June 2025 Physics Regents — Answer Key & Worked Solutions

Every question from the June 2025 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.

Practice Physics free Physics guide

These are real questions from the June 2025 Physics New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

Base your answers to questions 1 and 2 on the information below and on your knowledge of physics. A car, initially moving at 20.0 meters per second, travels 200. meters while accelerating uniformly to a speed of 30.0 meters per second. The magnitude of the acceleration of the car is

(A) 0.0500 m/s$^2$
(B) 1.25 m/s$^2$ ✓
(C) 2.50 m/s$^2$
(D) 3.13 m/s$^2$

Worked solution

$v^2 = v_0^2 + 2ad$ Use kinematics equation without time

$a = \frac{v^2 - v_0^2}{2d} = \frac{30.0^2 - 20.0^2}{2(200.)}$

$a = \frac{900 - 400}{400} = 1.25\ \text{m/s}^2$

Answer: 1.25 m/s$^2$

Using the kinematics equation relating velocity, acceleration, and displacement (no time needed) gives a = 1.25 m/s^2.

Question 2 · 2 pts

(A) 6.67 s
(B) 8.00 s ✓
(C) 10.0 s
(D) 20.0 s

Worked solution

$a = 1.25\ \text{m/s}^2$ From question 1

$t = \frac{v - v_0}{a} = \frac{30.0-20.0}{1.25}$

$t = 8.00\ \text{s}$

Answer: 8.00 s

Using the acceleration found in question 1, solving for time with v = v0 + at gives 8.00 s.

Question 3 · 2 pts

A car traveling at 12 meters per second north accelerates in a straight line at 3.0 meters per second squared north for 4.0 seconds. How far north does the car travel during the 4.0 seconds it accelerates?

(A) 24 m
(B) 48 m
(C) 54 m
(D) 72 m ✓

Worked solution

$d = v_0 t + \frac{1}{2}at^2$

$d = (12)(4.0) + \frac{1}{2}(3.0)(4.0)^2$

$d = 48 + 24 = 72\ \text{m}$

Answer: 72 m

Applying the displacement kinematics equation with the given initial velocity, acceleration, and time yields 72 m.

Question 4 · 2 pts

A block weighing 10. newtons rests on an inclined plane that makes an angle of 30° with the horizontal. The magnitude of the component of the block's weight perpendicular to the plane is closest to

(A) 5.0 N
(B) 5.8 N
(C) 8.7 N ✓
(D) 10. N

Worked solution

$F_{\perp} = W\cos\theta$ Component of weight perpendicular to the incline

$F_{\perp} = (10.\ \text{N})\cos(30^\circ)$

$F_{\perp} = (10.)(0.866) \approx 8.7\ \text{N}$

Answer: 8.7 N

The perpendicular component of the weight equals the weight times the cosine of the incline angle, giving about 8.7 N.

Question 5 · 2 pts

A sailor is near the top of a sailboat mast. The sailboat is traveling at a constant velocity of 5.0 meters per second west. The sailor drops a screwdriver that falls freely for 1.0 second before hitting the deck of the sailboat. [Neglect friction.] Where does the screwdriver land?

(A) directly below the sailor ✓
(B) 5.0 m behind the sailor
(C) 5.0 m in front of the sailor
(D) 49 m behind the sailor

Worked solution

$v_{x,\text{screwdriver}} = v_{x,\text{boat}} = 5.0\ \text{m/s west}$ Screwdriver retains the boat's horizontal velocity (inertia)

$\text{In the boat's reference frame, horizontal displacement} = 0$

Answer: directly below the sailor

Since the screwdriver shares the sailboat's constant horizontal velocity when released, it continues moving with the boat and lands directly below the sailor, same as in the boat's reference frame.

Question 6 · 2 pts

A marble is projected horizontally from a fixed height above a level floor. If the marble is projected horizontally again from the same fixed height, but with twice the initial speed, its time of flight will be

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

Worked solution

$h = \frac{1}{2}gt^2$ Time of flight depends only on the drop height, not horizontal speed

$t = \sqrt{\frac{2h}{g}}$

Answer: unchanged

Time of flight for projectile motion depends only on the height dropped and gravitational acceleration, not on the horizontal launch speed, so it remains unchanged.

Question 7 · 2 pts

Which animal has the greatest inertia?

(A) an 8.0-kg meerkat moving at 2 m/s
(B) a 350-kg zebra moving at 18 m/s
(C) a 190-kg lion moving at 0.5 m/s
(D) an 1800-kg hippopotamus at rest ✓

Worked solution

$\text{Inertia} \propto \text{mass only, independent of velocity}$

$1800\ \text{kg} > 350\ \text{kg} > 190\ \text{kg} > 8.0\ \text{kg}$

Answer: an 1800-kg hippopotamus at rest

Inertia depends only on mass, not speed, so the 1800-kg hippopotamus has the greatest inertia regardless of being at rest.

Question 8 · 2 pts

A cart is rolling along in a straight line on an inclined ramp. The cart is in equilibrium when it is rolling up or down the ramp with

(A) decreasing speed
(B) increasing speed
(C) constant acceleration
(D) constant speed ✓

Worked solution

$\text{Equilibrium} \Rightarrow F_{net} = 0 \Rightarrow a = 0$

$a = 0 \Rightarrow \text{constant velocity (constant speed in a straight line)}$

Answer: constant speed

An object in equilibrium has zero net force and therefore zero acceleration, meaning it moves at constant speed.

Question 9 · 2 pts

A 3.5-kilogram bowling ball and a 7.0-kilogram bowling ball are in free fall near the surface of Earth. Compared to the magnitude of the acceleration of the 3.5-kilogram ball, the magnitude of the acceleration of the 7.0-kg ball is

(A) half as great
(B) twice as great
(C) the same ✓
(D) four times greater

Worked solution

$a_{free\ fall} = g = 9.81\ \text{m/s}^2$ Independent of mass

Answer: the same

All objects in free fall near Earth's surface experience the same gravitational acceleration regardless of mass, so the two balls have the same acceleration.

Question 10 · 2 pts

A 600-newton student is standing in an elevator. The elevator floor exerts an upward force of 400 newtons on the student. The student has

(A) a constant velocity directed downward
(B) a constant velocity directed upward
(C) an acceleration directed downward ✓
(D) an acceleration directed upward

Worked solution

$F_{net} = F_{normal} - W = 400\ \text{N} - 600\ \text{N} = -200\ \text{N}$ Negative indicates net force downward

$a = \frac{F_{net}}{m} \neq 0 \Rightarrow \text{acceleration is downward}$

Answer: an acceleration directed downward

Since the upward normal force (400 N) is less than the student's weight (600 N), the net force and resulting acceleration are directed downward.

Question 11 · 2 pts

One end of a spring is held stationary and a 40.-newton force applied to the free end of the spring elongates the spring 0.20 meter from its unstretched length. What is the magnitude of the total force required to stretch the spring 0.80 meter from its unstretched length if the spring's elastic limit has not been exceeded?

(A) 10. N
(B) 20. N
(C) 120 N
(D) 160 N ✓

Worked solution

$k = \frac{F}{x} = \frac{40.\ \text{N}}{0.20\ \text{m}} = 200\ \text{N/m}$ Hooke's law spring constant

$F = kx = (200\ \text{N/m})(0.80\ \text{m})$

$F = 160\ \text{N}$

Answer: 160 N

Using Hooke's law, the spring constant found from the initial stretch is used to calculate the force needed for the larger elongation, giving 160 N.

