← Mechanical Systems

4 Linear momentum & collisions

Momentum usually rides alongside circular motion in Question One rather than owning its own question — do not let it feel optional because of that. Momentum p = m v A VECTOR — direction (sign) matters as much as size. Conservation (no external force) Σp<sub>before</sub> = Σp<sub>after</sub> True for every collision or explosion, elastic or not.

Elastic, inelastic, perfectly inelastic

  • Momentum is conserved in ALL three cases — always apply it first.
  • Elastic: kinetic energy is also conserved (rare in real collisions).
  • Inelastic: kinetic energy decreases (converted to heat, sound, deformation).
  • Perfectly inelastic: the objects stick together and move with one common velocity — the maximum possible KE loss consistent with momentum conservation.
Worked example A head-on collision (1-D) A 3.00 kg trolley moving at 5.00 m s⁻¹ collides head-on with a 2.00 kg trolley moving at 9.00 m s⁻¹ in the opposite direction. They stick together. Find their combined velocity. Take the first trolley's direction as positive: p_before = (3.00 × 5.00) + (2.00 × −9.00) = 15.0 − 18.0 = −3.00 kg m s⁻¹. Perfectly inelastic: combined mass = 3.00 + 2.00 = 5.00 kg. p_after = p_before, so v = −3.00 / 5.00 = −0.600 m s⁻¹. ✓ v = 0.600 m s⁻¹, in the direction the 2.00 kg trolley was originally travelling: the lighter trolley had the larger momentum. Worked example A collision at right angles (2-D) A 1.20 kg object moving east at 4.00 m s⁻¹ collides and sticks to a 0.800 kg object moving north at 3.00 m s⁻¹. Find the size and direction of their combined velocity. Momentum is conserved in EACH direction separately. East: p_x = 1.20 × 4.00 = 4.80 kg m s⁻¹. North: p_y = 0.800 × 3.00 = 2.40 kg m s⁻¹. Combined mass = 1.20 + 0.800 = 2.00 kg. The two components are at right angles, so the resultant momentum is Pythagorean: p = √(4.80² + 2.40²) = √(23.04 + 5.76) = √28.8 = 5.37 kg m s⁻¹. v = p/m = 5.37 / 2.00 = 2.68 m s⁻¹. Direction: θ = tan⁻¹(2.40/4.80) = 26.6° north of east. ✓ v ≈ 2.68 m s⁻¹, 26.6° north of east.

Centre of mass of a two-body system

  • For two masses m₁ and m₂ a distance r apart, the centre of mass sits a distance d = m₂r / (m₁ + m₂) from the centre of m₁.
  • The centre of mass sits closer to whichever mass is LARGER.
  • If m₂ ≪ m₁ (e.g. a moon orbiting a planet), d is very small — the centre of mass sits almost exactly at the centre of the larger body.
Worked example Centre of mass — "show that" Saturn (5.68×10²⁶ kg) and its moon Titan (1.35×10²³ kg) are 1.22×10⁹ m apart. Show that their centre of mass is 2.90×10⁵ m from the centre of Saturn. d = m_Titan × r / (m_Saturn + m_Titan). d = (1.35×10²³ × 1.22×10⁹) / (5.68×10²⁶ + 1.35×10²³). d = (1.647×10³²) / (5.680135×10²⁶) = 2.899×10⁵ m. ✓ d ≈ 2.90×10⁵ m from Saturn's centre.

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