Let's break down the problem step-by-step: --- ## **Problem Restatement** - A mass \( m_1 \) hangs from a spring (spring constant \( k \)), at static equilibrium. - A second mass \( m_2 \) is dropped from height \( h \) and sticks to \( m_1 \) **(perfectly inelastic collision, no rebound)**. - Find the **subsequent motion** (i.e., the motion of the combined mass after the collision). --- ## **Step 1: Initial Static Equilibrium** At static equilibrium, the spring force balances the weight of \( m_1 \): \[ k x_ = m_1 g \implies x_ = \frac{m_1 g}{k} \] where: - \( x_ \) = initial stretch of spring (from natural length) - \( g \) = acceleration due to gravity --- ## **Step 2: Mass \( m_2 \) Drops and Sticks** - Mass \( m_2 \) is dropped from height \( h \). - Just before collision, its velocity: \( v_2 = \sqrt{2gh} \). - It sticks to \( m_1 \): **perfectly inelastic collision**. **Let’s find the velocity of the combined mass immediately after collision:** ### **Conservation of Momentum** Let \( v_1 = \) (since \( m_1 \) was at rest at equilibrium): \[ (m_1 + m_2) v' = m_1 \cdot + m_2 \cdot v_2 \] \[ v' = \frac{m_2}{m_1 + m_2} \sqrt{2gh} \] where \( v' \) is the velocity of the combined mass immediately after sticking. --- ## **Step 3: Motion After Collision (Oscillation Analysis)** Now, the combined mass (\( m_1 + m_2 \)) moves as a mass-spring system. ### **New Equilibrium Position** \[ k x_{\text{eq}} = (m_1 + m_2)g \implies x_{\text{eq}} = \frac{(m_1 + m_2)g}{k} \] ### **Displacement from New Equilibrium** Immediately after the collision: - The position is still at \( x_ \) (old equilibrium). - The new equilibrium is at \( x_{\text{eq}} \). - So, initial displacement from new equilibrium: \[ y_ = x_ - x_{\text{eq}} = \frac{m_1 g}{k} - \frac{(m_1 + m_2)g}{k} = -\frac{m_2 g}{k} \] (Negative: above new equilibrium.) - Initial velocity: \( v' \) (downward, so positive). --- ## **Step 4: Equation of Motion (Simple Harmonic Motion)** The system now undergoes **SHM** about the new equilibrium. For mass \( M = m_1 + m_2 \): \[ y(t) = A \cos(\omega t) + B \sin(\omega t) \] where: - \( \omega = \sqrt{\frac{k}{m_1 + m_2}} \) - \( y \) is measured from new equilibrium position. **Initial Conditions:** - At \( t = \): \( y() = y_ = -\frac{m_2 g}{k} \) - \( \dot{y}() = v' \) \[ y() = A = -\frac{m_2 g}{k} \] \[ \dot{y}() = -A \omega \sin() + B \omega \cos() = B \omega = v' \implies B = \frac{v'}{\omega} \] Plug in \( v' \): \[ B = \frac{1}{\omega} \cdot \frac{m_2}{m_1 + m_2} \sqrt{2gh} \] Therefore, the **subsequent motion** is: --- ## **Final Answer: Subsequent Motion** \[ \boxed{ y(t) = -\frac{m_2 g}{k} \cos\left(\sqrt{\frac{k}{m_1 + m_2}}\, t\right) + \frac{m_2}{m_1 + m_2} \sqrt{\frac{2gh}{k/(m_1 + m_2)}} \sin\left(\sqrt{\frac{k}{m_1 + m_2}}\, t\right) } \] where \( y(t) \) is measured **downward from the new equilibrium position**. --- ### **Summary of Steps** 1. **Static equilibrium**: Find initial and new spring stretches. 2. **Collision**: Use momentum to get velocity after sticking. 3. **Oscillation**: Write SHM equation about the new equilibrium, use initial displacement and velocity to solve for constants. Let me know if you'd like a worked numerical example or further clarification!