Two-Dimensional Collisions & Explosions
- Conserve momentum independently along each axis in a two-dimensional collision
- Recombine component momenta into a resultant magnitude and direction
- Analyze an explosion as a collision run backward, starting from zero momentum
Momentum is a vector, so it conserves component by component
Conservation of momentum is not one equation in two dimensions — it is two independent equations, one for x and one for y. That independence is the whole method: set up the x-equation and the y-equation separately, solve each, then recombine at the end if the question asks for a magnitude and direction. Mixing the axes in a single equation is the fastest way to a wrong answer.
An explosion is a collision in reverse
When an object at rest breaks apart — a firework, a cannon firing, a person stepping off a raft — the total momentum before is zero, so the total after must also be zero. The fragments' momenta must cancel as vectors. For two fragments this means they fly apart along the same line in opposite directions with equal momentum magnitudes, so the lighter piece moves faster in exact proportion: m₁v₁ = m₂v₂.
A 2.0 kg puck moving east at 5.0 m/s strikes a 3.0 kg puck at rest. After the collision the 2.0 kg puck moves north at 2.0 m/s. Find the velocity of the 3.0 kg puck.
- 1.Take east as +x and north as +y. Initial momentum: pₓ = (2.0)(5.0) = 10 kg·m/s, pᵧ = 0.
- 2.After, the 2.0 kg puck carries pₓ = 0 and pᵧ = (2.0)(2.0) = 4.0 kg·m/s.
- 3.So the 3.0 kg puck must carry pₓ = 10 − 0 = 10 kg·m/s and pᵧ = 0 − 4.0 = −4.0 kg·m/s.
- 4.Its velocity components: vₓ = 10 ÷ 3.0 = 3.33 m/s, vᵧ = −4.0 ÷ 3.0 = −1.33 m/s.
- 5.Magnitude: √(3.33² + 1.33²) = √(11.1 + 1.8) = 3.6 m/s, directed tan⁻¹(1.33/3.33) = 22° south of east.
Why "the system" has to include everything that pushes
Momentum is conserved only when the net external force is zero. Choose your system so that the forces you care about are internal. In a collision between two carts, the contact forces are internal (a third-law pair) and cancel — so momentum is conserved even though each cart's momentum changes wildly. But if friction with the track is significant over the interval, that is an external force and momentum is not conserved. Collisions get away with ignoring friction because they are over so quickly that the friction impulse is negligible against the collision impulse.
Momentum conservation is per-axis, and a common trap is conserving the total speed instead. Speeds do not add — 3 m/s east and 4 m/s north combine to 5 m/s northeast, not 7 m/s. Always work in components and recombine only at the very end.
A 60 kg person stands on a stationary 120 kg raft on frictionless water and walks east at 2.0 m/s relative to the water. What happens to the raft?
A 1.0 kg object moving at 6.0 m/s east collides with a 1.0 kg object moving at 8.0 m/s north. They stick together. What is the magnitude of their combined momentum after the collision?
A stationary object explodes into exactly two fragments. Which must be true?
Draw and label a before/after sketch with your axes marked before writing anything. On two-dimensional momentum free-responses, the two component equations are separate rubric points, and setting them up correctly earns credit even if the arithmetic goes wrong.
Answer the 3 checkpoints as you read.
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