AP Physics 1 Practice Exam: Section I & II Notes

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

General Instructions

• The section has 50 questions to be answered in 90 minutes.
• Scratch work can be done in the test booklet, but only answers on the answer sheet will be scored.
• A calculator, equation sheet, and table of information are allowed.

Questions 1-45: Single-Choice Items

• Choose the single best answer from the four choices provided.

Question 1

• Concept: Conservation of charge in a circuit.
• Scenario: A circuit with a battery and light bulb is connected.
• Answer: (C) The net charge has not changed.
• Explanation: Charge is conserved in a closed circuit. Connecting the circuit does not create or destroy net charge.

Question 2

• Concept: Current in a series circuit.
• Scenario: A circuit with resistors in series.
• Answer: (A) A
• Explanation: The current is the same at all points in a series circuit.

Question 3

• Concept: Newton's Second Law.
• Scenario: Spring scales measure net force, and a sonic motion detector measures acceleration. A graph of net force vs. acceleration is constructed.
• Answer: (D) Inertial mass
• Explanation: According to Newton's Second Law (F = ma), the slope of a net force vs. acceleration graph is the inertial mass.

Questions 4 and 5

• Scenario: A 0.5-kg cart collides with a fixed wall. The collision is recorded at 20 frames per second.

Question 4

• Concept: Kinematics and measurement.
• Answer: (A) 0.25 m/s
• Explanation: The cart moves 1 meter between frames which are taken at 20 frames per second. So, the time for one frame is 1/20 = 0.05 seconds. The velocity is approximately 1 \, \text{m} / 0.05 \, \text{s} = 20 \, \text{m/s}. This is not one of the possible answers, so calculating the average, the cart travels approximately .5 \, \text{m} in 0.05 \, \text{s}, giving a velocity of 0.5 \, \text{m} / 0.05 \, \text{s} = 10 \, \text{m/s}.

Question 5

• Concept: Change in momentum during collision.
• Answer: (A) 27 N.s
• Explanation:

Question 6

• Concept: Standing waves and wavelength measurement.
• Scenario: A 60-Hz generator is connected to a string fixed at both ends, producing a standing wave.
• Answer: (B) B to D
• Explanation: The wavelength is the distance between two consecutive points in phase. In the diagram, the distance between B and D represents half a wavelength, so measuring from B to D allows determination of the wavelength.

Question 7

• Concept: Rotational inertia.
• Scenario: Four identical lead balls connected by rigid rods in a square configuration are rotated about different axes.
• Answer: (C) Ic>IA > IB
• Explanation: Rotational inertia depends on the distribution of mass relative to the axis of rotation. I = \sum mr^2.

Question 8

• Concept: Work done by a force.
• Scenario: A cart experiences a single horizontal force.
• Answer: (A) The magnitude of the force, the cart's initial speed, and the cart's final speed.
• Explanation: The work-energy theorem states that the work done on an object is equal to its change in kinetic energy: W = \Delta KE = \frac{1}{2}mvf^2 - \frac{1}{2}mvi^2. To determine the work, we need the initial and final speeds to calculate the change in kinetic energy, as well as the magnitude of the force and the distance over which it is applied.

Question 9

• Concept: Superposition and destructive interference.
• Scenario: Superimposing a wave pulse with another to produce complete destructive interference.
• Answer: (B)
• Explanation: Complete destructive interference occurs when two waves with equal amplitude and opposite phase (i.e., an inverted pulse) are superimposed.

Question 10

• Concept: Impulse and momentum change.
• Scenario: A 3-kg cart experiences a varying net force over time, displayed on a graph.
• Answer: (A) 5 N.s
• Explanation: The change in momentum (impulse) is equal to the area under the force vs. time graph. From t=0 to t=2 s. The area is 5 N.s

Question 11

• Concept: Energy in simple harmonic motion with damping.
• Scenario: A block attached to a vertical spring is pulled down and released.
• Answer: (B) Less than the distance A below equilibrium
• Explanation: Due to energy loss from damping (e.g., air resistance), the block will not reach the same height or depth on each oscillation.

