Study Notes on Linear Momentum and Related Concepts

LESSON 7-1: INTRODUCTION TO LINEAR MOMENTUM

Definition of Momentum

  • Momentum is defined as mass in motion.

  • It is the product of mass and velocity of an object.

  • Momentum is a vector quantity whose direction is the same as that of velocity.

  • Mathematical representation:
    P = mV

    Where:

    • P = linear momentum (kg-m/s)
    • m = mass (kg)
    • V = velocity (m/s)

Momentum of a System of Objects

  • The momentum of a system of objects is the vector sum of the momenta of all the individual objects in the system:
    P = m1 V1 + m2 V2 + … + mn Vn

Example 1: Calculation of Linear Momentum

  1. 60 kg halfback moving eastward at 9 m/s:

    • P = (60 ext{ kg})(9 ext{ m/s}) = 540 ext{ kg-m/s}

  2. 1000 kg car moving northward at 20 m/s:

    • P = (1000 ext{ kg})(20 ext{ m/s}) = 20000 ext{ kg-m/s N}

  3. 40 kg freshman moving southward at 2 m/s:

    • P = (40 ext{ kg})(2 ext{ m/s}) = 80 ext{ kg-m/s}

Example 2: Change in Momentum Based on Velocity Changes

  • Given: A car possesses 20000 units of momentum.
  • What would be the car's new momentum if:
    1. Velocity was doubled:
      • P' = m(2V)
      • P' = 2(20000) = 40000 units of momentum.
    2. Velocity was tripled:
      • P' = 3(mV)
      • P' = 3(20000) = 60000 units of momentum.
    3. Mass was doubled:
      • P' = 2m(V)
      • P' = 2(20000) = 40000 units of momentum.
    4. Both mass and velocity were doubled:
      • P' = (2m)(2V)
      • P' = 4(20000) = 80000 units of momentum.
    5. Mass was halved and velocity doubled:
      • P' = (1/2 m)(2V)
      • P' = 20000 units of momentum (remains the same).

Example 3: Calculation of Momentum and Ratios

  1. Momentum of a 110 kg football player running at 8 m/s:

    • P = mv = (110)(8) = 880 ext{ kg-m/s}

  2. Ratio with a football of 0.410 kg at 25 m/s:

    • P_{ball} = mv = (0.410)(25) = 10.25 ext{ kg-m/s}
    • Ratio: ext{Ratio} = rac{P{player}}{P{ball}} = rac{880}{10.25} o 85.85

Impulse - Momentum Theorem

  • The impulse applied to an object will be equal to the change in its momentum:
    ext{Impulse} = ext{Change in Momentum} = m riangle v = F riangle t

Impulse

  • Linear Impulse (J) is defined as the product of force and time interval:
    J = F riangle t
  • It is also a vector quantity whose direction is the same as that of force.

Example 4: Calculation of Impulse

  • If a football halfback experiences a force of 800 N for 0.9 seconds to the north:
    J = (800)(0.9) = 720 ext{ N-s}

Example 5: Average Force from Impulse

  • For a 10 kg model rocket's engine delivering 6.0 N-s of impulse for 0.755 s:
    J = F riangle t
    ightarrow F = rac{J}{ riangle t} = rac{6.0}{0.755} = 7.94 ext{ N}

Example 6: Momentum of a Supersonic Bomber

  • Mass: 21000 kg, Velocity: 400 m/s.
    • P = mv = (21000)(400) = 8.4 imes 10^6 ext{ kg-m/s}
    • If the bomber comes to a halt:
    • Impulse is given by:
      J = m(v2 - v1) = (21000)(0-400) = -8.4 imes 10^6 ext{ N-s}

Newton's Second Law of Motion

  • A change in momentum occurs due to force and time:
    • F = rac{ riangle P}{ riangle t}

Applications of Impulse and Momentum

  • Airbags in Cars: Designed based on impulse-momentum principles. They decrease the force over a longer time, reducing injury in collisions.
  • Baseball: A ball struck with a bat varies in momentum change and impulse based on contact time. A longer contact time leads to greater impulse and ball travel distance.

Example 7: Batsman and Ball Impulse Calculation

  • A batsman hits a ball of mass 0.15 kg:
    • Before: V1 = -12 ext{ m/s}, After: V2 = 12 ext{ m/s}
    • Impulse:
      J = m(v2 - v1) = (0.15)(12 - (-12)) = 0.15 imes 24 = 3.6 ext{ N-s}

Additional Examples

  • A golfer hits a 45g ball at 40 m/s, with contact for 0.8 seconds: The average force applied is:
  • Calculate impulses and forces for varied scenarios based on dynamics of motion and interactions between different objects.

Application in Collisions and Conservation Principle

  • Principle of conservation of momentum states:
    • Total momentum before a collision equals total momentum after the collision (P before = P after).
  • Isolated system momentum remains constant in absence of external forces.

Example 15: Skater's Recoil Velocity Calculation

  • Woman's mass: 54 kg, Man's mass: 88 kg, Woman's velocity after push: 2.5 m/s. Calculate man's recoil velocity.
  • Use conservation of momentum:
    • Initial: 0 = m1 v1 + m2 v2, Solve for v_2.

Additional Views on Collisions

  1. Elastic Collisions: Both momentum and kinetic energy conserved.
  2. Inelastic Collisions: Momentum conserved; some kinetic energy transformed to other forms.
  3. Perfectly Inelastic: Objects stick together post-collision; maximum kinetic energy lost.

Coefficient of Restitution (e)

  • A measure of how much kinetic energy remains after a collision.
  • Defined as the ratio of relative velocities after and before collisions:
    • e = rac{V{after}}{V{before}}
    • Special cases:
    • Perfectly elastic: e = 1
    • Perfectly inelastic: e = 0

Application in Explosions

  • Explosions are scenarios of conservation laws applied under instantaneous momentums.

Lesson 7.2: Collisions and Laws of Conservation of Momentum

  • Principle verification through example applications and problem-solving techniques utilizing momentum conservation equations and impulse formulas.