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SCIENCE · INVESTIGATE AND EXPLAIN

Locate Newton’s third-law partners on different bodies

An interaction produces a pair: the force of A on B and the force of B on A. Their magnitudes are equal and their directions opposite at the same instant. The labels reverse agent and recipient. The pair is not “a force and the motion it causes”; acceleration is a response, not the partner force.

Learning outcome: Compare equal masses and a 3:1 mass ratio. List each interaction force and the acceleration of its recipient. Multiply accelerations by their masses and verify cancellation for the whole system. Explain why this is not cancellation within either individual diagram.

Starting recap

Review signed vectors, components, acceleration, graph slopes and system boundaries. Every force must name its agent and recipient.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

At a selected instant two objects attract with magnitude 12 N. With masses 2 kg and 4 kg, their accelerations are +6 and−3 m/s² along the chosen axis. Their mass-weighted acceleration sum is 2×6+4×(−3)=0, consistent with zero net external horizontal force on the pair.

Open the experiment

Predict

If only the right mass doubles while the stipulated interaction force is fixed, which force and which acceleration changes?

Experiment

Compare equal masses and a 3:1 mass ratio. List each interaction force and the acceleration of its recipient. Multiply accelerations by their masses and verify cancellation for the whole system. Explain why this is not cancellation within either individual diagram.

Compare mutual forces and accelerations

Worked teaching model · synthetic values · Two separated objects experience a stipulated mutual attractive interaction of magnitude 12 N at one instant; no other horizontal forces. The model does not assume that the force stays constant as separation changes.

Reset keeps your prediction, captures and practice history.

Interaction arrows are grouped by recipient. For horizontal-only cart/block diagrams, balanced vertical interactions are omitted explicitly; this is a component diagram, not a claim that gravity vanished. For the spring case only the selected spring interaction is drawn; the opposite external holding force is tabulated but omitted from that diagram. For circular cases, the arrow is the required tension, not a claim it is available when the feasibility check fails. All arrows in one view share a scale; scale changes with controls. A zero force has no visible arrow. The force table supplies exact signs and units.

Left objectF1: (12, 0) NRight objectF1: (-12, 0) N
Proposed-model interaction components — right and up are positive
RecipientInteractionFx (N)Fy (N)
Left objectRight object on left120
Right objectLeft object on right-120
Bounded synthetic force-model results
QuantityValueUnit
Force on left object12N
Force on right object-12N
Left acceleration6m/s²
Right acceleration-3m/s²
Net internal force on pair0N

Mutual forces have equal magnitudes; accelerations can differ because masses differ.

The pair acts on different objects. Cancel the pair only when forming the combined-system external-force sum, not in either individual free-body diagram.

Mutual forces have equal magnitudes; accelerations can differ because masses differ.

The pair acts on different objects. Cancel the pair only when forming the combined-system external-force sum, not in either individual free-body diagram.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Orient: Locate Newton’s third-law partners on different bodies

An interaction produces a pair: the force of A on B and the force of B on A. Their magnitudes are equal and their directions opposite at the same instant. The labels reverse agent and recipient. The pair is not “a force and the motion it causes”; acceleration is a response, not the partner force.

2. Explain and derive

Equal interaction forces do not require equal accelerations. For one-dimensional motion with no other horizontal forces, a_left=F/m_left and a_right=−F/m_right. The less massive object changes velocity more rapidly. A large body can experience the same gravitational force as a small body yet have a much smaller acceleration.

A third-law pair cannot cancel on one object’s free-body diagram because its members act on different recipients. When both bodies belong to the chosen system, adding their force equations cancels that internal pair. Remaining external forces then determine center-of-mass acceleration.

Balanced forces on the same body are not automatically a third-law pair. A book’s upward table force and downward Earth force can balance, but their partners are the book’s force on the table and the book’s force on Earth. Matching magnitudes is insufficient; name the interaction and reverse agent/recipient.

3. Work an example

At a selected instant two objects attract with magnitude 12 N. With masses 2 kg and 4 kg, their accelerations are +6 and−3 m/s² along the chosen axis. Their mass-weighted acceleration sum is 2×6+4×(−3)=0, consistent with zero net external horizontal force on the pair.

4. Predict and investigate

Predict first: If only the right mass doubles while the stipulated interaction force is fixed, which force and which acceleration changes?

Compare equal masses and a 3:1 mass ratio. List each interaction force and the acceleration of its recipient. Multiply accelerations by their masses and verify cancellation for the whole system. Explain why this is not cancellation within either individual diagram. Use the lesson controls and record the settings with your evidence.

5. Transfer and explain

Try a new practice case

A 2 kg body experiences an isolated mutual force of 18 N toward its partner. Find the magnitude of its acceleration.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

Name the third-law partner of the track’s upward normal force on a cart. Why is the cart’s downward weight not that partner?

Retain your first answer, force diagram, equations, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Supplied synthetic outputs do not certify actual laboratory skill.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 2 topic mapping: 2.3; 2.5. Checked 30 October 2026. This is a content cross-reference, not AP authorization.

Apply

A light object and a heavy object pull on each other. With no other forces, which statement is correct?

Public practice checks a specific claim. It does not establish independent mastery.

Name the third-law partner of the track’s upward normal force on a cart. Why is the cart’s downward weight not that partner?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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