Two-Dimensional Collisions

You’ll be able to

Apply conservation of momentum independently to perpendicular components
Solve perfectly inelastic collisions in two dimensions
Use kinetic energy to classify a collision and check a solution for consistency

Momentum conserves component by component

Momentum is a vector, and vector conservation means each component is conserved separately and independently. A two-dimensional collision therefore gives you two equations — Σp_x before = Σp_x after, and Σp_y before = Σp_y after — which you solve simultaneously. There is no need for a new principle: the entire content of two-dimensional collisions is resolving each velocity into components before you start and recombining at the end.

Two-dimensional momentum conservation

Σm v_x (before) = Σm v_x (after) Σm v_y (before) = Σm v_y (after)

Choose axes to make the algebra easy — usually along the initial velocity of one object. A well-chosen axis often makes one of the two initial components zero.

Perfectly inelastic collisions are the tractable case

When two objects stick together, the two unknown final velocity components collapse into one unknown vector shared by the combined mass. Two equations then determine it completely, with no additional assumption required. Any other two-dimensional collision has four unknown final components and only two momentum equations, so it cannot be solved without extra information — typically a stated final direction, or the assumption that the collision is elastic, which supplies a third equation through kinetic energy.

Kinetic energy as a consistency check

Momentum is conserved in every collision; kinetic energy is not. Elastic collisions conserve K exactly; inelastic collisions lose some to deformation and heat; perfectly inelastic collisions lose the maximum consistent with momentum conservation. This gives a free and highly valuable check on any answer: compute K before and after, and if your final kinetic energy exceeds the initial value, your solution is physically impossible and contains an arithmetic error. A collision cannot manufacture energy.

Worked example

A 2.0 kg puck moving east at 3.0 m/s collides with a 3.0 kg puck moving north at 4.0 m/s. They stick together. Find the velocity of the combined mass and the kinetic energy lost.

1.x-momentum: p_x = (2.0)(3.0) + 0 = 6.0 kg·m/s. y-momentum: p_y = 0 + (3.0)(4.0) = 12.0 kg·m/s.
2.Combined mass is 5.0 kg, so v_x = 6.0/5.0 = 1.2 m/s and v_y = 12.0/5.0 = 2.4 m/s.
3.Speed: v = √(1.2² + 2.4²) = √(1.44 + 5.76) = √7.20 ≈ 2.7 m/s, at arctan(2.4/1.2) ≈ 63° north of east.
4.Energy: K_i = ½(2.0)(3.0)² + ½(3.0)(4.0)² = 9.0 + 24.0 = 33.0 J; K_f = ½(5.0)(7.20) = 18.0 J; lost = 15.0 J.

Answer: The pair moves at about 2.7 m/s, 63° north of east, having lost 15.0 J of the original 33.0 J to deformation and heat

Checkpoint

You solve a two-dimensional collision and find the total kinetic energy after the collision is greater than before. This means —

On the exam

Set up a small table with columns for x and y and rows for before and after. Filling it in before doing any algebra prevents the dominant error on these problems, which is not conceptual but bookkeeping — a dropped component or a sign flipped on one object.

Checkpoint

A two-dimensional collision in which the objects do NOT stick together generally cannot be solved from momentum conservation alone because —

Answer the 2 checkpoints as you read.