5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)
Determine Your Test Readiness
Test Yourself: AP Physics 1 Question Types
IN THIS CHAPTER
Summary: In the previous chapter you tested yourself on physics fundamentals. In this chapter you will test yourself on the types of questions found in the AP Physics 1, Algebra-Based Exam. This preview of question types will provide insight into which types of questions will be the most problematic for you. Complete explanations for all the questions in this diagnostic test are included at the end of this chapter.
In addition to reviewing AP Physics 1, Algebra-Based content and skills, you will need to become familiar with the question types on the exam and assess which types of questions will be most difficult for you.
This self-assessment, along with the one in Chapter 4 , should be used to help you develop a personalized test-prep plan based on your needs (see Chapter 3 ).
Self-Assessment: Question Types
In the self-assessment that follows, you’ll be introduced to the question types found on the AP Physics 1, Algebra-Based Exam. Sure, you’ve taken multiple-choice and essay tests before, but on the AP Physics 1 Exam, you’ll find some multiple-choice and free-response questions unlike anything you’ve probably seen before. Your test score will improve if you not only review physics subject areas but also become familiar with the types of questions you will encounter and practice responding to these. Adjust your study plan (see Chapter 3 ) to focus on any weaknesses you identify as a result of this diagnostic test. In Chapters 7 through 9 you’ll find proven strategies for attacking each of the question types on the AP Physics 1 Exam. Be sure to familiarize yourself with these, especially for those types of questions that proved difficult for you.
Descriptive Problems (Like Those You’ve Probably Seen Before)
1 . In the preceding diagram, forces F 1 and F 2 are acting on box M which is on a frictionless table. F 1 has a greater magnitude than F 2 . Of the following statements about the motion of box M , which is correct?
(A) Box M is accelerating in the direction of F 1 .
(B) Box M is accelerating in the direction of F 2.
(C) Box M is moving with a constant speed in the direction of F 1 .
(D) Box M is moving with a constant speed in the direction of F 2 .
2 . Four objects with mass m are rigidly connected and free to rotate in the plane of the page about the center point C . The objects experience two forces, F 1 and F 2 , as shown in the preceding diagram. Which of the following statements correctly analyzes how F 1 and F 2 will affect the angular velocity of the objects?
(A) Both forces apply torque in the same sense, so the angular velocity must increase.
(B) If the forces F 1 and F 2 are equal in magnitude, the angular velocity will not change.
(C) The angular velocity could decrease if the objects were initially rotating clockwise.
(D) F 2 does not apply any torque to the objects, so only F 1 can cause a change in angular velocity.
Calculation Problems (Like Those You’ve Probably Seen Before)
3 . A 2.0-kg ball is dropped such that its speed upon hitting the ground is 3.0 m/s. It rebounds, such that its speed immediately after collision is 2.0 m/s. What is the magnitude of the ball’s change in momentum?
(A) 10 N·s
(B) 5 N·s
(C) 2 N·s
(D) 1 N·s
4 . Two carts, each of mass 0.5 kg, move to the right at different speeds as shown in the diagram. Next, a student stops cart B with his hand. By how much has the linear momentum of the two-cart system changed after the student stops cart B?
(A) 0.4 N·s
(B) 0.1 N·s
(C) 0.3 N·s
(D) 0.7 N·s
5 . Three identical rods experience a single, identical force F , as shown in the diagrams. Each rod is initially rotating differently about its left edge, as described. Which of the following correctly ranks the magnitude of each rod’s change in angular speed during the first 0.5 s that the force is applied?
(A) 2 = 3 > 1
(B) 1 = 2 = 3
(C) 1 > 2 > 3
(D) 3 > 2 > 1
6 . A diver leaps upward and forward off of a diving board. As the diver is in the air, he twists his body such that his rotational speed when he hits the water is four times his rotational speed when he left the diving board. When he hits the water, what is the diver’s angular momentum about his center of mass?
(A) Greater than his angular momentum when he left the diving board by a factor of four
(B) Greater than his angular momentum when he left the diving board, but not by a factor of four
(C) Less than his angular momentum when he left the diving board
(D) Equal to his angular momentum when he left the diving board
Description of an Experiment
7 . A sprinter running the 100-meter dash is known to accelerate for the first few seconds of the race and then to run at constant speed the rest of the way. It is desired to design an experimental investigation to determine the sprinter’s maximum speed v . Which of the following procedures could correctly make that determination?
(A) Place poles 90 m and 100 m from the race’s start. Measure with a stopwatch the time t for the sprinter to travel between the poles. To find v , divide 10 m by t .
