5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

STEP 5

Build Your Test-Taking Confidence

AP PHYSICS 1 Practice Exam 1: Section I (Multiple-Choice)

AP PHYSICS 1 Practice Exam 1: Section II (Free-Response)

SOLUTIONS: AP Physics I Practice Exam 1, Section I (Multiple-Choice)

SOLUTIONS: AP Physics I Practice Exam 1, Section II (Free-Response)

AP PHYSICS 1 Practice Exam 2: Section I (Multiple-Choice)

AP PHYSICS 1 Practice Exam 2: Section II (Free-Response)

SOLUTIONS: AP Physics I Practice Exam 2, Section I (Multiple-Choice)

SOLUTIONS: AP Physics I Practice Exam 2, Section II (Free-Response)

Scoring the Practice Exams

Practice Exam 1

ANSWER SHEET FOR SECTION I

AP Physics 1 Practice Exam 1: Section I (Multiple-Choice)

Directions: The multiple-choice section consists of 50 questions to be answered in 90 minutes . You may write scratch work in the test booklet itself, but only the answers on the answer sheet will be scored. You may use a calculator, the equation sheet, and the table of information.

Questions 1–45: Single-Choice Items

Directions: Choose the single best answer from the four choices provided and grid the answer with a pencil on the answer sheet.

  1. A circuit consists of a battery and a light bulb. At first, the circuit is disconnected. Then, the circuit is connected, and the light bulb lights. After the light bulb has been lit for a few moments, how has the net charge residing on the circuit elements changed?

(A)   The net charge has become more positive.

(B)   The net charge has become more negative.

(C)   The net charge has not changed.

(D)   Whether the net charge becomes more positive or more negative depends on the initial net charge residing on the circuit elements before the bulb was lit.

  1. At which position in the above circuit will the charge passing that position in one second be largest?

(A)   A

(B)   B

(C)   C

(D)   D

  1. Spring scales are used to measure the net force applied to an object; a sonic motion detector is used to measure the object’s resulting acceleration. A graph is constructed with the net force on the vertical axis and the acceleration on the horizontal axis. Which of the following quantities is directly measured using the slope of this graph?

(A)   Gravitational mass

(B)   Weight

(C)   Velocity

(D)   Inertial mass

Questions 4 and 5 refer to the information below:

In the laboratory, a 0.5-kg cart collides with a fixed wall, as shown in the preceding diagram. The collision is recorded with a video camera that takes 20 frames per second. A student analyzes the video, placing a dot at the center of mass of the cart in each frame. The analysis is shown above.

  1. About how fast was the cart moving before the collision?

(A)   0.25 m/s

(B)   4.0 m/s

(C)   0.20 m/s

(D)   5.0 m/s

  1. Which of the following best estimates the change in the cart’s momentum during the collision?

(A)   27 N·s

(B)   13 N·s

(C)   1.3 N·s

(D)   2.7 N·s

  1. In the laboratory, a 60-Hz generator is connected to a string that is fixed at both ends. A standing wave is produced, as shown in the preceding figure. In order to measure the wavelength of this wave, a student should use a meterstick to measure from positions

(A)   B to C

(B)   B to D

(C)   D to E

(D)   A to F

  1. Four identical lead balls with large mass are connected by rigid but very light rods in the square configuration shown in the preceding figure. The balls are rotated about the three labeled axes. Which of the following correctly ranks the rotational inertia Iof the balls about each axis?

(A)   I B > I A = I C

(B)   I A > I C = I B

(C)   I C > I A > I B

(D)   I C > I A = I B

  1. In the laboratory, a cart experiences a single horizontal force as it moves horizontally in a straight line. Of the following data collected about this experiment, which is sufficient to determine the work done on the cart by the horizontal force?

(A)   The magnitude of the force, the cart’s initial speed, and the cart’s final speed

(B)   The mass of the cart and the distance the cart moved

(C)   The mass of the cart, the cart’s initial speed, and the cart’s final speed

(D)   The mass of the cart and the magnitude of the force

  1. A wave pulse on a string is shown above. Which pulse, when superimposed with the one above, will produce complete destructive interference?

(A)   

(B)   

(C)   

(D)   

  1. In the laboratory, a 3-kg cart experiences a varying net force. This net force is measured as a function of time, and the data collected are displayed in the graph above. What is the change in the cart’s momentum during the interval t= 0 to t = 2 s?

(A)   5 N·s

(B)   10 N·s

(C)   15 N·s

(D)   30 N·s

  1. A block is attached to a vertical spring. The block is pulled down a distance Afrom equilibrium, as shown above, and released from rest. The block moves upward; the highest position above equilibrium reached by the mass is less than A , as shown. When the mass returns downward, how far below the equilibrium position will it reach?

(A)   Greater than the distance A below equilibrium

(B)   Less than the distance A below equilibrium

(C)   Equal to the distance A below equilibrium

(D)   No distance—the block will fall only to the equilibrium position.

Questions 12 and 13 refer to the following information:

Two charged Styrofoam balls are brought a distance d from each other, as shown. The force on Ball B is 2 μN to the right. When the distance between the balls is changed, the force on Ball B is 8 μN to the right.

  1. Which of the following can indicate the sign of the charges of balls Aand B ?
  2. When the force on Ball Bis 8 μN, what is the distance between the centers of the two balls?

(A)   d /4

(B)   d /2

(C)   d /16

(D)   d /

  1. A disk of radius 1 m and rotational inertia I= 0.5 kg.m2 is free to rotate, but initially at rest. A blob of putty with mass 0.1 kg is traveling toward the disk with a speed of 10 m/s, as shown in the preceding figure. The putty collides with the outermost portion of the disk and sticks to the disk. What is the angular momentum of the combined disk-putty system after the collision?

(A)   5 kg·m2 /s

(B)   1 kg·m2 /s

(C)   0.5 kg·m2 /s

(D)   0 kg·m2 /s

  1. A 1-kg object is released from rest at the top of a rough-surfaced incline. The object slides without rotating to the bottom of the incline. The object’s kinetic energy at the bottom must be

(A)   Equal to the block’s gravitational potential energy when it was released, because total mechanical energy must be conserved.

(B)   Equal to the block’s gravitational potential energy when it was released, because the gain in kinetic energy compensates for the mechanical energy lost to thermal energy on the rough incline.

(C)   Less than the block’s gravitational potential energy when it was released, because the gravitational potential energy was converted both to thermal energy and to kinetic energy.

(D)   Less than the block’s gravitational potential energy when it was released, because the work done by the friction force must be greater than the block’s gain in kinetic energy.

  1. What is the current in the 4 Ω resistor in the circuit in the preceding illustration?

(A)   1.5 A

(B)   2.0 A

(C)   3.0 A

(D)   6.0 A

  1. Two sounds are played in the laboratory. A microphone is connected to an oscilloscope, which displays the traces shown above for each sound. On these traces, the horizontal axis is time; the vertical axis is related to the distance the microphone’s diaphragm is displaced from its resting position. The scales are identical for each diagram. Which of the following is correct about the sounds that produce the traces above?

(A)   Sound 1 is louder, and sound 2 is higher pitched.

(B)   Sound 2 is louder, and sound 2 is higher pitched.

(C)   Sound 1 is louder, and sound 1 is higher pitched.

(D)   Sound 2 is louder, and sound 1 is higher pitched.

  1. The radius of Mars is about half that of Earth; the mass of Mars is about one-tenth that of Earth. Which of the following is closest to the gravitational field at the surface of Mars?

(A)   10 N/kg

(B)   4 N/kg

(C)   2 N/kg

(D)   0.5 N/kg

  1. In an experiment, a marble rolls to the right at speed v, as shown in the top diagram. The marble rolls under a canopy, where it is heard to collide with marbles that were not initially moving. Such a collision is known to be elastic. After the collision, two equal-mass marbles are observed leaving the canopy with velocity vectors directed as shown. Which of the following statements justifies why the experimenter believes that a third marble was involved in the collision under the canopy?

(A)   Before collision, the only marble momentum was directed to the right. After the collision, the combined momentum of the two visible marbles is still to the right. Another marble must have a leftward momentum component to conserve momentum.

(B)   Before collision, the only marble momentum was directed to the right. After the collision, the combined momentum of the two visible marbles has a downward component; another marble must have an upward momentum component to conserve momentum.

(C)   Before collision, the only marble kinetic energy was directed to the right. After the collision, the combined kinetic energy of the two visible marbles is still to the right. Another marble must have a leftward kinetic energy component to conserve kinetic energy.

(D)   Before collision, the only marble kinetic energy was directed to the right. After the collision, the combined kinetic energy of the two visible marbles has a downward component; another marble must have an upward kinetic energy component to conserve kinetic energy.

  1. In the laboratory, two carts on a track collide in the arrangement shown in the preceding figure. Before the collision, the 2-kg cart travels through photogate 1, which measures its speed; the 0.25-kg cart is initially at rest. After the collision, the carts bounce off one another. Photogate 2 measures the speed of each cart as it passes.

A student is concerned about his experimental results. When he adds the momentum of both carts after collision, he gets a value greater than the momentum of the 2-kg cart before collision. Which of the following is a reasonable explanation for the discrepancy?

(A)   The track might have been slanted such that the carts were moving downhill.

(B)   Human error might have been involved in reading the photogates.

(C)   Friction might not have been negligible.

(D)   The collision might not have been elastic.

  1. Three wagons each have the same total mass (including that of the wheels) and four wheels, but the wheels are differently styled. The structure, mass, and radius of each wagon’s wheels are shown in the preceding chart. In order to accelerate each wagon from rest to a speed of 10 m/s, which wagon requires the greatest energy input?

(A)   Wagon A

(B)   Wagon B

(C)   Wagon C

(D)   All require the same energy input

  1. A swimmer is able to propel himself forward through the water by moving his arms. Which of the following correctly states the applicant and recipient of the force responsible for the swimmer’s forward acceleration?

(A)   The force of the surrounding water on the swimmer’s arms

(B)   The force of the swimmer’s arms on the swimmer’s torso

(C)   The force of the swimmer’s arms on the surrounding water

(D)   The force of the swimmer’s torso on the swimmer’s arms

  1. A 10-kg wagon moves horizontally at an initial speed of 5 m/s. A 30-N force is applied to the wagon by pulling the rigid handle, which is angled 60° above the horizontal. The wagon continues to move horizontally for another 20 m. A negligible amount of work is converted into thermal energy. By how much has the wagon’s kinetic energy increased over the 20 m?

(A)   300 J

(B)   600 J

(C)   125 J

(D)   63 J

  1. A moving 1.5-kg cart collides with and sticks to a 0.5-kg cart which was initially at rest. Immediately after the collision, the carts each have the same ______ as each other.

(A)   Velocity

(B)   Kinetic energy

(C)   Mass

(D)   Linear momentum

Questions 25 and 26 refer to the information below:

Four identical resistors are connected to a battery in the configuration shown in the preceding figure.

  1. Which of the following ranks the current Ithrough each resistor?

(A)   I 1 = I 4 > I 2 > I 3

(B)   I 1 = I 4 > I 2 = I 3

(C)   I 1 = I 2 = I 3 = I 4

(D)   I 1 > I 2 = I 3 > I 4

  1. Which graph represents the electric potential with respect to the negative end of the battery as a function of the location on a loop of wire starting from the positive end of the battery, going through resistors 1, 2, and 4, and ending back on the negative end of the battery?

(A)   

(B)   

(C)   

(D)   

  1. A force probe is used to stretch a spring by 20 cm. The graph of the force as a function of distance shown in the preceding figure is produced and used to determine the amount of work done in stretching the spring 20 cm. The experimenter reports the result as 3,000 N·cm. Which of the following is a reasonable estimate of the experimental uncertainty on this measurement?

(A)   3,000 ± 3 N.cm

(B)   3,000 ± 30 N.cm

(C)   3,000 ± 300 N.cm

(D)   3,000 ± 3,000 N.cm

  1. A string of fixed tension and linear mass density is attached to a vibrating speaker. It is observed that a speaker frequency of 60 Hz does not produce standing waves in the string. Which explanation for this phenomenon is correct?

(A)   The string length is not a multiple of half the wavelength of the wave.

(B)   The wave speed on the string is fixed.

(C)   60 Hz is in the lowest range of audible sound.

(D)   The wavelength of the wave produced by the speaker is equal to the speed of waves on the string divided by 60 Hz.

  1. In the laboratory, a long platform of negligible mass is free to rotate on a fulcrum. A force probe is placed a fixed distance xfrom the fulcrum, supporting the platform. An object of fixed mass is placed a variable distance dfrom the fulcrum. For each position d , the force probe is read. It is desired to determine the mass of the object from a graph of data. Which of the following can determine the object’s mass?

(A)   Plot the reading in the force probe times x on the vertical axis; plot the gravitational field times d on the horizontal axis. The mass is the slope of the line.

(B)   Plot the reading in the force probe on the vertical axis; plot the distance d on the horizontal axis. The mass is the area under the graph.

(C)   Plot the reading in the force probe on the vertical axis; plot the distance d multiplied by the distance x on the horizontal axis. The mass is the y-intercept of the graph.

(D)   Plot the reading in the force probe times d on the vertical axis; plot the distance x on the horizontal axis. The mass is the slope of the line divided by the gravitational field.

  1. In Collision A, two carts collide and bounce off each other. In Collision B, a ball sticks to a rigid rod, which begins to rotate about the combined center of mass. Which of the following statements about quantities in each collision is correct?

(A)   Collision A: each cart experiences the same force, time of collision, and change in kinetic energy. Collision B: the ball and the rod each experience the same torque, time of collision, and change in rotational kinetic energy.

(B)   Collision A: each cart experiences the same force, time of collision, and change in linear momentum. Collision B: the ball and the rod each experience the same torque, time of collision, and change in angular momentum.

(C)   Collision A: each cart experiences the same force, time of collision, and change in kinetic energy. Collision B: the ball and the rod each experience the same torque, time of collision, and change in angular momentum.

(D)   Collision A: each cart experiences the same force, time of collision, and change in velocity. Collision B: the ball and the rod each experience the same torque, time of collision, and change in angular velocity

  1. It is known that a lab cart is moving east at 25 cm/s at time t1 = 0.10 s, and then moving east at 15 cm/s at t 2 = 0.20 s. Is this enough information to determine the direction of the net force acting on the cart between t 1 and t 2 ?

