Recover center motion from an internal-force audit

Choose a fixed material boundary

For constant masses, xCM=(sum mi xi)/M. Differentiating gives M vCM=sum mi vi=P. Differentiating again gives M aCM=sum Fi. Reciprocal internal forces cancel in this last sum; only net external force changes total momentum. Internal interactions still change individual motions and can change relative kinetic energy.

Write both recipient equations

Our two point particles have masses 2 and 3 kg, initial positions−1 and 2 m and velocities 2 and−1 m/s. A prescribed internal interaction gives forces+3 and−3 N. Only particle 2 also receives external force A+bt. Thus a 1=3/2 and a 2=(A+bt−3)/3. No rigid separation or taut connector is imposed; equal acceleration is not assumed.

Integrate and then average

The individual positions are x 1=−1+2t+3t²/4 and x 2=2−t+(A−3)t²/6+bt³/18. Weighting them gives xCM=.8+.2t+At²/10+bt³/30. Alternatively, integrate aCM=(A+bt)/5 with xCM(0)=.8 and vCM(0)=.2. Agreement is an algebraic consistency check between two accounts of the same supplied model.

Test internal redistribution

For A=2,b=1,t=2, x 1=6 and x 2=−2/9 m. Their weighted center is 34/15 m. Total momentum is initial 1 plus external impulse 6, or 7 kg m/s. Different internal force choices could change both individual positions while leaving the same center history, provided the reciprocal forces and external history remain consistent.

Handle increasing carried mass explicitly

A conveyor holds collected material at constant horizontal speed V=3 m/s while receiving material with zero horizontal velocity at rate 2 kg/s. The drive must deliver horizontal momentum 6 N s each second, so its required horizontal force is 6 N. The stored carried momentum grows at V dm/dt, despite zero acceleration of the already-moving part. If incoming material instead already moves at 3 m/s, it carries that momentum into the boundary and no such horizontal transfer is needed. This changed-input case shows why an open-system inventory derivative alone is not a universal force rule.

Worked original case

A=2,b=1 gives external impulse 6 N s and total final momentum 7 kg m/s. Center velocity 7/5 m/s follows independently from the weighted particle velocities.

Investigation

Compare A=0,b=0 with A=2,b=1. Derive individual and center trajectories, retain both initial center conditions, and identify why internal forces cannot supply the external impulse. Analyze the two conveyor inflow velocities as a written boundary check.

Public numerical transfer practice

Try a new practice case

A 5 kg fixed-mass system receives external impulse 10 N s from initial center velocity−1 m/s. What is final center velocity?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original content checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax University Physics momentum. Packets are supplied/generated, not performed collisions or instrument measurements. No physical impact, release, explosive materials or moving machinery is requested. Public solutions and virtual captures do not certify independent mastery or supervised laboratory work.