Law of Conservation of Momentum

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🔹 Conservation of Momentum

Conservation of Momentum: The total momentum of an isolated system remains constant if no external force acts on it.

Mathematical Statement:
Total momentum before collision = Total momentum after collision
p₁ + p₂ = p₁’ + p₂’

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Requirements:

  1. Isolated system: No external forces acting
  2. Internal forces only: Forces between objects in the system
  3. Newton’s Third Law: Action-reaction pairs are internal

Examples of Isolated Systems:

  • Two colliding balls on a smooth surface
  • Gun and bullet system during firing
  • Rocket and expelled gases in space
rocket

1. Elastic Collision:

  • Momentum and kinetic energy are conserved
  • Objects bounce back after collision
  • Example: Collision between two steel balls

2. Inelastic Collision:

  • Only momentum conserved, kinetic energy not conserved
  • Objects may stick or deform
  • Example: Car crash, clay balls sticking together

3. Perfectly Inelastic Collision:

  • Objects stick together after collision
  • Maximum loss of kinetic energy
  • Example: Bullet embedding in a wood block
collision

Gun Recoil:

  • Initially: Gun and bullet at rest (Total momentum = 0)
  • After firing: Bullet forward, gun backward
  • Formula: m₁v₁ + m₂v₂ = 0

Rocket Propulsion:

  • Gases expelled downward at high speed
  • Rocket gains upward momentum
  • Works even in vacuum due to conservation of momentum

Walking and Swimming:

  • Person pushes ground or water backward
  • Body moves forward with equal momentum
pool
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Head-on Collision:

  • Two objects move toward each other
  • After collision: velocities change but total momentum same
  • Formula: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Explosion Problems:

  • Initially at rest; after explosion, parts move in opposite directions
  • Total momentum remains zero

Basic Sign Convention Rules:

  • Choose a positive direction (e.g., right or forward)
  • Opposite directions are taken as negative

Fundamental Principle:

  • One of the core conservation laws in physics
  • Applies from atoms to galaxies
  • Predicts motion outcomes in collisions and explosions
explosion

Engineering Applications:

  • Vehicle safety (airbags, crumple zones)
  • Sports gear and design
  • Spacecraft propulsion and maneuvering
  • Industrial machine analysis (impact, vibration)
  • Conservation of Momentum: p₁ + p₂ = p₁’ + p₂’
  • Gun Recoil: m₁v₁ + m₂v₂ = 0

Solution:

  • Initial momentum = 0
  • Final momentum = 4 × v + 0.02 × 400 = 0
  • Recoil velocity = -2 m/s (backward)

Solution:

  • Initial momentum = 2 × 5 + 3 × 2 = 16 kg⋅m/s
  • Final momentum = (2 + 3) × v = 5v
  • Final velocity = 16/5 = 3.2 m/s

Solution:

  • Taking right as positive direction: Initial: m₁ = 5 kg, u₁ = +8 m/s; m₂ = 3 kg, u₂ = -6 m/s Final: m₁ = 5 kg, v₁ = -2 m/s; m₂ = 3 kg, v₂ = ?
  • Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ 5(8) + 3(-6) = 5(-2) + 3v₂ 40 – 18 = -10 + 3v₂ 22 = -10 + 3v₂ 3v₂ = 32 v₂ = 10.67 m/s (to the right)

Solution:

Taking car’s direction as positive: Initial: m₁ = 1000 kg, u₁ = +20 m/s; m₂ = 1500 kg, u₂ = -15 m/s Final: Combined mass = 2500 kg, v = ?

Conservation of momentum: 1000(20) + 1500(-15) = 2500v 20000 – 22500 = 2500v -2500 = 2500v v = -1 m/s

The vehicles move at 1 m/s in the truck’s original direction.