Elastic Collisions in One Dimension — Question 7

Question 7

An intermediate mass can improve energy transfer

A mass $m$ moving at $u$ hits a stationary mass $x$, which then hits a stationary mass $M$. Each of these first two encounters is elastic and one dimensional. Consider the speed of $M$ immediately after its first encounter, before any later recollision.

1. Derive this speed as a function of $x$.

2. Find the intermediate mass that maximizes it and evaluate the speed for $M=9m$.

Question 7 - Solution

An intermediate mass can improve energy transfer

(a) Derivation and calculation

After the first collision, $v_x=2mu/(m+x)$. After the second, $$\boxed{v_M=\frac{2x}{x+M}v_x=\frac{4mxu}{(m+x)(x+M)}}.$$ For positive $x$, maximize this by minimizing $$\frac{(m+x)(x+M)}x=m+M+x+\frac{mM}{x}\geq m+M+2\sqrt{mM}.$$ Thus $\boxed{x=\sqrt{mM}}$.

(b) Result and interpretation

For $M=9m$, the optimum is $x=3m$: $$\boxed{v_M=\frac{4m(3m)}{(4m)(12m)}u=\frac u4}.$$ Direct impact gives $$v_{M,\mathrm{direct}}=\frac{2m}{m+9m}u=\frac u5.$$ These results describe the stated first two impacts, before any later recollisions.

Physical check

The optimal intermediate mass is the geometric mean of the endpoint masses.