(A)   Yes, since we know the cart is slowing down, its momentum change is opposite the direction of movement, and the net force is in the direction of momentum change.

(B)   No, because we don’t know whether forces such as friction or air resistance might be acting on the cart.

(C)   No, because we don’t know the mass of the cart.

(D)   Yes, since we know the cart keeps moving to the east, the net force must be in the direction of motion.

  1. A rigid rod is pivoted at its right end. Three forces of identical magnitude but different directions are applied at the positions 1, 2, and 3 as shown. Which of the following correctly ranks the torques τ1 , τ 2 , and τ 3 provided by the forces F 1 , F 2 , and F 3 ?

(A)   τ 1 > τ 2 > τ 3

(B)   τ 3 > τ 2 > τ 1

(C)   τ 2 > τ 1 > τ 3

(D)   τ 2 > τ 1 = τ 3

  1. A block hanging vertically from a spring undergoes simple harmonic motion. Which of the following graphs could represent the acceleration aas a function of position x for this block, where x = 0 is the midpoint of the harmonic motion?

(A)   

(B)   

(C)   

(D)   

  1. The preceding diagram represents a photograph of three transverse waves, each of which is moving to the right in the same material as the others. Which of the following ranks the waves by their amplitudes?

(A)   A = B > C

(B)   B > C > A

(C)   A > C > B

(D)   A = B = C

  1. The mass of the Earth is 5.97 × 1024kg. The Moon, whose center is 3.84 × 108 m from the Earth’s center, has mass 7.35 × 1022 kg. Which of the following is the best estimate of the gravitational force of the Earth on the Moon?

(A)   1039 N

(B)   1029 N

(C)   1019 N

(D)   109 N

  1. A children’s toy consists of a cart whose very light wheels are attached to a rubber band. This rubber band can wind and unwind around the axle supporting the wheels.

This toy is given a shove, after which the toy rolls across a flat surface and up a ramp. It is observed that the toy does not go a consistent distance up the ramp—in some trials it ends up higher than in other trials, even though the shove imparts the same kinetic energy to the cart each time. Which of the following is a reasonable explanation for this phenomenon?

(A)   Depending on how the rubber band is initially wound, more or less potential energy can be transferred from the rubber band to the kinetic energy of the car’s motion.

(B)   The normal force on the cart’s wheels will be different depending on how much the rubber band winds or unwinds.

(C)   How much energy is transferred from kinetic energy to gravitational potential energy depends on the vertical height at which the cart ends up.

(D)   Some of the cart’s initial kinetic energy will be dissipated due to work done by friction.

  1. A man stands on a platform scale in an elevator. The elevator moves upward, speeding up. What is the action-reaction force pair to the man’s weight?

(A)   The force of the elevator cable on the man

(B)   The force of the man on the scale

(C)   The force of the elevator cable on the elevator

(D)   The force of the man on the Earth

  1. The preceding diagram shows a speaker mounted on a cart that moves to the right at constant speed v. Wave fronts for the constant-frequency sound wave produced by the speaker are indicated schematically in the diagram. Which of the following could represent the wave fronts produced by the stationary speaker playing the same note?

(A)   

(B)   

(C)   

(D)   

  1. A table supports a wooden block placed on the tabletop. Which fundamental force of nature is responsible for this interaction, and why?

(A)   The electric force, because the protons in the nuclei of the top atomic layer of the table repel the nuclei in the bottom atomic layer of the wood.

(B)   The gravitational force, because by F = GMm/r 2 , the force of the table on the wood at that close range is sufficient to balance the force of the Earth on the wood.

(C)   The electric force, because the outer electrons in the top atomic layer of the table repel the outer electrons in the bottom atomic layer of the wood.

(D)   The strong nuclear force, because the protons in the nuclei of the top atomic layer of the table repel the nuclei in the bottom atomic layer of the wood.

  1. A solid sphere (I= 0.06 kg·m2 ) spins freely around an axis through its center at an angular speed of 20 rad/s. It is desired to bring the sphere to rest by applying a friction force of magnitude 2.0 N to the sphere’s outer surface, a distance of 0.30 m from the sphere’s center. How much time will it take the sphere to come to rest?

(A)   4 s

(B)   2 s

(C)   0.06 s

(D)   0.03 s

  1. Which of the following force diagrams could represent the forces acting on a block that slides to the right while slowing down?

(A)   

(B)   

(C)   

(D)   

  1. Standing waves are produced by a 100-Hz generator in a string of fixed length. The tension in the string is increased until a new set of standing waves is produced. Will the wavelength of the new standing waves be greater than or less than the wavelength of the original standing waves?

(A)   Less, because the tension in the string varies directly with the wave speed, which varies inversely with the wavelength.

(B)   Greater, because the tension in the string varies directly with the wave speed, which varies inversely with the wavelength.

(C)   Greater, because the tension in the string varies directly with the wave speed, which varies directly with the wavelength.

(D)   Less, because the tension in the string varies directly with the wave speed, which varies directly with the wavelength.

  1. Two electrically charged balls are separated by a short distance, producing a force of 50 μN between them. Keeping the charge of each ball the same, the mass of one of the balls but not the other is doubled. What is the new electric force between the balls?

(A)   50 μN

(B)   100 μN

(C)   200 μN

(D)   400 μN

  1. A block of mass mis attached to a spring of force constant k . The mass is stretched a distance A from equilibrium and released from rest. At a distance x from the equilibrium position, which of the following represents the kinetic energy of the block?

(A)   

(B)   

(C)   

(D)   

  1. A man stands with his hands to his sides on a frictionless platform that is rotating. Which of the following could change the angular momentum of the man-platform system?

(A)   The man catches a baseball thrown to him by a friend.

(B)   The man thrusts his arms out away from his body

(C)   The man thrusts his arms out away from his body, and then quickly brings his arms back to his side again.

(D)   The man jumps straight up in the air and lands back on the platform.

Questions 46–50: Multiple-Correct Items

Directions: Identify exactly two of the four answer choices as correct and grid the answers with a pencil on the answer sheet. No partial credit is awarded; both of the correct choices, and none of the incorrect choices, must be marked for credit.

  1. The distance between the centers of two objects is d. Each object has identical mass m and identical charge –q . Choose all of the correct statements about the similarities and differences between the electric and gravitational force between the two objects. Select two answers.

(A)   Both the electric and the gravitational force depend inversely on the square of the distance d .

(B)   Just as the gravitational force depends on the sum of the two masses m and m , the electric force depends on the sum of the two charges –q and –q .

(C)   For any measureable m and q in the laboratory, the electric force is many orders of magnitude larger than the gravitational force.

(D)   Both the electric and gravitational forces are attractive.

  1. Which placement of voltmeters will allow for determination of the voltage across resistor R2 in the circuit diagrammed in the preceding figure? Select two answers.

(A)   

(B)   

(C)   

(D)   

  1. A student of mass 50 kg stands on a scale in an elevator. The scale reads 800 N. Which of the following could describe how the elevator is moving? Select two answers.

(A)   Moving downward and slowing down

(B)   Moving downward and speeding up

(C)   Moving upward and speeding up

(D)   Moving upward and slowing down

  1. A 1-m-long pipe is closed at one end. The speed of sound in the pipe is 300 m/s. Which of the following frequencies will resonate in the pipe? Select two answers.

(A)   75 Hz

(B)   150 Hz

(C)   225 Hz

(D)   300 Hz

  1. The device shown in the preceding figure consists of two wheels connected by a thick axle. A force can be applied to the axle by pulling a rope at several positions along the axle. Assuming the spool does not slip on the table, which of the pictured applied forces would cause rotation to the right of the device’s wheels? Select two answers.

(A)   F 1

(B)   F 2

(C)   F 3

(D)   F 4

STOP. End of Physics 1 Practice Exam 1—Multiple-Choice Questions

AP Physics 1 Practice Exam 1: Section II (Free-Response)

Directions: The free-response section consists of five questions to be answered in 90 minutes. Budget approximately 20 to 25 minutes each for the first two longer questions; the next three shorter questions should take about 12 to 17 minutes each. Explain all solutions thoroughly, as partial credit is available. On the actual test, you will write the answers in the test booklet; for this practice exam, you will need to write your answers on a separate sheet of paper.

  1. (12 points)

In Experiment 1, two carts collide on a negligible-friction track: Cart A with mass 500 g, and Cart B with unknown mass. Before the collision, Cart B is at rest. Adhesive is attached to the carts such that after the collision, the carts stick together. The speeds of Cart A before collision and after collision are measured using the sonic motion detector, as shown in the diagram.

(a)   In one trial, the motion detector is turned on, Cart A is given a shove, the carts collide, and then the detector is turned off. The detector produces the velocity-time graph shown as follows. On the graph, indicate with a circle the portion of the graph that represents the collision occurring. Explain how you figured this out.

(i)   Use the graph to estimate the speed of Cart A before the collision.

(ii)   Use the graph to estimate the speed of Cart A after the collision.

(b)   In numerous trials, the speeds of Cart A before and after the collision are measured. You are asked to construct a graph of this data whose slope can be used to calculate the mass of Cart B.

(i)   What should you graph on each axis?

(ii)   Explain in several sentences how you will use the slope of this graph to calculate the mass of Cart B. Be specific both about the calculations you will perform, and about why those calculations will produce the mass of Cart B.

(c)   In Experiment 2, the adhesive is removed such that the carts bounce off of one another. The motion detector is again positioned to read the speed of Cart A before and after collision.

(i)   Describe an experimental procedure by which the speed of Cart B after collision can be measured. You may use any equipment available in your physics laboratory, but you may not use a second sonic motion detector.

(ii)   The masses of Carts A and B are now both known; the speeds of both carts before and after collision have been measured. Explain how you could determine whether the collision in Experiment 2 was elastic. Be sure to describe specifically the calculations you would perform, as well as how you would use the results of those calculations to make the determination.

  1. (12 points)

A uniform meterstick, which weighs 1.5 N, is supported by two spring scales. One scale is attached 20 cm from the left-hand edge; the other scale is attached 30 cm from the right-hand edge, as shown in the preceding diagram.

(a)   Which scale indicates a greater force reading? Justify your answer qualitatively, with no equations or calculations.

(b)   Calculate the reading in each scale.

(c)   Now the right-hand scale is moved closer to the center of the meterstick but is still hanging to the right of center. Explain your answers to the following in words with reference to your calculations in (b).

(i)   Will the reading in the left-hand scale increase, decrease, or remain the same?

(ii)   Will the reading in the right-hand scale increase, decrease, or remain the same?

(d)   Now the scales are returned to their original locations, as in the diagram. Where on the meterstick could a 0.2-N weight be hung so as to increase the reading in the right-hand spring scale by the largest possible amount? Justify your answer.

  1. (7 points)

Space Probe A orbits in geostationary orbit directly above Jupiter’s equator, 90,000 km above the surface. Identical Space Probe B sits on the surface of Jupiter.

(a)   Which probe, if either, has a greater speed? Justify your answer

(b)   Which probe, if either, has the greatest acceleration toward the center of Jupiter? Justify your answer.

(c)   Consider the list of gravitational forces below:

  1. The force of Jupiter on Space Probe A
  2. The force of Jupiter on Space Probe B
  3. The force of Space Probe A on Jupiter
  4. The force of Space Probe B on Jupiter
  5. The force of Space Probe A on Space Probe B
  6. The force of Space Probe B on Space Probe A

(i)    Rank the magnitudes of the six gravitational forces listed above from greatest to least. If two or more quantities are the same, indicate so clearly in your ranking.

Greatest ______ ______ ______ ______ ______ ______ Least

(ii)   Justify your ranking.

  1. (7 points)

Two blocks, Block A of mass m and Block B of mass 2m , are attached together by a spring. The blocks are free to move on a level, frictionless surface. The spring is compressed and then the blocks are released from rest.

Consider two different systems. One system consists only of Block A; the other system consists of both blocks and the connecting spring. In a clear, coherent, paragraph-length response, explain whether kinetic energy, total mechanical energy, and/or linear momentum is conserved in each of the systems described.

  1. (7 points)

Three identical light bulbs are connected in the circuit shown above. The switch S is initially in the open position and then is closed at time t .

(a)   Describe any changes that occur in the current through each bulb when the switch is closed. Justify your answers.

(b)   Describe any changes that occur in the brightness of each bulb when the switch is closed. Justify your answers.

(c)   When the switch is closed, does the power output of the battery increase, decrease, or remain the same? Justify your answer.

(d)   When the switch is closed, the current in Bulb 1 changes. Explain why this change in current does not violate the law of conservation of charge.

