Construct separate component ledgers for a planar encounter
Momentum is a vector, so conservation applies separately to each component in a fixed inertial frame. A northward contribution cannot be added to an eastward contribution as though both were positive numbers on one axis. Draw components, sum them by direction, and only then calculate a resultant magnitude if needed.
Learning outcome: Compare inputs(3,4),(6,4) and(6,2) m/s. Sketch the component addition and predict direction changes before using output. Verify each component equation and distinguish a change in resultant direction from a change in total momentum magnitude.
Starting recap
Review signed velocity components, force-time areas, kinetic energy and system boundaries. Keep momentum vectors distinct from scalar energy.
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For u=3 m/s east and w=4 m/s north, the incoming momentum components are 6 and 12 kg m/s. Shared velocity components are 1.2 and 2.4 m/s; speed is√7.2≈2.68 m/s and direction about 63.4° north of east. Initial kinetic energy is 9+24=33 J; final translational energy is 18 J.
Open the experimentPredict
If only the eastward input speed doubles, should the final northward component also double?
Experiment
Compare inputs(3,4),(6,4) and(6,2) m/s. Sketch the component addition and predict direction changes before using output. Verify each component equation and distinguish a change in resultant direction from a change in total momentum magnitude.
Set up perpendicular momentum ledgers
Worked teaching model · synthetic values · Two objects moving along perpendicular axes meet and stick. No external impulse. Components are known separately, so no coupled unknown-angle equations are needed. This is a simple semiquantitative AP1 vector construction, not a general two-dimensional collision solver.
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Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.
| Quantity | Value | Unit |
|---|---|---|
| East momentum before and after | 6 | kg m/s |
| North momentum before and after | 12 | kg m/s |
| Shared east velocity | 1.2 | m/s |
| Shared north velocity | 2.4 | m/s |
| Shared speed | 2.68328 | m/s |
| Direction north of east | 63.43495 | degrees |
| Initial kinetic energy | 33 | J |
| Final kinetic energy | 18 | J |
Accessible plot coordinates
| east momentum (kg m/s) | north momentum (kg m/s) |
|---|---|
| 0 | 0 |
| 6 | 0 |
| 6 | 12 |
| east momentum (kg m/s) | north momentum (kg m/s) |
|---|---|
| 0 | 0 |
| 6 | 12 |
Conserve east and north components separately; do not add momentum magnitudes as scalars.
5vx=2u_east and 5vy=3u_north. Changing eastward input alone changes vx, not vy. Solid path adds the component arrows tip-to-tail; dashed segment is the resultant momentum.
Conserve east and north components separately; do not add momentum magnitudes as scalars.
5vx=2u_east and 5vy=3u_north. Changing eastward input alone changes vx, not vy. Solid path adds the component arrows tip-to-tail; dashed segment is the resultant momentum.
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If the objects did not stick, how would you write the two component conservation equations? Explain what additional information would be needed without trying to solve every possible final angle.
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