5 Steps to a 5: AP Physics C - Greg Jacobs 2019

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Physics C—Mechanics Practice Exam 1—Multiple-Choice Questions

ANSWER SHEET

Physics C—Mechanics Practice Exam 1—Multiple-Choice Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

1. A cannon is mounted on a truck that moves forward at a speed of 5 m/s. The operator wants to launch a ball from a cannon so the ball goes as far as possible before hitting the level surface. The muzzle velocity of the cannon is 50 m/s. At what angle from the horizontal should the operator point the cannon?

(A) 5°

(B) 41°

(C) 45°

(D) 49°

(E) 85°

2. A car moving with speed v reaches the foot of an incline of angle θ. The car coasts up the incline without using the engine. Neglecting friction and air resistance, which of the following is correct about the magnitude of the car’s horizontal acceleration a_x and vertical acceleration a_y?

(A) a_x = 0; a_y < g

(B) a_x = 0; a_y = g

(C) a_x < g; a_y < g

(D) a_x < g; a_y = g

(E) a_x < g; a_y > g

3. A bicycle slows down with an acceleration whose magnitude increases linearly with time. Which of the following velocity—time graphs could represent the motion of the bicycle?

(A)

(B)

(C)

(D)

(E)

4. A cart is sliding down a low friction incline. A device on the cart launches a ball, forcing the ball perpendicular to the incline, as shown above. Air resistance is negligible. Where will the ball land relative to the cart, and why?

(A) The ball will land in front of the cart, because the ball’s acceleration component parallel to the plane is greater than the cart’s acceleration component parallel to the plane.

(B) The ball will land in front of the cart, because the ball has a greater magnitude of acceleration than the cart.

(C) The ball will land in the cart, because both the ball and the cart have the same component of acceleration parallel to the plane.

(D) The ball will land in the cart, because both the ball and the cart have the same magnitude of acceleration.

(E) The ball will land behind the cart, because the ball slows down in the horizontal direction after it leaves the cart.

5. The quantity “jerk,” j, is defined as the time derivative of an object’s acceleration,

What is the physical meaning of the area under a graph of jerk vs. time?

(A) The area represents the object’s acceleration.

(B) The area represents the object’s change in acceleration.

(C) The area represents the object’s change in velocity.

(D) The area represents the object’s velocity.

(E) The area represents the object’s change in position.

6. A particle moves along the x-axis with a position given by the equation x(t) = 5 + 3t, where x is in meters, and t is in seconds. The positive direction is east. Which of the following statements about the particle is FALSE.

(A) The particle is east of the origin at t = 0.

(B) The particle is at rest at t = 0.

(C) The particle’s velocity is constant.

(D) The particle’s acceleration is constant.

(E) The particle will never be west of position x = 0.

7. A mass hangs from two ropes at unequal angles, as shown above. Which of the following makes correct comparisons of the horizontal and vertical components of the tension in each rope?

(A)

(B)

(C)

(D)

(E)

8. The force of air resistance F on a mass is found to obey the equation F = bv², where v is the speed of the mass, for the range of speeds investigated in an experiment. A graph of F vs. v² is shown above. What is the value of b?

(A) 0.83 kg/m

(B) 1.7 kg/m

(C) 3.0 kg/m

(D) 5.0 kg/m

(E) 1.0 kg/m

9. A box sits on an inclined plane without sliding. As the angle of the plane (measured from the horizontal) increases, the normal force

(A) increases linearly

(B) decreases linearly

(C) does not change

(D) decreases nonlinearly

(E) increases nonlinearly

10. Which of the following conditions are necessary for an object to be in static equilibrium?

I. The vector sum of all torques on the object must equal zero.

II. The vector sum of all forces on the object must equal zero.

III. The sum of the object’s potential and kinetic energies must be zero.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I, II, and III

11. A student pushes a big 16-kg box across the floor at constant speed. He pushes with a force of 50 N angled 35° from the horizontal, as shown in the diagram above. If the student pulls rather than pushes the box at the same angle, while maintaining a constant speed, what will happen to the force of friction?

(A) It must increase.

(B) It must decrease.

(C) It must remain the same.

(D) It will increase only if the speed is greater than 3.1 m/s.

(E) It will increase only if the speed is less than 3.1 m/s.

12. Consider a system consisting only of the Earth and a bowling ball, which moves upward in a parabola above Earth’s surface. The downward force of Earth’s gravity on the ball, and the upward force of the ball’s gravity on the Earth, form a Newton’s third law force pair. Which of the following statements about the ball is correct?

(A) The ball must be in equilibrium since the upward forces must cancel downward forces.

(B) The ball accelerates toward the Earth because the force of gravity on the ball is greater than the force of the ball on the Earth.

(C) The ball accelerates toward the Earth because the force of gravity on the ball is the only force acting on the ball.

(D) The ball accelerates away from Earth because the force causing the ball to move upward is greater than the force of gravity on the ball.

(E) The ball accelerates away from Earth because the force causing the ball to move upward plus the force of the ball on the Earth are together greater than the force of gravity on the ball.

13. A mass m is attached to a mass 3m by a rigid bar of negligible mass and length L. Initially, the smaller mass is located directly above the larger mass, as shown above. How much work is necessary to flip the rod 180° so that the larger mass is directly above the smaller mass?

(A) 4mgL

(B) 2mgL

(C) mgL

(D) 4πmgL

(E) 2πmgL

14. A ball rolls horizontally with speed v off of a table a height h above the ground. Just before the ball hits the ground, what is its speed?

(A)

(B)

(C)

(D) v

(E)

15. A pendulum is launched into simple harmonic motion in two different ways, as shown above, from a point that is a height h above its lowest point. During both launches, the bob is given an initial speed of 3.0 m/s. On the first launch, the initial velocity of the bob is directed upward along the pendulum’s path, and on the second launch it is directed downward along the pendulum’s path. Which launch will cause the pendulum to swing with the larger amplitude?

(A) the first launch

(B) the second launch

(C) Both launches produce the same amplitude.

(D) The answer depends on the initial height h.

(E) The answer depends on the length of the supporting rope.

16. The mass M is moving to the right with velocity v₀ at position x = x₀. Neglect friction. The spring has force constant k. What is the total mechanical energy of the block at this position?

(A)

(B)

(C)

(D)

(E)

17. A sphere, a cube, and a cylinder, all of equal mass, are released from rest from the top of a short incline. The surface of the incline is extremely slick, so much so that the objects do not rotate when released, but rather slide with negligible friction. Which reaches the base of the incline first?

(A) the sphere

(B) the cube

(C) the cylinder

(D) All reach the base at the same time.

(E) The answer depends on the relative sizes of the objects.

18. Block B is at rest on a smooth tabletop. It is attached to a long spring, which is in turn anchored to the wall. Block A slides toward and collides with block B. Consider two possible collisions:

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

Collision II: Block A sticks to block B.

Which of the following is correct about the speed of block B immediately after the collision?

(A) It is faster in case II than in case I ONLY if block B is heavier.

(B) It is faster in case I than in case II ONLY if block B is heavier.

(C) It is faster in case II than in case I regardless of the mass of each block.

(D) It is faster in case I than in case II regardless of the mass of each block.

(E) It is the same in either case regardless of the mass of each block.

19. A 0.30-kg bird is flying from right to left at 30 m/s. The bird collides with and sticks to a 0.50-kg ball that is moving straight up with speed 6.0 m/s. What is the magnitude of the momentum of the ball/bird combination immediately after collision?

(A) 12.0 N·s

(B) 9.5 N·s

(C) 9.0 N·s

(D) 6.0 N·s

(E) 3.0 N·s

20. The force F on a mass is shown above as a function of time t. Which of the following methods can be used to determine the impulse experienced by the mass?

I. multiplying the average force by t_max

II. calculating the area under the line on the graph

III. taking the integral

(A) II only

(B) III only

(C) II and III only

(D) I and II only

(E) I, II, and III

21. A projectile is launched on level ground in a parabolic path so that its range would normally be 500 m. When the projectile is at the peak of its flight, the projectile breaks into two pieces of equal mass. One of these pieces falls straight down, with no further horizontal motion. How far away from the launch point does the other piece land?

(A) 250 m

(B) 375 m

(C) 500 m

(D) 750 m

(E) 1000 m

Questions 22 and 23

A rigid rod of length L and mass M is floating at rest in space far from a gravitational field. A small blob of putty of mass m < M is moving to the right, as shown above. The putty hits and sticks to the rod a distance 2L/3 from the top end.

22. How will the rod/putty contraption move after the collision?

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

(B) The contraption will have no translational motion, but will rotate about the center of mass of the rod and putty combined.

(C) The contraption will move to the right and rotate about the position of the putty.

(D) The contraption will move to the right and rotate about the center of mass of the rod and putty combined.

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

23. What quantities are conserved in this collision?

(A) linear and angular momentum, but not kinetic energy

(B) linear momentum only

(C) angular momentum only

(D) linear and angular momentum, and linear but not rotational kinetic energy

(E) linear and angular momentum, and linear and rotational kinetic energy

24. A car rounds a banked curve of uniform radius. Three forces act on the car: a friction force between the tires and the road, the normal force from the road, and the weight of the car. Which provides the centripetal force which keeps the car in circular motion?

(A) the friction force alone

(B) the normal force alone

(C) the weight alone

(D) a combination of the normal force and the friction force

(E) a combination of the friction force and the weight

25. A ball of mass m anchored to a string swings back and forth to a maximum position A, as shown above. Point C is partway back to the vertical position. What is the direction of the mass’s acceleration at point C?

(A) along the mass’s path toward point B

(B) toward the anchor

(C) away from the anchor

(D) between a line toward the anchor and a line along the mass’s path

(E) along the mass’s path toward point A

26. In a carnival ride, people of mass m are whirled in a horizontal circle by a floorless cylindrical room of radius r, as shown in the diagram above. If the coefficient of friction between the people and the tube surface is μ, what minimum speed is necessary to keep the people from sliding down the walls?

(A)

(B)

(C)

(D)

(E)

Questions 27 and 28

The uniform, rigid rod of mass m, length L, and rotational inertia I shown above is pivoted at its left-hand end. The rod is released from rest from a horizontal position.

27. What is the linear acceleration of the rod’s center of mass the moment after the rod is released?

(A)

(B)

(C)

(D)

(E)

28. What is the linear speed of the rod’s center of mass when the mass passes through a vertical position?

(A)

(B)

(C)

(D)

(E)

29. The 1.0-m-long nonuniform plank, shown above, has weight 1000 N. It is to be supported by two rods, A and B, as shown above. The center of mass of the plank is 30 cm from the right edge. Each support bears half the weight of the plank. If support B is 10 cm from the right-hand edge, how far from the left-hand edge should support A be?

(A) 0 cm

(B) 10 cm

(C) 30 cm

(D) 50 cm

(E) 70 cm

30. A mass m on a spring oscillates on a horizontal surface with period T. The total mechanical energy contained in this oscillation is E. Imagine that instead a new mass 4m oscillates on the same spring with the same amplitude. What is the new period and total mechanical energy?

(A)

(B)

(C)

(D)

(E)

31. A mass m is attached to a horizontal spring of spring constant k. The spring oscillates in simple harmonic motion with amplitude A. What is the maximum speed of this simple harmonic oscillator?

(A)

(B)

(C)

(D)

(E)

32. An empty bottle goes up and down on the surface of the ocean, obeying the position function x = Acos(ωt). How much time does this bottle take to travel once from its lowest position to its highest position?

(A)

(B)

(C)

(D)

(E)

33. The Space Shuttle orbits 300 km above the Earth’s surface; the Earth’s radius is 6400 km. What is the acceleration due to Earth’s gravity experienced by the Space Shuttle?

(A) 4.9 m/s²

(B) 8.9 m/s²

(C) 9.8 m/s²

(D) 10.8 m/s²

(E) zero

34. An artificial satellite orbits Earth just above the atmosphere in a circle with constant speed. A small meteor collides with the satellite at point P in its orbit, increasing its speed by 1%, but not changing the instantaneous direction of the satellite’s velocity. Which of the following describes the satellite’s new orbit?

(A) The satellite now orbits in an ellipse, with P as the farthest approach to Earth.

(B) The satellite now orbits in an ellipse, with P as the closest approach to Earth.

(C) The satellite now orbits in a circle of larger radius.

(D) The satellite now orbits in a circle of smaller radius.

(E) The satellite cannot maintain an orbit, so it flies off into space.

35. Mercury orbits the sun in about one-fifth of an Earth year. If 1 AU is defined as the distance from the Earth to the sun, what is the approximate distance between Mercury and the sun?

(A) (1/25) AU

(B) (1/9) AU

(C) (1/5) AU

(D) (1/3) AU

(E) (1/2) AU

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

Physics C—Mechanics Practice Exam 1—Free-Response Questions

Time: 45 minutes. You may refer to the constants sheet and the equation, both of which are found in the appendix. You may use a calculator.

CM 1

Two 5-kg masses are connected by a light string over two massless, frictionless pulleys. Each block sits on a frictionless inclined plane, as shown above. The blocks are released from rest.

(a) Determine the magnitude of the acceleration of the blocks.

Now assume that the 30° incline provides a resistive force which depends on speed v. This resistive force causes the entire system’s acceleration to be given by the expression

a = 1.8 − 0.03v

where a speed v in m/s gives an acceleration in m/s². The blocks are again released from rest.

(b) i. On the axes below, sketch a graph of the speed of the 5 kg block as a function of time. Label important values, including any asymptotes and intercepts.

ii. Explain how the expression for acceleration leads to the graph you drew.

(c) Explain how to figure out the terminal speed of the 5 kg block.

(d) The terminal speed is 60 m/s — that’s a typical speed in automobile racing. Explain briefly why this result is physically reasonable, even though the blocks are on a track in a physics laboratory.

CM 2

A hollow glass sphere of radius 8.0 cm rotates about a vertical diameter with frequency 5 revolutions per second. A small wooden ball of mass 2.0 g rotates inside the sphere, as shown in the diagram above.

(a) Draw a free-body diagram indicating the forces acting on the wooden ball when it is at the position shown in the picture above.

(b) Calculate the angle θ, shown in the diagram above, to which the ball rises.

(c) Calculate the linear speed of the wooden ball as it rotates.

(d) The wooden ball is replaced with a steel ball of mass 20 g. Describe how the angle θ to which the ball rises will be affected. Justify your answer.

CM 3

A heavy ball of mass m is attached to a light but rigid rod of length L. The rod is pivoted at the top and is free to rotate in a circle in the plane of the page, as shown above.

(a) The mass oscillates to a maximum angle θ. On the picture of the mass m below, draw a vector representing the direction of the NET force on the mass while it is at angle θ. Justify your choice of direction.

(b) Is the magnitude of the net force at the maximum displacement equal to mg sinθ or mg cosθ? Choose one and justify your choice.

(c) Derive an expression for the ball’s potential energy U as a function of the angle θ. Assume that a negative angle represents displacement from the vertical in the clockwise direction.

(d) On the axes below, sketch a graph of the mass’s potential energy U as a function of the angle θ for angles between −90° and +360°. Label maximum and minimum values on the vertical axis.

(e) The pendulum is considered a classic example of simple harmonic motion when it undergoes small-amplitude oscillation. With specific reference to the graph you made in part (d), explain why the assumption of simple harmonic motion is valid.

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

Physics C—Electricity and Magnetism Practice Exam 1—Multiple-Choice Questions

ANSWER SHEET

Physics C—Electricity and Magnetism Practice Exam 1—Multiple-Choice Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

1. Experimenter A uses a very small test charge q_o, and experimenter B uses a test charge 2q_o to measure an electric field produced by two parallel plates. A finds a field that is

(A) greater than the field found by B

(B) the same as the field found by B

(C) less than the field found by B

(D) either greater or less than the field found by B, depending on the accelerations of the test charges

(E) either greater or less than the field found by B, depending on the masses of the test charges

2. A solid conducting sphere has radius R and carries positive charge Q. Which of the following graphs represents the electric field E as a function of the distance r from the center of the sphere?

(A)

(B)

(C)

(D)

(E)

3. An electron moving at constant velocity enters the region between two charged plates, as shown above. Which of the paths above correctly shows the electron’s trajectory after leaving the region between the charged plates?

(A) A

(B) B

(C) C

(D) D

(E) E

4. Two isolated particles, A and B, are 4 m apart. Particle A has a net charge of 2Q, and B has a net charge of Q. The ratio of the magnitude of the electric force on A to that on B is

(A) 4:1

(B) 2:1

(C) 1:1

(D) 1:2

(E) 1:4

5. A uniform electric field points to the left. A small metal ball charged to −2 mC hangs at a 30° angle from a string of negligible mass, as shown above. The tension in the string is measured to be 0.1 N. What is the magnitude of the electric field? (sin 30° = 0.50; cos 30° = 0.87; tan 30° = 0.58.)

