Solve coupled carts and test the system boundary

1. Orient: Solve coupled carts and test the system boundary

When objects share a motion constraint, choose whether to analyze each object or the whole system. Two carts connected by a taut inextensible massless string share an acceleration along a straight track. The leading cart is pulled by an external 20 N force; the trailing cart is pulled only by the string. Friction is absent in the stated model.

2. Explain and derive

Let leading and trailing masses be m₁ and m₂, rightward positive. The leading equation is 20−T=m₁a and the trailing equation isT=m₂a. Adding cancels the internal tension, giving a=20/(m₁+m₂). Then substitute into either individual equation to recover T=20 m₂/(m₁+m₂). The external force is not automatically the string tension.

Tension equality depends on the ideal string model. A massive string accelerating along its length would need unequal end forces. A stretching string could allow different cart accelerations. A practical investigation must therefore record connection properties and test the shared-motion assumption rather than treating the equations as unconditional identities.

Use the supplied model as a worked investigation, then design an independent comparison: keep total mass fixed while swapping which cart is heavier. Predict shared acceleration and tension before revealing outputs. The whole-system equation predicts unchanged acceleration, while the individual equation predicts changed tension. This paired intervention distinguishes the two claims.

3. Work an example

With leading mass 2 kg and trailing mass 3 kg, acceleration is 20/5=4 m/s². Tension is 3×4=12 N. Leading net force 20−12=8 N equals 2×4; trailing net force 12 N equals 3×4. Swapping masses leaves acceleration 4 m/s² but lowers tension to 2×4=8 N.

4. Predict and investigate

Predict first: At fixed total mass, does exchanging leading and trailing masses change acceleration, tension, both or neither?

Compare(2,3) kg with(3,2) kg and retain both free-body diagrams. Then add mass only to the trailing cart and predict changes in acceleration and tension separately. Verify both individual equations and the combined equation numerically. State what independent observation could falsify the shared-acceleration assumption. Use the lesson controls and record the settings with your evidence.

5. Transfer and explain

Try a new practice case

A 4 kg leading cart pulls a 1 kg trailing cart on a frictionless track under a 20 N external force. Find string tension.

Your numerical result (N) Check number

Reveal worked check (practice, not mastery)

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

Create a new case with a constant 4 N opposing force on the trailing cart. Derive both equations and the new acceleration before calculating tension; explain why friction remains external to the two-cart system.

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.

Ideal pulleys and a massive-string contrast

An ideal string is inextensible and has negligible mass. An ideal pulley has negligible mass and axle friction and rotates about an axis through its center of mass. These assumptions allow the same string-tension magnitude along the modeled path; changing them may require separate segment forces or rotational dynamics.

A massive hanging chain is a useful qualitative counterexample: a section near its top supports more chain below than a section near its bottom, so its tension is greater. The required AP Physics 1 treatment of a massive string or chain is qualitative, not a demanded continuum calculation. Predict and check: mark upper, middle and lower sections of a stationary uniform hanging chain and rank tension upper greater than middle greater than lower. Reverse the drawing on the page without changing gravity: the physical ranking follows supported material, not page position. Contrast this with a massless taut horizontal string between the carts, whose equal-tension approximation is one reason the two-cart equations simplify.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 2 topic mapping: 2.1; 2.2; 2.3; 2.5. Checked 30 October 2026. This is a content cross-reference, not AP authorization.