Two elastic balls of masses m_1 = 10 g and m_2 = 20 g hang…
Two elastic balls of masses $m_1 = 10\ \mathrm{g}$ and $m_2 = 20\ \mathrm{g}$ hang side by side, just touching, on threads of equal length $l = 18\ \mathrm{cm}$. The ball $m_1$ is pulled aside until its thread makes $\alpha = 60^\circ$ with the vertical and released; it collides elastically with $m_2$. Find the largest angles the threads make after the collision.
Explanation of Method
Elastic collision of two pendulum balls.
Step 1: Speed before the collision
\[ h = l(1 - \cos 60^\circ) = 0.09\ \mathrm{m}, \]
\[ v_1 = \sqrt{2gh} \approx 1.33\ \mathrm{m/s}. \]
Step 2: Elastic collision
\[ v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1 = -\frac{v_1}{3}, \]
\[ v_2' = \frac{2m_1}{m_1 + m_2}\,v_1 = \frac{2v_1}{3}. \]
Step 3: Heights and angles
Heights scale with the square of the speed: $h_1 = h/9 = 1\ \mathrm{cm}$, $h_2 = 4h/9 = 4\ \mathrm{cm}$.
\[ \cos\theta_1 = 1 - \frac{h_1}{l} = 1 - \frac{1}{18} = \frac{17}{18}, \]
\[ \cos\theta_2 = 1 - \frac{h_2}{l} = 1 - \frac{4}{18} = \frac{7}{9}. \]
\[\theta_1 \approx 19.2^\circ\ \text{(back)}, \quad \theta_2 \approx 38.9^\circ\]
Check
Energy: $ m_1 h_1 + m_2 h_2 = 10 \cdot 1 + 20 \cdot 4 = 90\ \mathrm{g\,cm} = m_1 h\ \checkmark $
Result
\[\boxed{\,\theta_1 \approx 19.2^\circ\ \text{(back)}, \quad \theta_2 \approx 38.9^\circ\,}\]
Remark
About the original solution. The collision speeds and the heights (1 cm and 4 cm) are right, but the angles were evaluated wrongly: $\arccos(17/18) = 19.2^\circ$, not 22°, and $\arccos(7/9) = 38.9^\circ$, not 35°.