(A) 25 N/C

(B) 50 N/C

(C) 2,500 N/C

(D) 5,000 N/C

(E) 10,000 N/C

6. A thin semicircular conductor of radius R holds charge +Q. What is the magnitude and direction of the electric field at the center of the circle?

(A)

(B)

(C)

(D)

(E) The electric field is zero at the center.

7. Above an infinitely large plane carrying charge density σ, the electric field points up and is equal to σ/2ε_o. What is the magnitude and direction of the electric field below the plane?

(A) σ/2ε_o, down

(B) σ/2ε_o, up

(C) σ/ε_o, down

(D) σ/ε_o, up

(E) zero

8. Three charges are arranged in an equilateral triangle, as shown above. At which of these points is the electric potential smallest?

(A) A

(B) B

(C) C

(D) D

(E) E

9. The diagram shows a set of equipotential surfaces. At point P, what is the direction of the electric field?

(A) left

(B) right

(C) up the page

(D) down the page

(E) either left or right, which one cannot be determined

10. A metal sphere carries charge Q; a nonconducting sphere of equal size carries the same charge Q, uniformly distributed throughout the sphere. These spheres are isolated from each other. Consider the electric field at the center of the spheres, within the spheres, and outside the spheres. Which of these electric fields will be the same for both spheres, and which will be different?

(A)

(B)

(C)

(D)

(E)

11. Under what conditions is the net electric flux through a closed surface proportional to the enclosed charge?

(A) under any conditions

(B) only when the enclosed charge is symmetrically distributed

(C) only when all nearby charges are symmetrically distributed

(D) only when there are no charges outside the surface

(E) only when enclosed charges can be considered to be point charges

12. A hollow metal ring of radius r carries charge q. Consider an axis straight through the center of the ring. At what point(s) along this axis is/are the electric field equal to zero?

(A) only at the center of the ring

(B) only at the center of the ring, and a very long distance away

(C) only a very long distance away

(D) only at the center of the ring, a distance r away from the center, and a very long distance away

(E) everywhere along this axis

13. A parallel plate capacitor consists of identical rectangular plates of dimensions a × b, separated by a distance c. To cut the capacitance of this capacitor in half, which of these quantities should be doubled?

(A) a

(B) b

(C) c

(D) ab

(E) abc

14. Two identical capacitors are hooked in parallel to an external circuit. Which of the following quantities must be the same for both capacitors?

I. the charge stored on the capacitor

II. the voltage across the capacitor

III. the capacitance of the capacitor

(A) I only

(B) II only

(C) II and III only

(D) I and III only

(E) I, II, and III

15. A 2 μF capacitor is connected directly to a battery. When the capacitor is fully charged, it stores 600 μC of charge. An experimenter replaces the 2 μF capacitor with three 18 μF capacitors in series connected to the same battery. Once the capacitors are fully charged, what charge is stored on each capacitor?

(A) 100 μC

(B) 200 μC

(C) 600 μC

(D) 1200 μC

(E) 1800 μC

16. A spherical conductor carries a net charge. How is this charge distributed on the sphere?

(A) The charge is evenly distributed on the surface.

(B) The charge resides on the surface only; the distribution of charge on the surface depends on what other charged objects are near the sphere.

(C) The charge moves continually within the sphere.

(D) The charge is distributed uniformly throughout the sphere.

(E) The charge resides within the sphere; the distribution of charge within the sphere depends on what other charged objects are near the sphere.

17. Three resistors are connected to a battery as shown in the diagram above. The switch is initially open. When the switch is closed, what happens to the total voltage, current, and resistance in the circuit?

(A)

(B)

(C)

(D)

(E)

18. In the circuit shown above, the 0.5-F capacitor is initially uncharged. The switch is closed at time t = 0. What is the time constant (the time for the capacitor to charge to 63% of its maximum charge) for the charging of this capacitor?

(A) 5 s

(B) 10 s

(C) 20 s

(D) 30 s

(E) 40 s

19. In the circuit shown above, what is the current through the 3 Ω resistor?

(A) 0 A

(B) 0.5 A

(C) 1.0 A

(D) 1.5 A

(E) 2.0 A

20. A light bulb rated at 100 W is twice as bright as a bulb rated at 50 W when both are connected in parallel directly to a 100-V source. Now imagine that these bulbs are instead connected in series with each other. Which is brighter, and by how much?

(A) The bulbs have the same brightness.

(B) The 100-W bulb is twice as bright.

(C) The 50-W bulb is twice as bright.

(D) The 100-W bulb is four times as bright.

(E) The 50-W bulb is four times as bright.

21. A uniform magnetic field B is directed into the page. An electron enters this field with initial velocity v to the right. Which of the following best describes the path of the electron while it is still within the magnetic field?

(A) It moves in a straight line.

(B) It bends upward in a parabolic path.

(C) It bends downward in a parabolic path.

(D) It bends upward in a circular path.

(E) It bends downward in a circular path.

22. Wire is wound around an insulated circular donut, as shown above. A current I flows in the wire in the direction indicated by the arrows. The inner, average, and outer radii of the donut are indicated by r₁, r₂, and r₃, respectively. What is the magnitude and direction of the magnetic field at point P, the center of the donut?

(A) zero

(B)

(C)

(D)

(E)

23. A wire carries a current toward the top of the page. An electron is located to the right of the wire, as shown above. In which direction should the electron be moving if it is to experience a magnetic force toward the wire?

(A) into the page

(B) out of the page

(C) toward the bottom of the page

(D) toward the top of the page

(E) to the right

24. Which of the following statements about electric and magnetic fields is FALSE:

(A) A charge moving along the direction of an electric field will experience a force, but a charge moving along the direction of a magnetic field will not experience a force.

(B) All charges experience a force in an electric field, but only moving charges can experience a force in a magnetic field.

(C) A positive charge moves in the direction of an electric field; a positive charge moves perpendicular to a magnetic field.

(D) All moving charges experience a force parallel to an electric field and perpendicular to a magnetic field.

(E) A negative charge experiences a force opposite the direction of an electric field; a negative charge experiences a force perpendicular to a magnetic field.

25. Which of these quantities decreases as the inverse square of distance for distances far from the objects producing the fields?

(A) the electric field produced by a finite-length charged rod

(B) the electric field produced by an infinitely long charged cylinder

(C) the electric field produced by an infinite plane of charge

(D) the magnetic field produced by an infinitely long, straight current-carrying wire

(E) the magnetic field produced by a wire curled around a torus

26. A proton enters a solenoid. Upon entry, the proton is moving in a straight line along the axis of the solenoid. Which of the following is a correct description of the proton’s motion within the solenoid?

(A) The proton will be bent in a parabolic path.

(B) The proton will be bent in a circular path.

(C) The proton will continue in its straight path at constant velocity.

(D) The proton will continue in its straight path and slow down.

(E) The proton will continue in its straight path and speed up.

27. A uniform magnetic field points into the page. Three subatomic particles are shot into the field from the left-hand side of the page. All have the same initial speed and direction. These particles take paths A, B, and C, as labeled in the diagram above. Which of the following is a possible identity for each particle?

28. The electric dipole shown above consists of equal-magnitude charges and has an initial leftward velocity v in a uniform magnetic field pointing out of the page, as shown above. The dipole experiences

(A) a clockwise net torque, and a net force to the left

(B) a counterclockwise net torque, and a net force to the left

(C) no net torque, and a net force to the left

(D) a counterclockwise net torque, and no net force

(E) a clockwise net torque, and no net force

29. A beam of electrons has speed 10⁷ m/s. It is desired to use the magnetic field of the Earth, 5 × 10⁻⁵ T, to bend the electron beam into a circle. What will be the radius of this circle?

(A) 1 nm

(B) 1 μm

(C) 1 mm

(D) 1 m

(E) 1 km

30. A very small element of wire of length dL carries a current I. What is the direction of the magnetic field produced by this current element at point P, shown above?

(A) to the right

(B) toward the top of the page

(C) into the page

(D) out of the page

(E) there is no magnetic field produced at point P by this element.

31. A loop of wire surrounds a hole in a table, as shown above. A bar magnet is dropped, north end down, from far above the table through the hole. Let the positive direction of current be defined as counterclockwise as viewed from above. Which of the following graphs best represents the induced current I in the loop?

(A)

(B)

(C)

(D)

(E)

32. A rectangular loop of wire has dimensions a × b and includes a resistor R. This loop is pulled with speed v from a region of no magnetic field into a uniform magnetic field B pointing through the loop, as shown above. What is the magnitude and direction of the current through the resistor?

(A) Bav/R, left-to-right

(B) Bbv/R, left-to-right

(C) Bav/R, right-to-left

(D) Bbv/R, right-to-left

(E) Bba/R, right-to-left

33. A conducting wire sits on smooth metal rails, as shown above. A variable magnetic field points out of the page. The strength of this magnetic field is increased linearly from zero. Immediately after the field starts to increase, what will be the direction of the current in the wire and the direction of the wire’s motion?

34. A uniform magnetic field B points parallel to the ground. A toy car is sliding down a frictionless plane inclined at 30°. A loop of wire of resistance R and cross-sectional area A lies in the flat plane of the car’s body, as shown above. What is the magnetic flux through the wire loop?

(A) zero

(B) BA cos 30°

(C) BA cos 60°

(D) BA

(E) (BA cos 60°)/R

35. If the two equal resistors R₁ and R₂ are connected in parallel to a 10-V battery with no other circuit components, the current provided by the battery is I. In the circuit shown above, an inductor of inductance L is included in series with R₂. What is the current through R₂ after the circuit has been connected for a long time?

(A) zero

(B) (1/4) I

(C) (1/2) I

(D) I

(E)

STOP. End of Physics C—Electricity and Magnetism Practice Exam 1—Multiple-Choice Questions

Physics C—Electricity and Magnetism Practice Exam 1—Free-Response Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

E&M 1

A metal sphere of radius R₁ carries charge +Q. A concentric spherical metal shell, of inner radius R₂ and outer radius R₃, carries charge +2Q.

(a) Let r represent the distance from the center of the spheres. Calculate the electric field as a function of r in each of the following four regions:

1. between r = 0 and r = R₁

2. between r = R₁ and r = R₂

3. between r = R₂ and r = R₃

4. between r = R₃ and r = 0

(b) How much charge is on each surface of the outer spherical shell? Justify your answer.

(c) Determine the electric potential of the outer spherical shell.

(d) Determine the electric potential of the inner metal sphere.

E&M 2

A 1 MΩ resistor is connected to the network of capacitors shown above. The circuit is hooked to a 10-V battery. The capacitors are initially uncharged. The battery is connected, and the switch is closed at time t = 0.

(a) Determine the equivalent capacitance of C₁, C₂, and C₃.

(b) Determine the charge on and voltage across each capacitor after a long time has elapsed.

(c) On the axes below, sketch the total charge on C₃ as a function of time.

(d) After the capacitors have been fully charged, the switch is opened, disconnecting C₁ and C₂ from the circuit. What happens to the voltage across and charge on C₃? Justify your answer.

E&M 3

In the laboratory, far from the influence of other magnetic fields, the Earth’s magnetic field has a value of 5.00 × 10⁻⁵ T. A compass in this lab reads due north when pointing along the direction of Earth’s magnetic field.

A long, straight current-carrying wire is brought close to the compass, deflecting the compass to the position shown above, 48° west of north.

(a) Describe one possible orientation of the wire and the current it carries that would produce the deflection shown.

(b) Calculate the magnitude B_wire of the magnetic field produced by the wire that would cause the deflection shown.

(c) The distance d from the wire to the compass is varied, while the current in the wire is kept constant; a graph of B_wire vs. d is produced. On the axes below, sketch the shape of this graph.

(d) It is desired to adjust this plot so that the graph becomes a straight line. The vertical axis is to remain B_wire, the magnetic field produced by the wire. How could the quantity graphed on the horizontal axis be adjusted to produce a straight-line graph? Justify your answer.

(e) The current carried by the wire is 500 mA. Determine the slope of the line on the graph suggested in part (d).

STOP. End of Physics C—Electricity and Magnetism Practice Exam 1—Free-Response Questions

Physics C—Mechanics Practice Exam 1—Multiple-Choice Solutions

1. D—A projectile has its maximum range when it is shot at an angle of 45° relative to the ground. The cannon’s initial velocity relative to the ground in this problem is given by the vector sum of the man’s 5 m/s forward motion and the cannon’s 50 m/s muzzle velocity. To get a resultant velocity of 45°, the man must shoot the cannon at only a slightly higher angle, as shown in the diagram below.

2. C—The car stays on the plane, and slows down as it goes up the plane. Thus, the net acceleration is in the direction down the plane, which has both a nonzero horizontal and vertical component. The car is not in free fall, so its vertical acceleration is less than g.

3. E—Acceleration is the slope of the v—t graph. Because acceleration increases, the slope of the v—t graph must get steeper, eliminating choices A and B. The bike slows down, so the speed must get closer to zero as time goes on, eliminating choices C and D.

4. C—The cart’s acceleration is g sinθ, down the plane, the ball’s acceleration is g, straight down. (So the magnitudes of acceleration are different and choice D is wrong.) The component of the ball’s acceleration along an axis parallel to the plane is also g sinθ, equal to the ball’s acceleration component.

5. B—The area under a jerk—time graph is the quantity jdt. The derivative

can be interpreted as a change in acceleration over a time interval,

Solving algebraically, jΔt is Δa, meaning the change in acceleration.

6. B—At t = 0, x = +5 m, so the particle is east of the origin to start with. The velocity is given by the derivative of the position function, v(t) = 3 m/s. This is a constant velocity; the acceleration is thus zero (and constant), but at t = 0 the velocity is also 3 m/s, so choice B is false.

7. B—Consider the horizontal and vertical forces separately. The only horizontal forces are the horizontal components of the tensions. Because the block is in equilibrium, these horizontal tensions must be equal, meaning only choices B and D can be right. But the ropes can’t have equal horizontal AND vertical tensions, otherwise they’d hang at equal angles. So D can’t be the right choice, and B must be right.

8. B—The equation F = bv² is of the form y = mx, the equation of a line. Here F is the vertical axis, v² is the horizontal axis, so b is the slope of the line. Looking at the graph, the slope is 5.0 N/3.0 m²/s² = 1.7 kg/m.

9. D—Because no forces act perpendicular to the incline except for the normal force and the perpendicular component of weight, and there is no acceleration perpendicular to the incline, the normal force is equal to the perpendicular component of weight, which is mg cosθ. As the angle increases, the cosine of the angle decreases. This decrease is nonlinear because a graph of F_N vs. θ would show a curve, not a line.

10. D—In equilibrium, the net force and the net torque must both be zero. Static equilibrium means the object is stationary, so kinetic energy must be zero. However, potential energy can take on any value—a sign suspended above a roadway is in static equilibrium, yet has potential energy relative to Earth’s surface.

11. B—The friction force is equal to the coefficient of friction times the normal force. The coefficient of friction is a property of the surfaces in contact, and thus will not change here. However, the normal force decreases when the cart is pulled rather than pushed—the surface must apply more force to the box when there is a downward component to the applied force than when there is an upward component. Speed is irrelevant because equilibrium in the vertical direction is maintained regardless.

12. C—The ball accelerates toward the Earth because, although it is moving upward, it must be slowing down. The only force acting on the ball is Earth’s gravity. Yes, the ball exerts a force on the Earth, but that force acts on the Earth, not the ball. According to Newton’s third law, force pairs always act on different objects, and thus can never cancel.

13. B—The work done on an object by gravity is independent of the path taken by the object and is equal to the object’s weight times its vertical displacement. Gravity must do 3mgL of work to raise the large mass, but must do mg(−L) of work to lower the small mass. The net work done is thus 2mgL.

14. C—Use conservation of energy. Position 1 will be the top of the table; position 2 will be the ground. PE₁ + KE₁ = PE₂ + KE₂. Take the PE at the ground to be zero. Then . The ms cancel. Solving for v₂, you get choice C. (Choice E is wrong because it’s illegal algebra to take a squared term out of a square root when it is added to another term.)

15. C—Consider the conservation of energy. At the launch point, the potential energy is the same regardless of launch direction. The kinetic energy is also the same because KE depends on speed alone and not direction. So, both balls have the same amount of kinetic energy to convert to potential energy, bringing the ball to the same height in every cycle.

