Newton’s law of impact: Two bodies move in the same direction with approach velocities u1 and u2; after a 1D elastic collision their separation velocities are v1 and v2. What is the coefficient of restitution e?

Introduction / Context:

In one-dimensional collisions along the line of impact, Newton’s law of restitution connects the relative speeds before and after impact. The coefficient of restitution e ranges from 0 (perfectly plastic) to 1 (perfectly elastic) and quantifies how ‘‘bouncy’’ the collision is.

Given Data / Assumptions:

• Motion is collinear (same straight line).
• u1 and u2 are pre-impact velocities of approach (same direction so u1 > u2 for approach).
• v1 and v2 are post-impact velocities of separation (v2 > v1 if they separate).

Concept / Approach:

Definition: e = (relative speed of separation) / (relative speed of approach). With our sign convention (same direction along a line), the relative approach speed is u1 - u2 and the relative separation speed is v2 - v1.

Step-by-Step Solution:

Verification / Alternative check:

When e = 1, v2 - v1 = u1 - u2 (perfectly elastic). When e = 0, v2 = v1 (no rebound), matching physical intuition.

Why Other Options Are Wrong:

• Swapping sums for differences or reversing numerator/denominator breaks the standard definition.

Common Pitfalls:

• Sign confusion: always use relative speeds (magnitudes along the line), not algebraic velocities with arbitrary signs for this simple formula.

Final Answer:

e = (v2 - v1) / (u1 - u2)