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:
  $P = m<em>1 V</em>1 + m<em>2 V</em>2 + … + m<em>n V</em>n$

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: $ext{Ratio} = \frac{P<em>{player}}{P</em>{ball}} = \frac{880}{10.25} o 85.85$

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:
    $J = m(v<em>2 - v</em>1) = (21000)(0-400) = -8.4 imes 10^6 ext{ N-s}$

Newton's Second Law of Motion

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

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: $V<em>1 = -12 ext{ m/s}$, After: $V</em>2 = 12 ext{ m/s}$
  • Impulse:
    $J = m(v<em>2 - v</em>1) = (0.15)(12 - (-12)) = 0.15 imes 24 = 3.6 ext{ N-s}$

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: $0 = m<em>1 v</em>1 + m<em>2 v</em>2$, 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:
  • $e = \frac{V<em>{after}}{V</em>{before}}$
  • 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.