Questions 12 and 13

• Scenario: Two charged Styrofoam balls are brought near each other.

Question 12

• Concept: Signs of electric charges.
• Answer: (C) negative, negative
• Explanation: Like charges repel.

Question 13

• Concept: Coulomb's Law.
• Answer: (B) d/2
• Explanation: Coulomb's law states F = k \frac{q1 q2}{r^2}. If the force quadruples (8 \mu N is 4 times 2 \mu N), the distance must be halved since force is inversely proportional to the square of the distance.

Question 14

• Concept: Conservation of angular momentum.
• Scenario: A putty blob collides and sticks to a rotating disk.
• Answer: (B) 1 kg.m²/s
• Explanation: The angular mometum of the putty is L = r \times p = rmv = 10.1*10= 1 \text{kg m}^2/s Because the disk is initally at rest, the total angular momentum after the collision is 1 \text{kg m}^2/s

Question 15

• Concept: Energy conservation on a rough incline.
• Scenario: A block slides down a rough incline.
• Answer: (C) Less than the block's gravitational potential energy when it was released, because the gravitational potential energy was converted both to thermal energy and to kinetic energy.
• Explanation: Due to the rough surface, some of the initial potential energy is converted to thermal energy via friction, so the kinetic energy at the bottom will be less than the initial potential energy.

Question 16

• Concept: Current in a parallel circuit.
• Scenario: A circuit with resistors in parallel.
• Answer: (C) 3.0 A
• Explanation: The total resistance through both branches is 1/R = 1/12 + 1/4 = 1/3, so R = 3. Therefore, the equivalent resistance of the combination is also 3 \Omega. Because the batteries are in series, the power is 12V. So the total resistance in the circuit is 6 \Omega. Therefore, the current is I = V/R = 12/6 = 2A. Then, the voltage drop across each of the parallel branches is IR = 23 = 6V Then, applying ohm's law to the 4 \Omega resistor, 6V = I * 4 \Omega, so I = 1.5 A

Question 17

• Concept: Sound wave properties and oscilloscope traces.
• Scenario: Two sounds are displayed on an oscilloscope.
• Answer: (D) Sound 2 is louder, and sound 1 is higher pitched.
• Explanation: Amplitude represents loudness, and frequency represents pitch. Sound 2 has a larger amplitude (louder), and Sound 1 has a higher frequency (higher pitch).

Question 18

• Concept: Gravitational field strength.
• Scenario: Comparing gravitational field on Mars and Earth.
• Answer: (B) 4 N/kg
• Explanation: The gravitational field is g = \frac{GM}{r^2}. If the radius is halved (\frac{1}{2}) and the mass is one-tenth (\frac{1}{10}), then g{Mars} = g{Earth} * \frac{1/10}{(1/2)^2} = g{Earth} * \frac{4}{10}. Thus, g{Mars} = 10 * \frac{4}{10} = 4 N/kg

Question 19

• Concept: Conservation of momentum in collisions.
• Scenario: Elastic collision of marbles under a canopy.
• Answer: (B) Before collision, the only marble momentum was directed to the right. After the collision, the combined momentum of the two visible marbles has a downward component; another marble must have an upward momentum component to conserve momentum.
• Explanation: Momentum must be conserved in both the x and y directions. If the visible marbles have a downward momentum component after the collision, a third marble must have an upward component to balance it.

Question 20

• Concept: Momentum conservation in cart collisions.
• Scenario: Two carts collide on a track; momentum after collision appears greater than before.
• Answer: (A) The track might have been slanted such that the carts were moving downhill.
• Explanation: External forces, such as a slanted track providing a component of gravity in the direction of motion, would cause the momentum to increase, violating the conservation principle.

