Two balls of equal mass move toward each other on a smooth…

Two balls of equal mass move toward each other on a smooth horizontal surface at $v_1 = 10\ \mathrm{m/s}$ and $v_2 = 20\ \mathrm{m/s}$. What are their velocities after a head-on elastic collision?

Explanation of Method

Head-on elastic collision of equal balls.

Step 1: Equal masses exchange velocities

With $m_1 = m_2$, the momentum and energy equations

\[ v_1 + v_2 = v_1' + v_2', \]

\[ v_1^2 + v_2^2 = v_1'^2 + v_2'^2 \] have the non-trivial solution $v_1' = v_2$, $v_2' = v_1$. Taking ball 1's direction as positive: $v_1 = +10$, $v_2 = -20\ \mathrm{m/s}$.

\[v_1' = -20\ \mathrm{m/s}, \qquad v_2' = +10\ \mathrm{m/s}\]

Each ball bounces back with the speed the other one had: ball 1 at 20 m/s, ball 2 at 10 m/s.

$\text{before}$ $\text{after}$ $1$ $2$ $10\ \mathrm{m/s}$ $20\ \mathrm{m/s}$ $20\ \mathrm{m/s}$ $10\ \mathrm{m/s}$

Equal masses in a head-on elastic collision exchange their velocities: each ball bounces back with the speed the other one had (arrows to scale).

Result

\[\boxed{\,v_1' = -20\ \mathrm{m/s}, \qquad v_2' = +10\ \mathrm{m/s}\,}\]

Remark

Check. The original speeds are correct (20 and 10 m/s), but it did not say that both balls reverse direction; for a head-on collision the exchanged velocities point back the way each ball came.