Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Internal and External Forces — The Two-Block Train

Assessments aligned to 2026 AP Physics 1 standards

Experimental Design and Analysis (LAB)  |  MID-LEVEL  |  10 points  |  25 min

▤ Scenario

Two blocks sit at rest on a long, level track whose friction is negligible. Block A has mass 2.0 kg and Block B has mass 4.0 kg. Block A is on the left and Block B is directly to its right, so that the flat vertical faces of the two blocks touch. The blocks are not fastened to each other in any way — they simply touch.

A battery-powered fan unit is clamped to the outer (left-hand) face of Block A, as shown in Figure 1. It drives air backwards and pushes horizontally on Block A with a constant force of magnitude P = 12 N directed to the right; the thrust has been measured with a force probe. The fan unit’s mass is negligible compared with that of either block. When the fan is switched on, the two blocks remain in contact and travel down the track together.

Figure 1 — The two blocks at rest on the track, with the fan unit clamped to Block A.

All motion is one-dimensional and horizontal. Take the ground as the reference frame and the +x direction to the right. Air resistance is negligible.

✎ Free Response Questions

Part A

Calculate the magnitude of the acceleration of the two-block train. Then calculate the magnitude of the force that Block A exerts on Block B, and state its direction.

Part B

A student makes the following claim:

“The push is 12 N and it drives the whole train, so when I apply Newton’s second law to Block B by itself I should still use the full 12 N as the force exerted on Block B.”

Indicate whether the student’s claim is correct or incorrect. Justify your response. Your justification should refer to the forces exerted on each block.

Then describe a change to the arrangement that would change the force exerted on the front block while leaving both the push and the acceleration of the train unchanged, and calculate the new value of that force.

Part C

The student now wants to test in the laboratory whether it is the total mass being pushed — rather than the mass of the block the fan is clamped to — that sets the acceleration of the train. The question to be answered is: how does the acceleration of the train depend on the total mass being pushed, when the push is held constant?

The student has the two blocks, the fan unit and the track described in the Scenario, together with a metre stick, a stopwatch, a balance, and a set of known masses that can be placed on top of the blocks.

Describe an experimental procedure to collect data that would answer this question. Include any steps necessary to reduce experimental uncertainty. Then describe how the data collected could be graphed and how that graph would be analysed to answer the question.

Part D

A second group works with a different pair of blocks on the same track, with the fan unit clamped to the outer face of the rear block. A light force sensor is fixed between the two blocks so that it reads the magnitude F_c of the contact force between them. The group measures the total mass of the pair on a balance and obtains M = 5.0 kg. Leaving both blocks unchanged, they run the fan at five different settings, recording the thrust P with a force probe and the contact force F_c at each setting. The results are shown in Figure 2.

Figure 2 — The push and the contact force measured at each of the five fan settings.

By the same reasoning as in Part A, for this pair of blocks the contact force F_c is directly proportional to the push P.

On the grid in Figure 3, create a graph of the contact force F_c against the push P.

•   Clearly label the vertical axis, including units as appropriate.

•   Plot the five data points from Figure 2.

Draw a straight best-fit line for the data you plotted. Then, using the best-fit line that you drew and the measured total mass M, calculate an experimental value for the mass m_front of the front block.

Figure 3 — Grid for graphing the contact force against the push.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A — Model Answer

Take the two blocks together as one system. The blocks are not fastened, but that does not matter here: the forces they exert on each other act between two parts of the chosen system, they are equal in magnitude and opposite in direction, and together they contribute nothing to the net force on it. Vertically, the weight of each block is balanced by the normal force from the track, so the only unbalanced external force on the system is the push.

M = m_A + m_B = 2.0 kg + 4.0 kg = 6.0 kg

a = F_net / M = P / M = (12 N) / (6.0 kg) = 2.0 m/s²

The train accelerates at 2.0 m/s² in the +x direction. The forces the blocks exert on each other are internal to the two-block system, so they cannot change the motion of the train as a whole.