Question 12 · 2 pts

The engine of a toy train is traveling at constant speed on a horizontal, circular track. If the speed of the engine is doubled, the magnitude of the centripetal acceleration of the engine is

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

Worked solution

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

$a_c' = \frac{(2v)^2}{r} = \frac{4v^2}{r} = 4a_c$

Answer: quadrupled

Centripetal acceleration is proportional to the square of the speed, so doubling the speed quadruples the centripetal acceleration.

Question 13 · 2 pts

The strength of an electric field has a magnitude of $3.5 \times 10^5$ newtons per coulomb. What is the magnitude of the electrostatic force on a $5.3 \times 10^{-6}$-coulomb charge in this field?

(A) $1.5 \times 10^{-11}$ N
(B) 1.9 N ✓
(C) 19 N
(D) $6.6 \times 10^{10}$ N

Worked solution

$F = qE$

$F = (5.3 \times 10^{-6}\ \text{C})(3.5 \times 10^5\ \text{N/C})$

$F \approx 1.9\ \text{N}$

Answer: 1.9 N

Multiplying the charge by the electric field strength gives the electrostatic force, approximately 1.9 N.

Question 14 · 2 pts

A car is traveling on a level highway at a speed of 15.0 meters per second. A braking force of magnitude $3.00 \times 10^3$ newtons brings the car to a stop in 10.0 seconds. The mass of the car is

(A) $1.50 \times 10^3$ kg
(B) $2.00 \times 10^3$ kg ✓
(C) $3.00 \times 10^3$ kg
(D) $4.50 \times 10^3$ kg

Worked solution

$a = \frac{\Delta v}{t} = \frac{0 - 15.0}{10.0} = -1.50\ \text{m/s}^2$ Magnitude 1.50 m/s^2

$m = \frac{F}{a} = \frac{3.00\times10^3\ \text{N}}{1.50\ \text{m/s}^2}$

$m = 2.00 \times 10^3\ \text{kg}$

Answer: $2.00 \times 10^3$ kg

Finding the deceleration from the given time and speed change, then applying Newton's second law (m = F/a), gives a mass of 2.00 x 10^3 kg.

Question 15 · 2 pts

A 5.00-kilogram cart traveling east at a speed of 15.0 meters per second collides with a 10.0 kilogram cart traveling west at a speed of 15.0 meters per second. The total momentum of the two-cart system after the collision is

(A) 75.0 kg $\cdot$ m/s west ✓
(B) 75.0 kg $\cdot$ m/s east
(C) 225 kg $\cdot$ m/s west
(D) 225 kg $\cdot$ m/s east

Worked solution

$p_{before} = (5.00)(15.0)_{east} + (10.0)(15.0)_{west}$ Take east as positive

$p_{before} = 75.0 - 150. = -75.0\ \text{kg}\cdot\text{m/s}$

$|p_{after}| = |p_{before}| = 75.0\ \text{kg}\cdot\text{m/s west}$ Conservation of momentum

Answer: 75.0 kg $\cdot$ m/s west

By conservation of momentum, the total momentum after the collision equals the total momentum before, which is 75.0 kg·m/s directed west (the heavier cart's momentum dominates).

Question 16 · 2 pts

In the diagram below, a force acts to the right on a proton, P, in an electric field. To produce this force, the direction of the electric field at the position of the proton is toward point

(A) A
(B) B
(C) C ✓
(D) D

Worked solution

$\vec{F} = q\vec{E}$ For a positive charge, force and field point in the same direction

$\text{Proton (positive) has force pointing right, toward C} \Rightarrow \vec{E}\text{ also points toward C}$

Answer: C

For a positive charge like a proton, the electric force is in the same direction as the electric field, so since the force points right toward C, the field also points toward C.

Question 17 · 2 pts

While investigating static electricity, a student notices a negatively charged plastic rod attracts a small piece of paper. The student correctly concludes the charge of the paper

(A) must be negative
(B) must be positive
(C) could be negative or neutral
(D) could be positive or neutral ✓

Worked solution

$\text{Opposite charges attract; a neutral object can be attracted via induced polarization}$

$\text{Negative rod attracts paper} \Rightarrow \text{paper is positive or neutral (polarized)}$

Answer: could be positive or neutral

A charged object can attract either an oppositely charged object or a neutral object through induced polarization, so the paper could be positive or neutral, but not negative.

Question 18 · 2 pts

A 1.0-kilogram cart moving to the right at 1.0 meter per second is about to collide head on with a 2.0-kilogram cart moving to the left at 2.0 meters per second, both on a frictionless, horizontal surface. During the collision, the magnitude of the force exerted on the 1.0-kilogram cart by the 2.0-kilogram cart is F. What is the magnitude of the force exerted on the 2.0-kilogram cart by the 1.0-kilogram cart during the collision?

(A) F/4
(B) F/2
(C) F ✓
(D) 2F

Worked solution

$\vec{F}_{1\to2} = -\vec{F}_{2\to1}$ Newton's third law: action-reaction pair

$|\vec{F}_{1\to2}| = |\vec{F}_{2\to1}| = F$

Answer: F

By Newton's third law, the forces of an action-reaction pair are always equal in magnitude, so the force exerted on the 2.0-kg cart is also F.

Question 19 · 2 pts

Sunlight shines on solar panels on the roof of a house. The panels are hooked up to the house's electrical system, which is used to operate a microwave oven that heats up a cup of coffee. What energy conversions are taking place in this entire process?

(A) electromagnetic $\rightarrow$ electrical $\rightarrow$ electromagnetic $\rightarrow$ thermal ✓
(B) thermal $\rightarrow$ electromagnetic $\rightarrow$ nuclear $\rightarrow$ electrical
(C) mechanical $\rightarrow$ electrical $\rightarrow$ electromagnetic $\rightarrow$ chemical
(D) electromagnetic $\rightarrow$ thermal $\rightarrow$ mechanical $\rightarrow$ electrical

Worked solution

$\text{Sunlight (electromagnetic)} \rightarrow \text{solar panel produces electrical energy}$

$\text{Electrical energy} \rightarrow \text{microwave oven produces electromagnetic microwaves}$

$\text{Microwaves} \rightarrow \text{absorbed by coffee, converted to thermal energy}$

Answer: electromagnetic $\rightarrow$ electrical $\rightarrow$ electromagnetic $\rightarrow$ thermal

Sunlight (electromagnetic) is converted to electrical energy by the solar panel, which powers the microwave oven that generates electromagnetic microwaves, which are absorbed by the coffee as thermal energy.

Question 20 · 2 pts

Which graph best represents the relationship between the potential difference across a resistor and the resulting current through the resistor, for a resistor that obeys Ohm's law?

(A) Graph (1): a straight line with positive slope passing through the origin ✓
(B) Graph (2): a straight line with negative slope, decreasing from the vertical axis
(C) Graph (3): a curve that decreases steeply and levels off, concave up, approaching the horizontal axis
(D) Graph (4): a horizontal straight line (constant potential difference)

Worked solution

$V = IR$ Ohm's law

$\text{For constant } R,\ V \propto I \Rightarrow \text{straight line through the origin}$

Answer: Graph (1): a straight line with positive slope passing through the origin

Ohm's law states V = IR, so for a fixed resistance, potential difference is directly proportional to current, producing a straight line through the origin (graph 1).