Question 21

• Concept: Energy and rotational inertia.
• Scenario: Three wagons with different wheel styles are accelerated to the same speed.
• Answer: (C) Wagon C
• Explanation: The wagon will require the greatest amount of energy that has the highest rotational inertia. Wagon A: I = 1/2 * 0.5 * .1^2 = .0025 \text{kg m}^2. Wagon B: I = 1/2 * 0.2 * .2^2 = .004 \text{kg m}^2. Wagon C: I = 0.2 * .1^2 = .002 \text{kg m}^2 However, Wagon C also has translational energy as. All the energy in their wheels is rotational, and rotational inertia depends on the mass and radius distribution. Since Wagon C uses a hollow hoop, this will use significantly more energy.

Question 22

• Concept: Newton's Third Law (action-reaction pairs).
• Scenario: A swimmer propels forward through water.
• Answer: (C) The force of the swimmer's arms on the surrounding water
• Explanation: The swimmer exerts a force on the surrounding water (action), and the water exerts an equal and opposite force back on the swimmer (reaction), propelling them forward.

Question 23

• Concept: Work-energy theorem.
• Scenario: A wagon is pulled by a force at an angle.
• Answer: (A) 300 J
• Explanation: The work done by the force is W = Fd \cos(\theta) = 30 * 20 * \cos(60) = 30 * 20 * (1/2) = 300 J. The increase in kinetic energy is equal to the amount of work done.

Question 24

• Concept: Inelastic collisions.
• Scenario: A moving cart collides and sticks to a stationary cart.
• Answer: (A) Velocity
• Explanation: Momentum is conserved in a collision. Velocity is the same. Momentum is equal to mass * velocity. Kinetic energy is not conserved in an inelastic collision where the carts stick together. Mass is fixed.

Questions 25 and 26

• Scenario: Four identical resistors connected to a battery in a specific configuration.

Question 25

• Concept: Current in series and parallel circuits.
• Answer: (D) I₁ > I2 = I3 > I₄
• Explanation: Resistors 2 and 3 are in series, so they have the same current. Resistors 2, 3, and 4 are a parallel branch with Resistor 1. Hence total current through Resistor 1 is larger.

Question 26

• Concept: Electric potential along a circuit loop.
• Answer: (B)
• Explanation: Electric potential drops across each resistor. Since \DeltaV must be the same across each branch, the branch with 1 resistor will drop more sharply.

Question 27

• Concept: Uncertainty in experimental measurements.
• Scenario: Stretching a spring with a force probe; the work done is reported with uncertainty.
• Answer: (C) 3,000 ± 300 J
• Explanation: The uncertainty should be about 10% which is roughly 300.

Question 28

• Concept: Standing waves and resonance.
• Scenario: A string attached to a vibrating speaker does not produce standing waves at 60 Hz.
• Answer: (A) The string length is not a multiple of half the wavelength of the wave.
• Explanation: Standing waves occur when the string length is an integer multiple of half-wavelengths. If this condition is not met, resonance will not occur.

Question 29

• Concept: Torque and equilibrium.
• Scenario: A platform rotates on a fulcrum; a force probe supports the platform with a variable weight.
• Answer: (A) Plot the reading in the force probe times x on the vertical axis; plot the gravitational field times d on the horizontal axis. The mass is the slope of the line.
• Explanation:

Question 30

• Concept: Conservation laws in collisions.
• Scenario: Comparing two types of collisions.
• Answer: (B) Collision A: each cart experiences the same force, time of collision, and change in linear momentum. Collision B: the ball and the rod each experience the same torque, time of collision, and change in angular momentum.
• Explanation: In Collision A, momentum is conserved and Newton's Third Law applies. In Collision B, angular momentum is conserved, and the rotational analogue of Newton's Third Law applies.

