Rigid-body collision

A rigid-body collision changes both translation and rotation through a short contact impulse. For a freely moving body of mass $M$, an impulse $\mathbf{J}$ applied at displacement ${\mathbf{r}}_c$ from its center of mass giveswhere $\mathbf{L}$ is angular momentum about the center of mass. An off-centre hit can therefore create spin even when the incoming body had none.

The relevant collision velocity is the relative contact-point velocity, including rotational motion. A normal restitution law and a tangential contact or friction law are needed to determine the impulse; conservation of total momentum alone does not specify the outcome. A ball struck through its centre and the same ball struck off-centre illustrate this distinction. The rigid approximation describes the before-and-after motion while replacing the short deformation process by an impact law.

An ideal elastic collision between a sphere and a stationary plane reverses normal approach velocity. On the fully rough, tangentially elastic branch, the tangential contact-point velocity reverses too. If $K_t$ is the signed tangential impulse, translation and spin obey $M\Delta v=K_t$ and $I\Delta \Omega =a{K}_t$, with positive spin anticlockwise. Restoring tangential translational plus rotational kinetic energy givesThe nonzero root is the displayed impulse; the other root is the smooth-contact branch. For $I=2M{a}^2/5$, the map is $v^{\prime }=(3v-4a\Omega )/7$ and $a{\Omega }^{\prime }=-(10v+3a\Omega )/7$. It preserves kinetic energy and the angular momentum about the contact point, $I\Omega -Mav$. An unspecified rough surface need not realize this idealization: finite friction or a dissipative tangential contact law requires additional constitutive data. Setting the outgoing contact slip to zero instead loses $M(v+a\Omega {)}^2/7$ of kinetic energy for this uniform sphere.

A short transverse impulse $J$ at lever arm $b$ gives a pivoted rigid body angular velocity $\Omega =Jb/I$ and contact-point velocity $b\Omega =J/m_{\mathrm{eff}}$, where $I$ is its pivot moment of inertia. Thus $m_{\mathrm{eff}}=I/b^2$ plays the role of a translational mass in the impact direction. A point projectile elastically striking an initially stationary body stops immediately after impact precisely when its mass equals this effective value. The pivot can exchange linear momentum while doing no work.

A normal impulse $j$ on a free planar rigid body of mass $M$ and moment of inertia $I$, at perpendicular lever arm $b$, changes the contact point's normal velocity by $j(1/M+b^2/I)$. Its effective mass in that direction is therefore $(1/M+b^2/I{)}^{-1}$. This follows by adding the translational change $j/M$ to the rotational contribution $b(jb/I)$.

For an incoming point mass $m$, the inverse reduced contact mass is $1/m+1/M+b^2/I$. A smooth elastic collision with initial relative approach speed $u$ has impulse $2u/(1/m+1/M+b^2/I)$, since the relative normal velocity reverses. This separates effective inertia at a contact from the body's total mass and explains the lever-arm dependence in an equal-speed elastic impact on a free rod.