STOP. End of Physics 1 Practice Exam 1—Free-Response Questions

Solutions: AP Physics 1 Practice Exam 1, Section I (Multiple-Choice)

Questions 1–45: Single-Choice Items

  1. C—A circuit allows charge to flow and changes the electrical energy of the charges in the wires; however, by conservation of charge, the net charge of the circuit can’t change unless some path is provided for charge to flow out of the circuit.
  2. A—Kirchoff’s junction rule, a statement of conservation of electric charge in a circuit, says that current entering a junction equals current leaving a junction. Current is charge flowing per second. The charge passing point A in one second must be equal to the sum of the charge passing B, C, and D in one second; A must have the greatest charge flow per second.
  3. D—The relevant equation connecting force and mass is F = ma . The slope is the vertical axis divided by the horizontal axis, or F/a = m . So the slope is mass—but why inertial not gravitational mass? Inertia is defined as an object’s resistance to acceleration. If acceleration is involved, you’re talking inertial mass. Gravitational mass would involve the weight of an object in a gravitational field.
  4. B—The dots divide the 1-meter distance into five parts. In the time between dots, the cart travels 1/5 of a meter, or 0.2 m. The time between dots is 1/20 of a second, or 0.05 s. At constant speed, the speed is given by distance/time: 0.20 m/0.05 s = 4 m/s.
  5. D—Initially, the cart’s mass is 0.5 kg and speed is 4 m/s, so the cart’s momentum is mv = 2 N·s. In the collision, the cart loses that 2 N·s in order to stop briefly and then gains more momentum in order to speed up again. So the momentum change must be more than 2 N·s. How much more? After collision, the cart is moving slower than 4 m/s because the dots are closer together, so the cart’s momentum is less than 2 N·s. The cart’s momentum change is (2 N·s) + (something less than 2 N·s); the only possible answer is 2.7 N·s.
  6. D—On a standing wave, the wavelength is measured from node-to-node-to-node (i.e., across two “humps”).
  7. D—The rotational inertia of a point mass is MR 2 , where R is the distance from the mass to the axis of rotation. Pretend the side of the square is of length 2 m, and that each mass is 1 kg. For axis A, each mass has rotational inertia (1 kg) (1 m)2 = 1 kg·m2 . With four masses total, that’s 4 kg·m2 . For axis B, each mass is  m from the axis (the diagonal of the square is 2 m, each mass is half a diagonal from the axis). Each mass has (1 kg)( m)2 = 2 kg·m2 . Two masses make a total of 4 kg·m2 . And for axis C, the masses are each 2 m from the axis, so they each have (1 kg) × (2 m)2 = 4 kg·m2 . With two masses, that’s a total of 8 kg·m2 . So this would be ranked axis C, followed by equal axes A and B.
  8. C—The work done by the force is force times distance; but that’s not an option. The other option to find work is that work done by a nonconservative force is equal to the change in an object’s potential and kinetic energy. There’s no potential energy change, because the surface is horizontal. The kinetic energy ( mv 2 ) change can be determined by knowing the mass and the speed change.
  9. B—When two pulses superimpose, the amplitudes add algebraically. That means that a part of the wave pulse on the top of the rope is canceled by a part of the wave pulse on the bottom of the rope. Since the middle part of the wave pulse is closer to the rope than the two humps, the middle hump of Choice B will cancel this pulse—its middle part is also closer to the rope than its humps.
  10. A—Change in momentum is also known as impulse and is equal to force times time interval. On this graph, the multiplication of the axes means to take the area under the graph. Each segment of the data looks like it represents a straight line, making a big triangle. The area of a triangle is ½(base)(height). That’s ½(5 N)(2 s) = 5 N·s.
  11. B—If friction and air resistance are negligible, a mass on a spring oscillates about the equilibrium position, reaching the same maximum distance above and below. In this case, since the mass doesn’t get all the way to position A at the top, mechanical energy was lost (to friction or air resistance or some nonconservative force). Thus, without some external energy input, the mass won’t reach its maximum position at the bottom, either—at the bottom it will have no kinetic energy, so all the energy will be potential, and we’ve already established that some total mechanical energy was lost.
  12. C—If Ball B is forced to the right by Ball A, then Ball A must be forced to the left by Ball B—that’s Newton’s third law. These balls repel—only like-signed charges repel.
  13. B—Coulomb’s law gives the force of one charge on another,  . We haven’t changed the charges Q ; it’s the force of A on B that’s quadrupled. Since the force F has increased, the distance between charges r has decreased. Since the force has quadrupled, and since the r in the denominator is square, the distance has been cut in half (1/2)2 = 1/4; and making the denominator four times smaller makes the whole fraction four times bigger.
  14. B—In a collision, momentum—including angular momentum—is conserved. The question might as well be asking, “What is the angular momentum of the two objects before the collision?” And since the disk is at rest initially, the question is asking the even easier question, “What is the angular momentum of the putty before collision?” The axis of rotation is the center of the disk. The putty is a point mass; the angular momentum of a point mass is mvr with r the distance of closest approach to the axis. That’s (0.1 kg)(10 m/s)(1 m) = 1 kg·m2 /s.
  15. C—Mechanical energy is not conserved because of the work done by the nonconservative friction force provided by the rough-surfaced incline. The object starts with only gravitational potential energy, because it is higher than its lowest position and at rest. This gravitational energy is converted to thermal energy via the friction force, and to kinetic energy because the object speeds up.
  16. A—First simplify the 4 Ω and 12 Ω parallel combination to a 3 Ω equivalent resistance. In series with the other 3 Ω resistance, that gives a total resistance for the circuit of 6 Ω. By Ohm’s law used on the whole circuit, 12 V = I (6 Ω). The total current in the circuit is thus 2 A. This current must split between the two parallel resistors. The only possible answer, then, is 1.5 Ω—the current in the 4 Ω resistor must be less than the total current.
  17. C—The amplitude of a sound is related to the loudness. Sound 1 has higher amplitude, so it is louder. The pitch of a sound is related to the frequency. Since the horizontal axis is time, the peak-to-peak distance corresponds to a period, which is the inverse of the frequency. Sound 1 has a smaller period, so it has a larger frequency and a larger pitch.
  18. B—The gravitational field at the surface of a planet is  . The numerator for Mars is 1/10 that of Earth, reducing the gravitational field by 1/10. The denominator for Mars is (1/2)2 = 1/4 that of Earth, increasing the gravitational field by a factor of 4 (because a smaller denominator means a bigger fraction). The overall gravitational field is multiplied by 4/10. On Earth, the gravitational field is 10 N/kg, so on Mars, g = 4 N/kg.
  19. B—Choices C and D are wrong because kinetic energy doesn’t have a direction. Choice A is wrong because momentum conservation does not require a leftward momentum component—since the initial momentum was all to the right, the final momentum should be to the right. It’s the vertical momentum that’s the problem. Since the vertical momentum was zero to start with, any vertical momentum after collision must cancel out.
  20. A—Choice B is ridiculous—scientists should never refer generically to “human error.” Significant friction should reduce, not increase, the speed (and thus the momentum) measured by Photogate 2. The elasticity of a collision refers to kinetic energy conservation, not momentum conservation—even inelastic collisions must conserve momentum. If the track is slanted downhill to the right, then the carts speed up; conservation of momentum won’t be valid between Photogates 1 and 2 because the downhill component of the gravitational force is a force external to the two-cart system.
  21. B—The energy input must be enough to change the translational kinetic energy of the cart and to change the rotational kinetic energy of the wheels. Since all carts have the same mass and change speed by the same amount, they all require the same energy input to change the translational KE. Whichever wheels have the largest rotational inertia will require the largest energy input to get to the same speed. Calculating, wagon B has the largest rotational inertia of 0.004 kg·m2 .
  22. A—Choice C is not correct because if the swimmer is accelerating, the responsible force must act on the swimmer, not on something else. That force can act on any part of the swimmer’s body. But a force provided by the swimmer himself on the swimmer himself won’t accelerate him—that’s like pulling yourself up by your own bootstraps. Choices B and D do not consider a force external to the swimmer. The answer is A: the Newton’s third law force pair to the force of the swimmer’s arms on the water.
  23. A—The work-energy theorem says that the work done by a nonconservative force is equal to the change in potential energy plus the change in kinetic energy. Since the wagon is on a horizontal surface, the potential energy change is zero; the work done by the 30-N pulling force is the change in the wagon’s KE. Work is force times parallel displacement, so we don’t use 30 N in this formula, we use the component of the 30-N force parallel to the 20-m displacement. That’s (30 N)(cos 60)(20 m) = 300 N.
  24. A—Yes, momentum is conserved. That means the carts combined have the same momentum as they did in sum before the collision. And sure, if we were solving for the speed after collision, we’d combine the masses together for the calculation, but that doesn’t mean that the carts both have the same mass—one cart is 1.5 kg, the other 0.5 kg. The problem asks what is the same for each cart in comparison to the other. Since the carts are stuck together, they must move together. They have the same velocity as one another.
  25. B—Current can only travel through a wire. At the junction after Resistor 1, the current splits; the current comes back together after resistors 2 and 3. Thus, resistors 1 and 4 have the same current that’s equal to the total coming from the battery. Since resistors 2 and 3 are identical, they split the current evenly.
  26. D—Since resistors 1 and 4 take the same current and are identical, the voltage drops the same amount across them. Since Resistor 2 takes less current than the others but has the same resistance, by V = IR the voltage drops less across Resistor 2 than the others. Choice A is wrong because all voltage drops are the same. Choices B and C have Resistor 2 dropping the voltage by more than at least one of the other resistors. Choice D includes the same voltage drop across resistors 1 and 4, but a smaller drop across 2.
  27. C—The work done by the spring is the area under a force-distance graph because work = force times distance. Using the best-fit line drawn as the top of a triangle, the area is (1/2) × (300 N)(20 cm) = 3,000 N.cm.* Now, put your ruler along the data points. Try to draw another line that’s still a reasonable best fit, but is a bit shallower. Where does that line intersect the 20-cm position? It intersects at a point probably not much below 280 N, maybe even 290 N. The smallest possible work done, given this data, would be area = (1/2)(280 N)(20 cm) = 2,800 N.cm, which is 200 N.cm short of the 3,000 N.cm original estimate. That’s closest to Choice C. If you’ve done a lot of in-class lab work, you might have noticed that your data often look about as scattered as shown in the graph; and that anything you calculate is never much closer to a known value or to your classmates’ calculations than 5 or 10 percent. Here, Choice B works out to an uncertainty of 1 percent; Choice D is 100 percent. So C is the reasonable choice.
  28. A—Standing waves on a string only exist when the wave pattern produces a node at each end. A wavelength is measured node-to-node-to-node; so half a wavelength is node-to-node. If the string length isn’t a multiple of this node-to-node distance, then some fraction of a wave pattern would be left on one end rather than a node. That can’t happen on a fixed string, so no standing waves exist.
  29. A—Since the platform itself is of negligible mass, only two torques act on the platform: counterclockwise by the force of the force probe (F P ), and clockwise by the downward force of the object (mg ). Torque is force times distance from a fulcrum, so set these torques equal: F P (x ) = mg (d ). Solving for the mass, we get  . Plot the numerator on the vertical, the denominator on the horizontal, and the slope is the mass m .
  30. B—The problem says nothing about the collision being “elastic”; therefore, the change in kinetic energy of any sort does not have to be the same for each object. Velocity is never conserved in a collision—momentum is—so Choice D is ridiculous. Newton’s third law demands that the force of one cart on another is the same, and so also the torque on each object about the center of mass.
  31. A—Net force includes the contributions of all forces, including friction or air resistance; and net force is in the direction of acceleration , not of motion.
  32. A—Torque is force times distance from a fulcrum; but that force must be perpendicular to the rod, so in this case the force used in the equation will be the vertical component, which includes a sin 45o term for F 1 and F 3 . The sine of 45o is 0.7; call the length of the rod L , so F 2 is a distance L /2 from the pivot, and F 3 is about L /4 from the pivot. So τ 1 = 0.7FL. τ 2 = 0.5FL. τ 3 = (0.7·0.25)FL .
  33. D—The net force of a spring is kx , and so changes linearly with distance (because the x is not squared or square rooted). Acceleration is related to the net force, so acceleration also changes linearly with distance. The force is always toward the equilibrium position: when the block is pulled down, it is forced (and thus accelerates) up; when the block is pushed up, it accelerates down. So a negative x gives a positive a , as in (D).
  34. A—The amplitude is measured from the midpoint to the peak or trough of a wave. Waves A and B are each two “bars” above the midpoint, while C is only one bar above the midpoint.
  35. C—Use the equation  but just use the power of 10 associated with the value—the answer choices are so far separated that more precision would be useless.  . Add exponents in the numerator because everything is multiplied together:  . Now subtract exponents for the division problem: F = 1019 N.
  36. A—Choice B is wrong—the normal force on the flat surface is equal to the cart’s weight, regardless of the rubber band. Choice C is true but does not explain different heights in each trial—the problem said that the kinetic energy provided to the cart was the same every time. Choice D may or may not be true but is irrelevant in any case—even if kinetic energy is lost to work done by friction, neither the force of nor the coefficient of friction changes in different trials, so that can’t explain different heights. Now, the rubber band, though, that can change things. If it’s initially wound and able to unwind as the cart moves, it can transfer some of its elastic potential energy to kinetic energy of the cart. Or, if it’s initially unwound, it will require some kinetic energy in order to wind up again and store elastic potential energy.
  37. D—The man’s weight is the force of the Earth on the man. The Newton’s third law force pair is then the force of the man on the Earth.
  38. C—The Doppler effect states that a moving speaker will cause a stationary observer to hear a higher frequency. This is because the wave fronts will be emitted closer together than if the speaker were stationary—the waves will seem “scrunched” together. Well, for the wave fronts to be “scrunched” in the diagram for the moving cart, they’d have to be farther apart in the diagram for the stationary cart.
  39. C—Choice B is ridiculous because not only is gravity the weakest of the fundamental forces, it makes no sense that the gravitational force of the maybe 100-kg table is similar to the gravitational force of the 1024 kg Earth in the equation  . Choices A and D are ridiculous because the protons are about 105 (100,000) times smaller than the diameter of an atom—any interactions between atoms have to involve the electrons at the outside of the atoms, not protons far away in the center.
  40. B—This is a calculation using τ net =  . The net torque on the sphere is force times distance from the center, or (2.0 N)(0.30 m) = 0.60 m·N. Now the angular acceleration can be calculated: (0.60 m·N) = (0.06 kg·m2 )(α ), so α = 10 rad/s per second. Use the definition of angular acceleration: the sphere loses 10 rad/s of speed each second. It started with 20 rad/s of speed, so after 1 s it has 10 rad/s of speed; after 2 s it will have lost all its speed.
  41. A—There’s no such thing as the “force of motion,” so motion does not belong on a force diagram. The block slows down, so the block has acceleration in the opposite direction of motion, or left. That means there must be a leftward force.
  42. C—The speed of waves on a string is given by  . All we really need to know is that the bigger the tension, the bigger the wave speed—we know that because tension is in the numerator. Then use v = λf : wavelength is in the numerator, so wave speed varies directly with wavelength.
  43. A—The electric force of one charge on another is given by  . The mass is not in this equation; doubling the mass does nothing to the electric force.
  44. A—Here use the work-energy theorem, W NC = ΔPE + ΔKE . The change in kinetic energy is what we’re looking for, since the block began at rest. The only force doing work is the spring force, and that’s a conservative force, so W NC = 0. What we want, then, is the change in the block’s potential energy, or the initial and final potential energies subtracted from one another. The potential energy of a block on a spring is ½kx 2 . What values do we use for x here? The initial and final positions are used. At first the block is at position x = A . At the end, the block is at a distance called x from equilibrium.
  45. A—Angular momentum can only change when a torque that’s external to the man-platform system acts; that is, when a torque is applied by something that isn’t the man or the platform. The force of the baseball on the man acting anywhere but precisely at the center of the man’s rotational motion will provide a torque, and thus change the man-platform’s angular momentum. The other choices all involve interactions only between the man and the platform.

Questions 46–50: Multiple-Correct Items (You must indicate both correct answers; no partial credit is awarded.)

