6.03i–6.03k - Restitution and direct impact

What restitution measures

An identical pair of smooth spheres gives a striking question: can a moving sphere hit a stationary one and stop dead, while the other leaves at exactly the incoming speed? Momentum permits it. Restitution tells you when it happens.

Imagine $A$ to the left of $B$, with right positive. Just before a direct impact, $A$ must be catching $B$, so $u_A-u_B>0$. After impact the pair must separate or have equal velocity, so $v_B-v_A\ge0$. Use these relative speeds, not the speed of one sphere in isolation.

The limiting values have particular names. A perfectly elastic collision has $e=1$ and no loss of kinetic energy. An inelastic collision in this terminology has $e=0$: in a direct two-body impact, the bodies coalesce and there is the maximum loss of kinetic energy compatible with momentum conservation. With $0<e<1$, the collision loses some kinetic energy and the bodies separate. These names, the coefficient range and the energy consequences are not supplied in the formula booklet.

For identical masses with initial velocities $u$ and $0$, conservation gives $v_A+v_B=u$. If $e=1$, restitution gives $v_B-v_A=u$; adding and subtracting gives $v_A=0$, $v_B=u$. This answers the opening question. If $e=0$, the common velocity is instead $u/2$. Initial kinetic energy is $\tfrac12 mu^2$, while final kinetic energy is $2(\tfrac12 m(u/2)^2)=\tfrac14 mu^2$: half the initial kinetic energy has become other forms of energy.

Why is the loss greatest at $e=0$ for any two positive masses? Let $M=m_A+m_B$, let $c$ be their conserved momentum divided by $M$, and let $w$ be their relative velocity. Expanding the two squares in the kinetic-energy expression gives

The first term stays fixed because momentum does. Restitution changes the relative velocity from $w$ to $-ew$, so the second term is multiplied by $e^2$. It vanishes at $e=0$, giving the smallest final kinetic energy and therefore the greatest loss. At $e=1$, it is unchanged. This identity is not in the formula booklet; you can establish energy changes directly from $\tfrac12 mv^2$, rather than memorise another collision formula.

Check the order of the two relative speeds in this short solution.

Two smooth spheres in direct impact

A direct impact has both centres and both velocities on the line of impact, so one signed coordinate is enough.

For smooth spheres the contact impulse acts along the line joining their centres. On a smooth horizontal surface, other horizontal impulses are absent. Write conservation of momentum and Newton's experimental law separately, then solve simultaneously. Avoid guessing which sphere reverses: signed unknowns let the equations decide.

Change only $B$'s initial velocity: it now moves towards $A$. Both equations change, but the sign convention does not.

These methods also find an unknown coefficient when a final velocity is supplied: use momentum to find the other final velocity, then take the separation-to-approach ratio.

The next pair has different masses, so it also checks that restitution is not a mass-weighted average.

A smooth fixed wall

A fixed wall has zero velocity before and after impact, so the restitution equation becomes especially short.

Choose a signed coordinate perpendicular to the wall. The sphere approaches with one sign and leaves with the other. The support supplies an external impulse to the sphere–wall system; conserve momentum for the sphere alone only if you include that impulse.