Now isolate Block B. The fan never touches Block B, so the only horizontal force exerted on Block B is the force F_AB exerted on it by Block A. Block B has the same acceleration as the train, because the two blocks stay in contact.

F_AB = m_B·a = (4.0 kg)(2.0 m/s²) = 8.0 N

Block A exerts a force of 8.0 N on Block B, directed in the +x direction.

Figure 4 — Horizontal forces on the two-block system and on Block B alone, to one scale.

Scoring (2 points):

▸ A1 — 1 point: For obtaining a = 2.0 m/s² by applying Newton’s second law with the push as the only unbalanced external force on the combined 6.0 kg of the two blocks, or by any equivalent route. This point does not depend on any other part of the response.

Scoring Note: A1 is a content criterion, not a route criterion. Writing Newton’s second law separately for each block, with a common acceleration and forces between the blocks that are equal in magnitude and opposite in direction, and solving the two equations together satisfies A1 on the same terms.

▸ A2 — 1 point: For obtaining 8.0 N as the magnitude of the force Block A exerts on Block B and stating that it is directed in the +x direction — from Newton’s second law applied to Block B alone, F_AB = m_B·a, or by any equivalent route. A magnitude equal to m_B·a or to P − m_A·a, where a is the response’s own acceleration, also earns A2, provided the direction is stated.

⚠︎ Common error (partial credit): Obtains both magnitudes correctly but never states the direction of the force exerted on Block B — earns 1 of 2 points, A1.

⚠︎ Common error (partial credit): Obtains the acceleration correctly but multiplies it by Block A’s mass, giving 4.0 N as the force exerted on Block B — earns 1 of 2 points, A1.

⚠︎ Common error (partial credit): Divides the push by Block A’s mass alone, a = (12 N)/(2.0 kg) = 6.0 m/s², then isolates Block B correctly, F_AB = (4.0 kg)(6.0 m/s²) = 24 N in the +x direction — earns 1 of 2 points, A2. A force on Block B of 24 N, twice the 12 N push, shows that the acceleration is wrong: the push is the only external horizontal force on the two blocks together, so it must be divided by their combined 6.0 kg.

Part B — Model Answer

The claim is incorrect.

Isolate each block in turn. Horizontally, Block B is touched only by Block A, so the single horizontal force exerted on Block B is the 8.0 N force from Block A found in Part A. The 12 N push is exerted on Block A, not on Block B. Along the track, Block A is touched by two things: the fan, which pushes it forward with P = 12 N, and Block B, which by Newton’s third law pushes back on Block A with a force F_BA of the same magnitude as F_AB, in the −x direction.

F_net on A = P − F_BA = 12 N − 8.0 N = 4.0 N

a_A = (4.0 N) / (2.0 kg) = 2.0 m/s²

Each block has an acceleration of 2.0 m/s², which is what moving together requires. If a 12 N force acted on Block B, Block B would accelerate at (12 N)/(4.0 kg) = 3.0 m/s² — faster than Block A — and the blocks would separate, contradicting the fact that they stay in contact.

Figure 5 — Horizontal forces on each block; the dashed arrow is the 12 N the claim assumes.

The claim goes wrong by carrying a force from one system to another. Newton’s second law for a chosen system uses only the external forces exerted on that system. For the two-block system, the forces between the blocks are internal and cancel, so the push is the only external horizontal force and Newton’s second law gives a = P/M for the 6.0 kg system as a whole. For Block B alone, the force from Block A is external, and the push, which is exerted on Block A, is not a force on Block B at all.

The 8.0 N depends on this particular arrangement. Swap the two blocks: unclamp the fan unit and clamp it to the outer face of the 4.0 kg block, with that block now at the rear and the 2.0 kg block riding in front. The push is still 12 N and the total mass is still 6.0 kg, so the acceleration is still 2.0 m/s², but the force on the front block is now

F_front = (2.0 kg)(2.0 m/s²) = 4.0 N

At a fixed acceleration, the force on the front block is only what that block needs to share the train’s acceleration, so it depends on which mass rides in front.

