A body of mass m_1 = 4 kg moving at v_1 = 10 m/s hits a body…

A body of mass $m_1 = 4\ \mathrm{kg}$ moving at $v_1 = 10\ \mathrm{m/s}$ hits a body of twice its mass at rest; they stick together. By how much does the internal energy of the system increase?

Explanation of Method

Internal energy released in a sticking collision.

Step 1: Common speed

\[ m_1 v_1 = 3m_1 v \;\Longrightarrow\; v = \frac{v_1}{3}. \]

Step 2: Lost kinetic energy

\[ \Delta E = \tfrac12 m_1 v_1^2 - \tfrac12 (3m_1)\left(\frac{v_1}{3}\right)^2 = \tfrac12 m_1 v_1^2 \cdot \tfrac23. \]

\[\Delta E = \tfrac13 m_1 v_1^2 \approx 133.3\ \mathrm{J}\]

$E_k = 200\ \mathrm{J}$ $\text{before}$ $66.7$ $\text{heat } 133.3\ \mathrm{J}$ $\text{after}$ $v_2 = \tfrac{v_1}{3},\quad E_{k2} = \tfrac13 E_{k1}$

In the perfectly inelastic collision two thirds of the kinetic energy turn into internal energy (heat and deformation).

Result

\[\boxed{\,\Delta E = \tfrac13 m_1 v_1^2 \approx 133.3\ \mathrm{J}\,}\]

Remark

Check. The original result is correct.