Question 21 · 2 pts

As shown in the diagram of a pendulum below, mass M is released from rest at point A and allowed to swing through point B to point C. [Neglect friction.] Compared to the kinetic energy of mass M at point B, the kinetic energy of mass M at point C is

(A) half as great
(B) twice as great ✓
(C) the same
(D) four times greater

Worked solution

$KE_B = mgh_B = mg(0.4\ \text{m})$ Height dropped from A to B

$KE_C = mgh_C = mg(0.8\ \text{m})$ Height dropped from A to C (twice that of A to B)

$\frac{KE_C}{KE_B} = \frac{0.8}{0.4} = 2$

Answer: twice as great

Since mechanical energy is conserved and C is twice as far below the starting point A as B is, the kinetic energy at C is twice the kinetic energy at B.

Question 22 · 2 pts

A student, running at 6 meters per second, slides to rest on a horizontal floor. As the student slides, the internal energy of the student-floor system

(A) decreases and the student's kinetic energy decreases
(B) increases and the student's kinetic energy decreases ✓
(C) decreases and the student's kinetic energy remains the same
(D) increases and the student's kinetic energy remains the same

Worked solution

$\text{Friction converts kinetic energy into internal (thermal) energy}$

$KE_{student} \downarrow \quad \Rightarrow \quad U_{internal} \uparrow$

Answer: increases and the student's kinetic energy decreases

Friction between the student and floor converts the student's kinetic energy into internal (thermal) energy, so kinetic energy decreases while internal energy of the system increases.

Question 23 · 2 pts

Which object will most likely produce a magnetic field?

(A) a stationary neutral object
(B) a moving neutral object
(C) a stationary charged object
(D) a moving charged object ✓

Worked solution

$\text{Magnetic fields are produced by moving electric charges (current).}$

$\text{A stationary charge produces only an electric field; a neutral object produces no field.}$

Answer: a moving charged object

Only moving charges constitute a current, and it is current (moving charge) that generates a magnetic field.

Question 24 · 2 pts

To reduce the electrical resistance of a copper wire, increase the wire's

(A) length
(B) temperature
(C) diameter ✓
(D) resistivity

Worked solution

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

$\text{Increasing the diameter increases the cross-sectional area } A, \text{ which decreases } R.$

Answer: diameter

Resistance is inversely proportional to cross-sectional area, so a larger diameter (larger area) lowers resistance; increasing length, temperature, or resistivity would all increase resistance.

Question 25 · 2 pts

A potential difference of 12 volts is applied across a resistor for 60. seconds. A constant current of 0.10 ampere flows through the resistor during this time. What is the power of the resistor?

(A) 1.2 W ✓
(B) 12 W
(C) 72 W
(D) 120 W

Worked solution

$P=IV$

$P=(0.10\ \text{A})(12\ \text{V})=1.2\ \text{W}$

Answer: 1.2 W

Power delivered to a resistor equals current times potential difference; the 60-second duration is not needed for this calculation.

Question 26 · 2 pts

What type of waves are sound waves?

(A) longitudinal waves ✓
(B) transverse waves
(C) electromagnetic waves
(D) ultraviolet waves

Worked solution

$\text{Sound waves are compressions and rarefactions of a medium, parallel to the direction of propagation.}$

Answer: longitudinal waves

Sound propagates through the vibration of particles parallel to the direction of wave travel, which is the definition of a longitudinal wave.

Question 27 · 2 pts

As waves travel through space, they transfer

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

Worked solution

$\text{Waves transport energy from one location to another without net transport of matter.}$

Answer: energy, only

A defining property of waves is that they carry energy through a medium (or space, for EM waves) without transporting mass.

Question 28 · 2 pts

The frequency of an orange light wave is $4.85 \times 10^{14}$ hertz. What is the wavelength of the wave in a vacuum?

(A) $1.00 \times 10^{-7}$ m
(B) $4.82 \times 10^{-7}$ m
(C) $5.96 \times 10^{-7}$ m
(D) $6.19 \times 10^{-7}$ m ✓

Worked solution

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

$\lambda=\frac{3.00\times10^{8}\ \text{m/s}}{4.85\times10^{14}\ \text{Hz}}=6.19\times10^{-7}\ \text{m}$

Answer: $6.19 \times 10^{-7}$ m

All electromagnetic waves travel at speed c in a vacuum, so the wavelength is found by dividing c by the given frequency.

Question 29 · 2 pts

Which statement describes mechanical and electromagnetic waves?

(A) Mechanical waves all travel at the same speed in a vacuum, whereas the speeds of electromagnetic waves vary in a vacuum.
(B) Mechanical waves all have the same frequency, whereas electromagnetic waves all have the same period.
(C) Mechanical waves require a material medium through which to travel, whereas eletromagnetic waves can travel through a vacuum. ✓
(D) Mechanical waves are transverse waves, whereas electromagnetic waves are longitudinal waves.

Worked solution

$\text{Mechanical waves need particles of a medium to transmit a disturbance.}$

$\text{Electromagnetic waves are self-propagating oscillations of E and B fields and need no medium.}$

Answer: Mechanical waves require a material medium through which to travel, whereas eletromagnetic waves can travel through a vacuum.

The key distinguishing feature is that mechanical waves cannot travel through a vacuum since they require matter to propagate, while electromagnetic waves can.

Question 30 · 2 pts

Which diagram represents a parallel circuit with an equivalent resistance of 4.0 ohms?

(A) A single-loop (series) circuit with a battery and two 2.0 Ω resistors connected one after another in the loop.
(B) A single-loop (series) circuit with a battery and two 8.0 Ω resistors connected one after another in the loop.
(C) A circuit with a battery connected to two 2.0 Ω resistors in separate parallel branches.
(D) A circuit with a battery connected to two 8.0 Ω resistors in separate parallel branches. ✓

Worked solution

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{1}{8.0}+\frac{1}{8.0}=\frac{2}{8.0} \Rightarrow R_{eq}=4.0\ \Omega$

Answer: A circuit with a battery connected to two 8.0 Ω resistors in separate parallel branches.

Only the parallel combination of two 8.0-ohm resistors gives an equivalent resistance of 4.0 ohms; the series combinations give 4.0 or 16 ohms and the parallel 2.0-ohm resistors give 1.0 ohm.

Question 31 · 2 pts

The diagram below shows four waves. Which two waves, when superimposed, would produce complete destructive interference at every point along the waves?

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

Worked solution

$\text{Complete destructive interference at every point requires two waves of equal amplitude that are exactly 180}^\circ\text{ out of phase.}$

$\text{Of the four waves shown, B and D have equal amplitude and are inverted (180}^\circ\text{) relative to each other.}$

Answer: B and D

For total cancellation everywhere along the waves, the two waves must have identical amplitude and be perfectly out of phase; only waves B and D meet both conditions.

Question 32 · 2 pts

A musician plays a flute by blowing across the opening in the mouth piece of the flute. This causes the air inside the flute to vibrate and produce sound waves. The phenomenon that causes the creation of these sound waves in a flute is

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

Worked solution

$\text{Air blown across the flute opening sets up a standing wave in the air column at its natural (resonant) frequency.}$

Answer: resonance

The air column inside the flute vibrates at its natural frequency in response to the driving disturbance, which is the phenomenon of resonance.

Question 33 · 2 pts

Compared to FM radio waves, AM radio waves

(A) travel faster in a vacuum
(B) travel slower in a vacuum
(C) have a higher frequency
(D) have a longer wavelength ✓

Worked solution

$c=f\lambda, \text{ and all EM waves (including AM and FM radio) travel at the same speed } c \text{ in a vacuum.$

$\text{AM radio has a lower frequency than FM radio, so since } \lambda=\frac{c}{f}, \text{ AM waves have a longer wavelength.}$

Answer: have a longer wavelength

AM and FM radio waves are both electromagnetic and travel at the same speed in a vacuum; AM's lower carrier frequency corresponds to a longer wavelength.

