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Minimum speed of ball in a box

Short answer

Treating as a particle in a box, minimum speed v_min = h/(2mL). With m=0.5 kg, L=100 m, v_min = 6.626×10⁻³⁴/(2×0.5×100) ≈ 6.6×10⁻³⁶ m/s.

A football of 500 gm is confined between two impenetrable walls of Dasharath Rangashala that can be treated as a box of length 100m. Calculate the minimum speed of the ball.

Treat the ball as a particle in a one-dimensional box of length L. The allowed energies are Eₙ = n²h²/(8mL²). The minimum energy is for n = 1:

E₁ = h²/(8mL²) = ½ m v² → v_min = h/(2mL)

Given: m = 500 g = 0.5 kg, L = 100 m, h = 6.626×10⁻³⁴ J·s.

E₁ = (6.626×10⁻³⁴)² / (8 × 0.5 × 100²)
   = 4.39×10⁻⁶⁷ / 4×10⁴
   = 1.10×10⁻⁷¹ J

v_min = h / (2 m L)
      = 6.626×10⁻³⁴ / (2 × 0.5 × 100)
      = 6.63×10⁻³⁶ m/s

Answer: v_min ≈ 6.6×10⁻³⁶ m/s.

This speed is far too small to detect, so the ball behaves classically and appears to be at rest. Quantisation is not noticeable for large masses in large boxes.

✓ Answer checked by Claude Opus 5.5 (high effort)

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