  1. Aand C —A is right because of the equations  and  . B is incorrect—the numerators involve the product, not the sum, of masses and charges. C is correct; you could memorize that fact, or you could try plugging in the powers of 10 for measurable masses and charges. Try charges as small as nanocoulombs, or 10−9 C, and milligrams, or 10−6 kg. Since k for the electric force equation is 109 N·m2 /C2 , and G for the gravitational force equation is 10−11 N·m2 /kg2 , the electric force will always come out much bigger. D is incorrect because while gravity always attracts, only opposite-signed charges attract.
  2. Aand C —This is essentially Kirchoff’s loop rule, which boils down to “items in parallel take the same voltage.” It’s pretty easy to see that the voltmeter in (A) is in parallel with R 2 , but in (C) it’s harder. Note that the current must divide between R 2 and the voltmeter, and then the current comes immediately back together after going through the voltmeter and R 2 . That’s parallel. The combination of R 3 and R 4 are also in parallel with the voltmeter here, but that’s not relevant to the problem. Choice (B) is wrong because the voltmeter is in series with R 2 . Choice (D) is wrong because the voltmeter is in parallel with the battery, which is not in parallel with R 2 , because R 1is in the way.
  3. Aand C —A free-body diagram would include the student’s 500-N (50 kg·g ) weight downward, and the scale’s 800-N force upward. (No, there’s no “force of the elevator.” The elevator isn’t in contact with the student.) So the net force is upward; and the acceleration is in the direction of the net force, also upward. Upward acceleration could mean one of two things: speeding up and moving upward, or slowing down and moving downward.
  4. Aand C —The fundamental frequency of a standing wave on a pipe closed at one end is  . Here, with v = 300 m/s and L = 1 m, the fundamental frequency is 75 Hz. On a closed pipe, odd multiples of the fundamental frequency will resonate. So 75 Hz, 3·75 Hz = 225 Hz, and 5·75 Hz = 375 Hz, with nothing in between.
  5. Aand D —The key is knowing where the “fulcrum,” or the pivot for rotation, is. Here, that’s the contact point between the surface and the wheel. Now look at the line of the force. In F 1 , the line of force is pulling right above the pivot point, and that causes clockwise, or forward, rotation. For F 2 , the line of force is pulling up to the right of the pivot, and that causes counterclockwise, or backward, rotation. F 3 ’s line of force is going right through the pivot point, so it will cause no rotation at all. F 4 pulls up with a line of force to the left of the pivot. This produces clockwise, or forward, rotation.

* The units are N·cm rather than joules because a joule is a newton times a meter, not a newton times a centimeter.

Solutions: AP Physics 1 Practice Exam 1, Section II (Free-Response)

Obviously your solutions will not be word-for-word identical to what is written below. Award points for your answer as long as it contains the correct physics, and as long as it does not contain incorrect physics.

Question 1

Part (a)

1 point for a correct circle on the diagram.

1 point for a correct explanation: When the carts collide, the moving cart must lose speed. The vertical axis of the velocity-time graph indicates speed. The circled portion of the graph is the only place where the vertical axis value drops rapidly, as the cart’s speed must drop in the collision.

(i)    1 point for answer: 0.60 m/s (or thereabouts—anywhere between, say, 0.55 m/s and 0.60 m/s is fine.)

(ii)   1 point for answer: 0.25 m/s (or thereabouts—anything between, say, 0.22 m/s and 0.29 m/s is fine.)

Part (b)

(i)   1 point: The easiest answer is to put the speed of Cart A before collision on the vertical axis; and to put the speed of Cart A after collision on the horizontal axis. There are other answers that will work.

(ii)   3 points: Award one or two points of partial credit for correct but incomplete physics. For example, writing and using an expression for momentum conservation should earn a point, even if the rest of the explanation doesn’t follow properly.

Conservation of momentum means the total momentum before the collision equals the total momentum after the collision. Before the collision, the total momentum is that of Cart A : m A v A . After the collision, the total momentum is (m A + m B )v ′ where v ′is the speed of Cart A (and, because they stick together, the speed of Cart B, too). Set these momentum expressions equal:

m A v A = (m A + m B )v ′.

This equation can be solved for the y- axis variable divided by the x -axis variable:

So, to get the mass of Cart B, I’d determine the slope of the line on the graph, and set that equal to  . The mass of Cart A is given as 500 g, so I’d plug that in and solve for m B .

Part (c)

(i)   2 points: Award one point for a partially correct description, two points for a complete and correct description.

Three ideas occur, but many are possible:

  • Measure the distance that Cart B has to travel to the end of the track. When the carts collide, start a stopwatch; when Cart B hits the end of the track, stop the stopwatch. The speed of Cart B is the distance you measured divided by the time on the stopwatch.
  • After the collision, when the detector has already read the speed of Cart A but before Cart B reaches the end of the track, lift up Cart A. Now the detector can read Cart B’s speed.
  • Let Cart B roll off the end of the track and fall to the floor as a projectile. Measure the vertical height yof the track off the ground; the time t that the cart was in the air is given by y = ½gt 2 , where g is 10 m/s per second. Measure the horizontal distance from the track’s edge to the spot where the cart landed. Then the speed of the cart is this horizontal distance divided by the calculated time of flight.

(ii)   2 points: Award one point for a partially correct description, two points for a complete and correct description.

“Elastic” means that the total kinetic energy of the two carts was the same before and after collision. Before the collision, the only kinetic energy is that of Cart A: ½m A v A 2 . After the collision, the total kinetic energy is the sum of the kinetic energy of both carts, where each cart’s kinetic energy is given by ½mv 2 . Compare the total kinetic energy after collision to Cart A’s kinetic energy before collision. If these values are equal, the collision was elastic. If the kinetic energy after the collision is less than the kinetic energy before collision, the collision was not elastic.

Question 2

Part (a)

2 points: Award one point for a partially correct description, two points for a complete and correct description. Consider the center of the meterstick as the fulcrum; then the weight of the meterstick provides no torque. The oppositely directed torques applied by each scale must be equal, because the meterstick is in equilibrium. Torque is force times distance from the fulcrum; since the right-hand scale’s torque calculation includes a smaller distance from the fulcrum, the right-hand scale must apply more force in order to multiply to the same torque.

Part (b)

4 points: Full credit for a complete and correct answer. Award three points partial credit for a correct approach with incorrect answers. Award two points for a correct approach and correct answer for one of the scales, but not the other. Award at least one point if the answer involved some use of torque equilibrium.

For this calculation, consider the left-hand scale as the fulcrum—that way, the left-hand scale provides no torque, and we only have to solve for one unknown variable. Set counterclockwise torques equal to clockwise torques, with T2 the reading in the right-hand scale.

The weight of the meterstick provides the clockwise torque; the right-hand scale provides the counterclockwise torque.

∗∗ (T 2 )(50 cm) = (1.5 N)(30 cm)

Solve for T 2 to get T 2 = 0.9 N

Next, the sum of the scale readings has to be the 1.5 N weight of the meterstick:

0.9 N + T 1 = 1.5 N

Giving T 1 = 0.6 N

Part (c)

(i)   2 points: Award one point for a partially correct description, two points for a complete and correct description.

Look at the starred calculation in Part (b). By moving the right-hand scale closer to the center, the scale will be less than 50 cm from the left-hand scale; but the meterstick’s center will still be 30 cm from the fulcrum. So when we solve for T 2 , we’re dividing (1.5 N) (30 cm) by a smaller value, giving a bigger T 2 reading.

But the question asks for the reading in the left-hand scale, which adds to T 2 to the same 1.5 N. A bigger T 2 adds to a smaller T 1 to get 1.5 N. Answer: decrease.

(ii)   2 points: Award one point for a partially correct description, two points for a complete and correct description.

See Part (i): T 2 , the reading in the right-hand scale, will increase.

Part (d)

2 points: Award one point for a partially correct description, two points for a complete and correct description.

Again, start from the equilibrium of torques using the left-hand scale as the fulcrum:

(T 2 )(50 cm) = (1.5 N)(30 cm)

Hanging a 0.2-N weight would provide a clockwise torque that would add to the torque applied by the meterstick’s weight on the right of this equation. Algebraically, T 2 is increased by adding to the numerator of the right side of this equation. We want to add the biggest possible torque.

Torque is force times distance from the fulcrum. We want, then, the largest possible distance from the fulcrum, which would be the right-hand edge of the meterstick, 80 cm from the left-hand scale.

Question 3

Part (a)

1 point for both a correct answer and a correct justification.

Probe A. The probe’s speed is the circumference of its circular motion divided by its period of revolution. We already established that the period is the same for each. Probe A has a bigger orbital radius, meaning a larger circumference of its circular motion, meaning a greater speed.

Part (b)

2 points : Award one point for the correct answer with a partially correct justification; award both points for the fully correct answer and justification.

Probe A. The centripetal acceleration is  . The problem is that Probe A has both a larger speed v and a larger orbital radius r . In order to answer the question, it’s necessary to replace the speed v by circumference over period,  . Now the acceleration is  . Okay, now we know: both probes have the same orbital period T , and r is in the numerator. The bigger-radius orbit—Probe A—has the greater acceleration.

Part (c)

(i)   1 point for correct ranking

Greatest      2 = 4 > 1 = 3 > 5 = 6     Least

(ii)   3 points: Award one point for justifying all three sets of force pairs set equal. Award one more point for justifying act least one correct portion of the ranking. Award the third point for justifying a second correct portion of the ranking.

By Newton’s Third Law, the three force pairs can be immediately set equal: that’s #1 with #3, #2 with #4, and #5 with #6. Next, we know that the force of Jupiter on either probe is given by  , where M and m are the masses of Jupiter and the probe, respectively. Since the probes are identical, the numerator is the same for both #1 and #2, but the distance of the probe from Jupiter’s center is smaller for Probe B. Therefore, Probe B experiences more force, and force #2 is greater than force #1. As for force #5, Jupiter is an enormous planet, many times more massive than Earth, even. There’s no way that the product of the space probes’ masses can ever approach Jupiter’s mass, meaning that the numerator of the force equation must be way smaller for force #5.

Question 4

The paragraph response must discuss kinetic energy, total mechanical energy, and linear momentum for each of the two systems. For each of these three quantities in each system, award one point for correctly explaining whether it is conserved and correctly justifying why it is or isn’t conserved. For example:

1 point: In system A, kinetic energy is not conserved. When the blocks are released, Block A speeds up away from Block B. Kinetic energy depends on mass and speed only. Since Block A’s speed increases without changing its mass, kinetic energy cannot remain constant.

1 point: In system A, total mechanical energy is not conserved. Since the system consists only of Block A, there is no interaction with another object that would allow for the storage of potential energy. The force of the spring on Block A would be a force external to the system, and the spring does work on Block A because Block A moves parallel to the spring force; when a net force external to the system does work, mechanical energy is not conserved.

1 point: In system A, linear momentum is not conserved. Either the reasoning for system A’s kinetic energy or total mechanical energy can be extended here. Linear momentum depends on mass and speed and Block A’s speed changes without changing mass. Or, the spring force is external to the system, and momentum is only conserved in systems for which no net external force acts.

1 point: In system B, kinetic energy is not conserved. Kinetic energy is a scalar, so kinetic energy of a system of objects is just the addition of the kinetic energies of all the objects in the system. Both blocks speed up, so both blocks are increasing their kinetic energy, increasing the system’s kinetic energy.

1 point: In system B, total mechanical energy is conserved. No force external to the spring-blocks system does work, so mechanical energy is conserved. The kinetic energy gained by the blocks was converted from potential energy stored in the spring.

1 point: In system B, linear momentum is conserved. No force external to the spring-blocks system acts, so linear momentum is conserved. Here even though Block B gains linear momentum, momentum is a vector—its gain of momentum is canceled by the momentum gained by Block A in the opposite direction.

Add 1 point if the paragraph correctly states whether each quantity is conserved in each system, regardless of whether the justifications are legitimate.

Question 5

Part (a)

1 point for correctly identifying and justifying Bulb 1’s current increase and 1 point for correctly identifying and justifying Bulb 3’s decreased current.

Initially, the circuit is just Bulbs 1 and 3 in series. When Bulb 2 is added, the voltage from the battery is unchanged. Yet the total resistance of the circuit decreases, because an additional parallel path is added. Therefore, by V = IR with constant V , the total current in the circuit increases.

Bulb 1 takes the total current, so Bulb 1’s current increases. For Bulb 1 only , the resistance is a property of the bulb and thus doesn’t change. So by V = IR with constant R , Bulb 1 takes an increased voltage, too.

Then by Kirchoff’s loop rule, an increase voltage across Bulb 1 means a decreased voltage across Bulb 3. And for Bulb 3 only, by V = IR with constant R , Bulb 3’s current also decreases. (Obviously Bulb 2’s current increases from nothing to something.)

Part (b)

1 point for either a correct answer with correct justification; or, for an answer consistent with the answers to Part (a) with reference to the power dissipated by the bulbs.

All bulbs have an unchanging resistance. Brightness depends on power, which is I 2 R . With constant R , a bigger current means more brightness; a smaller current means less brightness. So Bulb 1 gets brighter and Bulb 3 gets dimmer.

Part (c)

2 points for a fully correct answer with justification. One of these two points can be earned for a partially correct justification, or for an incorrect answer that is justified consistently with the answers to (A) or (B).

For the whole circuit, use power = V 2 /R . The voltage of the battery is unchanged because it’s still the same battery. The resistance of the circuit decreases because of the extra parallel path. So, decreasing the denominator increases the entire value of the equation, so power increases.

Part (d)

2 points for a complete and correct explanation. One of these two points can be earned by a partially correct, or an incomplete, justification.

Conservation of charge in circuits is expressed in Kirchoff’s junction rule—the current entering a junction equals the current leaving the junction. At any given moment of time, the junction rule holds. Now, when the switch is closed, more current flows from the battery than before. That’s not a violation of charge conservation, because the materials in the battery contain way more charged particles than are ever flowing through the wires. After the switch is closed, more current flows into the junction right before the switch than before, but more current also flows out of that junction than before. Charge conservation doesn’t mean that the same current must always flow in a circuit, it just says that whatever charge does flow in a circuit must flow along the wires.

Practice Exam 2

ANSWER SHEET FOR SECTION I

AP Physics 1 Practice Exam 2: Section I (Multiple-Choice)

Directions: The multiple-choice section consists of 50 questions to be answered in 90 minutes . You may write scratch work in the test booklet itself, but only the answers on the answer sheet will be scored. You may use a calculator, the equation sheet, and the table of information.