(B) Estimate that the sprinter accelerates for the first 2.5 s. Mark on the track the location of the sprinter after 2.5 s. Use a measuring tape to find the distance d the sprinter traveled in this time. Divide d by 2.5 s to get v .
(C) Measure with a stopwatch the time t for the sprinter to run the 100 m. To find v , divide 100 m by t .
(D) Measure with a stopwatch the time t for the sprinter to run the 100 m. Divide 100 m by t 2 to get the average acceleration a . Then since the sprinter starts from rest, v is given by .
8 . Two blocks of known masses M 1 and M 2 , with M 1 > M 2 , are connected by string over a freely rotating light pulley, as shown in the preceding diagram. A video camera records the motion of the blocks and pulley after the blocks are released from rest in the position shown. It is desired to use the video to measure the angular velocity of the pulley when block m 1 has fallen a known distance d . Which of the following approaches will best make this experimental determination?
(A) Treat the system as a single mass. The net force is the difference between the blocks’ weights, m 1 g – m 2 g ; Newton’s second law gives an acceleration of . Use the kinematic equation v f 2= v 0 2 + 2ad with v 0 = 0 to determine the final speed of the block; then the angular velocity is this speed divided by the radius of the pulley.
(B) Mark a spot on the edge of the pulley. Run the video until the blocks have gone the distance d . In that time, count the total revolutions that spot makes, including any partial revolutions measured using a protractor. Divide the total revolutions by the time the video ran to get the angular velocity.
(C) Pause the video when mass m 2 has just reached the distance d ; note the location of a position on the rim of the pulley. Advance the video one frame. Use a protractor to measure the angle through which the noted location on the pulley moved in that one frame. Divide that angle by the camera’s time between frames to get the angular velocity of the pulley.
(D) Make a graph of the position of mass m 1 as a function of time, determining the block’s position by pausing the video after every frame. The slope of this graph is the pulley’s angular velocity.
9 . Two hanging blocks, each attached to a different spring, undergo oscillatory motion. It is desired to determine, without stopping the motion, which block experiences the greater maximum acceleration. Which of the following procedures would accomplish that determination?
(A) Place a motion detector underneath each block. On the velocity-time graphs output by the detector, look at the maximum vertical axis value, indicating the highest speed that block attained. Whichever block attains the higher speed has the larger acceleration.
(B) Place a motion detector underneath each block. On the velocity-time graphs output by the detector, look at the steepest portion of the graph. Whichever block makes the steeper maximum slope on the velocity-time graph has the greater maximum acceleration.
(C) Place a motion detector underneath each block. On the position-time graphs output by the detector, look at the maximum vertical axis value, indicating the amplitude of the motion. Whichever block oscillates with the larger amplitude has the larger acceleration.
(D) Place a motion detector underneath each block. On the position-time graphs output by the detector, look at the steepest portion of the graph. Whichever block makes the steeper maximum slope on the position-time graph has the greater maximum acceleration.
Analysis of an Experiment
10 . In the laboratory, measured net torques τ are applied to an initially stationary pivoted bar. The resulting change in the bar’s angular speed after 1 s is measured and recorded as Δω . Which of the following graphs will produce a slope equal to the bar’s rotational inertia about the pivot point?
(A) τ versus Δω
(B) τ versus
(C) τ versus (Δω )2
(D) τ versus
Set of Questions Referring to the Same Stem
Questions 11 and 12: A 1-kg cart moves to the right at 2 m/s. This cart collides with a 0.5-kg cart that is initially at rest; the carts stick together after the collision. Friction on the surface is negligible.
11 . What is the kinetic energy of the two-cart system after the collision?
(A) 1.3 J
(B) 0 J
(C) 0.7 J
(D) 2 J
12 . If instead the carts collide elastically, which of the following is correct about the linear momentum and kinetic energy of the two-cart system after the collision compared to the collision in which the carts stuck together?
(A) The linear momentum and the kinetic energy will both be greater.
(B) The linear momentum will be greater, but the kinetic energy will be the same.
(C) The kinetic energy will be greater, but the linear momentum will be the same.
(D) The linear momentum and the kinetic energy will both be the same.
Multiple Correct Questions
Questions 13 through 15 are multiple correct : Mark the two correct answers.
13 . (multiple correct) An astrophysicist is modeling the behavior of stars orbiting the center of the Milky Way galaxy. In his model, he must consider the effect of each of the four fundamental natural forces. Which of the following correctly indicates a negligible force with a correct explanation for neglecting that force? Select two answers.
(A) The gravitational force is negligible, because the order of magnitude of the gravitational constant G (10−11 N·m2 /kg2 ) is extraordinarily small compared to the order of magnitude of the Coulomb’s law constant (k = 109N·m2 /C2 ).
