Newton’s law for collision of perfectly elastic bodies: The velocity of separation of the colliding bodies is related to their velocity of approach by which statement?

Introduction / Context:Collisions in one dimension are characterized by conservation laws and an impact law. Newton’s law of restitution introduces the coefficient of restitution e (0 ≤ e ≤ 1) connecting the relative speeds before and after impact. For perfectly elastic impacts, e = 1.

Given Data / Assumptions:

• Smooth, central impact (line of centers collinear with motion).
• Short impact duration; external impulses along line of motion negligible.
• Perfect elasticity implies no kinetic energy loss along line of impact.

Concept / Approach:Define relative speed of approach as (u1 − u2) along the line of centers before impact, and relative speed of separation as (v2 − v1) after impact (signs chosen consistently). Newton’s impact law states:(velocity of separation) = e * (velocity of approach).With e = 1 for perfectly elastic collisions.

Step-by-Step Solution:

Verification / Alternative check:For equal masses in a head-on elastic collision, the bodies exchange velocities, which satisfies e = 1 and momentum conservation.

Why Other Options Are Wrong:Direct or inverse proportionality without a defined constant is incomplete; the law specifies a constant ratio e (option C). Sum/difference formulations do not represent the restitution law.

Common Pitfalls:Confusing absolute with relative velocities; mixing sign conventions; applying restitution off the line of centers in oblique impact without resolving components.

Final Answer:Velocity of separation bears a constant ratio to velocity of approach (the coefficient of restitution)