Coefficient of Restitution

Definition

• Represented by the symbol e.

• Defined as the ratio of the difference of final velocities after a collision to the initial velocities before the collision:

  [ e = \frac{V_1' - V_2'}{V_2 - V_1} ]

Types of Collisions

1. Perfectly Elastic Collision (e = 1)

   • Kinetic energy is conserved.
   • No loss of kinetic energy.

2. Completely Inelastic Collision (e = 0)

   • Kinetic energy is not conserved.
   • Objects stick together after collision.

3. Inelastic Collision (0 < e < 1)

   • Kinetic energy is not conserved but some energy remains.
   • Objects do not stick together.

Practice Problems

Problem 1

• Scenario: A 5 kg block moving East at 8 m/s strikes a 10 kg block at rest.

Part A: Combined Final Velocity

• Initial momentum:
  • ( M_1 V_1 + M_2 V_2 = 5 imes 8 + 10 imes 0 = 40 ext{ kg m/s} )
• Final momentum:
  • ( (M_1 + M_2) V_f = 15 V_f )
• Set equal and solve:
  • ( 40 = 15 V_f )
  • ( V_f = \frac{40}{15} \approx 2.67 ext{ m/s} )

Part B: Coefficient of Restitution

• Since both blocks stick together, their final velocities are equal:
  • ( V_1' = V_2' = 2.67 ext{ m/s} )
• Coefficient of restitution calculation:
  • ( e = \frac{V_1' - V_2'}{V_2 - V_1} = \frac{2.67 - 2.67}{0 - 8} = 0 )
• Conclusion: Completely Inelastic Collision.

Problem 2

• Scenario: An 8 kg block moving East at 6 m/s strikes a 4 kg block moving East at 2 m/s. The 8 kg block continues East at 4 m/s.

Finding the Final Velocity of the 4 kg Block

• Use conservation of momentum:
  • ( M_1 V_1 + M_2 V_2 = M_1 V_1' + M_2 V_2' )
  • Solve:
  • ( 8 imes 6 + 4 imes 2 = 8 imes 4 + 4 V_2' )
  • ( 48 + 8 = 32 + 4 V_2' )
  • ( V_2' = \frac{24}{4} = 6 ext{ m/s} )

Coefficient of Restitution Calculation

• Calculate:
  • ( e = \frac{V_1' - V_2'}{V_2 - V_1} = \frac{4 - 6}{2 - 6} = \frac{-2}{-4} = 0.5 )
• Conclusion: Inelastic Collision.

Kinetic Energy Check

• Calculate before and after collision:
  • Before: ( 144 + 8 = 152 ext{ J} )
  • After: ( 64 + 72 = 136 ext{ J} )
• Energy loss: ( 152 - 136 = 16 ext{ J} ) confirms inelastic collision.

Problem 3

• Scenario: A 4 kg ball moving East at 5 m/s hits a 2 kg ball at rest. After the collision, the 4 kg ball moves East at 1.67 m/s and the 2 kg ball moves East at 6.67 m/s.

Coefficient of Restitution Calculation

• Calculate:
  • ( e = \frac{V_1' - V_2'}{V_2 - V_1} = \frac{1.67 - 6.67}{0 - 5} = \frac{-5}{-5} = 1 )
• Conclusion: Perfectly Elastic Collision.

Kinetic Energy Check

• Before collision:
  • ( 12 imes 4 imes 5^2 = 150 ext{ J} )
• After collision:
  • Check both:
    • 4 kg ball: ( 12 imes 4 imes 1.67^2 \approx 11.12 ext{ J} )
    • 2 kg ball: ( 12 imes 2 imes 6.67^2 \approx 80.10 ext{ J} )
• Total kinetic energy before and after is conserved, confirming elastic collision.