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AP® Physics 1: Algebra-Based Free-Response Questions Past Paper 2025 Solution

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AP® Physics 1: Algebra-Based 2025 FRQ Set 1: Detailed Solutions

A Note from Your AP Physics 1 Educator: Welcome! This guide will walk you through the 2025 Set 1 Free-Response Questions for AP Physics 1. Success on this exam requires more than just plugging numbers into formulas; it demands a deep conceptual understanding of physics principles, the ability to design experiments, analyze data, and construct clear, evidence-based arguments. For each question, we'll break down the prompt, strategize an approach, and review a model answer that exemplifies the kind of thinking and justification required for a top score.

Question 1: Momentum and Energy in Collisions

Collision of a Block and Cart

A. i. On the axes shown in Figure 2, sketch a graph of the magnitude px of the x-component of the momentum of the block-cart system...

Thinking Process: The system is the block and the cart. The principle to apply is conservation of momentum. We must check for external horizontal forces on the system. The problem describes a horizontal surface and a block being dropped vertically. Gravity and the normal force from the surface are vertical. The forces during the collision between block and cart are internal. Therefore, there are no net external horizontal forces on the system. This means the total horizontal momentum of the block-cart system is conserved, i.e., constant, throughout the entire process. The graph of a constant positive value versus time is a horizontal line.

The graph is a single horizontal line at a constant positive value of px, starting from t=0 and continuing for all t > t2.

A graph of momentum vs. time, showing a horizontal line at a positive momentum value across the entire time axis.

A. ii. Derive an expression for the speed vf... in terms of mc, vc, and physical constants...

Thinking Process: The collision is perfectly inelastic. Since there are no external horizontal forces, linear momentum in the x-direction is conserved. I will set the total momentum before the collision equal to the total momentum after the collision and solve for the final velocity, vf.

By the principle of conservation of linear momentum for an isolated system in the horizontal direction:

Σpinitial = Σpfinal

The initial momentum is only from the cart, as the block is released from rest and has no horizontal velocity.

pinitial = mcvc + mb(0) = mcvc

The final momentum is the total mass of the system moving with a common final velocity vf.

pfinal = (mc + mb)vf = (mc + (1/5)mc)vf = (6/5)mcvf

Setting them equal:

mcvc = (6/5)mcvf

The mass of the cart, mc, cancels out. Solving for vf:

vf = (5/6)vc

A. iii. Derive an expression for the change in the kinetic energy ΔK...

Thinking Process: The change in kinetic energy is ΔK = Kfinal - Kinitial. I need to write expressions for the initial and final kinetic energies of the system. The initial kinetic energy is just that of the cart. The final kinetic energy is that of the combined block-cart mass. I'll use the expression for vf derived in the previous part.

The change in kinetic energy of the system is given by ΔK = Kf - Ki.

The initial kinetic energy (at t=0) is due only to the cart's motion:

Ki = (1/2)mcvc2

The final kinetic energy (at t > t2) is of the combined mass moving at speed vf. Using vf = (5/6)vc from part (ii):

Kf = (1/2)(mc + mb)vf2 = (1/2)(6/5)mc * ((5/6)vc)2

Kf = (1/2)(6/5)mc * (25/36)vc2 = (1/2)mcvc2 * (6/5 * 25/36)

Kf = (1/2)mcvc2 * (5/6) = (5/6)Ki

Now, calculate the change, ΔK:

ΔK = Kf - Ki = (5/6)Ki - Ki = -(1/6)Ki

ΔK = -(1/6)(1/2)mcvc2 = -(1/12)mcvc2

B. Indicate whether the x-component of the momentum of the new block-cart system... remains constant during Δt.

Thinking Process: The system is the block and the cart. The force of friction between the block and the cart during sliding is an *internal* force to this system. By Newton's third law, the force on the block is equal and opposite to the force on the cart. These internal forces cannot change the total momentum of the system. We assume there are no external horizontal forces (like friction from the ground). Therefore, the system's total horizontal momentum must be conserved.

     Increases

     Decreases

 X  Remains constant

Justify your response:

The frictional force between the new block and the cart is an internal force to the block-cart system. According to the law of conservation of momentum, the total momentum of a system cannot be changed by internal forces. Since there are no net external horizontal forces acting on the system (gravity and normal force are vertical, and surface friction is assumed to be negligible), the total x-component of the momentum of the block-cart system must remain constant during the time interval Δt.

