Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 2 / ii

Table of contents
- Solution ii

Solution

The half-target mass assigned to equal fragments contains

N_{f} = \frac{\frac{1}{2} ( π ρ D _{t}^{3} /6 )}{π ρ D _{f}^{3} /6} = \frac{D _{t}^{3}}{2 D _{f}^{3}}

fragments. Their total geometric cross-section is consequently

σ_{tot, 2} = N_{f} \frac{π D _{f}^{2}}{4} = \frac{π}{8} \frac{D _{t}^{3}}{D _{f}} .

Ignoring the negligible single largest fragment in

σ_{tot, 1}

,

\frac{σ _{tot, 2}}{σ _{tot, 1}} ≃ \frac{α _{f} - 3}{4 - α _{f}} \frac{D _{m i n, f}^{α_{f} - 3} D _{m a x, f}^{4 - α_{f}}}{D _{f}} .

For example, if

D_{f} = D_{m i n, f}

, this ratio is

(α_{f} - 3) (D_{m a x, f} / D_{m i n, f})^{4 - α_{f}} / (4 - α_{f})

and is large for a broad size range. This reflects the inverse-size cross-section per unit mass of equal-density spherical fragments: concentrating the mass at small

D_{f}

creates more area.

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