﻿ ﻿Build Your Test-Taking Confidence - 5 Steps to a 5: AP Physics C (2016)

## 5 Steps to a 5: AP Physics C (2016)

### STEP 5

PHYSICS C—Mechanics Practice Exam—Multiple-Choice Questions

PHYSICS C—Mechanics Practice Exam—Free-Response Questions

PHYSICS C—Electricity and Magnetism Practice Exam—Multiple-Choice Questions

PHYSICS C—Electricity and Magnetism Practice Exam—Free-Response Questions

PHYSICS C—Mechanics Practice Exam—Multiple-Choice Solutions

PHYSICS C—Electricity and Magnetism Practice Exam—Multiple-Choice Solutions

PHYSICS C—Practice Exams—Free-Response Solutions

Physics C—Mechanics Practice Exam—Multiple-Choice Questions

Physics C—Mechanics Practice Exam—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 = bv 2 , where v is the speed of the mass, for the range of speeds investigated in an experiment. A graph of F vs. v 2 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?

1. The vector sum of all torques on the object must equal zero.
2. 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 v 0 at position x = x 0 . 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?

1. multiplying the average force bytmax
2. 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 = A cos(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—Multiple-Choice Questions

Physics C—Mechanics Practice Exam—Free-Response Questions

Time: 45 minutes. You may refer to the Constants sheet and Equations sheet in the Appendixes. You may also use a calculator on this portion of the exam.

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.

(b) Determine the tension in the string.

Now assume that the 30° incline is rough, so that the coefficient of friction between the block and the plane is 0.10. The 60° incline is still frictionless.

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

(d) Determine the tension in the string.

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—Free-Response Questions

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

Physics C—Electricity and Magnetism Practice Exam—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?

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

(A) 100 μ C

(B) 200 μ C

(C) 600 μ C

(D) 1200 μ C

(E) 1800 μ C

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?

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 r 1 , r 2 , and r 3 , 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 μ m

(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 R 1 and R 2 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 R 2 . What is the current through R 2 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—Multiple-Choice Questions

Physics C—Electricity and Magnetism Practice Exam—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 R 1 carries charge +Q . A concentric spherical metal shell, of inner radius R 2 and outer radius R 3 , 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. betweenr= 0 and r = R 1
2. betweenr= R 1 and r = R 2
3. betweenr= R 2 and r = R 3
4. betweenr= R 3 and r = o

(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 C 1 , C 2 , and C 3 .

(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 C 3 as a function of time.

(d) After the capacitors have been fully charged, the switch is opened, disconnecting C 1 and C 2 from the circuit. What happens to the voltage across and charge on C 3 ? 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—Free-Response Questions

Physics C—Mechanics Practice Exam—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 vt graph. Because acceleration increases, the slope of the vt 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, j Δt 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 = bv 2 is of the form y = mx , the equation of a line. Here F is the vertical axis, v 2 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 FN 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 m s cancel. Solving for v 2 , 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 1 / 2 mv 0 2 . The potential energy is provided entirely by the spring—gravitational potential energy requires a verticaldisplacement, which doesn”t occur here. The PE of the spring is 1 / 2 kx 0 2 .

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 Abounces. 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 FN and Ff 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 FN toward the center of the circle, mg down, and the force of friction up:

Because the person is not sliding down, mg = Ff . And because the motion of the person is circular, the normal force is a centripetal force, so FN = mv 2 /r . The force of friction by definition is μFN . Combining these equations, we have mg = μmv 2 /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 = . 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 = , 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 ) = mgL 2 /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 / 2 2 . 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 / 2 I (v 2 /[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 / 2 kx 2 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 / 2 kA 2 . 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 /r 2 ! 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—Electricity and Magnetism Practice Exam—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/r 2 , 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 = k 2QQ /(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 /R 2 .

