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FRQ: Centre of Mass of a Composite Object — The L-Shaped Cut-Out Plate

Assessments aligned to 2026 AP Physics 1 standards

Experimental Design and Analysis (LAB)  |  CHALLENGE  |  10 points  |  25 min

▤ Scenario

A workshop cuts parts from uniform square steel plates. Each uncut plate has side length L = 0.400 m and mass M = 1.20 kg, and has uniform thickness and uniform density, so the mass of any piece of a plate is proportional to its area.

Coordinates are fixed to the plate. The origin O is at the geometric centre of the uncut plate, the x-axis runs to the right along the plate and the y-axis runs upwards along the plate. The uncut plate is symmetric about both of these axes, so the centre of mass of the uncut plate is at O.

From one plate a square piece of side 0.100 m is removed, as shown in Figure 1. The centre of the removed square is at (0.150 m, 0.150 m), so the removed square sits flush against the top and right edges of the plate. The removed piece is discarded, and the remaining L-shaped plate is the object of interest.

Figure 1 — The uncut plate with a 0.100 m square removed from its upper-right corner.

All positions are measured in this plate-fixed x–y coordinate system, with rightward and upward taken as positive. Each plate is treated as a two-dimensional object of negligible thickness, and each piece of a plate may be modelled as a single object located at that piece’s own centre of mass.

✎ Free Response Questions

Part A

Without performing any calculation, indicate whether the centre of mass of the remaining L-shaped plate lies in the upper-right quadrant, at O, or in the lower-left quadrant, by writing one of the following.

______ In the upper-right quadrant

______ At O

______ In the lower-left quadrant

Justify your answer using qualitative reasoning beyond referencing equations.

Part B

Calculate the mass of the removed square and the mass of the remaining L-shaped plate. Then calculate the x- and y-coordinates of the centre of mass of the remaining L-shaped plate.

Part C

The workshop now wants to answer this question: how does the distance of a plate’s centre of mass from O depend on the side length of the square removed from it? On every plate the removed square is centred at the same fixed point (0.100 m, 0) on the x-axis.

Describe an experimental procedure that could be used to answer this question, using a supply of uncut plates identical to the one above, a ruler, a pin, a pencil, and a length of string with a small weight tied to one end. State which quantity is varied and which is measured, and describe how each measurement is made, including any steps necessary to reduce experimental uncertainty. Then describe how the resulting data would be analysed to answer the question.

Part D

A technician investigates a different question, using five uncut plates of the same side length L = 0.400 m. From each plate a square of the same side length s is removed, but on each plate the removed square is centred at a different distance d from O along the positive x-axis. The value of s was not recorded. For each plate the technician measures the distance |x_cm| by which the centre of mass has shifted from O. The results are shown in Figure 2.

Figure 2 — Shift of the centre of mass measured on each of the five plates.

For a plate of side L with a square of side s removed at a distance d from O along the x-axis, the centre of mass of the remaining plate lies at

x_cm = −s²d / (L² − s²)

Plot the five data points on the grid in Figure 3 and draw a best-fit straight line. Then use the slope of your line, together with the relationship above, to calculate the side length s of the removed square.

Figure 3 — Grid for plotting the shift of the centre of mass against the distance d.

❖ 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

The centre of mass of a system is the mass-weighted average position of the objects that make it up. The uncut plate is symmetric about both axes, so that average sits at O. Removing material from the upper-right corner takes away part of what was pulling the average towards the upper right, so the average position of the material that is left shifts away from the region that was removed.

The centre of mass of the remaining L-shaped plate therefore lies in the lower-left quadrant, at negative x and negative y — mass taken out of one region moves the mass-weighted average away from that region, so the shift is always opposite to the cut.

Figure 4 — The cut removes mass from the upper right, so the centre of mass moves along y = x.

It also lies on the line y = x, because the uncut plate and the removed square are both symmetric about that line, so the remaining plate is symmetric about it too.

Scoring (2 points):

▸ A1 — 1 point: For indicating that the centre of mass of the remaining plate lies in the lower-left quadrant.

▸ A2 — 1 point: For justifying that direction by establishing that mass removed from one region shifts the mass-weighted average position away from the region removed.

Scoring Note: A2 is a content criterion, not a route criterion. It requires the response to establish that the average position moves away from the material taken out — by appealing directly to the centre of mass as a mass-weighted average position, by comparing the mass now lying on either side of O and concluding that the average must lie towards the heavier side, or by any equivalent qualitative route.

⚠︎ Common error (partial credit): Selects the lower-left quadrant but justifies it only by observing that the upper-right side of the plate is now lighter, without drawing from that any conclusion about where the average position must lie — earns 1 of 2 points, A1.

✗ Common error (no credit): Selects the upper-right quadrant, on the reasoning that the centre of mass moves towards the gap left by the removed piece — earns 0 of 2 points. This treats the cut as something that attracts the centre of mass, rather than as the loss of the mass that was holding it there.

