FRQ: Spring Force and Stretch — Weighing a Rock Without a Balance

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

Students are asked to determine the mass m_R of a small rock sample, but the laboratory balance is out of order. They have a spring whose spring constant is not known, with a light pointer fixed to its hook; a stand from which the spring can be hung vertically, with a clamp that can hold a meterstick upright beside it; a meterstick; eight slotted masses of 100 g each that can be hung from the spring’s hook singly or stacked on a hanger of negligible mass; and a string bag of negligible mass in which the rock can be hung from the hook, as shown in Figure 1.

Figure 1 — The equipment for weighing the rock; no balance is available.

The rock’s mass is known to lie between 200 g and 600 g, and the spring is not damaged by any hanging load up to 1.0 kg. The students are asked to take measurements that will allow them to create a linear graph that could be used to determine m_R. Use the laboratory as the reference frame, and use g = 10 N/kg.

✎ Free Response Questions

Part A

Describe an experimental procedure to collect data that would allow the students to determine m_R. Include any steps necessary to reduce experimental uncertainty.

Part B

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

The following information applies to parts C and D.

In a different experiment, a student determines the spring constant k of a second spring. One end of the spring is attached to a post at the end of a level air track, and the other end is attached to a glider that moves along the track with negligible friction. A scale printed along the track gives the position x of a mark on the glider, measured from the post. The student pulls the glider away from the post with a hand-held force sensor, holds the glider at rest while the sensor reads each of five chosen values F, and records the glider’s position x each time, as shown in Figure 2.

Figure 2 — The glider held at rest by the force sensor; x is read from the scale on the track.

The results are shown in Figure 3. The student creates a graph with F plotted on the horizontal axis.

Figure 3 — The force sensor reading and the glider’s position in each of the five trials.

Part C

On the grid in Figure 4, create a graph that can be used to determine k.

•   Label the vertical axis with a measured or calculated quantity, including units as appropriate. The graphed quantities should yield a linear graph that can be used to determine k.

•   Clearly label the vertical axis with a numerical scale.

•   Plot the corresponding data points on the grid.

•   Draw a best-fit line for the data you plotted.

Figure 4 — Grid for graphing the data in Figure 3.

Part D

Using the best-fit line that you drew in part C, calculate an experimental value for k.

❖ 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

Setup. Hang the spring from the stand and clamp the meterstick upright beside it, so that the pointer on the hook lies against its scale. With nothing hanging from the spring, record the pointer’s reading y_0, the reading at the spring’s relaxed length.

Figure 5 — Each stretch is the change in the pointer’s reading from its no-load reading y_0.

Trials. Hang 100 g on the hanger, wait until it hangs at rest, and record the pointer’s reading y with your eye level with the pointer; the stretch is the size of the change in reading, Δx = |y − y_0|. Repeat for loads of 200 g, 300 g, and so on up to 800 g, a range that covers the rock’s expected 200 g to 600 g. Remove and rehang each load and take its reading three times, averaging the three. Finally hang the rock in the string bag and record its reading several times in the same way, averaging to find its stretch Δx_R. Every stretch is the change in the pointer’s reading from the relaxed reading, not the length of the spring.

Scoring (2 points):

▸ A1 — 1 point: For describing a procedure that includes both of the following: hanging several different known loads from the spring and, for each, measuring the spring’s stretch (or a length or position reading from which the stretch can be found); and hanging the rock and measuring the spring’s stretch (or the corresponding reading) with the rock.

▸ A2 — 1 point: For a procedure that indicates a reasonable method of reducing experimental uncertainty — for example repeating each reading and averaging, choosing known loads on both sides of the rock’s expected mass and stating that this lets its stretch be read between calibration points, reading the meterstick at eye level against a fixed pointer, or waiting until each load hangs at rest before reading.

Scoring Note: Hanging several known loads, which A1 requires, does not by itself satisfy A2, even when they happen to span the rock’s expected mass.

⚠︎ Common error (partial credit): Hangs only one known mass and the rock, taking each reading several times and averaging — earns 1 of 2 points, A2. One known load gives a single point, not the linear graph the task asks for, and rests the whole calibration on a single load.

✗ Common error (no credit): Hangs the rock alone, measures the spring’s length once, and uses none of the known masses — earns 0 points. With no known load the spring constant is never found, so the rock’s stretch cannot be turned into a mass, and a single reading offers no way to reduce uncertainty.

Part B — Model Answer

Plot the weight of each known load, mg, on the vertical axis against its stretch Δx on the horizontal axis, and draw the best-fit straight line. Each load hangs at rest, so the spring’s upward force balances its weight, kΔx = mg, and the points lie on a straight line through the origin whose slope is k. The rock’s weight is then k·Δx_R, and

m_R = k·Δx_R / g

Figure 6 — Schematic: the rock’s stretch, read against the calibration line, gives its weight.

Equivalently, read the weight at which the best-fit line reaches the rock’s stretch and divide it by g; or plot the stretch against the load’s mass, a straight line of slope g/k, and find m_R = Δx_R ÷ slope. A third route hangs the rock together with added known loads, keeping the total under 1.0 kg: the stretch against the added mass is a straight line that crosses the horizontal axis at an added mass of −m_R, though reading that intercept is an extrapolation and so less precise. Every one of these graphs works because the known loads calibrate the spring.

Scoring (2 points):

▸ B1 — 1 point: For indicating quantities that could be plotted to give a straight line from which m_R can be found — for example the weight of the known loads against the stretch, or the known mass against the stretch, with either quantity on either axis; the pointer reading against the load; or, with the rock hung together with added known loads, the stretch against the added mass.

