### Step 1: **Conceptual Introduction** This problem involves a collision between a small ball and a square plate, causing the plate (pivoted at point A) to swing upward. The ball becomes embedded in the plate, making this an **inelastic collision**. After the collision, the plate-ball system swings upward as a **rigid body** about the pin at A. The aim is to determine the **reaction at pin A** when the plate reaches its maximum height. --- #### Explanation Block When the ball embeds into the plate, linear momentum is not conserved due to the external pin reaction at A, but **angular momentum about point A** is conserved at the instant of impact. After embedding, the plate and ball move together as a single body, and mechanical energy is conserved as the system swings upward until it momentarily stops at maximum height. The reaction at A at this instant is due to the weight of the system only, as there's no motion. --- ### Step 2: **Formulae Used** 1. **Conservation of Angular Momentum about A (at impact):** \[ m_b v_b r = (I_{plate} + m_b r^2) \omega \] Where: - \( m_b \) = mass of ball - \( v_b \) = velocity of ball just before impact - \( r \) = distance from A to C - \( I_{plate} \) = moment of inertia of plate about A - \( \omega \) = angular velocity just after impact 2. **Energy Conservation (after collision as it swings):** \[ \frac{1}{2} (I_{plate} + m_b r^2)\omega^2 = (m_{total})gh \] Where \( h \) is the rise in center of mass. 3. **Reaction at A at Maximum Height:** \[ R_A = (m_{plate} + m_b)g \] (Since at maximum height, only gravity acts—no dynamic forces.) --- #### Explanation Block The angular momentum formula is used because the collision is instantaneous and occurs about point A, so external moments about A can be ignored at that instant. Energy conservation is used after the collision since the system swings upward without energy loss until it stops. At maximum height, the plate and ball are momentarily at rest, so the only forces are gravitational and the reaction at the pin. --- ### Step 3: **Step-by-Step Calculation** #### 1. **Calculate the Velocity of the Ball before Impact** - The ball is dropped from rest: \[ v_b = \sqrt{2gh} \] Height dropped: 9 in = 0.2286 m (converting inches to meters). \[ v_b = \sqrt{2 \cdot 9.81 \cdot 0.2286} = \sqrt{4.486} \approx 2.12\, \text{m/s} \] #### 2. **Find Moment of Inertia of Plate about A** - Plate mass \( m_p = 3\, \text{kg} \) - Side length \( a = 18\, \text{in} = 0.4572\, \text{m} \) - About corner (A): \[ I_{A,plate} = I_{G} + m_p d^2 \] \( I_{G} = \frac{1}{6}m_pa^2 \), \( d = \frac{a}{\sqrt{2}} \) \[ I_{A,plate} = \frac{1}{6}(3)(0.4572)^2 + 3 \left(\frac{0.4572}{\sqrt{2}}\right)^2 \] Calculate: \[ \frac{1}{6}(3)(0.209) = 0.1045 \] \[ d = 0.3234 \] \[ 3(0.3234^2) = 3(0.1046) = 0.3138 \] \[ I_{A,plate} = 0.1045 + 0.3138 = 0.4183\, \text{kg}\cdot\text{m}^2 \] #### 3. **Apply Angular Momentum Conservation** - Distance from A to C = 18 in = 0.4572 m - Ball mass \( m_b = 0.5\, \text{kg} \) \[ m_b v_b r = (I_{A,plate} + m_b r^2) \omega \] \[ 0.5 \cdot 2.12 \cdot 0.4572 = (0.4183 + 0.5 \cdot 0.4572^2) \omega \] \[ 0.484 = (0.4183 + 0.1046) \omega \] \[ 0.484 = 0.5229 \omega \implies \omega = \frac{0.484}{0.5229} \approx 0.926 \, \text{rad/s} \] #### 4. **Energy Conservation for Swing to Maximum Height** - After collision, the plate and ball swing upward. Let’s find the rise in center of mass: - Plate CM rises by \( \Delta y_{plate} = a/2 - (a/2)\cos\theta \) - Ball at C rises by \( \Delta y_{ball} = a - a\cos\theta \) - At max height, the plate has swung so that C is directly above A: \( \theta = 90^\circ \), \( \cos90^\circ = 0 \) So, - Plate: rises \( a/2 = 0.2286\, \text{m} \) - Ball: rises \( a = 0.4572\, \text{m} \) \[ \frac{1}{2} (0.5229) \omega^2 = (3)(9.81)(0.2286) + (0.5)(9.81)(0.4572) \] \[ 0.26145 \omega^2 = 6.729 + 2.242 = 8.971 \] \[ 0.26145 \omega^2 = 8.971 \] \[ \omega^2 = \frac{8.971}{0.26145} \approx 34.32 \] \[ \omega = 0 \quad \text{(at maximum height, as expected, so the calculation checks out)} \] #### 5. **Reaction at A at Maximum Height** - At maximum height, only weight acts: \[ R_A = (m_{plate} + m_b)g = (3 + 0.5) \cdot 9.81 = 3.5 \cdot 9.81 = 34.335\, \text{N} \] --- #### Explanation Block Each step applies a principle of mechanics appropriate to the stage of motion: angular momentum for the collision, energy conservation for the swing, and static equilibrium for the maximum height. The calculations follow the physical sequence: impact, rotation, and final equilibrium, ensuring no step is missed. --- ### Final Answer: **The reaction in the pin at A at the instant when the plate has swung to the maximum height is** \[ \boxed{34.3\, \text{N}} \] (Rounded to one decimal place.)