Questions 1–45: Single-Choice Items

Directions: Choose the single best answer from the four choices provided and grid the answer with a pencil on the answer sheet.

Questions 1 and 2 refer to the following information:

A person pulls on a string, causing a block to move to the left at a constant speed. The free body diagram shows the four forces acting on the block: the tension (T ) in the string, the normal force (F n ), the weight (W ), and the friction force (F f ). The coefficient of friction between the block and the table is 0.30.

  1. Which is the Newton’s third law force pair to T?

(A)   The force of the block on the string

(B)   The force of the block on the table

(C)   The force of the table on the block

(D)   The force of friction on the block

  1. Which of the following correctly ranks the four forces shown?

(A)   T > F f W = F n

(B)   W = F n T = F f

(C)   W = F n T > F f

(D)   W = F n T = F f

  1. Which of the following circuits shows a placement of meters and an observation of their readings that would allow researchers to experimentally demonstrate whether energy is conserved in the circuit?

(A)   Researchers look to see whether the readings on voltmeters 1 and 2 add to the reading in voltmeter 3.

(B)   Researchers look to see whether the voltmeter readings are equal.

(C)   Researchers look to see whether the readings on ammeters 1 and 2 add to the reading in ammeter 3.

(D)   Researchers look to see whether the ammeter readings are equal.

Questions 4 and 5 refer to the circuit shown in the figure, which includes a 9 V battery and three resistors.

  1. Which of the following ranks the resistors by the charge that flows through each in a given time interval?

(A)   300 Ω > 100 Ω = 50 Ω

(B)   50 Ω > 100 Ω > 300 Ω

(C)   300 Ω > 100 Ω > 50 Ω

(D)   50 Ω = 100 Ω > 300 Ω

  1. What is the voltage across the 50 Ω resistor?

(A)   9.0 V

(B)   6.0 V

(C)   3.0 V

(D)   1.0 V

  1. The charge on an oil drop is measured in the laboratory. Which of the following measurements should be rejected as highly unlikely to be correct?

(A)   6.4 × 10–19 C

(B)   8.0 × 10–19 C

(C)   4.8 × 10–19 C

(D)   2.4 × 10–19 C

  1. A cart attached to a spring initially moves in the xdirection at a speed of 0.40 m/s. The spring is neither stretched nor compressed at the cart’s initial position (x = 0.5 m). The figure shows a graph of the magnitude of the net force experienced by the cart as a function of x , with two areas under the graph labeled. Is it possible to analyze the graph to determine the change in the cart’s kinetic energy as it moves from its initial position to x = 1.5 m?

(A)   No, the cart’s mass must be known.

(B)   Yes, subtract area 2 from area 1.

(C)   Yes, add area 2 to area 1.

(D)   Yes, determine area 1.

  1. A horse is attached to a cart that is at rest behind it. Which force, or combination of forces, explains how the horse-cart system can accelerate from rest?

(A)   The forward static friction force of the ground on the horse is greater than any friction forces acting backward on the cart, providing a forward acceleration.

(B)   The forward force of the horse on the cart is greater than the backward force of the cart on the horse, providing a forward acceleration.

(C)   The force of the horse’s muscles on the rest of the horse-cart system provides the necessary acceleration.

(D)   The upward normal force of the ground on the horse is greater than the horse’s weight, providing an upward acceleration.

  1. A pipe full of air is closed at one end. A standing wave is produced in the pipe, causing the pipe to sound a note. Which of the following is a correct statement about the wave’s properties at the closed end of the pipe?

(A)   The pressure is at a node, but the particle displacement is at an antinode.

(B)   The pressure is at an antinode, but the particle displacement is at a node.

(C)   The pressure and the particle displacement are both at nodes.

(D)   The pressure and the particle displacement are both at antinodes.

  1. In the laboratory, a cart of mass mis held in place on a smooth incline by a rope attached to a spring scale, as shown in the figure. The angle of the incline from the horizontal θ varies between 0° and 90°. A graph of the reading in the spring scale as a function of the angle θ is produced. Which of the following will this graph look like?

(A)   

(B)   

(C)   

(D)   

Questions 11 and 12 refer to the following information: A bicycle wheel of known rotational inertia is mounted so that it rotates clockwise around a vertical axis, as shown in the first figure. Attached to the wheel’s edge is a rocket engine, which applies a clockwise torque τ on the wheel for a duration of 0.10 s as it burns. A plot of the angular position θ of the wheel as a function of time t is shown in the graph.

  1. In addition to the wheel’s rotational inertia and the duration of time the engine burns, which of the following information from the graph would allow determination of the net torque the rocket exerts on the wheel?

(A)   The area under the graph between t = 0 s and t = 3 s

(B)   The change in the graph’s slope before and after t = 2 s

(C)   The vertical axis reading of the graph at t = 3 s

(D)   The vertical axis reading of the graph at t = 2 s

  1. Which of the following graphs sketches the angular acceleration α of the wheel as a function of time?

(A)   

(B)   

(C)   

(D)   

  1. A rock drops onto a pond with a smooth surface. A few moments later, the wave produced by the rock’s impact arrives at the shore, touching the ankles of a wading child. Which of the following observations provides evidence that the portion of the wave hitting the child’s ankles carries less energy than the wave did when it was first created by the rock’s impact?

(A)   The wave is moving more slowly.

(B)   The wave pulse’s width has become greater.

(C)   The wave pulse’s width has become smaller.

(D)   The wave’s height has become smaller.

Questions 14 and 15 refer to the following information:

Two metal balls of equal mass (100 g) are separated by a distance (d ). In state 1, shown in the figure, the left ball has a charge of +20 µC, while the right ball has a charge of –20 µC. In state 2, both balls have an identical charge of –40 µC.

  1. Which of the following statements about the magnitudes of the electrostatic and gravitational forces between the two balls is correct?

(A)   The electrostatic force is greater in state 2 than in state 1. The electrostatic force is greater than the gravitational force in state 1, but the gravitational force is greater than the electrostatic force in state 2.

(B)   The electrostatic force is greater in state 1 than in state 2. The electrostatic force is greater than the gravitational force in state 1, but the gravitational force is greater than the electrostatic force in state 2.

(C)   The electrostatic force is greater in state 2 than in state 1. The electrostatic force is greater than the gravitational force in both states.

(D)   The electrostatic force is greater in state 1 than in state 2. The electrostatic force is greater than the gravitational force in both states.

  1. An experimenter claims that he took the balls from state 1 to state 2 without causing them to contact anything other than each other. Which of the following statements provides correct evidence for the reasonability of this claim?

(A)   The claim is not reasonable, because there was more net charge in the two-ball system in state 2 than in state 1.

(B)   The claim is not reasonable, because there was identical charge on each ball in state 2.

(C)   The claim is reasonable, because in both states, each ball carried the same magnitude of charge as the other.

(D)   The claim is reasonable, because the positive charge in state 1 could have been canceled out by the available negative charge.

  1. A guitar string creates a sound wave of known frequency. Which of the following describes a correct and practical method of measuring the wavelength of the sound wave with a meterstick?

(A)   Lightly touch the guitar string in the middle such that a single node is created. Measure the length of the string; this is the wavelength.

(B)   Measure the length of the guitar string; this is half the wavelength.

(C)   Adjust the length of a pipe placed near the string so that resonances are heard. Measure the difference between the pipe lengths for consecutive resonances; this is half the wavelength.

(D)   Measure the peak-to-peak distance of the wave as it passes; this is the wavelength.

  1. When a 0.20 kg block hangs at rest vertically from a spring of force constant 4 N/m, the spring stretches 0.50 m from its unstretched position, as shown in the figure. Subsequently, the block is stretched an additional 0.10 m and released such that it undergoes simple harmonic motion. What is the maximum kinetic energy of the block in its harmonic motion?

(A)   0.50 J

(B)   0.02 J

(C)   0.72 J

(D)   0.20 J

Questions 18 and 19 refer to the following information:

A student pushes cart A toward a stationary cart B, causing a collision. The velocity of cart A as a function of time is measured by a sonic motion detector, with the resulting graph shown in the figure.

  1. At which labeled time did the collision begin to occur?

(A)   A

(B)   B

(C)   C

(D)   D

  1. What additional measurements, in combination with the information provided in the graph, could be used to verify that momentum was conserved in this collision?

(A)   The mass of each cart and cart B’s speed after the collision

(B)   The force of cart A on cart B, and cart B’s speed after the collision

(C)   The mass of each cart only

(D)   The force of cart A on cart B only

  1. A car of mass minitially travels at speed v . The car brakes to a stop on a road that slants downhill, such that the car’s center of mass ends up a vertical height h below its position at the start of braking. Which of the following is a correct expression for the increase in the internal energy of the road-car system during the braking process?

(A)    mv – mgh

(B)    mv 2

(C)   0

(D)    mv 2 + mgh

  1. The circuits shown in the figure contain the same three resistors connected in different configurations. Which of the following correctly explains which configuration (if either) produces the larger current coming from the battery?

(A)   Circuit 1; the larger resistor is closer to the battery.

(B)   Circuit 2; the equivalent resistance of the three resistors is smaller than in circuit 1.

(C)   Neither; the batteries are both the same.

(D)   Neither; the individual resistors are the same in each circuit, even if they are in a different order.

  1. A student pushes a puck across a table, moving it from position x= 0 to position x = 0.2 m. After he lets go, the puck continues to travel across the table, coming to rest at position x = 1.2 m. When the puck is at position x = 1.0 m, which of the following is a correct assertion about the net force on the puck?

(A)   The net force is in the negative direction, because the puck is moving in the positive direction but slowing down.

(B)   The net force is down, because the puck is near the Earth, where gravitational acceleration is 10 m/s2 downward.

(C)   The net force is in the positive direction, because the student’s push caused the puck to speed up in the positive direction.

(D)   The net force is zero, because the student’s push in the positive direction must equal the force of friction in the negative direction.

  1. In an experiment, a cart is placed on a flat, negligible-friction track. A light string passes over a nearly ideal pulley. An object with a weight of 2.0 N hangs from the string. The system is released, and the sonic motion detector reads the cart’s acceleration. Can this setup be used to determine the cart’s inertial mass?

(A)   Yes, by dividing 2.0 N by the acceleration, and then subtracting 0.2 kg.

(B)   No, because only the cart’s gravitational mass could be determined.

(C)   Yes, by dividing 2.0 N by the acceleration.

(D)   No, because the string will have different tensions on either side of the pulley.

  1. An object hangs by a string from a car’s rearview mirror, as shown in the figure. The car is speeding up and moving to the right. Which of the following diagrams correctly represents the forces acting on the object?

(A)   T: Force of string on object

W: Force of Earth on object

Fo : Force of object on string

(B)   T: Force of string on object

W: Force of Earth on object

Fo : Force of object on string

(C)   T: Force of string on object

W: Force of Earth on object

Fo : Force of object on string

(D)   T: Force of string on object

W: Force of Earth on object

  1. The diagram shown here represents the particles in a longitudinal standing wave. Which of the following is an approximate measure of the standing wave’s maximum amplitude?

(A)   Half the distance from 1 to 4

(B)   The distance from 2 to 3

(C)   The distance from 1 to 4

(D)   Half the distance from 2 to 3

Questions 26 and 27 refer to the following information:

A small ball of mass m moving to the right at speed v collides with a stationary rod, as shown in the figure. After the collision, the ball rebounds to the left with speed v 1 , while the rod’s center of mass moves to the left at speed v 2 . The rod also rotates counterclockwise.

  1. Which of the following equations determines the rod’s change in angular momentum about its center of mass during the collision?

(A)    where I is the rod’s rotational inertia about its center of mass, and ω is its angular speed after collision.

(B)   Iv 2 /r , where I is the rod’s rotational inertia about its center of mass, and r is half the length of the rod.

(C)   mv 1 d , where d is the distance between the line of the ball’s motion and the rod’s center of mass.

(D)   mv 1 r , where r is half the length of the rod.

  1. Is angular momentum about the rod’s center of mass conserved in this collision?

(A)   No, the ball always moves in a straight line and thus does not have angular momentum.

(B)   No, nothing is spinning clockwise after the collision to cancel the rod’s spin.

(C)   Yes, the only torques acting are the ball on the rod and the rod on the ball.

(D)   Yes, the rebounding ball means the collision was elastic.

Questions 28 and 29 refer to the following information:

A model rocket with a mass of 100 g is launched straight up. Eight seconds after launch, when it is moving upward at 110 m/s, the force of the engine drops as shown in the force-time graph.

  1. Which of the following is the best estimate of the impulse applied by the engine to the rocket after the t= 8 s mark?

(A)   100 N·s

(B)   20 N·s

(C)   5 N·s

(D)   50 N·s

  1. Which of the following describes the motion of the rocket between t= 8 s and t = 10 s?

(A)   The rocket moves upward and slows down.

(B)   The rocket moves downward and speeds up.

(C)   The rocket moves upward at a constant speed.

(D)   The rocket moves upward and speeds up.

  1. The velocity-time graph shown here represents the motion of a 500 g cart that initially moved to the right along a track. It collided with a wall at approximately time (t) = 1.0 s. Which of the following is the best estimate of the impulse experienced by the cart in this collision?

(A)   3.6 N·s

(B)   0.5 N·s

(C)   0.2 N·s

(D)   1.8 N·s

  1. The Space Shuttle orbits 300 km above Earth’s surface; Earth’s radius is 6,400 km. What is the gravitational acceleration experienced by the Space Shuttle?

(A)   Zero

(B)   4.9 m/s2

(C)   9.8 m/s2

(D)   8.9 m/s2

  1. A person stands on a scale in an elevator. He notices that the scale reading is less than his usual weight. Which of the following could possibly describe the motion of the elevator?

(A)   It is moving downward and slowing down.

(B)   It is moving upward and slowing down.

(C)   It is moving upward at a constant speed.

(D)   It is moving downward at a constant speed.

  1. A textbook weighs 30 N at sea level. Earth’s radius is 6,400 km. Which of the following is the best estimate of the textbook’s weight on a mountain peak located 6,000 m above sea level?

(A)   60 N

(B)   15 N

(C)   30 N

(D)   7.5 N

  1. A satellite orbits the moon in a circle of radius R. For the satellite to double its speed but maintain a circular orbit, what must the new radius of its orbit be?

