Momentum and Collisions Review

Collision Types

  • Elastic Collision:
    • A collision where the total kinetic energy before the collision is conserved after the collision.
  • Inelastic Collision:
    • A collision where the total kinetic energy before the collision is not conserved after the collision.
  • Perfectly Inelastic Collision:
    • A collision where the total kinetic energy before the collision is not conserved after the collision, and the objects stick together.
  • Conservation of Momentum:
    • Momentum is conserved (meaning it does not change) for ALL collisions. (True)

Kinetic Energy of Collisions

  • Scenario:
    • Two dogs collide: a 4 kg dog and a 35 kg dog, both charging at 10 m/s.
  • Calculations:
    • Momentum of Each Dog Before Collision:
      • Dog 1 (4 kg): p<em>1=m</em>1∗v1=4 kg∗10 m/s=40 kg m/sp<em>1 = m</em>1 * v_1 = 4 \text{ kg} * 10 \text{ m/s} = 40 \text{ kg m/s}
      • Dog 2 (35 kg): p<em>2=m</em>2∗v2=35 kg∗(−10) m/s=−350 kg m/sp<em>2 = m</em>2 * v_2 = 35 \text{ kg} * (-10) \text{ m/s} = -350 \text{ kg m/s}
    • Kinetic Energy of Each Dog Before Collision:
      • Dog 1 (4 kg): KE<em>1=(1/2)∗m</em>1∗v12=0.5∗4 kg∗(10 m/s)2=200 JKE<em>1 = (1/2) * m</em>1 * v_1^2 = 0.5 * 4 \text{ kg} * (10 \text{ m/s})^2 = 200 \text{ J}
      • Dog 2 (35 kg): KE<em>2=(1/2)∗m</em>2∗v22=0.5∗35 kg∗(−10 m/s)2=1750 JKE<em>2 = (1/2) * m</em>2 * v_2^2 = 0.5 * 35 \text{ kg} * (-10 \text{ m/s})^2 = 1750 \text{ J}
    • Kinetic Energy of the Small Dog After Collision:
      • Small dog (4 kg) is "slingshotted" at 30 m/s in the opposite direction.
      • KE<em>1′=(1/2)∗m</em>1∗v1′2=0.5∗4 kg∗(30 m/s)2=1800 JKE<em>1' = (1/2) * m</em>1 * v_1'^2 = 0.5 * 4 \text{ kg} * (30 \text{ m/s})^2 = 1800 \text{ J}
    • Type of Collision:
      • Big dog stops, small dog rebounds. To determine the collision type, compare total kinetic energy before and after.
      • Before: KEtotal=200 J+1750 J=1950 JKE_{total} = 200 \text{ J} + 1750 \text{ J} = 1950 \text{ J}
      • After: KEtotal′=1800 J+0 J=1800 JKE_{total}' = 1800 \text{ J} + 0 \text{ J} = 1800 \text{ J}
      • Since kinetic energy is not conserved, this is an inelastic collision.

Momentum of Collisions

  • Scenario:
    • Two steel spheres collide: Sphere 1 (8 kg) at 5 m/s, Sphere 2 (12 kg) at 3 m/s. After the collision, Sphere 2 travels at 2 m/s, and Sphere 1's speed is unknown.
  • Conservation of Momentum:
    • m<em>1∗v</em>1+m<em>2∗v</em>2=m<em>1∗v</em>1′+m<em>2∗v</em>2′m<em>1 * v</em>1 + m<em>2 * v</em>2 = m<em>1 * v</em>1' + m<em>2 * v</em>2'
    • (8 kg∗5 m/s)+(12 kg∗3 m/s)=(8 kg∗v1′)+(12 kg∗2 m/s)(8 \text{ kg} * 5 \text{ m/s}) + (12 \text{ kg} * 3 \text{ m/s}) = (8 \text{ kg} * v_1') + (12 \text{ kg} * 2 \text{ m/s})
    • 40 kg m/s+36 kg m/s=8 kg∗v1′+24 kg m/s40 \text{ kg m/s} + 36 \text{ kg m/s} = 8 \text{ kg} * v_1' + 24 \text{ kg m/s}
    • 76 kg m/s=8 kg∗v1′+24 kg m/s76 \text{ kg m/s} = 8 \text{ kg} * v_1' + 24 \text{ kg m/s}
    • 52 kg m/s=8 kg∗v1′52 \text{ kg m/s} = 8 \text{ kg} * v_1'
    • v1′=6.5 m/sv_1' = 6.5 \text{ m/s}

Impulse and Force

  • Airbag Impact:
    • Airbags increase collision time, reducing force.
  • Impulse:
    • Impulse (J) is the change in momentum: J=F∗ΔtJ = F * \Delta t
    • Given impulse J=10 N sJ = 10 \text{ N s}
  • Calculations:
    • Without Airbag:
      • Δt=0.02 s\Delta t = 0.02 \text{ s}
      • F=J/Δt=(10 N s)/(0.02 s)=500 NF = J / \Delta t = (10 \text{ N s}) / (0.02 \text{ s}) = 500 \text{ N}
    • With Airbag:
      • Δt=0.1 s\Delta t = 0.1 \text{ s}
      • F=J/Δt=(10 N s)/(0.1 s)=100 NF = J / \Delta t = (10 \text{ N s}) / (0.1 \text{ s}) = 100 \text{ N}