🪐FRQ: Three Spheres on a Line Aboard an Orbiting Laboratory

Assessments aligned to 2026 AP Physics 1 standards

📋Scenario

Inside a large free-falling research laboratory, a robotic arm holds three uniform spheres so that their centres of mass lie on a single straight line.

1. Sphere X has mass 800 kg and is held at position x = 0.
2. Sphere Y has mass 5.00 kg and is held at position x = 2.00 m.
3. Sphere Z has mass 1600 kg and is held at position x = 6.00 m.

Because the laboratory is in free fall, no supporting forces act on the spheres, and the gravitational influence of the laboratory structure itself has been measured and found to be negligible.

All three centres of mass lie on the x-axis and all motion is one-dimensional.

Define the positive direction as the +x direction, from X toward Z. Use G = 6.67 × 10⁻¹¹ N·m²/kg².

📝 Free Response Questions

(a) On a diagram, represent Sphere Y as a single labelled dot. Draw and label every gravitational force exerted on Sphere Y by the other two spheres, using arrow lengths that correctly represent the relative magnitudes of those forces. 

(b) Calculate the magnitude and state the direction of the net gravitational force exerted on Sphere Y in its position at x = 2.00 m. 

(c) Sphere X and Sphere Z are held fixed. Determine the position on the x-axis between X and Z at which Sphere Y would experience zero net gravitational force. State whether your answer would change if Sphere Y were replaced by a sphere of different mass, and justify that statement. 

(d) The robotic arm now slides a thick lead plate into the gap between Sphere X and Sphere Y, without touching any sphere. Student M claims: “The force that X exerts on Y will now be smaller, because the lead plate blocks some of the gravity travelling from X to Y.” Student N replies: “Lead cannot block gravity, so nothing about the forces on Y changes at all.” Evaluate both claims. Identify what is correct and what is incorrect in each, and state what actually happens to the force exerted on Y by X and to the net gravitational force on Y. 

🗝️Answer Key & Scoring Guide

Part (a) — Model Answer

One dot, labelled Y (see the middle of the diagram). Two arrows originate on that dot:

1. An arrow pointing in the −x direction (toward X), labelled F_(on Y by X).
2. An arrow pointing in the +x direction (toward Z), labelled F_(on Y by Z).

The arrow toward X must be drawn approximately twice as long as the arrow toward Z, because F_(on Y by X) = 6.67 × 10⁻⁸ N while F_(on Y by Z) = 3.34 × 10⁻⁸ N.

Sphere Y is 2.00 m from X and 4.00 m from Z. Although Z is twice as massive as X, it is also twice as far away, and the distance enters as a square — so the factor of 2 in mass is defeated by the factor of 4 in r², leaving X's pull twice as strong.

Scoring (2 points):

✔️1 point: Exactly two gravitational force arrows drawn on the single dot representing Y, one directed toward X and one directed toward Z, each correctly labelled with the object exerting it.

✔️1 point: The arrow toward X drawn clearly longer than the arrow toward Z (a ratio of approximately 2:1).

⚠️Common error: Drawing the arrow toward Z longer because Z is more massive — earns 1 point for directions only; this is the failure to weigh the inverse-square distance factor against the mass factor.

⚠️Common error: Drawing a third arrow for the laboratory or for a supporting force — earns 0 points for the first criterion; the laboratory is in free fall and its influence is stated to be negligible.

Part (b) — Model Answer

Compute each pairwise force separately, then add them as vectors along the x-axis.

(1) Force exerted on Y by X, with r = 2.00 m:

F_(on Y by X) = (6.67 × 10⁻¹¹)(800)(5.00) / (2.00)² = (2.668 × 10⁻⁷) / 4.00 = 6.67 × 10⁻⁸ N (in the −x direction)

(2) Force exerted on Y by Z, with r = 6.00 − 2.00 = 4.00 m:

F_(on Y by Z) = (6.67 × 10⁻¹¹)(1600)(5.00) / (4.00)² = (5.336 × 10⁻⁷) / 16.0 = 3.34 × 10⁻⁸ N (in the +x direction)

 Adding with signs:

 ΣF = (−6.67 × 10⁻⁸) + (+3.34 × 10⁻⁸) = −3.34 × 10⁻⁸ N

The net gravitational force exerted on Sphere Y has magnitude 3.34 × 10⁻⁸ N and is directed in the −x direction, that is, toward Sphere X.

Scoring (3 points):

✔️1 point: Both pairwise forces set up correctly, each using the correct centre-to-centre separation — 2.00 m for X and 4.00 m for Z.

✔️1 point: Both magnitudes correct — 6.67 × 10⁻⁸ N and 3.34 × 10⁻⁸ N.

