---
title: "Minimum speed of ball in a box"
subject: "Engineering Physics"
chapter: 10
key: "semester_1/engineering_physics:football-box-minimum-speed"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/football-box-minimum-speed
site: BCT Engineering
---

# Minimum speed of ball in a box

Engineering Physics · Chapter 10: Photon and Matter Waves

- Asked in: 2071 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/football-box-minimum-speed
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-10

## Question

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.

## 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.

## Answer

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.

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the football treated?
   A. As a free particle
   B. As a particle in a 1D infinite box of length L
   C. As a hydrogen atom
   D. As a harmonic oscillator
   Answer: B. The walls act as impenetrable boundaries. (Key point: Model)

2. Which energy gives the minimum speed?
   A. n = 0, E = 0
   B. n = 2, E = h²/(2mL²)
   C. n = 1, E₁ = h²/(8mL²)
   D. n = 1, E = h²/(4mL²)
   Answer: C. n = 0 is not allowed, so the lowest level is n = 1. (Key point: Minimum energy)

3. Setting E₁ = ½mv² gives:
   A. v_min = h/(mL)
   B. v_min = h/(8mL)
   C. v_min = 2h/(mL)
   D. v_min = h/(2mL)
   Answer: D. v² = h²/(4m²L²). (Key point: Speed formula)

## Related in Engineering Physics

- [Compound pendulum centres and minimum period](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acompound-pendulum-centers-minimum) — Engineering Physics, chapter 1
- [Minimum lines to resolve sodium lines](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Agrating-minimum-lines-resolve-sodium) — Engineering Physics, chapter 4
- [Separation of two achromatic lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-equivalent-focal-fifty) — Engineering Physics, chapter 5
- [Cardinal points and equivalent focal length](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acardinal-points-equivalent-focal-length) — Engineering Physics, chapter 5
- [Equivalent focal length and principal points](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-ten-four-principal) — Engineering Physics, chapter 5
- [Drift speed of electrons, 20 A](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adrift-speed-copper-twenty-amp) — Engineering Physics, chapter 8, asked 2×

## Related in other subjects

- [Ball kicked, find launch speed](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Aball-kicked-speed) — Applied Mechanics (Dynamics), chapter 1
- [Plate towed between two parallel walls](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aplate-between-walls-force) — Fluid Mechanics, chapter 1
- [Minimum impervious floor length, 10000 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abarrage-min-floor-length-q10000) — Irrigation and Drainage Engineering, chapter 5, asked 2×
- [Minimum impervious floor length, 11000 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Abarrage-min-floor-length-q11000) — Irrigation and Drainage Engineering, chapter 5, asked 2×