Three balls of masses m_1 , m_2 , m_3 lie at rest along a…

Three balls of masses $m_1$, $m_2$, $m_3$ lie at rest along a line. The first is given speed $v_1$ and hits the second, which then hits the third; both collisions are elastic and head-on. What mass $m_2$ gives the third ball the greatest speed?

Explanation of Method

Best middle mass for passing on speed.

Step 1: Two elastic collisions

A ball moving at $u$ that hits a ball at rest gives it the speed $\tfrac{2m_{\text{incoming}}}{m_{\text{incoming}} + m_{\text{target}}}\,u$: $v_2 = \frac{2m_1}{m_1 + m_2}\,v_1$.

\[ v_3 = \frac{2m_2}{m_2 + m_3}\,v_2 = \frac{4m_1 m_2\,v_1}{(m_1 + m_2)(m_2 + m_3)}. \]

Step 2: Maximum

Maximise $f(m_2) = \dfrac{m_2}{(m_1 + m_2)(m_2 + m_3)}$: $f'(m_2) = 0 \;\Longleftrightarrow\; m_1m_3 - m_2^2 = 0$.

\[m_2 = \sqrt{m_1 m_3}\]

The numerator of $f'$ is

\[ (m_1 + m_2)(m_2 + m_3) - m_2(m_1 + m_3 + 2m_2) = m_1m_3 - m_2^2. \]

The third ball then gets

\[ v_{3,\max} = \frac{4m_1 v_1}{\left(\sqrt{m_1} + \sqrt{m_3}\right)^2}. \]

$v_3/v_1$ $m_2/m_1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $0.44$ $m_2 = \sqrt{m_1m_3} = 2m_1$ $m_3 = 4m_1$

Example with $m_3 = 4m_1$: the third ball's speed as a function of the middle mass. It peaks at $m_2 = \sqrt{m_1 m_3} = 2m_1$, where $v_3 = 4v_1/9$.

Result

\[\boxed{\,m_2 = \sqrt{m_1 m_3}\,}\]

Remark

About the original solution. The final answer $m_2 = \sqrt{m_1m_3}$ is correct, but its formula for the second collision had $2m_3$ instead of $2m_2$ in the numerator. With that formula $v_3$ would only grow as $m_2$ shrinks and there would be no maximum; the derivative step silently assumed the right formula.