Question 34 · 2 pts

A star approaches Earth. Compared to the frequency and wavelength of the light emitted by the star, the light detected by an observer on Earth would have a

(A) lower frequency and shorter wavelength
(B) lower frequency and longer wavelength
(C) higher frequency and shorter wavelength ✓
(D) higher frequency and longer wavelength

Worked solution

$\text{A source approaching an observer produces a blueshift (Doppler effect for light).}$

$\text{Blueshift means the observed frequency increases and the observed wavelength decreases.}$

Answer: higher frequency and shorter wavelength

Since the star is moving toward Earth, the light is Doppler blueshifted, meaning the observed frequency is higher and the wavelength shorter than emitted.

Question 35 · 2 pts

A beam of electrons directed at a polycrystalline nickel target scatters to form a diffraction pattern on a detecting screen. The diffraction of electrons is an example of the

(A) particle nature of light
(B) Doppler effect
(C) action-reaction principle
(D) wave nature of matter ✓

Worked solution

$\text{Electrons, which are particles of matter, produce a diffraction pattern, a behavior characteristic of waves.}$

$\text{This demonstrates de Broglie's wave nature of matter.}$

Answer: wave nature of matter

Diffraction is a wave phenomenon; observing it with an electron beam (the Davisson-Germer type experiment) demonstrates that matter exhibits wave-like behavior.

Question 36 · 2 pts

What is the approximate width of a typical five-passenger automobile?

(A) $2 \times 10^{-1}$ m
(B) $2 \times 10^{0}$ m ✓
(C) $2 \times 10^{1}$ m
(D) $2 \times 10^{2}$ m

Worked solution

$\text{A typical car width is about 2 meters, i.e., } 2\times10^0\ \text{m}.$

Answer: $2 \times 10^{0}$ m

An everyday car is roughly 2 meters wide, matching the order-of-magnitude estimate of $2 \times 10^{0}$ m.

Question 37 · 2 pts

Two identical metal spheres with charges of +3.0 microcoulombs and -1.0 microcoulomb, respectively, are brought into contact and then separated. Which statement describes the electrostatic force between the spheres?

(A) It is attractive before contact and repulsive after contact. ✓
(B) It is attractive both before contact and after contact.
(C) It is repulsive before contact and attractive after contact.
(D) It is repulsive both before contact and after contact.

Worked solution

$\text{Before contact: charges are opposite (+3.0 } \mu C \text{ and } -1.0\ \mu C), \text{ so the force is attractive.}$

$q_{final}=\frac{(+3.0)+(-1.0)}{2}\ \mu C=+1.0\ \mu C \text{ on each identical sphere after contact.}$

$\text{Since both spheres now carry the same sign charge, the force after contact is repulsive.}$

Answer: It is attractive before contact and repulsive after contact.

Opposite charges attract before contact; touching identical conducting spheres shares the total charge equally, leaving both spheres positively charged and thus repelling each other afterward.

Question 38 · 2 pts

Which graph best represents the relationship between velocity, $v$, and time, $t$, for an object moving with constant positive acceleration?

(A) A horizontal line showing constant velocity (zero slope) on a v vs. t graph.
(B) A straight line with positive slope through the origin on a v vs. t graph. ✓
(C) An upward-curving (increasing slope) line on a v vs. t graph.
(D) A straight line with negative slope on a v vs. t graph.

Worked solution

$a=\frac{dv}{dt}=\text{constant} \Rightarrow v=v_0+at$

$\text{A constant slope on a } v\text{ vs. } t \text{ graph corresponds to a straight line, and constant positive acceleration gives a positive slope.}$

Answer: A straight line with positive slope through the origin on a v vs. t graph.

Constant acceleration means velocity changes linearly with time, which is graphically a straight line with constant (positive) slope, not a curve or horizontal/negative-slope line.

Question 39 · 2 pts

A soccer ball is kicked from point $P_i$ at an angle above a horizontal field and lands on the field at point $P_f$. [Neglect friction.] As the ball travels from $P_i$ to $P_f$ in air, the horizontal component of its velocity

(A) decreases, only
(B) increases, only
(C) decreases and then increases
(D) remains the same ✓

Worked solution

$\text{With no friction (air resistance), there is no horizontal force acting on the projectile.}$

$a_x=0 \Rightarrow v_x=\text{constant throughout the flight.}$

Answer: remains the same

Neglecting friction means the only force on the ball in flight is gravity, which acts vertically; with no horizontal force, the horizontal velocity component stays constant.

Question 40 · 2 pts

Which graph best represents the relationship between the forces of kinetic friction and the normal force exerted on wooden boxes of various weights as they are moved at constant velocity across a level, wooden floor?

(A) A horizontal line showing kinetic friction force constant regardless of normal force.
(B) A straight line with negative slope showing kinetic friction force decreasing as normal force increases.
(C) An upward-curving line showing kinetic friction force increasing faster than normal force.
(D) A straight line with positive slope through the origin showing kinetic friction force directly proportional to normal force. ✓

Worked solution

$F_f=\mu_k F_N$

$\text{Since } \mu_k \text{ is constant for a given pair of surfaces, } F_f \text{ is directly proportional to } F_N, \text{ giving a straight line through the origin.}$

Answer: A straight line with positive slope through the origin showing kinetic friction force directly proportional to normal force.

The kinetic friction force is directly proportional to the normal force through the coefficient of kinetic friction, so the correct graph is a straight line with positive slope passing through the origin.

Question 41 · 2 pts

A 75-kilogram student runs up a set of stairs a vertical distance of 13 meters in 25 seconds. The average power developed by the student is

(A) 3.0 W
(B) 39 W
(C) 380 W ✓
(D) 9600 W

Worked solution

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

$P=\frac{(75\ \text{kg})(9.81\ \text{m/s}^2)(13\ \text{m})}{25\ \text{s}}\approx 382\ \text{W}\approx 380\ \text{W}$

Answer: 380 W

Average power is the work done against gravity (mgh) divided by the time taken, giving approximately 380 watts.

Question 42 · 2 pts

The diagram below represents an apparatus consisting of two conducting rails connected to a resistor, $R$, and a copper wire of length $d$, connected to each rail in a uniform magnetic field directed into the page. Electrons will flow through resistor $R$ if the copper wire is moving

(A) to the right, only
(B) to the left, only
(C) either right or left ✓
(D) neither right nor left

Worked solution

$\varepsilon=BLv$

$\text{Whenever the conducting wire moves through the magnetic field (right or left), it cuts field lines and an EMF is induced, driving current through } R.$

$\text{The direction of current reverses depending on the direction of motion, but current (electron) flow occurs for either direction as long as the wire is moving.}$

Answer: either right or left

Motional EMF is induced any time the conducting rod moves through the magnetic field, whether to the right or left, so electrons flow through R for motion in either direction, just reversing direction of current.

Question 43 · 2 pts

The current through an incandescent lamp is 0.500 ampere. The number of elementary charges that pass through the lamp in 5.00 seconds is

(A) $8.00 \times 10^{-20}$
(B) $4.00 \times 10^{-19}$
(C) $3.13 \times 10^{18}$
(D) $1.56 \times 10^{19}$ ✓

Worked solution

$q=It=(0.500\ \text{A})(5.00\ \text{s})=2.50\ \text{C}$

$n=\frac{q}{e}=\frac{2.50\ \text{C}}{1.60\times10^{-19}\ \text{C}}=1.56\times10^{19}$

Answer: $1.56 \times 10^{19}$

The total charge is found from q = It, and dividing by the elementary charge e gives the number of electrons that passed through the lamp.

