§ 1. Conservation of Linear Momentum (Grade 9)
Elastic & inelastic collision of spheres, momentum exchange. In a closed mechanical system, the vector sum of momenta of all bodies remains constant under any mutual interactions.
Theory Summary & Scientific Mechanism
In an isolated mechanical system where external forces cancel out (∑ Fₑxt = 0), Newton’s third law guarantees that internal impact forces between two colliding bodies are equal in magnitude and opposite in direction at every instant: F₁₂ = -F₂₁. Consequently, the total linear momentum p = m₁v₁ + m₂v₂ remains strictly conserved across all collision regimes, from perfectly elastic to completely plastic.
The degree of elasticity is governed by the coefficient of restitution e = (|v'₂ - v'₁|)/(|v₁ - v₂|) ∈ [0, 1]. Relative to the center-of-mass velocity v_C = (m₁ v₁ + m₂ v₂)/(m₁ + m₂), the post-impact velocities are v'₁ = v_C - e (m₂)/(m₁ + m₂)(v₁ - v₂) and v'₂ = v_C + e (m₁)/(m₁ + m₂)(v₁ - v₂). The dissipated mechanical energy converted into internal heat and plastic deformation is Q = (1)/(2)μ (v₁ - v₂)² (1 - e²), where μ = (m₁ m₂)/(m₁ + m₂) is the reduced mass.
By varying the mass ratio m₁/m₂ and restitution e, observe the two fundamental limits: for e = 1 and m₁ = m₂, the spheres completely exchange velocities (Q = 0), whereas for e = 0 they coalesce and move at the center-of-mass velocity v_C. This formalism supports ballistic pendulums, neutron moderation in nuclear reactors, and particle collider kinematics.
Everyday Analogy
Think of two billiard balls or train cars coupling on a track. Even if the spheres are soft like clay and warm up on impact, their combined forward push (linear momentum mv) never vanishes because no outside force pushed them!
Interactive Question & Derivation
A sphere of mass m₁ = 2 kg moving at velocity v₁ = 4 m/s undergoes a head-on perfectly elastic collision (e = 1) with an identical stationary sphere of mass m₂ = 2 kg. What will their velocities be after impact?
- A: Both spheres will move together at velocity u = 2 m/s
- B: The first sphere stops (v₁' = 0), and the second receives all its velocity (v₂' = 4 m/s)
- C: The first sphere rebounds backward at -4 m/s while the second remains stationary
Study Hint: Ask yourself: during impact, the spheres push only on each other (internal forces F₁₂ = -F₂₁). Can internal forces change total momentum, and where does lost kinetic energy go when e < 1?
Proof & Explanation: By conservation of momentum and mechanical energy, in a head-on perfectly elastic collision (e=1) of two spheres of equal mass (m₁ = m₂), a complete exchange of velocities occurs. When the collision is inelastic (e < 1), part of the kinetic energy is converted into internal thermal energy Δ Q = (1)/(2)μᵣed vᵣel² (1 - e²).
Key Equations & 3D Simulation Parameters
v₁' = ((m₁ - e m₂)v₁ + (1+e)m₂ v₂)/(m₁ + m₂), v₂' = ((m₂ - e m₁)v₂ + (1+e)m₁ v₁)/(m₁ + m₂)
- Mass of 1st sphere m₁: 0.5 … 6 kg (default: 2 kg)
- Mass of 2nd sphere m₂: 0.5 … 6 kg (default: 2 kg)
- Initial velocity v₁: 1 … 8 m/s (default: 4 m/s)
- Initial velocity v₂: -4 … 3 m/s (default: 0 m/s)
- Restitution coeff. e: 0 … 1 (default: 1)
- Impact offset Δz: -0.6 … 0.6 m (default: 0 m)