Internal kinetic energy of two balls

Problem

Two identical balls of mass $m = 0{,}10\;\text{kg}$ move along the same line: the first at $v_1 = +3{,}0\;\text{m/s}$, the second at $v_2 = -2{,}0\;\text{m/s}$. Using König’s theorem, calculate the internal kinetic energy of the system.

Solution

CM velocity: $v_\text{CM} = \frac{m v_1 + m v_2}{2m} = \frac{3{,}0 + (-2{,}0)}{2} = 0{,}50\;\text{m/s}$

Velocities relative to the CM: $v_1' = 3{,}0 - 0{,}50 = 2{,}5\;\text{m/s}, \qquad v_2' = -2{,}0 - 0{,}50 = -2{,}5\;\text{m/s}$

Internal kinetic energy: $E_\text{cin,int} = \tfrac{1}{2}m v_1'^2 + \tfrac{1}{2}m v_2'^2 = \tfrac{1}{2}\cdot 0{,}10\cdot 2{,}5^2 \cdot 2 = 0{,}10\cdot 6{,}25$ $\ev{E_\text{cin,int} \approx 0{,}63\;\text{J}}$

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Topics: Centro di massa Concepts: Energia cinetica · Energia interna Skills: Cambio di sistema di riferimento