 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-316/2/vi/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 316 / 2 / vi / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-05

Combining constant mass flux in a collisional cascade with the rate in part (iv) gives

\dot{M}_{loss} (D) \propto D^{6 + p - α + (1 - α) (3 + b - 2 p) /3} .

Set the exponent to zero and multiply by three:

3 (6 + p - α) + (1 - α) (3 + b - 2 p) = 0,

or

21 + b + p - α (6 + b - 2 p) = 0

. Therefore the velocity-dependent collisional cascade has

α = \frac{21 + b + p}{6 + b - 2 p} .

For

b = p = 0

, this recovers the Dohnanyi collisional cascade slope

α = 7/2

. The denominator must be nonzero, and the resulting slope and destructive cutoff must satisfy the assumptions used for the collision-rate integral. A size-dependent vertical belt thickness would supply an additional volume factor and change this result.

 Read the full article

 New to topics? Read the docs here!