Question 44 · 2 pts

Base your answer on the following information and diagram: The diagram below represents an electrical circuit. The reading of $A_1$ is 3.0 amperes. The circuit consists of a battery connected to an ammeter $A_1$ in series with a 3.0-ohm resistor, which is in turn connected in series to a 6.0-ohm resistor with a voltmeter $V$ connected across part of the circuit. The equivalent resistance of the circuit is

(A) 0.5 Ω
(B) 2.0 Ω
(C) 3.0 Ω
(D) 9.0 Ω ✓

Worked solution

$\text{The 3.0 } \Omega \text{ and 6.0 } \Omega \text{ resistors are connected in series (single current path, single ammeter reading).}$

$R_{eq}=R_1+R_2=3.0\ \Omega+6.0\ \Omega=9.0\ \Omega$

Answer: 9.0 Ω

Since the circuit is a single loop (series circuit) with only one ammeter reading for the whole circuit, the equivalent resistance is simply the sum of the two resistors.

Question 45 · 2 pts

The diagram below represents an electrical circuit. The reading of $A_1$ is 3.0 amperes. Compared to the potential difference across the 6-ohm resistor, the potential difference across the 3-ohm resistor is

(A) the same
(B) twice as much
(C) one half as much ✓
(D) one quarter as much

Worked solution

$I = 3.0\ \text{A through both resistors (series circuit)}$

$V_3 = IR_3 = (3.0\ \text{A})(3.0\ \Omega) = 9.0\ \text{V}$

$V_6 = IR_6 = (3.0\ \text{A})(6.0\ \Omega) = 18\ \text{V}$

$\dfrac{V_3}{V_6} = \dfrac{9.0}{18} = \dfrac{1}{2}$

Answer: Choice (3) one half as much

The 3.0-ohm and 6.0-ohm resistors are in series (the voltmeter carries no current), so the same 3.0 A flows through both. Since V = IR, the potential difference is directly proportional to resistance, so the drop across the 3-ohm resistor is half that across the 6-ohm resistor.

Question 46 · 2 pts

Which expression is a unit of work?

(A) $\dfrac{V^2 \cdot s}{\Omega}$ ✓
(B) $\dfrac{V^2}{\Omega}$
(C) $\dfrac{kg \cdot m}{s^2}$
(D) $\dfrac{kg \cdot m^3}{s^3}$

Worked solution

$P = \dfrac{V^2}{R} \rightarrow \text{units of } \dfrac{V^2}{\Omega} \text{ are watts (power)}$

$W = Pt \rightarrow \text{units of } \dfrac{V^2}{\Omega}\cdot s \text{ are watts} \times \text{seconds} = \text{joules}$

Answer: Choice (1) $\dfrac{V^2 \cdot s}{\Omega}$

Work (and energy) is measured in joules, which equal watts multiplied by seconds. Since V^2/Ω has the units of power (watts), multiplying by seconds gives joules, a unit of work.

Question 47 · 2 pts

The diagram below represents a wave propagating to the right through a uniform medium. Point A represents a particle of the medium. At the instant shown, the particle at point A is moving

(A) right
(B) left
(C) up
(D) down ✓

Worked solution

$\text{For a transverse wave, particle motion is perpendicular to the direction of wave travel.}$

$\text{Shifting the waveform slightly to the right shows point A will next take on the displacement just to its left (a lower value), so A is moving down.}$

Answer: Choice (4) down

Particles in a transverse wave oscillate perpendicular to the direction the wave travels. Using the 'shift the wave slightly and see where the point moves to' method, point A, located on the descending part of the curve just ahead of the wave's motion, is moving downward at this instant.

Question 48 · 2 pts

A ray traveling through air strikes a boundary with crown glass, as shown in the diagram below. Which path would the reflected ray take?

(A) A
(B) B
(C) C
(D) D ✓

Worked solution

$\text{Law of reflection: } \theta_i = \theta_r$

$\text{The reflected ray stays in the original medium (air) and is symmetric to the incident ray about the normal.}$

Answer: Choice (4) D

The reflected ray remains in air (above the boundary) and makes an angle equal to the angle of incidence, measured on the opposite side of the normal, which corresponds to path D.

Question 49 · 2 pts

The diagram below shows wave fronts going through an opening and spreading into the area behind the barrier. Which wave phenomenon is represented in the diagram?

(A) resonance
(B) Doppler effect
(C) diffraction ✓
(D) reflection

Worked solution

$\text{Waves bending/spreading around an obstacle or through an opening is diffraction.}$

Answer: Choice (3) diffraction

The spreading of straight wave fronts into circular wave fronts after passing through a narrow opening is the defining characteristic of diffraction.

Question 50 · 2 pts

What occurs as an electron in a mercury atom moves from energy level b to energy level a?

(A) a photon is emitted with an energy of 4.64 J
(B) a photon is absorbed with an energy of 4.64 J
(C) a photon is emitted with an energy of $7.42 \times 10^{-19}$ J ✓
(D) a photon is absorbed with an energy of $7.42 \times 10^{-19}$ J

Worked solution

$\text{An electron dropping from a higher energy level (b) to a lower one (a) emits a photon.}$

$E_{photon} = |E_b - E_a| = 4.64\ \text{eV}$

$E_{photon} = (4.64\ \text{eV})(1.60\times10^{-19}\ \text{J/eV}) = 7.42\times10^{-19}\ \text{J}$

Answer: Choice (3) a photon is emitted with an energy of $7.42 \times 10^{-19}$ J

Moving from a higher energy level to a lower one releases energy as an emitted photon. Converting the 4.64 eV energy difference between mercury's levels b and a into joules gives 7.42 x 10^-19 J.

Question 51 · 1 pt · constructed response

Base your answers to questions 51 through 53 on the information below and on your knowledge of physics. A 750-watt toaster is operating at 120 volts and is being used to heat a blueberry bagel for 2.5 minutes. Determine the resistance of the toaster.

Model answer: 19.2 ohms

Worked solution

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

$R = \dfrac{V^2}{P} = \dfrac{(120\ \text{V})^2}{750\ \text{W}}$

$R = \dfrac{14400}{750} = 19.2\ \Omega$

Answer: 19.2 ohms

Using the power equation P = V^2/R and solving for resistance gives R = V^2/P, which yields 19.2 ohms for this toaster.

Question 52 · 2 pts · constructed response

A 750-watt toaster is operating at 120 volts and is being used to heat a blueberry bagel for 2.5 minutes. Calculate the total amount of electrical energy consumed by the toaster during the heating of the bagel. [Show all work, including the equation and substitution with units.]

Model answer: $1.125 \times 10^5$ J (112,500 J)

Worked solution

$t = 2.5\ \text{min} \times 60\ \text{s/min} = 150\ \text{s}$

$E = Pt$

$E = (750\ \text{W})(150\ \text{s}) = 1.125\times10^{5}\ \text{J}$

Answer: $1.125 \times 10^5$ J (112,500 J)

Electrical energy consumed equals power multiplied by time. Converting 2.5 minutes to 150 seconds and multiplying by the 750 W rating gives 112,500 joules.

Question 54 · 1 pt · constructed response

In a linear accelerator, a proton is accelerated from rest through a potential difference of $4.40 \times 10^6$ volts. Determine the total kinetic energy, in joules, gained by this proton as it travels through this linear accelerator.

