Elastic Collisions in One Dimension — Question 3

Question 3

Why matching masses maximizes single-collision transfer

A mass $m$ at speed $u$ collides elastically with a stationary mass $M$. Define $r=M/m>0$.

1. Derive the fraction of incident kinetic energy transferred to the target.

2. Find its maximum and compare $r$ with $1/r$.

Question 3 - Solution

Why matching masses maximizes single-collision transfer

(a) Derivation and calculation

With $r=M/m$ and $v_2=2u/(1+r)$, $$\boxed{\eta=\frac{Mv_2^2}{mu^2}=\frac{4r}{(1+r)^2}}.$$ $$(1+r)^2-4r=(r-1)^2\geq0\Rightarrow\eta\leq1.$$ Equality occurs only at $\boxed{r=1}$.

(b) Result and interpretation

Under a reciprocal mass ratio, $$\eta(1/r)=\frac{4/r}{(1+1/r)^2}=\frac{4r}{(1+r)^2}=\eta(r).$$ Thus equally mismatched reciprocal ratios transfer the same energy fraction, while equal masses transfer all the incident energy.

Physical check

The transfer tends to zero as either $r\to0$ or $r\to\infty$.