Integrate momentum and energy over a moving distribution

Define what is being summed

For particles, total momentum is the vector sum of mi vi. For a continuous material distribution, replace each small mass by dm and integrate v dm. Density and velocity must describe the same instant and boundary. Momentum is frame dependent and signed by components; summing speed magnitudes would discard cancellation.

Declare a deformable instantaneous packet

The original line occupies 0≤x≤2 m with lambda(x)=2(1+beta x/2) kg/m, where the denominator is 2 m. Its velocity is v(x)=1+qx m/s with q in inverse seconds. A nonzero velocity gradient means different material elements move differently; the packet is not a rigid rod moving without deformation. The model calculates an instant, not its later density evolution.

Integrate with the same differential mass

Using dm=lambda dx, define M=integral dm, Q=integral x dm and R=integral x² dm. Here M=4+2beta kg, Q=4+8beta/3 kg m and R=16/3+4beta kg m². Momentum is P=uM+qQ with u=1 m/s. The center velocity equals P/M for this fixed material set; the geometric midpoint does not replace the mass weighting.

Compare two different kinetic inventories

Total kinetic energy is one-half integral (u+qx)² dm, giving (u²M+2uqQ+q²R)/2. Center translation contributes P²/(2M). Their difference is the nonnegative kinetic energy of motion relative to the center. It vanishes for uniform velocity, not merely for zero total momentum. This integral decomposition is an extended calculation supporting the core system-momentum concept.

Test cancellation with a changed case

At beta=0,q=−1, velocity varies from+1 to−1 across a uniform distribution. P=0 but K=2/3 J: oppositely moving material has not become stationary. At beta=1,q=.5, M=6 kg and P=28/3 kg m/s, so center velocity 14/9 m/s. A measured P alone cannot uniquely identify every local velocity or density.

Worked original case

Uniform beta=0,q=−1 gives zero total momentum and positive total K=2/3 J. The same system cannot be represented by one resting point if internal motion matters.

Investigation

Compare q=0 and−1 at beta=0, then beta=2. Reconstruct M,P and K from integrals; retain the center-relative energy and explain why this instantaneous packet cannot predict later density without further dynamics.

Public numerical transfer practice

Try a new practice case

Two masses 2 kg have velocities+3 and−3 m/s. What is their total momentum in kg m/s?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

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.