(A)   ½R

(B)   4R

(C)   ¼R

(D)   2R

  1. A 0.5 kg cart begins at rest at the top of an incline, 0.06 m vertically above its end position. It is released and allowed to travel down the smooth incline, where it compresses a spring. Between the positions labeled “Begin” and “End” in the figure, the work done on the cart by the earth is 0.30 J; the work done on the cart by the spring is –0.20 J. What is the cart’s kinetic energy at the position labeled “End”?

(A)   0.80 J

(B)   0.10 J

(C)   0.50 J

(D)   0.40 J

Questions 36 and 37 refer to the following information:

A rigid rod of length L and mass M sits at rest on an air table with negligible friction. A small blob of putty with a mass of m moves to the right on the same table, as shown in overhead view in the figure. The putty hits and sticks to the rod, a distance of 2L /3 from the top end.

  1. How will the rod-putty system move after the collision?

(A)   The system will have no translational motion, but it will rotate about the rod’s center of mass.

(B)   The system will move to the right and rotate about the rod-putty system’s center of mass.

(C)   The system will move to the right and rotate about the rod’s center of mass.

(D)   The system will have no translational motion, but it will rotate about the rod-putty system’s center of mass.

  1. What quantities, if any, must be conserved in this collision?

(A)   Linear momentum only

(B)   Neither linear nor angular momentum

(C)   Angular momentum only

(D)   Linear and angular momentum

  1. Bob and Tom hold a rod with a length of 8 m and weight of 500 N. Initially, Bob and Tom each hold the rod 2 m from the its ends, as shown in the figure. Next, Tom moves slowly toward the right edge of the rod, maintaining his hold. As Tom moves to the right, what happens to the torque about the rod’s midpoint exerted by each person?

(A)   Bob’s torque decreases, and Tom’s torque increases.

(B)   Bob’s torque increases, and Tom’s torque decreases.

(C)   Both Bob’s and Tom’s torque increases.

(D)   Both Bob’s and Tom’s torque decreases.

  1. An object of mass mhangs from two ropes at unequal angles, as shown in the figure. Which of the following makes correct comparisons between the horizontal and vertical components of the tension in each rope?

(A)   Horizontal tension is equal in both ropes, but vertical tension is greater in rope A.

(B)   Both horizontal and vertical tension are equal in both ropes

(C)   Horizontal tension is greater in rope B, but vertical tension is equal in both ropes.

(D)   Both horizontal and vertical tension are greater in rope B

Questions 40 and 41 refer to the following information:

Block B is at rest on a smooth tabletop. It is attached to a long spring, which in turn is anchored to the wall. Identical block A slides toward and collides with block B . Consider two collisions, each of which occupies a duration of about 0.10 s:

Collision I : Block A bounces back off of block B .

Collision II : Block A sticks to block B .

  1. In which collision, if either, does block Bmove faster immediately after the collision?

(A)   In collision I, because block A experiences a larger change in momentum, and conservation of momentum requires that block B does as well.

(B)   In collision I, because block A experiences a larger change in kinetic energy, and conservation of energy requires that block B does as well.

(C)   In neither collision, because conservation of momentum requires that both blocks must have the same momentum as each other in each collision.

(D)   In neither collision, because conservation of momentum requires that both blocks must change their momentum by the same amount in each collision.

  1. In which collision, if either, is the period and frequency of the ensuing oscillations after the collision larger?

(A)   Period and frequency are the same in both.

(B)   Period is greater in collision II, and frequency is greater in collision I.

(C)   Period and frequency are both greater in collision I.

(D)   Period and frequency are both greater in collision II.

  1. A string is fixed at one end but free to move at the other end. The lowest frequency at which this string will produce standing waves is 10 Hz. Which of the following diagrams represents how the string will look when it is vibrating with a frequency of 30 Hz?

(A)   

(B)   

(C)   

(D)   

  1. Three equal-mass objects (A, B, and C) are each initially at rest horizontally on a pivot, as shown in the figure. Object A is a 40 cm long, uniform rod, pivoted 10 cm from its left edge. Object B consists of two heavy blocks connected by a very light rod. It is also 40 cm long and pivoted 10 cm from its left edge. Object C consists of two heavy blocks connected by a very light rod that is 50 cm long and pivoted 20 cm from its left edge. Which of the following correctly ranks the objects’ angular acceleration about the pivot point when they are released?

(A)   A = B > C

(B)   A > B = C

(C)   A < B < C

(D)   A > B > C

  1. A stationary wave source emits waves at a constant frequency. As these waves move to the right, they are represented by the wave front diagram shown here. Sometime later, the wave source is moving at a constant speed to the right. Which of the following wave front diagrams could represent the propagation of the waves produced by the source?

(A)   

(B)   

(C)   

(D)   

  1. An object of mass mis attached to an object of mass 3m by a rigid bar of negligible mass and length L . Initially, the smaller object is at rest directly above the larger object, as shown in the figure. How much work is necessary to flip the object 180°, such that the larger mass is at rest directly above the smaller mass?

(A)   2πmgL

(B)   4mgL

(C)   4πmgL

(D)   2mgL

Questions 46–50: Multiple-Correct Items

Directions: Identify exactly two of the four answer choices as correct, and grid the answers with a pencil on the answer sheet. No partial credit is awarded; both of the correct choices, and none of the incorrect choices, must be marked for credit.

  1. Which of the experiments listed here measure inertial mass? Select two answers.

(A)   Attach a fan that provides a steady 0.2 N force to a cart. Use a sonic motion detector to produce a velocity-time graph as the cart speeds up. The cart’s mass is 0.2 N divided by the slope of the velocity-time graph.

(B)   Hang an object from a spring and cause the object to oscillate in simple harmonic motion. Use a stopwatch to determine the period of the motion. The object’s mass is this period squared, divided by 4π 2 , divided by the spring’s force constant.

(C)   Place an object on one side of a two-pan balance scale. On the other pan, place objects with previously calibrated masses. The object’s mass is equal to the amount of calibrated mass on the other plate when the plates balance.

(D)   Hang a spring from a clamp, then hook an object on the spring. Use a ruler to measure how far the spring stretched once the object was attached. This distance multiplied by the force constant of the spring determines the object’s weight. Use 10 N/kg to convert this weight to mass.

  1. An object on a spring vibrates in simple harmonic motion. A sonic motion detector is placed under the object. Which of the following determines the period of the object’s oscillation? Select two answers.

(A)   The maximum slope on a position-time graph divided by the maximum slope on a velocity-time graph

(B)   The maximum vertical axis value on a position-time graph divided by the maximum vertical axis value on a velocity-time graph

(C)   The time between positive maxima on a velocity-time graph

(D)   The time between positive maxima on a position-time graph

  1. Two carts on a negligible-friction surface collide with each other. Which of the following is a correct statement about an elastic collision between the carts? Select two answers.

(A)   Some of the kinetic energy of the two-cart system is converted to thermal and sound energy.

(B)   The carts bounce off of each other.

(C)   Linear momentum is conserved.

(D)   The carts stick together.

  1. The resistance of a sample of circular cross-section wire is known. A measurement of which two of the following would allow for a calculation of the resistivity of the wire? Select two answers.

(A)   The wire’s diameter

(B)   The wire’s density

(C)   The wire’s length

(D)   The wire’s temperature

  1. The net torque τon an object of rotational inertia I is shown as a function of time t . At time t max , the object has speed ω and angular acceleration α . Which of the following methods correctly determines the change in the object’s angular momentum? Select two answers.

(A)   Multiply I by α /t max .

(B)   Multiply the average torque by t max .

(C)   Calculate the area under the line on the graph.

(D)   Multiply I by ω .

STOP. End of Physics 1 Practice Exam 2—Multiple-Choice Questions

AP Physics 1 Practice Exam 2: Section II (Free-Response)

Directions: The free-response section consists of five questions to be answered in 90 minutes . Budget approximately 20 to 25 minutes each for the first two longer questions; the next three shorter questions should take about 12 to 17 minutes each. Explain all solutions thoroughly, as partial credit is available. On the actual test, you will write the answers in the test booklet; for this practice exam, you will need to write your answers on a separate sheet of paper.

  1. (7 points)

In the laboratory, you are given three resistors: R 1 = 30 kΩ, R 2 = 60 kΩ, and R 3 = 120 kΩ. You are to use these three resistors in a circuit with a 12 V battery such that one resistor is in series with the battery, and the other two are parallel with each other.

(a)   Diagram a circuit with these three resistors such that the resistor in series with the battery takes the largest possible voltage across it. Justify your answer.

(b)   For these three given resistors, is it possible for the resistor in series with the battery to take a smaller voltage than either of the two parallel resistors? Justify your answer.

(c)   Now the resistors are connected in series with the battery, as shown in the figure. On the axes provided, sketch a graph of the electric potential (V ) measured along the circuit, starting at position A, going in the direction shown, and ending at position B. Consider the electric potential at position B to be zero.

  1. (12 points)

In the laboratory, a student connects a toy vehicle to a hollow block with a string. The hollow block has a mass of 100 g and contains an additional 300 g object. The vehicle is turned on, causing it to move forward along a table at a constant speed. A force probe records the tension in the string as a function of time, and a sonic motion detector reads the position of the cart as a function of time. The positioning of the probes is shown in the diagram. The data collected are shown below.

(a)   Explain why the force reading marked point A on the graph is significantly different from the reading marked point B on the graph

(b)   i.   Calculate the impulse applied by the string on the block.

  1. A student in the lab contends, “The block moved at a constant speed, so it has no change in momentum and should thus experience no impulse.” Evaluate the validity of this student’s statement with reference to the answer to part (i).

Now the student is asked to determine whether the coefficient of kinetic friction between the block and the table depends on the block’s speed.

(c)   Describe an experimental procedure that the student could use to collect the necessary data, including all the equipment he or she would need.

(d)   How should the student analyze the data to determine whether the coefficient of friction depends on the block’s speed? What evidence from the analysis would be used to make the determination?

  1. (12 points)

An object of mass m is attached via a rope to the stem of the device, as shown in the figure. The top portion of the device includes two rocks, each with a mass of M , located near the end of a hollow horizontal pipe of length L . The rotational inertia of the pipe itself and the cylindrical support is assumed to be negligible compared to that of the rocks inside the pipe. As the object falls from rest, the device begins to rotate.

(a)   Explain why the tension in the hanging rope is not equal to mg .

(b)   Can the angular acceleration of the device be calculated given the tension in the rope T and the other information provided in the description? If so, explain in several sentences how the calculation could be performed. If not, explain in several sentences why not, including what additional measurements would be necessary and how those measurements could be performed. In either case, you should not actually do the calculations, but provide complete instructions so that another student could use them.

(c)   Derive an expression for the rotational inertia of the device. If you need to define new variables to represent easily measurable quantities, do so clearly.

(d)   How would replacing the hanging object with a new object with a mass of 2m affect the angular acceleration of the device? Answer with specific reference to the equation you derived in (c).

(e)   Explain how the reasonability of the assumption that the rotational inertia of the pipe and the support is negligible in the calculation of the device’s rotational inertia could be justified. You may use either a theoretical or an experimental approach to your justification.

  1. (7 points)

A spring is attached vertically to a table. Undisturbed, the spring has a length of 20 cm. In procedure A, a block with a mass of 400 g is placed gently on top of the spring, compressing it so its length is 15 cm. In procedure B, a student pushes the block farther down, such that the spring has a length of 10 cm, and then releases the block from rest. The block is projected to height h.

Consider a system consisting of the block, the spring, and Earth. Define the potential energy of this system as zero when the spring has a length of 15 cm.

(a)   i.   Calculate the work done by the student on the block-spring-Earth system in procedure B.

  1. Justify your choice of equations in your calculation, as well as any distance values you used.

(b)   Now the same spring is used with a block whose mass is 800 g. In procedure C, the 800 g block is placed gently on top of the spring, compressing it to a length of 10 cm. In procedure D, a student pushes the block 5 cm farther down and releases it from rest. In terms of h , how high will this 800 g block be projected? Justify your answer.

  1. [paragraph response] (7 points)

An object can be supported by two symmetric strings in one of two ways, as shown above. In which configuration are the strings more likely to break? Justify your answer, referencing physics principles in a clear, coherent, paragraph-length explanation.

STOP. End of Physics 1 Practice Exam 2—Free-Response Questions

Solutions: AP Physics 1 Practice Exam 2, Section I (Multiple-Choice)

