Use system boundaries without inventing a pulling contact

Define the recipient and model

Two initially resting carts touch on a frictionless line. The rear cart has mass 2 kg and the front has selected mass M. A force F acts only on the rear cart. Take rightward, toward the front cart, as positive. The contact can push the carts apart but cannot pull them together: there is no string, glue or magnetic attachment.

Build the algebraic relationship

For a positive push that maintains contact, both carts share acceleration A=F/(2+M). The front cart receives contact N=MA to the right. The rear receives F to the right and N to the left, so F−N=2A. The equal-and-opposite contact pair acts on different recipients. It cancels when both carts are included in the system, but not on either individual free-body diagram.

Check the changed state

If F becomes negative, forcing a common acceleration would give a negative contact N. That result rejects the assumed contact branch; it is not a new attractive surface force. The rear cart instead accelerates F/2 leftward while the front remains at rest. Starting together from rest, their gap after one second is −F/4 metres. At F=0 both remain at rest and the contact force is zero.

Interpret evidence and limits

The center-of-mass acceleration remains total external force divided by total mass even after separation. Compute (2a_rear+M a_front)/(2+M)=F/(2+M). A valid system law does not mean each component shares that acceleration. The interaction diagram should show a recipient-specific arrow for each force; acceleration arrows belong in a separate kinematic annotation.

Draw each recipient separately

Only horizontal interactions are drawn here. On each cart, its own upward floor normal and downward weight balance vertically. The contact arrows form a reciprocal pair across the two diagrams; they do not cancel on either single cart. This diagram is the positive-push branch; in the separating branch both contact arrows are removed. The arrows are schematic and not a scale comparison of F and N.

Worked original packet

For M=3 kg and F=8 N, common acceleration is 1.6 m/s² and contact is 4.8 N. For the same M and F=−8 N, rear acceleration is −4 m/s², front acceleration is zero and the gap at 1 s is 2 m.

Bounded investigation

Compare F=8,0,−8 N at front mass 3 kg. Solve recipient equations and the pair equation independently. Retain the failed negative-contact candidate before replacing it with the separated branch.

Changed-case numerical practice

Try the changed case before feedback

A 2 kg rear cart pushes a 4 kg front cart with 12 N on a frictionless line. Find the contact force in newtons.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original prose, data and diagrams checked 30 October 2026 against OpenStax College Physics 2e. This is an algebra-based virtual evidence exercise, not a physical experiment, equipment certification, transport recommendation or independent mastery result. Save assumptions and a competing explanation with the course evidence; supervised practical requirements remain external.