This is my impression. Maybe I have this wrong.
In the following, $\mathbb D^3$ is the free 3D linear space (or free module) over the dual numbers. We may use the terms infinitesimal part and standard part to refer to the two parts of a dual number. An element of this space is referred to as a screw.
A screw is a representation of a linear velocity and an angular velocity, together. Given a screw $\mathbf v \in \mathbb D^3$, the expression $\exp(t \mathbf v)$ represents the affine transformation which given a rigid body $b$, subjects it to a rigid body motion at constant linear+angular velocity $\mathbf v$ for $t$ units of time. More formally, if $\mathbb D^3$ is the algebra of screws, then there is a mapping $\exp : \mathbb D^3 \to SE(3)$, where $SE(3)$ is the special Euclidean group in 3 dimensions.
Above, the "algebra" $\mathbb D^3$ can either be understood as:
The Lie algebra $\mathfrak{se}_3$. In which case, $\exp$ is simply the usual exponential map sending a Lie algebra to a corresponding Lie group.
The six-dimensional linear subspace of the dual quaternions of the form $ai + bj + ck + a'\epsilon i + b'\epsilon j + c' \epsilon j$. The mapping $\exp$ is then surjective over the unit dual quaternions. The unit dual quaternions are isomorphic to $2SE(3)$, which is the double cover of $SE(3)$. The dual quaternions provide a useful formalism for representing rigid body motions, especially in computing applications, in spite of the unintuitive fact that they double-cover the desired group.
The logarithms of those $4 \times 4$ affine matrices which represent elements of $SE(3)$.
Now I think that if $\mathbf v$ is a velocity screw, and $M$ is a $3 \times 3$ dual number matrix with standard part a multiple of the identity matrix and infinitesimal part symmetric, then $\mathbf p = M \mathbf v$ is a momentum screw. This represents a linear and angular momentum together. A force screw can be defined as well (force and torque together). Then Newton's second and third law are the same as the usual ones, verbatim.
In order to go from the screw velocities $\mathbf v(t)$ at any time $t$, and the initial position+orientation of a rigid body $x(0)$ (as an element of $SE(3)$ and not a screw), to the position+orientation $x(t)$ of the body at time $t$, one would need to perform a kind of product integration (as opposed to the usual summation integral) using the idea that $x(t + dt) = \exp(\mathbf v(t)\, dt) x(t)$. If $x(t)$ is understood to be either a $4 \times 4$ affine matrix or a dual quaternion, then this becomes the fairly concrete differential equation $\frac{d}{dt}x(t) = \mathbf v(t) x(t)$. Can somebody confirm this?
Note: If $p(0)$ is a point on the rigid body at time $t=0$, then $p(t) = x(t) \cdot p(0)$ is the same point at time $t$, where we have used the notation for a group action to show that $x(t)$ (which is a member of either the group $SE(3)$ or $2SE(3)$) acts on $p(0)$ as a rigid body motion producing $p(t)$.
