Integrate a vector force and differentiate the resulting momentum

Use the momentum form of Newton’s law

For a fixed material system, the derivative of total momentum equals the net external force. Integrating between two times gives the net impulse and momentum change. The initial momentum is an independent condition. A component force can be zero at one instant while momentum remains nonzero because previous forces have accumulated effects.

Integrate each component

For the 2 kg point object, Fx=A t(2−t), Fy=−B over 0 to 2 s; the time constant 2 carries seconds. Initial velocity(−1,1) gives momentum(−2,2). Horizontal impulse is A[t²−t³/3] from 0 to 2, or 4A/3. Vertical impulse is−2B. Add these components to the initial momentum before dividing by mass.

Recover the local relation

The integrated functions are px=−2+A(t²−t³/3) and py=2−Bt. Their derivatives reproduce Fx and Fy. A rising momentum graph has positive force, a flat segment zero force and a falling segment negative force in that coordinate. Average force over the full interval is impulse divided by 2 s, not generally peak or endpoint force.

Check direction and differentials

For A=3,B=1, impulse(4,−2) changes momentum to(2,0), giving final velocity(1,0). The final direction does not equal the direction of every earlier force. For a variable-direction force history, impulse is the vector integral and need not point along the instantaneous force at an arbitrary chosen time.

Avoid an open-boundary shortcut

If mass crosses a boundary, the momentum carried by that material must be included. At constant velocity the product derivative d(mv)/dt=v dm/dt is not zero, but identifying that inventory derivative with external force requires accounting for incoming and outgoing momentum. A force-free stream can enter and leave a region while its stored inventory stays constant. State whether the system follows fixed material or is an open control region.

Worked original case

A=3,B=1 gives J=(4,−2) N s and final v=(1,0) m/s. Both horizontal pulse endpoints have Fx=0, yet its total impulse is 4 N s.

Investigation

Compare A=1 and 4 at B=1. Integrate before opening feedback; sketch momentum slopes and force areas. Then change B and explain how final momentum direction differs from the horizontal pulse direction.

Public numerical transfer practice

Try a new practice case

A constant-mass 2 kg object receives horizontal impulse 6 N s from initial vx=−1 m/s. What is final vx in m/s?

Numerical result in the requested units Check this practice result

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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.