(B) The electric force is negligible, because it cannot act beyond atomic (∼10−10 m) distances.
(C) The weak force is negligible because although it is responsible for a star’s energy release via nuclear fusion, the nuclear nature of the weak force means that it is irrelevant for star-to-star interactions at a distance.
(D) The strong force is negligible because it cannot act beyond nuclear (~10−15 m) distances.
14 . (multiple correct) A car tire initially rotates clockwise with a rotational speed of 20 rad/s. The rotation gradually slows, such that 2 s later the tire rotates clockwise with a rotational speed of 10 rad/s. Considering clockwise as the positive direction, which of the following vectors is positive? Select two answers.
(A) the tire’s angular acceleration
(B) the tire’s angular momentum
(C) the net torque on the tire
(D) the tire’s angular velocity
15 . (multiple correct) An electron in a vacuum chamber is moving at constant velocity in the direction shown in the preceding diagram. Which of the following force vectors F applied to the electron would increase the electron’s kinetic energy? Select two answers.
Note: The following question is part of a free-response question.
16 . Two blocks are connected over a light, frictionless pulley, as shown. Block A of mass 2 kg is on a frictionless surface; block B of mass 1 kg hangs freely. The blocks are released from rest, with block A 50 cm from the end of the table.
(a) Calculate the speed of block A when it reaches the end of the table.
(b) The mass of block A is now increased. Explain in words but with specific reference to your calculation in (c) how, if at all, block A’s speed at the edge of the table will change.
Lab-Based Free-Response Question with Graphical Analysis
17 . In the laboratory, a constant-frequency generator creates standing waves on a rope, as shown in the preceding diagram. The speed of waves on the string is varied by changing the tension in the rope.
(a) Next to the picture above, draw and label a line indicating one wavelength’s distance on the standing wave.
(b) A student makes the measurements recorded in the table that follows. On the axes provided, label an appropriate scale and graph the wave speed versus the wavelength.
(c) Use a best-fit line to determine the frequency of the generator.
(d) A different lab group used a frequency generator that provided a frequency of 150 Hz. On the graph, draw and label what a best-fit line for this other group’s data might look like.
Solutions for the AP Physics 1 Question Types Assessment
1 . (A) The force F 1 is bigger than the force F 2 , meaning the net force is in the direction of F 1 . Newton’s second law says that the direction of acceleration must be the same as the direction of net force.
2 . (C) Both F 1 and F 2 apply a counterclockwise torque. So these torques add together to give a net counterclockwise torque. That means the angular speed must change, but not necessarily increase. If the objects were rotating clockwise to start with, then a counterclockwise net torque would slow the angular speed.
3 . (A) The ball has a momentum of 6 N·s downward before the collision. In the collision, the ball has to stop, losing all 6 N·s; then the ball has to gain 4 N·s to go in the other direction. That’s a total change of 10 N·s. If you’d prefer to call the downward direction the negative direction, then the change in momentum is the final minus initial momentum: that’s (+4 N·s) – (–6 N·s) = +10 N·s.
4 . (A) The initial momentum of the system is (0.5 kg)(0.6 m/s) + (0.5 kg)(0.8 m/s) = 0.7 N·s. After Cart B is stopped, the momentum of the system is just (0.5 kg)(0.6 m/s) = 0.3 N·s. So, the change in momentum is 0.4 N·s.
5 . (B) The torque provided by F is identical for the three rods, because the force and distance from the fulcrum are the same for all. The rods are identical, so their rotational inertias are the same. By Newton’s second law for rotation, τ net = Iα , because both τ net and I are the same, the angular acceleration α must also be the same for all rods. Then angular acceleration is change in angular speed per second and all change speeds by the same amount in the same amount of time.
6 . (D) Since the only torques acting on the diver are due to the forces applied by his own muscles, no torques external to the diver act. Thus, angular momentum is conserved.
7 . (A) By the last 10 m, the sprinter will be moving at a steady speed, so just dividing distance by time is valid. Choice (B) also divides distance by time, but during the part of the race in which the sprinter is speeding up—invalid. Choice (C) assumes a constant speed for the whole race, which is incorrect according to the problem statement. Choice (D) not only assumes acceleration for the entire race, but uses a horrendously bogus method for finding acceleration. (There’s no equation in the world that says acceleration equals distance divided by time squared.)
