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LESSON 11 / 14 · TOPIC 4.3

Momentum in two directions

You will be able to: Set up separate x and y momentum balances and reason about direction changes.

Free study resourceReview editionTeacher review pending

How do momentum components balance on a tabletop?

A puck moving east hits another puck on a smooth tabletop. Afterward, one moves partly north. If the initial northward momentum was zero and outside horizontal impulses are negligible, the other puck must carry a matching southward momentum component.

A useful starting point: Vector components →

Words and symbols before equations

Component
The signed part of a vector along a chosen axis.
x and y axes
Here +x is east and +y is north, both in the tabletop plane.
Angle θ
Measured counterclockwise from +x for the first outgoing momentum.
Sine/cosine
Ratios giving perpendicular/parallel components: p_y=p sinθ and p_x=p cosθ.
Add outgoing momentum vectors head to tail+x east+y northAB0246123Start: tail of AEnd: A + B = (6, 0)
Read this model snapshot. A=(1.5, 2.6); B=(4.5, -2.6) kg·m/s. x total=6; y total=0. Arrow locations construct a sum; they are not puck positions.
What this picture assumes

Initial total P=(6,0) kg·m/s; final A has magnitude 3 kg·m/s. Axis ticks show momentum components in kg·m/s. This is component accounting, not a trajectory or a kinetic-energy model.

Connect the picture to the physics

Draw the initial and final vectors in the same frame. Momentum is conserved separately along any axis with negligible net external impulse: P_xi=P_xf and P_yi=P_yf. Never add perpendicular magnitudes as if they were along one line.

Suppose initial total momentum is (6,0) kg·m/s and final momentum of A is (2,3). Then B must have (6−2,0−3)=(4,−3) kg·m/s. The y components cancel while the x components add to 6.

For an outgoing vector of magnitude q at angle θ, A contributes (q cosθ,q sinθ). B must make up the remaining components. Increasing θ at fixed q up to 90° reduces A’s eastward component and increases its northward component.

This lesson uses component setup and simple known-component subtraction. Full two-dimensional simultaneous collision solving is outside this AP Physics 1 treatment. Component balance alone also does not establish the kinetic-energy outcome.

A worked example, step by step

An initial total momentum of (8,0) kg·m/s becomes two outgoing momenta. A has components (3,4) kg·m/s. Find B’s components and explain the north–south balance.

  1. x balance: 8=3+p_Bx, so p_Bx=+5 kg·m/s.
  2. y balance: 0=4+p_By, so p_By=−4 kg·m/s.
  3. B points southeast: positive eastward component, negative northward component.
  4. The two y components cancel. Its full vector is (5,−4), not the scalar sum 5−4=1.
Common mix-up

Conserve each component separately. A zero total y momentum does not mean both outgoing y components are zero.

CHECK THE IDEA

If A gains positive y momentum, what must happen elsewhere in an isolated system initially with P_y=0?

Compare with an explanation

Other parts must gain an equal total negative y momentum.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Initial total momentum is (6,0). A’s final momentum magnitude is 3 kg·m/s. Rotate A between 0° and 90° and predict B’s components. The display is a vector constraint, not a complete collision trajectory or energy model.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Add outgoing momentum vectors head to tail+x east+y northAB0246123Start: tail of AEnd: A + B = (6, 0)

A=(1.5, 2.6); B=(4.5, -2.6) kg·m/s. x total=6; y total=0. Arrow locations construct a sum; they are not puck positions.

Components balance independently0+A: x1.5B: x4.5A: y2.6B: y-2.6Momentum component (kg·m/s) · same scale for every bar · full half-axis 4.5

Initial total P=(6,0) kg·m/s; final A has magnitude 3 kg·m/s. Axis ticks show momentum components in kg·m/s. This is component accounting, not a trajectory or a kinetic-energy model.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use a momentum or impulse 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.

1. Initial P=(5,0); final p_A=(2,1). Final p_B is…

Show answer and reasoning

(3,−1). Subtract component by component: (5−2,0−1).

2. Rotate a fixed outgoing momentum from east toward north. Its x component…

Show answer and reasoning

Decreases. p_x=p cosθ decreases as θ rises from 0° to 90°.

Original written challenge

4 points · self-check · not an official AP question

Initial P=(10,0) kg·m/s. Final p_A=(6,2) kg·m/s. (a) Write the x balance. (b) Write the y balance. (c) Find p_B. (d) Predict how p_By changes if p_Ay increases to +3 while initial P_y remains zero.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: 10=6+p_Bx.
  2. 1 point: 0=2+p_By.
  3. 1 point: p_B=(4,−2) kg·m/s.
  4. 1 point: p_By becomes −3 kg·m/s to preserve zero total y momentum.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1How many balances for a planar collision?

One for x and one for y.

RECALL 2Where does θ start in this lesson?

At +x, increasing counterclockwise.

RECALL 3Does a vector balance specify energy conservation?

No. Kinetic energy needs a separate check.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Momentum in two directions

  • P_xi=P_xf and P_yi=P_yf if external impulses in those axes are negligible.
  • p_x=p cosθ; p_y=p sinθ for θ from +x.

Remember: Conserve each component separately. A zero total y momentum does not mean both outgoing y components are zero.

Conditions: Initial total P=(6,0) kg·m/s; final A has magnitude 3 kg·m/s. Axis ticks show momentum components in kg·m/s. This is component accounting, not a trajectory or a kinetic-energy model.

Refresh Kid · 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. Fall-2026 corrections also checked. 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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