§ 5. Newton's Cradle (Grade 9)
Transmission of momentum and energy through a chain of steel spheres. Simultaneous satisfaction of momentum and mechanical energy conservation uniquely dictates not only the total momentum, but also the exact number of rebounding spheres.
Theory Summary & Scientific Mechanism
Newton’s cradle shows that conservation of momentum alone is insufficient to uniquely determine the outcome of a collision, both linear momentum and kinetic energy must be conserved simultaneously. When N identical spheres of mass m strike a stationary row at velocity v₀, a rapid elastic compression pulse (Hertzian contact wave) propagates through the chain.
Suppose k spheres were ejected at velocity u after N spheres struck at speed v₀. Enforcing momentum conservation Nmv₀ = kmu and kinetic energy conservation N(mv₀²)/(2) = k(mu²)/(2) and dividing the two equations immediately yields u = v₀ and therefore k = N. Exactly the same number of spheres must swing out on the far side!
When the restitution coefficient e < 1 (accounting for internal acoustic losses and imperfect steel elasticity), each impact dissipates a small fraction of mechanical energy into heat Q, gradually exciting small collective oscillations across all spheres.
Everyday Analogy
The famous desk toy! Pull back one steel ball on the left, and after the click, exactly one ball flies out on the right. Why not two balls at half speed? Because both momentum AND kinetic energy must be conserved simultaneously!
Interactive Question & Derivation
In a Newton's cradle of 5 identical steel spheres, N = 2 spheres on the left are pulled back and released together with impact speed v. Why cannot just 1 sphere fly out on the right with double speed 2v?
- A: Because that would violate conservation of linear momentum (2mv ≠ m(2v))
- B: Momentum would be conserved (2mv = m· 2v), but kinetic energy would double ((1)/(2)m(2v)² = 2mv² > mv²), violating energy conservation!
- C: Because the outermost sphere has greater string tension than the middle spheres
Study Hint: If one ball of speed v launched two balls at speed v/2, momentum would match (mv = 2m(v/2)), but calculate the kinetic energy (1)/(2)mv², would it match?
Proof & Explanation: If a single sphere rebounded at speed 2v, its momentum would be 2mv (momentum conserved), but its kinetic energy would be (1)/(2)m(2v)² = 2mv², whereas the initial kinetic energy of the two incoming spheres was 2·(1)/(2)mv² = mv². The unique solution to both conservation equations is that Nₒut = Nᵢn spheres rebound with speed u = v!
Key Equations & 3D Simulation Parameters
Nₒut = Nᵢn, u = v (for e = 1)
- Lifted spheres N: 1 … 3 pcs (default: 2 pcs)
- Release angle θ₀: 15 … 55 ° (default: 40 °)
- Steel restitution e: 0.7 … 1 (default: 0.99)
- Suspension length L: 2.4 … 4.2 m (default: 3.4 m)