16. B—Total mechanical energy is defined as kinetic energy plus potential energy. The KE here is . The potential energy is provided entirely by the spring—gravitational potential energy requires a vertical displacement, which doesn’t occur here. The PE of the spring is .

17. D—When an object rotates, some of its potential energy is converted to rotational rather than linear kinetic energy, and thus it moves more slowly than a non-rotating object when it reaches the bottom of the plane. However, here none of the objects rotate! The acceleration does not depend on mass or size.

18. D—Momentum must be conserved in the collision. If block A bounces, it changes its momentum by a larger amount than if it sticks. This means that block B picks up more momentum (and thus more speed) when block A bounces. The mass of the blocks is irrelevant because the comparison here is just between bouncing and not bouncing. So B goes faster in collision I regardless of mass.

19. B—The momentum of the bird before collision is 9 N·s to the left; the momentum of the ball is initially 3 N·s up. The momentum after collision is the vector sum of these two initial momentums. With a calculator you would use the Pythagorean theorem to get 9.5 N·s; without a calculator you should just notice that the resultant vector must have magnitude less than 12 N·s (the algebraic sum) and more than 9 N·s.

20. E—Impulse is defined on the equation sheet as the integral of force with respect to time, so III is right. The meaning of this integral is to take the area under a F vs. t graph, so II is right. Because the force is increasing linearly, the average force will be halfway between zero and the maximum force, and the rectangle formed by this average force will have the same area as the triangle on the graph as shown, so I is right.

21. D—The center of mass of the projectile must maintain the projectile path and land 500 m from the launch point. The first half of the projectile fell straight down from the peak of its flight, which is halfway to the maximum range, or 250 m from the launch point. So the second half of equal mass must be 250 m beyond the center of mass upon hitting the ground, or 750 m from the launch point.

22. D—By conservation of linear momentum, there is momentum to the right before collision, so there must be momentum to the right after collision as well. A free-floating object rotates about its center of mass; because the putty is attached to the rod, the combination will rotate about its combined center of mass.

23. A—Linear and angular momentum are conserved in all collisions (though often angular momentum conservation is irrelevant). Kinetic energy, though, is only conserved in an elastic collision. Because the putty sticks to the rod, this collision cannot be elastic. Some of the kinetic energy must be dissipated as heat.

24. D.

The centripetal force must act toward the center of the car’s circular path. This direction is NOT down the plane, but rather is purely horizontal. The friction force acts down the plane and thus has a horizontal component; the normal force acts perpendicular to the plane and has a horizontal component. So BOTH F_N and F_f contribute to the centripetal force.

25. D—The mass’s acceleration has two components here. Some acceleration must be centripetal (i.e., toward the anchor) because the mass’s path is circular. But the mass is also speeding up, so it must have a tangential component of acceleration toward point B. The vector sum of these two components must be in between the anchor and point B.

26. B—The free-body diagram for a person includes F_N toward the center of the circle, mg down, and the force of friction up:

Because the person is not sliding down, mg = F_f. And because the motion of the person is circular, the normal force is a centripetal force, so F_N = mv²/r. The force of friction by definition is μF_N. Combining these equations, we have mg = μmv²/r; solve for v to get answer choice B. Note: Without any calculation, you could recognize that only choices A and B have units of speed, so you would have had a good chance at getting the answer right just by guessing one of these two!

27. B—Use Newton’s second law for rotation, τ_net = Iα. The only torque about the pivot is the rod’s weight, acting at the rod’s center; this torque is thus mgL/2. So the angular acceleration, α, of the rod is mgL/2I. But the question asks for a linear acceleration, a = rα, where r is the distance to the center of rotation. That distance here is L/2. So combining, you get a = (L/2)(mgL/2I) = mgL²/4I.

28. D—We cannot use rotational kinematics here because the net torque, and thus the angular acceleration, is not constant. Use conservation of energy instead. The potential energy at the release point is mg(L/2) (L/2 because the rod’s center of mass is that far vertically above its lowest point). This potential energy is converted entirely into rotational kinetic energy ¹/₂Iω². The rod’s angular velocity ω is equal to v/(L/2), where v is the linear speed of the center of mass that we’re solving for. Plugging in, you get mgL/2 = ¹/₂I(v²/[L/2]²). Solving for v, choice D emerges from the mathematics.

29. D—Choose any point at all as the fulcrum; say, the center of mass. Rod B supports 500 N, and is located 20 cm from the fulcrum, producing a total counterclockwise torque of 10,000 N·cm. Rod A also supports 500 N; call its distance from the fulcrum “x”. So 10,000 = 500x, and x = 20 cm. This means Rod A is located 20 cm left of the center of mass, or 50 cm from the left edge.

30. B—The period of a mass on a spring is

with the mass under the square root. So when the mass is quadrupled, the period is only multiplied by two. The total mechanical energy is the sum of potential plus kinetic energy. At the greatest displacement from equilibrium (i.e., at the amplitude), the mass’s speed is zero and all energy is potential; potential energy of a spring is ¹/₂kx² and does not depend on mass. So, because the amplitude of oscillation remains the same, the total mechanical energy does not change.

31. D—The maximum potential energy of the mass is at the amplitude, and equal to ¹/₂kA². This is entirely converted to kinetic energy at the equilibrium position, where the speed is maximum. So set . Solving for v_max, you get choice D. (Note: Only choices C and D have units of velocity! So guess between these if you have to!)

32. B—The bottle’s lowest position is x = −A, and its highest position is x = +A. When t = 0, cos (0) = 1 and the bottle is at x = +A. So, find the time when the cosine function goes to −1. This is when ωt = π, so t = π/ω.

33. B—Don’t try to calculate the answer by saying mg = GMm/r²! Not only would you have had to memorize the mass of the Earth, but you have no calculator and you only have a minute or so, anyway. So think: the acceleration must be less than 9.8 m/s², because that value is calculated at the surface of the Earth, and the Shuttle is farther from Earth’s center than that. But the added height of 300 km is a small fraction (~5%) of the Earth’s radius. So the gravitational acceleration will not be THAT much less. The best choice is thus 8.9 m/s². (By the way, acceleration is not zero—if it were, the Shuttle would be moving in a straight line, and not orbiting.)

34. B—The orbit can no longer be circular—circular orbits demand a specific velocity. Because the satellite gains speed while at its original distance from the planet, the orbit is now elliptical. Because the direction of the satellite’s motion is still tangent to the former circular path, in the next instant the satellite will be farther from Earth than at point P, eliminating answer choice A. The satellite will not “fly off into space” unless it reaches escape velocity, which cannot be 1% greater than the speed necessary for a low circular orbit.

35. D—Kepler’s third law states that for all planets in the same system, their period of orbit squared is proportional to the average distance from the sun cubed. Using units of years and AU, for Earth, (1 year)² = (1 AU)³. For Mercury, we have (¹/₅ year)² = (? AU)³. Solving for the question mark, you find that the distance from Mercury to the sun is the cube root of ¹/₂₅ AU, which is closest to ¹/₃ AU.

Physics C—Mechanics Practice Exam 1—Free-Response Solutions

Notes on grading your free-response section

For answers that are numerical, or in equation form:

*For each part of the problem, look to see if you got the right answer. If you did, and you showed any reasonable (and correct) work, give yourself full credit for that part. It’s okay if you didn’t explicitly show EVERY step, as long as some steps are indicated and you got the right answer. However:

*If you got the WRONG answer, then look to see if you earned partial credit. Give yourself points for each step toward the answer as indicated in the rubrics below. Without the correct answer, you must show each intermediate step explicitly in order to earn the point for that step. (See why it’s so important to show your work?)

*If you’re off by a decimal place or two, not to worry—you get credit anyway, as long as your approach to the problem was legitimate. This isn’t a math test. You’re not being evaluated on your rounding and calculator-use skills.

*You do not have to simplify expressions in variables all the way. Square roots in the denominator are fine; fractions in nonsimplified form are fine. As long as you’ve solved properly for the requested variable, and as long as your answer is algebraically equivalent to the rubric’s, you earn credit.

*Wrong, but consistent: Often you need to use the answer to part (a) in order to solve part (b). But you might have the answer to part (a) wrong. If you follow the correct procedure for part (b), plugging in your incorrect answer, then you will usually receive full credit for part (b). The major exceptions are when your answer to part (a) is unreasonable (say, a car moving at 10⁵ m/s, or a distance between two cars equal to 10⁻¹⁰⁰ meters), or when your answer to part (a) makes the rest of the problem trivial or irrelevant.

For answers that require justification:

*Obviously your answer will not match the rubric word-for-word. If the general gist is there, you get credit.

*But the reader is not allowed to interpret for the student. If your response is vague or ambiguous, you will NOT get credit.

*If your response consists of both correct and incorrect parts, you will usually not receive credit. It is not possible to try two answers, hoping that one of them is right. (See why it’s so important to be concise?)

CM 1

(a)

1 pt: Write Newton’s second law for the direction along the plane for each block.

For the right block, T − mg sin 30° = ma

For the left block mg sin 60° − T = ma

1 pt: for including consistent directions for both equations

1 pt: for including a trig function in both equations, even if it’s the wrong function

1 pt: solve these equations simultaneously to get the acceleration

1 pt: answer is a = 1.8 m/s². (An answer of a = −1.8 m/s² is incorrect because the magnitude of a vector cannot be negative.)

(Alternatively you can just recognize that mg sin 60° pulls left, while mg sin 30° pulls right, and use Newton’s second law directly on the combined system. Be careful, though, because the mass of the ENTIRE system is 10 kg, not 5 kg!)

(b)

i.

1 pt: for a graph that starts at v = 0, has exponential form, and sort of levels off to a maximum speed.

1 pt: for indicating that the asymptote represents the terminal speed, or 60 m/s.

ii.

1 pt: since acceleration is dv/dt, we have a differential equation: dv/dt = 1.8 − 0.03v. We need to find the expression for v.

1 pt: In a differential equation where the derivative is proportional to the function itself, the solution has an exponential term.

1 pt: In the solution e is raised to the −0.03t power because 0.03 is what multiplies the v.

1 pt: Since the blocks start at rest and speed up, the solution includes (1−e^−0.03t) not just e^−0.03t. [Those previous three points could be earned for a direct algorithmic solution for v, using separation of variables and integration.]

1 pt: For large t, the exponential term goes to zero, and so (1−e^−0.03t) should be multiplied by the terminal velocity. That’s why there’s an asymptote.

(c)

1 pt: The terminal velocity is when acceleration equals zero.

1 pt: solve the equation 0 = 1.8 − 0.03v. (You get 60 m/s.)

(d)

1 pt: The blocks only reach this terminal speed after a long time. An incline in a physics lab can’t be more than a couple of meters long, meaning the blocks don’t have the time and space to reach 60 m/s.

CM 2

(a)

1 pt: The weight of the ball acts down.

1 pt: The normal force acts up and left, perpendicular to the surface of the glass.

1 pt: No other forces act.

(b)

1 pt: The normal force can be broken into vertical and horizontal components, where the vertical is F_N cos θ and the horizontal is F_N sin θ. (The vertical direction goes with cosine here because θ is measured from the vertical.)

1 pt: The net vertical force is zero because the ball doesn’t rise or fall on the glass. Setting up forces equal to down, F_N cos θ = mg.

1 pt: The horizontal force is a centripetal force, so F_N sin θ = mv²/r sin θ.

1 pt: For using r sin θ and not just r. (Why? Because you need to use the radius of the actual circular motion, which is not the same as the radius of the sphere.)

1 pt: The tangential speed “v” is the circumference of the circular motion divided by the period. Since period is 1/f, and because the radius of the circular motion is r sin θ, this speed v = 2πr sin θf.

1 pt: Now divide the vertical and horizontal force equations to get rid of the F_N term:

sin θ/cos θ = v²/r sin θ g.

1 pt: Plug in the speed and the sin θ terms cancel, leaving cos θ = g/4π²rf².

1 pt: Plugging in the given values (including r = 0.08 m), θ = 83°.

(c)

1 pt: From part (a), the linear speed is 2πr sin θf.

1 pt: Plugging in values, the speed is 2.5 m/s. (If you didn’t get the point in part (a) for figuring out how to calculate linear speed, but you do it right here, then you can earn the point here.)

(d)

1 pt: The angle will not be affected.

1 pt: Since the mass of the ball does not appear in the equation to calculate the angle in part (b), the mass does not affect the angle.

CM 3

(a)

1 pt: The net force is at an angle down and to the left, perpendicular to the rod.

1 pt: Because the ball is instantaneously at rest, the direction of the velocity in the next instant must also be the direction of the acceleration; this direction is along the arc of the ball’s motion.

(b)

1 pt: The magnitude of the net force is mg sin θ.

1 pt: It’s easiest to use a limiting argument: When θ = 90°, then the net force would be simply the weight of the ball, mg. mg sin 90° = mg, while mg cos 90° = zero; hence the correct answer.

(c)

1 pt: The force on the mass is −mg sin θ, the negative arising because the force is always opposite the displacement.

1 pt: Potential energy is derived from force by U = — ∫ Fdx.

1 pt: The distance displaced x = Lθ.

1 pt: The differential dx becomes L dθ.

1 pt: The integral becomes ∫mgL sin θ dθ, which evaluates to −mgL cos θ. (Here the constant of integration can be taken to be any value at all because the zero of potential energy can be chosen arbitrarily.)

[Alternate solution: Using geometry, it can be found that the height of the bob above the lowest point is L − L cos θ. Thus, the potential energy is mgh = mg(L − L cos θ). This gives the same answer, but has defined the arbitrary constant of integration as mgL.]

(d)

1 pt: The graph should look like some sort of sine or cosine function, oscillating smoothly. The graph may be shifted up or down and still receive full credit.

1 pt: The graph should have an amplitude of mgL, though the graph can be shifted arbitrarily up or down on the vertical axis.

1 pt: The graph should have a minimum at θ = 0.

1 pt: The graph should have a maximum at θ = 180°.

(e)

1 pt: For simple harmonic motion, the restoring force must be linearly proportional to the displacement, like F = −kx. This yields an energy function that is quadratic: −(−kx)·dx integrates to give U = ¹/₂kx². The graph of the energy of a simple harmonic oscillator is, thus, parabolic.

1 pt: Near the θ = 0 position, the graph in part (e) is shaped much like a parabola, only deviating from a parabolic shape at large angles; so the pendulum is a simple harmonic oscillator as long as the energy graph approximates a parabola.

AP Physics C—Mechanics

Full Exam Scoring

Multiple Choice: Number Correct______(35 max)

Total Free Response______(45 max)

1.286 × Multiple Choice + Free Response = Raw Score______(90 max)

Physics C—Electricity and Magnetism Practice Exam 1—Multiple-Choice Solutions

1. B—An electric field exists regardless of the amount of charge placed in it, and regardless of whether any charge at all is placed in it. So both experimenters must measure the same field (though they will measure different forces on their test charges).

2. D—You could use Gauss’s law to show that the field outside the sphere has to decrease as 1/r², eliminating choices B and E. But it’s easier just to remember that an important result of Gauss’s law is that the electric field inside a conductor is always zero everywhere, so D is the only possibility.

3. B—While in the region between the plates, the negatively charged electron is attracted to the positive plate, so bends upward. But after leaving the plates, there is no more force acting on the electron. Thus, the electron continues in motion in a straight line by Newton’s first law.

4. C—This is a Newton’s third law problem! The force of A on B is equal (and opposite) to the force of B on A. Or, we can use Coulomb’s law: The field due to A is k(2Q)/(4 m)². The force on B is QE = k2QQ/(4 m)². We can do the same analysis finding the field due to B and the force on A to get the same result.

5. A—The charge is in equilibrium, so the horizontal component of the tension must equal the electric force. This horizontal tension is 0.1 N times sin 30° (not cosine because 30° was measured from the vertical), or 0.05 N. The electric force is qE, where q is 0.002 C. So the electric field is 0.050 N/0.002 C. Reduce the expression by moving the decimal to get 50/2, or 25 N/C.

6. D—The answer could, in principle, be found using the integral form of Coulomb’s law. But you can’t do that on a one-minute multiple-choice problem. The electric field will point down the page—the field due to a positive charge points away from the charge, and there’s an equal amount of charge producing a rightward field as a leftward field, so horizontal fields cancel. So, is the answer B or D? Choice B is not correct because electric fields add as vectors. Only the vertical component of the field due to each little charge element contributes to the net electric field, so the net field must be less than kQ/R².

7. A—Use the symmetry of the situation to see the answer. Because the infinitely large plane looks the same on the up side as the down side, its electric field must look the same, too—the field must point away from the plane and have the same value.

