Sometimes it is convenient to construct a system of units so that...

$\left[\dfrac{momentum}{Force}\right] = {X}^{p - r}$

$⇒ \left[\dfrac{mv}{ma}\right] = {X}^{p - r}$

$⇒ \left[\dfrac{v}{a}\right] = {X}^{p - r}$

but given that $\dfrac{v}{a} = {X}^{\beta - p}$

So,

$p - r = \beta - p ⇒ 2p = \beta + r . . . \left(1\right)$

\left[\dfrac{spee{d}^{2}}{acceleration}\right] = \left[position\right] ⇒ \dfrac{{X}^{2\beta}}{{X}^{p}} = {X}^{\alpha} ⇒ 2\beta - p = \alpha ⇒ p + \alpha = 2\beta . . . \left(2\right)$</p><p>and </p><p>\left[\dfrac{force}{linearmomentum}\right] = \left[\dfrac{acceleration}{speed}\right] ⇒ \left[\dfrac{{X}^{r}}{{X}^{q}}\right] = \left[\dfrac{{X}^{p}}{{X}^{\beta}}\right] ⇒ r - q = p - \beta ⇒ p + q - r = \beta . . . \left(3\right)$