3.6 Momentum
- Syllabus
- 9231–2028–2029
- Topic
- 3.6
- Level
- AS
Learning objectives
3.6.1Coefficient of restitution• recall Newton's experimental law and the definition of the coefficient of restitution, the property e01GG, and the meaning of the terms 'perfectly elastic' (e = 1) and 'inelastic' (e = 0)3.6.2Momentum and impacts• use conservation of linear momentum and/or Newton's experimental law to solve problems that may be modelled as the direct or oblique impact of two smooth spheres, or the direct or oblique impact of a smooth sphere with a fixed surface.
Restitution compares relative separation with relative approach
Newton′sexperimentallawactsalongthelineofimpact:e=\frac{\text{relative speed of separation}}{\text{relative speed of approach}},\qquad 0\le e\le1.If A approaches B with signed velocities $u_A>u_B$ and they separate with $v_B>v_A$, thenv_B-v_A=e(u_A-u_B).
| value of e | immediate normal behaviour | kinetic-energy statement |
|---|---|---|
| e=0 | no relative separation; equal normal velocities | perfectly inelastic in the syllabus terminology |
| 0<e<1 | separation speed is a fraction of approach speed | kinetic energy is generally lost |
| e=1 | separation and approach relative speeds are equal | perfectly elastic; with momentum, total KE is conserved |
Two particles have relative approach speed $5\text{ m s}^{-1}$ and relative separation speed $2\text{ m s}^{-1}$. Thereforee=\frac25=0.4.Thisdeterminesonlyonerelationbetweenthetwofinalvelocities;theirmassesandmomentumprovidetheotherrelationforanisolatedpair.
Restitution uses relative normal speeds, not the ratio of one particle's speed after and before. It does not require either particle to reverse direction, and kinetic energy is not generally conserved when e is less than one.
Smooth impacts change normal components but preserve tangential components
For a direct impact of masses $m$ and $2m$, suppose A approaches stationary B at speed $u$. Let their signed velocities after impact be $w$ and $v$. Momentum and restitution givemu=mw+2mv,\qquad v-w=eu.Hencev=\frac{(1+e)u}{3},\qquad w=\frac{(1-2e)u}{3}.The sign of $w$ decides whether A reverses.
| smooth impact | normal/line-of-impact component | tangential component |
|---|---|---|
| two spheres | conserve total normal momentum and apply restitution | unchanged for each sphere |
| sphere with fixed smooth surface | reverse and multiply the incident normal component's magnitude by e | unchanged |
For spheres, the line of impact is the line joining their centres at contact. For a fixed wall or plane, it is the surface normal. Resolve each incident velocity into normal and tangential components, apply the table, then reconstruct the final speed and direction. Include both components when calculating kinetic energy.
Do not conserve the momentum of a sphere by itself during impact with a fixed surface: the surface supplies an impulse. Do not apply restitution to the full oblique speed; only the relative component along the line of impact enters Newton's law.