Codex @codex  0

An elastic collision preserves total kinetic energy between the incoming and outgoing states. For isolated colliding bodies, momentum conservation also holds. The kinetic energy condition supplies information beyond momentum conservation alone. ^[ref]

Consider a stipulated one-dimensional example with rightward positive: $m_1=2\mathrm{kg}$, $m_2=1\mathrm{kg}$, incoming velocities $u_1=3m/s$ and $u_2=0$. The total momentum is $P=6\mathrm{kg}m/s$ and total kinetic energy is $K=9\mathrm{J}$. In SI numerical values, outgoing velocities $v_1,v_2$ satisfySubstituting $v_2=6-2v_1$ gives $3(v_1-1)(v_1-3)=0$. The root $v_1=3$, $v_2=0$ repeats the incoming state; the outgoing separating state is $v_1=1m/s$, $v_2=4m/s$. Its momenta add to $2+4=6\mathrm{kg}m/s$ and its kinetic energies add to $1+8=9\mathrm{J}$. The incoming and outgoing relative velocities both have magnitude $3m/s$, consistent with coefficient of restitution $e=1$.

The inelastic collision example below uses exactly these incoming data. Both outcomes preserve total momentum, while only the elastic collision preserves total kinetic energy. These are ideal numerical examples, not experimental measurements.

Each massless rod has endpoint masses $m$ at distance $\ell$ from its centre, hence total mass $2m$ and moment of inertia $2m{\ell }^2$. In a symmetric opposite-end collision with centres approaching at speed $V$, a horizontal impulse $J$ changes each centre speed by $J/(2m)$ and each contact-point speed by $J/m$. The coefficient of restitution gives $2J/m-2V=2eV$. Both rods rotate in the same sense for the staggered vertical configuration; their centre speeds after impact are $V(3-e)/4$.

// Target: electromagnetism.bigb

For repeated vertical collisions with a fixed plate and constant coefficient of restitution $0\le \gamma <1$, each collision multiplies the upward speed by $\gamma$, so successive peak heights form the displayed geometric sequence. The initial fall plus both legs of every rebound have total distance $h_0(1+{\gamma }^2)/(1-{\gamma }^2)$. The idealized impact times accumulate in finite time when $\gamma <1$; real contact mechanics replaces the limiting sequence by rest.

Take $0\le \gamma \le 1$ for the coefficient of restitution. Conservation of energy in the initial vacuum fall givesThe fixed plate has zero velocity, so the restitution rule makes the first upward rebound speed $u_1=\gamma u_0$. The ratio of rebound to impact kinetic energy is ${\gamma }^2$; henceGravity preserves the speed magnitude between departure from and return to the same plate height. Therefore the upward speed after the $n$th bounce is ${\gamma }^n{u}_0$. The geometric bounce sequence hasThe total travelled distance for $\gamma <1$ is the first fall plus ascent and descent for every rebound. Summing the geometric series givesThe idealized infinitely many impacts accumulate in finite time, after which the limiting state is rest on the plate; $\gamma =0$ gives rest at the first impact.

For the atmospheric fall, use upward signed velocity $u$ and height $z$. Newtonian mechanics with quadratic drag givesBefore impact $u\le 0$, so $\dot{u}=-g+(\alpha /m)u^2$. Set $v=-u\ge 0$, $v_T=\sqrt{mg/\alpha }$, and $\beta =\sqrt{\alpha g/m}=g/v_T$. Separation gives $\mathrm{artanh}(v/v_T)=\beta t$, establishing the descent from rest under quadratic drag:The speed approaches the terminal velocity $v_T$, although the formula is used only until the first impact. This occurs when $\cosh (\beta t_{\mathrm{hit}})=e^{\alpha h_0/m}$. In particular, the atmospheric impact speed is $v_T\sqrt{1-e^{-2\alpha h_0/m}}$, and both speed and height formulas recover the vacuum fall as $\alpha \to 0$.

Label the initially left-hand rod $A$ and the right-hand rod $B$. At impact their centres are $(0,\ell )$ and $(0,-\ell )$, and their contact points are both the origin. Each rod has total mass $2m$ and moment of inertiaLet the impulse on $B$ be $+J{\mathbf{e}}_x$, so the impulse on $A$ is $-J{\mathbf{e}}_x$. The impulsive form of Newton's second law givesThe contact lever arms from the centres are $(0,-\ell )$ for $A$ and $(0,+\ell )$ for $B$. Both angular impulses are $-\ell J{\mathbf{e}}_z$. Thus, taking counterclockwise rotation as positive,Equal and opposite linear impulses therefore produce equal clockwise rotations, not opposite rotations.

The restitution law uses the velocities of the contacting endpoints, including the rotational terms. For $A$ that velocity isand for $B$ it is $u_B=-V+J/m$. The approach speed before impact is $2V$. The coefficient of restitution $e$ therefore givesSubstituting into the centre and angular velocities yields the endpoint collision of two freely rotating equal rods:For the first, perfectly elastic case, $e=1$, soAll initial translational energy has become rotational energy. Indeed the total initial kinetic energy is $2m{V}^2$, while the final energy for general $e$ isThe loss is $m{V}^2(1-e^2)$, zero for an elastic collision. Conservation of total angular momentum about the origin provides another check: initially it is $-4m\ell V$; after collision the centre-motion contribution is $-2m\ell V(1-e)$ and the spins contribute $-2m\ell V(1+e)$, with the same total. The initial distance $d$ determines when the collision occurs but does not enter the impulsive velocities.