Irodov Problem 1.205

Problem Statement

A proton (mass m) elastically scatters off a stationary proton. After collision, one proton moves at angle θ₁ and the other at θ₂ to the original direction. Show that θ₁ + θ₂ = 90°.

Given

Two equal masses m, elastic collision. One at rest initially.

Concepts & Formulas

For equal-mass elastic collisions in 2D: the velocity vectors of the two particles after collision are always perpendicular. This follows from conservation of both momentum and KE with m₁ = m₂.

Step-by-Step Solution

Step 1: Conservation: mv₀ = mv₁’ + mv₂’ (vectors), ½mv₀² = ½mv₁’² + ½mv₂’².
Step 2: v₀² = v₁’² + v₂’² + 2v₁’·v₂’ (from momentum squared) and v₀² = v₁’² + v₂’² (from KE).
Step 3: Therefore 2v₁’·v₂’ = 0 → v₁’ ⊥ v₂’ → θ₁ + θ₂ = 90°.

Worked Calculation

v₁’·v₂’ = 0 → θ₁ + θ₂ = 90°.

Boxed Answer

θ₁ + θ₂ = 90° (for equal-mass elastic collision)

Physical Interpretation

This is the famous 90° rule for equal-mass elastic scattering — billiard balls always part at right angles, which snooker players know intuitively.