Balance momentum separately along two axes
You will be able to: Use x and y conservation equations to find an unknown final velocity vector.
How can objects moving in different directions share one conserved momentum?
One object carries 3 kg·m/s east and another carries 4 kg·m/s north. Together their momentum is (3, 4) kg·m/s, with magnitude 5—not 7. The vector triangle keeps both directions visible.
A useful starting point: Recoil trades momentum between separating parts →
Words and symbols before equations
- Momentum component Pₓ or Pᵧ
- Signed total momentum along the selected x or y axis.
- Before and after vectors
- Each total is formed by adding all included object components.
- Direction angle θ
- An angle measured from a stated axis; signs determine the quadrant.
- External impulse vector
- Its x and y components separately determine whether each momentum component changes.
What this picture assumes
Two 1 kg particles: A initially at (4,0) m/s, B initially at rest. Net external impulse is zero. A chosen outgoing velocity determines B by momentum balance; an energy check flags outcomes that require energy release and cannot represent a passive collision.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- A: (2, 1) m/s; B: (2, -1) m/s. Total momentum stays (4,0) kg·m/s. K_f = 5 J. K is reduced: an inelastic outcome is possible energetically.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Write Pₓ,i = Pₓ,f and Pᵧ,i = Pᵧ,f only when the corresponding external impulses vanish. These are independent equations, but they describe one vector conservation statement.
For two outgoing objects with one known, p₂,f = P_i − p₁,f component by component. Divide by m₂ to get v₂,f. Magnitude and direction can then be recovered using a right triangle and the correct quadrant.
Momentum conservation alone does not classify a collision or determine every unknown velocity. Additional conditions, such as a measured outgoing direction, common sticking velocity, or conserved K in an elastic collision, may be needed. A result obeying momentum can still require an impossible energy increase for a passive collision.
A worked example, step by step
Two equal 1 kg pucks have initial total momentum (4, 0) kg·m/s. After their interaction, puck A has velocity (2, 1) m/s. Find puck B’s velocity and final total K. Assume negligible external impulse.
- Puck A’s momentum is (2, 1) kg·m/s because its mass is 1 kg.
- Puck B must have p_B = (4, 0) − (2, 1) = (2, −1) kg·m/s.
- Divide by 1 kg: v_B = (2, −1) m/s. Its speed is √5 = 2.24 m/s, directed below +x.
- Final K is ½(1)(2²+1²) + ½(1)(2²+1²) = 5 J. Initial total momentum alone does not specify initial K, so elasticity cannot be determined without initial velocities.
Conserve vector components, not the sum of speeds or momentum magnitudes.
Does conserving Pₓ ensure that Pᵧ is conserved?
Compare with an explanation
No. Each component needs its own balance and external-impulse check.
Predict. Change one thing. Explain.
The model begins with a 1 kg puck at (4,0) m/s striking an identical resting puck. Change A’s outgoing components. Predict B’s components, then check the energy label to see whether that outcome can describe a passive collision.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
A: (2, 1) m/s; B: (2, -1) m/s. Total momentum stays (4,0) kg·m/s. K_f = 5 J. K is reduced: an inelastic outcome is possible energetically.
Two 1 kg particles: A initially at (4,0) m/s, B initially at rest. Net external impulse is zero. A chosen outgoing velocity determines B by momentum balance; an energy check flags outcomes that require energy release and cannot represent a passive collision.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant impulse, system boundary, momentum or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAn isolated pair has initial total P = (6, 2) kg·m/s. A 2 kg object leaves at velocity (1, 2) m/s. Find the momentum, velocity and speed of the other 1 kg object.
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Compare with the answer and four-point rubric
- 1 point: First final momentum is 2(1,2) = (2,4) kg·m/s.
- 1 point: Second momentum is (6,2) − (2,4) = (4,−2) kg·m/s.
- 1 point: Second velocity is (4,−2) m/s.
- 1 point: Its speed is √(4²+2²) = √20 = 4.47 m/s; it travels in +x and −y.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1How many momentum equations are available in a plane?
Two independent scalar component equations.
RECALL 2Why can a momentum solution still need an energy check?
It may demand more final kinetic energy than a passive collision can supply.
RECALL 3What information is lost by adding momentum magnitudes?
The directions and cancellation between components.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Balance momentum separately along two axes
- Pₓ,f = Pₓ,i + Jₓ,ext; Pᵧ,f = Pᵧ,i + Jᵧ,ext.
- For an isolated pair: p₂,f = P_i − p₁,f.
- Speed = √(vₓ² + vᵧ²); keep component signs for direction.
Remember: Conserve vector components, not the sum of speeds or momentum magnitudes.
Conditions: Two 1 kg particles: A initially at (4,0) m/s, B initially at rest. Net external impulse is zero. A chosen outgoing velocity determines B by momentum balance; an energy check flags outcomes that require energy release and cannot represent a passive collision.
Refresh Kid · AP Physics C: Mechanics Unit 4 (official Unit 4) · Objectives 4.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 4.3, objectives 4.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 4: Linear Momentum. The unit covers Topics 4.1–4.4. Calculus connects force to momentum derivatives and impulse integrals. Collision calculations use one or two dimensions; the optional spatial fragment diagram is qualitative. Changing-mass examples explicitly account for momentum carried across a boundary, so dp/dt is not used blindly for an open system. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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