1. Effective Length

We have $n$ identical balls of radius $r$ arranged in a line of length $L$. The physical space occupied by the matter of the balls is $n \times (2r)$. The “free volume” (or free length in 1D) available for the motion of the centers of the balls is: $$ L_{eff} = L – 2nr $$

2. Dynamic Equivalence

Since all balls have equal mass $m$ and collisions are perfectly elastic, when two balls collide, they simply exchange velocities. $$ v_1′ = v_2, \quad v_2′ = v_1 $$ Dynamically, this is indistinguishable from the balls passing through each other without interacting. Therefore, we can model the system as $n$ non-interacting point particles moving in a box of length $L_{eff}$.

3. Force Calculation

For a single particle moving with velocity $v$, the time between collisions with the same wall is $\Delta t = \frac{2 L_{eff}}{v}$. Momentum change per collision: $\Delta p = 2mv$. Average Force by one particle: $f = \frac{\Delta p}{\Delta t} = \frac{2mv}{2L_{eff}/v} = \frac{mv^2}{L_{eff}}$.

Total Force exerted by all $n$ balls: $$ F = \sum_{i=1}^n f_i = \sum_{i=1}^n \frac{m v_i^2}{L_{eff}} = \frac{1}{L_{eff}} \sum m v_i^2 $$ We are given the total Kinetic Energy $K = \sum \frac{1}{2} m v_i^2$. Therefore, $\sum m v_i^2 = 2K$.