Question 2: Energy Conservation

Block on a Ramp with a Spring

A. Draw shaded bars that represent K, Ug, and Us to complete the energy bar charts in Figure 2 and Figure 3...

Thinking Process: The system is block-spring-Earth, and friction is negligible, so total mechanical energy E is conserved. From the given chart at x=10D (Figure 4), the total energy is E = K + Ug + Us = 8E0 + 2E0 + 4E0 = 14E0. This value is constant.
  • At x=0 (Figure 2): The block is at rest, so K=0. The spring is uncompressed, so Us=0. All energy must be gravitational potential, so Ug = E = 14E0.
  • At x=6D (Figure 3): The spring is still uncompressed (contact at x=8D), so Us=0. The gravitational potential energy Ug is proportional to the vertical height. The total vertical drop corresponds to a distance of 12D along the ramp. At x=6D, the block has descended half the total distance, so its height is halfway between the start and the zero point. Thus, Ug(6D) = (12D-6D)/(12D) * Ug(0) = (6/12) * 14E0 = 7E0. By conservation of energy, K = E - Ug - Us = 14E0 - 7E0 - 0 = 7E0.

For Figure 2 (x = 0): A bar for K at 0. A bar for Ug up to 14E0. A bar for Us at 0.

For Figure 3 (x = 6D): A bar for K up to 7E0. A bar for Ug up to 7E0. A bar for Us at 0.

Completed energy bar charts for x=0 and x=6D as described in the solution.

B. Starting with conservation of energy, derive an equation for the spring constant k...

Thinking Process: I will apply conservation of mechanical energy between the initial point (x=0) and the final point (x=12D). At x=0, all the energy is gravitational potential. At x=12D, all the energy is spring potential. The key is to correctly express the initial height and the final spring compression in terms of the given variables.

We apply the principle of conservation of mechanical energy for the block-spring-Earth system between x=0 and x=12D.

Ei = Ef => Ki + Ug,i + Us,i = Kf + Ug,f + Us,f

The initial state is at x=0, and the final state is at x=12D.

At x=0: Ki = 0 (from rest). Us,i = 0 (spring uncompressed). The vertical height above the zero-point (x=12D) is h = (12D)sin(θ). So, Ug,i = Mg(12D)sin(θ).

At x=12D: Kf = 0 (momentarily at rest). Ug,f = 0 (defined as zero). The spring compression is Δx = (12D - 8D) = 4D. So, Us,f = (1/2)k(Δx)2 = (1/2)k(4D)2.

Setting the initial and final energies equal:

Mg(12D)sin(θ) = (1/2)k(16D2) = 8kD2

Now, we solve for the spring constant k:

k = (12MgD sin(θ)) / (8D2)

k = (3Mg sin(θ)) / (2D)

C. On the axes shown in Figure 6, do the following...

Thinking Process:
  1. i. Sketch Total Mechanical Energy E: Energy is conserved, so E is constant. From Part A, we know E = 14E0. I'll draw a horizontal line at the 14E0 level across the graph from 8D to 12D.
  2. ii. Sketch Gravitational Potential Energy Ug: Ug = Mgh. Height `h` decreases linearly as the block slides down the ramp (h = (12D - x)sinθ). Therefore, Ug is a decreasing linear function of x. We know Ug(12D) = 0 (from the problem definition) and we can find Ug(8D). From the bar charts, Ug(10D)=2E0. The change in Ug is proportional to the change in vertical height. The drop from 10D to 12D corresponds to 2E0. So the drop from 8D to 10D (same distance) also corresponds to 2E0. This means Ug(8D) must be 4E0. I will draw a straight line connecting the point (8D, 4E0) to the point (12D, 0).

The completed graph is shown below.
(i) The total mechanical energy E is a constant horizontal line at E = 14E0.
(ii) The gravitational potential energy Ug is a straight, downward-sloping line starting at (8D, 4E0) and ending at (12D, 0).

Energy graph showing a constant total energy E at 14E0, a given parabolic spring potential energy Us, and a derived linear gravitational potential energy Ug.

D. Indicate whether the speed v9D... is greater than, less than, or equal to the speed v8D... Justify...

Thinking Process: Speed is related to kinetic energy (K = 1/2Mv2). I will compare the kinetic energy at x=8D and x=9D using the energy conservation equation K = E - Ug - Us and the graphs from part C.
  • At x=8D: K8D = E - Ug(8D) - Us(8D) = 14E0 - 4E0 - 0 = 10E0.
  • At x=9D: From the Ug line I drew, Ug(9D) is halfway between Ug(8D)=4E0 and Ug(10D)=2E0, so Ug(9D)=3E0. From the given Us curve, Us(9D) is approximately 1E0. Therefore, K9D = 14E0 - 3E0 - 1E0 = 10E0.
Since K8D = K9D, their speeds must be equal.

     v9D > v8D

     v9D < v8D

 X  v9D = v8D

Justify your response:

The speed of the block is directly related to its kinetic energy K. Using the energy graphs from part C and the principle of energy conservation (K = E - Ug - Us), we can compare the kinetic energy at both points.

