`v_C = \frac{2m_A}{m_A + m_C} v_0`

`v_B = \frac{2m_C}{m_C + m_B} v_C`

`v_B = \frac{2m_C}{m_C + m_B} \cdot \frac{2m_A}{m_A + m_C} v_0 = \frac{4m_A m_C}{(m_A + m_C)(m_C + m_B)} v_0`

`y = \frac{m_A m_C + m_A m_B + m_C^2 + m_B m_C}{m_C} = m_A + \frac{m_A m_B}{m_C} + m_C + m_B`

`y = (m_A + m_B) + ( m_C + \frac{m_A m_B}{m_C} )`

Áp dụng bất đẳng thức Cauchy:

`m_C + \frac{m_A m_B}{m_C} \geq 2\sqrt{m_C \cdot \frac{m_A m_B}{m_C}} = 2\sqrt{m_A m_B}`

Dấu "=" xảy ra khi:`m_C = \frac{m_A m_B}{m_C} \Rightarrow m_C = \sqrt{m_A m_B}`

`m_C = \sqrt{40 \cdot 90} = \sqrt{3600} = 60 \text{ g}`

`v_{B(max)} = \frac{4 \cdot 40 \cdot 60}{(40 + 60)(60 + 90)} \cdot 5,20`

`v_{B(max)} = \frac{9600}{100 \cdot 150} \cdot 5,20 = \frac{9600}{15000} \cdot 5,20 = 0,64 \cdot 5,20`

`->` $v_{B(max)} = 3,328 \text{ m/s}$