8. C—Another way to look at this question is, “Where would a small positive charge end up if released near these charges?” because positive charges seek the smallest potential. The positive charge would be repelled by the +2Q charge and attracted to the −Q charges, so would end up at point C. Or, know that potential due to a point charge is kq/r. Point C is closest to both −Q charges, so the r terms will be smallest, and the negative contribution to the potential will be largest; point C is farthest from the +2Q charge, so the r term will be large, and the positive contribution to the potential will be smallest.

9. A—A positive charge is forced from high to low potential, which is generally to the left; and the force on a positive charge is in the direction of the electric field. At point P itself the electric field is directly to the left because an electric field is always perpendicular to equipotential surfaces.

10. C—The charge on the metal sphere distributes uniformly on its surface. Because the nonconducting sphere also has a uniform charge distribution, by symmetry the electric fields will cancel to zero at the center. Outside the spheres we can use Gauss’s law: E·A = Q_enclosed/ε_o. Because the charge enclosed by a Gaussian surface outside either sphere will be the same, and the spheres are the same size, the electric field will be the same everywhere outside either sphere. But within the sphere? A Gaussian surface drawn inside the conducting sphere encloses no charge, while a Gaussian surface inside the nonconducting sphere does enclose some charge. The fields inside must not be equal.

11. A—That’s what Gauss’s law says: net flux through a closed surface is equal to the charge enclosed divided by ε_o. Though Gauss’s law is only useful when all charge within or without a Gaussian surface is symmetrically distributed, Gauss’s law is valid always.

12. B—The electric field at the center of the ring is zero because the field caused by any charge element is canceled by the field due to the charge on the other side of the ring. The electric field decreases as 1/r² by Coulomb’s law, so a long distance away from the ring the field goes to zero. The field is nonzero near the ring, though, because each charge element creates a field pointing away from the ring, resulting in a field always along the axis.

13. C—Capacitance of a parallel-plate capacitor is ε_oA/d, where A is the area of the plates, and d is the separation between plates. To halve the capacitance, we must halve the area or double the plate separation. The plate separation in the diagram is labeled c, so double distance c.

14. E—We are told that the capacitors are identical, so their capacitances must be equal. They are hooked in parallel, meaning the voltages across them must be equal as well. By Q = CV, the charge stored by each must also be equal.

15. E—First determine the voltage of the battery by Q = CV. This gives V = 600 μC/2 μF = 300 V. This voltage is hooked to the three series capacitors, whose equivalent capacitance is 6 μF (series capacitors add inversely, like parallel resistors). So the total charge stored now is (6 μF)(300 V) = 1800 μC. This charge is not split evenly among the capacitors, though! Just as the current through series resistors is the same through each and equal to the total current through the circuit, the charge on series capacitors is the same and equal to the total.

16. B—The charge does reside on the surface, and, if the conductor is alone, will distribute evenly. But, if there’s another nearby charge, then this charge can repel or attract the charge on the sphere, causing a redistribution.

17. D—The voltage must stay the same because the battery by definition provides a constant voltage. Closing the switch adds a parallel branch to the network of resistors. Adding a parallel resistor reduces the total resistance. By Ohm’s law, if voltage stays the same and resistance decreases, total current must increase.

18. C—The time constant for an RC circuit is equal to RC. The resistance used is the resistance encountered by charge that’s flowing to the capacitor; in this case, 40 Ω. So RC = 20 s.

19. E—Assume that the current runs clockwise in the circuit, and use Kirchoff’s loop rule. Start with the 7-V battery and trace the circuit with the current: + 7V − I(3Ω) + 3V − I(2Ω) = 0. Solve for I to get 2.0 A.

20. C—The intrinsic property of the light bulb is resistance; the power dissipated by a bulb depends on its voltage and current. When the bulbs are connected to the 100-V source, we can use the expression for power P = V²/R to see that the bulb rated at 50 watts has twice the resistance of the other bulb. Now in series, the bulbs carry the same current. Power is also I²R; thus the 50-watt bulb with twice the resistance dissipates twice the power, and is twice as bright.

21. E—The electron bends downward by the right-hand rule for a charge in a B field—point to the right, curl fingers into the page, and the thumb points up the page. But the electron’s negative charge changes the force to down the page. The path is a circle because the direction of the force continually changes, always pointing perpendicular to the electron’s velocity. Thus, the force on the electron is a centripetal force.

22. A—This is one of the important consequences of Ampère’s law. The magnetic field inside the donut is always along the axis of the donut, so the symmetry demands of Ampère’s law are met. If we draw an “Amperian Loop” around point P but inside r₁, this loop encloses no current; thus the magnetic field must be zero.

23. C—The magnetic field due to the wire at the position of the electron is into the page. Now use the other right-hand rule, the one for the force on a charged particle in a magnetic field. If the charge moves down the page, then the force on a positive charge would be to the right, but the force on a (negative) electron would be left, toward the wire.

24. C—A positive charge experiences a force in the direction of an electric field, and perpendicular to a magnetic field; but the direction of a force is not necessarily the direction of motion.

25. A—The electric field due to any finite-sized charge distribution drops off as 1/r² a long distance away because if you go far enough away, the charge looks like a point charge. This is not true for infinite charge distributions, though. The magnetic field due to an infinitely long wire is given by

not proportional to 1/r²; the magnetic field produced by a wire around a torus is zero outside the torus by Ampère’s law.

26. C—The magnetic field produced by a single loop of wire at the center of the loop is directly out of the loop. A solenoid is a conglomeration of many loops in a row, producing a magnetic field that is uniform and along the axis of the solenoid. So, the proton will be traveling parallel to the magnetic field. By F = qvB sin θ, the angle between the field and the velocity is zero, producing no force on the proton. The proton continues its straight-line motion by Newton’s first law.

27. E—By the right-hand rule for the force on a charged particle in a magnetic field, particle C must be neutral, particle B must be positively charged, and particle A must be negatively charged. Charge B must be more massive than charge A because it resists the change in its motion more than A. A proton is positively charged and more massive than the electron; the neutron is not charged.

28. E—The force on the positive charge is upward; the force on the negative charge is downward. These forces will tend to rotate the dipole clockwise, so only A or E could be right. Because the charges and velocities are equal, the magnetic force on each = qvB and is the same. So, there is no net force on the dipole. (Yes, no net force, even though it continues to move to the left.)

29. D—The centripetal force keeping the electrons in a circle is provided by the magnetic force. So set qvB = mv²/r. Solve to get r = (mv)/(qB). Just look at orders of magnitude now: r = (10⁻³¹ kg) (10⁷m/s)/(10⁻¹⁹C)(10⁻⁵T). This gives r = 10²⁴/10²⁴ = 10⁰ m ~ 1 m.

30. E—An element of current produces a magnetic field that wraps around the current element, pointing out of the page above the current and into the page below. But right in front (or anywhere along the axis of the current), the current element produces no magnetic field at all.

31. D—A long way from the hole, the magnet produces very little flux, and that flux doesn’t change much, so very little current is induced. As the north end approaches the hole, the magnetic field points down. The flux is increasing because the field through the wire gets stronger with the approach of the magnet; so, point your right thumb upward (in the direction of decreasing flux) and curl your fingers. You find the current flows counterclockwise, defined as positive. Only A or D could be correct. Now consider what happens when the magnet leaves the loop. The south end descends away from the loop. The magnetic field still points down, into the south end of the magnet, but now the flux is decreasing. So point your right thumb down (in the direction of decreasing flux) and curl your fingers. Current now flows clockwise, as indicated in choice D. (While the magnet is going through the loop, current goes to zero because the magnetic field of the bar magnet is reasonably constant near the center of the magnet.)

32. A—You remember the equation for the induced EMF in a moving rectangular loop, ε = Blv. Here l represents the length of the wire that doesn’t change within the field; dimension a in the diagram. So the answer is either A or C. To find the direction of induced current, use Lenz’s law: The field points out of the page, but the flux through the loop is increasing as more of the loop enters the field. So, point your right thumb into the page (in the direction of decreasing flux) and curl your fingers; you find the current is clockwise, or left to right across the resistor.

33. D—Start by finding the direction of the induced current in the wire using Lenz’s law: the magnetic field is out of the page. The flux increases because the field strength increases. So point your right thumb into the page, and curl your fingers to find the current flowing clockwise, or south in the wire. Now use the right-hand rule for the force on moving charges in a magnetic field (remembering that a current is the flow of positive charge). Point down the page, curl your fingers out of the page, and the force must be to the west.

34. C—There is clearly nonzero flux because the field does pass through the wire loop. The flux is not BA, though, because the field does not go straight through the loop—the field hits the loop at an angle. So is the answer BA cos 30°, using the angle of the plane; or BA cos 60°, using the angle from the vertical? To figure it out, consider the extreme case. If the incline were at zero degrees, there would be zero flux through the loop. Then the flux would be BA cos 90°, because cos 90° is zero, and cos 0° is one. So don’t use the angle of the plane, use the angle from the vertical, BA cos 60°.

35. C—The inductor resists changes in current. But after a long time, the current reaches steady state, meaning the current does not change; thus the inductor, after a long time, might as well be just a straight wire. The battery will still provide current I, of which half goes through each equal resistor.

Physics C—Electricity and Magnetism Practice Exam 1—Free-Response Solutions

E&M 1

(a)

1 pt: Inside a conductor, the electric field must always be zero. E = 0.

1 pt: Because we have spherical symmetry, use Gauss’s law.

1 pt: The area of a Gaussian surface in this region is 4πr². The charge enclosed by this surface is Q.

1 pt: So, E = Q_enclosed/ε_oA = Q/4π ε_or².

1 pt: Inside a conductor, the electric field must always be zero. E = 0.

2 pts: Just as in part 2, use Gauss’s law, but now the charge enclosed is 3Q. E = 3Q/4π ε_or².

(b)

1 pt: −Q is on the inner surface.

1 pt: +3Q is on the outer surface.

1 pt: Because E = 0 inside the outer shell, a Gaussian surface inside this shell must enclose zero charge, so −Q must be on the inside surface to cancel the +Q on the small sphere. Then to keep the total charge of the shell equal to +2Q, +3Q must go to the outer surface.

(c)

1 pt: Because we have spherical symmetry, the potential due to both spheres is 3Q/4π ε_or, with potential equal to zero an infinite distance away.

1 pt: So at position R₃, the potential is 3Q/4π ε_oR₃. (Since E = 0 inside the shell, V is the same value everywhere in the shell.)

(d)

1 pt: Integrate the electric field between R₁ and R₂ to get V = Q/4π ε_or + a constant of integration.

1 pt: To find that constant, we know that V(R₂) was found in part (c), and is 3Q/4π ε_oR₃. Thus, the constant is

1 pt: Then, potential at R₁ = Q/4π ε_oR₁ + the constant of integration.

E&M 2

(a)

1 pt: The series capacitors add inversely,

so C_eq for the series capacitors is 3 μF.

1 pt: The parallel capacitor just adds in algebraically, so the equivalent capacitance for the whole system is 5 μF.

(b)

1 pt: After a long time, the resistor is irrelevant; no current flows because the fully charged capacitors block direct current.

1 pt: The voltage across C₃ is 10 V (because there’s no voltage drop across the resistor without any current).

1 pt: By Q = CV the charge on C₃ is 20 μC.

1 pt: Treating C₁ and C₂ in series; the equivalent capacitance is 3 μF, the voltage is 10 V (in parallel with C₃).

1 pt: The charge on the equivalent capacitance of C₁ and C₂ is 30 μC; thus the charge on C₁ = 30 μC, and the charge on C₂ is also 30 μC (charge on series capacitors is the same).

1 pt: Using Q = CV, the voltage across C₁ is 7.5 V.

1 pt: Using Q = CV, the voltage across C₂ is 2.5 V.

(c)

1 pt: For a graph that starts at Q = 0.

1 pt: For a graph that asymptotically approaches 20 μC (or whatever charge was calculated for C₃ in part b).

1 pt: For calculating the time constant of the circuit, RC = 5 s.

1 pt: For the graph reaching about 63% of its maximum charge after one time constant.

(d)

1 pt: For recognizing that the voltage does not change.

1 pt: For explaining that if voltage changed, then Kirchoff’s voltage rule would not be valid around a loop including C₃ and the battery (or explaining that voltage is the same across parallel components, so if one is disconnected the other’s voltage is unaffected).

E&M 3

(a)

1 pt: For placing the wire along a north—south line.

1 pt: The wire could be placed above the compass, with the current traveling due north. (The wire also could be placed underneath the compass, with current traveling due south.) (Points can also be earned for an alternative correct solution: for example, the wire could be placed perpendicular to the face of the compass (just south of it), with the current running up.)

(b)

1 pt: The B field due to Earth plus the B field caused by the wire, when added together as vectors, must give a resultant direction of 48° west of north.

1 pt: Placing these vectors tail-to-tip, as shown below, tan 48° = B_wire/B_Earth.

1 pt: So B_wire = B_Earthtan 48° = 5.6 × 10⁻⁵ T.

(c)

1 pt: The magnetic field due to a long, straight, current-carrying wire is given by

where r is the distance from the wire to the field point, represented in this problem by d.

1 pt: So B is proportional to 1/d; this results in a hyperbolic graph.

1 pt: This graph should be asymptotic to both the vertical and horizontal axes.

(d)

1 pt: Place 1/d on the horizontal axis.

2 pts: The equation for the field due the wire can be written

Everything in the first set of parentheses is constant. So, this equation is of the form y = mx, which is the equation of a line, if 1/d is put on the x-axis of the graph. (1 point can be earned for a partially complete explanation. On this problem, no points can be earned for justification if the answer is incorrect.)

(e)

1 pt: The slope of the graph, from the equation above, is

1 pt: For plugging in values correctly, including 0.5 A or 500 mA.

1 pt: For units on the slope equivalent to magnetic field times distance (i.e., T·m, T·cm, mT·m, etc.).

1 pt: For a correct answer, complete with correct units: 1.0 × 10⁻⁷ Tm, or 1.0 × 10⁻⁴ mT·m.

AP Physics C—Electricity and Magnetism

Full Exam Scoring

Multiple-Choice: Number Correct______(35 max)

Total Free-Response______(45 max)

1.286 × Multiple-Choice + Free-Response = Raw Score______(90 max)

Physics C—Mechanics Practice Exam 2—Multiple-Choice Questions

ANSWER SHEET

Physics C—Mechanics Practice Exam 2—Multiple-Choice Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

1. A jet airplane has mass 300,000 kg. In order to take off, it accelerates uniformly over a horizontal distance of 3000 m, reaching a top speed of 80 m/s. Neglecting drag forces, which of the following is closest to the average horizontal force provided by the airplane’s engines?

(A) 10⁶ N

(B) 10⁴ N

(C) 10³ N

(D) 10⁵ N

(E) 10⁰ N

2. On a one-dimensional track, a large cart in motion collides with a small cart at rest. The carts bounce off each other. Which of the following is true about linear momentum and kinetic energy?

(A) Momentum may or may not be conserved; kinetic energy may or may not be conserved.

(B) Momentum must be conserved; kinetic energy must be conserved.

(C) Momentum may or may not be conserved; kinetic energy must be conserved.

(D) Momentum must be conserved; kinetic energy may or may not be conserved.

(E) Momentum must be conserved; kinetic energy must not be conserved.

3. The escape velocity for a rocket from Earth’s surface is 11 km/s. What is the escape velocity for a rocket on the surface of a planet whose mass is the same as Earth’s but whose radius is half that of Earth?

(A) 2.8 km/s

(B) 6.5 km/s

(C) 7.8 km/s

(D) 11 km/s

(E) 15.4 km/s

4. A 1-kg block hangs from a string at rest. A person holds a spring scale attached vertically to the string, pulling upward so that the scale reads 10 N. What is the tension in the string?

(A) 19 N

(B) 9 N

(C) 0 N

(D) 10 N

(E) 20 N

5. Three blocks A, B, and C are connected by taut light strings on a frictionless surface, as shown earlier. A rightward force F = 24 N is applied to block C. Which of the following correctly ranks the accelerations a_A, a_B, and a_C of each block?

(A) a_A > a_B > a_C

(B) a_C > a_B > a_A

(C) a_A = a_B > a_C

(D) a_A = a_B = a_C

(E) a_C > a_B = a_A

6. A 250-g cart rolls to the right at 100 cm/s and collides with a 500-g cart, which is initially rolling left at 75 cm/s. The carts stick together. What is the speed of the carts right after they collide?

