Translational Kinetic Energy — Question 7

Question 7

A single elastic collision need not share energy equally

In a one-dimensional elastic collision, a molecule of mass $m$ and initial speed $u$ strikes a stationary molecule of mass $3m$. Use conservation of momentum and kinetic energy; take the incoming direction positive.

1. Find both final velocities and the fraction of initial kinetic energy transferred to the target.

2. For general target mass $M$, derive the transferred fraction $4mM/(m+M)^2$ and determine which mass ratio maximizes it.

Question 7 - Solution

A single elastic collision need not share energy equally

(a) Method and calculation

$$mu=mv+3mV,\quad\tfrac12mu^2=\tfrac12mv^2+\tfrac12(3m)V^2.$$$$v=-u/2,\quad V=u/2,\quad\frac{K_{\mathrm{target}}}{K_i}=\frac{3m(u/2)^2}{mu^2}=\frac34.$$

(b) Interpretation and further calculation

$$V=\frac{2m}{m+M}u\ \Rightarrow\ \frac{K_{\mathrm{target}}}{K_i}=\frac{4mM}{(m+M)^2}.$$$$\frac{4mM}{(m+M)^2}=1-\frac{(M-m)^2}{(M+m)^2}\le1.$$Equality occurs at $M=m$. A single event transfers an amount fixed by its initial geometry and velocities; equilibrium equipartition is an ensemble statement.

Physical check

The lighter projectile rebounds here, while the heavier target takes 75% of the initial energy.