Part B — Model Answer

The plate is uniform, so mass is proportional to area.

area of the uncut plate = L² = (0.400 m)² = 0.160 m²
area of the removed square = (0.100 m)² = 0.0100 m²

m_s = M × (0.0100 m² / 0.160 m²) = (1.20 kg)/16 = 0.0750 kg
m_L = M − m_s = 1.20 kg − 0.0750 kg = 1.125 kg

The uncut plate can be treated as a system of two objects: the remaining L-shaped plate, of mass m_L, whose centre of mass is at the unknown point (x_L, y_L); and the removed square, of mass m_s, whose centre of mass is at its own geometric centre (0.150 m, 0.150 m). The centre of mass of that two-object system is the centre of mass of the uncut plate, which is at O.

Figure 5 — The uncut plate as two objects; the detail near O is magnified five times.

x_cm = (m_L x_L + m_s x_s) / (m_L + m_s) = 0
m_L x_L + m_s x_s = 0
(1.125 kg) x_L + (0.0750 kg)(0.150 m) = 0
x_L = −(0.0750 kg)(0.150 m) / (1.125 kg) = −0.0100 m

The whole arrangement is symmetric about the line y = x, and x_s = y_s, so the same working gives y_L = −0.0100 m.

The centre of mass of the remaining L-shaped plate is at (−0.0100 m, −0.0100 m) — 1.00 cm to the left of O and 1.00 cm below it, in the lower-left quadrant as predicted in Part A, and an offset of L/40 along each axis. The two pieces are treated as a system whose centre of mass is already known, so the unknown piece is found by addition and never by giving the removed square a negative mass.

Scoring (3 points):

▸ B1 — 1 point: For using the uniform density of the plate — that the mass of a piece is proportional to its area — to obtain the mass of the removed square and the mass of the remaining plate. The expected values are m_s = 0.0750 kg and m_L = 1.125 kg.

▸ B2 — 1 point: For writing a mass-weighted centre-of-mass relation for the uncut plate treated as a system of two objects — the remaining plate and the removed square — with the centre of mass of that system placed at O. Award this point wherever such a relation is written, whatever values the response goes on to substitute into it.

▸ B3 — 1 point: For obtaining both coordinates of the centre of mass of the remaining plate, each negative and with units, from a relation in which the remaining plate’s own mass multiplies its own position. The expected value is x_L = y_L = −0.0100 m. A response that carries its own value of that mass forward from B1 earns this point wherever the substitution into its own relation is correct.

Scoring Note: B2 and B3 are content criteria, not route criteria. A response that instead writes the uncut plate’s centre of mass as the mass-weighted average of the whole plate and the removed square, and rearranges that to isolate the remaining plate’s position, satisfies B2 and B3 on the same terms — it is the same relation rearranged, and no mass is given a negative value. What B2 requires is that the two pieces and the known position of their combined centre of mass both appear; what B3 requires is that the position obtained is the remaining plate’s, carries its sign and its units, and follows from the response’s own masses.

Scoring Note: B1, B2 and B3 may each be earned independently of the response in Part A.

⚠︎ Common error (partial credit): Sets up the two-object relation correctly but drops the minus sign when solving, placing the centre of mass at (+0.0100 m, +0.0100 m) — earns 2 of 3 points, B1 and B2.

⚠︎ Common error (partial credit): Obtains both masses correctly but then multiplies the unknown position by the full mass M = 1.20 kg rather than by the 1.125 kg of the remaining plate, obtaining −0.00938 m for each coordinate — earns 2 of 3 points, B1 and B2.

✗ Common error (no credit): Skips the mass calculation entirely and averages the position of O and the position of the removed square, giving (0.0750 m, 0.0750 m) — earns 0 of 3 points. No mass is ever obtained and no mass-weighted relation is ever written, so nothing in the response depends on the quantity a centre of mass is an average over, and the result sits on the wrong side of O.

Part C — Model Answer

Quantity varied: the side length s of the removed square. Quantity measured: the distance of the plate’s centre of mass from O. Everything else is held constant — the side length, thickness and material of the plate, and the point (0.100 m, 0) at which every removed square is centred.

1.  Take several identical uncut plates. From each one remove a square of a different side length — for example 0.05 m, 0.10 m, 0.15 m and 0.20 m — with every square centred at (0.100 m, 0). Measure each side length s with the ruler.

2.  Push the pin through a point P_1 near the edge of one plate and suspend the plate from the pin so that it hangs freely and can swing. Wait until it stops swinging.

3.  Hang the string with the weight on it from the same pin so that it hangs down in front of the plate. The string hangs vertically, so the centre of mass lies somewhere on that vertical line. Rule a pencil line along the string across the plate.

4.  Repeat steps 2 and 3 with the plate suspended from a second point P_2 well away from the first. The two pencil lines cross at the centre of mass of that plate. Ruling a third line from a third suspension point is a useful check: the three lines should meet at one point, and the size of any triangle they leave is the uncertainty in it.

5.  Lay the plate flat and use the ruler to measure the distance from O to the crossing point.

6.  Repeat steps 2 to 5 for every plate, so that each value of s has a measured centre-of-mass distance.

A balance method is equally acceptable: rest the plate on the edge of the ruler, mark the line along which it balances, repeat with the plate turned through roughly a right angle, and take the crossing of the two balance lines as the centre of mass. The two determinations do the same work as the two suspension lines.