8 . (C) Choice (A) is wrong because it describes a calculational prediction, not an experimental measurement. Choice (D) determines linear velocity, not angular velocity. Choices (B) and (C) both discuss angular velocity experimentally, but Choice (B) gives the average angular velocity, while Choice (C) explicitly describes an instantaneous angular velocity because it is finding the angular displacement over a very short time period when the block is essentially at the position d .
9 . (B) Acceleration is the slope of a velocity-time graph. Choices (A) and (D) give speed rather than acceleration. Choice (C) is a distance, not an acceleration.
10 . (A) The relevant equation relates change in angular speed per second—that is, angular acceleration—to net torque. That’s τ net = Iα . The rotational inertia is I , and plotting τ net versus α will give the slope equal to I . Which choice is that? Since the angular speed change was measured over 1 s, Δω is the angular acceleration in this case.
11 . (A) The initial kinetic energy is possessed only by the 1-kg cart and is equal to ½mv 2 = 2 J. We don’t know the speed of the carts after the collision, so we have to calculate that via momentum conservation. (Kinetic energy is not conserved because the carts stick together—this is an inelastic collision.) The total momentum before collision is the sum of mv for each cart = 2 N·s + 0 = 2 N·s. The total momentum after collision must also be 2 N·s by momentum conservation. (Momentum is always conserved in a collision.) The carts’ combined mass is 1.5 kg, so the speed must be 1.3 m/s in order to multiply to the 2 N·s total. Kinetic energy of the combined masses is now ½(1.5 kg)(1.3 m/s)2 = 1.3 J.
12 . (C) Linear momentum is conserved in all collisions, regardless of the elasticity of the collision. The value of the total momentum will be 2 N·s either way. Kinetic energy was lost in the first collision, because kinetic energy is conserved only in an elastic collision. In the second collision, though, no kinetic energy was lost. Therefore, kinetic energy is bigger after the second collision, but smaller after the first collision.
13 . (C) and (D) Choice A is wrong: While it’s true that G is much smaller than k , so what? It’s the gravitational force of the mass of the stars in the galaxy’s center that pulls on the mass of the other stars. Without gravity, stars and planets would not orbit. Choice B is not correct, either. The electric force is, in fact, negligible, but the reasoning presented is wrong. Electric forces dang well can act beyond atomic distances; otherwise, clothes out of the dryer could never experience static cling. The electric force is negligible because electrical interactions can be both attractive and repulsive; in sum, virtually all of the attractive and repulsive forces between fundamental particles will cancel. The other statements are correct.
14 . (B) and (D) The tire is rotating in the positive direction. Angular velocity and angular momentum are always in the direction of rotation. Since the tire is slowing its rotation, though, angular acceleration must be opposite the direction of angular velocity. And by Newton’s second law for rotation, net torque is in the same direction as angular acceleration.
15 . (A) and (C) Any force that does positive work on the electron will increase the electron’s kinetic energy. Positive work means that a component of the force is in the same direction as the electron’s motion. Choice (D) is perpendicular to the electron’s motion, so that force does no work; choices (A) and (C) all have a component up and to the right, so those forces do work to increase the electron’s kinetic energy. Choice B has a component in the direction opposite the electron’s motion, so choice B does work to decrease the electron’s kinetic energy.
16 . (a) Treating the two blocks as a single system, the net force is the 10-N hanging weight. The mass of this two-block system is 3 kg, so by F net = ma , the system’s acceleration is 10 N/3 kg = 3.3 m/s per second.
Now, Block A starts from rest, travels 0.50 m, and accelerates at 3.3 m/s per second. Using the kinematics equation , we get .
(b) The net force on the system does not change, because the 10-N weight is still the same. However, the mass of the system increases. The solution in (a) for the acceleration of the system includes the system’s mass in the denominator, so the acceleration of the system will decrease.
Then, since Block A still travels the same distance to the edge of the table from rest, the same kinematics equation as in (a) works, but this time with a smaller acceleration. With that acceleration in the numerator of the final speed equation, the final speed will also decrease.
17 . (a)
(c) The slope of the line is the frequency of the generator, because the relevant equation is v = λ f . That matches up with the equation for a line y = mx with the wave speed v on the y -axis and the wavelength λ on the x -axis. The slope value m corresponds with the frequency f .
To calculate the slope, use two points on the line that are not data points. I’ve used the ones that are marked with open circles. Then the slope is the rise over the run (5,000–3,000) cm/s/(78–50) cm. That gives a slope of 71 Hz.
(d) A bigger frequency means a bigger slope. The graph will still go through the origin (because there’s no y- intercept term in v = λf ), and the origin is not pictured in the graph. I’ve drawn the line such that the slope is about 150 Hz, up 1,500 cm/s on the vertical axis and over 10 cm on the horizontal axis.