Questions 1–45: Single-Choice Items

  1. A—The tension is the force of the string acting on the block. Newton’s third law says that since the string pulls on the block, the block pulls equally on the string. (The answer is NOT for force of friction: Newton’s third law force pairs can never act on the same object.)
  2. B—Since the block has no acceleration, left forces equal right forces, and up forces equal down forces. The equation for a force of friction is F f μF n . Since the coefficient of friction μ is less than 1, the friction force must be less than the normal force.
  3. A—The energy gained or lost by a charge as it goes through a circuit element is voltage, and a voltmeter measures voltage. In this circuit loop, any energy gained by charges across the battery must equal the energy lost by the charges through the two resistors; this is known as Kirchoff’s Loop Rule.
  4. D—“Charge that flows through each in a given time interval” is a complicated way of saying “current.” Current through series resistors must always be the same through each, so the 50 Ω resistor and the 100 Ω resistor should rank equally. Then simplify the circuit to two parallel branches. The left branch has an equivalent resistance of 150 Ω and the right branch of 300 Ω. With the same voltage across each branch, the larger current goes through the path with smaller resistance by Ohm’s law.
  5. C—Consider just the two series resistors, which have an equivalent resistance of 150 Ω and are connected to the 9 V battery. The current through that branch of the circuit is  . Now consider just the 50 Ω resistor. The voltage across it is (0.06 A)(50 Ω) = 3 V. As a sanity check, we know that for two resistors in series (which take the same current), V = IR says that the smaller resistor takes less voltage; 3 V across the 50 Ω resistor leaves 6 V across the 100 Ω resistor.
  6. D—All experimental evidence in existence shows that the smallest unit of isolated charge is the electron charge e = 1.6 × 10–19 C. The charge measured in A is equivalent to 4 fundamental charges; in B, 5 fundamental charges; and in C, 3 fundamental charges. But choice D would require an isolated charge of 1.5 e , which is not possible.
  7. D—The work done by the net force is the area under the graph. Since the cart only moved from position x = 0.5 m to x = 1.5 m, area 1 is the work done by the net force. By the work-energy theorem, work done by the net force is the change in an object’s kinetic energy. (Yes, the mass must be known to determine the values of the initial and final kinetic energy; however, the question asks only for the change in kinetic energy.)
  8. A—To evaluate the acceleration of the horse-cart system, you can only consider the forces applied by objects external to the system. This eliminates choices B and C, which discuss forces of the system on itself. Choice D is ridiculous because the horse remains on the ground. Choice A is correct: the ground pushes forward on the horse’s hooves because the horse’s hooves push backward on the ground.
  9. B—The closed end of the pipe makes it impossible for the air particles right next to the end to vibrate at all, which is why a particle displacement node must exist there. It’s easiest, I think, just to remember that in a sound wave, pressure is at an antinode when particle displacement is at a node and vice versa. Or remember that the pressure must be equal to atmospheric pressure at the open end, which is exposed to the atmosphere, and greater inside because the air is compressed, not sucked out of the pipe.
  10. C—The cart is in equilibrium, so the spring scale’s force is equal to the component of the cart’s weight that acts parallel to the plane. That component is mg sinθ. Thus, the graph of spring scale reading vs. θ should be a sine function, as in choice C. Choice B is wrong because that’s what a sine function looks like all the way to 180°; here, the angle of the incline only goes to 90°.
  11. B—Newton’s second law for rotation says τ net =  , where α is the angular acceleration, or change in the wheel’s angular velocity per time. To find the wheel’s angular velocity, look at the slope of the angular position versus time graph. The slope changes after the torque is applied; so the change in the slope is the change in the angular velocity, which (when divided by the 0.10 s duration of the rocket firing) gives the angular acceleration.
  12. B—Angular acceleration is the change in angular speed in each second. To find angular speed, take the slope of the angular position versus time graph shown. This slope is constant for two seconds, then it changes to another constant slope. Therefore, there is no angular acceleration during the time when the slope doesn’t change—with no change in angular speed, there’s no acceleration. The only change in angular speed comes at the moment the slope changes, so that’s the only time when there’s any angular acceleration.
  13. D—The energy carried by a wave depends on the wave’s a mplitude, meaning its height. The speed and pulse width or wavelength have no impact on the energy of a wave.
  14. C—The magnitude of the electrostatic force depends on the product of the charges, regardless of sign. In state 2, the charges are both bigger than in state 1, giving a bigger electrostatic force in state 2. Without even doing the calculation, you can recognize that since Newton’s gravitational constant G is orders of magnitude less than the coulomb’s law constant k , the electrostatic force will be much greater than the gravitational force between two objects in virtually any laboratory situation. The only time these forces become comparable is when objects the size of planets exert gravitational forces.
  15. A—Charge is conserved, meaning that while negative charge can neutralize positive charge, the net charge of a system cannot change. The net charge of state 1 is zero; thus, the net charge of state 2 must also be zero, regardless of whether the balls touched or not. State 2 has a net charge of –80 µC, not zero; thus, the claim is not reasonable due to this violation of charge conservation.
  16. C—You want the wavelength of the sound wave, not the wave on the guitar. Choices A and B both correctly determine the wavelength of a wave on a guitar string, but these are not the waves you are looking for. Choice D does discuss the sound wave in air, but when is the last time you visually “saw” a sound wave passing you in the air, let alone saw any peaks that you could practically measure with a meterstick? So do the experiment in choice C. Going from one resonance to the next adds half a wavelength to the standing wave in the pipe, regardless of whether the pipe is open or closed.
  17. B—You can look at this two ways. The hard way is to consider the spring energy gained and the gravitational energy lost in stretching the spring the additional 0.10 m separately. The block-earth system loses mgh = (0.20 kg)(10 N/kg)(0.10 m) = 0.20 J of gravitational energy; but the block-spring system gains ½kx 2 2 – ½kx 1 2 = {½(4 N/m)(0.60 m)2 – ½(4 N/m)(0.50 m)2 } = 0.22 J of spring energy. Thus, the net work done on the block in pulling it the additional 0.10 m is 0.02 J. That’s what is converted into the block’s maximum kinetic energy.

You can also look at it the easy way. With a vertical spring, consider the block-earth-spring system as a whole. Define the hanging equilibrium as the zero of the whole system’s potential energy; then the potential energy of the whole system can be written as ½kx 2 , where x is the distance from this hanging equilibrium position. That’s ½(4 N/m)(0.10 m)2 = 0.02 J.

  1. C—The graph represents the speed of cart A, the one that’s initially moving. So right before the collision, the vertical axis of the graph must be nonzero. Right after the collision, the vertical axis must quickly either decrease or perhaps become negative if the cart changed directions. That’s what happens in the tenth of a second or so after the time labeled C.
  2. A—Conservation of momentum requires that the total momentum of the two-cart system be the same before and after the collision. You already know cart A’s speed and direction of motion before and after the collision by looking at the vertical axis of the graph; so the mass of cart A will give us cart A’s momentum before and after the collision. You know cart B has no momentum before the collision. But you need both cart B’s mass AND its velocity after the collision to finish the momentum conservation calculation.
  3. D—“Increase in the internal energy of the road-car system” is a fancy way of saying “work done by the car’s brakes to stop the car.” Without the brakes, the car would have gained mgh of mechanical energy in dropping the height h , giving it a total mechanical energy of ½mv 2 + mgh . The brakes convert all of that mechanical energy to internal energy.
  4. B—The equivalent resistance of the parallel resistors in circuit 1 is 12 kΩ; adding that to the 100 kΩ resistor gives a total resistance in circuit 1 of 112 kΩ. Circuit 2’s parallel combination has an equivalent resistance of 23 kΩ, giving a total resistance in that circuit of 43 kΩ. By V = IR applied to both circuits in their entirety with the same total voltage, the circuit with smaller total resistance will produce the larger current. That’s circuit 2.
  5. A—The motion after the push is finished is toward more positive x , so it is in the positive direction. The puck is slowing down. When an object slows down, its acceleration and its net force are in the direction opposite the motion. (Choice C would be correct at a position at which the student were still pushing the puck, but at x = 1.0 m, the student had already let go, so the puck was slowing down.)
  6. A—The net force on the cart-object system is 2.0 N. Dividing 2.0 N by the acceleration gives the mass of the entire cart-object system, not just the mass of the cart; so subtract the 0.2 kg mass of the object from the whole system mass to get the cart mass. This is actually inertial mass, because it uses the equation F net ma ; Newton’s second law defines inertial mass as resistance to acceleration.
  7. D—All of the diagrams have the object’s weight correct. Any other force acting on the object must be provided by something in contact with that object. The only thing making contact with the object is the string, so add the force of the string on the object. That’s it. The force of the object on the string doesn’t go on this diagram, because this diagram only includes forces acting ON the object, not exerted by the object.
  8. D—Positions 1 and 4 show minimum particle displacement, so these are the nodes. The maximum amplitude occurs at the antinode, which is somewhere near the positions labeled 2 and 3. The particles are moving left and right in this longitudinal wave. The amplitude is twice the peak-to-peak displacement; since the particle displacement near the antinode is about the distance from 2 to 3, the amplitude is half that.
  9. A—The rod starts from rest, so its final angular momentum is the same as its change in angular momentum. Although the equation ω = v/r is valid for a point object moving in a circle, it does not apply to a rotating rod; thus, choice B is wrong. Choice C gives the final angular momentum of the ball, which is not the same as the angular momentum change of the rod because the ball does NOT start from rest.
  10. C—Choice C states the fundamental condition for angular momentum conservation, which is correct here. The ball does have angular momentum about the rod’s center of mass before and after the collision, because its line of motion does not go through the rod’s center of mass. Whether or not the collision is elastic has to do with conservation of mechanical energy, not angular or linear momentum.
  11. C—Impulse is the area under a force-time graph. From t = 8 s to t = 10 s, the area is an approximate rectangle of 2 N times 2 s, giving 4 N·s. Add an approximate triangle from t = 10 s to t = 11 s, which has the area ½(2 N)(1 s) = 1 N·s. That gives a total of 5 N·s.
  12. D—The force of the engine on the rocket during this time is 2 N upward. The weight of the rocket is 1 N (that is, 0.1 kg times the gravitational field of 10 N/kg). So the net force is still upward during this time. Since the rocket was already moving upward, it will continue to move upward and speed up.
  13. D—Impulse is change in momentum. The initial momentum was something like (0.5 kg)(1.6 m/s) = 0.8 N·s to the right. The cart came to rest, changing its momentum by 0.8 N·s, then sped back up, again changing momentum by (0.5 kg)(2 m/s) = 1.0 N·s. Thus, the total momentum change is about 1.8 N·s.
  14. D—The gravitational acceleration is given by GM/d 2 , where d is the Space Shuttle’s distance to Earth’s center. You don’t know values for G and M , nor do you need to know them. You do know that at Earth’s surface, 6,400 km from the center, the gravitational acceleration is 9.8 m/s2 . To calculate using this equation at the height of the Space Shuttle, the numerator remains the same; however, the denominator increases from (6,400 km)2 to (6,700 km)2 , a difference of about 8%. (Try it in your calculator if you don’t believe me.) Thus, the gravitational acceleration will decrease by about 8%, giving choice D. (Yes, things seem “weightless” in the Space Shuttle. That’s not because g = 0 there, but because everything inside the shuttle, including the shuttle itself, is in free fall, accelerating at 8.9 m/s/s toward the center of Earth.)
  15. B—The scale reading is less than the man’s weight; that means that the net force is downward. By Newton’s second law, the person’s acceleration is downward too. Downward acceleration means either moving down and speeding up, or moving up and slowing down. Only choice B works.
  16. C—The weight of the textbook is GMm/d 2 , where M and m are the masses of Earth and the textbook. These don’t change on a mountain. The d term will change, because d represents the distance between the textbook and Earth’s center. But that will change the denominator of the weight equation from (6,400 km)2 to (6,406 km)2 ; in other words, not to the two digits expressed in the answers. No need to use the calculator. You can see that the choices require the weight to either stay the same, double, or be cut in half. The weight of the textbook remains 30 N.
  17. C—In circular motion around a planet, the centripetal force is provided by gravity:  . Solving for the radius of the satellite’s circular orbit, we get  . The numerator of this expression doesn’t change, because the mass of the planet M and Newton’s gravitation constant G don’t change. The satellite’s speed v is doubled. Since the v term is squared, that increases the denominator by a factor of 4. So the radius of orbit d is now ¼ as much as before.
  18. B—Consider the cart alone, which as a single object has no internal energy or potential energy. Thus, any work done on the cart will change the cart’s kinetic energy. The cart began with no kinetic energy at all. Earth increased the cart’s kinetic energy by 0.30 J, as stated in the problem; the spring decreased the cart’s kinetic energy by 0.20 J. A gain of 0.30 J and a loss of 0.20 J leaves 0.10 J of kinetic energy.
  19. B—Conservation of linear momentum requires that the center of mass of the system continue to move to the right after the collision. The rotation will be about the combined rod-putty center of mass. To understand that, imagine if the putty were really heavy. Then after the collision, the rod would seem to rotate about the putty, because the center of mass of the rod-putty system would be essentially at the putty’s location. In this case, you don’t know whether the putty or the rod is more massive, but you do know that when the two objects stick together, they will rotate about wherever their combined center of mass is located.
  20. D—No unbalanced forces act here other than the putty on the rod and the rod on the putty. (The weight of these objects is canceled by the normal force.) Thus, linear momentum is conserved. No torques act on the rod-putty system except those due to each other; thus, angular momentum is conserved. It’s essentially a fact of physics that in a collision between two objects, both linear and angular momentum must be conserved.
  21. C—The net torque on the rod must be zero, because the rod doesn’t rotate. Therefore, whatever happens to Bob’s torque must also happen to Tom’s torque—these torques must cancel each other out. That eliminates choices A and B. The forces provided by Bob and Tom must add up to 500 N, the weight of the rod. Try doing two quick calculations: In the original case, Bob and Tom must each bear 250 N of weight and are each 2 m from the midpoint, for 500 N·m of torque each. Now put Tom farther from the midpoint—say, 3 m away. For the torques to balance, Fbob (2 m) = Ftom (3 m). The only way to satisfy this equation and get both forces to add up to 500 N is to use 300 N for Fbob and 200 N for Ftom . Now, the torque provided by each is (300 N)(2 m) = 600 N·m. The torque increased for both people.
  22. A—The object is in equilibrium, so left forces equal right forces. Thus, the horizontal tensions must be the same in each rope. Rope A pulls at a steeper angle than rope B, but with the same amount of horizontal force as rope B. To get to that steeper angle, the vertical component of the tension in rope A must be larger than in rope B.
  23. A—There’s no indication that energy must be conserved in collision 1. However, momentum is always conserved in a collision. When block A bounces, its momentum has to change to zero and then change even more to go back the other way. Since block A changes momentum by more in collision I, block B must as well because conservation means that any momentum change by block A must be picked up by block B. Choices C and D are wrong because, among other things, they use conservation of momentum to draw conclusions about two separate collisions; momentum conservation means that total momentum remains the same before and after a single collision, not in all possible collisions.
  24. B—The period of a mass-on-a-spring oscillator is  . The important part here is that the mass term is in the numerator—a larger mass means a larger period. More mass oscillates on the spring in collision II, so collision II has a greater period. Frequency is the inverse of the period, so the period is smaller in collision II.
  25. D—The fundamental will have a node at one edge, an antinode at the other, and no nodes in between. The next allowable harmonic will have the same end conditions as the fundamental plus one additional node. A string fixed at one end and free to move at the other can only produce standing waves whose frequency is an odd multiple of the fundamental. So 30 Hz is the next available frequency after the fundamental with one additional node.
  26. D—Angular acceleration is net torque divided by rotational inertia,  . Imagine each object has total mass of 2 kg. Begin by comparing objects A and B: To find the net torque on object A, assume the entire 20 N weight is concentrated at the dot representing the rod’s center of mass. That’s located 10 cm from the pivot, giving a net torque of 200 N·cm. For object B, consider the torques provided by each block separately. The right block provides a torque of (10 N)(30 cm) = 300 N·cm clockwise; the left block provides a torque of (10 N)(10 cm) = 100 N·cm counterclockwise. That makes the net torque 200 N·cm, the same as for object A. But object B has more rotational inertia, since its 2 kg of mass are concentrated farther away from the pivot than object A’s mass. So the denominator of the angular acceleration equation is bigger for object B with the same numerator, which means A > B.