(A) 75 cm/s

(B) 83 cm/s

(C) 25 cm/s

(D) 17 cm/s

(E) 50 cm/s

7. An object of mass m is constrained to move along the x-axis only. It experiences a potential energy function , where U is in joules and x is in meters. Which of the following represents the net force F on the object?

(A) b − kx

(B) kx − b

(C)

(D)

(E) kx + b

8. Three objects, of mass 2 kg, 2 kg, and 4 kg, are arranged as shown in the diagram. Which of the following is closest to the y-coordinate of the three objects’ center of mass?

(A) 2.0 cm

(B) 3.5 cm

(C) 1.5 cm

(D) 0.5 cm

(E) 1.0 cm

9. A block is released from rest at the top of a ramp on which friction is negligible. At the bottom of the ramp, the block encounters a rough, level surface on which the coefficient of kinetic friction with the block is 0.20. The block slides to rest after traveling 45 cm on the rough surface, as shown in the diagram. From what height vertically above the rough surface was the block released?

(A) 9 cm

(B) 22.5 cm

(C) The answer cannot be determined without knowing the mass of the block.

(D) 225 cm

(E) 90 cm

Questions 10 and 11: A 0.20-kg object is attached to a spring, which hangs vertically. A motion detector placed underneath the object records the position-time data shown earlier.

10. At which of the following times is the object’s acceleration downward and at its maximum magnitude?

(A) 1.5 s

(B) 2.0 s

(C) 0.0 s

(D) 1.0 s

(E) 0.5 s

11. What is the spring constant of the spring?

(A) 0.9 N/m

(B) 5.0 N/m

(C) 2.5 N/m

(D) 2.0 N/m

(E) 3.9 N/m

12. Which of the following quantities, if any, is a scalar quantity?

(A) A velocity of −2.0 m/s

(B) A potential energy of −2.0 J

(C) A momentum of −2.0 N·s

(D) A displacement of −2.0 m

(E) None of the above is a scalar quantity.

13. A ball of mass m is attached to a light string, as shown earlier. The ball is whirled counterclockwise in a vertical circle at constant speed. At which labeled position is the tension in the string largest?

(A) A

(B) B

(C) C

(D) D

(E) None of the above; the tension is constant throughout.

14. An object moves along the x-axis according to the potential energy function U(x) = ax² − b, where a and b are positive constants. Which of the following lists all of the positions in which the particle is in stable equilibrium?

(A) x = 0 only

(B) x = 0, x = b, and x = −b

(C) x = b and x = −b

(D) x = 0, x = a/b, and x = −a/b

(E) x = a/b and x = −a/b

15. A 200-W motor lifts an object at a constant speed of 0.5 m/s. The object ends up 100 m east of where it started and elevated 100 m vertically from its starting point after 20 seconds of elapsed time. What is the mass of the object?

(A) 2 kg

(B) 4 kg

(C) 0.2 kg

(D) 4 kg

(E) 8 kg

16. The velocity of a particle moving along the x-axis is given as a function of time by the expression v(t) = 4t² − 3t + 2, where v is in units of m/s and t in s. About how far did the particle move between t = 0 and t = 2 s?

(A) 9 m

(B) 12 m

(C) 24 m

(D) 19 m

(E) 13 m

Questions 17—19: A moon executes an elliptical orbit about its planet. Point P is where the moon is closest to the planet; point A is where the moon is farthest from the planet.

17. As the moon moves from point P to point A and then back to its original position, what happens to its angular momentum about the central planet?

(A) It increases, then decreases

(B) It decreases, then increases

(C) It decreases the entire way.

(D) It remains constant.

(E) It increases the entire way.

18. As the moon moves from point P to point A and then back to its original position, what happens to the potential energy of the moon-planet system?

(A) It increases, then decreases

(B) It decreases, then increases

(C) It decreases the entire way.

(D) It remains constant.

(E) It increases the entire way.

19. As the planet moves from point P to point A and then back to its original position, what happens to the magnitude of the force of the moon on the planet?

(A) It increases, then decreases

(B) It decreases, then increases

(C) It decreases the entire way.

(D) It remains constant.

(E) It increases the entire way.

20. Block A has mass 4 kg; block B has mass 2 kg. Each block experiences an identical but unknown net force for a time of 3 s. Which of the following correctly compares block A’s change in linear momentum Δp_A to block B’s change in linear momentum Δp_B?

(A) Δp_A = Δp_B ≠ 0

(B) Δp_A = (1/12)Δp_B

(C) Δp_A = Δp_B = 0

(D) Δp_A = (1/2)Δp_B

(E) Δp_A = (1/6)Δp_B

21. A meterstick of nonnegligible mass m is pivoted at its center, as shown. Two blocks, block 1 and block 2, are placed on top of the meterstick. When the meterstick is released from the rest at the position shown, it begins to rotate clockwise. Which of the following statements about the masses of the two blocks is correct?

(A) It cannot be determined which block has greater mass, because the lever arms associated with the torques provided by each block are different.

(B) The blocks are of equal mass, because the magnitude of the angular acceleration of each block is the same.

(C) Block 1 has greater mass, because the lever arm associated with the torque it provides is larger than that of block 2.

(D) Block 2 has greater mass, because more torque is provided by block 2 with a smaller lever arm.

(E) The blocks are of equal mass, because the meterstick’s mass is not negligible.

22. In an amusement park ride, people of mass m stand with their backs to the inside of a large cylinder of radius R. The maximum coefficient of static friction between a person’s clothes and the cylinder wall is μ. The cylinder rotates fast enough that the floor can drop away and the riders remain “stuck” to the wall without falling. What is the minimum period of rotation for which the riders will “stick” to the wall?

(A)

(B) μmg

(C)

(D)

(E) 2πμmg

23. The solid disk shown has mass M and radius R. The rotational inertia of the disk about axis 1, which passes through the disk’s center, is (1/2)MR². What is the rotational inertia of the disk about axis 2, which is tangential to the disk’s edge?

(A) (3/2)MR²

(B) (2/5)MR²

(C) MR²

(D) (1/3)MR²

(E) (2/3)MR²

24. A 0.50-kg lab cart on a frictionless surface is attached to a spring, as shown in the figure. The rightward direction is considered positive. The spring is neither compressed nor stretched at position x = 0. The cart is released from rest at position x = +0.25 m at time t = 0. Which of the following graphs represents the force applied by the spring as a function of x?

(A)

(B)

(C)

(D)

(E)

Questions 25 and 26: A ball is thrown off a cliff, with an initial velocity directed at an angle θ downward and to the right.

25. Which of the following is correct about the ball’s acceleration from when the ball is released until it hits the ground?

(A) It remains constant and is directed at the angle θ.

(B) It remains constant and is directed downward.

(C) It increases and its direction changes.

(D) It increases and is directed downward.

(E) It increases and is directed at the angle θ.

26. Which of the following graphs represents the vertical position y of the ball as a function of time, where upward is the positive direction?

(A)

(B)

(C)

(D)

(E)

27. A double-pulley consists of pulleys of two different radii attached together such that they rotate together. The radius r₁ of the inner pulley is 1 cm, and the radius r₂ of the outer pulley is 3 cm. A 1-kg object and a 2-kg object are hung from the pulley as shown in the diagram. How much torque must be applied in order to hold the pulley at rest?

(A) 50 N·cm

(B) 7 N·cm

(C) 70 N·cm

(D) 5 N·cm

(E) 30 N·cm

28. Identical balls A and B are thrown straight upward with initial speeds v_A and v_B, respectively. Ball A goes three times higher than ball B. What is the relationship between v_A and v_B?

(A) v_A = 6v_B

(B) v_A = 9v_B

(C) v_A = 1.7v_B

(D) v_A = 2.4v_B

(E) v_A = 3v_B

29. Two identical balls, A and B, fall vertically and hit the floor with speed v_1. Ball A rebounds with speed v₂; ball B does not rebound. Consider the impulse exerted on the floor and the impulse exerted on the ball during these collisions. Which case(s) cause greater magnitude of impulse?

30. A cart of mass m moves to the right on a level surface along the x-axis. The only horizontal force on the cart is that of air resistance, which can be modeled as being linearly proportional to the cart’s speed v. Which of the following differential equations could describe Newton’s second law for this cart?

(A)

(B)

(C)

(D)

(E)

31. Train A is initially traveling east with speed 10 m/s, slowing down with acceleration 5 m/s². Train B is initially at rest a distance of 200 m from train A’s starting location. Train B speeds up to the west with identical acceleration 5 m/s². How long does it take for the trains to pass each other?

(A) 80 s

(B) 5 s

(C) 40 s

(D) 10 s

(E) 20 s

32. An object moves along the x-axis with a velocity v given as a function of time t as shown in the graph. During which section(s) of the graph is the object’s acceleration constant, but not zero?

(A) AB and BC only

(B) AB, BC, and DE only

(C) CD only

(D) DE only

(E) AB only

33. A particle of mass m is released from rest at position x = 0 at time t = 0. The velocity v of the particle is given by the equation . Which of the following functions could represent the net force on the particle?

(A) 6t³

(B) t³

(C) 3t

(D) 6t

(E) (3/2)t

34. A ball can move to the left or right along a track, where right is considered positive. A graph of the ball’s velocity as a function of time is shown. Which of the following describes the ball’s motion?

(A) The ball slows down to the right, then speeds up to the right.

(B) The ball slows down to the right, then speeds up to the left.

(C) The ball speeds up and then slows down.

(D) The ball slows down to the left, then speeds up to the left.

(E) The ball slows down to the left, then speeds up to the right.

35. A block of mass m experiences a force F, which pushes down and to the right at an angle θ, as shown. The block encounters a rough surface of coefficient of kinetic friction μ. As the block moves a distance d, how much work is done by the net force on the block?

(A) (F cosθ − μmg)d

(B) (F − μmg)d

(C) Fd

(D) Fd cosθ

(E) (F cosθ − μ[mg + F sinθ])d

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

Physics C—Mechanics Practice Exam 2—Free-Response Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

1. An end of a uniform rod of length d and mass M is attached to a frictionless horizontal surface by a frictionless pivot, as shown. Point C marks the midpoint of the rod. The rod is initially motionless, but is free to rotate about the pivot.

(a) Derive an expression for the rotational inertia of the rod about the pivot in terms of given variables and fundamental constants.

(b) Student 1 slides a clay disk of mass m₁ perpendicular to the rod with speed v₁, as shown. The disk sticks to the rod a distance x from the pivot. Derive an expression for the angular speed ω of the rod-disk system immediately after the collision.

(c) Student 2 slides a rubber disk of mass m₁ perpendicular to the rod, also with speed v₁. However, this rubber disk bounces off the rod. After which collision, if either, does the rod rotate with greater angular speed?

____ after colliding with the clay disk

____ after colliding with the rubber disk

____ both collisions result in the same angular speed for the rod

____ which collision produces greater angular speed for the rod cannot be determined

Justify your answer.

For the following two questions, consider collisions involving the clay disk and assume M >> m₁.

(d) In order to maximize the angular speed of the rod-disk system after collision, should the disk hit the rod left of, right of, or at point C? Briefly justify your answer.

(e) How, if at all, would your answer to (d) change if the rod’s mass distribution were nonuniform, with linear mass density getting larger farther from the pivot? Justify your answer.

2. An object of mass m is attached to a spring of spring constant k, as shown in the diagram. The object rests on a horizontal frictionless surface. The spring is compressed a distance d from the equilibrium position, where x = 0. The object is then released from rest to oscillate in simple harmonic motion.

(a) Derive a differential equation that can be used to solve for the block’s position x as a function of time t.

(b) Write an expression for the block’s position x as a function of time that satisfies the differential equation in (a) and the conditions in the problem statement.

(c) The block loses its attachment to the spring when it is at the equilibrium position while moving to the right in its oscillation. The block falls off the table a height y above the ground. Derive an expression for the horizontal distance R that the block lands from the table.

(d) Instead consider that the block lost its attachment to the spring when it was moving to the right in its oscillation, but a distance ½d to the right of the equilibrium position. How, if at all, does the distance R change?

Justify your answer.

(e) On the axes provided, sketch a graph of horizontal position x versus time for the block that loses attachment a distance ½d to the right of the equilibrium position. Let t = 0 be the moment when the spring is fully compressed; continue the graph until the block hits the floor.

(f) On the axes provided, sketch a graph of speed versus time for the block that loses attachment a distance ½d to the right of the equilibrium position. Let t = 0 be the moment when the spring is fully compressed; continue the graph until the block hits the floor.

3. In the laboratory, a block of unknown mass m₁ is attached via a light rope and pulleys to a hanger of known mass m₂. The pulleys are of negligible mass and turn with negligible friction. The block and hanger are released from rest in the configuration shown.

(a) Describe a procedure for measuring the acceleration of the hanger. You should use equipment usually found in a high school laboratory.

(b) Is the magnitude of the acceleration of the block greater than, less than, or equal to the magnitude of the acceleration of the hanger? Justify your answer.

Now m₂ is varied by placing disks of different masses on the hanger. For each value of m₂, the acceleration a of the hanger is measured. A graph is produced with m₂g on the vertical axis and a on the horizontal axis.

(c) Is this graph linear?

___ yes ___ no

• If yes, explain how you know and then explain how you would use a line of best fit to determine the mass m₁.

• If no, explain why not and then explain whether the graph would be concave-up or concave-down.

(d) In a final experiment, the disks are removed from the hanger, and the pulleys are replaced with pulleys that do experience significant friction as they turn. Will the acceleration of the hanger in this final experiment be measured to be greater than, less than, or equal to the acceleration of the hanger in the original experiment? Justify your answer.

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

Physics C—Electricity and Magnetism Practice Exam 2—Multiple-Choice Questions

ANSWER SHEET

Physics C—Electricity and Magnetism Practice Exam 2—Multiple-Choice Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

1. A object of charge q moves at constant speed v in a uniform magnetic field B in a circular path of radius R. How much work is done by the field on the object in each revolution?

(A) qvBR

(B) pR²qvB

(C) 2pR²qvB

(D) zero

(E) 2pqvBR

2. A uniform magnetic field B points into the page and is limited to the region shown. A loop of wire is pulled to the right such that it maintains a steady speed. At which labeled position is the current in the wire at its largest counterclockwise value?

(A) A

(B) B

(C) C

(D) D

(E) E

Questions 3 and 4: Two point objects produce electric potentials at the origin. Object 1, located along the positive x-axis, produces an electric potential of +100 V at the origin; Object 2, located along the negative y-axis, produces an electric potential of +100 V at the origin.

3. What is the net electric potential at the origin?

(A) +200 V

(B) 0 V

(C) +100 V

(D) The answer cannot be determined without knowing the relative distances of the charges from the origin.

(E) (+200 V) (cos 45°)

4. The magnitude of the charge on object 1 is twice the magnitude of the charge on object 2. What is the ratio of object 1’s distance from the origin to object 2’s distance from the origin?

(A) 2:1

(B) 1:1

(C) 1:4

(D) 4:1

(E) 1:2

Questions 5—7:

Two isolated and identical long, straight, parallel wires carry currents I₁ and I₂ in the directions shown in the top view. The wires are separated by a distance d. The magnitude of current in wire 2 initially is twice the magnitude of current in wire 1. The top wire experiences a force F₁ due to the bottom wire.

5. In the laboratory, I₂ is held constant while I₁ is changed to different values both greater and less than I₂. A force probe measures the force F₁ on the top wire for each different I₁. Which of the following graphs represents F₁ as a function of I₁?

(A)

(B)

(C)

(D)

(E)

6. Which of the following indicates the magnitude and direction of the force F₂ on wire 2 due to wire 1?

(A) Half F₁ and downward

(B) Twice F₁ and downward

(C) Equal to F₁ and upward

(D) Equal to F₁ and downward

(E) Half F₁ and upward

7. Now the distance d is doubled while the currents I₁ and I₂ remain the same. In terms of F₁, what is the new force of wire 2 on wire 1?

(A) 2F₁

(B) (1/8) F₁

(C) (1/4) F₁

(D) (1/2) F₁

(E) F₁

8. A very long, straight wire carries a current of I₁ in a direction toward the top of the page, as shown in the diagram. A rectangular coil with two sides parallel to the straight wire carries a current I₂. The near side of the rectangular coil is 2 cm away from the straight wire; the far side is 7 cm from the straight wire. Which of the following correctly sketches the magnetic field magnitude B as a function of the distance x along the top segment of the rectangular wire loop?

(A)

(B)

(C)

(D)

(E)

9. A magnetic field that is aimed perpendicular to the plane of a wire loop varies as a function of time given by the equation B(t) = 4t. Which of the following sketches the emf produced in the loop as a function of time?

