Newton’s experimental law

The key idea. When two bodies collide, they do not always bounce apart at the same speed they came together. Newton's experimental law (NEL) captures exactly how much they bounce with a single number, the coefficient of restitution $e$:

$e = \frac{\text{speed of separation}}{\text{speed of approach}}$

The speed of approach is how fast the gap between the bodies is closing before impact; the speed of separation is how fast the gap is opening after impact. Because real bodies never bounce apart faster than they came together (energy would have to come from nowhere), $e$ always satisfies:

$0 \leq e \leq 1$

This formula is NOT in MF19 — you must memorise it. Examiners assume instant recall.

Two extreme cases you must know:

• $e = 1$ — perfectly elastic. The bodies separate exactly as fast as they approached. Kinetic energy is conserved.
• $e = 0$ — perfectly inelastic. There is no separation: the bodies move together (they coalesce) after impact.

Cambridge tip. $e$ is a ratio of speeds, so it is always positive. Build it from the relative speeds (gap-closing, gap-opening) and you will never get the sign wrong.

When a smooth sphere hits a fixed surface (a wall or the ground), the surface does not move, so there is no momentum equation to write. NEL alone gives the rebound.

Direct (head-on) impact. If the sphere strikes the wall along the normal with speed $u$, the speed of approach is $u$ and the speed of separation is the rebound speed $v$. Since the wall is fixed:

$e = \frac{v}{u} \quad\Rightarrow\quad v = eu$

The sphere bounces straight back with its speed reduced by the factor $e$.

Oblique impact — resolve into two components. When the sphere meets a smooth wall at an angle, split its velocity into:

• the normal component (perpendicular to the wall), and
• the tangential component (parallel to the wall).

Then apply two rules:

• Normal component: reversed in direction and multiplied by $e$ (this is NEL — the wall is the second "body").
• Tangential component: unchanged — the wall is smooth, so it exerts no force parallel to itself.

Putting it together. If the sphere arrives at angle $\alpha$ to the wall with speed $u$:

• tangential speed after $= u\sin\alpha$ (unchanged),
• normal speed after $= eu\cos\alpha$ (reversed),
• rebound speed $= \sqrt{(u\sin\alpha)^2 + (eu\cos\alpha)^2}$, at a new angle to the wall.

Cambridge tip. Draw the wall first, mark the normal, and resolve before touching any numbers. Mixing up which component reverses is the single biggest source of lost marks here.

Many questions hand you speeds before and after a collision and ask you to find $e$. The method is mechanical: identify the approach speed, identify the separation speed, divide.

Recovering $e$ from a wall bounce. A ball dropped from rest falls and hits the floor at speed $u$, then rebounds to speed $v$. Since the floor is fixed:

$e = \frac{v}{u}$

If you are given heights instead of speeds, use $v^2 = u^2 + 2as$ (or $g$-energy) to convert: a ball dropped from height $h_1$ strikes at $\sqrt{2gh_1}$ and, rebounding to height $h_2$, leaves at $\sqrt{2gh_2}$, so:

$e = \frac{\sqrt{2gh_2}}{\sqrt{2gh_1}} = \sqrt{\frac{h_2}{h_1}}$

Classifying the impact:

• $e = 1$ → perfectly elastic → KE conserved → bounce height equals drop height.
• $0 < e < 1$ → partially elastic → some KE lost as heat/sound → bounces lower each time.
• $e = 0$ → perfectly inelastic → no bounce → bodies coalesce / ball stays put.

Kinetic energy is generally lost. Unless $e = 1$, every real impact converts some KE into heat, sound and deformation. The KE loss is:

$\text{KE loss} = \text{KE before} - \text{KE after}$

For a ball of mass $m$ bouncing off a fixed wall, $\text{KE loss} = \tfrac12 m u^2 - \tfrac12 m v^2 = \tfrac12 m u^2 (1 - e^2)$, since $v = eu$. Notice the loss depends on $1 - e^2$, so it is always $\geq 0$.

Cambridge tip. "Perfectly elastic" is a technical phrase meaning $e = 1$ and KE conserved — not just "bouncy". If a question states an impact is perfectly elastic, write $e = 1$ immediately; if perfectly inelastic, write $e = 0$ and set the two final velocities equal.