Coefficient of restitution and collisionsAQA A-Level Further Maths: Revision notes

For two bodies in a direct collision (along the line of motion), Newton's experimental law says $\text{speed of separation}=e\times\text{speed of approach},$ where $e$ is the coefficient of restitution, a dimensionless constant that depends on the materials. With velocities $u_A,u_B$ before and $v_A,v_B$ after, along the same positive direction: $v_B-v_A=e(u_A-u_B)$. The values of $e$ satisfy $0\le e\le1$. If $e=1$ the collision is perfectly elastic (no kinetic energy lost). If $e=0$ the bodies do not separate (they coalesce or move together). For $0<e<1$ some kinetic energy is lost.

To find two unknown velocities after a collision you need two equations:

1. Conservation of momentum: $m_1u_1+m_2u_2=m_1v_1+m_2v_2$.
2. Newton's experimental law: $v_2-v_1=e(u_1-u_2)$. Solve them simultaneously. Example: equal masses $m$, $A$ at $5$ m s⁻¹ chasing $B$ at $1$ m s⁻¹, $e=0.5$. Momentum: $v_A+v_B=6$. Newton: $v_B-v_A=0.5(5-1)=2$. So $v_B=4$ and $v_A=2$ m s⁻¹. For equal masses and a target at rest, the results are $v_A=\frac{u(1-e)}{2}$ and $v_B=\frac{u(1+e)}{2}$.

When a particle hits a fixed smooth surface directly, the surface has infinite effective mass and does not move, so Newton's law gives $\text{speed after}=e\times\text{speed before},$ with the direction reversed. Momentum is not conserved (the wall provides an impulse). Example: a ball at $8$ m s⁻¹ rebounds at $5$ m s⁻¹, so $e=\frac58$. Hitting a second wall of the same material, it rebounds at $\frac58\times5=3.125$ m s⁻¹. After $n$ impacts the speed is $e^n$ times the original.

Kinetic energy lost in a collision $=$ KE before $-$ KE after. For a fixed surface, KE after $=e^2\times$ KE before, so the fraction lost is $1-e^2$. For $e=\frac58$ the fraction lost is $1-\frac{25}{64}=\frac{39}{64}$. For two particles, calculate KE before and after using the actual speeds. In the example above the KE before is $13m$ and after is $10m$, so $3m$ is lost. If $e=1$, KE is conserved. Kinetic energy never increases in a collision, which is why $e\le1$.

In problems with three or more bodies, work through the collisions in order, using the speeds from the first collision as the starting speeds for the next. A further collision between two bodies occurs only if the one behind is moving faster than the one in front, or if they are moving towards each other (for example after rebounding from a wall). Example: spheres of equal mass $A$ then $B$ then $C$ in a line. After $A$ hits $B$ at speed $u$, $v_A=\frac{u(1-e)}{2}$ and $v_B=\frac{u(1+e)}{2}$. After $B$ hits $C$, $v_B'=\frac{u(1-e^2)}{4}$ and $v_C=\frac{u(1+e)^2}{4}$. $A$ catches $B$ again if $v_A>v_B'$, which is true for $e<1$.