(A)

(B)

(C)

(D)

(E)

10. A positively charged object moves to the left at constant speed v. The object enters a region containing uniform magnetic and electric fields and continues to move in a straight line at constant speed. The electric field E is directed down toward the bottom of the page, as shown. What is the direction of the magnetic field?

(A) Into the page

(B) Down toward the bottom of the page

(C) Up toward the top of the page

(D) To the right

(E) Out of the page

Questions 11—13: Four objects with charge +Q, −Q, and —q are placed at the corners of a square of side x, as shown in the diagram.

11. What is the direction of the net electric force on the object of charge —q?

12. Which of the following represents the magnitude of the electric field due to the other three charges at the location of the —q charge?

(A)

(B)

(C)

(D)

(E)

13. Which of the following expressions gives the electric potential at the position of the —q charge due to the other two charges?

(A)

(B)

(C)

(D)

(E)

Questions 14—16: A battery with voltage V, two resistors R₁ and R₂, two switches S₁ and S₂, and an inductor L are connected in a circuit as shown. The resistance of R₂ is twice that of R₁. At time t = 0, the switches have been open for a long time and the inductor stores no energy. At time t = 0, switch S₁ is closed. At time t = t₁ switch S₁ is opened and, simultaneously, switch S₂ is closed.

14. Which of the following expressions represents the voltage across the resistor R₁ as a function of time before t₁?

(A)

(B)

(C) V

(D)

(E)

15. Which of the following graphs represents the current I through the inductor as a function of time?

(A)

(B)

(C)

(D)

(E)

16. What is the maximum energy stored in the inductor?

(A)

(B) zero

(C)

(D)

(E)

17. Two identical resistors are connected in parallel to a battery of voltage V, dissipating power P. Next, the same resistors are instead connected in series to the same battery. What is the new total voltage and total power in the new circuit?

(A) 4V, P/4

(B) V, P

(C) 4V, 4P

(D) V, 4P

(E) V, P/4

18. In a region of space, the electric field is zero everywhere. Which of the following MUST be true about the electric potential in this region?

(A) It must be the same value, either positive or negative, everywhere.

(B) It must be zero everywhere.

(C) It must be the same positive value everywhere.

(D) It must be zero at at least one point in the region.

(E) It must be the same negative value everywhere.

Questions 19 and 20: Two parallel plate capacitors, each with capacitance C, are connected in series with one another as part of a larger circuit. Each stores charge Q and has voltage difference V between its plates. These two capacitors are then replaced by a single equivalent capacitor and reconnected to the same larger circuit.

19. What is the capacitance of the replacement capacitor?

(A) C

(B) C/2

(C) 2C

(D) 3C/2

(E) C/

20. What is the charge on and voltage across the replacement capacitor?

21. Two large oppositely charged parallel plates are initially separated by a distance d. The uniform electric field between the plates has magnitude E. Next, the magnitude of the charge on each plate is doubled, while the separation distance d is doubled. What is the new electric field magnitude?

(A) E

(B) E/4

(C) E/2

(D) 4E

(E) 2E

22. Four objects with charge +Q are arranged on the corners of a square, as shown. Three of the objects are fixed in place; the one at the top right of the square is free to move and is released from rest. No external forces act on the four-object system. Which of the following correctly describes the motion of the free object?

(A) It moves up and to the right with increasing acceleration, always speeding up.

(B) It moves down and left until it hits the bottom-right object.

(C) It moves up and to the right with decreasing acceleration until it eventually comes to rest.

(D) It oscillates about its original position.

(E) It moves up and to the right with decreasing acceleration, always speeding up.

23. Two hollow metal spheres of different radii initially carry different charges: the larger sphere 1 carries charge +q₁, and the smaller sphere 2 carries charge −q₂. When the spheres are connected by a long conducting wire, charge will flow until:

(A) The electric field at each sphere’s surface is the same

(B) The charge residing on the surface of each sphere is equal

(C) The electric field at each sphere’s center is the same

(D) The charge residing on the surface of each sphere is zero

(E) The electric potential at each sphere’s surface is the same

Questions 24 and 25: A parallel-plate capacitor is filled with air. Each plate has area 10 cm². The separation between plates is 2.0 cm. The top place stores +200 nC of charge, and the bottom plate stores −200 nC of charge.

24. An insulating material of dielectric constant k is inserted between the plates. What happens to the capacitance of the capacitor?

(A) (0.44 pF) / κ

(B) (0.44 pF)·κ

(C) (0.44 pF)

(D) (0.44 pF)·κ²

(E) (0.44 pF) / κ²

25. The original capacitor is again filled with air. Now the plates are replaced with new plates of area 20 cm² but the same charge and separation. How is the potential difference across the plates affected?

(A) The potential difference is quadrupled.

(B) The potential difference is cut in half.

(C) The potential difference is doubled.

(D) The potential difference is unchanged.

(E) The potential difference is reduced to ¼ of its previous value.

26. A square loop of wire with side length 0.10 m is situated in the plane of the page. A uniform magnetic field B points into the page, along the z-axis, as shown in the diagram. Which of the following manipulations would increase the magnetic flux through the wire loop?

(A) Increase the side length of the square loop

(B) Rotate the loop about the z-axis

(C) Decrease the magnetic field strength

(D) Rotate the loop about the x-axis

(E) Rotate the loop about the y-axis

27. A solid, nonconducting sphere carries a uniform volume charge distribution r. Which of the following is a correct statement?

(A) All positions in the sphere are at the same electric potential.

(B) All charge that is not bound to an atom resides on the surface of the sphere.

(C) The net charge enclosed by any Gaussian surface entirely within the conductor must be zero.

(D) The electric field just outside the sphere is directed perpendicular to the sphere’s surface.

(E) The electric field magnitude inside the sphere is zero.

28. A proton moving to the right encounters an electric field that points up. What is the direction of the electric force on the proton once the proton enters the electric field?

(A) At an angle down and to the right

(B) Down

(C) At an angle up and to the right

(D) Right

(E) Up

29. A solid conducting sphere of radius R contains a distribution of charge —Q along its surface. Although the electric field at the sphere’s center is zero, the electric potential is not. Why not?

(A) Because the electric potential is defined as equal to zero only at a position very far away from the sphere, the potential cannot be zero anywhere else.

(B) Because the electric field is only zero right at the sphere’s center, a changing electric field nearby must create a changing electric potential nearby, too.

(C) Because in the formula for electric potential R is in the denominator; when R = 0 the potential is not zero, it is undefined.

(D) Because a negative charge would be forced away from the sphere toward V = 0 far away, the potential at the sphere’s surface is nonzero; and in a conductor, the potential is uniform throughout.

(E) Because a positive test charge placed at the sphere’s center would be attracted to the surface, and positive charges are forced high-to-low potential.

30. Can a negatively charged conductor apply an attractive force on an uncharged insulator?

(A) YES — If the uncharged insulator is small enough, the r term in the denominator of Coulomb’s law would become small, making the whole expression large and creating a measurable force.

(B) NO — Because charge can’t move freely along the insulator’s surface, charge separation by induction cannot occur.

(C) NO — Because one of the q terms in Coulomb’s law would be zero, it’s impossible for a negatively charged object to apply an electric force on an uncharged object.

(D) YES — The positive charges can induce a local negative charge on the near surface of the insulator.

(E) YES — The electrons in the negatively charged conductor can attract the protons in the insulator.

31. Two objects, each with charge +Q, are a fixed distance d apart. Each applies an electric force F to the other. Next, an object with charge —(1/8)Q is placed on the midpoint between the first two objects. The new electric force experienced by each of the positively charged objects is:

(A) (3/4)F

(B) 3F

(C) (1/2)F

(D) F

(E) zero

32. The electric potential along the y-axis is given by V = —3y² — 2y — 3, where y is in meters and V is in volts. At what position on the y-axis is the electric field equal to zero?

(A) +0.33 m

(B) —3 m

(C) 0 m

(D) —0.33 m

(E) +3 m

33. A long straight wire of circular cross-section with radius a carries a uniform areal current density j. What is the value of the line integral of the magnetic field around the circumference of the wire?

(A) μ₀jπα

(B) 2μ₀jπα

(C) 2μ₀jπα²

(D) μ₀j

(E) μ₀jπα²

Questions 34 and 35: A 120-V source is connected to three resistors, as shown in the diagram. The resistance of R₃ is twice the resistance of R₂, which is twice the resistance of R₁.

34. Which of the following ranks the power P₁, P₂, and P₃ dissipated by each resistor?

(A) P₁ > P₃ > P₂

(B) P₂ > P₃ > P₁

(C) P₁ = P₂ = P₃

(D) P₃ > P₂ > P₁

(E) P₁ > P₂ > P₃

35. Which of the following is correct about the equivalent resistance of this circuit?

(A) It is between the resistances of R₂ and R₃.

(B) It is greater than the resistance of R₃.

(C) It is between the resistances of R₁ and R₂.

(D) It is less than the resistance of R₁.

(E) It is equal to the resistance of R₂.

STOP. End of Physics C—Electricity and Magnetism Practice Exam 2—Multiple-Choice Questions

Physics C—Electricity and Magnetism Practice Exam 2—Free-Response Questions

Time: 45 minutes. You may refer to the constants sheet and the equation sheet, both of which are found in the appendix. You may use a calculator.

1. (15 points)

A point object carrying positive charge +Q sits in the hollow center of a spherical conducting shell, as shown in the diagram. The inner radius of the shell is r₁, and the outer radius is r₂. Positions A, B, and C are located distances a, b, and c from the shell’s center, respectively.

(a) Determine the magnitude of the electric field at each of the following positions:

i. Position C

ii. Position B

iii. Position A

(b) Draw a vector indicating the direction and relative magnitude of the electric field at positions D, E, and F. If the electric field is zero, indicate so explicitly.

(c) Determine the electric potential at each of the following positions:

i. Position C

ii. Position B

iii. Position A

(d) i. What charge resides on the inner surface of the shell? Justify your answer.

ii. What charge resides on the outer surface of the shell? Justify your answer.

(e) Now the charged point object is relocated such that it is off-center, as shown in the diagram.

i. Describe any changes to the magnitude or direction of the electric fields at positions B and C with the relocated central object. Justify your answer.

ii. Describe any changes to the amount, sign, or distribution of charge residing on the inner and outer surface of the conducting shell. Justify your answer.

2. An uncharged capacitor is connected to the circuit shown, where R_A > R_B. Initially the switch is open, connected neither to position 1 nor to position 2. At time t = 0, the switch is thrown to position 1.

(a) i. What is the current through resistor R₁ immediately after t = 0?

ii. What is the voltage across the capacitor a long time after t = 0?

Now, at time t_2, which is a long time after t = 0, the switch is thrown to position 2.

(b) i. Which resistor, if either, takes a larger current immediately after time t₂? Justify your answer briefly.

ii. Which resistor, if either, takes a larger voltage immediately after time t₂? Justify your answer briefly.

iii. What is the voltage across the capacitor a long time after time t₂?

(c) i. Show how to use Kirchhoff’s rules to write (but do NOT solve) a differential equation for the charge Q on the capacitor as a function of time for all times after t₂.

ii. On the axes provided, sketch a graph of the charge on the capacitor Q as a function of time, beginning at t₂ and ending a long time afterward.

iii. On the axes provided, sketch the magnitude of current through R_A as a function of time, beginning at t₂ and ending a long time afterward.

(d) A student in the laboratory is assigned to verify the current graph in (c) iii. experimentally. The student proposes the following procedure:

I’ll hook a digital multimeter set to measure current in series with R_A. I’ll start a stopwatch simultaneously with throwing the switch to position 2. I’ll call out every 10 s, when my partner will record the ammeter reading.

Explain under what conditions, if any, this procedure will effectively provide evidence for the time-dependent current of the circuit as predicted in (c) iii.

3. (15 points)

A very long wire carries current I₁. Nearby, a square wire loop of side x carries a current I₂, which is produced by a very small battery within the loop as shown. The top side of the square loop is located a distance a from the long wire.

(a) i. In terms of given variables and fundamental constants, calculate the magnitude of the total force applied by the long wire on the wire loop.

ii. Draw a vector indicating the direction of the magnetic force calculated in (a) i.

Now the battery in the square loop is removed, and the loop is placed at the same location. For the following, use values a = 0.02 m, x = 0.05 m, and I₁ = 0.50 A.

(b) Calculate the magnetic flux through the square loop.

(c) Calculate the induced current in the square loop.

(d) Finally, the square loop is moved from its initial position to a new position, a distance a + x from the long wire. It takes 2 s for the wire to change locations. Using the values I₁ = 0.50 A, a = 0.020 m, and x = 0.050 m, calculate the average EMF induced in the square loop while it changes positions.

STOP. End of Physics C—Electricity and Magnetism Practice Exam 2—Free-Response Questions

Physics C—Mechanics Practice Exam 2—Multiple-Choice Solutions

1. D—The airplane’s acceleration can be found from kinematics, using v² = v₀² + 2aΔx with v₀ = 0. This gives a = (80 m/s)²/2(3000 m) = 1.1 m/s/s. The net force is ma, so multiply by the 300,000-kg mass to get about 310,000 N. That’s 3 × 10⁵ N.

2. D—In collisions, momentum generally must be conserved because no net external force acts. But kinetic energy could be converted into internal energy in the collision. Without data to check or an explicit statement of an elastic collision, kinetic energy may or may not be conserved.

3. E—For an object of mass m launched with escape velocity, its total mechanical energy at the surface of the planet is kinetic plus potential, or . When the object gets very far from the planet, its kinetic energy is zero (because escape velocity is the minimum speed necessary to leave the planet) and its potential energy is also zero (by definition, any position a long way from the planet has zero potential energy). So set the previous expression equal to zero and solve for v to get . Now apply semi-quantitative reasoning—the numerator does not change, but the planet’s radius gets cut in half. The radius is in the denominator and under the square root, meaning the speed is multiplied by . Or, you can intuit that the gravitational field at the surface of the planet will be stronger because the planet’s radius is smaller. That would increase, not decrease, the escape velocity. Probably it’s easier the conceptual way.

4. D—The block is in equilibrium, at rest. Thus the weight equals the tension, which is 10 N. The scale simply reads this tension. (If you wanted to do a free body diagram of the scale, it would include a 10 N tension pulling down, and the 10 N force of the hand pulling up.)

5. D—The blocks are attached together. Acceleration is the change in speed every second. In 1 second, all the blocks must change speed by the same amount; if they didn’t, one of the blocks would become unattached. The objects experience different net forces because they are of different mass, but the accelerations must be the same.

6. D—The total momentum before collision is (250 g)(100 m/s) = 25,000 g·cm/s to the right and (500 g)(75 cm) = 37,500 g·cm/s to the left. Subtracting, this gives 12,500 g·cm/s total momentum to the left. That’s the same momentum as after collision, by conservation. After collision, because the carts stick, the total momentum of 12,500 g·cm/s is the total mass of 750 g times the speed. Dividing, the speed is 17 cm/s.

7. A—Take the negative derivative of a potential energy function to get force. Here, the force is —kx + b.

8. E—Ignore the x direction entirely. In the y direction, there’s 4 kg at y = 0 and 4 kg at y = 2 cm. These equal masses have a center of mass directly in between, at y = 1 cm.

9. A—The energy conversion here is gravitational potential energy (mgh) to work done by friction (equal to friction force times distance, where friction force is mu times the normal force): mgh=μmgd. Solving, h = μd. Here μ is 0.20 and d = 0.45 m. That’s 0.09 m, or 9 cm.

10. D—The vertical axis of the graph indicates that up is the positive direction. The slope of this graph is velocity. When an object slows down, its acceleration is in the opposite direction of its motion; when an object speeds up, its acceleration is in the same direction as its motion. So we want a negative slope that gets steeper (meaning downward, increasing velocity) or a positive slope that gets shallower (meaning upward, decreasing velocity). That’s near the top of the peaks of the graph, like at t = 1.0 s.

11. D—The period of an object oscillating on a spring is . The graph shows the period of oscillation—the time for one complete cycle—to be 2.0 s. The mass is given as 0.20 kg. Plug into the equation earlier to get k = 2.0 N/m.

12. C—Displacement, velocity, and momentum are vector quantities because their direction matters. When using a negative sign with these, the negative indicates a direction. However, all forms of energy are scalar quantities, just amounts with no direction. The negative sign on potential energy simply means an amount of potential energy less than zero.