Scoring Note: B1 requires naming what is plotted against what, but either quantity may go on either axis, and constant factors such as g are not required. Stating only that the stretch is proportional to the load does not satisfy B1. B1 may be earned independently of the response in part A.

▸ B2 — 1 point: For describing how m_R is found from a graph that yields a straight line, consistent with the quantities plotted — for example the slope of weight against stretch is k, so m_R = k·Δx_R/g; the slope of stretch against mass is g/k, so m_R = Δx_R ÷ slope; the rock’s mass is read from the best-fit line at the rock’s stretch; or, with the rock hung together with added known loads, m_R is the size of the horizontal intercept of the stretch against the added mass.

Scoring Note: B2 may be earned independently of the response in part A.

⚠︎ Common error (partial credit): Describes a correct straight-line graph but finds the rock’s mass from the spring’s total length with the rock hung, m_R = k·L_R/g, rather than from its stretch — earns 1 of 2 points, B1. The spring’s force depends on the change in its length from the relaxed length, not on its length.

⚠︎ Common error (partial credit): Plots the stretch on the vertical axis against the weight but treats the slope as k, so that m_R = slope × Δx_R/g — earns 1 of 2 points, B1. With the stretch on the vertical axis, the slope is 1/k.

✗ Common error (no credit): States only that the stretch is proportional to the load, so the data would be graphed and the rock’s mass “read from the graph”, naming no quantity for either axis and no feature of the graph — earns 0 points. A graph whose axes are never named cannot be analyzed.

Part C — Model Answer

Plot the glider’s position x, in centimeters, on the vertical axis. While the glider is at rest, the sensor’s pull balances the spring’s force, so F = k(x − x_0), where x_0 is the glider’s position when the spring is relaxed — a position the student never measured. Rearranged,

x = x_0 + (1/k)·F

so the graph of x against F is a straight line of slope 1/k whose vertical intercept is x_0. A vertical scale from 28 cm to 60 cm, at 1 cm per small square, spreads the five points across the grid. Plotting position rather than stretch shifts the line up by x_0 but leaves its slope unchanged.

Figure 7 — Position against force, with the best-fit line; its intercept is the relaxed position.

Scoring (4 points):

▸ C1 — 1 point: For indicating a vertical-axis quantity that, plotted against F, yields a linear graph whose slope can be used to determine k — for example the glider’s position x, or its change in position from any one reading.

Scoring Note: C1 may be earned independently of parts A and B.

▸ C2 — 1 point: For labelling the vertical axis with the quantity whose values are actually plotted, including units, with a linear numerical scale.

Scoring Note: The vertical scale need not start at zero.

▸ C3 — 1 point: For correctly plotting the five data points on the response’s own vertical scale.

▸ C4 — 1 point: For drawing a straight best-fit line that approximates the trend of the plotted data.

Scoring Note: C3 and C4 may be earned for data plotted consistently with the response’s own vertical scale, whatever quantity the axis names.

⚠︎ Common error (partial credit): Labels the vertical axis as the stretch, Δx, while plotting the position readings from Figure 3 — earns 3 of 4 points, C1, C3 and C4. The readings are positions measured from the post; a stretch would be each reading minus the unknown relaxed position.

⚠︎ Common error (partial credit): Plots and labels correctly but joins the points dot to dot, or forces the line through (0 N, 0 cm) or through the grid’s lower-left corner — earns 3 of 4 points, C1, C2 and C3. A best-fit line follows the trend of all the points.

Part D — Model Answer

With position on the vertical axis and force on the horizontal axis, the slope of the best-fit line, its rise divided by its run, equals 1/k. Two points on the line, well apart and not data points, are (1.0 N, 32.5 cm) and (11.0 N, 57.5 cm):

slope = rise ÷ run = (57.5 cm − 32.5 cm) / (11.0 N − 1.0 N)

slope = 2.50 cm/N = 0.0250 m/N

k = 1 / slope = 1 / (0.0250 m/N) = 40.0 N/m

Figure 8 — The slope from two well-separated points on the best-fit line, and k = 1/slope.

The line meets the vertical axis at 30.0 cm, the glider’s position with the spring relaxed, a value the slope method never needs. Dividing a single force by its position reading instead — 10.0 N ÷ 0.551 m = 18 N/m — treats the position as the stretch. The slope uses only changes in position, so the unknown relaxed position drops out of k.

Scoring (2 points):

▸ D1 — 1 point: For correctly relating the best-fit line to k — with position plotted against force, slope = 1/k, so k = 1/slope; or k = F/(x − x_0) for a point on the line, with x_0 read from the line’s vertical intercept; or an equivalent relation consistent with the response’s own axes, provided the graph is one that yields a straight line.

▸ D2 — 1 point: For a value of k from 36 N/m to 44 N/m, calculated from the best-fit line, with units.

Scoring Note: D2 requires the value itself to fall in the range, whatever the upstream work. The value must come from the response’s best-fit line, either its slope from two points on the line, which may include data points the line passes through, or its vertical intercept with a point on the line. A value in equivalent units, such as 0.40 N/cm, is accepted.

⚠︎ Common error (partial credit): Relates the slope correctly, k = 1/slope, but mislabels the units, reporting k = 0.40 N/m where 1/(2.50 cm/N) is 0.40 N/cm, or 40 N/m — earns 1 of 2 points, D1.

✗ Common error (no credit): With position plotted against force, takes k equal to the slope, reporting k = 2.5 N/m or 0.025 N/m — earns 0 points. With force on the horizontal axis, the slope is the change in position per newton, 1/k.

✗ Common error (no credit): Divides a single force by its position reading, such as 10.0 N ÷ 0.551 m = 18 N/m, treating the position as the stretch — earns 0 points. The position includes the unknown relaxed position x_0.

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

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