Combining <constant mass flux in a collisional cascade> with the rate in part (iv) gives
$$
\dot M_{\rm loss}(D)\propto
D^{6+p-\alpha+(1-\alpha)(3+b-2p)/3}.
$$
Set the exponent to zero and multiply by three:
$$
3(6+p-\alpha)+(1-\alpha)(3+b-2p)=0,
$$
or $21+b+p-\alpha(6+b-2p)=0$. Therefore the <velocity-dependent collisional cascade> has
$$
\boxed{\alpha=\frac{21+b+p}{6+b-2p}.}
$$
For $b=p=0$, this recovers the <Dohnanyi collisional cascade> slope $\alpha=7/2$. 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.

Solution

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

For <catastrophic disruption threshold> $Q_D^*=Q_bD^b$, encounter speed $v_{\rm rel}=v_pD^p$, fixed belt volume, and small impactors, the <catastrophic projectile diameter> scales as $D_{\rm cc}\propto D^{1+(b-2p)/3}$. If $\alpha>1$, small destructive impactors dominate the <catastrophic planetesimal collision rate>, giving
$$
R_{\rm cc}(D)\propto D^{2+p}D_{\rm cc}^{1-\alpha}.
$$
Mass per logarithmic bin scales as $D^{4-\alpha}$. <Constant mass flux in a collisional cascade> then gives
$$
\alpha=\frac{21+b+p}{6+b-2p}.
$$
Size-independent speed and strength recover the <Dohnanyi collisional cascade> slope $7/2$. A size-dependent belt thickness changes this derivation. The broader coupled size-and-velocity problem is studied by https://arxiv.org/abs/1111.0667[Pan and Schlichting].

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 Incoming links: Velocity-dependent collisional cascade