Question 31

• Concept: Net force and momentum change.
• Scenario: A lab cart slows down while moving east.
• Answer: (A) Yes, since we know the cart is slowing down, its momentum change is opposite the direction of movement, and the net force is in the direction of momentum change.
• Explanation: A decrease in speed indicates a net force opposing the motion. The net force is in the direction of the momentum change.

Question 32

• Concept: Torque.
• Scenario: Forces applied to a rigid rod pivoted at one end.
• Answer: (C) T₂ > T₁ > T₃
• Explanation: Torque is calculated as T = rF\sin(\theta). T1: T = Fd1. T2: T = F2d * \sin(90) = 2Fd. T3: T = F*2d * sin(\theta), where theta is less than 90.

Question 33

• Concept: Simple harmonic motion.
• Scenario: Acceleration vs. position graph for a block-spring system.
• Answer: (D)
• Explanation: In SHM, acceleration is proportional to the displacement from equilibrium but in the opposite direction: a = -(\frac{k}{m})x.

Question 34

• Concept: Wave properties.
• Scenario: Three transverse waves moving in the same material.
• Answer: (B) B>C>A
• Explanation: The visible amplitude is the radius. B has highest amplitude, C is next, and A is lowest.

Question 35

• Concept: Gravitational force.
• Scenario: Calculating the gravitational force between Earth and Moon.
• Answer: (B) 1020 N
• Explanation: Calculate F = G(Mm/r^2). F = 6.7 * 10^{-11} * (5.97 * 10^{24}) * (7.35 * 10^{22}) / (3.84 * 10^8)^2 \approx 2 * 10^{20}

Question 36

• Concept: Energy transfer and dissipation.
• Scenario: A toy car with a rubber band rolling up a ramp with inconsistent distances.
• Answer: (A) Depending on how the rubber band is initially wound, more or less potential energy can be transferred from the rubber band to the kinetic energy of the car's motion.
• Explanation: The internal state of the rubber band winding affects the conversion of potential energy to kinetic energy, influencing the car's motion.

Question 37

• Concept: Action-reaction pairs of forces.
• Scenario: A man stands on a scale in an accelerating elevator.
• Answer: (D) The force of the man on the Earth
• Explanation: Weight is the force of gravity of the Earth on the man. The reation force is the force of the man on the Earth.

Question 38

• Concept: Doppler effect.
• Scenario: Wave fronts from a speaker moving on a cart.
• Answer: (A)
• Explanation: The sound waves should be evenly spaced since there is no doppler effect.

Question 39

• Concept: Fundamental forces.
• Scenario: A table supporting a wooden block.
• Answer: (C) The electric force, because the outer electrons in the top atomic layer of the table repel the outer electrons in the bottom atomic layer of the wood.
• Explanation: Electromagnetic forces between atoms and molecules in the table and block prevent them from passing through each other.

Question 40

• Concept: Rotational motion and torque.
• Scenario: Bringing a spinning sphere to rest with friction.
• Answer: (B) 2 s
• Explanation: \tau = I\alpha. \alpha = \tau/I = Fr/I = 2*.3/.06 = 10 \text{rad}/s^2 So t = \Delta w/\alpha = (20-0)/10 = 2 s

Question 41

• Concept: Free-body diagrams.
• Scenario: Forces acting on a block sliding to the right while slowing down.
• Answer: (B)
• Explanation: The net force must oppose the motion if the object is slowing. Which means the horizontal for (Friction) must be on the left.

Question 42

• Concept: Standing waves and tension.
• Scenario: Changing standing waves by adjusting tension.
• Answer: (B) Greater, because the tension in the string varies directly with the wave speed, which varies inversely with the wavelength.
• Explanation: Wave speed on a string is v = \sqrt{\frac{T}{\mu}}, where T is tension and \mu is linear mass density. A greater tension will lead to a greater wavelength.

Question 43

• Concept: Coulomb's Law.
• Scenario: Changing the mass of charged balls.
• Answer: (A) 50 μN
• Explanation: Electric force depends only on charge and distance, not mass.