Now consider object C. It experiences less net torque than A and B. Calculate (10 N)(30 cm) = 300 N·cm clockwise, and (10 N)(20 cm) = 200 N·cm counterclockwise for a net torque of 100 N·cm. And object C has the same mass distributed even farther from the pivot point than either of the other two objects, giving C an even bigger rotational inertia. In the angular acceleration equation, object C gives a smaller numerator and a bigger denominator than object B, meaning C < B. Put it all together to get A > B > C.

  1. B—Since the source is still moving at constant speed, the wave fronts will be equally spaced. However, since the source is moving, it will have traveled a bit with the wave before emitting the next wave. Thus, the wave fronts will be closer together. As an alternate explanation, a stationary observer at the right of the page would hear a higher frequency by the Doppler effect, meaning that more waves should pass the observer in one second; that requires wave fronts to be closer together.
  2. D—Consider the system consisting of the objects and Earth, with the location of the 3m mass being the zero of gravitational energy. The initial gravitational energy of the system is mgL . After the rotation, the final gravitational energy of the system is 3mgL . That extra gravitational energy of 2mgL came from the work done on the system, meaning choice D. If you want instead to think of work on the objects as force times distance, remember that the force of Earth on the objects acts straight down, not along a circle. So the distance term to use here is just L , not πL .

Questions 46–50: Multiple-Correct Items (You must indicate both correct answers; no partial credit is awarded.)

  1. A and B—Choice A uses the equation F net = ma , which essentially defines inertial (as opposed to gravitational) mass. Choice B measures inertial mass, because it measures mass as resistance to the acceleration caused by the spring force. Choices C and D measure an object’s behavior in a gravitational field; by definition, that’s gravitational , not inertial, mass.
  2. C and D—Though both speed divided by acceleration (as in choice A) and position divided by speed (as in choice B) give units of time, the time indicated has no relation to the period of the motion. The period is defined as the time for one complete cycle to happen. One complete cycle can be defined as the time for the object to get back to its extreme position, as in choice D, or it can be defined as the time between the occurrences of maximum velocity in the same direction, as in choice C.
  3. B and C—Elastic means that mechanical energy is conserved in the collision, so the answer is not choice A, which describes a loss of mechanical energy. Mechanical energy conservation requires that carts bounce, so choice B is correct, but choice D is not. Linear momentum is conserved in all collisions, elastic or not elastic.
  4. A and C—The relevant equation here is  . The length of the wire is the variable L , so choice C is correct. The diameter measurement (choice A) can be used with the formula for the area of a circle to get A , the cross-sectional area of the wire.
  5. B and C—Change in angular momentum is τ Δt , which means the area under a torque-time graph; thus, choice C is correct. Since this graph is linear, the average torque multiplied by the maximum time is the same thing as the area under the graph, so choice B is also correct. While choice D describes a calculation of angular momentum, it does not correctly give the change in angular momentum, so you don’t know whether the object had an initial angular momentum or not.

Solutions: AP Physics 1 Practice Exam 2, Section I (Free-Response)

Obviously your solutions will not be word-for-word identical to what is written below. Award points for your answer as long as it contains the correct physics and as long as it does not contain incorrect physics.

Question 1

Part (a)

The resistor in series with the battery takes the largest current of any of the three resistors by Kirchoff’s junction rule. By Ohm’s law, V = IR , a resistor with both the largest current and the largest resistance will similarly take the largest voltage. So put the largest resistor R 3 in series with the battery, as shown.

1 point for a correct diagram

1 point for a correct justification

Part (b)

Yes. Consider the equivalent circuit drawn, where the second resistor is the equivalent resistor of the parallel combination. These two resistors each take the same current by Kirchoff’s junction rule; by Ohm’s law, with the same current, the larger of the two takes the larger voltage. If the equivalent resistance of the parallel combination is greater than the resistance of the resistor in series with the battery, then the parallel combination will take more voltage than the first resistor. In this case, putting R1 in series with the battery and the other two resistors in the parallel combination satisfies this condition; the equivalent resistance of R 2 and R 3 in parallel is greater than 30 kW.

1 point for comparing the resistance of the parallel combination to that of the series resistor

1 point for drawing the correct conclusion that the situation described is possible.

Part (c)

The total resistance of this circuit is the sum of the three resistors, 210 kΩ. The total current in the circuit is 12 V/210 kΩ = 57 μA (that is, 0.000057 A). Using that current in V = IR for each resistor gives a voltage drop of 3.4 V, 6.9 V, and 1.7 V across each resistor in turn. So start the graph at 12 V. The voltage drops by 3.4 V after the 60 kΩ resistor, down to 8.6 V, as shown in the diagram. (The voltage does not change along the wire itself, which has essentially zero resistance compared to the resistors.) When we get to the 120 kΩ resistor, drop the voltage another 6.9 V, down to 1.7 V left. The voltage drops the rest of the way to zero after the 30 kΩ resistor, as required by Kirchoff’s loop rule.

1 point for any graph having three distinct voltage drops ending at 0 V.

1 point for any graph that somehow indicates the largest voltage drop across the middle resistor, and the smallest voltage drop across the last resistor

1 point for any graph with three distinct horizontal segments representing the voltage in the wires not changing.

Question 2

Part (a)

The maximum coefficient of static friction is greater than the coefficient of kinetic friction. The graph shows the maximum force of static friction, when the block starts moving, to be 1.7 N; when the block is moving, the friction force drops to 1.0 N, which is appropriate for the lower kinetic friction force.

1 point for correctly explaining the difference between static and kinetic friction in this case.

Part (b)

(i)   Impulse is the area under a force-time graph. The graph has an average force of about 1.0 N, and a time interval of about 2 s (from t = 2 s to t = 4 s). That’s an impulse of 2 N·s.

1 point for using the area under the F vs. t graph

1 point for making reasonable estimates of time and force from the graph

(ii)   The student is incorrect. While it is true that the net impulse on an object is equal to the object’s change in momentum, the graph shown does not include the net force. The graph shows the force of the string only. If you were to subtract the impulse provided by the friction force, you would get zero net impulse and thus zero change in momentum.

1 point for discussing the difference between the force of the string and the net force on the cart

1 point for including no incorrect statements

Part (c)

The student could use a set of toy carts, each of which moves at a different constant speed. The student should connect the carts to the block via the force probe and produce force-time graphs for each cart as the cart moves across the table. The constant speed traveled by each cart can be measured by placing a sonic motion detector in front of the cart.

3 points for a complete and correct description. Of these points, 1 or 2 could be holistically awarded for a partially complete and/or partially correct description.

Part (d)

The student should make a plot of speed on the vertical axis versus the force of kinetic friction on the horizontal axis. The speed of each cart would be determined by the slope of the position-time graph produced by the sonic motion detector. The force of kinetic friction would be measured by the force probe.

The block is the same mass each time, which means it always experiences the same normal force; the coefficient of friction is the friction force divided by the normal force. So if the coefficient of friction changes, the force probe reading would change as well.

If this graph is horizontal, then the force of friction and the coefficient of friction do not change with speed. If the graph is sloped, then we can conclude that the coefficient of friction does change with speed.

1 point for using a graph with a significant number of data points, or perhaps statistical analysis of multiple trials.

1 point for correctly relating the friction coefficient to the force probe reading with constant F n .

2 points for a complete and correct analysis. One of these points can be holistically awarded for a partially complete and/or partially correct analysis.

Question 3

Part (a)

1 point: As the hanging object m falls, it speeds up. Its acceleration is therefore downward, and so is the net force acting on it. To get a downward net force, the downward gravitational force mg must be greater than the rope’s tension.

Part (b)

4 points:

  • 1 pointfor using an applicable fundamental relationship such as  (or a =  .)
  • 1 pointfor a correct discussion of how to get τ net (or α )
  • 1 pointfor a correct discussion of how to estimate I (or a )
  • 1 pointfor a correct conclusion to answer the question

The angular acceleration can be found by dividing the net torque on the device by its rotational inertia. The rotational inertia I of the device can be calculated based on the assumption that the pipe and the stem do not contribute; the device consists of two pointed masses, each a distance of L/ 2 from the center of rotation. The net torque on the device is Tr , where r is the radius of the support.

The only unknown information in all of this is the radius of the support r . So no, the angular acceleration cannot be calculated with the information provided unless r is measured with calipers.

(Alternate solution: The angular acceleration of the device α is related to the linear acceleration a of the falling mass by a =  . The radius of the support r must be measured with calipers. To get a , the distance the mass falls from rest d can be measured with a meterstick, and the time t to fall that distance can be measured with a stopwatch. Then kinematics can be used to calculate a . So no, the angular acceleration cannot be calculated with the information given here.)

Part (c)

1 point for clearly defining and correctly using any new variables

1 point for a correct expression for I (or a )

1 point for a correct final answer

As described in part (b), let r be the radius of the support.

By Newton’s second law for rotation,  .

Using the net torque and rotational inertia explained in part (b),  .

(Alternate solution, as described in part (b):  .

Falling from rest means zero initial velocity, so d = ½at 2 , with d and t defined as shown. Algebraically, this makes  .

Combining these two statements gives  .)

Part (d)

1 point for describing the quantity in the equation in (c) that would be affected by the new mass

1 point for describing with reference to the equation how that change would affect the acceleration

While the tension in the hanging rope is not equal to the weight of the hanging object, increasing the hanging object’s mass would increase the tension in the rope T . Since T is in the numerator of the equation for angular acceleration, and since the parameters in the denominator are unchanged, the angular acceleration would increase.

(Alternate solution: We would be able to measure that the hanging object falls the same distance in less time. Note that this is NOT because “heavier objects fall faster”; this is really a consequence of the reasoning already given, that the tension in the rope increases without changing the properties of the device. With a smaller t in the denominator and other variables unchanged, the angular acceleration would increase.)

Part (e)

2 points for a complete and correct solution. One of these points can be awarded for a partially complete or partially correct solution.

Using a calculational approach: You assumed that the rotational inertia of the device was wholly due to the two rocks as point objects, 2M(L/ 2)2 . The rotational inertia of a cylinder rotating horizontally and vertically can be looked up, and the mass of the pipe and support can be measured. You could calculate the additional rotational inertia provided by the cylinder and support. If this additional rotational inertia is substantially less than 2M (L/ 2)2 such that the calculation of α would still come out approximately the same, then this additional inertia is negligible.

Using an experimental approach: Measure the angular acceleration of the device directly. This can be done with frame-by-frame video analysis or with a photogate set to measure the increase in angular velocity. If the angular acceleration measured matches that predicted by the equation in (c), then the assumption that the rotational inertia is due wholly to the point masses is reasonable. However, if the angular acceleration is measured to be noticeably smaller than that predicted in (c), then the rotational inertia of the pipe and support do contribute meaningfully to the calculation and are not negligible.

Question 4

Part (a)

(i)   1 point for calculating the spring constant

1 point for using the calculated spring constant in a correct equation to determine the work done

The spring constant of the spring can be determined by procedure A. The rock applies a 4 N force on the spring, compressing it 0.05 m. By F = kx , that gives a spring constant k of (4 N)/(0.05 m) = 80 N/m.

The potential energy of the spring-block-Earth system is just ½kx 2 , where x is the distance from the position of the block after procedure A. (If you were talking about just the spring-block system, you would use the distance from the undisturbed position, but then you would have to consider the work done on the block-Earth system separately.) So in procedure B, the block-Earth system gains ½(80 N/m)(0.05 m)2 of potential energy, which is 0.10 J. That’s how much work was done by the student.

(ii)   1 point for correctly justifying the use of data from procedure A to get the spring constant

1 point for justifying use of ½kx 2 as a change in potential energy of the correct system

See above.

Part (b)

1 point for recognizing that the same potential energy is available to be converted to KE

1 point for words or equations showing that the mass only shows up in the denominator of an expression for the height

1 point for using an equation or conservation of energy reasoning to get ½h.

The new spring-block-Earth system stores the same 0.10 J of potential energy: ½kx 2 , where x is the 0.05 m distance from the position of the block after procedure C. That 0.10 J is converted into kinetic energy, then to purely gravitational energy. Gravitational energy is mgh ; the new height is h new = 0.10 J/mg . Since this new mass is twice as much as before, and since height is in the denominator, the new height is half as much as h .

Question 5

  • 1 pointfor appropriate free body diagram(s) or equivalent clear descriptions of forces acting on the object.
  • 1 pointfor relating the tension in the rope to the likelihood that the string will break.
  • 1 pointfor using horizontal and vertical components of the strings’ tensions.
  • 1 pointfor use of trigonometry (graphically, or with sines and cosines) relating the magnitude of the tension to the components of the tension.
  • 1 pointsfor an equivalence between the vertical tension components and the object’s weight.
  • 1 pointfor recognizing that the vertical components of tension must be the same in either configuration.
  • 1 pointfor connecting identical vertical tensions with different horizontal tensions to show that configuration 1 gives the larger magnitude of tension.

Example of a good paragraph : In each configuration, the forces acting on the object are the two equal tensions T , and the object’s weight. Equilibrium demands that the vertical tension components are together equal to the object’s weight; and that the horizontal tension components are equal to each other. No matter what the angle, the vertical components of tension must remain the same (because the weight of the object can’t change.) The vertical component is given by T sinθ , with angle θ measured from horizontal. To keep this vertical component the same no matter the angle, the tension must increase as the sine of the angle decreases. The angle θ is smaller in configuration 1; therefore, the tension is larger and the string is more likely to break.

Scoring the Practice Exams

As this book goes to press, the College Board has released the raw score cutoffs for just one AP Physics 1 exam. The chart below reflects approximately those cutoffs.

Remember that the raw percentage score necessary to obtain a 5, 4, 3, or 2 is not a fixed number. The scores are scaled each year so, regardless of whether the questions on that year’s exam are hard or easy, the meaning of each score is similar year after year after year. What you see below is merely my best educated guess at a reasonable score conversion. I’ll even use this very score conversion in my own classes. But I don’t guarantee the accuracy of the chart any more than I guarantee Arsenal to win the Premier League.

Multiple-Choice Raw Score: Number Correct______(50 points maximum)

Free response total:______(45 points maximum)

The final score is equal to (1.11 × the free response score) + (the multiple choice score)

Total score:______(100 points maximum)

Approximate Score Conversion Chart (only a guesstimate, see above)