13. D—The net force is ma, which in circular motion is mv²/r. Because speed, radius, and mass don’t change throughout the motion, the net force is constant. To find the net force, add or subtract the individual forces on the ball. At the peak, both the tension in the string and the weight pull down: F_net = T + mg. At the bottom, though, the tension pulls up while the weight pulls down: F_net = T — mg. With F_net and mg constants, the tension at the bottom must be greater—at the bottom, tension is F_net + mg, whereas at the top, tension is F_net — mg.

14. A—Stable equilibrium is where the U vs. x graph has a local minimum. This function is an upward-opening parabola, so anywhere where the derivative is zero is a local minimum. Set the derivative 2ax = 0; solving, x = 0.

15. D—Power P is work done per time. The motor must do work to change the object’s potential energy by mgh, where h is 100 m. (Not 100 m, because the h in this formula represents vertical height only. No work is done to carry the object horizontally.) The relevant equation becomes , where t is the 20-s time interval. Solving, . That’s (200 W)(20 s) / (10 m/s/s) (100 m) = 4 kg.

16. A—To get displacement, integrate the velocity function. The limits of the definite integral should be 0 to 2 s. The integral gives . This evaluates to 8.7 m, closest to A.

17. D—The only force on the moon is the force of the planet. This force acts on a line directly through the planet’s center; thus, this force provides no torque about the planet. Because no torque acts on the moon, the moon’s angular momentum is conserved.

18. A—The formula for the moon-planet potential energy is , where M is the planet’s mass, m is the moon’s mass, and d is the distance between them. The zero of potential energy is taken by convention to be when the moon is a very long way from the planet. As the moon gets closer to the planet, the d term in the denominator decreases. This increases the absolute value of the term. But that term is negative, that is, below zero; a smaller d means a smaller potential energy.

19. B—The magnitude of the force of the planet on the moon is . As the moon gets closer to the planet, the d term in the denominator decreases, increasing the amount of the force.

20. A—Force times time is impulse. Both objects experienced the same nonzero force for the same time, so experience the same impulse. Impulse is change in momentum.

21. D—Because the pivot is at the meterstick’s center of mass, the meterstick’s weight provides no torque. The angular acceleration is clockwise; thus, clockwise torques are greater than counterclockwise torques. These torques are equal to the weight of each object times the object’s distance from the pivot. Object 2 provides a greater torque, yet its lever arm (the distance from the pivot) is smaller. Thus, object 2 must be more massive.

22. D—The centripetal force mv²/r is the normal force F_n. The friction force, also equal to μF_n, balances the weight. The leads to the equation . The speed of an object in circular motion is the circumference over the period T, so replace v with . Now manipulate algebraically to solve for T.

23. A—Use the parallel axis theorem: I = I_cm + Mh², where h is the distance between parallel axes. The rotational inertia through the object’s center of mass is (1/2)MR². The distance between the two axes is R. So I = (1/2)MR² + MR² = (3/2)MR².

24. D—When the displacement is positive—that is, to the right—the force of the spring is negative, that is, to the left. The magnitude of the force of a spring is kx and so is linear in x. Only choice D is linear, with negative force for positive displacement.

25. B—Regardless of how it’s thrown, once released, the ball is in free-fall: no forces except gravity act on the ball. All objects in free-fall experience a constant acceleration of 10 m/s/s downward.

26. B—The slope of the position-time graph is the velocity. Initially, the slope must be nonzero and negative because the ball moves down. And then the graph curves; the slope gets steeper because the ball’s speed increases.

27. A—The torque provided by each string is force times lever arm. The lever arm in this case is the radius of the pulley. So clockwise torque is (20 N)(3 cm) = 60 N·cm. The counterclockwise torque is (10 N)(1 cm) = 10 N·cm. The net torque is thus (60 − 10) = 50 N·cm.

28. C—Use kinematics equation v_f² = v_o² + 2aΔx. The ball comes to rest at its peak, v_f = 0. So . Here the maximum height Δx is under a square root. So three times higher means times bigger speed, and the square root of 3 is 1.7.

29. B—Impulse is change in momentum. Both balls lose the same amount of momentum in hitting the floor and coming to rest; however, ball A changes its momentum even more in order to rebound. So ball A changes its momentum by a larger amount, therefore experiencing the greater impulse. Impulse is also force times time of collision. By Newton’s third law, the force of the ball on the floor is equal to the force of the floor on the ball, and the same collision takes the same amount of time. So if ball A experiences more impulse in its collision, so does the floor.

30. A—Translated to equations, the problem says F = −bv, where b is some positive constant. (The term bv is negative, because the force of air resistance must act opposite the direction of velocity.) We know net force F = ma, and a is the time derivative of velocity. So rewrite: . Rearrange algebraically to get.

31. E—The equation for train A’s position is derived from Δx = v₀t + ½at²: x = 10t — 2.5t², with x = 0 at its original location and calling positive x to the east. That makes train A’s initial velocity positive and its acceleration negative. Train B’s position is given by x = 200 − 2.5t². This works because at time t = 0 x = 200; then after 1 s, x = 197.5 m, so is moving west as expected. Now set both expressions for x equal to each other: 10t — 2.5t² = 200 − 2.5t². This solves to give t = 20 s.

32. A—Acceleration is the slope of a velocity-time graph. The acceleration is constant when the graph is a line: AB, BC, and CD. But CD’s section is horizontal, indicating zero acceleration.

33. D—Net force is ma, where a is the time derivative of the velocity function. This derivative gives . So ma is 6t.

34. A—Here the graph is always above the horizontal axis; this means the motion is always in the positive direction (i.e., right). The vertical axis values get closer to zero, reach zero, and then get farther from zero. Thus, the speed gets closer then farther from zero: the car slows down and speeds up.

35. E—The net force on the block includes the horizontal component of F to the right and the friction force to the left. That’s Fcosθ to the right. To the left, the force of friction is μF_n, where the normal force F_n in this case is the weight mg plus the vertical component of F in order to maintain vertical equilibrium. So the friction force to the left is μ(mg + Fsinθ). The net force is now Fcosθ — μ(mg + Fsinθ). And this net force is parallel to the displacement d, so multiply these to get the net work: [Fcosθ — μ(mg + Fsinθ)]d.

Physics C—Mechanics Practice Exam 2—Free-Response Solutions

Notes on grading your free-response section

1.

(a)

The rotational inertia is given by .

To evaluate the integral, set λ as the constant mass per unit length. Replace dm with λdx, giving . This evaluates to . Here, λ is m/d, making the final result I = (1/3)md².

1 point for using the correct general calculus expression for I

1 point for replacing dm with λdx

1 point for setting up an integral with an x²dx term and limits from 0 to d

1 point for correctly evaluating a correct integral

1 point for recognizing that λ is m/d

1 point for the correct answer

(b)

Angular momentum is conserved. Before, the total angular momentum is just that of the disk, m₁v₁x. Afterward, the total angular momentum is the sum of the disk and rod’s Iω, which is . Set the expressions equal and solve to get .

1 point for a correct expression for total momentum before collision

1 point for a correct expression for I of the rod-disk system

1 point for setting expressions for total momentum before collision equal to total momentum after.

1 point for the correct expression for ω

(c)

The rubber disk provides greater angular speed after collision. In both cases the angular momentum before the collision is the same. However, the rubber disk rebounds, which means it has negative momentum after the collision. That means that the rod must have even more positive momentum after the collision in order to subtract the rubber disk’s momentum and still get the same total momentum.

1 point for correct answer

1 point for correct justification

(d)

To the left of point C. Angular momentum for the disk is m₁v₁x, and the bigger x, the bigger the total angular momentum that can be transferred into the rod-disk after the collision.

1 point for correct answer and justification

(e)

The answer wouldn’t change. The total angular momentum of the system before collision is still m₁v₁x. That angular momentum is still transferred to the system after the collision. Because this angular momentum depends on x but not on properties of the rod, the rod’s linear density is irrelevant.

1 point for correct answer

1 point for correct justification

2.

(a)

The net force is ma, and also is equal to —kx because it’s provided by a spring. Acceleration . Combine to get .

1 point for setting ma equal to the force of the spring

1 point for substituting for a

(b)

The solution to a second-order differential equation is a cosine function. Initially, the compression is maximum and equal to d. So writex = d cos(ωt). But ω isn’t given in the problem statement. Take two derivatives to get . In order for this expression to match the one in (a), ω must be equal to . The final answer is thus .

1 point for writing an expression with a cosine or sine term

1 point for a correct answer

(c)

First, find the speed of the block when it loses contact with the spring. This requires energy conservation, because the block’s acceleration while attached to the spring is not constant. Spring potential energy is converted to kinetic energy: ½kd² = ½mv². The speed of the block is . Then when the block leaves the table, it falls for a time given by vertical kinematics with v₀ = 0: y = ½gt². Solve to get the time . Horizontally, the block’s speed doesn’t change; so multiply speed by time to get distance traveled in the air, .

1 point for using energy to get the block’s speed

1 point for using vertical kinematics to get time

1 point for the correct expression in terms of given variables

(d)

R becomes smaller. In the first case, all of the spring energy was transferred to the block’s kinetic energy. In the new case, the spring stores some energy at release. So the block has less kinetic energy, moves at a slower speed, and covers less horizontal distance in the same time of flight.

1 point for the correct answer

1 point for a correct justification

(e)

Initially the positon-time graph is sinusoidal, because the block was in simple harmonic motion. When the block is released, it moves at constant speed, represented by an unchanging slope (i.e., a straight line). That slope doesn’t change, even after the block leaves the table, because no further horizontal forces act on the block.

1 point for the correct sinusoidal shape before release from the spring

1 point for a straight line, tangent to the sinusoid, immediately after release

1 point for continuing the straight line until the end

(f)

Initially the speed is zero; the velocity-time graph is sinusoidal, because the block was in simple harmonic motion. When the block is released, its speed doesn’t change until the block leaves the table. Then, the block gains speed—the vertical velocity increases, so the Pythagorean sum of vertical and horizontal velocities must also increase even though horizontal velocity doesn’t change.

1 point for the correct sinusoidal shape before release from the spring

1 point for a horizontal line continuous with the sinusoid after release

1 point for increasing speed when the block leaves the table

3.

(a)

Many approaches. My favorite: Place a sonic motion detector beneath the hanger. Use the detector to make a velocity-time graph while the hanger is in motion. The slope of that graph is the hanger’s acceleration.

1 point for measuring quantities that could lead to a determination of acceleration

1 point for a plausible procedure

1 point for a detailed procedure

1 point for explaining how to get acceleration from measured quantities

1 point for multiple data points (note that any use of a graph satisfies this requirement)

(b)

Equal. Acceleration is the change in an object’s speed in some about of time. The objects are connected; if one speeds up, so does the other by the same amount. (Otherwise, they would become disconnected or the rope would go slack.)

1 point for the correct answer

1 point for a correct justification

(c)

The relevant equation here comes from applying Newton’s second law to the two-object system. The net force is m₂g — m₁g; the system mass is m₁ + m₂. Because the graph has m₂g on the vertical axis, solve for m₂g: m₂ g = (m₁ +m₂) a +m₁g. Because a appears linearly (i.e., not squared or square rooted) in the numerator, the graph is linear: yes. Identify each term with a term in the equation for a line, y = mx + b. The y variable is m₂g. The x variable is a. This leaves the slope of the graph as (m₁ + m₂). The b term is the y-intercept and is m₁g. To get m₁, find the vertical intercept of the best-fit line and divide by g.

1 point for recognizing that the graph is linear

1 point for explaining why the graph is linear

1 point for using Newton’s second law correctly to get an expression relating m₂g and a

1 point for using the vertical intercept to get m₁

1 point for explaining in words or mathematics why the vertical intercept is used to get m₁

(d)

The newly measured acceleration would be less. Now the net force on the system isn’t just m₂g — m₁g; there’s another force, the friction force, that acts opposite the direction of motion. Because the objects are released from rest, that’s also opposite the direction of acceleration. So the net force would be smaller than before. Because a = F_net / m and the system mass hasn’t changed, the acceleration would be smaller.

1 point for choosing “less”

1 point for explicitly explaining why the net force would be less

1 point for relating less net force to less acceleration using Newton’s second law

AP Physics C—Mechanics

Full Exam Scoring

Multiple Choice: Number Correct______(35 max)

Total Free-Response______(45 max)

1.286 × Multiple Choice + Free-Response = Raw Score______(90 max)

Physics C—Electricity and Magnetism Practice Exam 2—Multiple-Choice Solutions

1. D—The magnetic force on the object is indeed qvB. However, that force is always perpendicular to the direction of motion. (This is true for all magnetic forces on moving charged objects.) Work is equal to force times the distance traveled parallel to that force—because the force and motion are never parallel to one another, no work is done.

2. B—An emf is only induced in a wire when the flux through the wire is changing; thus the answer must be B or D, where the wire is entering or leaving the magnetic field. To find the direction of the induced current, use Lenz’s law—point the right thumb in the direction of the magnetic field, and ask if the flux is decreasing. At position D the flux is decreasing; so curl the fingers to find that at D the current is clockwise. At position B the flux is not decreasing; so flip the direction of the thumb and curl the fingers. That gives a counterclockwise current.

3. A—Electric potential is a scalar; thus, electric potentials at a point due to multiple objects are added together.

4. A—The potential due to a point charge is , where d is the distance from the charge. Both objects produce the same potential. So the equation with the bigger charge in the numerator must also have a bigger distance in the denominator and by the same factor because the equation has no squares or square roots.

5. C—The force of a magnetic field on a current carrying wire is F = ILB. Here the I is I₁ because wire 1 is experiencing the force; B is the magnetic field produced by wire 2 at the location of wire 1. That magnetic field is . Combining the equations, we find that . All values here are constant except I₁, which is in the numerator and to the first power; the graph is therefore linear.

6. C—Newton’s third law demands that the force of wire 1 on wire 2 be equal to the force of wire 2 on wire 1; so the magnitude of F₂ is equal to F₁. Wire 1 produces a magnetic field into the page at the location of wire 2 (by the second right-hand rule: point the thumb along I₁ and curl the fingers—the fingers point into the page at the location of wire 2). Then by the first right-hand rule (point in the direction of current, curl the fingers toward the magnetic field, the thumb gives the direction of force), the force on wire 2 is upward, toward wire 1.

7. D—Combining the equations for the magnetic field produced by wire 2 at the location of wire 1, and for the force exerted by that magnetic field on wire 1, we get . Here d is in the denominator and to the first power, so doubling d simply halves F₁.

8. D—The magnetic field produced by a current-carrying wire is given by , where r is the distance from the wire. So B goes inversely with x here, meaning only B or D could be right. The magnetic field does not go to zero just because a wire ends 7 cm away from another wire.

9. E—Emf is the time derivative of magnetic flux. Here, because the area of the loop and the angle of the magnetic field relative to the loop don’t change, the time derivative of the magnetic field determines the flux. The time derivative of 4t is simply a constant 4—the emf doesn’t vary in time.

10. E—The electric force on a positive charge is in the direction of the electric field, so downward. To create equilibrium allowing for constant speed, the magnetic force must be upward. Using the first right-hand rule, the fingers point left for the velocity and the thumb points up for the force, so the fingers indicating the magnetic field direction curl out of the page.

11. D—The force components are away from the negative charges and toward the positive charge. The negative charges produce equal components right and up, with a resultant 45 degrees up and right. The positive charge produces a force 45 degrees down and left. Because the positive charge is farther from —q than the other charges, the force it produces is smaller, and thus the net force is up and right.

12. C—The top-left and bottom-right charges each produce an electric field component of magnitude directed toward those charges. These perpendicular components add pythagoreanly to down and left, toward the +Q charge. That +Q charge produces an electric field pointing up and right of magnitude . Because this field is in the opposite direction of the field due to the other two charges, the two fields subtract. Finding a common denominator and subtracting yields choice.

13. D—Each of the —Q charges produces a potential of at the position in question. (Potentials add or subtract algebraically; negative charges produce negative potentials.) The +Q charge x away and so produces a potential of . Get a common denominator and add together to get choice.

14. B—An inductor resists change in current. Initially, when we close S₁, the current needs to change rapidly from zero to something; so the inductor stores energy initially, leaving zero voltage for the resistor. After a long time, though, when the current reaches a steady state, the circuit behaves as if the inductor weren’t there, meaning the voltage across R₁ is just V. Both choices B and D meet these limiting conditions—plug in t = 0 and you get a voltage of zero. But the time constant for an RL circuit is L/R, not RL. Thus, the correct choice is not D, but B.