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 +2Qcharge 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/r 2 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 o A /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 μ C/2 μ F = 300 V. This voltage is hooked to the three series capacitors, whose equivalent capacitance is 6 μ F (series capacitors add inversely, like parallel resistors). So the total charge stored now is (6 μ F)(300 V) = 1800 μ C. 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 = V 2 /Rto 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 I 2 R; 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 Ampére”s law. The magnetic field inside the donut is always along the axis of the donut, so the symmetry demands of Ampére”s law are met. If we draw an “Ampérean Loop” around point P but inside r 1 , 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/r 2 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/r 2 ; the magnetic field produced by a wire around a torus is zero outside the torus by Ampére”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 = mv 2 /r . Solve to get r = (mv )/(qB ). Just look at orders of magnitude now: r = (10−31 kg) (107 m/s)/(10−19 C)(10−5 T). 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 BAcos 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—Practice Exams—Free-Response Solutions

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.

*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. Call the mass of each identical block m .

1 pt: For the right block, Tmg sin 30 = ma .

1 pt: For the left block mg sin 60 − T = ma .

1 pt: Here the directions chosen are consistent, so that forces that accelerate the system to the left are positive. (However, you earn this point as long as directions are consistent.)

1 pt: Solve these equations simultaneously (it”s easiest just to add them together).

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

(Alternatively, you can just recognize that mg sin 60 pulls left, while mg cos 60 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)

1 pt: Just plug the acceleration back into one of the original Newton”s second law equations from part (a).

1 pt: You get T = 34 N.

(c)

For parts (c) and (d), points are awarded principally for showing the difference that friction makes in the solution. You earn credit for properly accounting for this difference, even if your overall solution is wrong, as long as you followed a similar process to parts (a) and (b).

1 pt: Following the solution for part (a), this time the right block”s equation becomes Tmg sin 30 − μFN , where μ is the coefficient of friction, given as 0.10.

1 pt: The normal force is equal to mg cos 30.

1 pt: The left block”s equation is the same as before, mg sin 60 − T = ma .

1 pt: Eliminating T and solving, we get 1.4 m/s2 . This is reasonable because we get a smaller acceleration when friction is included, as expected. [This answer point is awarded for ANY nonzero acceleration that is less than that calculated in part (a).]

(d)

1 pt: Plugging back into one of the equations in part (c), we find T = 36 N this time, or whatever tension is consistent with part (c).

1 pt: Awarded for ANY nonzero tension greater than that found in part (b).

1 pt: For proper units on at least one acceleration and one tension, and no incorrect units.

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 FN cos θ and the horizontal is FN 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, FN cos θ = mg .

1 pt: The horizontal force is a centripetal force, so FN sin θ = mv 2 /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 FN term:

sin θ /cos θ = v 2 /r sin θg .

1 pt: Plug in the speed and the sin θ terms cancel, leaving cos θ = g /4π2 rf 2 .

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 = .

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 LL cos θ . Thus, the potential energy is mgh = mg (LL 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: −(−kxdx integrates to give U = 1 / 2 kx 2 . 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)

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 4πr 2 . The charge enclosed by this surface is Q .

1 pt: So, E = Q enclosed /ε o A = Q /4π ε o r 2 .

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 o r 2 .

(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 o r , with potential equal to zero an infinite distance away.

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

(d)

1 pt: Integrate the electric field between R 1 and R 2 to get V = Q /4π ε o r + a constant of integration.

1 pt: To find that constant, we know that V (R 2 ) was found in part (c), and is 3Q /4π ε o R 3 . Thus, the constant is

1 pt: Then, potential at R 1 = Q /4π ε o R 1 + the constant of integration.

E&M 2

(a)

1 pt: The series capacitors add inversely,

so C eq for the series capacitors is 3 μ F.

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

(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 C 3 is 10 V (because there”s no voltage drop across the resistor without any current).

1 pt: By Q = CV the charge on C 3 is 20 μ C.

1 pt: Treating C 1 and C 2 in series; the equivalent capacitance is 3 μ F, the voltage is 10 V (in parallel with C 3 ).

1 pt: The charge on the equivalent capacitance of C 1 and C 2 is 30 μ C; thus the charge on C 1 = 30 μ C, and the charge on C 2 is also 30 μ C (charge on series capacitors is the same).

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

1 pt: Using Q = CV , the voltage across C 2 is 2.5 V.

(c)

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

1 pt: For a graph that asymptotically approaches 20 μ C (or whatever charge was calculated for C 3 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 C 3 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 Earth tan 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)

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