Question 44

• Concept: Energy in spring-mass systems.
• Scenario: Kinetic energy of a block attached to a spring.
• Answer: (A) \frac{1}{2}kA^2 - \frac{1}{2}kx^2
• Explanation: The total energy is 1/2 k A^2, where A represents the amplitude. Then KE can be found by KE = 1/2 k A^2 - 1/2 k x^2

Question 45

• Concept: Angular momentum conservation.
• Scenario: Changing angular momentum on a frictionless rotating platform.
• Answer: (A) The man catches a baseball thrown to him by a friend.
• Explanation: An external torque (provided by the baseball) is requird to change angular momentum.

Questions 46-50: Multiple-Correct Items

• Identify exactly two of the four answer choices as correct.

Question 46

• Concept: Similarities and differences between electric and gravitational forces.
• Answer: (A) and (C)
• (A): Both depend inversly on the square of the distance d.
• (C): For any measureable m and q, the electric force is many orders…

Question 47

• Concept: Voltmeter placement to measure voltage across a resistor.
• Answer: (A) and (D)
• Explanation:

Question 48

• Concept: Apparent weight in an elevator.
• Scenario: A student on scale reads 800 N (actual weight is 500N)
• Answer: (A) and (C)
• (A): The scale reads a higher value than force the force applied by the elevator on the student is grater due to their acceleration.
• (B): Elevator moving downward and slowing down
  (Elevator moving upward and speeding up

Question 49

• Concept: Resonance in a closed pipe.
• Scenario: Finding frequencies in a 1m long pipe closed at one end with Vsound 300 m/s
• Answer: (A) and (C)
• (A): The fundamental frequency F = 300/4 = 75
• (C): Only odd harmonics will resonate.

Question 50

• Concept: Forces causing forward rotation of a wheel.
• Scenario: Applied forces on an axle connecting two wheels
• Answer:
• Explanation:

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

General Instructions

• The section has five questions to be answered in 90 minutes.
• Budget approximately 20 to 25 minutes each for the first two longer questions.
• The next three shorter questions should take about 12 to 17 minutes each.
• Explain all solutions thoroughly, as partial credit is available.

Question 1: Experimental Analysis (12 points)

• Experiment 1: Two carts collide and stick together.
  • Cart A: mass 500 g
  • Cart B: unknown mass, initially at rest
• Experiment 2: Two carts colliding and bouncing off from each other.

(a) Identifying the Collision on the Graph

• The collision is represented by a sudden change in the velocity of Cart A.

(i) Speed of Cart A before the collision

• Estimate the velocity from the graph before the collision. (Example: 0.5 m/s)

(ii) Speed of Cart A after the collision

• Estimate the velocity from the graph after the collision. (Example: 0.2 m/s)

(b) Constructing a Graph to Calculate the Mass of Cart B

(i) Axes for the graph

• x-axis: Speed of Cart A before collision (v_{A,before})
• y-axis: Speed of Cart A after collision (v_{A,after})

(ii) Using the slope to calculate the mass of Cart B

• Conservation of momentum: mA v{A,before} = (mA + mB) v_{A,after}
• Rearrange the equation: v{A,after} = (\frac{mA}{mA + mB}) v_{A,before}
• The slope of the graph is (\frac{mA}{mA + m_B}).
• Calculate mB using the slope: mB = m_A (\frac{1}{slope} - 1)

(c) Experiment 2: Elastic Collision

(i) Procedure to measure the speed of Cart B after the collision

• Use a photogate to measure the time it takes for the Cart B to pass through after the collision.

(ii) Determination whether the collision was elastic

• Calculate KE before collision: calculate KE = 1/2 mv_A^2 for each cart.
• Collision is elastic if KE{before} = KE{after}

Question 2: Qualitative-Quantitative Translation (12 points)

• A uniform meterstick (weight = 1.5 N) is supported by two spring scales.
  • Left scale: 20 cm from the left-hand edge.
  • Right scale: 30 cm from the right-hand edge.