  • At x = 8D, the kinetic energy is K8D = E - Ug(8D) - Us(8D) = 14E0 - 4E0 - 0 = 10E0.
  • At x = 9D, the gravitational potential energy is Ug(9D) = 3E0 and the spring potential energy is Us(9D) = 1E0. Thus, the kinetic energy is K9D = 14E0 - 3E0 - 1E0 = 10E0.

Because the kinetic energy of the block is the same at both x = 8D and x = 9D, the speed of the block must also be the same at these two positions.

Question 3: Experimental Design

Balancing Systems and Rotational Motion

A. Describe an experimental procedure to collect data that would allow the students to determine m0.

Thinking Process: The goal is to find m0 using a linear graph. The principle is rotational equilibrium (Στ = 0). The setup involves balancing torques. I need to identify what I can vary (independent variable) and what I will measure (dependent variable). I can vary the position of the unknown mass (rm) and measure the force from the spring scale (FT) required to balance it.

1. Place the meterstick on the pivot at its center, the 50 cm mark, so the meterstick's own weight exerts no torque.
2. Attach the spring scale to a fixed hole on one side, for example, the 10 cm mark. This creates a constant lever arm for the scale's force.
3. Hang the unknown mass m0 from the hook in the hole at the 60 cm mark.
4. Pull vertically downward on the spring scale until the meterstick is perfectly horizontal. Use a level to ensure it is horizontal.
5. Record the force FT shown on the spring scale and the position of the mass m0.
6. Repeat steps 3-5 for at least four other positions of the mass m0 (e.g., at the 70 cm, 80 cm, and 90 cm marks), recording the corresponding force from the spring scale for each position.
7. To reduce uncertainty, for each position of the mass, take multiple readings of the force and calculate the average. Read the scale at eye level to avoid parallax error.

B. Describe how the data collected in part A could be graphed and how that graph would be analyzed to determine m0.

Thinking Process: From the procedure in A, I have pairs of data for the scale force (FT) and the position of the mass. I need to relate these using torques. Στ = FTrs - (m0g)rm = 0, where rs and rm are distances from the pivot. This rearranges to FT = (m0g / rs) * rm. This is a linear equation of the form y = (slope) * x. I will plot FT vs. rm. The slope will be m0g / rs.

The data could be analyzed by first calculating the lever arm, rm, for the unknown mass for each trial (rm = position - 50 cm). Then, create a graph with the measured force from the spring scale, FT, on the y-axis and the calculated lever arm, rm, on the x-axis. The plotted data should form a straight line that passes through the origin.

Draw a best-fit line through the data points and calculate its slope. According to the torque equation, FT * rs = m0g * rm, which can be rearranged to FT = (m0g / rs) * rm. Therefore, the slope of the graph is equal to m0g / rs. The unknown mass m0 can then be calculated by rearranging this relationship: m0 = (slope * rs) / g, where rs is the fixed lever arm of the spring scale and g is the acceleration due to gravity.

C. i. Indicate what measured or calculated quantity could be plotted on the vertical axis to yield a linear graph...

Thinking Process: The given equation is FT = 5Mg / (6 sinθ). The horizontal axis is 1/sinθ. I need to rearrange the equation to match y = (slope) * x. Let x = 1/sinθ. Then the equation becomes FT = (5Mg/6) * (1/sinθ). This fits the form y = m*x where y is FT and the slope is 5Mg/6. So the vertical axis should be FT.

Vertical axis:  FT     Horizontal axis: 1/sinθ

C. ii & iii. On the blank grid provided, create a graph... and draw a straight best-fit line.

Thinking Process: I need to first calculate the values for the horizontal axis, 1/sinθ, using the data from Table 1. Then I'll plot the (x, y) pairs, which are (1/sinθ, FT). Finally, I'll draw a best-fit line.
  • θ=22, 1/sin(22) ≈ 2.67. Point: (2.67, 21)
  • θ=31, 1/sin(31) ≈ 1.94. Point: (1.94, 17)
  • θ=36, 1/sin(36) ≈ 1.70. Point: (1.70, 13)
  • θ=45, 1/sin(45) ≈ 1.41. Point: (1.41, 12)
  • θ=80, 1/sin(80) ≈ 1.01. Point: (1.01, 8)
I will plot these points on the provided grid, label the axes with quantities and units, and then draw a single straight line that best represents the trend.

(The student would create a data table and plot the points on the grid.)

Calculated Data Table (Table 2):

1/sin(θ)FT (N)
2.6721
1.9417
1.7013
1.4112
1.018

Graph:

A graph of FT vs 1/sin(theta) with data points plotted and a best-fit line drawn through them.