Figure 6 — Locating the centre of mass: two ruled lines from different pins cross at it.

Analysis: plot the measured distance from O against the side length s and look at the shape of the graph. The points do not fall on a straight line — the distance grows more and more steeply as s increases, so the distance depends on s more strongly than in proportion to it. One determination fixes the centre of mass only to a line, so it takes a second, independent determination in a different direction to fix it to a point.

Scoring (3 points):

▸ C1 — 1 point: For a procedure in which several plates each have a square of a different side length removed, and in which that side length is measured on each plate with the ruler.

▸ C2 — 1 point: For describing a method that locates the centre of mass of a plate as a point rather than only as a line, by two independent determinations — for example suspending the plate freely from a pin, ruling a pencil line along the hanging string, and repeating from a second, well separated suspension point so that the two lines cross; or balancing the plate on the edge of the ruler in two different directions and taking the crossing of the two balance lines — and for reading the distance from O to that located point off the ruler.

Scoring Note: The second, well separated determination is the step that reduces the uncertainty in the located position, since a single line leaves the centre of mass undetermined along its whole length. A response that also rules a third line as a check, or that carries out the whole two-determination location a second time and averages the two located points, satisfies C2 on the same terms.

▸ C3 — 1 point: For stating how the collected data would be analysed to answer the question — by graphing the measured distance from O against the side length of the removed square and examining the shape of that graph, or by any equivalent comparison across the whole data set that establishes how the distance depends on the side length.

⚠︎ Common error (partial credit): Suspends each plate from a single point only, which fixes the centre of mass to a line rather than to a point and leaves no point to measure from O, but cuts and measures the plates correctly and describes a workable analysis — earns 2 of 3 points, C1 and C3.

⚠︎ Common error (partial credit): Removes a square of a different size from each plate but never measures those sizes, describing them only as small, medium and large, so the horizontal axis of the graph carries no values — but describes a workable location method and a workable analysis — earns 2 of 3 points, C2 and C3.

✗ Common error (no credit): Works the answer out instead of measuring it — writes down a set of side lengths on paper, calculates from each one where the centre of mass of that plate would lie, and reports those calculated positions as the results — earns 0 of 3 points. No plate is cut, no side length is read off the ruler, nothing is suspended or located and no distance is measured, so there is no procedure and no data, only the relationship the experiment was meant to test, assumed in advance.

Part D — Model Answer

The five points lie close to a straight line through the origin, as they should: L and s are the same for every plate, so s²/(L² − s²) is the same for every plate too, and |x_cm| is directly proportional to d.

Reading two widely separated points from the best-fit line, (2.0 cm, 3.3 mm) and (12.0 cm, 19.7 mm). The slope is a ratio of two lengths, so both are first expressed in the same unit:

slope = (19.7 mm − 3.3 mm) / (120.0 mm − 20.0 mm) = 16.4 mm / 100.0 mm = 0.164

From the given relationship, |x_cm| = [s²/(L² − s²)] d, so that slope is s²/(L² − s²) itself.

Figure 7 — The plotted data with a best-fit line; the marked rise and run give its slope.

s² / (L² − s²) = 0.164
s² (1 + 0.164) = 0.164 L²
s² = 0.164 (0.400 m)² / 1.164 = 0.02624 m² / 1.164 = 0.02254 m²
s = 0.150 m

The square removed from each plate has a side length of 0.150 m, or 15.0 cm. The line passes through the origin because s and L are the same on every plate, so the whole of the unknown sits in its slope.

Scoring (2 points):

▸ D1 — 1 point: For plotting all five data points correctly on the grid in Figure 3 and drawing a single straight best-fit line lying among them. This point is earned from the plotted graph alone and may be earned independently of the response in any earlier part.

▸ D2 — 1 point: For determining the slope of the drawn best-fit line, with the two axes expressed in a common unit, and using it with the given relationship to obtain a value for the side length that follows from that slope. The expected value is s ≈ 0.150 m; accept 0.14 m to 0.16 m.

Scoring Note: D2 is awarded from a slope read off the response’s own single straight best-fit line, whatever line that is, provided the two points used lie on that line — which may include plotted points the line happens to pass through. A slope taken between two plotted points with no single straight line drawn through the whole data set does not earn D2.

⚠︎ Common error (partial credit): Plots and fits correctly, then rearranges the relationship as s² = (slope) L², dropping the (1 + slope) denominator, and obtains s ≈ 0.162 m — earns 1 of 2 points, D1.

⚠︎ Common error (partial credit): Plots and fits correctly, then substitutes the slope as 1.64 read straight off a graph in millimetres against centimetres, without expressing the two axes in a common unit, and obtains s ≈ 0.315 m — a square too wide to fit on the plate at the largest value of d — earns 1 of 2 points, D1.

✗ Common error (no credit): Joins the plotted points dot to dot instead of drawing a single straight best-fit line, then reads a slope from one neighbouring pair — earns 0 of 2 points. With no fitted line there is nothing to take a slope from, so a value that happens to land in range has been read from two points rather than from the trend of all five.

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