 J.

Question 84 · 1 pt · constructed response

Base your answers to questions 81 through 85 on the information below and on your knowledge of physics. Alpha particles are stable decay products of many nuclear disintegrations. An alpha particle consists of two protons and two neutrons and has a mass of 6.644 × 10⁻²⁷ kilogram. When the individual masses of two protons and two neutrons are added, the sum is 6.695 × 10⁻²⁷ kilogram. The difference between the mass of an alpha particle and the total mass of its four individual component particles is called the mass defect. 84 Determine the total amount of energy that would result from the complete conversion of this mass defect to energy in megaelectronvolts (MeV).

Model answer: E ≈ 29 MeV

Worked solution

$1\ \text{MeV} = 1.60\times10^{-13}\ \text{J}$

$E = \frac{4.6\times10^{-12}\ \text{J}}{1.60\times10^{-13}\ \text{J/MeV}} \approx 29\ \text{MeV}$

Answer: E ≈ 29 MeV

Converting the energy from joules to megaelectronvolts using the conversion factor 1 MeV = 1.60 × 10⁻¹³ J.

Question 85 · 1 pt · constructed response

Base your answers to questions 81 through 85 on the information below and on your knowledge of physics. Alpha particles are stable decay products of many nuclear disintegrations. An alpha particle consists of two protons and two neutrons and has a mass of 6.644 × 10⁻²⁷ kilogram. When the individual masses of two protons and two neutrons are added, the sum is 6.695 × 10⁻²⁷ kilogram. The difference between the mass of an alpha particle and the total mass of its four individual component particles is called the mass defect. 85 What fundamental interaction is primarily responsible for holding protons and neutrons together in an alpha particle?

Model answer: The strong nuclear force (the strong interaction).

Worked solution

$\text{Strong nuclear force} \gg \text{Coulomb repulsion between protons at nuclear distances}$

Answer: The strong nuclear force (the strong interaction).

The strong nuclear force is the fundamental interaction that binds protons and neutrons together in a nucleus, overcoming the electrostatic repulsion between protons.

June 2024 Physics Regents — Answer Key & Worked Solutions

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