The graph has FT (N) on the vertical axis and 1/sin(θ) on the horizontal axis. The points are plotted according to the table, and a straight best-fit line is drawn through the points, passing through approximately (1.2, 10) and (2.5, 19).

D. Using the best-fit line..., calculate an experimental value for the mass M of the meterstick.

Thinking Process: I need to find the slope of my best-fit line, then use the relationship `slope = 5Mg/6` to solve for M. I must use points from my line, not the data points themselves. Let's pick two points far apart on the line I drew, for example, (1.2, 10 N) and (2.5, 19 N).

First, calculate the slope of the best-fit line using two points from the line, such as (1.2, 10 N) and (2.5, 19 N).

slope = Δy / Δx = (19 N - 10 N) / (2.5 - 1.2) = 9 N / 1.3 ≈ 6.92 N

From the linearized equation, we know that the slope is equal to 5Mg/6.

slope = 5Mg / 6

Now, we solve for M:

M = (6 * slope) / (5 * g)

M = (6 * 6.92 N) / (5 * 9.8 m/s2) = 41.52 / 49

M ≈ 0.85 kg

Question 4: Forces and Dynamics

Buoyancy and Newton's Second Law

A. Indicate whether a1 is greater than, less than, or equal to a2... Justify your answer...

Thinking Process:
  1. Identify forces: For the submerged block, there are two vertical forces: the downward force of gravity (Fg) and the upward buoyant force (FB).
  2. Compare forces between scenarios: The block's mass and volume are the same, so the gravitational force Fg is the same in both scenarios. The buoyant force is FB = ρfluidVg. We are given that saltwater is denser than freshwater (ρ2 > ρ1). Therefore, the buoyant force in saltwater (FB2) is greater than in freshwater (FB1).
  3. Compare net force: The net upward force is Fnet = FB - Fg. Since FB2 > FB1 and Fg is constant, the net upward force is greater in saltwater (Fnet2 > Fnet1).
  4. Apply Newton's Second Law: a = Fnet/m. Since the mass `m` is the same and Fnet2 > Fnet1, it follows that the acceleration is greater in saltwater (a2 > a1).

     a1 > a2

 X  a1 < a2

     a1 = a2

Justify your answer:

In both scenarios, two forces act on the block: the downward force of gravity and the upward buoyant force. The force of gravity is the same in both cases because the block's mass does not change. The buoyant force is equal to the weight of the fluid displaced, which depends on the fluid's density. Since saltwater (ρ2) is denser than freshwater (ρ1), the upward buoyant force on the block is greater in saltwater than in freshwater.

The net force on the block is the difference between the upward buoyant force and the downward gravitational force. Because the upward buoyant force is greater in saltwater, the net upward force on the block is greater in saltwater. According to Newton's second law, for a constant mass, a greater net force results in a greater acceleration. Therefore, the upward acceleration of the block will be greater in saltwater (a2) than in freshwater (a1).

B. Starting with Newton’s second law, derive an expression for the initial upward acceleration a...

Thinking Process: I will start with ΣF = ma. I will define the upward direction as positive. The forces are the buoyant force (FB) acting up and the gravitational force (Fg) acting down. I will substitute the expressions for these forces and solve for `a`.

We begin with Newton's second law, with the upward direction defined as positive.

ΣF = ma

The forces acting on the block are the upward buoyant force, FB, and the downward gravitational force, Fg.

FB - Fg = ma

The buoyant force is given by FB = ρVg, where ρ is the fluid density and V is the volume of the block. The gravitational force is Fg = mg.

Substituting these into the equation:

ρVg - mg = ma

Now, we solve for the acceleration, a:

a = (ρVg - mg) / m

a = g(ρV/m - 1)

C. Indicate whether the expression for the acceleration a you derived in part B is or is not consistent with the claim made in part A. Briefly justify...

Thinking Process: The claim in Part A was a1 < a2. My derived expression is a = g(ρV/m - 1). In this expression, g, V, and m are constants for the block. The acceleration `a` is therefore a linear function of the fluid density `ρ`. Since acceleration increases as fluid density increases, and we know ρ2 > ρ1, my derived expression predicts that a2 > a1. This is consistent.

The expression for acceleration derived in part B is consistent with the claim made in part A.

Justification:

The derived expression, a = g(ρV/m - 1), shows that the acceleration `a` is directly proportional to the density of the fluid, `ρ`, since g, V, and m are all positive constants for the two scenarios. Given that the density of saltwater is greater than the density of freshwater (ρ2 > ρ1), the derived equation mathematically requires that the acceleration in saltwater (a2) must be greater than the acceleration in freshwater (a1). This matches the claim made in part A.

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