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SCIENCE · INVESTIGATE AND EXPLAIN

Change system boundaries between capture and oscillation

Compound events require separate models for separate stages. Consider a horizontal spring mass that briefly captures a second object with zero horizontal velocity. During an idealized very short attachment, the spring’s horizontal impulse is negligible compared with the interaction impulse, so combined horizontal momentum is approximately conserved. Mechanical energy need not be conserved during this inelastic stage.

Learning outcome: Compare center and endpoint capture for the same added mass. Write momentum and energy ledgers for the brief event, then a separate energy ledger for the later oscillation. Change added mass and predict whether amplitude and period both change in each case.

Starting recap

Review Newton’s second law, radians, sine/cosine graphs, energy and momentum boundaries, and bounded measurement errors.

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Try

For m=1 kg, k=20 N/m, A=0.20 m and added mass1 kg, center-capture speed falls from0.8944 to0.4472 m/s. New amplitude is0.20/√2≈0.1414 m and energy drops from0.40 to0.20 J. At endpoint capture instead, amplitude stays0.20 m and horizontal mechanical energy stays0.40 J. Both cases have new period2π√(2/20)≈1.987 s.

Open the experiment

Predict

Can equal final masses and equal final periods have different final amplitudes?

Experiment

Compare center and endpoint capture for the same added mass. Write momentum and energy ledgers for the brief event, then a separate energy ledger for the later oscillation. Change added mass and predict whether amplitude and period both change in each case.

Separate a brief capture from the later oscillation

Worked teaching model · synthetic values · Original horizontal oscillator m=1 kg,k=20 N/m,A=.2 m. A second mass attaches instantaneously with zero horizontal velocity. External spring impulse is negligible during capture; subsequent motion is ideal. Vertical arrival energy and real attachment details are outside this horizontal model.

Reset keeps your prediction, captures and practice history.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Initial horizontal velocity0.89443m/s
Velocity immediately after capture0.44721m/s
New total mass2kg
New amplitude0.14142m
New period1.98692s
Mechanical energy before capture0.4J
Mechanical energy after capture0.2J
Horizontal mechanical energy converted internally0.2J
Accessible plot coordinates
Primary plotted points
coordinatevalue

At center capture, horizontal momentum is approximately conserved but mechanical energy decreases.

Use momentum only during the brief attachment, then energy for the later ideal oscillation. At center Anew=A√(m/(m+Δm)); at endpoint the unchanged spring extension fixes Anew=A. Spring momentum transfer over a full period is not negligible.

At center capture, horizontal momentum is approximately conserved but mechanical energy decreases.

Use momentum only during the brief attachment, then energy for the later ideal oscillation. At center Anew=A√(m/(m+Δm)); at endpoint the unchanged spring extension fixes Anew=A. Spring momentum transfer over a full period is not negligible.

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

Understand

1. Frame the question

Compound events require separate models for separate stages. Consider a horizontal spring mass that briefly captures a second object with zero horizontal velocity. During an idealized very short attachment, the spring’s horizontal impulse is negligible compared with the interaction impulse, so combined horizontal momentum is approximately conserved. Mechanical energy need not be conserved during this inelastic stage.

2. Build the mechanism

After attachment, the combined mass oscillates on the same ideal spring. Over this later interval the spring force changes momentum substantially, so momentum conservation for the mass alone no longer applies. Instead use the post-capture mechanical energy, provided drag and further dissipation are neglected. The new period follows the increased mass regardless of where capture occurred.

At equilibrium capture, the original speed is A√(k/m). Momentum gives vnew=m vold/(m+Δm). Converting the remaining kinetic energy to the next turning-point spring energy yields Anew=A√[m/(m+Δm)]. The lost horizontal mechanical energy becomes internal energy in the idealized attachment. It is incorrect to carry the original total mechanical energy unchanged across capture.

At endpoint capture, the original horizontal velocity and the incoming horizontal velocity are both zero. The spring extension is still A, so the post-capture amplitude remains A, although the larger mass lengthens the period. This is a horizontal model: vertical arrival energy, real attachment mechanics and a finite-duration spring impulse are excluded. It is not an instruction to drop material onto physical moving equipment.

3. Work an example

For m=1 kg, k=20 N/m, A=0.20 m and added mass1 kg, center-capture speed falls from0.8944 to0.4472 m/s. New amplitude is0.20/√2≈0.1414 m and energy drops from0.40 to0.20 J. At endpoint capture instead, amplitude stays0.20 m and horizontal mechanical energy stays0.40 J. Both cases have new period2π√(2/20)≈1.987 s.

4. Predict and investigate

Can equal final masses and equal final periods have different final amplitudes?

Compare center and endpoint capture for the same added mass. Write momentum and energy ledgers for the brief event, then a separate energy ledger for the later oscillation. Change added mass and predict whether amplitude and period both change in each case.

5. Transfer to a new case

Try a new practice case

An oscillator of mass1 kg and amplitude0.3 m captures an additional3 kg at rest horizontally when crossing equilibrium. Find its new amplitude.

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. Explain the boundary

Which assumption permits approximate momentum conservation during capture but not over the subsequent oscillation?

These are bounded virtual models and supplied numerical cases, not independent measurements or an observed physical investigation. Do not construct spring, pendulum or collision apparatus from these instructions. Practical evidence requires an appropriate supervised setting.

Sources

Original explanations, worked cases and numerical models. Framework checked 30 October 2026: AP Physics 1 Course and Exam Description, topics 7.4; 4.3. Mechanism reference: OpenStax University Physics 1, simple harmonic motion and pendulums. No official assessment items reproduced.

Apply

An equal stationary horizontal mass attaches to an oscillator at its positive endpoint. Its new amplitude is:

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

Which assumption permits approximate momentum conservation during capture but not over the subsequent oscillation?

Continue

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