Scoring (3 points):

▸ B1 — 1 point: For indicating that the claim is incorrect and supporting that verdict with the force actually exerted on Block B — that the only horizontal force exerted on Block B is the force from Block A, whose magnitude of 8.0 N is less than the 12 N push; or that a 12 N force on Block B would give it 3.0 m/s², not the acceleration it shares with Block A (2.0 m/s², or the response’s own Part A value); or any equivalent argument that the force on Block B cannot be 12 N, for example that Block B would then push back on Block A with 12 N, leaving Block A no net force. The magnitude of the force on Block B found in the response’s own Part A may be used, provided it is less than 12 N.

▸ B2 — 1 point: For the forces exerted on Block A — that Block B exerts on Block A a force equal in magnitude and opposite in direction to F_AB (Newton’s third law), so that the net force on Block A is 12 N − 8.0 N = 4.0 N, or 12 N minus the force on Block B from the response’s own Part A, provided that force is less than 12 N.

Scoring Note: Calculating the acceleration of Block A is not required for B2.

▸ B3 — 1 point: For describing a change that keeps the push and the total mass of the train unchanged but alters the mass riding in front — for example swapping the two blocks, so that the fan pushes the 4.0 kg block and the 2.0 kg block rides in front — and calculating the new force exerted on the front block: 4.0 N for the swap, or the product of the new front mass and the response’s own acceleration from Part A.

⚠︎ Common error (partial credit): Indicates that the claim is incorrect and quotes the 8.0 N force from Part A as the force on Block B, but never considers the forces exerted on Block A and describes no change to the arrangement — earns 1 of 3 points, B1.

⚠︎ Common error (partial credit): Indicates that the claim is incorrect and gives a complete argument about the forces on each block, including the 4.0 N net force on Block A, but describes the swap without calculating the new force on the front block — earns 2 of 3 points, B1 and B2.

✗ Common error (no credit): Agrees with the student on the grounds that the push is “the force on the train, so it acts on every part of it”, and concludes that no rearrangement can change the force on the front block while the push is unchanged — earns 0 of 3 points. A force is an interaction between two objects, and the fan interacts with Block A only.

Part C — Model Answer

Setup. Keep the fan clamped to Block A at one fixed setting throughout, so the push has the same value in every trial. Measure the mass of each block on the balance and add them to find the total mass M being pushed. Mark a start line on the track and a finish line a distance d further along, as far apart as the track allows, and measure d with the metre stick.

Figure 6 — One trial: known masses ride on Block B, and the fan setting never changes.

Trials. With the fan running at its setting, hold the train at rest with its front at the start line. Release it and start the stopwatch at the instant of release, then stop the stopwatch as the front of the train reaches the finish line, and record the time t. Repeat the timing at least three times for the same load and average the results, to reduce the effect of reaction time. Then place a known mass on top of Block B, add it to M, and repeat. Collect data for at least five different total masses, spanning as wide a range as the known masses allow.

Analysis. The train starts from rest and the push is constant, so its acceleration is constant and d = ½at², giving

a = 2d / t²

Calculate a for each total mass from its average time, and graph a against 1/M. If the acceleration is inversely proportional to the total mass, the points lie on a straight line through the origin, whose slope is the push. Because every load sits on Block B, the mass of Block A — the block the fan is clamped to — never changes, so a trend in the acceleration cannot be put down to it. At a constant push, the acceleration of the train is inversely proportional to the total mass being pushed.

Scoring (3 points):

▸ C1 — 1 point: For describing a procedure that changes the total mass of the train, and not the push, between trials — for example by placing known masses on the blocks — and that for each total mass measures the time the train takes to travel a measured distance from rest, or other quantities from which the acceleration can be found.

▸ C2 — 1 point: For a procedure that indicates a reasonable method of reducing experimental uncertainty — for example repeating the timing several times for each total mass and averaging, making the measured distance as long as the track allows so that reaction time is a smaller fraction of each time, or collecting data for several different total masses across a wide range.