✔️1 point: Correct vector addition giving 3.34 × 10⁻⁸ N with the direction stated as −x or “toward X”.

⚠️Common error: Using r = 6.00 m for the force exerted by Z, measuring from X rather than from Y — earns 1 point for the X calculation only.

⚠️Common error: Adding the two magnitudes to obtain 1.00 × 10⁻⁷ N — earns 2 points; the setup and magnitudes are right but the forces oppose each other and must be combined with signs.

✔️No credit for the third point if a magnitude is given without a direction; the net force is a vector.

Part (c) — Model Answer

Let the null position be a distance d from X, so that it is a distance (6.00 − d) from Z. Setting the two opposing pairwise forces equal in magnitude, and writing the mass of the middle sphere as m:

 G (800) m / d² = G (1600) m / (6.00 − d)²

 G and m appear on both sides and cancel:

 800 (6.00 − d)² = 1600 d²   ⇒   (6.00 − d)² = 2 d²   ⇒   6.00 − d = √2 · d

 d = 6.00 / (1 + √2) = 6.00 / 2.414 = 2.49 m

The net gravitational force on the middle sphere is zero at x = 2.49 m, which is 0.49 m further from X than Sphere Y's present position — consistent with part (b), where the net force at x = 2.00 m pointed back toward X.

The answer would not change if Sphere Y were replaced by a sphere of different mass. The mass m of the middle sphere appears as a factor in both pairwise forces and cancels identically when they are set equal. The null point is fixed entirely by the two outer masses and their separation, and is a property of the field created by X and Z rather than of whatever object is placed there.

Scoring (2 points):

✔️1 point: Correct equation setting the two opposing pairwise force magnitudes equal, with the two separations correctly expressed as d and (6.00 − d).

✔️1 point: Correct position x = 2.49 m (accept 2.5 m) AND the statement that the result is independent of the middle sphere's mass, justified by the cancellation of m.

⚠️Common error: Placing the null point at the midpoint, x = 3.00 m — earns 0 points; this ignores the unequal masses.

⚠️Common error: Solving 800/d² = 1600/(6−d)² but taking the negative root, giving a position outside the interval between X and Z — earns 1 point; the physical root is the one lying between the two spheres.

⚠️Common error: Answering that the null point would shift for a heavier sphere — loses the second point even if the position is correct; the cancellation of m is the substance of this part.

Part (d) — Model Answer

Both students are partly right and both are partly wrong, and the disagreement turns on the distinction between a single pairwise force and the net force.

Student M is wrong. There is no gravitational shielding. Inserting matter between two objects does not reduce the gravitational force they exert on each other — gravity is not absorbed, blocked or attenuated by intervening material. The force exerted on Y by X still depends only on the two masses m_X and m_Y and on the centre-to-centre separation between X and Y, none of which the plate has altered:

 F_(on Y by X) = G m_X m_Y / r² = 6.67 × 10⁻⁸ N, unchanged

Student N is right about the physics of shielding and wrong about the conclusion drawn from it. “Lead cannot block gravity” is correct, but “nothing about the forces on Y changes at all” does not follow, because the lead plate is itself an object with mass. That mass exerts its own gravitational force on Y, directed from Y toward the plate — that is, in the −x direction, since the plate lies between X and Y. This is a new, third pairwise force that did not exist before, and the net force on Y is the vector sum of all pairwise forces acting on it:

 ΣF_on Y = F_(on Y by X) + F_(on Y by Z) + F_(on Y by plate)

 So what actually happens is this.

1. The force exerted on Y by X is exactly unchanged at 6.67 × 10⁻⁸ N in the −x direction.
2. The net gravitational force on Y increases in magnitude in the −x direction, because the plate's attraction adds to X's pull rather than opposing it.
3. Student M reaches “something changes” by the wrong mechanism; Student N applies the right principle and then forgets that the shield has mass.

Scoring (3 points):

✔️1 point: States that Student M is incorrect and that the gravitational force exerted on Y by X is unchanged, explicitly because gravitational forces are not shielded or blocked by intervening matter.

✔️1 point: States that Student N's conclusion is incorrect because the lead plate itself has mass and therefore exerts its own gravitational force on Y.

✔️1 point: Correctly distinguishes the two quantities in the final answer — the pairwise force F_(on Y by X) is unchanged while the net gravitational force on Y changes — and identifies the change as an increase in the −x direction (toward X).

⚠️Common error: Agreeing fully with Student N — earns 1 point for the no-shielding principle only; this is the targeted error of treating the pairwise force and the net force as the same quantity.

⚠️Common error: Agreeing with Student M — earns 0 points; gravitational shielding is not a physical phenomenon.

No credit for the third criterion if the response states that the net force changes without specifying the direction, or states the direction as +x.