Impact mechanics – kinetic energy loss in a perfectly elastic collision In a perfectly elastic impact, the loss of kinetic energy is:

Difficulty: Easy

Correct Answer: Agree – zero (no loss of kinetic energy)

Explanation:

Introduction / Context:‘‘Elastic’’ and ‘‘inelastic’’ classify collisions based on kinetic energy behavior. In ideal elastic impacts, kinetic energy is conserved; in inelastic impacts, some kinetic energy transforms into other forms.

Given Data / Assumptions:

  • Coefficient of restitution e = 1 for a perfectly elastic impact.
  • System is isolated along the line of impact; no external impulses.
  • No deformation losses, heat, or sound energy losses in the idealization.

Concept / Approach:For e = 1, both linear momentum (along the line of impact) and kinetic energy are conserved. Thus the total kinetic energy before equals that after impact.

Step-by-Step Solution:

State conservation: Σp_before = Σp_after and ΣKE_before = ΣKE_after.With e = 1, relative velocity of separation equals relative velocity of approach, a hallmark of elastic behavior.Therefore, loss of kinetic energy = 0.

Verification / Alternative check:Classic 1D two-body formulas for elastic collision show KE_before − KE_after = 0; numerical examples with equal masses show a perfect velocity swap and identical total KE.

Why Other Options Are Wrong:

  • Saying some energy is always lost contradicts the ideal definition of ‘‘perfectly elastic’’.
  • ‘‘Depends only on masses’’ is incorrect; for e = 1, KE is conserved irrespective of specific masses.
  • ‘‘Cannot be determined’’ is false; it is determined by the definition of elastic impact.

Common Pitfalls:Confusing real collisions (often slightly inelastic) with the idealized, perfectly elastic case used in theory.

Final Answer:Agree – zero loss of kinetic energy

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