Model answer: $7.04 \times 10^{-13}$ J

Worked solution

$KE = qV$

$KE = (1.60\times10^{-19}\ \text{C})(4.40\times10^{6}\ \text{V})$

$KE = 7.04\times10^{-13}\ \text{J}$

Answer: $7.04 \times 10^{-13}$ J

All electrical potential energy lost by the proton as it is accelerated converts to kinetic energy, so KE = qV, using the elementary charge of a proton.

Question 55 · 2 pts · constructed response

Calculate the resistance of a 0.050-meter-long copper wire having a cross-sectional area of $5.73 \times 10^{-10}$ meter squared at 20°C. [Show all work, including the equation and substitution with units.]

Model answer: Approximately 1.5 $\Omega$

Worked solution

$R = \dfrac{\rho L}{A}, \quad \rho_{copper} = 1.72\times10^{-8}\ \Omega\cdot\text{m (Reference Table)}$

$R = \dfrac{(1.72\times10^{-8}\ \Omega\cdot\text{m})(0.050\ \text{m})}{5.73\times10^{-10}\ \text{m}^2}$

$R = \dfrac{8.6\times10^{-10}}{5.73\times10^{-10}} \approx 1.5\ \Omega$

Answer: Approximately 1.5 $\Omega$

Resistance of a wire is found from R = ρL/A, using copper's resistivity from the Reference Tables. Substituting the given length and cross-sectional area gives about 1.5 ohms.

Question 57 · 2 pts · constructed response

During a football game, player A, a 70.0-kilogram football player, attempts to prevent player B, a 90.0-kilogram player, from scoring a touchdown. Player B is running toward player A at a constant speed of 6.50 meters per second. Calculate the speed of player A such that the magnitude of the momentum of player A is equal to the magnitude of the momentum of player B. [Show all work, including the equation and substitution with units.]

Model answer: Approximately 8.36 m/s

Worked solution

$p_B = m_B v_B = (90.0\ \text{kg})(6.50\ \text{m/s}) = 585\ \text{kg}\cdot\text{m/s}$

$p_A = p_B \Rightarrow m_A v_A = 585\ \text{kg}\cdot\text{m/s}$

$v_A = \dfrac{585\ \text{kg}\cdot\text{m/s}}{70.0\ \text{kg}} \approx 8.36\ \text{m/s}$

Answer: Approximately 8.36 m/s

Momentum is mass times velocity. Setting player A's momentum magnitude equal to player B's calculated momentum and solving for A's speed gives about 8.36 m/s.

Question 59 · 2 pts · constructed response

A photon has a wavelength of $5.03 \times 10^{-7}$ meter. Calculate the energy of this photon. [Show all work, including the equation and substitution with units.]

Model answer: Approximately $3.95 \times 10^{-19}$ J

Worked solution

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

$E = \dfrac{(6.63\times10^{-34}\ \text{J}\cdot\text{s})(3.00\times10^{8}\ \text{m/s})}{5.03\times10^{-7}\ \text{m}}$

$E = \dfrac{1.989\times10^{-25}}{5.03\times10^{-7}} \approx 3.95\times10^{-19}\ \text{J}$

Answer: Approximately $3.95 \times 10^{-19}$ J

Photon energy is found using E = hc/λ. Substituting Planck's constant, the speed of light, and the given wavelength yields approximately 3.95 x 10^-19 J.

Question 61 · 1 pt · constructed response

Base your answers to questions 61 and 62 on the information below and on your knowledge of physics. A 53-newton box on a horizontal surface is pulled toward the right by a horizontal force of 27 newtons. The force of friction exerted on the box by the surface has a magnitude of 16 newtons. On the diagram in your answer booklet, use a scale of 1.0 cm = 10. N and start at point P to construct a vector representing the normal force exerted on the box by the floor.

Model answer: A 5.3 cm vector drawn straight up (perpendicular to and away from the floor) from point P, representing 53 N.

Worked solution

$\text{Box is in vertical equilibrium (no vertical acceleration): } F_N = F_g = 53\ \text{N, directed up}$

$\text{Length of vector} = \dfrac{53\ \text{N}}{10.\ \text{N/cm}} = 5.3\ \text{cm}$

Answer: A 5.3 cm vector drawn straight up (perpendicular to and away from the floor) from point P, representing 53 N.

Since the box does not accelerate vertically, the normal force must balance the box's weight, giving a normal force of 53 N directed straight up. Using the given scale, that corresponds to a 5.3 cm vector drawn upward from point P.

Question 62 · 1 pt · constructed response

Base your answers to questions 61 and 62 on the information below and on your knowledge of physics. A 53-newton box on a horizontal surface is pulled toward the right by a horizontal force of 27 newtons. The force of friction exerted on the box by the surface has a magnitude of 16 newtons. Determine the magnitude of the net force acting on the box.

Model answer: 11 N

Worked solution

$F_{net} = F_{applied} - F_{friction}$

$F_{net} = 27\ \text{N} - 16\ \text{N} = 11\ \text{N}$

Answer: 11 N

The applied force and friction force act in opposite horizontal directions, so the net horizontal force is their difference, 11 N.

Question 63 · 1 pt · constructed response

Base your answers to questions 63 through 65 on the information and diagram below and on your knowledge of physics. A ray of monochromatic light (f = $5.09 \times 10^{14}$ hertz) passes from air into a transparent medium, X, as shown in the diagram below. Using a protractor, determine the angle of refraction in medium X, to the nearest degree.

Model answer: Approximately 20° (measured directly from the printed diagram with a protractor; accepted values typically fall within about 18°-22°).

Worked solution

$\text{Angle of refraction is measured between the refracted ray inside medium X and the normal, using a protractor on the printed diagram.}$

Answer: Approximately 20° (measured directly from the printed diagram with a protractor; accepted values typically fall within about 18°-22°).

This question requires physically measuring the refracted ray's angle from the normal on the printed diagram with a protractor; based on the diagram's geometry the refracted ray bends toward the normal to roughly 20 degrees, since medium X is optically denser than air.

Question 64 · 2 pts · constructed response

Base your answers to questions 63 through 65 on the information and diagram below and on your knowledge of physics. A ray of monochromatic light (f = $5.09 \times 10^{14}$ hertz) passes from air into a transparent medium, X, as shown in the diagram below. Calculate the absolute index of refraction of medium X. [Show all work, including the equation and substitution with units.]

Model answer: Approximately 1.7 (using $n = \sin35°/\sin(\text{measured refraction angle})$; exact value depends on the protractor measurement from question 63).

Worked solution

$n_{air}\sin\theta_i = n_X\sin\theta_r, \quad n_{air}=1.00$

$n_X = \dfrac{\sin\theta_i}{\sin\theta_r} = \dfrac{\sin35°}{\sin(20°)}$

$n_X = \dfrac{0.574}{0.342} \approx 1.68$

Answer: Approximately 1.7 (using $n = \sin35°/\sin(\text{measured refraction angle})$; exact value depends on the protractor measurement from question 63).

Snell's Law relates the angle of incidence in air to the angle of refraction in medium X through the index of refraction. Using the 35 degree incidence angle and the protractor-measured refraction angle (about 20 degrees) from question 63 gives an index of refraction near 1.7.

Question 66 · 2 pts · constructed response

Base your answers to questions 66 through 68 on the information below and on your knowledge of physics. Three 4.0-kilogram spherical masses, A, B, and C, are positioned so that mass B is due south of mass A and mass C is due east of mass A. Mass B and mass C exert gravitational forces on mass A. The gravitational force, $F_B$, exerted by mass B on mass A is $5.08 \times 10^{-10}$ newton south. The force $F_C$ exerted by mass C on mass A points east, and the resultant gravitational force, R, exerted on mass A is $5.74 \times 10^{-10}$ newton at 27° east of south. Calculate the distance separating the centers of masses A and B. [Show all work, including the equation and substitution with units.]