(a) Which scale indicates a greater force reading?

• Consider the torques about the center of the meterstick. The right scale will indicate a greater force reading because it is closer to the center of the meterstick, and therefore must exert a stronger force.

(b) Calculate the reading in each scale.

• Let FL be the reading on the left scale and FR be the reading on the right scale.
• Equilibrium conditions:
  • \sum Fy = 0 \implies FL + F_R - 1.5 = 0
  • \sum \tau = 0 (about the center of the stick)
  • Larm = 0.5 - 0.2 = 0.3 then FL * 0.3 = F_R * 0.2
  • FR = 1.8 m and FL = .8 m
  • subbing in to FL + FR - 1.5 = 0
  • 0.8 + 1.8 m - 1.5 = 0
  • m = .7/2.6 = 0.27
• Solve the system of equations to find FL = 0.6 N and FR= 0.9 N.

(c) Moving the right-hand scale closer to the center of the meterstick

(i) Reading in the left-hand scale

• Increase: As the right-hand scale moves closer to the center, it bears less weight, causing the left-hand scale to bear more weight.

(ii) Reading in the right-hand scale

• Decrease: For the same reason as above.

(d) Hanging a 0.2 N weight

• To increase the right-hand scale reading by the largest amount: hang the weight at the left most side of the meter stick.

Question 3: Short Answer (7 points)

• Space Probe A: geostationary orbit above Jupiter's equator.
• Space Probe B: sitting on Jupiter's surface.

(a) Which probe has the greater speed?

• Space Probe A, orbiting in geostationary orbit, must have a higher orbital speed to maintain its orbit than Space Probe B, which is stationary on the surface relative to Jupiter.

(b) Which probe has the greater acceleration toward the center of Jupiter?

• Space probe B.

(c) Gravitational Forces Ranking

(i) Ranking

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

(ii) Justification

• Forces 1 (Jupiter on A) and 2 (Jupiter on B) are greatest because Jupiter's mass is much greater than the probes'.
• Forces 3 and 4 are equal to 1 and 2, respectively, due to Newton's third law.
• Forces 5 and 6 involve the product of two small masses (the probes) and are therefore much smaller.

Question 4: Short Answer (7 points)

• Two blocks (A and B) attached by a spring move on a frictionless surface.
• A has mass m and B has mass 2m.

Conservation Laws

• System A (Block A only):
  • Kinetic energy: Not conserved because the spring exerts a force on Block A, changing
  • Total mechanical energy: Not conserved, spring force does work.
  • Linear momentum: Not conserved, spring exerts an external force on Block A.
• System B (Both blocks and spring):
  • Kinetic energy: Not conserved, because the spring is compressed, and the blocks convert this potential to kinetic energy.
  • Total mechanical energy: Conserved. The system is isolates and spring is not doing any external work.
  • Linear momentum: Conserved. Since there are no external forces acting on the system, total linear momentum will not change.

Question 5: Short Answer (7 points)

• Three identical light bulbs connected in a circuit.

(a) Changes in current

• Bulb 1: Current decreases because the additional parallel path reduces the total resistance of the circuit, thus reducing the overall current through Bulb 1.
• Bulbs 2 and 3: When the switch is closed, they now have current; current increases from zero to some positive value

(b) Changes in brightness

• Bulb 1: Becomes dimmer because brightness depends on current. Since current decreases through the bulb, brightness decreases.
• Bulbs 2 and 3: Become lit because brightness depends on current. Since current increases from zero, brightness increases.

(c) Does the battery increase, decrease or remain the same

• Increase: the power output increases because the total resistance in a circuit decreases as you add branches

(d) Why does the law not violate conservation of charge

• There is no build up of charge at any given node. The same amount of charge flows in and flows out. So it follows the law of conservation.