▸ C3 — 1 point: For describing a graph of the data, or of quantities calculated from them, whose shape would show whether the acceleration is inversely proportional to the total mass, and what shape would show it — for example a against 1/M, or t² against M, with either quantity on either axis, lying on a straight line through the origin; or t² or 1/a against the added mass alone, lying on a straight line whose intercept corresponds to the 6.0 kg of the unloaded train.

Scoring Note: C1, C2 and C3 are independent, and each may be earned whether or not the others are. Placing the loads on Block B only is the most direct way to separate the total mass from the mass of the block the fan is clamped to, but it is not required for any point.

⚠︎ Common error (partial credit): Times the original 6.0 kg train several times and averages the results, and compares the measured acceleration with the 2.0 m/s² predicted from the total mass — earns 1 of 3 points, C2. That comparison can separate 2.0 m/s² from the 6.0 m/s² that Block A’s mass alone would give, but with a single load it cannot show how the acceleration depends on the total mass.

⚠︎ Common error (partial credit): Changes the load with the fan setting unchanged and repeats and averages each timing, but then stops at “compare the accelerations”, describing no graph and no test of how they depend on the total mass — earns 2 of 3 points, C1 and C2.

✗ Common error (no credit): Changes the fan setting once while keeping the load fixed, times a single run at each of the two settings over a short distance, and concludes that the acceleration grows with the push — earns 0 of 3 points. The push is not held constant and the total mass never changes, so the data can say nothing about how the acceleration depends on the mass being pushed.

Part D — Model Answer

With both blocks unchanged, the push gives the pair an acceleration a = P/M, and the front block — on which the only horizontal force is the contact force — has the same acceleration, so F_c = m_front·a. Together these give

F_c = (m_front / M)·P

so a graph of F_c against P is a straight line through the origin whose slope is m_front/M. The contact force is the only horizontal force on the front block and the push is the only external horizontal force on the pair, so the slope F_c/P equals m_front/M.

Reading two well-separated points from the best-fit line, (2.5 N, 1.5 N) and (7.5 N, 4.5 N):

slope = (4.5 N − 1.5 N) / (7.5 N − 2.5 N) = 3.0 N / 5.0 N = 0.60

Figure 7 — The contact force against the push, with the best-fit line and its slope triangle.

m_front = slope × M = (0.60)(5.0 kg) = 3.0 kg

The front block has a mass of 3.0 kg, a quantity the group never measured directly. The slope is a ratio of two forces and carries no unit.

Scoring (2 points):

▸ D1 — 1 point: For labelling the vertical axis with the contact force and its unit, on a linear scale, and plotting all five data points correctly. This point is earned from the graph alone and may be earned independently of the response in any earlier part.

Scoring Note: A point is plotted correctly if it lies within half a grid cell of its true position on the response’s own scale.

▸ D2 — 1 point: For drawing a single straight best-fit line through the plotted data and using its slope, related to the masses by slope = m_front/M, with M = 5.0 kg, to calculate a mass of the front block between 2.7 kg and 3.3 kg.

Scoring Note: The slope may be read from any two points on the drawn line, including plotted points that the line passes through.

⚠︎ Common error (partial credit): Labels, plots and fits correctly and reads the slope as 0.60, but reports 0.60 kg as the mass of the front block without relating the slope to the total mass — earns 1 of 2 points, D1.

⚠︎ Common error (partial credit): Labels and plots correctly but joins the points dot to dot instead of drawing one straight best-fit line, and takes the slope from a single neighbouring pair, for example 0.55 from the points at 2.0 N and 4.0 N, giving 2.75 kg — earns 1 of 2 points, D1.

✗ Common error (no credit): Draws no graph and divides a single data pair, 6.05 N by 10.0 N, to obtain a ratio — earns 0 of 2 points. With no plotted data and no best-fit line there is no slope to use, so neither criterion can be met.