15. D—Without the inductor, the current would go from zero to a maximum value instantaneously when S₁ was closed. With the inductor, the current increases to a maximum gradually with an exponential relationship. So only C and D get the first part of the current right. Then the current decreases back to zero exponentially—both C and D show that. What’s the difference? The time constant for changes in the current is given by L/R. The bigger the resistance, the shorter the time it takes to reach the steady-state current. Because R₂ is bigger than R₁, the drop in current after the second switch is closed is faster than the rise in current initially.

16. A—The energy stored by an inductor is . What current I is used? We need the maximum current. That’s V/R₁, when S₁ has been closed for a long time but before S₂ is closed. When S₂ is open, the resistor R₂ has no effect on the circuit. After S₂ is closed, the current in the inductor drops, so it cannot be the maximum.

17. E—Voltage is provided by a battery; with rare exceptions, this voltage does not depend on what the battery is attached to. Power is V²/R. The equivalent resistance of the circuit quadruples with the new arrangement. (Originally identical resistors R in parallel give equivalent resistance R/2; in series, the equivalent resistance is 2R.) So the numerator of the power equation doesn’t change while the denominator quadruples, causing power to be cut by ¼.

18. A—Pretend a positive charge is in fact located in this region. No electric field means no electric force on this charge. Because positive charges are forced high-to-low potential and there’s no force on this charge, there can’t be “high” and “low” potentials—the potential must be uniform. Uniform potential has the same effect whether it’s a uniform potential of zero, +5 V, −10 V, or whatever—only the difference in potential from one position to another is relevant here.

19. B—The equivalent resistance of series capacitors is given by . The shortcut is that the equivalent capacitance of two equal capacitors in series is half of each individual capacitance.

20. C—Even though each of the original capacitors stored charge Q, the equivalent capacitor also stores charge Q: capacitors in series each carry the same charge, which is equal to the total charge stored. In the initial situation, each capacitor took voltage V = Q/C. Now the equivalent capacitor takes voltage (Q)/(C/2) = 2Q/C = twice the voltage across each individual capacitor.

21. E—Start with the equation for capacitance of a parallel plate capacitor, . Doubling d causes the capacitance C to be cut in half. Next, look at the relationship between charge and voltage for a capacitor: Q = CV. Here, Q is doubled and C is halved, meaning that V must be increased by a factor of 4. Finally, the electric field between the plates is given by E = V/d. The numerator increases by a factor of 4, while the denominator only increases by a factor of 2. So E is doubled.

22. E—The acceleration depends on the net force. The net force on the free object decreases as it gets farther from the other objects, because the distance term in the denominator of Coulomb’s law gets larger. But the free object will always experience a force up and to the right. A net force in the direction of motion always causes an object to speed up, not to slow down.

23. E—Positive charge is forced high-to-low potential; negative charge is forced low-to-high potential. When there’s no difference in electric potential, then charge is not forced to flow.

24. B—The capacitance of a parallel plate capacitor filled with a vacuum (or air, which is close enough) is , which works out to 0.44 pF — 0.44 × 10^-12 F in this case. The capacitance is multiplied by the dielectric constant of the material filling the capacitor. Note that there’s no need to plug in values into the capacitance equation.

25. B—The potential difference (i.e., the voltage) across a capacitor is given by Q = CV. In this case the charge across the plates is unaffected. The capaci tance is given by ; because A is in the numerator and to the first power, the capacitance doubles when the area of the plates doubles. Back to Q = CV, the same Q with doubled C gives half the previous V.

26. A—Magnetic flux is BA, where B is the magnetic field strength and A is the area of the loop that is directly penetrated by the magnetic field. So we need to either increase the magnetic field strength or increase the area of the loop directly penetrated by the field. The rotations do not increase that area: rotating about the x- or y-axis would reduce the area of the loop available to the magnetic field penetration. Rotation about the z-axis wouldn’t change that area. But increasing the actual physical area of the loop would increase the A term.

27. D—Choices (A), (B), (C), and (E) are valid only for a conducting sphere. However, (D) is correct in this case, too, because the charge distribution is uniform; the symmetry of the charge distribution means that at any position on the surface, the charges in the sphere produce horizontal electric field components that cancel. Only the vertical (i.e., perpendicular to the surface) component remains.

28. E—A positively charged object is forced in the same direction as the electric field. The direction of motion has nothing to do with the direction of force.

29. D—It is correct that electric potential is zero a very long distance away from the sphere, but so what. The electric potential at the surface is given by V = −kQ/R; it is negative because it is produced by a negative charge. A negative charge near the surface of the sphere would be forced low-to-high potential, which would be from the negative-potential surface toward zero potential far away. But within the conducting sphere, the electric field is zero, meaning the potential cannot change. The potential at the sphere’s center is ALSO V = −kQ/R.

30. B—Charge separation by induction happens in conductors, in which charge can flow freely along the conductor’s surface. This phenomenon allows an uncharged conductor to experience attraction to another charged object. However, in an insulator, charges cannot flow freely, so there can’t even be a local area in which negative charges attract positive charges.

31. C—Consider the leftmost object. It experiences two electric forces, one from the +Q charge and one from the —(1/8)Q charge. Write Coulomb’s law for each of these forces, and subtract because the forces act in opposite directions: . Simplify the expression to get , which is half of the force of the two +Q charges on each other.

32. D—The electric field is related to the electric potential by . The negative derivative of the V function is 6y + 2; set that equal to zero to get y = −(1/3).

33. E—Ampere’s law says that the line integral of the magnetic field around a closed line is μ₀I_enclosed. How much current is enclosed by the circumference of the wire? The areal current density j times the area enclosed. The cross-sectional area enclosed here is just the area of the circular wire, πa².

34. E—The power dissipated by a resistor is given by V²/R. Here the voltage across each resistor is the same, because all are in parallel with each other. Mathematically, then, the smallest resistance will take the largest power. Distance from the battery is irrelevant.

35. D—You don’t need the equation for equivalent resistance of parallel resistors. No matter the values, the equivalent resistance of parallel resistors is always less than the smallest resistor.

Physics C—Electricity and Magnetism Practice Exam 2—Free-Response Solutions

1. (15 points)

A point object carrying positive charge +Q sits in the hollow center of an uncharged spherical conducting shell, as shown in the diagram. The inner radius of the shell is r₁, and the outer radius is r₂. Positions A, B, and C are located distances a, b, and c from the shell’s center, respectively.

(a) Determine the magnitude of the electric field at each of the following positions:

i. Position C

Because the conductor itself has no charge, the charge enclosed by a Gaussian surface of radius c is just Q. Then the reasoning in part i gives Q/4πε₀c² or kQ/c².

1 pt for correct answer

ii. Position B

Fact: The electric field inside a conductor is zero.

1 pt for correct answer

iii. Position A

Draw a spherical Gaussian surface with radius a. It encloses charge Q and has surface area 4πa2. By the symmetry of the problem, the integral in Gauss’s law becomes multiplication: EA = Q_enclosed/ε₀. The electric field is then Q/4πε₀a2; this is equivalent to kQ/a2 using k as the Coulomb’s law constant.

1 pt for correct answer

(b) Draw a vector indicating the direction and relative magnitude of the electric field at positions D, E, and F. If the electric field is zero, indicate so explicitly.

1 pt for each vector pointing away from the center of the sphere

1 pt for vectors at D and F being longer than the vector at E

(c) Determine the electric potential at each of the following positions:

i. Position C

The electric potential a long distance away is zero. Because the electric field outside the shell has the same form as that due to a point charge, so does the electric potential: +kQ/c.

1 pt for the correct answer

ii. Position B

The electric potential at the outer surface of the shell is +kQ/r₂. Because the electric field is zero within the shell, the potential cannot change within the shell—it’s still the same potential as at either surface. So +kQ/r_2.

1 pt for correct answer

iii. Position A

The electric potential inside a spherical shell of charge is equal to kq/r, where r is the radius of the shell. The electric potential due to a point charge is also kq/r, where this r represents the distance from the charge. At position A, we have three things contributing to the potential: the point charge +Q at the center, the ring of charge −Q on the inside surface, and the ring +Q on the outside surface. Superimpose these three to get a total potential of +kQ/a — kQ/r₁ + kQ/r₂^.

Note that this works out to give the same potential at the inside surface (+kQ/r₂) as at the outer surface.

1 pt for correct answer

(d)

i. A spherical Gaussian surface drawn inside the shell must enclose zero charge, because the electric field inside the shell is zero. We know such a shell encloses +Q from the point object, so the inner surface must carry charge —Q to make the total charge inside that Gaussian surface zero.

1 pt for correct answer

1 pt for a correct justification

ii. What charge resides on the outer surface of the shell? Justify your answer.

We know the shell is uncharged. With —Q on the inner surface, +Q must be on the outer surface to give a net charge of zero on the shell.

1 pt for an answer consistent with i and correct justification

(e) Now the charged point object is relocated such that it is off-center, as shown in the diagram.

i. Describe any changes to the magnitude or direction of the electric fields at positions B and C with the relocated central object. Justify your answer.

The electric field at B is still zero, because B is still inside a conductor. No change.

The electric field at C is unchanged, because the electric field outside a conductor is perpendicular to the conductor’s surface; the symmetry of the problem outside the shell is unchanged. A Gaussian sphere drawn through C still encloses the same charge, so no change.

1 pt for correct justification for no change in electric field at B

1 pt for correct justification that there is no change in the electric field at C

ii. Describe any changes to the amount, sign, or distribution of charge residing on the inner and outer surface of the conducting shell. Justify your answer.

The same amount and sign of charge must reside on each surface, according to Gauss’s law, in order to get an electric field of zero inside the conductor. However, whereas the outer surface’s charge will still be uniformly distributed, the charge on the inner surface will be differently distributed—the point object will induce a denser negative charge on the part of the surface closest to the object.

1 pt justifying the same amount and sign of charge on each surface with Gauss’s law

1 pt justifying the different distribution on the inner surface

2. An uncharged capacitor of capacitance C is connected to the circuit shown, where R_A > R_B. Initially the switch is open, connected neither to position 1 nor to position 2. At time t = 0, the switch is thrown to position 1.

(a)

i. What is the current through resistor R₁ immediately after t = 0?

The uncharged capacitor will initially allow current to flow as if the circuit consisted of just the battery and resistor. So by Ohm’s law the current is V/R_A.

1 pt for correct answer

ii. What is the voltage across the capacitor a long time after t = 0?

After a long time the capacitor will be charged, and thus will block current. Zero.

1 pt for correct answer

Now, at time t_2, which is a long time after t = 0, the switch is thrown to position 2.

(b)

i. Which resistor, if either, takes a larger current immediately after time t₂? Justify your answer briefly.

The resistors are in series—they take the same current by definition (or by Kirchhoff’s junction rule).

1 pt for answer and brief justification

ii. Which resistor, if either, takes a larger voltage immediately after time t₂? Justify your answer briefly.

In the instant after closing the switch, the capacitor is charged, meaning it has a voltage across it and will act briefly as a battery. Both resistors take the same current as each other. By Ohm’s law, then, the bigger resistor takes the bigger voltage. So R_A.

1 pt for correct answer

1 pt for justification

iii. What is the voltage across the capacitor a long time after time t₂?

The voltage is given by Q/C, where Q is the charge on the capacitor. This charge drains over time, so the voltage will also drop, reducing to zero after a long time.

1 pt for correct answer

(c)

i. Show how to use Kirchhoff’s rules to write (but do NOT solve) a differential equation for the charge Q on the capacitor as a function of time for all times after t₂.

The sum of the voltage changes will equal zero. Trace the circuit starting with the capacitor, which raises the voltage by Q/C. Each resistor drops the voltage by IR: Q/C — IR₁ — IR₂ = 0.

Current is dQ/dt; replace I to get

Q/C — (R₁ + R₂) dQ/dt = 0

And that is a differential equation for Q. All the variables except for Q and t are constants.

1 pt for writing any equation that tries to sum voltage changes to zero

1 pt for the correct Kirchhoff’s equation: Q/C — IR₁ — IR₂ = 0

1 pt replacing I with dQ/dt

ii. On the axes provided, sketch a graph of the charge on the capacitor Q as a function of time, beginning at t₂ and ending a long time afterward.

1 pt starting with nonzero Q and ending close to zero Q

1 pt showing something like exponential decay

iii. On the axes provided, sketch the magnitude of current through R_A as a function of time, beginning at t₂ and ending a long time afterward.

The current is initially large—see (d). The current drops to zero—see (e). In between, the current in any RC circuit changes by an exponential function.

1 pt starting with nonzero I and ending close to zero I

1 pt showing something like exponential decay

(d) A student in the laboratory is assigned to verify the current graph in (c) iii. experimentally. The student proposes the following procedure:

The time constant (R₁ + R₂)C for the capacitor’s discharge has to be comparable with the 10-s stopwatch call-out time.

1 pt for referencing the time constant of the circuit

1 pt for correctly relating the time constant to the 10-s stopwatch time, that is, they should be somewhat similar

3. (15 points)

A very long wire carries current I₁. Nearby, a square wire loop of side x carries a current I₂, which is produced by a very small battery within the loop as shown. The top side of the square loop is located a distance a from the long wire.

(a)

i. In terms of given variables and fundamental constants, calculate magnitude of the total force applied by the long wire on the wire loop.

The magnetic field produced by I₁ at the location of the wire loop is always into the page, by the right-hand rule associated with . Thus, by the right-hand rule associated with F = ILB, the force on the left-hand side of the loop is to the left; the force on the right-hand side of the loop is to the right. These forces are equal in magnitude, because each element of the loop is the same distance from the long wire. So the side forces cancel.

The force on the top wire is upward and equal to I₂xB, where . The force on the bottom wire is downward and equal to I₂xB, where . Subtract the two forces to get . No need to simplify—any equivalent form is acceptable.

1 pt for subtracting just force on top and bottom wire portions

1 pt for correct use of

1 pt for correct use of F = ILB

1 pt for combining equations and substituting a and x + a for d, and x for L

ii. Draw a vector indicating the direction of the magnetic force calculated in (a) i.

1 pt for the correct answer

Now the battery in the square loop is removed, and the loop is placed at the same location.

(b) Calculate the magnetic flux through the square loop.

Magnetic flux is . Here replace B with and dA with x·dr, where r is a variable representing distance from the long wire: . Take the integral to get , evaluated from a to x + a. That’s . Any form of answer is acceptable, as long as the integral is actually evaluated—no integral notation on the final answer.

1 pt for substituting into any form of the magnetic flux equation

1 pt for attempting to use calculus rather than straight multiplication because the magnetic field term is not constant

1 pt for evaluating the integral to some sort of logarithmic function, even if not entirely correct

1 pt for entirely correct solution, including correct substitution of integral limits

(c) Calculate the induced emf in the square loop.

Induced emf is the time derivative of the magnetic flux. This flux does not change in time, because the field and the area of the loop that the field penetrates don’t change in time. So zero induced current.

1 pt for using any form of the emf equation, either in words or variables (not just writing it, using it)

1 pt for explaining or showing that the ΔBA term is zero (i.e., the magnetic flux doesn’t change in time)

(d) Finally, the square loop is moved from its initial position to a new position, a distance a + x from the long wire. It takes 2 s for the wire to change locations. Using the values I₁ = 0.50 A, a = 0.020 m, and x = 0.050 m, calculate the average EMF induced in the square loop while it changes positions.

Using the solution in (c), we know the initial magnetic flux. Plugging into the result, the term gives 5.0 × 10^-9 in standard units. The natural log term becomes ln (7/2) = 1.3. So the initial flux is 6.3 × 10^-9 Tm².

The final flux is calculated using the same formula, just plugging in the different locations of the loop: x + x + a and x + a. That’s 0.12 m and 0.07 m. The natural log term this time becomes ln (12/7) = 0.53. So the final flux is 2.7 × 10^-9 Tm².

Subtract final minus initial flux to get the change in flux of 3.6 × 10^-9 Tm².

But induced emf is change in flux divided by the time interval over which the flux changed—divide by 2 s to get the final answer: the induced emf is 1.8 × 10^-9 V (i.e., 1.8 nV).

1 pt for plugging the correct values into whatever equation they came up with in (c) for the initial flux.

1 pt for plugging the correct values into whatever equation they came up with in (c) for the final flux

1 pt for dividing any defined change in flux by 2 s

1 pt for the correct numerical answer (i.e., somewhere between 1.5 and 2.0 nV); the answer must have correct units, but may be in any form

AP Physics C—Electricity and Magnetism

Full Exam Scoring

Multiple-Choice: Number Correct______(35 max)

Total Free-Response______(45 max)

1.286 × Multiple-Choice + Free-Response = Raw Score______(90 max)