Let's break down the problem step by step: ## Step 1: **Identify the Principle** There are **no external torques** acting on the system, so **angular momentum is conserved** about the center of the rod. --- ## Step 2: **Initial Angular Momentum (Before Collision)** - **Object 1**: Mass \(m_1\), velocity \(v_1\), sticks at a distance \(L/2\) from the center (bottom edge). - Perpendicular distance to axis of rotation: \(L/2\) - Angular momentum: \(L_1 = m_1 v_1 (L/2)\) - **Object 2**: Mass \(m_2\), velocity \(v_2\), sticks at a distance \(L/4\) from the center (top edge). - Perpendicular distance to axis of rotation: \(L/4\) - Angular momentum: \(L_2 = m_2 v_2 (L/4)\) - **Direction**: Check the direction of velocity relative to the rotation axis. Let's assume both velocities \(v_1\) and \(v_2\) are perpendicular to the rod and in the plane of rotation. The **total initial angular momentum** about the center: \[ L_{\text{initial}} = m_1 v_1 \left(\frac{L}{2}\right) + m_2 v_2 \left(\frac{L}{4}\right) \] --- ## Step 3: **Final Angular Momentum (After Collision)** After collision, the rod and both masses rotate together with angular speed \(\omega\). - **Moment of inertia of rod about center:** \(I_{\text{rod}} = \frac{1}{12} ML^2\) - **Moment of inertia of object 1:** \(I_1 = m_1 (L/2)^2 = m_1 \frac{L^2}{4}\) - **Moment of inertia of object 2:** \(I_2 = m_2 (L/4)^2 = m_2 \frac{L^2}{16}\) Total moment of inertia: \[ I_{\text{total}} = I_{\text{rod}} + I_1 + I_2 = \frac{1}{12} ML^2 + m_1 \frac{L^2}{4} + m_2 \frac{L^2}{16} \] Final angular momentum: \[ L_{\text{final}} = I_{\text{total}} \omega \] --- ## Step 4: **Set Initial and Final Angular Momentum Equal** \[ L_{\text{initial}} = L_{\text{final}} \] \[ m_1 v_1 \frac{L}{2} + m_2 v_2 \frac{L}{4} = \left( \frac{1}{12} M L^2 + m_1 \frac{L^2}{4} + m_2 \frac{L^2}{16} \right) \omega \] --- ## Step 5: **Solve for \(\omega\)** Divide both sides by \(L\): \[ m_1 v_1 \left(\frac{1}{2}\right) + m_2 v_2 \left(\frac{1}{4}\right) = \left( \frac{1}{12} M L + m_1 \frac{L}{4} + m_2 \frac{L}{16} \right) \omega \] Or, factor out \(L\): \[ m_1 v_1 \frac{L}{2} + m_2 v_2 \frac{L}{4} = \left( \frac{1}{12} M L^2 + m_1 \frac{L^2}{4} + m_2 \frac{L^2}{16} \right) \omega \] Divide both sides by \(L^2\): \[ m_1 v_1 \frac{1}{2L} + m_2 v_2 \frac{1}{4L} = \left( \frac{1}{12} M + m_1 \frac{1}{4} + m_2 \frac{1}{16} \right) \omega \] So, \[ \omega = \frac{ m_1 v_1 \frac{1}{2L} + m_2 v_2 \frac{1}{4L} }{ \frac{1}{12} M + m_1 \frac{1}{4} + m_2 \frac{1}{16} } \] Or, combine into a single fraction: \[ \boxed{ \omega = \frac{ m_1 v_1 \frac{L}{2} + m_2 v_2 \frac{L}{4} } { \frac{1}{12} M L^2 + m_1 \frac{L^2}{4} + m_2 \frac{L^2}{16} } } \] Or, simplify \(L\) out: \[ \boxed{ \omega = \frac{ 2 m_1 v_1 + m_2 v_2 }{ M \frac{1}{3} L + 4 m_1 L + m_2 L } } \] But the best, clear answer is: \[ \boxed{ \omega = \frac{ m_1 v_1 \frac{L}{2} + m_2 v_2 \frac{L}{4} } { \frac{1}{12} M L^2 + m_1 \frac{L^2}{4} + m_2 \frac{L^2}{16} } } \] --- **Summary of steps:** 1. Write initial angular momentum (sum for both masses). 2. Write final angular momentum (rotates as one object). 3. Set them equal (angular momentum conserved). 4. Solve for \(\omega\). Let me know if you need any substitutions or further simplifications!

The final angular velocity \(\omega\) of the system after both objects stick to the rod is given by: \[ \boxed{ \omega = \frac{ \displaystyle m_1 v_1 \frac{L}{2} + m_2 v_2 \frac{L}{4} }{ \displaystyle \frac{1}{12} M L^{2} + m_1 \frac{L^{2}}{4} + m_2 \frac{L^{2}}{16} } } \] **In words:** - Numerator: The sum of the initial angular momenta of the two objects (mass times velocity times their perpendicular distance from the center). - Denominator: The total moment of inertia of the system (rod plus the two masses at their respective distances). This expression allows calculation of the system’s angular velocity immediately after the collision, assuming no external torques and perfect sticking.