Model answer: Approximately 1.4 m (1.45 m)

Worked solution

$F_B = \dfrac{Gm_Am_B}{d^2}$

$d = \sqrt{\dfrac{Gm_Am_B}{F_B}} = \sqrt{\dfrac{(6.67\times10^{-11})(4.0)(4.0)}{5.08\times10^{-10}}}$

$d = \sqrt{\dfrac{1.067\times10^{-9}}{5.08\times10^{-10}}} = \sqrt{2.10} \approx 1.45\ \text{m}$

Answer: Approximately 1.4 m (1.45 m)

Newton's law of universal gravitation gives F = Gm1m2/d^2. Solving for the distance using the given gravitational force between A and B and their equal 4.0 kg masses yields about 1.4 to 1.45 meters.

Question 68 · 1 pt · constructed response

Mass $A$ is removed and a 7.0-kilogram mass is placed at the same position. Compare the magnitude of the resultant gravitational force on the 7.0-kilogram mass with the magnitude of the original resultant gravitational force, $R$.

Model answer: The new resultant gravitational force is greater than R; specifically it is 7.0/4.0 = 1.75 times R, or about 1.0 x 10^-9 N.

Worked solution

$F \propto m_{A}$ gravitational force on the mass at position A is directly proportional to that mass's value, since G, the other masses, and the distances are unchanged

$R_{new} = \dfrac{7.0}{4.0}\,R = 1.75 \times 5.74\times10^{-10}\ \text{N} \approx 1.0\times10^{-9}\ \text{N}$

Answer: The new resultant gravitational force is greater than R; specifically it is 7.0/4.0 = 1.75 times R, or about 1.0 x 10^-9 N.

Since the gravitational force from each source mass on the object at position A is directly proportional to the mass placed there, increasing the mass from 4.0 kg to 7.0 kg increases the resultant force proportionally, making it greater than the original R.

Question 69 · 1 pt · constructed response

Base your answers to questions 69 through 73 on the information and data table below and on your knowledge of physics. An electric train with a mass of 2.8 kilograms moves from rest down a long, straight track. The table shows the train's momentum during the first 4.0 seconds of its trip. Data Table: Time (s): 0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0. Momentum (kg·m/s): 0.0, 2.1, 4.2, 6.3, 8.4, 11.0, 13.0, 15.0, 18.0. Using information from the data table, construct a graph in your answer booklet, following the directions below. 69 Plot the momentum versus time for the first 4.0 seconds of the train's trip.

The best-fit line is drawn on the graph plotted in question 69, using this data for the 2.8 kg train: | Time (s) | Momentum (kg*m/s) | |---|---| | 0.0 | 0.0 | | 0.5 | 2.1 | | 1.0 | 4.2 | | 1.5 | 6.3 | | 2.0 | 8.4 | | 2.5 | 11.0 | | 3.0 | 13.0 | | 3.5 | 15.0 | | 4.0 | 18.0 |

Model answer: All nine data points (0.0, 0.0), (0.5, 2.1), (1.0, 4.2), (1.5, 6.3), (2.0, 8.4), (2.5, 11.0), (3.0, 13.0), (3.5, 15.0), (4.0, 18.0) should be plotted accurately on a grid with time (s) on the horizontal axis and momentum (kg·m/s) on the vertical axis, using an appropriate, evenly spaced scale that uses at least half the grid.

Worked solution

$\text{Choose scale for time axis: e.g., 0.5 s per interval}$

$\text{Choose scale for momentum axis: e.g., 2.0 kg·m/s per interval}$

$\text{Plot each (t, p) pair as a point}$

Answer: All nine data points (0.0, 0.0), (0.5, 2.1), (1.0, 4.2), (1.5, 6.3), (2.0, 8.4), (2.5, 11.0), (3.0, 13.0), (3.5, 15.0), (4.0, 18.0) should be plotted accurately on a grid with time (s) on the horizontal axis and momentum (kg·m/s) on the vertical axis, using an appropriate, evenly spaced scale that uses at least half the grid.

This is a data-plotting task; credit is earned for correctly plotting all nine points using a suitable, consistent scale on labeled axes.

Question 70 · 1 pt · constructed response

Draw a best-fit line, using a straight edge, to represent the relationship between momentum and time.

Model answer: A single straight line drawn with a straight edge that passes as close as possible to all plotted points, with roughly equal numbers of points above and below the line, starting near the origin.

Worked solution

$\text{Draw one straight line (not connect-the-dots) that best represents the overall trend of the plotted points}$

Answer: A single straight line drawn with a straight edge that passes as close as possible to all plotted points, with roughly equal numbers of points above and below the line, starting near the origin.

Since the momentum data increases roughly linearly with time, a single best-fit straight line through the plotted points represents the relationship, allowing the slope to be used to find the net force.

Question 71 · 1 pt · constructed response

Using your best-fit line, determine the average force that acted upon the train.

The best-fit line comes from the momentum versus time graph of this data for the 2.8 kg train: | Time (s) | Momentum (kg*m/s) | |---|---| | 0.0 | 0.0 | | 0.5 | 2.1 | | 1.0 | 4.2 | | 1.5 | 6.3 | | 2.0 | 8.4 | | 2.5 | 11.0 | | 3.0 | 13.0 | | 3.5 | 15.0 | | 4.0 | 18.0 | The average force equals the slope of the best-fit line (F = change in momentum / change in time).

Model answer: The average force is approximately 4.4 to 4.5 newtons (the slope of the best-fit momentum vs. time line).

Worked solution

$F_{avg} = \dfrac{\Delta p}{\Delta t} = \text{slope of best-fit line}$ Newton's second law in impulse-momentum form

$F_{avg} \approx \dfrac{18.0 - 0.0}{4.0 - 0.0} \approx 4.4\text{-}4.5\ \text{N}$ using approximate endpoints of the best-fit line

Answer: The average force is approximately 4.4 to 4.5 newtons (the slope of the best-fit momentum vs. time line).

Because F = dp/dt, the slope of a momentum vs. time graph gives the net average force acting on the object; reading the slope of the best-fit line yields about 4.4-4.5 N.

Question 72 · 2 pts · constructed response

Calculate the magnitude of the average acceleration of the train. [Show all work, including the equation and substitution with units.]

Model answer: The average acceleration of the train is approximately 1.6 m/s^2.

Worked solution

$v_f = \dfrac{p_f}{m} = \dfrac{18.0\ \text{kg·m/s}}{2.8\ \text{kg}} = 6.43\ \text{m/s}$ find final velocity from final momentum

$a = \dfrac{v_f - v_i}{t} = \dfrac{6.43\ \text{m/s} - 0}{4.0\ \text{s}}$ definition of average acceleration, train starts from rest

$a \approx 1.6\ \text{m/s}^2$

Answer: The average acceleration of the train is approximately 1.6 m/s^2.

Using the final momentum to find final velocity (v = p/m) and dividing the change in velocity by the elapsed time gives the average acceleration of the train.

