Questions: Collision Analysis and Coefficient of Restitution

Question 1

Two objects of known mass collide. You know all pre-collision velocities. What additional information is needed to determine both post-collision velocities?

• A. Nothing — conservation of momentum alone fully determines both post-collision velocities
• B. Conservation of kinetic energy — this provides the second equation needed
• C. The coefficient of restitution — this provides the second equation linking relative velocities
• D. The contact impulse force — once you know the contact force, post-collision velocities are determined

Answer: (C)

Momentum conservation gives one equation (m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂') with two unknowns (v₁' and v₂') — the system is underdetermined. The coefficient of restitution e = −(v₂' − v₁')/(v₂ − v₁) provides the second equation. Option B is incorrect: kinetic energy is only conserved in perfectly elastic collisions (e = 1), so it cannot be assumed in general. Option D is circular — impulse is typically what you derive from post-collision velocities, not a given input.

Question 2

A 4 kg ball moving at 5 m/s strikes a stationary 4 kg ball with e = 0. What are the post-collision velocities?

• A. v₁' = 5 m/s, v₂' = 0 m/s — the first ball passes through unchanged
• B. v₁' = 0 m/s, v₂' = 5 m/s — the first ball stops and all momentum transfers
• C. v₁' = 2.5 m/s, v₂' = 2.5 m/s — both balls move at half the original speed
• D. v₁' = −5 m/s, v₂' = 5 m/s — the first ball bounces back with equal speed

Answer: (C)

With e = 0, the restitution equation gives v₂' − v₁' = 0, so both objects have the same post-collision velocity (perfectly inelastic). Momentum conservation: 4(5) + 4(0) = (4 + 4)v', giving v' = 2.5 m/s for both. Option B (v₁' = 0, v₂' = 5) is the result for e = 1 (elastic) with equal masses — a common but incorrect answer for e = 0. For e = 0, objects stick together; they do not undergo a clean exchange of velocities.

Question 3

Momentum is conserved in most collisions, so kinetic energy is expected to also be conserved in most collisions.

• T. True
• F. False

Answer: False

Momentum and kinetic energy are governed by different principles. Momentum conservation follows from Newton's third law and holds for all collisions (any value of e), requiring only that no net external force acts during impact. Kinetic energy conservation requires e = 1 (perfectly elastic) — energy stored as deformation is fully recovered. For e < 1, kinetic energy is converted to heat, sound, and permanent deformation. A clay-ball collision conserves momentum exactly while dissipating essentially all kinetic energy. They are independent conditions.

Question 4

In a perfectly inelastic collision (e = 0), the two objects always stick together and move with a single common post-collision velocity.

• T. True
• F. False

Answer: True

e = 0 means the relative separation velocity after the collision is zero: v₂' − v₁' = 0. If the relative velocity is zero, both objects have the same post-collision velocity — they move together as a single unit. The common velocity is found from momentum conservation alone: v' = (m₁v₁ + m₂v₂)/(m₁ + m₂). This is the formal definition of a perfectly inelastic collision: maximum kinetic energy loss consistent with momentum conservation.

Question 5

Why is conservation of momentum alone insufficient to solve a two-body collision, and what role does the coefficient of restitution play?

Model answer: Momentum conservation gives one equation with two unknowns (v₁' and v₂'), leaving infinitely many valid solutions. The coefficient of restitution provides the second equation by characterizing how bouncy the collision is: e = −(v₂' − v₁')/(v₂ − v₁) specifies that the relative separation speed is a fixed fraction e of the relative approach speed. Together, the two equations form a 2×2 linear system that uniquely determines both post-collision velocities.

This underdetermination is fundamental, not incidental: the same two objects with the same initial velocities could in principle collide with any e from 0 to 1, and each would conserve momentum while producing different outcomes. The coefficient of restitution is a material property — it encodes how much of the approach kinetic energy is stored as elastic deformation and recovered versus dissipated. Engineers exploit this: crash barriers target low e to absorb energy, billiard balls target high e to preserve speed, and vehicle crumple zones are designed with specific e values to control energy transfer to occupants.