Question 74 · 1 pt · constructed response

Base your answers to questions 74 through 77 on the information and diagram below and on your knowledge of physics. The diagram below represents a medium as a periodic wave passes through it, traveling to the right. The wave requires 2.0 seconds to travel from point $A$ to point $B$. The scale used in the diagram is 1.0 centimeter = 0.10 meter. (The diagram shows a sinusoidal wave beginning at point A on the axis, rising to a crest, descending through the axis to a trough labeled point C, rising again to a second crest, then descending to an unlabeled point below the axis, with point B marked further to the right on the axis.) 74 Determine the wavelength of the wave in meters.

Model answer: The wavelength of the wave is approximately 0.40 meter, based on measuring one full crest-to-crest cycle on the diagram and converting using the given scale (1.0 cm = 0.10 m).

Worked solution

$\lambda = \text{(measured crest-to-crest distance in cm)} \times \dfrac{0.10\ \text{m}}{1.0\ \text{cm}}$ convert measured centimeters to meters using the given scale

$\lambda \approx 0.40\ \text{m}$ typical measured value for this diagram

Answer: The wavelength of the wave is approximately 0.40 meter, based on measuring one full crest-to-crest cycle on the diagram and converting using the given scale (1.0 cm = 0.10 m).

The wavelength is the horizontal distance for one complete wave cycle (crest to crest), found by measuring the diagram with a ruler in centimeters and converting to meters with the given scale.

Question 75 · 2 pts · constructed response

Calculate the speed of the wave in meters per second. [Show all work, including the equation and substitution with units.]

Model answer: The speed of the wave is approximately 0.40 m/s.

Worked solution

$v = \dfrac{d_{AB}}{t}$ speed equals total distance traveled divided by total time

$d_{AB} = \text{(measured A-to-B distance in cm)} \times \dfrac{0.10\ \text{m}}{1.0\ \text{cm}} \approx 0.80\ \text{m}$ convert measured distance using the given scale

$v = \dfrac{0.80\ \text{m}}{2.0\ \text{s}} = 0.40\ \text{m/s}$

Answer: The speed of the wave is approximately 0.40 m/s.

The wave speed is found by measuring the total horizontal distance from A to B on the scaled diagram, converting to meters, and dividing by the given travel time of 2.0 seconds.

Question 77 · 1 pt · constructed response

On the diagram in your answer booklet, place an X on the wave at a position that is 180 degrees out of phase with point C.

Model answer: The X should be placed at either crest of the wave (a point of maximum upward displacement), since point C is at a trough and a point 180 degrees out of phase with a trough is a crest.

Worked solution

$\text{Point C is at a trough (minimum displacement)}$

$\text{A point 180}^{\circ}\text{ out of phase is exactly one half-cycle away, i.e. a crest}$

$\text{Mark X at the nearest crest (peak) of the wave}$

Answer: The X should be placed at either crest of the wave (a point of maximum upward displacement), since point C is at a trough and a point 180 degrees out of phase with a trough is a crest.

Points 180 degrees out of phase move in exactly opposite directions at every instant; since C is at a trough, the point 180 degrees out of phase with it is at a crest of the same wave.

Question 78 · 1 pt · constructed response

Base your answers to questions 78 through 82 on the information and diagram below and on your knowledge of physics. A pogo stick is a popular children's toy with a spring on the bottom. When a 51.0-kilogram child stands at rest on a pogo stick, the spring is compressed 0.15 meter. (Diagram shows a child standing on a pogo stick with a spring at the bottom, labeled 'Child' and 'Spring'.) 78 Determine the magnitude of the weight of the child in newtons.

Model answer: The weight of the child is approximately 500. newtons.

Worked solution

$F_g = mg = (51.0\ \text{kg})(9.81\ \text{m/s}^2)$ weight equals mass times gravitational acceleration

$F_g \approx 500.\ \text{N}$

Answer: The weight of the child is approximately 500. newtons.

Weight is calculated by multiplying the child's mass by the acceleration due to gravity, g = 9.81 m/s^2.

Question 79 · 2 pts · constructed response

Calculate the spring constant of the spring on the pogo stick. [Show all work, including the equation and substitution with units.]

Model answer: The spring constant is approximately 3.3 x 10^3 N/m.

Worked solution

$F_s = kx$ Hooke's law, at equilibrium the spring force equals the child's weight

$k = \dfrac{F_s}{x} = \dfrac{500.\ \text{N}}{0.15\ \text{m}}$

$k \approx 3.3\times10^{3}\ \text{N/m}$

Answer: The spring constant is approximately 3.3 x 10^3 N/m.

At rest, the spring force balances the child's weight, so Hooke's law (F = kx) can be rearranged to solve for the spring constant using the weight and compression distance.

Question 81 · 2 pts · constructed response

Calculate the total energy stored in the pogo stick's spring when the child's weight has compressed the spring 0.15 meter. [Show all work, including the equation and substitution with units.]

Model answer: The total energy stored in the spring is approximately 38 joules (37.5 J).

Worked solution

$PE_s = \dfrac{1}{2}kx^2$ elastic potential energy formula

$PE_s = \dfrac{1}{2}(3.3\times10^{3}\ \text{N/m})(0.15\ \text{m})^2$

$PE_s \approx 37.5\ \text{J}$

Answer: The total energy stored in the spring is approximately 38 joules (37.5 J).

The elastic potential energy stored in a compressed spring is found using PE = (1/2)kx^2, using the spring constant found previously and the given compression distance.

Question 83 · 1 pt · constructed response

Base your answers to questions 83 through 85 on the information below and on your knowledge of physics. Scientists at the CERN Large Hadron Collider discovered a new subatomic particle. It is called the $X_b$ particle and is composed of one down quark, one strange quark, and one bottom quark. The mass of the $X_b$ particle is approximately 5950 MeV. 83 Determine both the sign and the magnitude of the charge of the $X_b$ particle, in elementary charges.

Model answer: The charge of the Xb particle is -1 elementary charge (negative, magnitude 1e).

Worked solution

$q_{down} = q_{strange} = q_{bottom} = -\dfrac{1}{3}e$ down, strange, and bottom are all down-type quarks, each with charge -1/3 e

$q_{X_b} = -\dfrac{1}{3}e - \dfrac{1}{3}e - \dfrac{1}{3}e = -1e$ sum the three quark charges

Answer: The charge of the Xb particle is -1 elementary charge (negative, magnitude 1e).

Down, strange, and bottom quarks are all down-type quarks, each carrying a charge of -1/3 elementary charge, so their sum gives a total particle charge of -1 elementary charge (negative).

Question 84 · 1 pt · constructed response

Identify the fundamental force that holds the quarks together in the $X_b$ particle.

Model answer: The strong force (strong nuclear force) holds the quarks together in the Xb particle.

Worked solution

$\text{Quarks are bound together by the strong (nuclear) interaction, mediated by gluons}$

Answer: The strong force (strong nuclear force) holds the quarks together in the Xb particle.

The strong nuclear force is the fundamental interaction responsible for binding quarks together to form composite particles such as baryons.

Question 85 · 1 pt · constructed response

Determine the mass of the $X_b$ particle, in universal mass units.

Model answer: The mass of the Xb particle is approximately 6.39 universal mass units (u).

Worked solution

$1\ \text{u} = 931\ \text{MeV}/c^2$ conversion factor from the Reference Tables for Physical Setting/Physics

$m_{X_b} = \dfrac{5950\ \text{MeV}}{931\ \text{MeV/u}} \approx 6.39\ \text{u}$

Answer: The mass of the Xb particle is approximately 6.39 universal mass units (u).

Dividing the given mass in MeV by the conversion factor of 931 MeV per universal mass unit converts the particle's mass into universal mass units.

June 2025 Physics Regents — Answer Key & Worked Solutions

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